![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2F54debde1de6ae475b0472f5951f4a9d0%2Fmermaid-ai-diagram-2026-04-21-111537.png?generation=1776770277109863&alt=media\")
Abstract. We build on openai/gpt-oss-120b in its base (released) form because the community has already shown it is strong on this task: the public baseline notebook 44-50 Let Me Over Cook—using the same family of model and tooling—achieved 44/50 on the public leaderboard, i.e. most problems are within reach of the unfine-tuned checkpoint with a good harness. In our own model comparisons, Qwen3 and Qwen3.5 did not match that bar and were discarded; we also set aside NVIDIA Nemotron (released later) after it looked weaker in our internal checks. Tinker SFT on gpt-oss-120b with a merged adapter, by contrast, reduced our public scores (37, 26 vs 38, 40 for the same submission family; see §7.3), so our final submission keeps the base weights. This submission solves AIMO-style problems with that fixed backbone served via vLLM, the same system prompt as 44-50 Let Me Over Cook, a short preference prompt that pushes step-by-step reasoning and heavy use of a stateful Python (Jupyter) tool, and parallel stochastic attempts with entropy-aware aggregation. The main methodological contribution is dynamic deadline allocation: a three-stage time budget that extends only when agreement among completed attempts is insufficient at stage boundaries—reducing wasted wall-clock on problems that already stabilize early. We also implemented a hybrid dense + BM25 retriever over a compact math corpus (techniques, example problems, and hints); it is disabled in the final submission when empirical checks showed net harm. We summarize other approaches we tried (continuations, layered developer prompts, global time-banking, and related SFT experiments) that did not match the stability of the final pipeline (finalnotebook). The shipped notebook is finalnotebook with base gpt-oss-120b; details on merged-SFT public scores are in §7. This writeup is methodology-first; we do not rely on extra plots or exhaustive numeric tables.
Reproducibility statement. All implementation details below are grounded in this submission’s Kaggle notebook finalnotebook. Tinker training runs are summarized in §7 (configs and metrics were saved with each run).
\nEDIT on May 13th, 2026. After the release of the private leaderboard, our final submission scored 41.0/50, which aligns with our arguments in this writeup about stability since we scored 40/50 on the public leaderboard. Also added a link to the code on GitHub.
\nAIMO3 requires producing a single integer answer in ({0,\\ldots,99999}) under a global notebook time limit and per-problem compute constraints.
\nBackbone and fine-tuning. The competition implementation standardizes on openai/gpt-oss-120b: a public AIMO3 notebook (44-50 Let Me Over Cook) already demonstrates 44/50 on the public leaderboard with this model, which motivates investing in inference-time design rather than replacing the base weights. We evaluated Qwen3 / Qwen3.5 and NVIDIA Nemotron as alternatives; they did not justify switching away from gpt-oss-120b under our resource and tooling constraints. We also fine-tuned gpt-oss-120b (merged SFT adapter) and found worse public results than the base model with the same harness (§7.3), so the primary submission uses the original checkpoint.
The approach treats each problem as a multi-attempt inference problem: several independent runs explore the space of reasoning trajectories; we then stop early when a strong consensus emerges, and otherwise allocate more time only when the evidence suggests the problem is still “open.”
\nKey Contributions (as implemented in the submission).
\nstdout can be extremely long—on the order of tens of thousands of model-side tokens in severe cases if passed through verbatim. Without truncation, tool returns alone can consume the context / generation budget, and otherwise valuable attempts may stop early (e.g., context exhaustion) before a final answer is produced.The competition submission is finalnotebook (38 / 40) (§7.3).
\ngpt-oss-120b (base checkpoint) from the Kaggle model path in the notebook (CFG.model_path)—the same line already shown to reach most problems in public (e.g. 44/50 in 44-50 Let Me Over Cook); we avoid merged SFT here because it hurt our public scores relative to this base (§1, §7.3).vLLM OpenAI-compatible API (python -m vllm.entrypoints.openai.api_server), with Harmony encoding for GPT-OSS (HarmonyEncodingName.HARMONY_GPT_OSS).print()-based output, SymPy/NumPy guidance, concise dumps, and timeout-aware splitting—plus a length cap on tool returns (§5.5).The end-to-end path is OpenAI-compatible client → Harmony template (GPT-OSS) → streaming completion with tool calls → Jupyter execution. We omit standalone architecture figures; the flow is summarized in Appendix E.
\nThe system_prompt in CFG is unchanged from the public AIMO3 starter notebook 44-50 Let Me Over Cook, by nihilisticneuralnet: it defines an IMO-level solver persona, a structured problem-solving loop (understand → explore → plan → execute → verify), verification requirements, and the required output format (\\boxed{} with an integer in ([0,99999])).
The tool_prompt is tuned for efficient computation in a stateful Python (Jupyter) notebook: it lists available libraries (math, numpy, sympy, etc.) and instructs the model to print() results (last expression is not auto-displayed), to use SymPy for exact symbolic work and NumPy for fast numeric work, to estimate complexity before large searches (prefer better-than-naive algorithms), to keep printed output concise (summarize large dumps), and to split or simplify work when a cell hits the per-call time limit. This steers tool use toward fewer wasteful megabyte-scale printouts and more targeted code.
A compact preference_prompt is appended to the user side of the conversation:
\n\nSolve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing
\n\\boxed{}.
This is intentionally short: it does not replace the system prompt; it steers sampling toward explicit computation and independent verification—aligned with competition scoring (correctness over eloquence).
\nThe notebook defines CorpusRetriever, a hybrid retriever:
topics, techniques, recognition_signals).sentence-transformers when the embedding model path exists; cosine similarity min-max normalized and fused with BM25:\n[\n\\text{score} = 0.5 \\cdot \\text{bm25}{\\text{norm}} + 0.5 \\cdot \\text{dense}{\\text{norm}}\n]\n(falls back to BM25-only if dense is unavailable).Retrieved items are formatted as short “strategy blocks” including technique/meta/similar-problem text, topics, insights, optional solution sketch, and computational hints.
\nCorpus size. The submission narrative uses a compact corpus on the order of ~200 entries (techniques, worked patterns, hints). The notebook logs len(self.retriever.entries) when a corpus file is loaded.
When retrieval is turned on, the top strategies are serialized into a [REFERENCE STRATEGIES] block (with closing [END REFERENCE]) and prepended to the user turn so the model sees structured hints, topics, techniques, and per-strategy relevance (BM25, dense, and fused combined scores). Below is a representative excerpt from logs (two retrieved entries; line breaks as in stdout):
[REFERENCE STRATEGIES]\n\n--- Strategy 1 [AIME-style] ---\nSimilar Problem: Triangle ABC inscribed in circle omega. Tangent to omega at B and C intersect at D. AD meets omega again at P. If AB=5, BC=9, AC=10, find AP (as m/n, answer m+n).\nTopics: geometry, circles\nTechniques: power_of_a_point, similar_triangles\nRecognition Signals:\n - tangent to circle\n - inscribed triangle\n - tangent intersection\nKey Insights:\n - D is the intersection of tangents at B,C, so DB=DC (tangent lengths)\n - Power of point D: DB^2 = DA * DP\n - D lies on the perpendicular bisector of BC\nStrategy: Place D as pole of BC w.r.t. omega. DB^2=DC^2=power of D. By power of a point: DA*DP = DB^2. Compute: use coordinates with BC on x-axis, or use the formula DB^2 = (circumradius)^2 * ... Alternatively: by the tangent-secant relation and similar triangles DAB ~ DPA (since angle DBA = inscribed angle = angle DPA). Then DA/DB = DB/DP gives DP = DB^2/DA. AP = DA - DP or DA + DP depending on configuration.\nCode Hint: Use circumradius R = abc/(4*Area). Area by Heron's formula. Coordinate geometry for exact computation.\nRelevance: combined=1.000, bm25=1.000, dense=1.000\n\n--- Strategy 2 [Competition] ---\nSimilar Problem: Point P lies on the circumcircle of square ABCD such that PA*PC = 56 and PB*PD = 90. Find the area of ABCD.\nTopics: geometry, circles\nTechniques: radical_axis, power_of_a_point\nRecognition Signals:\n - circumcircle of square\n - PA * PC\n - PB * PD\nKey Insights:\n - For a point on the circumcircle of a square with side s: use Ptolemy or power of a point\n - By Ptolemy on cyclic quadrilateral PABC: PA*BC + PC*AB = PB*AC\n - The square's circumradius is s*sqrt(2)/2\nStrategy: Let side = s. Circumradius R = s/sqrt(2). Place square with center at origin. For P on circumcircle: PA*PC and PB*PD relate to the position of P. Using complex numbers with vertices at R*e^{i*k*pi/2}: compute products. By Ptolemy on ABPC: PA*s + PC*s = PB*s*sqrt(2). Combined with PA*PC=56, PB*PD=90 and symmetry relations, solve for s^2 = area.\nCode Hint: Use complex number representation: place square vertices at s/sqrt(2) * exp(i*k*pi/2 + i*theta) and parameterize P.\nRelevance: combined=0.819, bm25=0.797, dense=0.841\n\n[END REFERENCE]\nEmpirical observation: injecting retrieved context sometimes reduced reliability—e.g., misleading analogies, overcrowding the context window, or steering the model away from the cleanest path for a specific problem.
\nIn the notebook’s default _load_corpus configuration, the corpus path is set to a non-existent placeholder (\"NOFILE\"), so the retriever is not loaded and solve_problem uses:
user_input = f'{problem} {self.cfg.preference_prompt}'.
To enable RAG in a fork, point corpus_candidates to your JSON corpus and ensure the Qwen3 embedding model path exists on Kaggle.
Stability, not just mean score. With hybrid retrieval turned on, we observed worse stability of end-to-end performance: for example, the over-engineered prompts + RAG public-leaderboard line in §7.3 spans 34 to 42 out of 50—a 8-point spread across submits—whereas runs without RAG (or with retrieval off in the shipped stack) were typically tighter for us. We saw similar high-variance behavior in other notebooks that enabled RAG but are not detailed in this report. We did not run a dedicated offline A/B at fixed seeds; still, the combination of unreliable peaks and heavy tail risk led us to ship without retrieval and invest budget in dynamic deadlines and voting instead.
\n
Parallel attempts have heterogeneous completion times. On some problems, several attempts finish quickly with a stable answer (we observed this pattern anecdotally for problems whose final answers included 57447, 160, and 520). Empirically, on one representative run the eight parallel attempts showed Response Length (assistant stream size, in tokens) from about 26k to 51k and Completion Time from about 292s to 557s—same harness, large spread—so some trajectories finish much earlier than others. Remaining attempts may still consume tail latency without changing the decision. A static “always run until the global max” policy wastes time that is better reallocated across the full contest slate.
\nDesign goal: treat “difficulty” operationally—not with a separate classifier—but via whether the attempt ensemble has converged by each stage boundary.
\nFor each problem, the solver computes a remaining notebook budget and derives a max_budget for that problem (function of time left, problems remaining, and configured caps). It then defines three checkpoints:
\nfirst_deadline = start + first_budget, with first_budget = min(first_problem_timeout, max_budget).second_deadline = start + second_budget, with second_budget = min(second_problem_timeout, max_budget).final_deadline = start + final_budget where final_budget = max_budget.Configured constants in CFG include (see notebook): first_problem_timeout, second_problem_timeout, final_problem_timeout, base_problem_timeout, notebook_limit, and agreement thresholds first_agreement_count, second_agreement_count.
A background loop polls completed attempts. When real time crosses the current stage deadline (with a small grace window in is_past_deadline), the policy compares the top vote count among valid numeric answers to thresholds:
| Stage | \nIf condition holds | \nElse | \n
|---|---|---|
| 1 | \nFinalize when the leading answer has ≥ first_agreement_count votes and len(valid_answers) > first_agreement_count. | \nExtend to stage 2 (or toward final if the stage-2 checkpoint is not later than the current deadline). | \n
| 2 | \nFinalize when the leading answer has ≥ second_agreement_count votes. | \nExtend to stage 3 (final budget). | \n
Intuition mapping (informal labels):
\nThis matches the motivation: do not pay for long tail attempts when the vote mass has already stabilized.
\n![\"Problem](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2F5a44a1a95f9ed7e326aeecc08ae29fab%2FProblem%20Retrieval%20Strategy-2026-04-21-112241.png?generation=1776770635752584&alt=media\")
\nCFG.attempts parallel threads (with workers pool size).early_stop copies, the solver sets a stop event and cancels pending futures.\\boxed{} format; this recovers trajectories that solved most of the problem but failed to emit the final box cleanly.stdout can be extremely long—on the order of tens of thousands of model-side tokens in severe cases if passed through verbatim. Without truncation, tool returns alone can consume the context / generation budget, and otherwise valuable attempts may stop early (e.g., context exhaustion) before a final answer is produced. The notebook’s AIMO3Tool therefore applies _truncate_output: if a tool return exceeds CFG.max_tool_output_chars (default 8192 characters), it keeps the head and tail and inserts a marker ...[python output truncated]... so the model still sees the beginning and end of the run without flooding the context._select_answer ranks answers by a sum of inverse-entropy weights across attempts (lower entropy ⇒ higher weight), then picks the top score.We did not produce separate distribution plots of time-to-consensus or stage-finalization rates; the design rationale above is the primary evidence for this component.
\nBeyond finalnotebook, we implemented and submitted several alternative designs that explored richer prompting, continuation-based search, global time-banking, and a merged SFT checkpoint. Public leaderboard scores were not consistently strong across reruns and configurations. Empirically, the main failure modes were over-long or over-specified preference / system-adjacent text (which diluted the stable baseline persona and steered tool use inefficiently) and time-management schemes that were harder to tune than the simpler dynamic per-problem deadline policy in the shipped solution.
\nChosen solution. We selected finalnotebook as the primary submission: it was submitted twice and scored 38 and 40 (50 points max), which is both more stable and better than the ranges below. More importantly, this approach holds enormous potential due to its time saving strategy.
\n| Approach | \nApprox. score range (of 50) | \nMain idea (brief) | \n
|---|---|---|
| Two-phase continuations | \n33–40 | \nInitial wave of attempts, then forked continuation trajectories (doubling the number of attempts 8==>16) with a dedicated continuation user message; optional compute-heavy system variant; long structured preference text; tooling penalties and deadline-aware forced \\boxed{} extraction. | \n
| Over-engineered prompts + RAG | \n34–42 | \nLayered developer_prompt, rewritten system copy, verbose preference steering; hybrid retrieval on (strategy blocks in the user turn). §7.3 table. | \n
| Merged SFT model (Tinker) | \n26–37 | \nSFT on gpt-oss-120b, adapter merged into full weights; CFG.model_path points at the merged HF directory. Public leaderboard: 37, 26 (two submits); §7.3. | \n
| Banked time + hard reserve | \n30–40 | \nGlobal time bank and hard-problem reserve: adaptive per-problem budgets draw from a bank while protecting a reserve—more complex scheduling than fixed or stage-wise caps. | \n
Why these did not work. The headline pattern was variance, not a clear ceiling lift: preference prompts that were long or multi-part often hurt consistency (either by distracting the model from the baseline reasoning contract, or by encouraging expensive tool loops without a matching gain in agreement). Time policies that depended on global bank state or multi-phase continuation schedules were sensitive to problem order and tail latency, so they did not reliably beat the simpler agreement-gated, stage-wise deadline extension in finalnotebook. The shipped solution keeps the short preference prompt and focuses innovation on deadline allocation and parallel voting—a combination that still has considerable headroom (e.g., tighter thresholds, retrieval gating, or calibrated early stopping), but already matches or exceeds the best outcomes from the exploratory line above.
\nWe trained openai/gpt-oss-120b with Tinker using LoRA (rank 32 in saved run configs) and the gpt_oss_no_sysprompt renderer. Each run produced configs, metrics, checkpoints, and timing metadata (typical Tinker / Tinker Cookbook layout). The code is here Code on GitHub. Representative settings:
| Run (label) | \nRegime | \nNotes | \n
|---|---|---|
| SFT (on mixture of AIMO3 Dataset and Astral Math ), 100k sample size | \nSFT | \nlearning_rate 1e-4, linear schedule, 1 epoch, batch 8, max length 32768, JSONL SFT dataset, train_on_what: last assistant message | \n
| SFT (on 10k examples of AIMO3 Dataset) | \nSFT | \nSame LR / rank; 4000 steps; JSONL SFT dataset | \n
SFT loss (from logs). Our long SFT run on Tinker (1 epoch on 100k samples) shows training NLL generally trending downward (about 0.77 → 0.35, with noisy batch-level fluctuations). A compact view from sampled steps:
\n![\"Figure](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2Ff91361cbbee80463ebc962383e1a6d3c%2Ffig4_sft_loss_train_nll.png?generation=1776882386983144&alt=media\")
These runs document what we tried; they are not the primary story of the final submission.
\nMerged SFT model submission. For supervised fine-tuning we merged the trained adapter into base gpt-oss-120b using the Tinker Cookbook path (tinker_cookbook.weights.build_hf_model). That yields a different checkpoint; serving it means CFG.model_path points at the merged directory. §7.3 and Figures 1–3 list merged SFT as its own row/category (orange in Figure 1).
§7.3 aggregates repeated public leaderboard submits: base-model stacks (continuation, over-engineered prompts + RAG, banked time), merged SFT (different weights), and our two finalnotebook runs (38, 40). Qualitatively, continuation-enabled can reach 40 but also 33–34; merged SFT shows a wide two-run spread (26–37) and a lower mean than shipped; banked-time includes very low scores (30–32). We do not claim SOTA against all open-weight or commercial systems; the relevant comparison is within our own runs and the AIMO3 leaderboard context.
\nTable: raw public leaderboard scores (out of 50). Base-model rows use gpt-oss-120b in CFG.model_path (not the merged SFT directory). Merged SFT is a different checkpoint (Tinker adapter merged into full weights)—listed on its own row. Multiple submits reflect iterative tuning and harness variance—not i.i.d. experiments—so we emphasize ranges, means, and visuals rather than (p)-values.
| Approach | \nScores (each submit) | \n(n) | \nMin | \nMax | \nMean | \n
|---|---|---|---|---|---|
| Continuation-enabled | \n40, 37, 38, 39, 33, 34, 34 | \n7 | \n33 | \n40 | \n36.4 | \n
| Over-engineered prompts + RAG | \n37, 35, 42, 34, 41, 38, 38, 38 | \n8 | \n34 | \n42 | \n37.9 | \n
| Banked time + reserve | \n30, 39, 40, 35, 34, 32, 33, 40 | \n8 | \n30 | \n40 | \n35.4 | \n
| Merged SFT (Tinker) | \n37, 26 | \n2 | \n26 | \n37 | \n31.5 | \n
| Shipped: finalnotebook (base) | \n38, 40 | \n2 | \n38 | \n40 | \n39 | \n
Figure 1 — Distribution per approach. Box plots (median in red, mean as blue dashed line) with jittered points for every submit. Merged SFT (Tinker) uses an orange box (different model family). The shipped finalnotebook row is green; with two submits it is a short, high band.
\n![\"Figure](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2Fc0fd1da3cee96524efc8b0b17bd1df7d%2Ffig1_boxplot_public_lb.png?generation=1776769279836016&alt=media\")
Figure 2 — Mean score with approximate 95% CI for the mean (more submits ⇒ narrower intervals). The shipped mean is 39 (two submits at 38 and 40).
\n![\"Figure](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2Fc612e2412cd6e02bf338a7ea1cc8e5cd%2Ffig2_mean_ci_public_lb.png?generation=1776769481185523&alt=media\")
Figure 3 — Range (min–max bar) and mean (dot). Highlights risk: banked time spans 30–40; merged SFT spans 26–37 on only two submits; over-engineered + RAG spans 34–42; continuation-enabled spans 33–40.
