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One-Token Verification for Reasoning Correctness Estimation
Zhan Zhuang, Xiequn Wang, Zebin Chen, Feiyang Ye, Ying Wei, Kede Ma, Yu Zhang
Intelligence
Status: succeeded | Model: Gemma-4-26B-A4B | Prompt: intel-v1 | Confidence: 91%
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Summary
The paper introduces One-Token Verification (OTV), a method for estimating reasoning correctness in Large Language Models (LLMs) using a single forward pass. OTV utilizes a learnable special token ([ToT]) and a LoRA-based verifier module to probe the model's key-value cache, providing token-level confidence scores. This approach enables early termination of reasoning traces, reducing token usage by up to 90% while outperforming existing internal and external verifiers on mathematical reasoning benchmarks.
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Zhan Zhuang → authored → One-Token Verification for Reasoning Correctness Estimation
confidence 95% · One-Token Verification for Reasoning Correctness Estimation Zhan Zhuang
Yu Zhang → authored → One-Token Verification for Reasoning Correctness Estimation
confidence 95% · Yu Zhang
Kede Ma → authored → One-Token Verification for Reasoning Correctness Estimation
confidence 95% · Kede Ma
Ying Wei → authored → One-Token Verification for Reasoning Correctness Estimation
confidence 95% · Correspondence to: Ying Wei
Xiequn Wang → authored → One-Token Verification for Reasoning Correctness Estimation
confidence 95% · Xiequn Wang 1
One-Token Verification → reducestokenusageby → 90%
confidence 95% · OTV reduces token usage by up to 90%
One-Token Verification → uses → Key-Value Cache
confidence 95% · probe internal reasoning signals through the key-value cache
One-Token Verification → uses → LoRA
confidence 95% · integrated into the LLM via low-rank adaptation
One-Token Verification → →
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Abstract
Abstract:Recent breakthroughs in large language models (LLMs) have led to notable successes in complex reasoning tasks, such as mathematical problem solving. A common strategy for improving performance is parallel thinking, in which multiple reasoning traces are generated and the final prediction is made using aggregation schemes like majority voting or best-of-$N$ decoding. However, two key challenges persist. First, multi-sample decoding incurs substantial inference latency, especially for long-form outputs. Second, effective mechanisms for reliably assessing the correctness of individual reasoning traces are still limited. To address these challenges, we introduce One-Token Verification (OTV), a computational method that estimates reasoning correctness in a single forward pass during generation. OTV is activated by a learnable token and integrated into the LLM via low-rank adaptation to probe internal reasoning signals through the key-value cache, supporting token-level correctness estimation at any stage of generation without disrupting primary reasoning. Experiments on mathematical reasoning benchmarks demonstrate that OTV consistently surpasses existing verifiers. Additionally, OTV reduces token usage by up to $90\%$ through correctness-guided early termination, prioritizing shorter, more reliable solutions.
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- Source: https://arxiv.org/abs/2603.01025v1
- Canonical: https://arxiv.org/abs/2603.01025v1
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One-Token Verification for Reasoning Correctness Estimation Zhan Zhuang 1 2 Xiequn Wang 1 Zebin Chen 1 Feiyang Ye 1 Ying Wei 3 Kede Ma 2 Yu Zhang 1 Abstract Recent breakthroughs in large language models (LLMs) have led to notable successes in complex reasoning tasks, such as mathematical problem solving. A common strategy for improving per- formance is parallel thinking, in which multiple reasoning traces are generated and the final pre- diction is made using aggregation schemes like majority voting or best-of-Ndecoding. However, two key challenges persist. First, multi-sample decoding incurs substantial inference latency, es- pecially for long-form outputs. Second, effec- tive mechanisms for reliably assessing the correct- ness of individual reasoning traces are still lim- ited. To address these challenges, we introduce One-Token Verification (OTV), a computational method that estimates reasoning correctness in a single forward pass during generation. OTV is activated by a learnable token and integrated into the LLM via low-rank adaptation to probe internal reasoning signals through the key-value cache, supporting token-level correctness estima- tion at any stage of generation without disrupting primary reasoning. Experiments on mathematical reasoning benchmarks demonstrate that OTV con- sistently surpasses existing verifiers. Additionally, OTV reduces token usage by up to90%through correctness-guided early termination, prioritizing shorter, more reliable solutions. 1. Introduction Large language models (LLMs) such as OpenAI o1 (Jaech et al., 2024), DeepSeek-R1 (Guo et al., 2025), and the Qwen3 series (Yang et al., 2025a) have recently demon- strated strong multi-step reasoning capabilities on challeng- ing tasks like mathematical problem solving. These ad- vances are largely driven by training pipelines that combine 1 Southern University of Science and Technology, Shenzhen, China 2 City University of Hong Kong, Hong Kong SAR 3 Zhejiang University, Hangzhou, China. Correspondence to: Ying Wei <ying.wei@zju.edu.cn>, Kede Ma<kede.ma@cityu.edu.hk>, Yu Zhang <yu.zhang.ust@gmail.com>. Preprint. March 3, 2026. 풕 ①5+3 = 8,then8-2=6. ②5+3 = 9? No, it’s 8.Then8-2=6. ③5+3 = 8,then8-2=5. ④5*3 = 15,then15-2=13. Query: Sally has 5 apples, buys 3, eats 2. How many are left? Token index 0.5 0 1.0 ① ③ ② ④ Estimated confidence Reasoner ... Verifier KV cache 푪 풕−ퟏ from reasoner Reuse Token sequence [ToT] Position 풕 LoRA-based Module + = Figure 1. Conceptual illustration of the proposed OTV. By reusing the KV cache and activating a LoRA-based verifier via a special token[ToT], OTV reliably estimates the correctness of reasoning traces in a single forward pass. supervised fine-tuning from human-supplied long chains of thought (Wei et al., 2022; Suzgun et al., 2023) and re- inforcement learning from outcome- or process-level feed- back (Ouyang et al., 2022; Shao et al., 2024; Team et al., 2025). In parallel, test-time scaling has emerged as a com- plementary paradigm that improves accuracy by allocating additional computation during inference (Brown et al., 2024; Venktesh et al., 2025). A simple yet effective instance is par- allel thinking, where the model generates multiple candidate solution traces and aggregates them into a final answer. A central challenge in parallel thinking is to reliably as- sess the correctness of individual reasoning traces. Existing approaches can be broadly grouped into internal and exter- nal verification. Internal methods rely on the model’s own token-level uncertainty or calibration 1 to derive trace-level confidence 2 scores, which are then used to perform majority voting over diverse traces (Wang et al., 2023b; Kang et al., 2025; Fu et al., 2025; Zhang et al., 2025a; Huang et al., 2025). These techniques are model-native, requiring no auxiliary models, but they often suffer from miscalibration and struggle to separate correct from incorrect reasoning traces, particularly for long-form solutions (Huang et al., 1 Calibration refers to the agreement between predicted proba- bilities and empirical outcome frequencies—for example, among predictions with probability 0.7, roughly 70% should be correct. 2 Here, confidence denotes the model’s predicted probability that its generated answer is correct. 