\n![\"Figure](\"https://www.googleapis.com/download/storage/v1/b/kaggle-user-content/o/inbox%2F3937164%2Fe80fecb20c9e3933784184bbd22244c1%2Ffig3_range_mean_public_lb.png?generation=1776769529217792&alt=media\")
Reading. Among base-model exploratory lines, continuation-enabled reaches 40 on some submits but 33–34 on others; over-engineered + RAG (mean ~37.9) and banked-time show wide variance. Merged SFT (different weights), with only two public scores (37, 26), mean ~31.5, includes the lowest observed point in the whole panel (26)—so our SFT model submission did not beat finalnotebook (38, 40) under this evaluation budget. We preferred finalnotebook as the primary submission for strong scores, a clean inference story (§3–6), and avoidance of the worst tails under banked-time (30–32), continuation (33–34), and merged SFT (26).
\nWe intentionally do not include exhaustive ablation tables (fixed vs dynamic deadlines, grid over first_agreement_count / second_agreement_count, early_stop sweeps) or per-topic error breakdowns: those would require a frozen offline harness and labels beyond the scope of this narrative. The methodology in §3–6 is the reproducible artifact readers can build on.
The system prompt follows the public AIMO3 notebook text in https://www.kaggle.com/code/nihilisticneuralnet/44-50-let-me-over-cook. The submission implementation (dynamic deadlines, tooling, aggregation) is documented in finalnotebook. Tooling and inference patterns build on the competition’s reference stack (kaggle_evaluation, vLLM, Harmony).
| Item | \nNotebook reference / value | \n
|---|---|
| Model path | \nCFG.model_path (gpt-oss-120b Kaggle input path) | \n
| Served name | \nCFG.served_model_name | \n
| Seeds | \nCFG.seed; per-attempt seed derived in _process_attempt | \n
| Time limits | \nnotebook_limit, first_problem_timeout, second_problem_timeout, final_problem_timeout, base_problem_timeout | \n
| Attempts / workers | \nattempts, workers | \n
| Tool limits | \njupyter_timeout, max_tool_output_chars, etc. | \n
| Retrieval | \nCorpusRetriever + _load_corpus path list; default disables retriever | \n
| Stop / consensus | \nearly_stop, unanimous rule, _select_answer weighting | \n
| Primary submission | \nfinalnotebook — two public leaderboard submissions scored 38 and 40 (of 50) | \n
| Exploratory approaches (negative / unstable) | \nContinuations; layered developer prompts; merged SFT model submission; banked time — see §6 | \n
| Tinker SFT | \nRun configs and metrics as in §7.1 (saved alongside each training job) | \n
| Merge SFT adapter → full HF | \nTinker Cookbook build_hf_model; merged directory as CFG.model_path for merged SFT submissions (§7) | \n
Exact Kaggle notebook revision IDs and pip freeze outputs may change per session; reproduce from the published finalnotebook and the competition Docker / input snapshot at submit time.
\n§7.3 includes three figures: (1) box plot + jittered submits (merged SFT in orange, shipped in green), (2) mean with uncertainty on the mean, (3) min–max range with mean dot.
\nFull tool transcripts are long and environment-specific. Readers should run finalnotebook on Kaggle and inspect stdout / displayed attempt tables for end-to-end behavior (retrieval off in the default configuration).
\nsystem_prompt we reuse unchanged: https://www.kaggle.com/code/nihilisticneuralnet/44-50-let-me-over-cook.openai/gpt-oss-120b as referenced in Tinker configs.Adding to the comments by @sambealing below, I have a few additional questions:
\nRAG Efficiency: Do you think the RAG failure could partly be due to the model already having much of the relevant knowledge internalized, given its capacity at roughly 120B total parameters? One could argue that RAG may flood the context with additional information, making the actual problem statement less salient and reducing inference speed due to much longer input sequences before reasoning even begins.
Time Budget: You mention using a max_budget per problem. Is this constant empirically motivated, estimated by a predictor, or determined in some other way? In the write-up, you briefly mention a function that calculates it. Since time management is a crucial part of AIMO3, it would be helpful to describe this component in more detail.
SFT: Why did you choose to run SFT on an already RL-tuned model such as gpt-oss-120b? Was this mainly due to simplicity, time constraints, or another motivation? Methods such as GRPO seem more aligned with the model’s current stage of training. The reported results suggest that SFT may not have been the best fit and may have overfit to the training data format, potentially weakening some of the reasoning behavior learned during earlier RL stages. Do you agree with this interpretation? Additionally, did you encounter any issues when merging the LoRA weights, such as the vanishing-weight problem reported by other contestants?
Reproducibility: Please add the code required to reproduce your fine-tuning and experiments, so others can reproduce or adapt your workflow where possible.
Hi @philipvonderlind. Thanks for your questions. Below are my answers.
\nYes, I think that is a plausible part of the whole story. gpt-oss-120b is already strong on many IMO/AIME-style patterns without retrieval (the public baseline reaches 44/50), so naive RAG can add redundant context rather than missing knowledge. That matches what I saw: strategy blocks sometimes made the problem statement less salient, increased prefill cost (longer inputs before the first reasoning token), and sometimes steered the model off a path it would otherwise have taken. So I do not read the RAG failure as \"retrieval cannot help\", but as \"default-on retrieval is often net harmful on a model that already internalizes much of the corpus.\" That is why I disabled it in the final notebook due to time constraints in deeper investigations. I now believe conditional RAG (e.g. only at stage 2/3) could help on some hard/rare problems.
max_budget is not learned by a predictor and not fixed globally—it is recomputed per problem from remaining notebook time and problems left.
At each problem, the notebook computes:
\ntime_left = notebook_limit − elapsed_global\nreserved_time = (problems_remaining − 1) × base_problem_timeout (default 280 s per future problem)\nmax_budget = time_left − reserved_time \nthen clamped to [base_problem_timeout, final_problem_timeout] (defaults 280–900 s)
\nStage deadlines are derived from that cap:
\nStage 1: min(first_problem_timeout, max_budget) → 400 s
Stage 2: min(second_problem_timeout, max_budget) → 640 s
Stage 3: max_budget itself (up to 900 s)
notebook_limit is set from the competition’s remaining wall-clock budget (initialized around 17 400 s in CFG, adjusted at session start). The constants (280, 400, 640, 900) are empirically chosen heuristics based on notebook local runs on an upsampled version of the reference set: reserve a minimum slice for every remaining problem, allow extension only when votes are insufficient, and cap any single problem so later ones are not starved. It is simpler than my banked-time variant (§6), which was harder to tune.
Why SFT? Mainly practical constraints: I already had tool-integrated SFT traces (My Math Corpus Dataset) and a working Tinker SFT pipeline, while GRPO (which I also explored in side scripts) would have needed more rollout/tool infrastructure and compute before the submission deadline. SFT was the fastest path to test whether imitation on I traces would help—not a claim that SFT is theoretically optimal post-RL.
\nSo I partly agree with your interpretation. Lower public scores (37, 26; mean 31.5) are consistent with overfitting to the training data (and its format) and/or hurting behaviors that matter at inference (multi-attempt diversity, tool use under the full system prompt). But I would not confidently attribute everything to “RL capabilities erased” via SFT as it could also be possible that I did not merge Tinker adapters properly.
\nMerge / LoRA issues: Yes. Tinker adapters are not standard PEFT LoRA; build_hf_model from Tinker Cookbook did not work reliably on my side (including MXFP4 / key-naming issues on gpt-oss-120b). I used a two-stage merge with manual adapter weight renaming before merging. I did not fully debug whether there was a vanishing-weight issue that other teams reported, but a bad merge is my leading hypothesis for the weak SFT public scores. With only two public SFT submissions, I treat the low mean of scores as inconclusive and unreliable until I can serve a checkpoint merged through a validated procedure (and compare against GRPO checkpoints).
\nThe code is on GitHub.
", "votes": 0, "replies": [] } ] }, { "id": 3457729, "author_name": "Sam Bealing", "author_url": "", "post_date": "2026-05-14T08:39:08.940000", "content": "Hi, thank you for your submission to the Writeup Prize. @philipvonderlind and I will post some questions below that help us in assessing your submission more closely, as described in the updated discussion post.
\nLook forward to reading your responses on these.
", "votes": 0, "replies": [ { "id": 3457889, "author_name": "JK-Piece", "author_url": "", "post_date": "2026-05-14T15:00:13.177000", "content": "Thanks @sambealing for your questions. Below are answers.
\nYes. I reran the final notebook locally on the full AIME 2026 labeled set (30 problems). All 30 finalized at stage 1 (400 s budget); none extended to stage 2 or 3. Accuracy was 30/30 (100%). In every case, parallel attempts reached unanimous agreement before the stage-1 deadline, so the extension policy never triggered. See logs at the bottom of this reply.
\nI do not think the stability loss with RAG is universal. In the final submission I disabled RAG because naive injection hurt end-to-end reliability in my earlier runs (§4.3), especially alongside the heavy system prompt used.
\nWhen I did enable RAG, I added instructions to use retrieved strategies and hints but not to copy them blindly, with verification required first. On the 10 reference problems provided, retrieval sometimes surfaced useful techniques and shortened reasoning; on other problems, hints steered the model onto wrong paths (contamination) and wrong answers.
\nSo I believe some problem types can benefit from RAG; the issue is unreliable net effect, not RAG always hurting. A natural extension would be conditional RAG---e.g. activate retrieval only when a problem reaches stage 2 or 3, appending references to ongoing conversations for the hardest cases rather than injecting them on every problem upfront.
\nI can see two main explanations. First (most probable hypothesis): the merged checkpoint may not reflect a correct Tinker→HF conversion—Tinker adapters are not standard PEFT LoRA, and I may not have applied the right merge procedure before serving. In fact the build_hf_model method provided by Tinker CookBook did not work on my side. So I had to perform a two stage merging procedure which involved renaming some specific adapter weight names first.
Second: overfitting to the training data, combined with limited offline evaluation before my two public submissions (with scores 37, 26). The decreasing NLL (Fig. 4) only shows that the model somehow learned the training data, but it does not guarantee a strong end model. With only 2 submissions, I treat the low score mean as inconclusive and unreliable until I validate a properly merged checkpoint.
", "votes": 0, "replies": [ { "id": 3457892, "author_name": "JK-Piece", "author_url": "", "post_date": "2026-05-14T15:08:04.650000", "content": "ID: 8742f9\nQuestion: The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex…
\nThe figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Let $N$ be the number of paths the bug can take from the lower left corner ($A$) to the upper right corner ($B$). One such path from $A$ to $B$ is shown by the thick line segments in the figure. Find $\\sqrt{N}$.
\n[asy]\nsize(12cm);\ndraw((0, 0)--(0, 1));\nfor (int i = 0; i <= 9; i += 1)\n{\ndraw((i, 1)--(i + 1, 1)--(i + 1, 0)--(i, 0)--(i + 1, 1));\n}\ndraw((0, 0)--(0, 1)--(2, 1)--(2, 0)--(3, 1)--(3, 0)--(4, 1)--(4, 0)--(5, 0)--(5, 1)--(6, 1)--(6, 0)--(7, 1)--(8, 1)--(8, 0)--(10, 0)--(10, 1), red + linewidth(2));\nlabel(\"$A$\", (0, 0), dir(180));\nlabel(\"$B$\", (10, 1), dir(0));\n[/asy] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n5024 | \n4 | \n0 | \n0.811 | \n243 | \nanswered | \n59.13 | \n
| 1 | \n7 | \n6260 | \n2 | \n0 | \n0.903 | \n243 | \nanswered | \n72.26 | \n
| 2 | \n3 | \n6497 | \n2 | \n0 | \n0.798 | \n243 | \nanswered | \n74.67 | \n
| 3 | \n8 | \n7256 | \n5 | \n1 | \n0.794 | \n243 | \nanswered | \n82.55 | \n
UNANIMOUS ANSWER: 243
\nAnswer: 243 | Ground Truth: 243 | ✅
\nID: 5cc894\nQuestion: Let $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the…
\nLet $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the $2a + 2b - 4$ cells that surround an $(a - 2) \\times (b - 2)$ (possibly empty) rectangle of cells in the grid. For example, the following diagram shows a way to partition a $6 \\times 6$ grid of cells into $4$ cell loops.
\n| P P P P | Y Y |\n| P | R R | P | Y | Y |\n| P | R R | P | Y | Y |\n| P P P P | Y | Y |\n| G G G G | Y | Y |\n| G G G G | Y Y |
\nFind the number of ways to partition a $10 \\times 10$ grid of cells into $5$ cell loops so that every cell of the grid belongs to exactly one cell loop. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n8177 | \n9 | \n0 | \n0.891 | \n83 | \nanswered | \n98.10 | \n
| 1 | \n4 | \n9962 | \n4 | \n0 | \n0.844 | \n179 | \nanswered | \n109.96 | \n
| 2 | \n1 | \n9566 | \n6 | \n1 | \n0.784 | \n83 | \nanswered | \n114.32 | \n
| 3 | \n8 | \n10538 | \n8 | \n1 | \n0.849 | \n83 | \nanswered | \n128.12 | \n
| 4 | \n5 | \n10843 | \n21 | \n3 | \n0.779 | \n83 | \nanswered | \n130.85 | \n
UNANIMOUS ANSWER: 83
\nAnswer: 83 | Ground Truth: 83 | ✅
\nID: 744366\nQuestion: Let $S$ denote the value of the infinite sum\n[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$…
\nLet $S$ denote the value of the infinite sum\n[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n4410 | \n8 | \n0 | \n0.737 | \n669 | \nanswered | \n50.35 | \n
| 1 | \n8 | \n4493 | \n12 | \n1 | \n0.663 | \n669 | \nanswered | \n51.79 | \n
| 2 | \n3 | \n4585 | \n6 | \n0 | \n0.625 | \n669 | \nanswered | \n51.88 | \n
| 3 | \n2 | \n5249 | \n5 | \n0 | \n0.670 | \n669 | \nanswered | \n58.13 | \n
UNANIMOUS ANSWER: 669
\nAnswer: 669 | Ground Truth: 669 | ✅
\nID: b80bba\nQuestion: \nFind the number of functions $\\pi$ mapping the set $A ={1,2,3,4,5,6}$ onto $A$ such that for every $a \\in A,$\n[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n]…
\nFind the number of functions $\\pi$ mapping the set $A ={1,2,3,4,5,6}$ onto $A$ such that for every $a \\in A,$\n[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n1651 | \n2 | \n0 | \n0.660 | \n396 | \nanswered | \n18.17 | \n
| 1 | \n5 | \n1773 | \n3 | \n0 | \n0.723 | \n396 | \nanswered | \n19.55 | \n
| 2 | \n1 | \n2056 | \n2 | \n0 | \n0.669 | \n396 | \nanswered | \n22.35 | \n
| 3 | \n3 | \n2127 | \n3 | \n0 | \n0.607 | \n396 | \nanswered | \n23.11 | \n
UNANIMOUS ANSWER: 396
\nAnswer: 396 | Ground Truth: 396 | ✅
\nID: bce9ef\nQuestion: Let $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D…
\nLet $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D$ lie on the same side of line $AE.$ Suppose further that $AB$ is an integer with $AB < 2026$ and the area of pentagon $ABCDE$ is an integer multiple of $16.$ Find the number of possible values of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n12334 | \n3 | \n0 | \n0.755 | \n503 | \nanswered | \n136.30 | \n
| 1 | \n4 | \n12760 | \n13 | \n1 | \n0.761 | \n1002 | \nanswered | \n143.73 | \n
| 2 | \n5 | \n13789 | \n5 | \n0 | \n0.754 | \n503 | \nanswered | \n153.05 | \n
| 3 | \n8 | \n13974 | \n7 | \n0 | \n0.751 | \n503 | \nanswered | \n155.26 | \n
| 4 | \n6 | \n15984 | \n9 | \n2 | \n0.729 | \n503 | \nanswered | \n172.94 | \n
UNANIMOUS ANSWER: 503
\nAnswer: 503 | Ground Truth: 503 | ✅
\nID: cba4eb\nQuestion: Find the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b….
\nFind the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n2144 | \n4 | \n0 | \n0.736 | \n70 | \nanswered | \n24.08 | \n
| 1 | \n2 | \n2339 | \n3 | \n0 | \n0.794 | \n70 | \nanswered | \n26.17 | \n
| 2 | \n1 | \n2653 | \n5 | \n0 | \n0.764 | \n70 | \nanswered | \n29.58 | \n
| 3 | \n5 | \n2703 | \n4 | \n0 | \n0.780 | \n70 | \nanswered | \n30.02 | \n
UNANIMOUS ANSWER: 70
\nAnswer: 70 | Ground Truth: 70 | ✅
\nID: cda722\nQuestion: For integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n[\na_1 + …
\nFor integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n[\na_1 + a_2 + a_3 + \\cdots + a_n = 12 \\quad \\text{and} \\quad a_1 \\circ a_2 \\circ a_3 \\circ \\cdots \\circ a_n = 0\n]\nwhere the operations are performed from left to right; that is, $a_1 \\circ a_2 \\circ a_3$ means $(a_1 \\circ a_2) \\circ a_3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n2209 | \n3 | \n0 | \n0.875 | \n157 | \nanswered | \n25.53 | \n
| 1 | \n3 | \n2249 | \n4 | \n0 | \n0.837 | \n157 | \nanswered | \n26.15 | \n
| 2 | \n8 | \n2406 | \n3 | \n0 | \n0.845 | \n157 | \nanswered | \n27.62 | \n
| 3 | \n7 | \n2502 | \n3 | \n0 | \n0.897 | \n157 | \nanswered | \n28.55 | \n
UNANIMOUS ANSWER: 157
\nAnswer: 157 | Ground Truth: 157 | ✅
\nID: f7b4a9\nQuestion: Let $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC…
\nLet $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC}$ at $D.$ Let $E \\neq A$ and $F \\neq A$ be the intersections of $\\omega$ with segments $\\overline{AB}$ and $\\overline{AC},$ respectively. Suppose that $AB = 200, AC = 225,$ and all of $AE, AF, BD,$ and $CD$ are positive integers. Find the sum of all possible values of $BC.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n5727 | \n2 | \n0 | \n0.758 | \n850 | \nanswered | \n60.41 | \n
| 1 | \n2 | \n6207 | \n2 | \n0 | \n0.708 | \n850 | \nanswered | \n65.95 | \n
| 2 | \n6 | \n6343 | \n3 | \n0 | \n0.540 | \n850 | \nanswered | \n67.47 | \n
| 3 | \n3 | \n6727 | \n2 | \n0 | \n0.712 | \n850 | \nanswered | \n71.45 | \n
UNANIMOUS ANSWER: 850
\nAnswer: 850 | Ground Truth: 850 | ✅
\nID: f3d7b7\nQuestion: Find the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$…
\nFind the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 3 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n2711 | \n5 | \n0 | \n0.687 | \n50 | \nanswered | \n29.22 | \n
| 1 | \n4 | \n3372 | \n1 | \n0 | \n0.698 | \n50 | \nanswered | \n35.80 | \n
| 2 | \n2 | \n3659 | \n8 | \n0 | \n0.672 | \n50 | \nanswered | \n39.63 | \n
| 3 | \n3 | \n3806 | \n2 | \n0 | \n0.701 | \n50 | \nanswered | \n40.41 | \n
UNANIMOUS ANSWER: 50
\nAnswer: 50 | Ground Truth: 50 | ✅
\nID: f969f6\nQuestion: Call finite sets of integers $S$ and $T$ cousins if
\n$S$ and $T$ have the same number of elements,
$S$ and $T$ are disjoint, and
the elements of $S$ can be paired with the elements of $T$ so that…
\nCall finite sets of integers $S$ and $T$ cousins if
$S$ and $T$ have the same number of elements,
$S$ and $T$ are disjoint, and
the elements of $S$ can be paired with the elements of $T$ so that the elements in each pair differ by exactly $1$.\nFor example, ${1,2,5}$ and ${0,3,4}$ are cousins. Suppose that the set $S$ has exactly $4040$ cousins. Find the least number of elements the set $S$ can have. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
Stage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n11691 | \n6 | \n0 | \n0.868 | \n107 | \nanswered | \n130.24 | \n
| 1 | \n4 | \n14143 | \n16 | \n2 | \n0.859 | \n107 | \nanswered | \n167.23 | \n
| 2 | \n3 | \n16430 | \n6 | \n0 | \n0.893 | \n107 | \nanswered | \n184.39 | \n
| 3 | \n7 | \n19128 | \n22 | \n3 | \n0.821 | \n107 | \nanswered | \n220.86 | \n
UNANIMOUS ANSWER: 107
\nAnswer: 107 | Ground Truth: 107 | ✅
\nID: f3d797\nQuestion: A plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the p…
\nA plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the plane clockwise through angle $\\theta$ around point $A^\\prime$ to point $B^\\prime$. Suppose that $AB^\\prime = \\frac{4}{3}$. The value of $\\cos \\theta$ can be written as $\\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n1909 | \n1 | \n0 | \n0.559 | \n65 | \nanswered | \n20.36 | \n
| 1 | \n7 | \n1900 | \n4 | \n0 | \n0.601 | \n65 | \nanswered | \n20.50 | \n
| 2 | \n6 | \n2084 | \n2 | \n0 | \n0.654 | \n65 | \nanswered | \n22.18 | \n
| 3 | \n5 | \n2500 | \n2 | \n0 | \n0.425 | \n65 | \nanswered | \n25.94 | \n
UNANIMOUS ANSWER: 65
\nAnswer: 65 | Ground Truth: 65 | ✅
\nID: 0e674f\nQuestion: A hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk suc…
\nA hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk such that a sphere of radius $42$ can be placed on top of the disk at $P$ and lie completely inside the hemisphere. The area of $\\mathcal T$ divided by the area of the disk is $\\tfrac pq,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n2501 | \n2 | \n0 | \n0.778 | \n79 | \nanswered | \n27.00 | \n
| 1 | \n7 | \n2796 | \n2 | \n0 | \n0.622 | \n79 | \nanswered | \n31.03 | \n
| 2 | \n5 | \n2956 | \n4 | \n0 | \n0.784 | \n79 | \nanswered | \n32.15 | \n
| 3 | \n1 | \n3050 | \n6 | \n0 | \n0.789 | \n79 | \nanswered | \n33.80 | \n
UNANIMOUS ANSWER: 79
\nAnswer: 79 | Ground Truth: 79 | ✅
\nID: f2f2d9\nQuestion: The integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and le…
\nThe integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and let $M$ be the sum of the absolute differences between adjacent cells. That is,\n[\nM = \\sum^8_{i=1} \\sum^7_{j=1} (|a_{i,j+1} - a_{i,j}| + |a_{j+1,i} - a_{j,i}|).\n]\nFind the remainder when the maximum possible value of $M$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 24 times…\n🐍 Executing Python code 13 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 17 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n9031 | \n12 | \n2 | \n0.745 | \n896 | \nanswered | \n104.73 | \n
| 1 | \n8 | \n15183 | \n9 | \n0 | \n0.739 | \n896 | \nanswered | \n176.24 | \n
| 2 | \n4 | \n16049 | \n19 | \n4 | \n0.613 | \n896 | \nanswered | \n190.60 | \n
| 3 | \n2 | \n15551 | \n16 | \n2 | \n0.738 | \n896 | \nanswered | \n196.55 | \n
UNANIMOUS ANSWER: 896
\nAnswer: 896 | Ground Truth: 896 | ✅
\nID: d5d04c\nQuestion: Find the greatest integer $n$ such that the cubic polynomial\n[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta…
\nFind the greatest integer $n$ such that the cubic polynomial\n[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta$, and $\\gamma$ are complex numbers, and there are exactly seven different possible values for $\\alpha + \\beta + \\gamma$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 8 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n8782 | \n13 | \n1 | \n0.715 | \n132 | \nanswered | \n98.54 | \n
| 1 | \n6 | \n9559 | \n13 | \n1 | \n0.625 | \n132 | \nanswered | \n107.40 | \n
| 2 | \n5 | \n12093 | \n14 | \n3 | \n0.682 | \n132 | \nanswered | \n146.25 | \n
| 3 | \n3 | \n13308 | \n14 | \n1 | \n0.703 | \n132 | \nanswered | \n147.31 | \n
UNANIMOUS ANSWER: 132
\nAnswer: 132 | Ground Truth: 132 | ✅
\nID: e6ea64\nQuestion: Let $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$….