1 arXiv:2603.01025v1 [cs.LG] 1 Mar 2026 One-Token Verification for Reasoning Correctness Estimation 2023; Xiong et al., 2024). External methods instead train dedicated verifiers (Cobbe et al., 2021; Hosseini et al., 2024; Zhang et al., 2024c; Lightman et al., 2024; Wang et al., 2024; Yang et al., 2024; Zhao et al., 2025) to score inter- mediate steps or final answers. Even though these auxiliary models can deliver richer feedback, they treat the base LLM as a black box and often add substantial inference over- head, while also risking domain mismatch. As a result, their achieved accuracy still remains well below the upper bound suggested by the Pass@kmetric (Chen et al., 2021). More- over, existing approaches typically defer decision making until complete reasoning traces are generated, because their scoring rules are defined over final answers or full trajecto- ries. When early termination is enabled, performance may degrade noticeably if intermediate signals are unreliable. Consequently, decoding cost, as exacerbated by “System- 2”-style overthinking (Chen et al., 2024a), raises significant efficiency concerns. These limitations call for a deeper form of internal verifica- tion that goes beyond surface-level, logit-based heuristics. Such verification should 1) explicitly leverage the internal states of the ongoing reasoning process (Burns et al., 2022; Azaria & Mitchell, 2023; Zhang et al., 2025a; Li et al., 2025), 2) provide fine-grained, token-level estimations of reason- ing correctness to unlock practical early termination (Fu et al., 2025; Zhang et al., 2025a; Lee et al., 2025), and 3) be inexpensive to invoke during generation. To this end, we introduce One-Token Verification (OTV), a computational method that augments a reasoning LLM with a LoRA-based verifier (Hu et al., 2022), operating directly on the model’s key-value (KV) cache (see Figure 1). Con- cretely, OTV introduces a special token of truth ([ToT]). When inserted at inference time, this token leverages LoRA- enhanced cross attention to access the cached KV states ac- cumulated during the ongoing reasoning process. A small re- gression head maps the last-layer hidden state of[ToT]to a scalar estimate of reasoning correctness, providing token- level verification of the partial trace. Crucially, the LoRA updates are gated so that the base LLM behaves identically to the original reasoner in its default mode, and adopts a ver- ifier role only conditioned on the presence of[ToT]in the input. OTV thus reuses the LLM’s full internal computation, and incurs only a single forward pass per verification query. Training OTV relies solely on cheap pseudo-scores assigned to every token in a reasoning trace, derived from its final correctness label. We further show that OTV training ad- mits parallelization: by inserting[ToT]simultaneously at all candidate positions and reusing a cached prefix, OTV computes correctness scores for the entire trace in a single forward pass. This preserves the token-level parallelism of standard Transformer training. We evaluate OTV on multi- ple reasoning LLMs (i.e., Qwen3-4B-Instruct, Qwen3-8B, and DAPO-Qwen-32B) and a suite of math benchmarks. Across diverse parallel thinking strategies and several effi- cient variants, OTV consistently outperforms prior internal and external verifiers, substantially narrowing the gap to- ward Pass@kwhile reducing token usage by up to90%via confidence-guided early termination. Beyond post-trained reasoning LLMs, we show that calibrating OTV on a pre- trained base model also yields sizable gains in raw mathe- matical reasoning accuracy and robustness. 2. Related Work Parallel thinking Test-time scaling has become a cen- tral paradigm for strengthening LLM reasoning, and it is typically pursued along three axes: enlarging the effective input context (e.g., retrieval-augmented generation (Lewis et al., 2020) and tool use (Schick et al., 2023)), increasing sequential reasoning depth (Jaech et al., 2024; Muennighoff et al., 2025), and widening generation through parallel sam- pling (Comanici et al., 2025; Wen et al., 2025; Yang et al., 2025b; Hsu et al., 2025; Zheng et al., 2025b). Among these, parallel thinking stands out as one of the most widely adopted approaches. Representative instances include ma- jority voting (Wang et al., 2023b), Best-of-Ndecoding (Sti- ennon et al., 2020), and tree- or graph-structured search (Yao et al., 2023; Zhang et al., 2024a). These methods explore and evaluate multiple (partial) solutions before committing to a final answer, which generally achieve a higher per- formance ceiling than single-trace decoding (Ghosal et al., 2025), but at the cost of substantial token overhead. To control this budget, recent work investigates trajectory prun- ing (Lee et al., 2025; Wang et al., 2025; Fu et al., 2025; Huang et al., 2025) to terminate low-promise traces early. Reasoning correctness estimationAssessing the correct- ness of individual reasoning traces (Lee & Hockenmaier, 2025) is critical for reliable early termination of LLMs in parallel thinking. Existing approaches can be grouped into internal and external verification. Internal methods exploit the model’s own representations (Lin et al., 2022; Fadeeva et al., 2024). For example, self-consistency (Wang et al., 2023b) aggregates diverse traces via majority voting, while self-certainty (Kang et al., 2025) derives confidence from output distributions. DeepConf (Fu et al., 2025) prunes low-confidence traces, and self-calibration (Huang et al., 2025) distills confidence scores into a single-pass estima- tor. Probing-based methods further train prediction heads on last-layer hidden states (Zhang et al., 2025a; Lee et al., 2025), showing that models internally encode correctness signals that can support calibrated early exits. External methods instead introduce auxiliary verifiers that operate on the generated text. Outcome reward models as- sess only the final answer (Cobbe et al., 2021; Yu et al., 2 One-Token Verification for Reasoning Correctness Estimation 2023; Chen et al., 2024b; Liu et al., 2024; Lu et al., 2024; Zhang et al., 2025b), whereas process reward models score intermediate steps and aggregate them into an overall cor- rectness estimate (Uesato et al., 2022; Lightman et al., 2024; Wang et al., 2023a; Zhang et al., 2025c). Subsequent work extends verification to broader domains (Zeng et al., 2025) and richer criteria (Golovneva et al., 2022; Wang et al., 2024), or reframes verification as a reasoning-induced pre- diction task (Ankner et al., 2024; Zhang et al., 2024c). Other lines develop critic-style feedback models (Zheng et al., 2023; 2025a; Ye et al., 2025) or ensembles over multiple verifiers and solutions (Lifshitz et al., 2025; Zhong et al., 2025; Zhao et al., 2025) to improve reliability. Although external verifiers often provide stronger feedback, they ignore the base LLM’s internal dynamics and introduce extra inference cost, with potential domain mismatch. In contrast, internal methods are model-native and efficient, but remain susceptible to miscalibration, especially for long- form solutions. OTV lies between these two extremes. By attaching a LoRA- based module that is activated only in verification mode, OTV allows the same reasoning LLM to also operate as a verifier. It provides token-level correctness estimates with negligible additional computation, thereby combining the model-specificity and efficiency of internal approaches with explicit scoring capabilities of external verifiers. 3. Proposed Method: OTV In this section, we introduce OTV, a computational method for reasoning correctness estimation. OTV integrates a rea- soning LLM with 1) a LoRA-based module that is activated only during verification and 2) a special verification token that probes the model’s KV cache. We now describe the four components: the LoRA module design (Sec. 3.1), KV cache probing (Sec. 3.2), pseudo-labeling (Sec. 3.3), and parallelized implementation (Sec. 3.4). 3.1. LoRA-based Verification LoRA (Hu et al., 2022) is a widely used parameter-efficient fine-tuning method that augments pretrained weight matri- ces with low-rank updates. Given a weight matrixW ∈ R d out ×d in , LoRA introduces two trainable matricesR ∈ R d out ×r andS ∈ R r×d in with rankr ≪ mind in ,d out , and defines the updated weight asW LoRA = W + RS. LoRA only trains the adapters, which substantially reduces the trainable-parameter count and optimizer-state memory. In OTV, we attach a LoRA-based verification module to selected layers of the base LLM. To preserve the original reasoning ability, we adopt a gating mechanism (Samragh et al., 2025), which adds the LoRA pathway in parallel to each linear layer and activates it only in verification mode. For an inputx t ∈ R d in ×1 at positiont, the corresponding output is z t = (W + m t RS)x t ,(1) wherem t ∈0, 1is a binary gate. Whenm t = 0, the LLM behaves identically to the original reasoner; whenm t = 1, the LoRA update is applied, and the model assumes the verifier role. The gating provides a clean separation between reasoning and verification, and minimizes the risk that fine- tuning degrades the base model’s reasoning capabilities. 3.2. KV Cache-based Internal Representation Probing During autoregressive decoding, Transformer-based LLMs maintain a KV cache at layerland positiont, denoted asC (l) t = K (l) t , V (l) t whereK (l) t = [k (l) 1 ,...,k (l) t ], V (l) t = [v (l) 1 ,...,v (l) t ], andk (l) t ,v (l) t ∈ R D×1 areD- dimensional key and value vectors, withC (l) 0 =∅ . For anL- layer LLM, the forward pass at positiont + 1is functionally determined by the current inputx t+1 and the accumulated KV cacheC t = S L l=1 C (l) t . Compared with the last-layer hidden states (Zhang et al., 2025a; Lee et al., 2025), which represent a lossy summary of the preceding context and dis- card its token-wise and layer-wise structure, the KV cache serves as a sufficient statistic of the prefix for the underlying LLM, allowing richer, task-specific pooling over the entire reasoning trajectory. OTV is designed to explicitly exploit this structure. We introduce a special token of truth ([ToT]), which is inserted only in verification mode (i.e., whenm t+1 = 1in Eq. (1)) at an arbitrary positiont + 1to estimate the correctness of the partial trace up to tokent. Instead of recomputing the prefix, the model reuses the KV cacheC t at every layer and performs a single forward pass augmented with the LoRA-based verification module, described in Sec. 3.1. Letx (l) t+1 be the input to layerl, withx (0) t+1 representing the initial embedding of[ToT]. The LoRA-augmented query, key, and value vectors are given by ̃ q (l) t+1 = W (l) q + R (l) q S (l) q x (l) t+1 , ̃ k (l) t+1 = W (l) k + R (l) k S (l) k x (l) t+1 , ̃ v (l) t+1 = W (l) v + R (l) v S (l) v x (l) t+1 . (2) whereW (l) ∗ ,∗ ∈ q,k,v, denotes the frozen pretrained weights at layerlandR (l) ∗ , S (l) ∗ are the corresponding train- able LoRA matrices. Using the cached prefixC t and the updated vectors ̃ q (l) t+1 , ̃ k (l) t+1 , ̃ v (l) t+1 , the attention output for [ToT] at layer l is ̃ h (l) t+1 = h V (l) t , ̃ v (l) t+1 i softmax K (l) t , ̃ k (l) t+1 ⊺ ̃ q (l) t+1 √ D , (3) wheresoftmax(·)is applied column-wise. After[ToT]is propagated through allLlayers, we obtain its final hidden 3 One-Token Verification for Reasoning Correctness Estimation state ̃ h (L) t+1 ∈ R D×1 . OTV applies a three-layer perceptron g(·) to map ̃ h (L) t+1 to a scalar prediction: ˆc t = g ̃ h (L) t+1 ∈ [0, 1],(4) which estimates the likelihood (i.e., confidence) that the reasoning trajectory is correct up to positiont. Predicting a continuous score rather than decoding a token from the vo- cabulary avoids entangling verification with the model’s linguistic prior over specific tokens (e.g., “correct” and “wrong”) and yields a signal that is easier to calibrate and threshold for routing. 3.3. Token-level Pseudo-confidence Labeling OTV requires token-level supervision to train the veri- fier, but collecting such process-supervision signals (Light- man et al., 2024) or generating search-based rollouts is costly (Wang et al., 2023a; Luo et al., 2024; Zhang et al., 2024b; Feng et al., 2024; Setlur et al., 2024; Guan et al., 2025). Instead, we derive pseudo-confidence targets solely from outcome-level supervision.Concretely, given a training dataset, we sample a reasoning tracex 1:T = [x 1 ,...,x T ]and assign a scalar targetc t ∈ [0, 1]at each to- ken positiont. We interpretc t = 0as confidently incorrect, c t = 1as confidently correct, andc t = 0.5as maximally uncertain. Lety ∈ 0, 1indicate the final correctness of the full trace. In the default setting, we initializec 0 = 0.5 as a neutral prior and progressively bias confidence toward yas evidence accrues. We implement this using a Linear ramp that enforces a monotone, linear interpolation from uncertainty to the final outcome: c t = linear(t) = 0.5 + (y− 0.5) t T .(5) After assigning token-level pseudo-confidence labels, we train the LoRA-based verification module and the regression head by minimizing the mean squared error (MSE) over response tokens: ℓ = 1 T T X t=1 (c t − ˆc t ) 2 .(6) Although we adopt the linear ramp as our default pseudo- confidence prior, we also explore several alternative labeling rules (e.g., constant, sigmoid, noise-perturbed, and stepwise ramps). In addition, we present detailed ablation studies to evaluate performance and provide a theoretical analysis of our pseudo-confidence labeling rules in Sec. 4.4 and Appendix C, respectively. 3.4. Parallelization A key advantage of Transformers (Vaswani et al., 2017) is their ability to train with token-level parallelism. OTV preserves this property: although the confidence score at positiontis obtained by one-token verification (in Eq. (3)), we can compute scores for all positions in a trace with a single forward pass by probing all prefixes in parallel. Consider a reasoning trace of lengthT. For each prefix ending att ∈ 1,...,T, we want a verifier query that attends only to the cached KVC t associated with that prefix (i.e., the firstttokens). To this end, we construct a probe sequence ofT +1truth tokens,[ToT] 1:T+1 . In verification mode, we collect their LoRA-augmented projections at layer las e Q (l) T+1 , e K (l) T+1 , e V (l) T+1 ∈ R D×(T+1) . To enforce that probe tokent + 1only “sees” the prefix up tot, we use a triangular mask M∈ R T×(T+1) defined as: M i,j = ( 0, i < j, −∞, i≥ j, (7) wherei ∈ 1,...,Tandj ∈ 1,...,T + 1. At layerl, the cache-to-probe attention logits are A (l) ctp = K (l) T ⊺ e Q (l) T+1 + M.(8) To match the single-token computation (in Eq. (3)), each probe token shall attend only to itself within the set of probe tokens. Accordingly, we compute the per-probe self-logits as the diagonal of the probe-to-probe dot-product matrix: a (l) self = diag e K (l) T+1 ⊺ e Q (l) T+1 ∈ R (T+1)×1 ,(9) wherediag(·)extracts the diagonal entries of a square ma- trix and returns them as a column vector. We then append these self-logits as an additional row and apply a column- wise softmax to obtain the attention probabilities: A (l) T+1 = " A (l) ctp a (l) self ⊺ # ∈ R (T+1)×(T+1) , P (l) T+1 = " P (l) ctp p (l) self ⊺ # = softmax A (l) T+1 √ D ! . (10) The resulting attention outputs for all probes at layer l are e H (l) T+1 = V (l) T P (l) ctp + e V (l) T+1 diag p (l) self ∈ R D×(T+1) . (11) Here,diag(·)converts a column vector into a square diago- nal matrix. Finally, stacking these computations across all Llayers yieldsT + 1final[ToT]hidden states, each of which is mapped to a scalar confidence (in Eq. (4)). This construction is mathematically equivalent to runningT + 1 independent single-token verification queries. However, it consolidates these queries in a single forward pass by reusing the same cached prefix. We apply the MSE loss only at probe positions corresponding to response tokens, excluding those in the prompt and query segment. Algorith- mic descriptions of the training and inference procedures are given in Appendix A. 4 One-Token Verification for Reasoning Correctness Estimation 4. Experiments In this section, we test OTV for parallel thinking across multiple open-source reasoning LLMs in both standard and advanced mathematical reasoning settings. We consider common aggregation schemes, including self-consistency, weighted self-consistency, and best-of-N, along with early- termination variants. We report accuracy and efficiency metrics (i.e., token cost, and verification overhead) and provide a qualitative analysis of OTV confidence trajectories over the course of generation. 4.1. Experimental Setups Models and datasets We evaluate OTV on three repre- sentative open-source reasoning LLMs: Qwen3-4B-Instruct, Qwen3-8B (Yang et al., 2025a), and DAPO-Qwen-32B (Yu et al., 2025). In all cases, the base LLM is kept frozen and used as the reasoner. To cover a range of reasoning difficulty, we consider two experimental regimes. For standard mathe- matical reasoning, we train OTV on MetaMathQA (Yu et al., 2024) and test on GSM8K (Cobbe et al., 2021), a widely adopted benchmark of grade-school math problems. For ad- vanced mathematical reasoning, we train on DAPO17K (Yu et al., 2025), a more challenging dataset that provides only final answers (without human-written reasoning traces), and evaluate on the AIME24 (MAA, 2024) and AIME25 (MAA, 2025) benchmarks, which comprise contest-style problems from the American Invitational Mathematics Examination. Our main experiments focus on this advanced regime, as it rigorously tests long-horizon, multi-step logical reasoning. Training protocol For each training instance, we sam- ple the base LLM8times to obtain diverse reasoning traces, and derive token-level pseudo-confidence targets from the final correctness label to supervise OTV. At infer- ence time, for each test question, we first generate a pool of256complete reasoning traces, setting the temperature parameter to1.0, and repeatedly sample subsets from this pool for downstream aggregation. Unless otherwise stated, we fine-tune LoRA modules (applied to query, key, and value projections) together with the regression head using LlamaFactory(Zheng et al., 2024) for3epochs, with a learning rate of 10 −4 and a batch size of 128. Competing methods We categorize competing meth- ods into internal and external verifiers. Internal methods are model-native, including DeepConf (Fu