\nLet $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 8 times…\n🐍 Executing Python code 4 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n1675 | \n5 | \n0 | \n0.506 | \n244 | \nanswered | \n18.56 | \n
| 1 | \n1 | \n2406 | \n2 | \n0 | \n0.478 | \n244 | \nanswered | \n26.29 | \n
| 2 | \n8 | \n2524 | \n4 | \n0 | \n0.620 | \n244 | \nanswered | \n27.75 | \n
| 3 | \n2 | \n2953 | \n9 | \n0 | \n0.486 | \n244 | \nanswered | \n32.03 | \n
UNANIMOUS ANSWER: 244
\nAnswer: 244 | Ground Truth: 244 | ✅
\nID: 8ff9e9\nQuestion: Find the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:
\n$a_k \\in {1,2,3}$ for all $k.$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$
$a_1a_2 a…
\nFind the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:
$a_k \\in {1,2,3}$ for all $k.$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$
$a_1a_2 a_4 + a_2a_3a_5 + a_3a_4 a_6 + a_4 a_5 a_7 + a_5 a_6 a_1 + a_6 a_7 a_2 + a_7 a_1 a_3$ is a multiple of $3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
Stage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n1200 | \n2 | \n0 | \n0.638 | \n393 | \nanswered | \n13.87 | \n
| 1 | \n3 | \n1429 | \n2 | \n0 | \n0.748 | \n393 | \nanswered | \n16.41 | \n
| 2 | \n5 | \n1593 | \n3 | \n0 | \n0.702 | \n393 | \nanswered | \n18.12 | \n
| 3 | \n1 | \n1709 | \n2 | \n0 | \n0.740 | \n393 | \nanswered | \n19.08 | \n
UNANIMOUS ANSWER: 393
\nAnswer: 393 | Ground Truth: 393 | ✅
\nID: ac19f7\nQuestion: A standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The…
\nA standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The probability that Alice and Bob each receive at least two coins before Carol receives any coins can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $100m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n2868 | \n2 | \n0 | \n0.546 | \n754 | \nanswered | \n33.70 | \n
| 1 | \n8 | \n3703 | \n7 | \n0 | \n0.696 | \n754 | \nanswered | \n43.71 | \n
| 2 | \n7 | \n3781 | \n4 | \n0 | \n0.731 | \n754 | \nanswered | \n44.75 | \n
| 3 | \n5 | \n5266 | \n7 | \n0 | \n0.552 | \n754 | \nanswered | \n57.80 | \n
UNANIMOUS ANSWER: 754
\nAnswer: 754 | Ground Truth: 754 | ✅
\nID: d47975\nQuestion: Isosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both …
\nIsosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both triangles have integer lengths. Find the minimum possible value of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n6887 | \n4 | \n0 | \n0.634 | \n245 | \nanswered | \n74.73 | \n
| 1 | \n1 | \n7072 | \n6 | \n1 | \n0.730 | \n245 | \nanswered | \n77.36 | \n
| 2 | \n4 | \n7834 | \n9 | \n0 | \n0.626 | \n245 | \nanswered | \n85.49 | \n
| 3 | \n3 | \n7305 | \n6 | \n1 | \n0.667 | \n245 | \nanswered | \n85.61 | \n
UNANIMOUS ANSWER: 245
\nAnswer: 245 | Ground Truth: 245 | ✅
\nID: 0015eb\nQuestion: Let $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpend…
\nLet $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpendicular $\\overline{BC},$ with $A'$ and $B$ not on the same side of line $B'C'.$ Find the integer closest to the area of hexagon $AA'CC'BB'.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 8 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 9 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n6640 | \n10 | \n1 | \n0.654 | \n156 | \nanswered | \n76.62 | \n
| 1 | \n6 | \n8349 | \n20 | \n2 | \n0.639 | \n156 | \nanswered | \n98.50 | \n
| 2 | \n2 | \n9434 | \n14 | \n1 | \n0.663 | \n156 | \nanswered | \n109.25 | \n
| 3 | \n4 | \n9341 | \n20 | \n1 | \n0.672 | \n156 | \nanswered | \n109.50 | \n
UNANIMOUS ANSWER: 156
\nAnswer: 156 | Ground Truth: 156 | ✅
\nID: 1eaadd\nQuestion: Triangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four sph…
\nTriangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four spheres, all on the same side of $\\mathcal P,$ have radii $1, 2, 3,$ and $r$ and are tangent to $\\mathcal P$ at points $A, B, C,$ and $D,$ respectively. The four spheres are also each tangent to a second plane $\\mathcal T$ and are all on the same side of $\\mathcal T.$ The value of $r$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 17 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n7008 | \n9 | \n0 | \n0.716 | \n161 | \nanswered | \n79.21 | \n
| 1 | \n7 | \n7594 | \n10 | \n0 | \n0.655 | \n161 | \nanswered | \n86.17 | \n
| 2 | \n1 | \n8101 | \n11 | \n1 | \n0.592 | \n161 | \nanswered | \n91.21 | \n
| 3 | \n2 | \n9021 | \n9 | \n1 | \n0.526 | \n161 | \nanswered | \n99.51 | \n
UNANIMOUS ANSWER: 161
\nAnswer: 161 | Ground Truth: 161 | ✅
\nID: 4d8666\nQuestion: Patrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after…
\nPatrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after Tanya left, Jose started bicycling along the same road from school to the park. Tanya ran at a constant rate of $2$ miles per hour faster than Patrick walked, Jose bicycled at a constant rate of $7$ miles per hour faster than Tanya ran, and all three arrived at the park at the same time. The distance from the school to the park is $\\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n1231 | \n1 | \n0 | \n0.487 | \n277 | \nanswered | \n12.75 | \n
| 1 | \n2 | \n1423 | \n1 | \n0 | \n0.514 | \n277 | \nanswered | \n14.91 | \n
| 2 | \n8 | \n1436 | \n1 | \n0 | \n0.468 | \n277 | \nanswered | \n14.99 | \n
| 3 | \n5 | \n1518 | \n2 | \n0 | \n0.551 | \n277 | \nanswered | \n15.80 | \n
UNANIMOUS ANSWER: 277
\nAnswer: 277 | Ground Truth: 277 | ✅
\nID: 7968ea\nQuestion: A real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$?…
\nA real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$? Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 3 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n2005 | \n5 | \n0 | \n0.534 | \n441 | \nanswered | \n22.34 | \n
| 1 | \n4 | \n2857 | \n5 | \n1 | \n0.491 | \n441 | \nanswered | \n31.46 | \n
| 2 | \n6 | \n2808 | \n8 | \n1 | \n0.528 | \n441 | \nanswered | \n31.46 | \n
| 3 | \n8 | \n3186 | \n5 | \n1 | \n0.454 | \n441 | \nanswered | \n34.74 | \n
UNANIMOUS ANSWER: 441
\nAnswer: 441 | Ground Truth: 441 | ✅
\nID: 282256\nQuestion: Find the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a numb…
\nFind the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a number whose representation reads the same from left to right as from right to left. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n3449 | \n2 | \n0 | \n0.777 | \n62 | \nanswered | \n39.15 | \n
| 1 | \n6 | \n3395 | \n4 | \n1 | \n0.793 | \n62 | \nanswered | \n44.15 | \n
| 2 | \n1 | \n3957 | \n7 | \n0 | \n0.749 | \n62 | \nanswered | \n44.98 | \n
| 3 | \n4 | \n4078 | \n2 | \n0 | \n0.644 | \n62 | \nanswered | \n45.28 | \n
UNANIMOUS ANSWER: 62
\nAnswer: 62 | Ground Truth: 62 | ✅
\nID: 51697b\nQuestion: Consider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The…
\nConsider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 11 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 8 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n8545 | \n13 | \n0 | \n0.645 | \n223 | \nanswered | \n98.52 | \n
| 1 | \n1 | \n8768 | \n9 | \n0 | \n0.585 | \n223 | \nanswered | \n100.38 | \n
| 2 | \n7 | \n9027 | \n12 | \n1 | \n0.623 | \n223 | \nanswered | \n104.21 | \n
| 3 | \n5 | \n9244 | \n10 | \n0 | \n0.782 | \n223 | \nanswered | \n105.55 | \n
UNANIMOUS ANSWER: 223
\nAnswer: 223 | Ground Truth: 223 | ✅
\nID: 652b17\nQuestion: An urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability …
\nAn urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability that exactly $4$ of them are red equals the probability that exactly $5$ of them are red. Find the sum of the five least values of $n$ for which this is possible. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n1686 | \n3 | \n0 | \n0.578 | \n190 | \nanswered | \n17.93 | \n
| 1 | \n2 | \n2172 | \n3 | \n0 | \n0.591 | \n190 | \nanswered | \n23.36 | \n
| 2 | \n7 | \n2213 | \n2 | \n0 | \n0.543 | \n190 | \nanswered | \n23.74 | \n
| 3 | \n4 | \n2391 | \n3 | \n0 | \n0.574 | \n190 | \nanswered | \n25.48 | \n
UNANIMOUS ANSWER: 190
\nAnswer: 190 | Ground Truth: 190 | ✅
\nID: 7a1a55\nQuestion: For each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, the…
\nFor each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, then $b=8$, and $72$ as a numeral in base $8$ equals $7\\cdot 8+2=58$; therefore $f(72)=58$. Find the number of positive integers $n$ less than $1000$ such that $f(n)=n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n2361 | \n3 | \n0 | \n0.877 | \n279 | \nanswered | \n25.78 | \n
| 1 | \n1 | \n2679 | \n5 | \n0 | \n0.780 | \n279 | \nanswered | \n29.39 | \n
| 2 | \n7 | \n2998 | \n4 | \n0 | \n0.707 | \n279 | \nanswered | \n32.58 | \n
| 3 | \n4 | \n3046 | \n2 | \n0 | \n0.635 | \n279 | \nanswered | \n32.90 | \n
UNANIMOUS ANSWER: 279
\nAnswer: 279 | Ground Truth: 279 | ✅
\nID: c6d3cf\nQuestion: In an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be…
\nIn an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be expressed as $m \\sqrt n,$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 11 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 17 times…\n🐍 Executing Python code 16 times…\n🐍 Executing Python code 14 times…\n🐍 Executing Python code 21 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 22 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n8395 | \n12 | \n2 | \n0.545 | \n681 | \nanswered | \n96.43 | \n
| 1 | \n4 | \n15102 | \n17 | \n3 | \n0.574 | \n681 | \nanswered | \n185.96 | \n
| 2 | \n6 | \n15529 | \n20 | \n6 | \n0.588 | \n681 | \nanswered | \n186.80 | \n
| 3 | \n5 | \n16184 | \n17 | \n2 | \n0.572 | \n681 | \nanswered | \n195.80 | \n
UNANIMOUS ANSWER: 681
\nAnswer: 681 | Ground Truth: 681 | ✅
\nID: 5e99c1\nQuestion: Joanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then …
\nJoanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then rolls the die again, places the sticker labeled 2 on the top face, and continues this process to place the rest of the stickers in order. If the die ever lands with a sticker already on its top face, the new sticker is placed to cover the old sticker. Let $p$ be the conditional probability that at the end of the process exactly one face has been left blank, given that all the even-numbered stickers are visible on faces of the die. Then $p$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 7 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n6043 | \n3 | \n0 | \n0.765 | \n29 | \nanswered | \n66.51 | \n
| 1 | \n5 | \n6129 | \n5 | \n1 | \n0.885 | \n29 | \nanswered | \n74.45 | \n
| 2 | \n1 | \n7787 | \n1 | \n0 | \n0.792 | \n29 | \nanswered | \n84.05 | \n
| 3 | \n2 | \n8489 | \n6 | \n0 | \n0.838 | \n29 | \nanswered | \n91.10 | \n
UNANIMOUS ANSWER: 29
\nAnswer: 29 | Ground Truth: 29 | ✅
\nID: 77bffe\nQuestion: Find the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms….
\nFind the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n1256 | \n2 | \n0 | \n0.803 | \n178 | \nanswered | \n13.77 | \n
| 1 | \n2 | \n1629 | \n2 | \n0 | \n0.761 | \n178 | \nanswered | \n17.64 | \n
| 2 | \n3 | \n1974 | \n2 | \n0 | \n0.786 | \n178 | \nanswered | \n21.36 | \n
| 3 | \n1 | \n2330 | \n3 | \n0 | \n0.727 | \n178 | \nanswered | \n24.74 | \n
UNANIMOUS ANSWER: 178
\nAnswer: 178 | Ground Truth: 178 | ✅
\nID: c4ae02\nQuestion: For each positive integer $r$ less than $502,$ define\n[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all bino…
\nFor each positive integer $r$ less than $502,$ define\n[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all binomial coefficients of the form $\\binom{10000}{k}$ for which $0\\le k\\le 10000$ and $k-r$ is a multiple of $502.$ Find the number of integers in the list $S_0,S_1,\\dots,S_{501}$ that are multiples of the prime number $503.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 9 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n5037 | \n6 | \n0 | \n0.682 | \n39 | \nanswered | \n62.10 | \n
| 1 | \n1 | \n5244 | \n7 | \n1 | \n0.644 | \n39 | \nanswered | \n64.32 | \n
| 2 | \n8 | \n5498 | \n15 | \n2 | \n0.702 | \n39 | \nanswered | \n67.36 | \n
| 3 | \n3 | \n5896 | \n7 | \n0 | \n0.590 | \n39 | \nanswered | \n69.41 | \n
UNANIMOUS ANSWER: 39
\nAnswer: 39 | Ground Truth: 39 | ✅
\n==================================================
\nCorrect: 30/30
\nDetails:\n 8742f9: pred=243, gt=243 ✅\n 5cc894: pred=83, gt=83 ✅\n 744366: pred=669, gt=669 ✅\n b80bba: pred=396, gt=396 ✅\n bce9ef: pred=503, gt=503 ✅\n cba4eb: pred=70, gt=70 ✅\n cda722: pred=157, gt=157 ✅\n f7b4a9: pred=850, gt=850 ✅\n f3d7b7: pred=50, gt=50 ✅\n f969f6: pred=107, gt=107 ✅\n f3d797: pred=65, gt=65 ✅\n 0e674f: pred=79, gt=79 ✅\n f2f2d9: pred=896, gt=896 ✅\n d5d04c: pred=132, gt=132 ✅\n e6ea64: pred=244, gt=244 ✅\n 8ff9e9: pred=393, gt=393 ✅\n ac19f7: pred=754, gt=754 ✅\n d47975: pred=245, gt=245 ✅\n 0015eb: pred=156, gt=156 ✅\n 1eaadd: pred=161, gt=161 ✅\n 4d8666: pred=277, gt=277 ✅\n 7968ea: pred=441, gt=441 ✅\n 282256: pred=62, gt=62 ✅\n 51697b: pred=223, gt=223 ✅\n 652b17: pred=190, gt=190 ✅\n 7a1a55: pred=279, gt=279 ✅\n c6d3cf: pred=681, gt=681 ✅\n 5e99c1: pred=29, gt=29 ✅\n 77bffe: pred=178, gt=178 ✅\n c4ae02: pred=39, gt=39 ✅
", "votes": 0, "replies": [] } ] } ] } ], "comment_count": 5, "messages": [ { "id": 3446312, "postDate": "2026-04-21T11:48:07.240Z", "content": "Abstract. We build on openai/gpt-oss-120b in its base (released) form because the community has already shown it is strong on this task: the public baseline notebook 44-50 Let Me Over Cook—using the same family of model and tooling—achieved 44/50 on the public leaderboard, i.e. most problems are within reach of the unfine-tuned checkpoint with a good harness. In our own model comparisons, Qwen3 and Qwen3.5 did not match that bar and were discarded; we also set aside NVIDIA Nemotron (released later) after it looked weaker in our internal checks. Tinker SFT on gpt-oss-120b with a merged adapter, by contrast, reduced our public scores (37, 26 vs 38, 40 for the same submission family; see §7.3), so our final submission keeps the base weights. This submission solves AIMO-style problems with that fixed backbone served via vLLM, the same system prompt as 44-50 Let Me Over Cook, a short preference prompt that pushes step-by-step reasoning and heavy use of a stateful Python (Jupyter) tool, and parallel stochastic attempts with entropy-aware aggregation. The main methodological contribution is dynamic deadline allocation: a three-stage time budget that extends only when agreement among completed attempts is insufficient at stage boundaries—reducing wasted wall-clock on problems that already stabilize early. We also implemented a hybrid dense + BM25 retriever over a compact math corpus (techniques, example problems, and hints); it is disabled in the final submission when empirical checks showed net harm. We summarize other approaches we tried (continuations, layered developer prompts, global time-banking, and related SFT experiments) that did not match the stability of the final pipeline (finalnotebook). The shipped notebook is finalnotebook with base gpt-oss-120b; details on merged-SFT public scores are in §7. This writeup is methodology-first; we do not rely on extra plots or exhaustive numeric tables.