et al., 2025), a training-free confidence estimator derived from token- level log-likelihood, and GenRM (Zhang et al., 2024c), a reasoning-induced verifier that prompts the model it- self with “Is the answer correct?” and uses the predicted “Yes/No” probability as a confidence score. External meth- ods adopt off-the-shelf reward models, including AceMath- RM-7B (Liu et al., 2024), an outcome-level reward model that scores the full solution, and several process reward models—VersaPRM (Zeng et al., 2025), Math-Shepherd- Mistral-7B (Wang et al., 2023a), and Qwen2.5-Math PRM variants (Zhang et al., 2025c)—that assign step-level re- wards along the reasoning trace. Following prior practice, we take the reward predicted at the last token as the overall verifier score. Inference-time aggregation strategiesWe evaluate three families of parallel-thinking decoders. •Self-consistency (i.e., majority voting) and its weighted variant (Wang et al., 2023b), where each trace’s final answer is weighted by its estimated confidence score. •Best-of-N, which selects the highest-scoring trace among N candidates. • Efficient Best-of-Nvariants, which reduce computa- tion via pruning or early termination: –Drop@10: Periodic score-based pruning. For every 10generated tokens, it drops the currently lowest- scoring trace and continues decoding until only one trace remains. – Stop@600: Fixed-length early commitment. Once traces reach600tokens, it terminates all but the highest-scoring trace and continues generation only for the surviving trace to completion. –Halve@300: Stage-wise halving. For every300gen- erated tokens, it removes the bottom half of traces according to the current scores, repeating this “halve- and-continue” procedure until one trace remains. We also report Pass@1(single-trace accuracy), Pass@k (oracle success amongktraces), and Maj@k(unweighted majority-voting accuracy overktraces). Notably, Pass@k serves as an upper bound for any aggregation method that must operate without ground-truth at inference time. While self-consistency and standard Best-of-Nrequire completing all traces with comparable token cost, efficient Best-of-N variants can substantially reduce decoding. For example, withN = 128and an average trace length of6, 000tokens, the pruning-based variants reduce total generation by nearly 90%. On the other hand, OTV introduces negligible compu- tation and token overhead because each verification query is implemented as a single forward pass (see Appendix B.4). 4.2. Main Results We first evaluate OTV on advanced mathematical reasoning in an offline setting, where the complete set of sampled reasoning traces is available and aggregation can be applied post hoc. Table 1 reports weighted majority-voting results using128traces per run. We weight votes by trace-level 5 One-Token Verification for Reasoning Correctness Estimation Table 1. Weighted majority-voting accuracy (%) on AIME. For process reward models, we use the final token score as the confidence for each reasoning trace. Following Fu et al. (2025), we use method-specific aggregation windows: confidence is computed as the mean score over the last tokens of each trace (100tokens for OTV;2, 048tokens for DeepConf). We then discard the bottom50%of traces by confidence before performing weighted majority voting. Each run samples128reasoning traces, and all results are averaged over64runs. We also report Pass@128as an oracle upper bound for parallel thinking. Within each column, the best result is highlighted in bold, and the second best is underlined. QWEN3-4BQWEN3-8BDAPO-QWEN-32B AIME24AIME25AIME24AIME25AIME24AIME25 PASS@12891.4683.1369.5856.9883.7568.75 Internal PASS@160.2946.6726.2919.3251.7736.42 MAJ@12875.42±1.6166.46±1.1644.22±2.1128.44±2.0666.72±0.9341.77±1.76 DEEPCONF77.76±2.4366.77±1.1845.73±1.5031.87±2.3566.77±1.1744.79±1.65 GENRM79.11±2.0666.72±1.8244.38±2.2732.66±2.3066.72±0.5942.55±1.66 External ACEMATH-RM-7B67.66±1.7460.00±2.3644.90±2.7026.72±2.1861.15±2.2242.76±2.00 VERSAPRM-8B75.52±1.6966.72±1.1044.48±2.3029.11±2.2266.98±0.9743.18±1.81 MATH-SHEPHERD-7B74.27±1.5066.56±1.0244.22±1.7926.04±1.9465.73±2.2443.65±1.93 QWEN2.5-PRM800K-7B75.16±1.6666.35±1.1444.22±2.3029.32±2.5767.03±1.4648.18 ±3.48 QWEN2.5-PRM-7B78.18±1.8563.49±2.3945.10±2.3529.48±2.6367.45±2.2646.77±3.06 OTV (OURS)83.33±1.5769.32±1.4646.56±1.2533.85±1.6970.83±1.8649.58±1.10 Table 2. Accuracy (%) and average output length (in parentheses) on AIME atN = 128under Best-of-Nand three compute-efficient variants (i.e., Drop@10, Stop@600, and Halve@300). All results are averaged over 64 runs. BEST-OF-NDROP@10STOP@600HALVE@300 AIME24AIME25AIME24AIME25AIME24AIME25AIME24AIME25 QWEN3-4B DEEPCONF64.95 (9664)43.07 (9322)62.86 (7044)40.78 (7260)59.90 (7150)40.89 (7084)61.98 (6742)43.54 (6318) VERSAPRM-8B54.48 (6560)43.28 (6132)60.52 (2589)43.44 (6334)63.75(3270)46.77 (6113)59.06 (2981)37.24 (6438) MATH-SHEPHERD-7B73.59 (5820)66.51 (5824)61.61 (5679)46.25(6116)57.45 (5989)47.45 (6218)54.64 (5722)45.36 (6118) QWEN2.5-PRM800K-7B69.90 (4891)45.10 (6202)66.77 (2196)42.86 (6445)60.31 (3434)49.53 (6409)62.24 (2919)45.21 (6619) QWEN2.5-PRM-7B71.77 (3720)53.33 (3948)63.80 (3040)45.83 (6588)65.73 (4211)44.95 (6304)66.46(3173)45.73(6416) OTV (OURS)73.44(5447)53.91(5416)63.39 (4427)46.46 (3225)63.75(4431)49.11(6542)67.03 (4132)49.02 (3170) DAPO-QWEN-32B DEEPCONF53.92 (5101)38.91 (3957)50.52 (6382)37.08 (4398)51.82 (7176)37.76 (4353)50.94 (3772)36.77 (4449) VERSAPRM-8B48.80 (5061)31.04 (4447)59.79 (5263)39.48 (4796)53.12 (5432)37.66 (5005)49.32 (5744)36.61 (5046) MATH-SHEPHERD-7B62.34(5051)42.40 (4570)55.52 (4819)39.22 (4475)55.21(5555)36.46 (5063)58.80 (4983)42.76(4919) QWEN2.5-PRM800K-7B54.17 (4722)47.81 (4426)53.28 (5585)41.20(4689)49.48 (5260)40.89(4523)47.40 (5732)40.00 (4528) QWEN2.5-PRM-7B57.03 (4888)47.24(4481)55.52 (5525)33.65 (5351)55.57 (4939)35.31 (5068)51.98 (5660)36.09 (4967) OTV (OURS)63.18 (4623)47.08 (4079)55.95(3397)50.68 (2926)53.54 (3211)48.23 (2577)55.05(3436)46.98 (2991) confidence scores and discard the lowest-confidence50% of traces prior to aggregation. Across all three backbone scales, OTV consistently delivers the best accuracy on both AIME24 and AIME25. Relative to unweighted majority voting, the gains are substantial (e.g., +7.9points for Qwen3-4B on AIME24 and+7.8points for DAPO-Qwen-32B on AIME25), indicating that OTV’s confidence estimates correlate more strongly with trace cor- rectness than heuristic confidence baselines. Compared with other internal methods, OTV is consistently superior, suggesting that a learned verifier that probes the model’s in- ternal states provides a more faithful correctness signal than logit-based confidence or self-queried “yes/no” verification. Moreover, external verifiers also fall short of OTV across all settings, despite extensive training and broad adoption, highlighting the difficulty of transferring generic reward models across backbones and shifting trace distributions. Table 2 compares OTV with standard Best-of-Nand three efficient variants (i.e., Drop@10, Stop@600, and Halve@300) atN = 128in an online setting. A key finding is that vanilla Best-of-Ncan be suboptimal in accuracy: av- eraged across tasks, it lags behind weighted majority voting (at the same decoding cost) by more than10points, reflect- ing the brittleness of selecting a single trace solely by the highest predicted confidence score. Across the efficient variants, OTV attains the best or near- best accuracy in most configurations. Moreover, unlike competing methods, it often yields shorter final traces than standard Best-of-N(approximately20%fewer tokens on average). This behavior is consistent with OTV’s linearly in- creasing pseudo-confidence target: given two correct traces, the shorter one accumulates confidence faster, reaches a higher score earlier, and is therefore more likely to sur- vive pruning or early-stopping criteria. Among the efficient 6 One-Token Verification for Reasoning Correctness Estimation 35 40 45 50 Score DeepConf 0.5 0.6 0.7 0.8 0.9 1.0 VersaPRM-8B 0.4 0.6 0.8 Math-Shepherd-7B 0.2 0.4 0.6 0.8 1.0 Qwen2.5-PRM-7B 0.2 0.4 0.6 0.8 OTV 40 45 50 Score DeepConf 0.2 0.4 0.6 0.8 1.0 VersaPRM-8B 0.2 0.4 0.6 0.8 Math-Shepherd-7B 0.2 0.4 0.6 0.8 1.0 Qwen2.5-PRM-7B 0.4 0.6 0.8 OTV 0.00.20.40.60.81.0 Normalized reasoning length 40 45 50 Score DeepConf 0.00.20.40.60.81.0 Normalized reasoning length 0.8 0.9 1.0 VersaPRM-8B 0.00.20.40.60.81.0 Normalized reasoning length 0.2 0.4 0.6 0.8 Math-Shepherd-7B 0.00.20.40.60.81.0 Normalized reasoning length 0.2 0.4 0.6 0.8 1.0 Qwen2.5-PRM-7B 0.00.20.40.60.81.0 Normalized reasoning length 0.2 0.4 0.6 0.8 OTV Figure 2. Confidence dynamics on three representative AIME24 problems (i.e., #3, #9, and #22). For each predictor, we plot the mean confidence trajectory over32sampled reasoning traces, shown separately for traces that end with correct (red) and incorrect (green) final answers. Shaded bands around each mean curve denote the inter-quantile range across traces, summarizing cross-trace variability. Figure 3. Effect of verifier capacity (i.e., LoRA rank) on training dynamics and downstream voting accuracy. Left: verifier training loss over optimization steps for the “probe” baseline, which trains only the regression head (no LoRA; no KV cache) and for OTV with varying LoRA ranks. Middle/Right: Weighted majority-voting accuracy on AIME, as a function of the number of sampled traces. All results are averaged over 64 runs. strategies, Halve@300offers the most favorable accuracy- efficiency trade-off (see also Appendix B.4), where OTV exhibits clear gains. Overall, these results indicate that model-specific verification provides a reliable and compute- efficient alternative to generic score-based selection. We further validate the generalizability of OTV across pretrained base models and diverse architectures (e.g., LLaMA), observing consistent improvements. Details are provided in Appendices B.2 and B.3. 