Reproducibility statement. All implementation details below are grounded in this submission’s Kaggle notebook finalnotebook. Tinker training runs are summarized in §7 (configs and metrics were saved with each run).
\nEDIT on May 13th, 2026. After the release of the private leaderboard, our final submission scored 41.0/50, which aligns with our arguments in this writeup about stability since we scored 40/50 on the public leaderboard. Also added a link to the code on GitHub.
\nAIMO3 requires producing a single integer answer in ({0,\\ldots,99999}) under a global notebook time limit and per-problem compute constraints.
\nBackbone and fine-tuning. The competition implementation standardizes on openai/gpt-oss-120b: a public AIMO3 notebook (44-50 Let Me Over Cook) already demonstrates 44/50 on the public leaderboard with this model, which motivates investing in inference-time design rather than replacing the base weights. We evaluated Qwen3 / Qwen3.5 and NVIDIA Nemotron as alternatives; they did not justify switching away from gpt-oss-120b under our resource and tooling constraints. We also fine-tuned gpt-oss-120b (merged SFT adapter) and found worse public results than the base model with the same harness (§7.3), so the primary submission uses the original checkpoint.
The approach treats each problem as a multi-attempt inference problem: several independent runs explore the space of reasoning trajectories; we then stop early when a strong consensus emerges, and otherwise allocate more time only when the evidence suggests the problem is still “open.”
\nKey Contributions (as implemented in the submission).
\nstdout can be extremely long—on the order of tens of thousands of model-side tokens in severe cases if passed through verbatim. Without truncation, tool returns alone can consume the context / generation budget, and otherwise valuable attempts may stop early (e.g., context exhaustion) before a final answer is produced.The competition submission is finalnotebook (38 / 40) (§7.3).
\ngpt-oss-120b (base checkpoint) from the Kaggle model path in the notebook (CFG.model_path)—the same line already shown to reach most problems in public (e.g. 44/50 in 44-50 Let Me Over Cook); we avoid merged SFT here because it hurt our public scores relative to this base (§1, §7.3).vLLM OpenAI-compatible API (python -m vllm.entrypoints.openai.api_server), with Harmony encoding for GPT-OSS (HarmonyEncodingName.HARMONY_GPT_OSS).print()-based output, SymPy/NumPy guidance, concise dumps, and timeout-aware splitting—plus a length cap on tool returns (§5.5).The end-to-end path is OpenAI-compatible client → Harmony template (GPT-OSS) → streaming completion with tool calls → Jupyter execution. We omit standalone architecture figures; the flow is summarized in Appendix E.
\nThe system_prompt in CFG is unchanged from the public AIMO3 starter notebook 44-50 Let Me Over Cook, by nihilisticneuralnet: it defines an IMO-level solver persona, a structured problem-solving loop (understand → explore → plan → execute → verify), verification requirements, and the required output format (\\boxed{} with an integer in ([0,99999])).
The tool_prompt is tuned for efficient computation in a stateful Python (Jupyter) notebook: it lists available libraries (math, numpy, sympy, etc.) and instructs the model to print() results (last expression is not auto-displayed), to use SymPy for exact symbolic work and NumPy for fast numeric work, to estimate complexity before large searches (prefer better-than-naive algorithms), to keep printed output concise (summarize large dumps), and to split or simplify work when a cell hits the per-call time limit. This steers tool use toward fewer wasteful megabyte-scale printouts and more targeted code.
A compact preference_prompt is appended to the user side of the conversation:
\n\nSolve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing
\n\\boxed{}.
This is intentionally short: it does not replace the system prompt; it steers sampling toward explicit computation and independent verification—aligned with competition scoring (correctness over eloquence).
\nThe notebook defines CorpusRetriever, a hybrid retriever:
topics, techniques, recognition_signals).sentence-transformers when the embedding model path exists; cosine similarity min-max normalized and fused with BM25:\n[\n\\text{score} = 0.5 \\cdot \\text{bm25}{\\text{norm}} + 0.5 \\cdot \\text{dense}{\\text{norm}}\n]\n(falls back to BM25-only if dense is unavailable).Retrieved items are formatted as short “strategy blocks” including technique/meta/similar-problem text, topics, insights, optional solution sketch, and computational hints.
\nCorpus size. The submission narrative uses a compact corpus on the order of ~200 entries (techniques, worked patterns, hints). The notebook logs len(self.retriever.entries) when a corpus file is loaded.
When retrieval is turned on, the top strategies are serialized into a [REFERENCE STRATEGIES] block (with closing [END REFERENCE]) and prepended to the user turn so the model sees structured hints, topics, techniques, and per-strategy relevance (BM25, dense, and fused combined scores). Below is a representative excerpt from logs (two retrieved entries; line breaks as in stdout):
[REFERENCE STRATEGIES]\n\n--- Strategy 1 [AIME-style] ---\nSimilar Problem: Triangle ABC inscribed in circle omega. Tangent to omega at B and C intersect at D. AD meets omega again at P. If AB=5, BC=9, AC=10, find AP (as m/n, answer m+n).\nTopics: geometry, circles\nTechniques: power_of_a_point, similar_triangles\nRecognition Signals:\n - tangent to circle\n - inscribed triangle\n - tangent intersection\nKey Insights:\n - D is the intersection of tangents at B,C, so DB=DC (tangent lengths)\n - Power of point D: DB^2 = DA * DP\n - D lies on the perpendicular bisector of BC\nStrategy: Place D as pole of BC w.r.t. omega. DB^2=DC^2=power of D. By power of a point: DA*DP = DB^2. Compute: use coordinates with BC on x-axis, or use the formula DB^2 = (circumradius)^2 * ... Alternatively: by the tangent-secant relation and similar triangles DAB ~ DPA (since angle DBA = inscribed angle = angle DPA). Then DA/DB = DB/DP gives DP = DB^2/DA. AP = DA - DP or DA + DP depending on configuration.\nCode Hint: Use circumradius R = abc/(4*Area). Area by Heron's formula. Coordinate geometry for exact computation.\nRelevance: combined=1.000, bm25=1.000, dense=1.000\n\n--- Strategy 2 [Competition] ---\nSimilar Problem: Point P lies on the circumcircle of square ABCD such that PA*PC = 56 and PB*PD = 90. Find the area of ABCD.\nTopics: geometry, circles\nTechniques: radical_axis, power_of_a_point\nRecognition Signals:\n - circumcircle of square\n - PA * PC\n - PB * PD\nKey Insights:\n - For a point on the circumcircle of a square with side s: use Ptolemy or power of a point\n - By Ptolemy on cyclic quadrilateral PABC: PA*BC + PC*AB = PB*AC\n - The square's circumradius is s*sqrt(2)/2\nStrategy: Let side = s. Circumradius R = s/sqrt(2). Place square with center at origin. For P on circumcircle: PA*PC and PB*PD relate to the position of P. Using complex numbers with vertices at R*e^{i*k*pi/2}: compute products. By Ptolemy on ABPC: PA*s + PC*s = PB*s*sqrt(2). Combined with PA*PC=56, PB*PD=90 and symmetry relations, solve for s^2 = area.\nCode Hint: Use complex number representation: place square vertices at s/sqrt(2) * exp(i*k*pi/2 + i*theta) and parameterize P.\nRelevance: combined=0.819, bm25=0.797, dense=0.841\n\n[END REFERENCE]\nEmpirical observation: injecting retrieved context sometimes reduced reliability—e.g., misleading analogies, overcrowding the context window, or steering the model away from the cleanest path for a specific problem.
\nIn the notebook’s default _load_corpus configuration, the corpus path is set to a non-existent placeholder (\"NOFILE\"), so the retriever is not loaded and solve_problem uses:
user_input = f'{problem} {self.cfg.preference_prompt}'.
To enable RAG in a fork, point corpus_candidates to your JSON corpus and ensure the Qwen3 embedding model path exists on Kaggle.
Stability, not just mean score. With hybrid retrieval turned on, we observed worse stability of end-to-end performance: for example, the over-engineered prompts + RAG public-leaderboard line in §7.3 spans 34 to 42 out of 50—a 8-point spread across submits—whereas runs without RAG (or with retrieval off in the shipped stack) were typically tighter for us. We saw similar high-variance behavior in other notebooks that enabled RAG but are not detailed in this report. We did not run a dedicated offline A/B at fixed seeds; still, the combination of unreliable peaks and heavy tail risk led us to ship without retrieval and invest budget in dynamic deadlines and voting instead.
\n
Parallel attempts have heterogeneous completion times. On some problems, several attempts finish quickly with a stable answer (we observed this pattern anecdotally for problems whose final answers included 57447, 160, and 520). Empirically, on one representative run the eight parallel attempts showed Response Length (assistant stream size, in tokens) from about 26k to 51k and Completion Time from about 292s to 557s—same harness, large spread—so some trajectories finish much earlier than others. Remaining attempts may still consume tail latency without changing the decision. A static “always run until the global max” policy wastes time that is better reallocated across the full contest slate.
\nDesign goal: treat “difficulty” operationally—not with a separate classifier—but via whether the attempt ensemble has converged by each stage boundary.
\nFor each problem, the solver computes a remaining notebook budget and derives a max_budget for that problem (function of time left, problems remaining, and configured caps). It then defines three checkpoints:
\nfirst_deadline = start + first_budget, with first_budget = min(first_problem_timeout, max_budget).second_deadline = start + second_budget, with second_budget = min(second_problem_timeout, max_budget).final_deadline = start + final_budget where final_budget = max_budget.Configured constants in CFG include (see notebook): first_problem_timeout, second_problem_timeout, final_problem_timeout, base_problem_timeout, notebook_limit, and agreement thresholds first_agreement_count, second_agreement_count.
A background loop polls completed attempts. When real time crosses the current stage deadline (with a small grace window in is_past_deadline), the policy compares the top vote count among valid numeric answers to thresholds:
| Stage | \nIf condition holds | \nElse | \n
|---|---|---|
| 1 | \nFinalize when the leading answer has ≥ first_agreement_count votes and len(valid_answers) > first_agreement_count. | \nExtend to stage 2 (or toward final if the stage-2 checkpoint is not later than the current deadline). | \n
| 2 | \nFinalize when the leading answer has ≥ second_agreement_count votes. | \nExtend to stage 3 (final budget). | \n
Intuition mapping (informal labels):
\nThis matches the motivation: do not pay for long tail attempts when the vote mass has already stabilized.
\n\nCFG.attempts parallel threads (with workers pool size).early_stop copies, the solver sets a stop event and cancels pending futures.\\boxed{} format; this recovers trajectories that solved most of the problem but failed to emit the final box cleanly.stdout can be extremely long—on the order of tens of thousands of model-side tokens in severe cases if passed through verbatim. Without truncation, tool returns alone can consume the context / generation budget, and otherwise valuable attempts may stop early (e.g., context exhaustion) before a final answer is produced. The notebook’s AIMO3Tool therefore applies _truncate_output: if a tool return exceeds CFG.max_tool_output_chars (default 8192 characters), it keeps the head and tail and inserts a marker ...[python output truncated]... so the model still sees the beginning and end of the run without flooding the context._select_answer ranks answers by a sum of inverse-entropy weights across attempts (lower entropy ⇒ higher weight), then picks the top score.We did not produce separate distribution plots of time-to-consensus or stage-finalization rates; the design rationale above is the primary evidence for this component.
\nBeyond finalnotebook, we implemented and submitted several alternative designs that explored richer prompting, continuation-based search, global time-banking, and a merged SFT checkpoint. Public leaderboard scores were not consistently strong across reruns and configurations. Empirically, the main failure modes were over-long or over-specified preference / system-adjacent text (which diluted the stable baseline persona and steered tool use inefficiently) and time-management schemes that were harder to tune than the simpler dynamic per-problem deadline policy in the shipped solution.
\nChosen solution. We selected finalnotebook as the primary submission: it was submitted twice and scored 38 and 40 (50 points max), which is both more stable and better than the ranges below. More importantly, this approach holds enormous potential due to its time saving strategy.
\n| Approach | \nApprox. score range (of 50) | \nMain idea (brief) | \n
|---|---|---|
| Two-phase continuations | \n33–40 | \nInitial wave of attempts, then forked continuation trajectories (doubling the number of attempts 8==>16) with a dedicated continuation user message; optional compute-heavy system variant; long structured preference text; tooling penalties and deadline-aware forced \\boxed{} extraction. | \n
| Over-engineered prompts + RAG | \n34–42 | \nLayered developer_prompt, rewritten system copy, verbose preference steering; hybrid retrieval on (strategy blocks in the user turn). §7.3 table. | \n
| Merged SFT model (Tinker) | \n26–37 | \nSFT on gpt-oss-120b, adapter merged into full weights; CFG.model_path points at the merged HF directory. Public leaderboard: 37, 26 (two submits); §7.3. | \n
| Banked time + hard reserve | \n30–40 | \nGlobal time bank and hard-problem reserve: adaptive per-problem budgets draw from a bank while protecting a reserve—more complex scheduling than fixed or stage-wise caps. | \n
Why these did not work. The headline pattern was variance, not a clear ceiling lift: preference prompts that were long or multi-part often hurt consistency (either by distracting the model from the baseline reasoning contract, or by encouraging expensive tool loops without a matching gain in agreement). Time policies that depended on global bank state or multi-phase continuation schedules were sensitive to problem order and tail latency, so they did not reliably beat the simpler agreement-gated, stage-wise deadline extension in finalnotebook. The shipped solution keeps the short preference prompt and focuses innovation on deadline allocation and parallel voting—a combination that still has considerable headroom (e.g., tighter thresholds, retrieval gating, or calibrated early stopping), but already matches or exceeds the best outcomes from the exploratory line above.
\nWe trained openai/gpt-oss-120b with Tinker using LoRA (rank 32 in saved run configs) and the gpt_oss_no_sysprompt renderer. Each run produced configs, metrics, checkpoints, and timing metadata (typical Tinker / Tinker Cookbook layout). The code is here Code on GitHub. Representative settings:
| Run (label) | \nRegime | \nNotes | \n
|---|---|---|
| SFT (on mixture of AIMO3 Dataset and Astral Math ), 100k sample size | \nSFT | \nlearning_rate 1e-4, linear schedule, 1 epoch, batch 8, max length 32768, JSONL SFT dataset, train_on_what: last assistant message | \n
| SFT (on 10k examples of AIMO3 Dataset) | \nSFT | \nSame LR / rank; 4000 steps; JSONL SFT dataset | \n
SFT loss (from logs). Our long SFT run on Tinker (1 epoch on 100k samples) shows training NLL generally trending downward (about 0.77 → 0.35, with noisy batch-level fluctuations). A compact view from sampled steps:
\n
These runs document what we tried; they are not the primary story of the final submission.
\nMerged SFT model submission. For supervised fine-tuning we merged the trained adapter into base gpt-oss-120b using the Tinker Cookbook path (tinker_cookbook.weights.build_hf_model). That yields a different checkpoint; serving it means CFG.model_path points at the merged directory. §7.3 and Figures 1–3 list merged SFT as its own row/category (orange in Figure 1).
§7.3 aggregates repeated public leaderboard submits: base-model stacks (continuation, over-engineered prompts + RAG, banked time), merged SFT (different weights), and our two finalnotebook runs (38, 40). Qualitatively, continuation-enabled can reach 40 but also 33–34; merged SFT shows a wide two-run spread (26–37) and a lower mean than shipped; banked-time includes very low scores (30–32). We do not claim SOTA against all open-weight or commercial systems; the relevant comparison is within our own runs and the AIMO3 leaderboard context.
\nTable: raw public leaderboard scores (out of 50). Base-model rows use gpt-oss-120b in CFG.model_path (not the merged SFT directory). Merged SFT is a different checkpoint (Tinker adapter merged into full weights)—listed on its own row. Multiple submits reflect iterative tuning and harness variance—not i.i.d. experiments—so we emphasize ranges, means, and visuals rather than (p)-values.
| Approach | \nScores (each submit) | \n(n) | \nMin | \nMax | \nMean | \n
|---|---|---|---|---|---|
| Continuation-enabled | \n40, 37, 38, 39, 33, 34, 34 | \n7 | \n33 | \n40 | \n36.4 | \n
| Over-engineered prompts + RAG | \n37, 35, 42, 34, 41, 38, 38, 38 | \n8 | \n34 | \n42 | \n37.9 | \n
| Banked time + reserve | \n30, 39, 40, 35, 34, 32, 33, 40 | \n8 | \n30 | \n40 | \n35.4 | \n
| Merged SFT (Tinker) | \n37, 26 | \n2 | \n26 | \n37 | \n31.5 | \n
| Shipped: finalnotebook (base) | \n38, 40 | \n2 | \n38 | \n40 | \n39 | \n
Figure 1 — Distribution per approach. Box plots (median in red, mean as blue dashed line) with jittered points for every submit. Merged SFT (Tinker) uses an orange box (different model family). The shipped finalnotebook row is green; with two submits it is a short, high band.
\n
Figure 2 — Mean score with approximate 95% CI for the mean (more submits ⇒ narrower intervals). The shipped mean is 39 (two submits at 38 and 40).
\n
Figure 3 — Range (min–max bar) and mean (dot). Highlights risk: banked time spans 30–40; merged SFT spans 26–37 on only two submits; over-engineered + RAG spans 34–42; continuation-enabled spans 33–40.
\n
Reading. Among base-model exploratory lines, continuation-enabled reaches 40 on some submits but 33–34 on others; over-engineered + RAG (mean ~37.9) and banked-time show wide variance. Merged SFT (different weights), with only two public scores (37, 26), mean ~31.5, includes the lowest observed point in the whole panel (26)—so our SFT model submission did not beat finalnotebook (38, 40) under this evaluation budget. We preferred finalnotebook as the primary submission for strong scores, a clean inference story (§3–6), and avoidance of the worst tails under banked-time (30–32), continuation (33–34), and merged SFT (26).
\nWe intentionally do not include exhaustive ablation tables (fixed vs dynamic deadlines, grid over first_agreement_count / second_agreement_count, early_stop sweeps) or per-topic error breakdowns: those would require a frozen offline harness and labels beyond the scope of this narrative. The methodology in §3–6 is the reproducible artifact readers can build on.
The system prompt follows the public AIMO3 notebook text in https://www.kaggle.com/code/nihilisticneuralnet/44-50-let-me-over-cook. The submission implementation (dynamic deadlines, tooling, aggregation) is documented in finalnotebook. Tooling and inference patterns build on the competition’s reference stack (kaggle_evaluation, vLLM, Harmony).
| Item | \nNotebook reference / value | \n
|---|---|
| Model path | \nCFG.model_path (gpt-oss-120b Kaggle input path) | \n
| Served name | \nCFG.served_model_name | \n
| Seeds | \nCFG.seed; per-attempt seed derived in _process_attempt | \n
| Time limits | \nnotebook_limit, first_problem_timeout, second_problem_timeout, final_problem_timeout, base_problem_timeout | \n
| Attempts / workers | \nattempts, workers | \n
| Tool limits | \njupyter_timeout, max_tool_output_chars, etc. | \n
| Retrieval | \nCorpusRetriever + _load_corpus path list; default disables retriever | \n
| Stop / consensus | \nearly_stop, unanimous rule, _select_answer weighting | \n
| Primary submission | \nfinalnotebook — two public leaderboard submissions scored 38 and 40 (of 50) | \n
| Exploratory approaches (negative / unstable) | \nContinuations; layered developer prompts; merged SFT model submission; banked time — see §6 | \n
| Tinker SFT | \nRun configs and metrics as in §7.1 (saved alongside each training job) | \n
| Merge SFT adapter → full HF | \nTinker Cookbook build_hf_model; merged directory as CFG.model_path for merged SFT submissions (§7) | \n
Exact Kaggle notebook revision IDs and pip freeze outputs may change per session; reproduce from the published finalnotebook and the competition Docker / input snapshot at submit time.
\n§7.3 includes three figures: (1) box plot + jittered submits (merged SFT in orange, shipped in green), (2) mean with uncertainty on the mean, (3) min–max range with mean dot.
\nFull tool transcripts are long and environment-specific. Readers should run finalnotebook on Kaggle and inspect stdout / displayed attempt tables for end-to-end behavior (retrieval off in the default configuration).
\nsystem_prompt we reuse unchanged: https://www.kaggle.com/code/nihilisticneuralnet/44-50-let-me-over-cook.openai/gpt-oss-120b as referenced in Tinker configs.\nAdding to the comments by @sambealing below, I have a few additional questions:
\nRAG Efficiency: Do you think the RAG failure could partly be due to the model already having much of the relevant knowledge internalized, given its capacity at roughly 120B total parameters? One could argue that RAG may flood the context with additional information, making the actual problem statement less salient and reducing inference speed due to much longer input sequences before reasoning even begins.
Time Budget: You mention using a max_budget per problem. Is this constant empirically motivated, estimated by a predictor, or determined in some other way? In the write-up, you briefly mention a function that calculates it. Since time management is a crucial part of AIMO3, it would be helpful to describe this component in more detail.
SFT: Why did you choose to run SFT on an already RL-tuned model such as gpt-oss-120b? Was this mainly due to simplicity, time constraints, or another motivation? Methods such as GRPO seem more aligned with the model’s current stage of training. The reported results suggest that SFT may not have been the best fit and may have overfit to the training data format, potentially weakening some of the reasoning behavior learned during earlier RL stages. Do you agree with this interpretation? Additionally, did you encounter any issues when merging the LoRA weights, such as the vanishing-weight problem reported by other contestants?
Reproducibility: Please add the code required to reproduce your fine-tuning and experiments, so others can reproduce or adapt your workflow where possible.
Hi @philipvonderlind. Thanks for your questions. Below are my answers.
\nYes, I think that is a plausible part of the whole story. gpt-oss-120b is already strong on many IMO/AIME-style patterns without retrieval (the public baseline reaches 44/50), so naive RAG can add redundant context rather than missing knowledge. That matches what I saw: strategy blocks sometimes made the problem statement less salient, increased prefill cost (longer inputs before the first reasoning token), and sometimes steered the model off a path it would otherwise have taken. So I do not read the RAG failure as \"retrieval cannot help\", but as \"default-on retrieval is often net harmful on a model that already internalizes much of the corpus.\" That is why I disabled it in the final notebook due to time constraints in deeper investigations. I now believe conditional RAG (e.g. only at stage 2/3) could help on some hard/rare problems.
max_budget is not learned by a predictor and not fixed globally—it is recomputed per problem from remaining notebook time and problems left.
At each problem, the notebook computes:
\ntime_left = notebook_limit − elapsed_global\nreserved_time = (problems_remaining − 1) × base_problem_timeout (default 280 s per future problem)\nmax_budget = time_left − reserved_time \nthen clamped to [base_problem_timeout, final_problem_timeout] (defaults 280–900 s)
\nStage deadlines are derived from that cap:
\nStage 1: min(first_problem_timeout, max_budget) → 400 s
Stage 2: min(second_problem_timeout, max_budget) → 640 s
Stage 3: max_budget itself (up to 900 s)
notebook_limit is set from the competition’s remaining wall-clock budget (initialized around 17 400 s in CFG, adjusted at session start). The constants (280, 400, 640, 900) are empirically chosen heuristics based on notebook local runs on an upsampled version of the reference set: reserve a minimum slice for every remaining problem, allow extension only when votes are insufficient, and cap any single problem so later ones are not starved. It is simpler than my banked-time variant (§6), which was harder to tune.
Why SFT? Mainly practical constraints: I already had tool-integrated SFT traces (My Math Corpus Dataset) and a working Tinker SFT pipeline, while GRPO (which I also explored in side scripts) would have needed more rollout/tool infrastructure and compute before the submission deadline. SFT was the fastest path to test whether imitation on I traces would help—not a claim that SFT is theoretically optimal post-RL.
\nSo I partly agree with your interpretation. Lower public scores (37, 26; mean 31.5) are consistent with overfitting to the training data (and its format) and/or hurting behaviors that matter at inference (multi-attempt diversity, tool use under the full system prompt). But I would not confidently attribute everything to “RL capabilities erased” via SFT as it could also be possible that I did not merge Tinker adapters properly.
\nMerge / LoRA issues: Yes. Tinker adapters are not standard PEFT LoRA; build_hf_model from Tinker Cookbook did not work reliably on my side (including MXFP4 / key-naming issues on gpt-oss-120b). I used a two-stage merge with manual adapter weight renaming before merging. I did not fully debug whether there was a vanishing-weight issue that other teams reported, but a bad merge is my leading hypothesis for the weak SFT public scores. With only two public SFT submissions, I treat the low mean of scores as inconclusive and unreliable until I can serve a checkpoint merged through a validated procedure (and compare against GRPO checkpoints).
\nThe code is on GitHub.
", "rawMarkdown": "Hi @philipvonderlind. Thanks for your questions. Below are my answers. \n\n## 1. RAG efficiency\n\nYes, I think that is a plausible part of the whole story. `gpt-oss-120b` is already strong on many IMO/AIME-style patterns without retrieval (the public baseline reaches 44/50), so naive RAG can add redundant context rather than missing knowledge. That matches what I saw: strategy blocks sometimes made the problem statement less salient, increased prefill cost (longer inputs before the first reasoning token), and sometimes steered the model off a path it would otherwise have taken. So I do not read the RAG failure as \"retrieval cannot help\", but as \"default-on retrieval is often net harmful on a model that already internalizes much of the corpus.\" That is why I disabled it in the final notebook due to time constraints in deeper investigations. I now believe conditional RAG (e.g. only at stage 2/3) could help on some hard/rare problems.\n\n## 2. Time budget (max_budget)\n\n`max_budget` is not learned by a predictor and not fixed globally—it is recomputed per problem from remaining notebook time and problems left.\n\nAt each problem, the notebook computes:\n```\ntime_left = notebook_limit − elapsed_global\nreserved_time = (problems_remaining − 1) × base_problem_timeout (default 280 s per future problem)\nmax_budget = time_left − reserved_time \n``` then clamped to [base_problem_timeout, final_problem_timeout] (defaults 280–900 s)\n\nStage deadlines are derived from that cap:\n\n- Stage 1: min(first_problem_timeout, max_budget) → 400 s\n\n- Stage 2: min(second_problem_timeout, max_budget) → 640 s\n\n- Stage 3: max_budget itself (up to 900 s)\n\n`notebook_limit` is set from the competition’s remaining wall-clock budget (initialized around 17 400 s in `CFG`, adjusted at session start). The constants `(280, 400, 640, 900)` are empirically chosen heuristics based on notebook local runs on an upsampled version of the reference set: reserve a minimum slice for every remaining problem, allow extension only when votes are insufficient, and cap any single problem so later ones are not starved. It is simpler than my banked-time variant (§6), which was harder to tune.\n\n## 3. SFT on an RL-tuned base\n\nWhy SFT? Mainly practical constraints: I already had tool-integrated SFT traces ([My Math Corpus Dataset](https://www.kaggle.com/datasets/jeannkouagou/aimo3-tool-integrated-reasoning)) and a working Tinker SFT pipeline, while GRPO (which I also explored in side scripts) would have needed more rollout/tool infrastructure and compute before the submission deadline. SFT was the fastest path to test whether imitation on I traces would help—not a claim that SFT is theoretically optimal post-RL.\n\nSo I partly agree with your interpretation. Lower public scores (37, 26; mean 31.5) are consistent with overfitting to the training data (and its format) and/or hurting behaviors that matter at inference (multi-attempt diversity, tool use under the full system prompt). But I would not confidently attribute everything to “RL capabilities erased” via SFT as it could also be possible that I did not merge Tinker adapters properly.\n\nMerge / LoRA issues: Yes. Tinker adapters are not standard PEFT LoRA; build_hf_model from Tinker Cookbook did not work reliably on my side (including MXFP4 / key-naming issues on gpt-oss-120b). I used a two-stage merge with manual adapter weight renaming before merging. I did not fully debug whether there was a vanishing-weight issue that other teams reported, but a bad merge is my leading hypothesis for the weak SFT public scores. With only two public SFT submissions, I treat the low mean of scores as inconclusive and unreliable until I can serve a checkpoint merged through a validated procedure (and compare against GRPO checkpoints).\n\n## 4 Reproducible SFT.\n\nThe [code is on GitHub](https://github.com/Jean-KOUAGOU/Tinker-SFT-AIMO3)." } ] }, { "id": 3457729, "postDate": "2026-05-14T08:39:08.940Z", "content": "Hi, thank you for your submission to the Writeup Prize. @philipvonderlind and I will post some questions below that help us in assessing your submission more closely, as described in the updated discussion post.
\nLook forward to reading your responses on these.
", "rawMarkdown": "Hi, thank you for your submission to the Writeup Prize. @philipvonderlind and I will post some questions below that help us in assessing your submission more closely, as described in the [updated discussion post](https://www.kaggle.com/competitions/ai-mathematical-olympiad-progress-prize-3/discussion/689703).\n\n1. **Stage transitions on public benchmarks:** Have you run the pipeline on any public benchmarks (e.g. AIME 2026), and if so, what's the distribution of problems across the three deadline stages (finalising at stage 1 vs extending to stage 2 or 3)?\n2. **RAG failure cases:** §4.3 disables RAG after observing worse stability. Do you think this reduction in stability is universal or might there be problem types that do benefit from RAG?\n3. **SFT mean score gap:** §7 shows your merged SFT submissions had a noticeably lower mean (31.5) than the shipped base model (39), despite training NLL trending downward smoothly (Fig 4). Do you have any hypotheses for why the SFT scores were quite a bit lower?\n\nLook forward to reading your responses on these.\n", "replies": [ { "id": 3457889, "postDate": "2026-05-14T15:00:13.177Z", "content": "Thanks @sambealing for your questions. Below are answers.
\nYes. I reran the final notebook locally on the full AIME 2026 labeled set (30 problems). All 30 finalized at stage 1 (400 s budget); none extended to stage 2 or 3. Accuracy was 30/30 (100%). In every case, parallel attempts reached unanimous agreement before the stage-1 deadline, so the extension policy never triggered. See logs at the bottom of this reply.
\nI do not think the stability loss with RAG is universal. In the final submission I disabled RAG because naive injection hurt end-to-end reliability in my earlier runs (§4.3), especially alongside the heavy system prompt used.
\nWhen I did enable RAG, I added instructions to use retrieved strategies and hints but not to copy them blindly, with verification required first. On the 10 reference problems provided, retrieval sometimes surfaced useful techniques and shortened reasoning; on other problems, hints steered the model onto wrong paths (contamination) and wrong answers.
\nSo I believe some problem types can benefit from RAG; the issue is unreliable net effect, not RAG always hurting. A natural extension would be conditional RAG---e.g. activate retrieval only when a problem reaches stage 2 or 3, appending references to ongoing conversations for the hardest cases rather than injecting them on every problem upfront.
\nI can see two main explanations. First (most probable hypothesis): the merged checkpoint may not reflect a correct Tinker→HF conversion—Tinker adapters are not standard PEFT LoRA, and I may not have applied the right merge procedure before serving. In fact the build_hf_model method provided by Tinker CookBook did not work on my side. So I had to perform a two stage merging procedure which involved renaming some specific adapter weight names first.
Second: overfitting to the training data, combined with limited offline evaluation before my two public submissions (with scores 37, 26). The decreasing NLL (Fig. 4) only shows that the model somehow learned the training data, but it does not guarantee a strong end model. With only 2 submissions, I treat the low score mean as inconclusive and unreliable until I validate a properly merged checkpoint.
", "rawMarkdown": "Thanks @sambealing for your questions. Below are answers.\n\n## 1. Stage transitions on public benchmarks\n\nYes. I reran the final notebook locally on the full AIME 2026 labeled set (30 problems). All 30 finalized at stage 1 (400 s budget); none extended to stage 2 or 3. Accuracy was 30/30 (100%). In every case, parallel attempts reached unanimous agreement before the stage-1 deadline, so the extension policy never triggered. See logs at the bottom of this reply.\n\n## 2. RAG failure cases\n\nI do not think the stability loss with RAG is universal. In the final submission I disabled RAG because naive injection hurt end-to-end reliability in my earlier runs (§4.3), especially alongside the heavy system prompt used.\n\nWhen I did enable RAG, I added instructions to use retrieved strategies and hints but not to copy them blindly, with verification required first. On the 10 reference problems provided, retrieval sometimes surfaced useful techniques and shortened reasoning; on other problems, hints steered the model onto wrong paths (contamination) and wrong answers.\n\nSo I believe some problem types can benefit from RAG; the issue is unreliable net effect, not RAG always hurting. A natural extension would be conditional RAG---e.g. activate retrieval only when a problem reaches stage 2 or 3, appending references to ongoing conversations for the hardest cases rather than injecting them on every problem upfront.\n\n## 3. SFT mean score gap (31.5 vs 39)\n\nI can see two main explanations. First (most probable hypothesis): the merged checkpoint may not reflect a correct Tinker→HF conversion—Tinker adapters are not standard PEFT LoRA, and I may not have applied the right merge procedure before serving. In fact the `build_hf_model` method provided by Tinker CookBook did not work on my side. So I had to perform a two stage merging procedure which involved renaming some specific adapter weight names first. \n\nSecond: overfitting to the training data, combined with limited offline evaluation before my two public submissions (with scores 37, 26). The decreasing NLL (Fig. 4) only shows that the model somehow learned the training data, but it does not guarantee a strong end model. With only 2 submissions, I treat the low score mean as inconclusive and unreliable until I validate a properly merged checkpoint.", "replies": [ { "id": 3457892, "postDate": "2026-05-14T15:08:04.650Z", "content": "ID: 8742f9\nQuestion: The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex…
\nThe figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Let $N$ be the number of paths the bug can take from the lower left corner ($A$) to the upper right corner ($B$). One such path from $A$ to $B$ is shown by the thick line segments in the figure. Find $\\sqrt{N}$.
\n[asy]\nsize(12cm);\ndraw((0, 0)--(0, 1));\nfor (int i = 0; i <= 9; i += 1)\n{\ndraw((i, 1)--(i + 1, 1)--(i + 1, 0)--(i, 0)--(i + 1, 1));\n}\ndraw((0, 0)--(0, 1)--(2, 1)--(2, 0)--(3, 1)--(3, 0)--(4, 1)--(4, 0)--(5, 0)--(5, 1)--(6, 1)--(6, 0)--(7, 1)--(8, 1)--(8, 0)--(10, 0)--(10, 1), red + linewidth(2));\nlabel(\"$A$\", (0, 0), dir(180));\nlabel(\"$B$\", (10, 1), dir(0));\n[/asy] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n5024 | \n4 | \n0 | \n0.811 | \n243 | \nanswered | \n59.13 | \n
| 1 | \n7 | \n6260 | \n2 | \n0 | \n0.903 | \n243 | \nanswered | \n72.26 | \n
| 2 | \n3 | \n6497 | \n2 | \n0 | \n0.798 | \n243 | \nanswered | \n74.67 | \n
| 3 | \n8 | \n7256 | \n5 | \n1 | \n0.794 | \n243 | \nanswered | \n82.55 | \n
UNANIMOUS ANSWER: 243
\nAnswer: 243 | Ground Truth: 243 | ✅
\nID: 5cc894\nQuestion: Let $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the…
\nLet $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the $2a + 2b - 4$ cells that surround an $(a - 2) \\times (b - 2)$ (possibly empty) rectangle of cells in the grid. For example, the following diagram shows a way to partition a $6 \\times 6$ grid of cells into $4$ cell loops.
\n| P P P P | Y Y |\n| P | R R | P | Y | Y |\n| P | R R | P | Y | Y |\n| P P P P | Y | Y |\n| G G G G | Y | Y |\n| G G G G | Y Y |
\nFind the number of ways to partition a $10 \\times 10$ grid of cells into $5$ cell loops so that every cell of the grid belongs to exactly one cell loop. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n8177 | \n9 | \n0 | \n0.891 | \n83 | \nanswered | \n98.10 | \n
| 1 | \n4 | \n9962 | \n4 | \n0 | \n0.844 | \n179 | \nanswered | \n109.96 | \n
| 2 | \n1 | \n9566 | \n6 | \n1 | \n0.784 | \n83 | \nanswered | \n114.32 | \n
| 3 | \n8 | \n10538 | \n8 | \n1 | \n0.849 | \n83 | \nanswered | \n128.12 | \n
| 4 | \n5 | \n10843 | \n21 | \n3 | \n0.779 | \n83 | \nanswered | \n130.85 | \n
UNANIMOUS ANSWER: 83
\nAnswer: 83 | Ground Truth: 83 | ✅
\nID: 744366\nQuestion: Let $S$ denote the value of the infinite sum\n[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$…
\nLet $S$ denote the value of the infinite sum\n[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n4410 | \n8 | \n0 | \n0.737 | \n669 | \nanswered | \n50.35 | \n
| 1 | \n8 | \n4493 | \n12 | \n1 | \n0.663 | \n669 | \nanswered | \n51.79 | \n
| 2 | \n3 | \n4585 | \n6 | \n0 | \n0.625 | \n669 | \nanswered | \n51.88 | \n
| 3 | \n2 | \n5249 | \n5 | \n0 | \n0.670 | \n669 | \nanswered | \n58.13 | \n
UNANIMOUS ANSWER: 669
\nAnswer: 669 | Ground Truth: 669 | ✅
\nID: b80bba\nQuestion: \nFind the number of functions $\\pi$ mapping the set $A ={1,2,3,4,5,6}$ onto $A$ such that for every $a \\in A,$\n[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n]…
\nFind the number of functions $\\pi$ mapping the set $A ={1,2,3,4,5,6}$ onto $A$ such that for every $a \\in A,$\n[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n1651 | \n2 | \n0 | \n0.660 | \n396 | \nanswered | \n18.17 | \n
| 1 | \n5 | \n1773 | \n3 | \n0 | \n0.723 | \n396 | \nanswered | \n19.55 | \n
| 2 | \n1 | \n2056 | \n2 | \n0 | \n0.669 | \n396 | \nanswered | \n22.35 | \n
| 3 | \n3 | \n2127 | \n3 | \n0 | \n0.607 | \n396 | \nanswered | \n23.11 | \n
UNANIMOUS ANSWER: 396
\nAnswer: 396 | Ground Truth: 396 | ✅
\nID: bce9ef\nQuestion: Let $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D…
\nLet $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D$ lie on the same side of line $AE.$ Suppose further that $AB$ is an integer with $AB < 2026$ and the area of pentagon $ABCDE$ is an integer multiple of $16.$ Find the number of possible values of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n12334 | \n3 | \n0 | \n0.755 | \n503 | \nanswered | \n136.30 | \n
| 1 | \n4 | \n12760 | \n13 | \n1 | \n0.761 | \n1002 | \nanswered | \n143.73 | \n
| 2 | \n5 | \n13789 | \n5 | \n0 | \n0.754 | \n503 | \nanswered | \n153.05 | \n
| 3 | \n8 | \n13974 | \n7 | \n0 | \n0.751 | \n503 | \nanswered | \n155.26 | \n
| 4 | \n6 | \n15984 | \n9 | \n2 | \n0.729 | \n503 | \nanswered | \n172.94 | \n
UNANIMOUS ANSWER: 503
\nAnswer: 503 | Ground Truth: 503 | ✅
\nID: cba4eb\nQuestion: Find the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b….