4.3. Visualization Figure 2 provides a qualitative view of how different con- fidence predictors evolve throughout generation on three representative AIME 24 problems (i.e., #3, #9, and #22). We plot, for each verifier, the mean confidence trajectory over32sampled traces, shown separately for correct (red) and incorrect (green) solutions. The shaded region around each mean curve reflects cross-trace variability (i.e., the 0.2–0.8inter-quantile range), facilitating comparisons of overall trends across verifiers. Additional OTV visualiza- tions for all AIME problems are provided in Appendix F. A consistent pattern emerges across all three cases: Deep- Conf and prior process reward models often produce highly entangled confidence curves, with substantial overlap be- tween correct and incorrect traces, which limits their ability to discriminate promising candidates from failures early in decoding. In contrast, OTV exhibits noticeably clearer stratification: confidence typically increases over the course of reasoning for correct traces, whereas incorrect traces remain comparatively suppressed. This enhanced separabil- ity aligns with the quantitative gains reported in Sec. 4.2, and helps explain OTV’s stronger performance under score- based selection and pruning and early-termination regimes. Appendix G provides additional token-level visualizations that align confidence curves with the generated text. This finer-grained analysis illustrates how OTV’s scores evolve within different solutions—often rising sharply after key computational steps—thereby improving interpretability at the granularity of individual reasoning steps. 7 One-Token Verification for Reasoning Correctness Estimation Table 3. Ablation of token-level pseudo-confidence labeling heuristics for training OTV. MAJ@128BEST-OF-NHALVE@300 AIME24AIME25AIME24AIME25AIME24AIME25 CONSTANT LABEL81.41±1.6568.49±2.0472.76 (4,340)54.27 (4,362)66.41 (3,951)49.27 (4,503) SIGMOID RAMP80.05±1.9166.51±0.9277.40 (4,617)56.41 (4,426)68.44 (4,210)46.30 (4,294) NOISE-PERTURBED RAMP81.87±1.6567.92±1.6172.60 (5,226)54.90 (5,152)71.30 (4,134)48.33 (5,736) STEPWISE RAMP82.40±1.5068.80±1.6077.29(5,351)55.36(5,452)70.63(4,231)37.60 (7,095) LINEAR RAMP (DEFAULT)83.33±1.5769.32±1.4673.44 (5,447)53.91 (5,416)67.03 (4,132)49.02 (3,170) Table 4. Effect of trace-level confidence aggregation and filtering on weighted majority voting. We convert OTV’s token-level confidence estimates into a single trace-level score by aggregating over the last few tokens, discard the lowest-confidenceρfraction of traces (i.e., 0%, 25%, 50%, and 75%), and evaluate three aggregation operators. maxmeanmin 0%25% 50% 75%0%25% 50% 75%0%25% 50% 75% ALL TOKENS75.7876.8280.4280.3675.9479.5382.1980.0080.1080.8381.2579.17 LAST 10% TOKENS76.7780.2182.7681.4682.1983.0283.2882.1983.0783.0783.1282.29 LAST 1, 600 TOKENS75.7878.6581.2582.1480.9982.7183.2382.2483.2383.2382.9779.95 LAST 400 TOKENS77.5080.3682.6682.8681.9382.8183.3383.2382.8683.2383.2883.75 LAST 100 TOKENS78.3980.2182.3481.8282.0882.9783.3382.7682.8682.6082.7683.96 4.4. Ablation Studies We conduct ablation studies to quantify how individual de- sign choices contribute to OTV’s performance. Specifically, we examine 1) the LoRA rank used in the verification mod- ule, 2) alternative pseudo-confidence labeling rules, and 3) the impact of trace-level confidence aggregation and filter- ing on weighted majority voting. Effect of LoRA rank We first vary the LoRA rankr while keeping the underlying reasoner fixed, and evaluate performance under weighted majority voting as the number of sampled traces increases. As shown in Figure 3, we observe a clear capacity-performance trade-off: increasingr reduces the verifier’s training loss and improves downstream accuracy, with substantial gains already at moderate ranks (e.g.,r = 16). Importantly, OTV consistently outperforms the “probe” baseline, which trains only the regression head (i.e., without LoRA fine-tuning or KV cache reuse). This suggests that the verifier benefits from both 1) additional adaptation capacity and 2) richer access to the reasoning trajectory via the KV cache, rather than relying solely on final-layer hidden states (Zhang et al., 2025a). Pseudo-confidence labelingWe next ablate the heuristic pseudo-confidence targetsc t defined in Sec. 3.3, while keep- ing the voting protocol fixed. Table 3 shows that constant la- bel, which applies a hindsight-style target uniformly over the entire trace, tends to bias the verifier toward shorter traces and slightly degrades the accuracy of weighted majority voting. In contrast, ramp-based alternatives achieve compa- rable performance while mitigating this uniform-hindsight bias by imposing a monotonic progression toward the final outcome. We thus adopt the linear ramp as the default due to its simplicity and stable performance. Trace-level confidence aggregation and filteringWe fur- ther examine how to map token-level confidence estimates to a single trace-level score for weighted majority voting. Following the DeepConf-style protocol (Fu et al., 2025), we aggregate confidence over the last segment of each trace and then remove the bottomρfraction of traces before voting. Table 4 reveals two consistent trends. First, aggregating over the tail of the trajectory (e.g., the last100tokens) is more re- liable than using the full trace, supporting the intuition that late-step verification better captures the fully formed reason- ing state and final answer. Second, moderate-to-aggressive filtering improves robustness by suppressing low-confidence candidates, with mean/min aggregation over roughly the last 100 tokens performing the best in our setting. 5. Conclusion In this work, we have introduced OTV, an efficient, model- specific computational method for estimating token-level reasoning correctness. OTV equips a reasoning LLM with a LoRA-gated verification pathway that is activated only when a dedicated[ToT]is inserted, allowing the model to enter verification mode without perturbing its default rea- soning behavior. By probing the model’s KV cache and producing a scalar confidence score through a small regres- sion head, OTV supports anytime verification at the cost of just a single forward pass per query. Experiments across multiple reasoning LLMs and math benchmarks show that OTV consistently improves parallel-thinking aggregation and pruning strategies relative to existing internal and ex- ternal verifiers, while delivering substantial efficiency gains via confidence-guided early termination. We further discuss limitations and future research directions in Appendix D. 8 One-Token Verification for Reasoning Correctness Estimation Impact Statement This paper presents work with the goal of advancing the field of machine learning. 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Solve-Detect-Verify: Inference-time scaling with flexible generative verifier. arXiv preprint arXiv:2505.11966, 2025. 