\nFind the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n2144 | \n4 | \n0 | \n0.736 | \n70 | \nanswered | \n24.08 | \n
| 1 | \n2 | \n2339 | \n3 | \n0 | \n0.794 | \n70 | \nanswered | \n26.17 | \n
| 2 | \n1 | \n2653 | \n5 | \n0 | \n0.764 | \n70 | \nanswered | \n29.58 | \n
| 3 | \n5 | \n2703 | \n4 | \n0 | \n0.780 | \n70 | \nanswered | \n30.02 | \n
UNANIMOUS ANSWER: 70
\nAnswer: 70 | Ground Truth: 70 | ✅
\nID: cda722\nQuestion: For integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n[\na_1 + …
\nFor integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n[\na_1 + a_2 + a_3 + \\cdots + a_n = 12 \\quad \\text{and} \\quad a_1 \\circ a_2 \\circ a_3 \\circ \\cdots \\circ a_n = 0\n]\nwhere the operations are performed from left to right; that is, $a_1 \\circ a_2 \\circ a_3$ means $(a_1 \\circ a_2) \\circ a_3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n2209 | \n3 | \n0 | \n0.875 | \n157 | \nanswered | \n25.53 | \n
| 1 | \n3 | \n2249 | \n4 | \n0 | \n0.837 | \n157 | \nanswered | \n26.15 | \n
| 2 | \n8 | \n2406 | \n3 | \n0 | \n0.845 | \n157 | \nanswered | \n27.62 | \n
| 3 | \n7 | \n2502 | \n3 | \n0 | \n0.897 | \n157 | \nanswered | \n28.55 | \n
UNANIMOUS ANSWER: 157
\nAnswer: 157 | Ground Truth: 157 | ✅
\nID: f7b4a9\nQuestion: Let $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC…
\nLet $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC}$ at $D.$ Let $E \\neq A$ and $F \\neq A$ be the intersections of $\\omega$ with segments $\\overline{AB}$ and $\\overline{AC},$ respectively. Suppose that $AB = 200, AC = 225,$ and all of $AE, AF, BD,$ and $CD$ are positive integers. Find the sum of all possible values of $BC.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n5727 | \n2 | \n0 | \n0.758 | \n850 | \nanswered | \n60.41 | \n
| 1 | \n2 | \n6207 | \n2 | \n0 | \n0.708 | \n850 | \nanswered | \n65.95 | \n
| 2 | \n6 | \n6343 | \n3 | \n0 | \n0.540 | \n850 | \nanswered | \n67.47 | \n
| 3 | \n3 | \n6727 | \n2 | \n0 | \n0.712 | \n850 | \nanswered | \n71.45 | \n
UNANIMOUS ANSWER: 850
\nAnswer: 850 | Ground Truth: 850 | ✅
\nID: f3d7b7\nQuestion: Find the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$…
\nFind the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 3 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n2711 | \n5 | \n0 | \n0.687 | \n50 | \nanswered | \n29.22 | \n
| 1 | \n4 | \n3372 | \n1 | \n0 | \n0.698 | \n50 | \nanswered | \n35.80 | \n
| 2 | \n2 | \n3659 | \n8 | \n0 | \n0.672 | \n50 | \nanswered | \n39.63 | \n
| 3 | \n3 | \n3806 | \n2 | \n0 | \n0.701 | \n50 | \nanswered | \n40.41 | \n
UNANIMOUS ANSWER: 50
\nAnswer: 50 | Ground Truth: 50 | ✅
\nID: f969f6\nQuestion: Call finite sets of integers $S$ and $T$ cousins if
\n$S$ and $T$ have the same number of elements,
$S$ and $T$ are disjoint, and
the elements of $S$ can be paired with the elements of $T$ so that…
\nCall finite sets of integers $S$ and $T$ cousins if
$S$ and $T$ have the same number of elements,
$S$ and $T$ are disjoint, and
the elements of $S$ can be paired with the elements of $T$ so that the elements in each pair differ by exactly $1$.\nFor example, ${1,2,5}$ and ${0,3,4}$ are cousins. Suppose that the set $S$ has exactly $4040$ cousins. Find the least number of elements the set $S$ can have. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
Stage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n11691 | \n6 | \n0 | \n0.868 | \n107 | \nanswered | \n130.24 | \n
| 1 | \n4 | \n14143 | \n16 | \n2 | \n0.859 | \n107 | \nanswered | \n167.23 | \n
| 2 | \n3 | \n16430 | \n6 | \n0 | \n0.893 | \n107 | \nanswered | \n184.39 | \n
| 3 | \n7 | \n19128 | \n22 | \n3 | \n0.821 | \n107 | \nanswered | \n220.86 | \n
UNANIMOUS ANSWER: 107
\nAnswer: 107 | Ground Truth: 107 | ✅
\nID: f3d797\nQuestion: A plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the p…
\nA plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the plane clockwise through angle $\\theta$ around point $A^\\prime$ to point $B^\\prime$. Suppose that $AB^\\prime = \\frac{4}{3}$. The value of $\\cos \\theta$ can be written as $\\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n1909 | \n1 | \n0 | \n0.559 | \n65 | \nanswered | \n20.36 | \n
| 1 | \n7 | \n1900 | \n4 | \n0 | \n0.601 | \n65 | \nanswered | \n20.50 | \n
| 2 | \n6 | \n2084 | \n2 | \n0 | \n0.654 | \n65 | \nanswered | \n22.18 | \n
| 3 | \n5 | \n2500 | \n2 | \n0 | \n0.425 | \n65 | \nanswered | \n25.94 | \n
UNANIMOUS ANSWER: 65
\nAnswer: 65 | Ground Truth: 65 | ✅
\nID: 0e674f\nQuestion: A hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk suc…
\nA hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk such that a sphere of radius $42$ can be placed on top of the disk at $P$ and lie completely inside the hemisphere. The area of $\\mathcal T$ divided by the area of the disk is $\\tfrac pq,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n2501 | \n2 | \n0 | \n0.778 | \n79 | \nanswered | \n27.00 | \n
| 1 | \n7 | \n2796 | \n2 | \n0 | \n0.622 | \n79 | \nanswered | \n31.03 | \n
| 2 | \n5 | \n2956 | \n4 | \n0 | \n0.784 | \n79 | \nanswered | \n32.15 | \n
| 3 | \n1 | \n3050 | \n6 | \n0 | \n0.789 | \n79 | \nanswered | \n33.80 | \n
UNANIMOUS ANSWER: 79
\nAnswer: 79 | Ground Truth: 79 | ✅
\nID: f2f2d9\nQuestion: The integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and le…
\nThe integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and let $M$ be the sum of the absolute differences between adjacent cells. That is,\n[\nM = \\sum^8_{i=1} \\sum^7_{j=1} (|a_{i,j+1} - a_{i,j}| + |a_{j+1,i} - a_{j,i}|).\n]\nFind the remainder when the maximum possible value of $M$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 24 times…\n🐍 Executing Python code 13 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 17 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n9031 | \n12 | \n2 | \n0.745 | \n896 | \nanswered | \n104.73 | \n
| 1 | \n8 | \n15183 | \n9 | \n0 | \n0.739 | \n896 | \nanswered | \n176.24 | \n
| 2 | \n4 | \n16049 | \n19 | \n4 | \n0.613 | \n896 | \nanswered | \n190.60 | \n
| 3 | \n2 | \n15551 | \n16 | \n2 | \n0.738 | \n896 | \nanswered | \n196.55 | \n
UNANIMOUS ANSWER: 896
\nAnswer: 896 | Ground Truth: 896 | ✅
\nID: d5d04c\nQuestion: Find the greatest integer $n$ such that the cubic polynomial\n[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta…
\nFind the greatest integer $n$ such that the cubic polynomial\n[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta$, and $\\gamma$ are complex numbers, and there are exactly seven different possible values for $\\alpha + \\beta + \\gamma$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 8 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n8782 | \n13 | \n1 | \n0.715 | \n132 | \nanswered | \n98.54 | \n
| 1 | \n6 | \n9559 | \n13 | \n1 | \n0.625 | \n132 | \nanswered | \n107.40 | \n
| 2 | \n5 | \n12093 | \n14 | \n3 | \n0.682 | \n132 | \nanswered | \n146.25 | \n
| 3 | \n3 | \n13308 | \n14 | \n1 | \n0.703 | \n132 | \nanswered | \n147.31 | \n
UNANIMOUS ANSWER: 132
\nAnswer: 132 | Ground Truth: 132 | ✅
\nID: e6ea64\nQuestion: Let $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$….
\nLet $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 8 times…\n🐍 Executing Python code 4 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n1675 | \n5 | \n0 | \n0.506 | \n244 | \nanswered | \n18.56 | \n
| 1 | \n1 | \n2406 | \n2 | \n0 | \n0.478 | \n244 | \nanswered | \n26.29 | \n
| 2 | \n8 | \n2524 | \n4 | \n0 | \n0.620 | \n244 | \nanswered | \n27.75 | \n
| 3 | \n2 | \n2953 | \n9 | \n0 | \n0.486 | \n244 | \nanswered | \n32.03 | \n
UNANIMOUS ANSWER: 244
\nAnswer: 244 | Ground Truth: 244 | ✅
\nID: 8ff9e9\nQuestion: Find the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:
\n$a_k \\in {1,2,3}$ for all $k.$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$
$a_1a_2 a…
\nFind the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:
$a_k \\in {1,2,3}$ for all $k.$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$
$a_1a_2 a_4 + a_2a_3a_5 + a_3a_4 a_6 + a_4 a_5 a_7 + a_5 a_6 a_1 + a_6 a_7 a_2 + a_7 a_1 a_3$ is a multiple of $3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
Stage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n2 | \n1200 | \n2 | \n0 | \n0.638 | \n393 | \nanswered | \n13.87 | \n
| 1 | \n3 | \n1429 | \n2 | \n0 | \n0.748 | \n393 | \nanswered | \n16.41 | \n
| 2 | \n5 | \n1593 | \n3 | \n0 | \n0.702 | \n393 | \nanswered | \n18.12 | \n
| 3 | \n1 | \n1709 | \n2 | \n0 | \n0.740 | \n393 | \nanswered | \n19.08 | \n
UNANIMOUS ANSWER: 393
\nAnswer: 393 | Ground Truth: 393 | ✅
\nID: ac19f7\nQuestion: A standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The…
\nA standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The probability that Alice and Bob each receive at least two coins before Carol receives any coins can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $100m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n4 | \n2868 | \n2 | \n0 | \n0.546 | \n754 | \nanswered | \n33.70 | \n
| 1 | \n8 | \n3703 | \n7 | \n0 | \n0.696 | \n754 | \nanswered | \n43.71 | \n
| 2 | \n7 | \n3781 | \n4 | \n0 | \n0.731 | \n754 | \nanswered | \n44.75 | \n
| 3 | \n5 | \n5266 | \n7 | \n0 | \n0.552 | \n754 | \nanswered | \n57.80 | \n
UNANIMOUS ANSWER: 754
\nAnswer: 754 | Ground Truth: 754 | ✅
\nID: d47975\nQuestion: Isosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both …
\nIsosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both triangles have integer lengths. Find the minimum possible value of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n6887 | \n4 | \n0 | \n0.634 | \n245 | \nanswered | \n74.73 | \n
| 1 | \n1 | \n7072 | \n6 | \n1 | \n0.730 | \n245 | \nanswered | \n77.36 | \n
| 2 | \n4 | \n7834 | \n9 | \n0 | \n0.626 | \n245 | \nanswered | \n85.49 | \n
| 3 | \n3 | \n7305 | \n6 | \n1 | \n0.667 | \n245 | \nanswered | \n85.61 | \n
UNANIMOUS ANSWER: 245
\nAnswer: 245 | Ground Truth: 245 | ✅
\nID: 0015eb\nQuestion: Let $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpend…
\nLet $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpendicular $\\overline{BC},$ with $A'$ and $B$ not on the same side of line $B'C'.$ Find the integer closest to the area of hexagon $AA'CC'BB'.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 8 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 9 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n6640 | \n10 | \n1 | \n0.654 | \n156 | \nanswered | \n76.62 | \n
| 1 | \n6 | \n8349 | \n20 | \n2 | \n0.639 | \n156 | \nanswered | \n98.50 | \n
| 2 | \n2 | \n9434 | \n14 | \n1 | \n0.663 | \n156 | \nanswered | \n109.25 | \n
| 3 | \n4 | \n9341 | \n20 | \n1 | \n0.672 | \n156 | \nanswered | \n109.50 | \n
UNANIMOUS ANSWER: 156
\nAnswer: 156 | Ground Truth: 156 | ✅
\nID: 1eaadd\nQuestion: Triangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four sph…
\nTriangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four spheres, all on the same side of $\\mathcal P,$ have radii $1, 2, 3,$ and $r$ and are tangent to $\\mathcal P$ at points $A, B, C,$ and $D,$ respectively. The four spheres are also each tangent to a second plane $\\mathcal T$ and are all on the same side of $\\mathcal T.$ The value of $r$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 6 times…\n🐍 Executing Python code 9 times…\n🐍 Executing Python code 17 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n7008 | \n9 | \n0 | \n0.716 | \n161 | \nanswered | \n79.21 | \n
| 1 | \n7 | \n7594 | \n10 | \n0 | \n0.655 | \n161 | \nanswered | \n86.17 | \n
| 2 | \n1 | \n8101 | \n11 | \n1 | \n0.592 | \n161 | \nanswered | \n91.21 | \n
| 3 | \n2 | \n9021 | \n9 | \n1 | \n0.526 | \n161 | \nanswered | \n99.51 | \n
UNANIMOUS ANSWER: 161
\nAnswer: 161 | Ground Truth: 161 | ✅
\nID: 4d8666\nQuestion: Patrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after…
\nPatrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after Tanya left, Jose started bicycling along the same road from school to the park. Tanya ran at a constant rate of $2$ miles per hour faster than Patrick walked, Jose bicycled at a constant rate of $7$ miles per hour faster than Tanya ran, and all three arrived at the park at the same time. The distance from the school to the park is $\\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n1231 | \n1 | \n0 | \n0.487 | \n277 | \nanswered | \n12.75 | \n
| 1 | \n2 | \n1423 | \n1 | \n0 | \n0.514 | \n277 | \nanswered | \n14.91 | \n
| 2 | \n8 | \n1436 | \n1 | \n0 | \n0.468 | \n277 | \nanswered | \n14.99 | \n
| 3 | \n5 | \n1518 | \n2 | \n0 | \n0.551 | \n277 | \nanswered | \n15.80 | \n
UNANIMOUS ANSWER: 277
\nAnswer: 277 | Ground Truth: 277 | ✅
\nID: 7968ea\nQuestion: A real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$?…
\nA real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$? Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 3 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n3 | \n2005 | \n5 | \n0 | \n0.534 | \n441 | \nanswered | \n22.34 | \n
| 1 | \n4 | \n2857 | \n5 | \n1 | \n0.491 | \n441 | \nanswered | \n31.46 | \n
| 2 | \n6 | \n2808 | \n8 | \n1 | \n0.528 | \n441 | \nanswered | \n31.46 | \n
| 3 | \n8 | \n3186 | \n5 | \n1 | \n0.454 | \n441 | \nanswered | \n34.74 | \n
UNANIMOUS ANSWER: 441
\nAnswer: 441 | Ground Truth: 441 | ✅
\nID: 282256\nQuestion: Find the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a numb…
\nFind the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a number whose representation reads the same from left to right as from right to left. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 2 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n3449 | \n2 | \n0 | \n0.777 | \n62 | \nanswered | \n39.15 | \n
| 1 | \n6 | \n3395 | \n4 | \n1 | \n0.793 | \n62 | \nanswered | \n44.15 | \n
| 2 | \n1 | \n3957 | \n7 | \n0 | \n0.749 | \n62 | \nanswered | \n44.98 | \n
| 3 | \n4 | \n4078 | \n2 | \n0 | \n0.644 | \n62 | \nanswered | \n45.28 | \n
UNANIMOUS ANSWER: 62
\nAnswer: 62 | Ground Truth: 62 | ✅
\nID: 51697b\nQuestion: Consider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The…
\nConsider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 11 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 8 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n8 | \n8545 | \n13 | \n0 | \n0.645 | \n223 | \nanswered | \n98.52 | \n
| 1 | \n1 | \n8768 | \n9 | \n0 | \n0.585 | \n223 | \nanswered | \n100.38 | \n
| 2 | \n7 | \n9027 | \n12 | \n1 | \n0.623 | \n223 | \nanswered | \n104.21 | \n
| 3 | \n5 | \n9244 | \n10 | \n0 | \n0.782 | \n223 | \nanswered | \n105.55 | \n
UNANIMOUS ANSWER: 223
\nAnswer: 223 | Ground Truth: 223 | ✅
\nID: 652b17\nQuestion: An urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability …
\nAn urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability that exactly $4$ of them are red equals the probability that exactly $5$ of them are red. Find the sum of the five least values of $n$ for which this is possible. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n1686 | \n3 | \n0 | \n0.578 | \n190 | \nanswered | \n17.93 | \n
| 1 | \n2 | \n2172 | \n3 | \n0 | \n0.591 | \n190 | \nanswered | \n23.36 | \n
| 2 | \n7 | \n2213 | \n2 | \n0 | \n0.543 | \n190 | \nanswered | \n23.74 | \n
| 3 | \n4 | \n2391 | \n3 | \n0 | \n0.574 | \n190 | \nanswered | \n25.48 | \n
UNANIMOUS ANSWER: 190
\nAnswer: 190 | Ground Truth: 190 | ✅
\nID: 7a1a55\nQuestion: For each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, the…
\nFor each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, then $b=8$, and $72$ as a numeral in base $8$ equals $7\\cdot 8+2=58$; therefore $f(72)=58$. Find the number of positive integers $n$ less than $1000$ such that $f(n)=n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n5 | \n2361 | \n3 | \n0 | \n0.877 | \n279 | \nanswered | \n25.78 | \n
| 1 | \n1 | \n2679 | \n5 | \n0 | \n0.780 | \n279 | \nanswered | \n29.39 | \n
| 2 | \n7 | \n2998 | \n4 | \n0 | \n0.707 | \n279 | \nanswered | \n32.58 | \n
| 3 | \n4 | \n3046 | \n2 | \n0 | \n0.635 | \n279 | \nanswered | \n32.90 | \n
UNANIMOUS ANSWER: 279
\nAnswer: 279 | Ground Truth: 279 | ✅
\nID: c6d3cf\nQuestion: In an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be…
\nIn an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be expressed as $m \\sqrt n,$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 3 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 11 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 12 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 17 times…\n🐍 Executing Python code 16 times…\n🐍 Executing Python code 14 times…\n🐍 Executing Python code 21 times…\n🐍 Executing Python code 15 times…\n🐍 Executing Python code 22 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n1 | \n8395 | \n12 | \n2 | \n0.545 | \n681 | \nanswered | \n96.43 | \n
| 1 | \n4 | \n15102 | \n17 | \n3 | \n0.574 | \n681 | \nanswered | \n185.96 | \n
| 2 | \n6 | \n15529 | \n20 | \n6 | \n0.588 | \n681 | \nanswered | \n186.80 | \n
| 3 | \n5 | \n16184 | \n17 | \n2 | \n0.572 | \n681 | \nanswered | \n195.80 | \n
UNANIMOUS ANSWER: 681
\nAnswer: 681 | Ground Truth: 681 | ✅
\nID: 5e99c1\nQuestion: Joanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then …
\nJoanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then rolls the die again, places the sticker labeled 2 on the top face, and continues this process to place the rest of the stickers in order. If the die ever lands with a sticker already on its top face, the new sticker is placed to cover the old sticker. Let $p$ be the conditional probability that at the end of the process exactly one face has been left blank, given that all the even-numbered stickers are visible on faces of the die. Then $p$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 5 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 7 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n6043 | \n3 | \n0 | \n0.765 | \n29 | \nanswered | \n66.51 | \n
| 1 | \n5 | \n6129 | \n5 | \n1 | \n0.885 | \n29 | \nanswered | \n74.45 | \n
| 2 | \n1 | \n7787 | \n1 | \n0 | \n0.792 | \n29 | \nanswered | \n84.05 | \n
| 3 | \n2 | \n8489 | \n6 | \n0 | \n0.838 | \n29 | \nanswered | \n91.10 | \n
UNANIMOUS ANSWER: 29
\nAnswer: 29 | Ground Truth: 29 | ✅
\nID: 77bffe\nQuestion: Find the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms….