12 One-Token Verification for Reasoning Correctness Estimation A. Algorithm Descriptions Algorithm 1 summarizes the parallelized OTV training procedure: for each sampled reasoning trace, we 1) cache the KV states produced by the frozen reasoner, 2) construct token-level pseudo-confidence targets, and 3) run a single parallel verification pass by inserting the truth token[ToT]at all probe positions. Algorithm 2 describes the OTV inference procedure: a single verification token[ToT]probes any cached prefix and returns a token-level correctness estimate. We use light purple shading to denote reasoning-mode steps (i.e., trace generation and KV caching) and light yellow shading to denote verification-mode steps (i.e., [ToT] probing). Algorithm 1 Parallelized OTV training Require: DatasetD 1: Trace sampling + KV caching. For each(q,a)∈D, sample a reasoning tracex 1:T ∼ LLM(q)and cache per-layer KV statesC T , whereC t =K (l) t , V (l) t L l=1 2: Pseudo-confidence targets. Compute token-level targetsc t T t=1 from the trace-level outcome labely ∈ 0, 1, where y is derived by comparing the model’s final answer to the ground-truth a 3: Single-pass parallel probing. Construct a probe sequence[ToT] 1:T+1 and apply a triangular probe maskMso probe t + 1 attends only to the cached prefix 1 : t; compute all token-level predictionsˆc t in one forward pass 4: Optimization. Minimize the MSE over response tokens and update LoRA and regression head parameters This implementation is equivalent to issuingT + 1separate one-token verification queries, but executes them in a single forward pass while reusing the same cached prefix states. Algorithm 2 OTV inference (one-token probing) Require: Prompt/question q, partial trace prefix x 1:t , cached KV statesC t 1: Prefix generation + caching. Generate (or reuse) a partial trace x 1:t and KV cacheC t 2: One-token verification query. Insert a single[ToT]at positiont + 1, reuseC t , and run a forward pass under the LoRA-gated verifier to obtain ˆc t ∈ [0, 1] 3:Trace-level scoring (optional). Aggregateˆc t (e.g., over the final segment) into a single score for ranking and pruning B. Additional Experimental Details B.1. Alternative Pseudo-Confidence Labeling Rules OTV uses outcome supervisiony ∈0, 1to define dense token-level targetsc t ∈ [0, 1]. Besides the default linear ramp, we evaluate four alternatives. • Constant label uses a hindsight-style target that assumes the model “knows” the final outcome across the trace: c t = y.(12) •Sigmoid ramp provides a smooth, parameterized alternative that can interpolate between gradual and abrupt transitions: c t = sigmoid (2y− 1)α t T ,(13) whereα > 0controls the shape: smallerαyields an almost-linear ramp, while largerαproduces a sharper transition toward the final label, approaching a step-like change (i.e., constant label). We set α = 4 by default. 13 One-Token Verification for Reasoning Correctness Estimation Table 5. Weighted majority-voting accuracy on GSM8K using MetaMath-LLaMA-7B and MetaMath-Mistral-7B as backbone models. Results are reported as mean± standard deviation over 64 runs. N148163264 LLaMA-7B PASS@k61.87±0.7475.01±0.7981.07±0.4985.36±0.4888.68±0.3591.27±0.38 MAJ@k61.87±0.7467.29±0.8869.57±0.4370.53±0.4471.47±0.3071.79±0.30 DEEPCONF61.87±0.7468.64±0.8070.63±0.6071.76±0.5872.06±0.2072.32±0.21 OTV (r = 2)61.87±0.7471.13±0.7473.50±0.4774.55±0.5274.80±0.3775.16±0.22 OTV (r = 16)61.87±0.7470.72±0.8172.82±0.3573.80±0.3274.37±0.2874.83±0.29 Mistral-7B PASS@k66.02±1.1883.16±0.6588.46±0.5091.58±0.4493.99±0.2895.87±0.24 MAJ@k66.02±1.1872.60±0.4977.33±0.3079.04±0.3780.21±0.4080.56±0.33 DEEPCONF66.02±1.1875.63±0.9078.62±0.8379.61±0.3480.81±0.3781.31±0.13 OTV (r = 2)66.02±1.1879.52±0.7382.05 ±0.4483.12±0.2584.06±0.2484.24±0.30 OTV (r = 16)66.02±1.1879.51±0.5782.27±0.3083.54±0.4284.18±0.3184.48±0.21 •Noise-perturbed ramp. To model local fluctuations in uncertainty while preserving a global monotonic trend, we perturb the linear baseline with additive sinusoid and Gaussian noise: c t = clip linear(t) + β sin 2πωt T + ε, 0, 1 ,(14) where β is the amplitude, ω is the frequency, and ε∼N (0,σ 2 ). We set β = 0.1 and ω = 3 by default. •Stepwise ramp (by reasoning steps). Some traces are naturally organized into discrete reasoning steps. We segment the trace intoΓsteps using “ ” delimiters and letγ(t)denote the index of the reasoning step to which tokentbelongs. We then define c t = 0.5 + (y− 0.5) γ(t) Γ ,(15) giving rise to a piecewise-constant trajectory that increases only when a new step begins. B.2. Evaluation on Pretrained Base Models 148163264 Number of traces 80 85 90 95 100 Accuracy (%) pass@k maj@k ours@k Figure 4. Evaluation on GSM8K using Qwen3-4B-Base. To further probe whether the observed gains are specific to post-trained reasoning models, we additionally evaluate OTV on a pretrained base model, Qwen3-4B-Base. In this setting, we train on MetaMathQA (Yu et al., 2024) and evaluate on GSM8K (Cobbe et al., 2021) to assess gains in raw mathematical reasoning capabilities. As shown in Figure 4, OTV substantially improves over Maj@kand narrows the gap toward Pass@k, indicating that the verification signal remains effective even without instruction tuning. B.3. Evaluation on Other Model Families To assess architectural generality beyond the Qwen family, OTV is applied to MetaMath- LLaMA-7B 3 and MetaMath-Mistral-7B 4 models on GSM8K under low-resource fine-tuning (using only 10k MetaMathQA subset). As shown in Table 5, OTV consistently outperforms majority voting and DeepConf across sample sizes, and increasing LoRA rank typically improves performance. B.4. Inference-time Verification Overhead OTV’s verification calls depend on the aggregation strategy (illustrated for N = 128). • Drop@10 (8, 256calls) prunes the single lowest-scoring trace every10generated tokens until only one trace remains. At the first checkpoint (t = 10), all128traces are verified and the worst is discarded; at the second (t = 20), the remaining 3 https://huggingface.co/meta-math/MetaMath-7B-V1.0 4 https://huggingface.co/meta-math/MetaMath-Mistral-7B 14 One-Token Verification for Reasoning Correctness Estimation 127 traces are verified and one is dropped, and so on. The total number of verification calls is the arithmetic sum: #Calls = N X n=1 n = N (N + 1) 2 = 128× 129 2 = 8, 256.(16) • Stop@600 (128calls) as a “verify-once” strategy, triggers the verifier only when a trace reaches the600-th token (or earlier if generation terminates). Consequently, each of the N candidate traces is verified exactly once: #Calls = N = 128.(17) • Halve@300 (254calls) halves the candidate pool every300tokens. That is, we verify128traces att = 300,64at t = 600, 32 at t = 900, and so forth. The total number of calls follows a geometric series with the upper bound: #Calls≈ ⌈log 2 (N−1)⌉ X k=0 N 2 k = 128 + 64 + 32 +· + 4 + 2 = 254.(18) C. Theoretical Analysis This section formalizes what OTV learns under different pseudo-confidence labeling rules by viewing verification as square-loss regression on a trace prefix s t = [q,x 1:t ], equivalently, its KV cache. C.1. Setups Let the base reasoner induce an autoregressive policyπ θ (x t | q,x 1:t )and sample tracesx 1:T = [x 1 ,...,x T ]with terminal correctness y ∈0, 1. For any pseudo-confidence labeling rule c t = c(t,T,y), OTV minimizes the per-token MSE: ℓ(φ) = E h (f φ (s t )− c(t,T,y)) 2 i ,(19) wheref φ (·)denotes the verifier, parameterized by vectorφand the expectation is taken over trajectoriesx 1:T ∼ π θ (·| q) sampled from the base reasoner. C.2. Optimal predictor under MSE Proposition C.1 (Risk Minimizer under MSE). For any fixedtand any states t , among all measurable functionsf φ (·), the minimizer of the conditional risk E (f φ (s t )− c(t,T,y)) 2 | s t is the conditional expectation f φ ⋆ (s t ) = E[c(t,T,y)| s t ].(20) Proof. Fix s t and consider any scalar a. By the bias-variance decomposition, E (a− c(t,T,y)) 2 | s t = (a− E[c(t,T,y)| s t ]) 2 + Var (c(t,T,y)| s t ),(21) which is minimized at a = E[c(t,T,y)| s t ]. This shows that the choice of pseudo-confidence labeling rulec t determines the conditional statistic that OTV would like to approximate during training. C.3. Constant Label Recovers a Monte-Carlo Correctness Value For the constant label rule c const (t,T,y) = y, the optimal predictor equals V MC (s t ) := P(y = 1| s t ) = E[y | s t ],(22) i.e., the probability that a continuation sampled from π θ (·| s t ) yields a correct final answer. 15 One-Token Verification for Reasoning Correctness Estimation C.4. Linear Ramp Induces an Inverse-Length Preference Consider the linear rampc lin (t,T,y) = 0.5 + (y− 0.5) t T , which satisfiesc lin (t,T,y)∈ [0, 1]fort≤ T, andc lin (T,T,y) = y. Define the expected inverse lengths of correct and incorrect completions from s t as μ + (s t ) := E 1 T s t , y = 1 , μ − (s t ) := E 1 T s t , y = 0 .(23) Proposition C.2. The MSE-optimal predictor under c lin is f φ ⋆ (s t ) = 0.5− t 2 μ − (s t ) | z baseline + V MC (s t )· t 2 μ + (s t ) + μ − (s t ) |z gain>0 ,(24) where V MC (s t ) = E[y | s t ]. In particular, for fixed s t and t, f φ ⋆ (s t ) is strictly increasing in V MC (s t ). Proof. By Proposition C.1 f φ ⋆ (s t ) = E 0.5 + (y− 0.5) t T s t = 0.5 + t· E y− 0.5 T s t .