\nFind the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 1 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n7 | \n1256 | \n2 | \n0 | \n0.803 | \n178 | \nanswered | \n13.77 | \n
| 1 | \n2 | \n1629 | \n2 | \n0 | \n0.761 | \n178 | \nanswered | \n17.64 | \n
| 2 | \n3 | \n1974 | \n2 | \n0 | \n0.786 | \n178 | \nanswered | \n21.36 | \n
| 3 | \n1 | \n2330 | \n3 | \n0 | \n0.727 | \n178 | \nanswered | \n24.74 | \n
UNANIMOUS ANSWER: 178
\nAnswer: 178 | Ground Truth: 178 | ✅
\nID: c4ae02\nQuestion: For each positive integer $r$ less than $502,$ define\n[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all bino…
\nFor each positive integer $r$ less than $502,$ define\n[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all binomial coefficients of the form $\\binom{10000}{k}$ for which $0\\le k\\le 10000$ and $k-r$ is a multiple of $502.$ Find the number of integers in the list $S_0,S_1,\\dots,S_{501}$ that are multiples of the prime number $503.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}.
\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds
\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 2 times…\n🐍 Executing Python code 1 times…\n🐍 Executing Python code 4 times…\n🐍 Executing Python code 7 times…\n🐍 Executing Python code 10 times…\n🐍 Executing Python code 9 times…
\n| Attempt | \nTurns | \nResponse Length | \nPython Calls | \nPython Errors | \nEntropy | \nAnswer | \nExit Reason | \nCompletion Time | \n
|---|---|---|---|---|---|---|---|---|
| 0 | \n6 | \n5037 | \n6 | \n0 | \n0.682 | \n39 | \nanswered | \n62.10 | \n
| 1 | \n1 | \n5244 | \n7 | \n1 | \n0.644 | \n39 | \nanswered | \n64.32 | \n
| 2 | \n8 | \n5498 | \n15 | \n2 | \n0.702 | \n39 | \nanswered | \n67.36 | \n
| 3 | \n3 | \n5896 | \n7 | \n0 | \n0.590 | \n39 | \nanswered | \n69.41 | \n
UNANIMOUS ANSWER: 39
\nAnswer: 39 | Ground Truth: 39 | ✅
\n==================================================
\nCorrect: 30/30
\nDetails:\n 8742f9: pred=243, gt=243 ✅\n 5cc894: pred=83, gt=83 ✅\n 744366: pred=669, gt=669 ✅\n b80bba: pred=396, gt=396 ✅\n bce9ef: pred=503, gt=503 ✅\n cba4eb: pred=70, gt=70 ✅\n cda722: pred=157, gt=157 ✅\n f7b4a9: pred=850, gt=850 ✅\n f3d7b7: pred=50, gt=50 ✅\n f969f6: pred=107, gt=107 ✅\n f3d797: pred=65, gt=65 ✅\n 0e674f: pred=79, gt=79 ✅\n f2f2d9: pred=896, gt=896 ✅\n d5d04c: pred=132, gt=132 ✅\n e6ea64: pred=244, gt=244 ✅\n 8ff9e9: pred=393, gt=393 ✅\n ac19f7: pred=754, gt=754 ✅\n d47975: pred=245, gt=245 ✅\n 0015eb: pred=156, gt=156 ✅\n 1eaadd: pred=161, gt=161 ✅\n 4d8666: pred=277, gt=277 ✅\n 7968ea: pred=441, gt=441 ✅\n 282256: pred=62, gt=62 ✅\n 51697b: pred=223, gt=223 ✅\n 652b17: pred=190, gt=190 ✅\n 7a1a55: pred=279, gt=279 ✅\n c6d3cf: pred=681, gt=681 ✅\n 5e99c1: pred=29, gt=29 ✅\n 77bffe: pred=178, gt=178 ✅\n c4ae02: pred=39, gt=39 ✅
", "rawMarkdown": "## Inference Logs on AIME 2026 (Question 1)\n\n------\nID: 8742f9\nQuestion: The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex...\n\n The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Let $N$ be the number of paths the bug can take from the lower left corner ($A$) to the upper right corner ($B$). One such path from $A$ to $B$ is shown by the thick line segments in the figure. Find $\\sqrt{N}$.\n\n[asy]\nsize(12cm);\ndraw((0, 0)--(0, 1));\nfor (int i = 0; i <= 9; i += 1)\n{\ndraw((i, 1)--(i + 1, 1)--(i + 1, 0)--(i, 0)--(i + 1, 1));\n}\ndraw((0, 0)--(0, 1)--(2, 1)--(2, 0)--(3, 1)--(3, 0)--(4, 1)--(4, 0)--(5, 0)--(5, 1)--(6, 1)--(6, 0)--(7, 1)--(8, 1)--(8, 0)--(10, 0)--(10, 1), red + linewidth(2));\nlabel(\"$A$\", (0, 0), dir(180));\nlabel(\"$B$\", (10, 1), dir(0));\n[/asy] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 2 | 5024 | 4 | 0 | 0.811 | 243 | answered | 59.13 |\n| 1 | 7 | 6260 | 2 | 0 | 0.903 | 243 | answered | 72.26 |\n| 2 | 3 | 6497 | 2 | 0 | 0.798 | 243 | answered | 74.67 |\n| 3 | 8 | 7256 | 5 | 1 | 0.794 | 243 | answered | 82.55 |\n\n\nUNANIMOUS ANSWER: 243\n\nAnswer: 243 | Ground Truth: 243 | ✅\n📊 Running Accuracy: 1/1 (100.0%)\n------\n\n------\nID: 5cc894\nQuestion: Let $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the...\n\n Let $a, b,$ and $n$ be positive integers with both $a$ and $b$ greater than or equal to $2$ and less than or equal to $2n$. Define an $a \\times b$ cell loop in a $2n \\times 2n$ grid of cells to be the $2a + 2b - 4$ cells that surround an $(a - 2) \\times (b - 2)$ (possibly empty) rectangle of cells in the grid. For example, the following diagram shows a way to partition a $6 \\times 6$ grid of cells into $4$ cell loops.\n\n| P P P P | Y Y |\n| P | R R | P | Y | Y |\n| P | R R | P | Y | Y |\n| P P P P | Y | Y |\n| G G G G | Y | Y |\n| G G G G | Y Y |\n\nFind the number of ways to partition a $10 \\times 10$ grid of cells into $5$ cell loops so that every cell of the grid belongs to exactly one cell loop. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 6 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 6 | 8177 | 9 | 0 | 0.891 | 83 | answered | 98.10 |\n| 1 | 4 | 9962 | 4 | 0 | 0.844 | 179 | answered | 109.96 |\n| 2 | 1 | 9566 | 6 | 1 | 0.784 | 83 | answered | 114.32 |\n| 3 | 8 | 10538 | 8 | 1 | 0.849 | 83 | answered | 128.12 |\n| 4 | 5 | 10843 | 21 | 3 | 0.779 | 83 | answered | 130.85 |\n\n\nUNANIMOUS ANSWER: 83\n\nAnswer: 83 | Ground Truth: 83 | ✅\n📊 Running Accuracy: 2/2 (100.0%)\n------\n\n------\nID: 744366\nQuestion: Let $S$ denote the value of the infinite sum\n\\[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n\\]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$...\n\n Let $S$ denote the value of the infinite sum\n\\[\n\\frac{1}{9} + \\frac{1}{99} + \\frac{1}{999} + \\frac{1}{9999} + \\cdots\n\\]\nFind the remainder when the greatest integer less than or equal to $10^{100} S$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 6 | 4410 | 8 | 0 | 0.737 | 669 | answered | 50.35 |\n| 1 | 8 | 4493 | 12 | 1 | 0.663 | 669 | answered | 51.79 |\n| 2 | 3 | 4585 | 6 | 0 | 0.625 | 669 | answered | 51.88 |\n| 3 | 2 | 5249 | 5 | 0 | 0.670 | 669 | answered | 58.13 |\n\n\nUNANIMOUS ANSWER: 669\n\nAnswer: 669 | Ground Truth: 669 | ✅\n📊 Running Accuracy: 3/3 (100.0%)\n------\n\n------\nID: b80bba\nQuestion: \nFind the number of functions $\\pi$ mapping the set $A =\\{1,2,3,4,5,6\\}$ onto $A$ such that for every $a \\in A,$\n\\[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n\\]...\n\n \nFind the number of functions $\\pi$ mapping the set $A =\\{1,2,3,4,5,6\\}$ onto $A$ such that for every $a \\in A,$\n\\[\n\\pi(\\pi(\\pi(\\pi(\\pi(\\pi(a)))))) = a.\n\\] Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 2 | 1651 | 2 | 0 | 0.660 | 396 | answered | 18.17 |\n| 1 | 5 | 1773 | 3 | 0 | 0.723 | 396 | answered | 19.55 |\n| 2 | 1 | 2056 | 2 | 0 | 0.669 | 396 | answered | 22.35 |\n| 3 | 3 | 2127 | 3 | 0 | 0.607 | 396 | answered | 23.11 |\n\n\nUNANIMOUS ANSWER: 396\n\nAnswer: 396 | Ground Truth: 396 | ✅\n📊 Running Accuracy: 4/4 (100.0%)\n------\n\n------\nID: bce9ef\nQuestion: Let $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D...\n\n Let $ABCDE$ be a nonconvex pentagon with internal angles $\\angle A = \\angle E = 90^\\circ$ and $\\angle B = \\angle D = 45^\\circ.$ Suppose that $DE < AB, AE = 20, BC = 14\\sqrt2,$ and points $B,C,$ and $D$ lie on the same side of line $AE.$ Suppose further that $AB$ is an integer with $AB < 2026$ and the area of pentagon $ABCDE$ is an integer multiple of $16.$ Find the number of possible values of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 6 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 2 | 12334 | 3 | 0 | 0.755 | 503 | answered | 136.30 |\n| 1 | 4 | 12760 | 13 | 1 | 0.761 | 1002 | answered | 143.73 |\n| 2 | 5 | 13789 | 5 | 0 | 0.754 | 503 | answered | 153.05 |\n| 3 | 8 | 13974 | 7 | 0 | 0.751 | 503 | answered | 155.26 |\n| 4 | 6 | 15984 | 9 | 2 | 0.729 | 503 | answered | 172.94 |\n\n\nUNANIMOUS ANSWER: 503\n\nAnswer: 503 | Ground Truth: 503 | ✅\n📊 Running Accuracy: 5/5 (100.0%)\n------\n\n------\nID: cba4eb\nQuestion: Find the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b....\n\n Find the number of integers less than or equal to 100 that are equal to $a+b+ab$ for some choice of distinct positive integers a and b. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 5 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 7 | 2144 | 4 | 0 | 0.736 | 70 | answered | 24.08 |\n| 1 | 2 | 2339 | 3 | 0 | 0.794 | 70 | answered | 26.17 |\n| 2 | 1 | 2653 | 5 | 0 | 0.764 | 70 | answered | 29.58 |\n| 3 | 5 | 2703 | 4 | 0 | 0.780 | 70 | answered | 30.02 |\n\n\nUNANIMOUS ANSWER: 70\n\nAnswer: 70 | Ground Truth: 70 | ✅\n📊 Running Accuracy: 6/6 (100.0%)\n------\n\n------\nID: cda722\nQuestion: For integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n\\[\na_1 + ...\n\n For integers $a$ and $b,$ let $a \\circ b = a - b$ if $a$ is odd and $b$ is even, and $a+b$ otherwise. Find the number of sequences $a_1, a_2, a_3, \\ldots, a_n$ of positive integers such that\n\\[\na_1 + a_2 + a_3 + \\cdots + a_n = 12 \\quad \\text{and} \\quad a_1 \\circ a_2 \\circ a_3 \\circ \\cdots \\circ a_n = 0\n\\]\nwhere the operations are performed from left to right; that is, $a_1 \\circ a_2 \\circ a_3$ means $(a_1 \\circ a_2) \\circ a_3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 5 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 5 | 2209 | 3 | 0 | 0.875 | 157 | answered | 25.53 |\n| 1 | 3 | 2249 | 4 | 0 | 0.837 | 157 | answered | 26.15 |\n| 2 | 8 | 2406 | 3 | 0 | 0.845 | 157 | answered | 27.62 |\n| 3 | 7 | 2502 | 3 | 0 | 0.897 | 157 | answered | 28.55 |\n\n\nUNANIMOUS ANSWER: 157\n\nAnswer: 157 | Ground Truth: 157 | ✅\n📊 Running Accuracy: 7/7 (100.0%)\n------\n\n------\nID: f7b4a9\nQuestion: Let $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC...\n\n Let $\\triangle ABC$ be a triangle with $D$ on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC.$ Let $\\omega$ be the circle that passes through $A$ and is tangent to segment $\\overline{BC}$ at $D.$ Let $E \\neq A$ and $F \\neq A$ be the intersections of $\\omega$ with segments $\\overline{AB}$ and $\\overline{AC},$ respectively. Suppose that $AB = 200, AC = 225,$ and all of $AE, AF, BD,$ and $CD$ are positive integers. Find the sum of all possible values of $BC.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 4 | 5727 | 2 | 0 | 0.758 | 850 | answered | 60.41 |\n| 1 | 2 | 6207 | 2 | 0 | 0.708 | 850 | answered | 65.95 |\n| 2 | 6 | 6343 | 3 | 0 | 0.540 | 850 | answered | 67.47 |\n| 3 | 3 | 6727 | 2 | 0 | 0.712 | 850 | answered | 71.45 |\n\n\nUNANIMOUS ANSWER: 850\n\nAnswer: 850 | Ground Truth: 850 | ✅\n📊 Running Accuracy: 8/8 (100.0%)\n------\n\n------\nID: f3d7b7\nQuestion: Find the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$...\n\n Find the sum of all real numbers $r$ such that there is at least one point where the circle with radius $r$ centered at $(4, 39)$ is tangent to the parabola with equation $2y = x^2 - 8x + 12.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 3 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 8 | 2711 | 5 | 0 | 0.687 | 50 | answered | 29.22 |\n| 1 | 4 | 3372 | 1 | 0 | 0.698 | 50 | answered | 35.80 |\n| 2 | 2 | 3659 | 8 | 0 | 0.672 | 50 | answered | 39.63 |\n| 3 | 3 | 3806 | 2 | 0 | 0.701 | 50 | answered | 40.41 |\n\n\nUNANIMOUS ANSWER: 50\n\nAnswer: 50 | Ground Truth: 50 | ✅\n📊 Running Accuracy: 9/9 (100.0%)\n------\n\n------\nID: f969f6\nQuestion: Call finite sets of integers $S$ and $T$ cousins if\n- $S$ and $T$ have the same number of elements,\n- $S$ and $T$ are disjoint, and\n- the elements of $S$ can be paired with the elements of $T$ so that...\n\n Call finite sets of integers $S$ and $T$ cousins if\n- $S$ and $T$ have the same number of elements,\n- $S$ and $T$ are disjoint, and\n- the elements of $S$ can be paired with the elements of $T$ so that the elements in each pair differ by exactly $1$.\nFor example, $\\{1,2,5\\}$ and $\\{0,3,4\\}$ are cousins. Suppose that the set $S$ has exactly $4040$ cousins. Find the least number of elements the set $S$ can have. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 12 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 5 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 8 | 11691 | 6 | 0 | 0.868 | 107 | answered | 130.24 |\n| 1 | 4 | 14143 | 16 | 2 | 0.859 | 107 | answered | 167.23 |\n| 2 | 3 | 16430 | 6 | 0 | 0.893 | 107 | answered | 184.39 |\n| 3 | 7 | 19128 | 22 | 3 | 0.821 | 107 | answered | 220.86 |\n\n\nUNANIMOUS ANSWER: 107\n\nAnswer: 107 | Ground Truth: 107 | ✅\n📊 Running Accuracy: 10/10 (100.0%)\n------\n\n------\nID: f3d797\nQuestion: A plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the p...\n\n A plane contains points $A$ and $B$ with $AB = 1$. Point $A$ is rotated in the plane counterclockwise through an acute angle $\\theta$ around point $B$ to point $A^\\prime$. Then $B$ is rotated in the plane clockwise through angle $\\theta$ around point $A^\\prime$ to point $B^\\prime$. Suppose that $AB^\\prime = \\frac{4}{3}$. The value of $\\cos \\theta$ can be written as $\\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 8 | 1909 | 1 | 0 | 0.559 | 65 | answered | 20.36 |\n| 1 | 7 | 1900 | 4 | 0 | 0.601 | 65 | answered | 20.50 |\n| 2 | 6 | 2084 | 2 | 0 | 0.654 | 65 | answered | 22.18 |\n| 3 | 5 | 2500 | 2 | 0 | 0.425 | 65 | answered | 25.94 |\n\n\nUNANIMOUS ANSWER: 65\n\nAnswer: 65 | Ground Truth: 65 | ✅\n📊 Running Accuracy: 11/11 (100.0%)\n------\n\n------\nID: 0e674f\nQuestion: A hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk suc...\n\n A hemisphere with radius $200$ sits on top of a horizontal circular disk with radius $200,$ and the hemisphere and disk have the same center. Let $\\mathcal T$ be the region of points P in the disk such that a sphere of radius $42$ can be placed on top of the disk at $P$ and lie completely inside the hemisphere. The area of $\\mathcal T$ divided by the area of the disk is $\\tfrac pq,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 6 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 3 | 2501 | 2 | 0 | 0.778 | 79 | answered | 27.00 |\n| 1 | 7 | 2796 | 2 | 0 | 0.622 | 79 | answered | 31.03 |\n| 2 | 5 | 2956 | 4 | 0 | 0.784 | 79 | answered | 32.15 |\n| 3 | 1 | 3050 | 6 | 0 | 0.789 | 79 | answered | 33.80 |\n\n\nUNANIMOUS ANSWER: 79\n\nAnswer: 79 | Ground Truth: 79 | ✅\n📊 Running Accuracy: 12/12 (100.0%)\n------\n\n------\nID: f2f2d9\nQuestion: The integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and le...\n\n The integers from $1$ to $64$ are placed in some order into an $8 \\times 8$ grid of cells with one number in each cell. Let $a_{i,j}$ be the number placed in the cell in row $i$ and column $j,$ and let $M$ be the sum of the absolute differences between adjacent cells. That is,\n\\[\nM = \\sum^8_{i=1} \\sum^7_{j=1} (|a_{i,j+1} - a_{i,j}| + |a_{j+1,i} - a_{j,i}|).\n\\]\nFind the remainder when the maximum possible value of $M$ is divided by $1000.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 24 times...\n🐍 Executing Python code 13 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 15 times...\n🐍 Executing Python code 17 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 3 | 9031 | 12 | 2 | 0.745 | 896 | answered | 104.73 |\n| 1 | 8 | 15183 | 9 | 0 | 0.739 | 896 | answered | 176.24 |\n| 2 | 4 | 16049 | 19 | 4 | 0.613 | 896 | answered | 190.60 |\n| 3 | 2 | 15551 | 16 | 2 | 0.738 | 896 | answered | 196.55 |\n\n\nUNANIMOUS ANSWER: 896\n\nAnswer: 896 | Ground Truth: 896 | ✅\n📊 Running Accuracy: 13/13 (100.0%)\n------\n\n------\nID: d5d04c\nQuestion: Find the greatest integer $n$ such that the cubic polynomial\n\\[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n\\]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta...\n\n Find the greatest integer $n$ such that the cubic polynomial\n\\[\nx^{3} - \\displaystyle\\frac{n}{6}x^{2} + (n - 11)x - 400\n\\]\nhas roots $\\alpha^{2}$, $\\beta^{2}$, and $\\gamma^{2}$, where $\\alpha$, $\\beta$, and $\\gamma$ are complex numbers, and there are exactly seven different possible values for $\\alpha + \\beta + \\gamma$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 9 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 9 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 8 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 2 | 8782 | 13 | 1 | 0.715 | 132 | answered | 98.54 |\n| 1 | 6 | 9559 | 13 | 1 | 0.625 | 132 | answered | 107.40 |\n| 2 | 5 | 12093 | 14 | 3 | 0.682 | 132 | answered | 146.25 |\n| 3 | 3 | 13308 | 14 | 1 | 0.703 | 132 | answered | 147.31 |\n\n\nUNANIMOUS ANSWER: 132\n\nAnswer: 132 | Ground Truth: 132 | ✅\n📊 Running Accuracy: 14/14 (100.0%)\n------\n\n------\nID: e6ea64\nQuestion: Let $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$....