(25) Conditioning on y ∈0, 1 yields E y− 0.5 T s t = P(y = 1| s t )· E 0.5 T s t ,y = 1 + P(y = 0| s t )· E −0.5 T s t ,y = 0 = 1 2 V MC (s t )μ + (s t )− 1 2 (1− V MC (s t ))μ − (s t ),(26) and substituting into Eq. (25) gives Eq. (24). Sinceμ + ,μ − > 0, the coefficient onV MC equals t 2 (μ + + μ − ) > 0, implying strict monotonicity. Eq. (24) shows that, beyond its monotonic dependence onV MC , the score is modulated byμ + andμ − ; holdingV MC fixed, larger μ + (i.e., shorter correct completions in expectation) increases f φ ⋆ , inducing a preference on shorter correct traces. D. Future Work Building on these observations, several directions appear promising. The first is to tighten model-verifier co-adaptation. Rather than treating the verifier as a fixed add-on, we may explore joint or continual training where the verifier tracks changes in the base model and the evolving decoding distribution. This naturally connects model-centric improvements (e.g., stronger reasoning backbones, better calibration, and decoding-time optimization) with data-centric improvements: OTV’s scores can be used to curate higher-quality training traces, perform hard-negative mining, and support active learning by prioritizing “high-uncertainty” cases for annotation or additional compute. Second, improving token-level pseudo-confidence labeling is a key opportunity. The current approach derives dense targets from final outcomes, but richer supervision could substantially improve performance. One direction is uncertainty-aware pseudo-labels that represent partial progress, reversible mistakes, or late-step slips rather than enforcing uniformly monotonic confidence trajectories. Another direction is hybrid bootstrapping for token-level supervision: augment outcome labels with auxiliary signals such as self-consistency/ensemble disagreement, step-boundary priors (e.g., transition markers between reasoning steps), and selective distillation from stronger but costlier process verifiers on a small subset. This can be implemented iteratively, where each improved verifier produces higher-fidelity pseudo-labels for subsequent training rounds, thereby reducing systematic bias introduced by weak initial targets. Third, it is desirable to extend the verifier output space and its role in decision-making, with the goal of unlocking broader applications. For example, a selective prediction formulation, e.g., moving from binary confidence to a ternary “correct/incorrect/unknown” signal, would allow for abstention and risk-controlled routing when the model is uncertain. Finally, it would be valuable to test OTV beyond math benchmarks, including code reasoning and tool-augmented tasks, and to study how verifier-guided compute allocation interacts with long-context settings (e.g., multi-step planning and multi-agent/ensemble decoding). 16 One-Token Verification for Reasoning Correctness Estimation E. Sensitivity to Solution Perturbations To test robustness to superficial edits vs. semantic errors, we construct controlled perturbations of the same solution text (see Table 6). Semantics-preserving edits (e.g., shortening/lengthening, formatting changes, or substituting the ground-truth answer) are contrasted with logic-breaking edits (e.g., changing numbers, altering logic, or corrupting the final step). Empirically, we observe that OTV’s confidence reliably increases for correct variants and drops for erroneous variants, suggesting it is driven by internal reasoning consistency rather than surface form. Table 6. Representative perturbed solution variants used in the sensitivity analysis. Each row applies a controlled edit to the same base solution to evaluate how the verifier’s confidence responds to semantics-preserving vs. logic-breaking modifications. TypeText Semantics-preserving modifications Ground-truthIt takes 2/2 = 1 bolt of white fiber the total amount of fabric is 2 + 1 = 3 bolts of fabric #### 3 OriginalTo solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. 1: Identify the amount of blue fiber needed. robe takes 2 bolts of blue fiber. 2: Calculate the amount of white fiber needed. robe takes half as much white fiber as blue fiber. Since it takes 2 bolts of blue fiber, it will take 2 / 2 = 1 bolt of white fiber. 3: Add the amounts of blue and white fiber to find the total number of bolts. bolts = Blue fiber + White fiber bolts = 2 + 1 bolts = 3. ShorteningTo find total bolts for the robe (blue + white fiber): 1: Blue fiber needed is 2 bolts. 2: White fiber is half blue, so 2/2=1 bolt. 3: Total = 2+1=3. Lengthening To solve the problem of determining the total number of bolts required for the robe, we must consider both the blue fiber and white fiber that are necessary for its construc- tion. 1: First, we need to identify the specific amount of blue fiber that is needed to make the robe. After checking the relevant specifications or requirements for the robe, it is determined that the robe takes 2 bolts of blue fiber. 2: Next, we need to calculate the amount of white fiber that is needed. According to the given information, the robe takes half as much white fiber as blue fiber. Since we already established that it takes 2 bolts of blue fiber, we can find the amount of white fiber by taking half of that number. To do this, we perform the calculation: 2 bolts (blue fiber) divided by 2, which equals 1 bolt. Therefore, the robe requires 1 bolt of white fiber. 3: Finally, to find the total number of bolts required for the robe, we need to add together the amounts of blue fiber and white fiber that we have determined. This means we take the 2 bolts of blue fiber and add the 1 bolt of white fiber to it. Performing this addition gives us: 2 + 1 = 3. Removing newline tokenTo solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. Step 1: Identify the amount of blue fiber needed. The robe takes 2 bolts of blue fiber. Step 2: Calculate the amount of white fiber needed. The robe takes half as much white fiber as blue fiber. Since it takes 2 bolts of blue fiber, it will take 2 / 2 = 1 bolt of white fiber. Step 3: Add the amounts of blue and white fiber to find the total number of bolts. Total bolts = Blue fiber + White fiber. Total bolts = 2 + 1. Total bolts = 3. Answer: 3. Logic-breaking modifications Changing numbersTo solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. 1: Identify the amount of blue fiber needed. robe takes 4 bolts of blue fiber. 2: Calculate the amount of white fiber needed. robe takes half as much white fiber as blue fiber. Since it takes 4 bolts of blue fiber, it will take 4 / 2 = 2 bolts of white fiber. 3: Add the amounts of blue and white fiber to find the total number of bolts. bolts = Blue fiber + White fiber bolts = 4 + 2 bolts = 6 : 6. Altering logic To solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. 1: Identify the amount of blue fiber needed. robe takes 2 bolts of white fiber. 2: Calculate the amount of white fiber needed. robe takes twice as much white fiber as blue fiber. Since it takes 2 bolts of blue fiber, it will take 2 * 2 = 4 bolts of white fiber. 3: Subtract the amounts of blue and white fiber to find the total number of bolts. bolts = Blue fiber - White fiber bolts = 2 - 1 bolts = 1 : 1. Adding extra stepsTo solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. 1: Identify the amount of blue fiber needed. robe takes 2 bolts of blue fiber. 2: Calculate the amount of white fiber needed. robe takes half as much white fiber as blue fiber. Since it takes 2 bolts of blue fiber, it will take 2 / 2 = 1 bolt of white fiber. 3: Add the amounts of blue and white fiber to find the total number of bolts. bolts = Blue fiber + White fiber bolts = 2 + 1 bolts = 3 4: Subtract 1 from the total because of a miscalculation. bolts = 3 - 1 bolts = 2 : 2. Corrupting final stepsTo solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. 1: Identify the amount of blue fiber needed. robe takes 2 bolts of blue fiber. 2: Calculate the amount of white fiber needed. robe takes half as much white fiber as blue fiber. Since it takes 2 bolts of blue fiber, it will take 2 / 2 = 1 bolt of white fiber. 3: Multiply the amounts of blue and white fiber to find the total number of bolts. bolts = Blue fiber × White fiber bolts = 2 × 1 bolts = 2 : 2. Repeating the question To solve this problem, we need to determine the total number of bolts required for the robe, which includes both blue and white fiber. A robe takes 2 bolts of blue fiber and half that much white fiber. How many bolts in total does it take? A robe takes 2 bolts of blue fiber and half that much white fiber. How many bolts in total does it take? A robe takes 2 bolts of blue fiber and half that much white fiber. How many bolts in total does it take? A robe takes 2 bolts of blue fiber and half that much white fiber. How many bolts in total does it take? A robe takes 2 bolts of blue fiber and half that much white fiber. How many bolts in total does it take? 17 One-Token Verification for Reasoning Correctness Estimation F. Trace-Level Confidence Dynamics on AIME Problems Figure 5 plots trace-level confidence trajectories for all AIME24/25 problems under Qwen3-4B-Instruct. Each curve corresponds to one sampled trace, with correct traces in red and incorrect traces in green. 