\n\n Let $N$ be the number of positive integer divisors of $17017^{17}$ that leave a remainder of $5$ when divided by $12$. Find the remainder when $N$ is divided by $1000$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 8 times...\n🐍 Executing Python code 4 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 4 | 1675 | 5 | 0 | 0.506 | 244 | answered | 18.56 |\n| 1 | 1 | 2406 | 2 | 0 | 0.478 | 244 | answered | 26.29 |\n| 2 | 8 | 2524 | 4 | 0 | 0.620 | 244 | answered | 27.75 |\n| 3 | 2 | 2953 | 9 | 0 | 0.486 | 244 | answered | 32.03 |\n\n\nUNANIMOUS ANSWER: 244\n\nAnswer: 244 | Ground Truth: 244 | ✅\n📊 Running Accuracy: 15/15 (100.0%)\n------\n\n------\nID: 8ff9e9\nQuestion: Find the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:\n- $a_k \\in \\{1,2,3\\}$ for all $k.$\n- $a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$\n- $a_1a_2 a...\n\n Find the number of ordered 7-tuples $(a_1, a_2, a_3, \\ldots, a_7)$ having the following properties:\n- $a_k \\in \\{1,2,3\\}$ for all $k.$\n- $a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3.$\n- $a_1a_2 a_4 + a_2a_3a_5 + a_3a_4 a_6 + a_4 a_5 a_7 + a_5 a_6 a_1 + a_6 a_7 a_2 + a_7 a_1 a_3$ is a multiple of $3.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 2 | 1200 | 2 | 0 | 0.638 | 393 | answered | 13.87 |\n| 1 | 3 | 1429 | 2 | 0 | 0.748 | 393 | answered | 16.41 |\n| 2 | 5 | 1593 | 3 | 0 | 0.702 | 393 | answered | 18.12 |\n| 3 | 1 | 1709 | 2 | 0 | 0.740 | 393 | answered | 19.08 |\n\n\nUNANIMOUS ANSWER: 393\n\nAnswer: 393 | Ground Truth: 393 | ✅\n📊 Running Accuracy: 16/16 (100.0%)\n------\n\n------\nID: ac19f7\nQuestion: A standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The...\n\n A standard fair six-sided die is rolled repeatedly. Each time the die reads 1 or 2, Alice gets a coin; each time it reads 3 or 4, Bob gets a coin; and each time it reads 5 or 6, Carol gets a coin. The probability that Alice and Bob each receive at least two coins before Carol receives any coins can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $100m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 6 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 4 | 2868 | 2 | 0 | 0.546 | 754 | answered | 33.70 |\n| 1 | 8 | 3703 | 7 | 0 | 0.696 | 754 | answered | 43.71 |\n| 2 | 7 | 3781 | 4 | 0 | 0.731 | 754 | answered | 44.75 |\n| 3 | 5 | 5266 | 7 | 0 | 0.552 | 754 | answered | 57.80 |\n\n\nUNANIMOUS ANSWER: 754\n\nAnswer: 754 | Ground Truth: 754 | ✅\n📊 Running Accuracy: 17/17 (100.0%)\n------\n\n------\nID: d47975\nQuestion: Isosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both ...\n\n Isosceles triangle $\\triangle ABC$ has $AB = BC.$ Let $I$ be the incenter of $\\triangle ABC.$ The perimeters of $\\triangle ABC$ and $\\triangle AIC$ are in the ratio $125:6,$ and all the sides of both triangles have integer lengths. Find the minimum possible value of $AB.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 6 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 5 | 6887 | 4 | 0 | 0.634 | 245 | answered | 74.73 |\n| 1 | 1 | 7072 | 6 | 1 | 0.730 | 245 | answered | 77.36 |\n| 2 | 4 | 7834 | 9 | 0 | 0.626 | 245 | answered | 85.49 |\n| 3 | 3 | 7305 | 6 | 1 | 0.667 | 245 | answered | 85.61 |\n\n\nUNANIMOUS ANSWER: 245\n\nAnswer: 245 | Ground Truth: 245 | ✅\n📊 Running Accuracy: 18/18 (100.0%)\n------\n\n------\nID: 0015eb\nQuestion: Let $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpend...\n\n Let $\\triangle ABC$ have side lengths $AB = 13, BC = 14,$ and $CA = 15.$ Triangle $\\triangle A'B'C'$ is obtained by rotating $\\triangle ABC$ about its circumcenter so that ${}\\overline{AC}$ is perpendicular $\\overline{BC},$ with $A'$ and $B$ not on the same side of line $B'C'.$ Find the integer closest to the area of hexagon $AA'CC'BB'.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 8 times...\n🐍 Executing Python code 9 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 12 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 15 times...\n🐍 Executing Python code 9 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 3 | 6640 | 10 | 1 | 0.654 | 156 | answered | 76.62 |\n| 1 | 6 | 8349 | 20 | 2 | 0.639 | 156 | answered | 98.50 |\n| 2 | 2 | 9434 | 14 | 1 | 0.663 | 156 | answered | 109.25 |\n| 3 | 4 | 9341 | 20 | 1 | 0.672 | 156 | answered | 109.50 |\n\n\nUNANIMOUS ANSWER: 156\n\nAnswer: 156 | Ground Truth: 156 | ✅\n📊 Running Accuracy: 19/19 (100.0%)\n------\n\n------\nID: 1eaadd\nQuestion: Triangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four sph...\n\n Triangle $\\triangle ABC$ lies in plane $\\mathcal P$ with $AB = 6, AC = 4,$ and $\\angle BAC = 90^\\circ.$ Let $D$ be the reflection across $\\overline{BC}$ of the centroid of $\\triangle ABC. {}$ Four spheres, all on the same side of $\\mathcal P,$ have radii $1, 2, 3,$ and $r$ and are tangent to $\\mathcal P$ at points $A, B, C,$ and $D,$ respectively. The four spheres are also each tangent to a second plane $\\mathcal T$ and are all on the same side of $\\mathcal T.$ The value of $r$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 6 times...\n🐍 Executing Python code 9 times...\n🐍 Executing Python code 17 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 5 | 7008 | 9 | 0 | 0.716 | 161 | answered | 79.21 |\n| 1 | 7 | 7594 | 10 | 0 | 0.655 | 161 | answered | 86.17 |\n| 2 | 1 | 8101 | 11 | 1 | 0.592 | 161 | answered | 91.21 |\n| 3 | 2 | 9021 | 9 | 1 | 0.526 | 161 | answered | 99.51 |\n\n\nUNANIMOUS ANSWER: 161\n\nAnswer: 161 | Ground Truth: 161 | ✅\n📊 Running Accuracy: 20/20 (100.0%)\n------\n\n------\nID: 4d8666\nQuestion: Patrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after...\n\n Patrick started walking at a constant rate along a straight road from school to the park. One hour after Patrick left, Tanya started running along the same road from school to the park. One hour after Tanya left, Jose started bicycling along the same road from school to the park. Tanya ran at a constant rate of $2$ miles per hour faster than Patrick walked, Jose bicycled at a constant rate of $7$ miles per hour faster than Tanya ran, and all three arrived at the park at the same time. The distance from the school to the park is $\\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 1 | 1231 | 1 | 0 | 0.487 | 277 | answered | 12.75 |\n| 1 | 2 | 1423 | 1 | 0 | 0.514 | 277 | answered | 14.91 |\n| 2 | 8 | 1436 | 1 | 0 | 0.468 | 277 | answered | 14.99 |\n| 3 | 5 | 1518 | 2 | 0 | 0.551 | 277 | answered | 15.80 |\n\n\nUNANIMOUS ANSWER: 277\n\nAnswer: 277 | Ground Truth: 277 | ✅\n📊 Running Accuracy: 21/21 (100.0%)\n------\n\n------\nID: 7968ea\nQuestion: A real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$?...\n\n A real number $x$ satisfies $\\sqrt[20]{x^{\\log_{2026}x}}=26x$. What is the number of positive divisors of the product of all possible positive values of $x$? Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 3 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 3 | 2005 | 5 | 0 | 0.534 | 441 | answered | 22.34 |\n| 1 | 4 | 2857 | 5 | 1 | 0.491 | 441 | answered | 31.46 |\n| 2 | 6 | 2808 | 8 | 1 | 0.528 | 441 | answered | 31.46 |\n| 3 | 8 | 3186 | 5 | 1 | 0.454 | 441 | answered | 34.74 |\n\n\nUNANIMOUS ANSWER: 441\n\nAnswer: 441 | Ground Truth: 441 | ✅\n📊 Running Accuracy: 22/22 (100.0%)\n------\n\n------\nID: 282256\nQuestion: Find the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a numb...\n\n Find the number of positive integer palindromes written in base $10$ with no zero digits, and whose digits add up to $13$. For example, $42124$ has these properties. Recall that a palindrome is a number whose representation reads the same from left to right as from right to left. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 2 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 7 | 3449 | 2 | 0 | 0.777 | 62 | answered | 39.15 |\n| 1 | 6 | 3395 | 4 | 1 | 0.793 | 62 | answered | 44.15 |\n| 2 | 1 | 3957 | 7 | 0 | 0.749 | 62 | answered | 44.98 |\n| 3 | 4 | 4078 | 2 | 0 | 0.644 | 62 | answered | 45.28 |\n\n\nUNANIMOUS ANSWER: 62\n\nAnswer: 62 | Ground Truth: 62 | ✅\n📊 Running Accuracy: 23/23 (100.0%)\n------\n\n------\nID: 51697b\nQuestion: Consider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The...\n\n Consider a tetrahedron with two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\\sqrt{10}, 5\\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent to a sphere with center $R.$ The distance $RS$ can be written as $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 12 times...\n🐍 Executing Python code 11 times...\n🐍 Executing Python code 15 times...\n🐍 Executing Python code 8 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 8 | 8545 | 13 | 0 | 0.645 | 223 | answered | 98.52 |\n| 1 | 1 | 8768 | 9 | 0 | 0.585 | 223 | answered | 100.38 |\n| 2 | 7 | 9027 | 12 | 1 | 0.623 | 223 | answered | 104.21 |\n| 3 | 5 | 9244 | 10 | 0 | 0.782 | 223 | answered | 105.55 |\n\n\nUNANIMOUS ANSWER: 223\n\nAnswer: 223 | Ground Truth: 223 | ✅\n📊 Running Accuracy: 24/24 (100.0%)\n------\n\n------\nID: 652b17\nQuestion: An urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability ...\n\n An urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability that exactly $4$ of them are red equals the probability that exactly $5$ of them are red. Find the sum of the five least values of $n$ for which this is possible. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 1 | 1686 | 3 | 0 | 0.578 | 190 | answered | 17.93 |\n| 1 | 2 | 2172 | 3 | 0 | 0.591 | 190 | answered | 23.36 |\n| 2 | 7 | 2213 | 2 | 0 | 0.543 | 190 | answered | 23.74 |\n| 3 | 4 | 2391 | 3 | 0 | 0.574 | 190 | answered | 25.48 |\n\n\nUNANIMOUS ANSWER: 190\n\nAnswer: 190 | Ground Truth: 190 | ✅\n📊 Running Accuracy: 25/25 (100.0%)\n------\n\n------\nID: 7a1a55\nQuestion: For each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, the...\n\n For each positive integer $n$ let $f(n)$ be the value of the base-ten numeral $n$ viewed in base $b$, where $b$ is the least integer greater than the greatest digit in $n$. For example, if $n=72$, then $b=8$, and $72$ as a numeral in base $8$ equals $7\\cdot 8+2=58$; therefore $f(72)=58$. Find the number of positive integers $n$ less than $1000$ such that $f(n)=n$. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 1 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 5 | 2361 | 3 | 0 | 0.877 | 279 | answered | 25.78 |\n| 1 | 1 | 2679 | 5 | 0 | 0.780 | 279 | answered | 29.39 |\n| 2 | 7 | 2998 | 4 | 0 | 0.707 | 279 | answered | 32.58 |\n| 3 | 4 | 3046 | 2 | 0 | 0.635 | 279 | answered | 32.90 |\n\n\nUNANIMOUS ANSWER: 279\n\nAnswer: 279 | Ground Truth: 279 | ✅\n📊 Running Accuracy: 26/26 (100.0%)\n------\n\n------\nID: c6d3cf\nQuestion: In an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be...\n\n In an equiangular pentagon, the sum of the squares of the side lengths equals $308,$ and the sum of the squares of the diagonal lengths equals $800.$ The square of the perimeter of the pentagon can be expressed as $m \\sqrt n,$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 3 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 11 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 12 times...\n🐍 Executing Python code 15 times...\n🐍 Executing Python code 17 times...\n🐍 Executing Python code 16 times...\n🐍 Executing Python code 14 times...\n🐍 Executing Python code 21 times...\n🐍 Executing Python code 15 times...\n🐍 Executing Python code 22 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 1 | 8395 | 12 | 2 | 0.545 | 681 | answered | 96.43 |\n| 1 | 4 | 15102 | 17 | 3 | 0.574 | 681 | answered | 185.96 |\n| 2 | 6 | 15529 | 20 | 6 | 0.588 | 681 | answered | 186.80 |\n| 3 | 5 | 16184 | 17 | 2 | 0.572 | 681 | answered | 195.80 |\n\n\nUNANIMOUS ANSWER: 681\n\nAnswer: 681 | Ground Truth: 681 | ✅\n📊 Running Accuracy: 27/27 (100.0%)\n------\n\n------\nID: 5e99c1\nQuestion: Joanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then ...\n\n Joanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then rolls the die again, places the sticker labeled 2 on the top face, and continues this process to place the rest of the stickers in order. If the die ever lands with a sticker already on its top face, the new sticker is placed to cover the old sticker. Let $p$ be the conditional probability that at the end of the process exactly one face has been left blank, given that all the even-numbered stickers are visible on faces of the die. Then $p$ can be written as $\\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 5 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 7 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 7 | 6043 | 3 | 0 | 0.765 | 29 | answered | 66.51 |\n| 1 | 5 | 6129 | 5 | 1 | 0.885 | 29 | answered | 74.45 |\n| 2 | 1 | 7787 | 1 | 0 | 0.792 | 29 | answered | 84.05 |\n| 3 | 2 | 8489 | 6 | 0 | 0.838 | 29 | answered | 91.10 |\n\n\nUNANIMOUS ANSWER: 29\n\nAnswer: 29 | Ground Truth: 29 | ✅\n📊 Running Accuracy: 28/28 (100.0%)\n------\n\n------\nID: 77bffe\nQuestion: Find the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms....\n\n Find the sum of the $10$th terms of all arithmetic sequences of integers that have first term equal to $4$ and include both $24$ and $34$ as terms. Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 1 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 7 | 1256 | 2 | 0 | 0.803 | 178 | answered | 13.77 |\n| 1 | 2 | 1629 | 2 | 0 | 0.761 | 178 | answered | 17.64 |\n| 2 | 3 | 1974 | 2 | 0 | 0.786 | 178 | answered | 21.36 |\n| 3 | 1 | 2330 | 3 | 0 | 0.727 | 178 | answered | 24.74 |\n\n\nUNANIMOUS ANSWER: 178\n\nAnswer: 178 | Ground Truth: 178 | ✅\n📊 Running Accuracy: 29/29 (100.0%)\n------\n\n------\nID: c4ae02\nQuestion: For each positive integer $r$ less than $502,$ define\n\\[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n\\]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all bino...\n\n For each positive integer $r$ less than $502,$ define\n\\[\nS_r=\\sum_{m\\ge 0}\\dbinom{10000}{502n+r},\n\\]\nwhere $\\binom{10000}{n}$ is defined to be $0$ when $n>10000.$ That is, $S_r$ is the sum of all binomial coefficients of the form $\\binom{10000}{k}$ for which $0\\le k\\le 10000$ and $k-r$ is a multiple of $502.$ Find the number of integers in the list $S_0,S_1,\\dots,S_{501}$ that are multiples of the prime number $503.$ Solve step-by-step. Use the python tool heavily: compute, don't guess. Verify your answer with a separate computation before writing \\boxed{}. \n\nStage 1 Budget: 400.00 seconds | Stage 2 Budget: 640.00 seconds | Final Budget: 900.00 seconds\n\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 2 times...\n🐍 Executing Python code 1 times...\n🐍 Executing Python code 4 times...\n🐍 Executing Python code 7 times...\n🐍 Executing Python code 10 times...\n🐍 Executing Python code 9 times...\n| Attempt | Turns | Response Length | Python Calls | Python Errors | Entropy | Answer | Exit Reason | Completion Time |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 0 | 6 | 5037 | 6 | 0 | 0.682 | 39 | answered | 62.10 |\n| 1 | 1 | 5244 | 7 | 1 | 0.644 | 39 | answered | 64.32 |\n| 2 | 8 | 5498 | 15 | 2 | 0.702 | 39 | answered | 67.36 |\n| 3 | 3 | 5896 | 7 | 0 | 0.590 | 39 | answered | 69.41 |\n\n\nUNANIMOUS ANSWER: 39\n\nAnswer: 39 | Ground Truth: 39 | ✅\n📊 Running Accuracy: 30/30 (100.0%)\n------\n\n\n==================================================\n📊 FINAL ACCURACY SUMMARY\n==================================================\nCorrect: 30/30\nAccuracy: 100.0%\n==================================================\n\nDetails:\n 8742f9: pred=243, gt=243 ✅\n 5cc894: pred=83, gt=83 ✅\n 744366: pred=669, gt=669 ✅\n b80bba: pred=396, gt=396 ✅\n bce9ef: pred=503, gt=503 ✅\n cba4eb: pred=70, gt=70 ✅\n cda722: pred=157, gt=157 ✅\n f7b4a9: pred=850, gt=850 ✅\n f3d7b7: pred=50, gt=50 ✅\n f969f6: pred=107, gt=107 ✅\n f3d797: pred=65, gt=65 ✅\n 0e674f: pred=79, gt=79 ✅\n f2f2d9: pred=896, gt=896 ✅\n d5d04c: pred=132, gt=132 ✅\n e6ea64: pred=244, gt=244 ✅\n 8ff9e9: pred=393, gt=393 ✅\n ac19f7: pred=754, gt=754 ✅\n d47975: pred=245, gt=245 ✅\n 0015eb: pred=156, gt=156 ✅\n 1eaadd: pred=161, gt=161 ✅\n 4d8666: pred=277, gt=277 ✅\n 7968ea: pred=441, gt=441 ✅\n 282256: pred=62, gt=62 ✅\n 51697b: pred=223, gt=223 ✅\n 652b17: pred=190, gt=190 ✅\n 7a1a55: pred=279, gt=279 ✅\n c6d3cf: pred=681, gt=681 ✅\n 5e99c1: pred=29, gt=29 ✅\n 77bffe: pred=178, gt=178 ✅\n c4ae02: pred=39, gt=39 ✅" } ] } ] } ], "complete": true, "source": "live", "fetched_at": "2026-10-20T04:22:09Z", "competition_id": 118448, "competition_slug": "ai-mathematical-olympiad-progress-prize-3", "team_id": 14801510, "solution_url": "/competitions/ai-mathematical-olympiad-progress-prize-3/writeups/aimo3-inference-time-reasoning-optional-rag-and" }