05001000150020002500 Token index 0.25 0.50 0.75 Score Problem 0 05001000150020002500 Token index 0.25 0.50 0.75 Score Problem 1 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 2 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 3 0200040006000 Token index 0.25 0.50 0.75 Score Problem 4 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 5 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 6 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 7 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 8 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 9 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 10 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 11 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 12 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 13 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 14 0500100015002000 Token index 0.25 0.50 0.75 Score Problem 15 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 16 0200040006000 Token index 0.25 0.50 0.75 Score Problem 17 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 18 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 19 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 20 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 21 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 22 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 23 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 24 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 25 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 26 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 27 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 28 050010001500 Token index 0.25 0.50 0.75 Score Problem 29 (a) AIME24 02004006008001000 Token index 0.25 0.50 0.75 Score Problem 0 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 1 0100020003000 Token index 0.25 0.50 0.75 Score Problem 2 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 3 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 4 05001000150020002500 Token index 0.25 0.50 0.75 Score Problem 5 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 6 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 7 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 8 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 9 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 10 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 11 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 12 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 13 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 14 050010001500 Token index 0.25 0.50 0.75 Score Problem 15 0100020003000 Token index 0.25 0.50 0.75 Score Problem 16 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 17 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 18 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 19 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 20 0200040006000 Token index 0.25 0.50 0.75 Score Problem 21 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 22 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 23 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 24 02000400060008000 Token index 0.25 0.50 0.75 Score Problem 25 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 26 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 27 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 28 0200040006000800010000 Token index 0.25 0.50 0.75 Score Problem 29 (b) AIME25 Figure 5. Trace-level confidence trajectories across problems in (a) AIME24 and (b) AIME25. 18 One-Token Verification for Reasoning Correctness Estimation G. Finer-Grained Visualization of Reasoning Correctness Estimation To complement the main results, we present token-level visualizations of OTV’s correctness estimates, where the generated reasoning text is color-coded from green (low confidence) to red (high confidence). We include six examples spanning three comparisons: 1) incorrect vs. correct traces for the same problem (Examples 1–2), demonstrating clear confidence separation; 2) shorter vs. longer correct solutions (Examples 3–4), where shorter traces sustain consistently higher confidence while longer traces show greater fluctuations, indicating intermediate uncertainty; and 3) correct vs. incorrect traces produced by a larger model on the same problem (Examples 5–6), where confidence declines near the end in both cases, but remains substantially higher overall for correct traces. Example 1 (AIME25, Problem 23): Qwen3-4B-Instruct-OTV There arenvalues ofxin the interval0 < x < 2πwheref(x) = sin (7π· sin(5x)) = 0. Fortof thesenvalues ofx, the graph of y = f(x) is tangent to the x-axis. Find n + t. (Correct Answer: 149) Wearegiventhefunction:f(x) = sin (7π· sin(5x))Wearetofind:-Thetotalnumbernofvaluesofxintheinterval 0x2πsuchthatf(x) = 0-Amongthesenvalues,howmanytaresuchthatthegraphis**tangent**tothex-axis(i.e., f(x) = 0andf ′ (x) = 0)-Thencomputen + t—Step1:Whenisf(x) = 0?Wehave:f(x) = sin (7π· sin(5x)) = 0The sinefunctioniszerowhenitsargumentisanintegermultipleofπ:7π· sin(5x) = kπfor some integer kDividebothsides byπ:7 sin(5x) = k ⇒ sin(5x) = k 7 Now,sincesin(5x)∈ [−1, 1],wemusthave:−1≤ k 7 ≤ 1 ⇒ −7≤ k ≤ 7 Sokrangesfrom−7to7,inclusive.That’s15integervalues.Foreachsuchk,wesolve:sin(5x) = k 7 Wewanttofind howmany**solutions**x∈ (0, 2π)satisfythisequation.Let’sdefineθ = 5x.Thenx∈ (0, 2π)⇒ θ ∈ (0, 10π)Soweare solvingsin(θ) = k 7 forθ ∈ (0, 10π)—Step2:Numberofsolutionstosin(θ) = cin(0, 10π)Forafixedc∈ (−1, 1),the equationsin(θ) = chas**twosolutionsperperiod**of2π,exceptwhenc =±1,whereithasonesolutionperperiod(atthe peak/trough).Soweanalyzebasedonthevalueofk:Case1:k = 0Thensin(5x) = 0Solutions:sin(θ) = 0,θ ∈ (0, 10π) Solutionsoccuratθ = π, 2π, 3π, 4π, 5π, 6π, 7π, 8π, 9πNote:θ = 0and10πareexcludedbecausex = 0andx = 2πare excluded.Soθ = π, 2π, 3π, 4π, 5π, 6π, 7π, 8π, 9π→9solutionsEachcorrespondstox = θ/5,sox = π/5, 2π/5,..., 9π/5 Allarein(0, 2π)So**9solutions**fork = 0—Case2:k ̸= 0Weconsiderk =±1,±2,...,±7Foreachsuchk, k 7 ∈ (−1, 1),sosin(θ) = k 7 has**twosolutionsperperiod**of2πin(0, 10π)Howmanyperiodsin(0, 10π)? 10π 2π = 5 fullperiodsSoineachperiod,twosolutions→totalof5× 2 = 10solutionspervalueofk ∈−7,..., 7\0But wait—arethereanyexceptions?Onlywhensin(θ) =±1,i.e.,k =±7Let’scheck:-k = 7:sin(5x) = 1-k =−7: sin(5x) =−1Forthese,sin(θ) =±1Ineach2πinterval,sin(θ) = 1occursonce:atθ = π 2 + 2πmSimilarly,sin(θ) =−1 atθ = 3π 2 + 2πmSoin(0, 10π),howmanysuchsolutions?-Forsin(θ) = 1:θ = π 2 , 5π 2 , 9π 2 , 13π 2 , 17π 2 Check: π 2 + 2πm10π π 2 + 2πm10πDividebyπ: 1 2 + 2m10⇒ 2m9.5⇒ m4.75Som = 0, 1, 2, 3, 4→5valuesSimilarlyforsin(θ) =−1: θ = 3π 2 + 2πm 3 2 + 2m10⇒ 2m9.5⇒ m4.75 →m = 0, 1, 2, 3, 4→5valuesSofork =±7,wehave**5solutions each**Forotherk =±1,±2,±3,±4,±5,±6,wehavesin(θ) = k 7 ∈ (−1, 1),so**twosolutionsperperiod**,5peri- ods→5× 2 = 10solutionseach—Step3:CounttotalnumberofsolutionsnListallkfrom−7to7:-k = 0:9solutions -k =±1,±2,±3,±4,±5,±6:6valuesofk,eachwith10solutions→6× 10 = 60-k =±7:eachwith5solutions→ 2× 5 = 10Totaln = 9 + 60 + 10 = 79Son = 79—Step4:Findt—numberofvalueswherethegraphis**tangent**to thex-axisThismeansf(x) = 0**and**f ′ (x) = 0Weneedtofindthevaluesofxwheref(x) = 0andf ′ (x) = 0First,recall: f(x) = sin (7π· sin(5x))Letu(x) = 7π· sin(5x),sof(x) = sin(u(x))Then:f ′ (x) = cos(u(x))· u ′ (x)Computeu ′ (x): u ′ (x) = 7π· cos(5x)· 5 = 35π cos(5x)So:f ′ (x) = cos(7π sin(5x))· 35π cos(5x)Wewantf ′ (x) = 0**and**f(x) = 0 Sotwoconditions:1.sin(7π sin(5x)) = 02.cos(7π sin(5x))· 35π cos(5x) = 0Notethat35π ̸= 0,sowecandivide:Socon- dition2becomes:cos(7π sin(5x)) = 0 or cos(5x) = 0Butfromcondition1:sin(7π sin(5x)) = 0Lets = sin(5x).Then: -sin(7πs) = 0-Eithercos(7πs) = 0orcos(5x) = 0Wearetofindvalueswhere**both**conditionshold.Soweconsider twocases:—CaseA:cos(5x) = 0Then5x = π 2 + mπ ⇒ x = π 10 + mπ 5 Weneedx∈ (0, 2π)So0 π 10 + mπ 5 2πDivideby π:0 1 10 + m 5 2Multiplyby10:01 + 2m202m− 1⇒ m− 0.5→m≥ 02m19⇒ m9.5→m≤ 9Som = 0, 1, 2,..., 9 →10valuesNow,foreachsuchx,wemustcheckwhetherf(x) = 0,i.e.,sin(7π sin(5x)) = 0Atsuchx,cos(5x) = 0, so5x = π 2 + mπThensin(5x) = sin π 2 + mπ = (−1) m Sosin(5x) = (−1) m ∈−1, 1Then7π sin(5x) =±7πSo sin(7π sin(5x)) = sin(±7π) = sin(7π) = 0Becausesin(7π) = sin(π) = 0Soyes,f(x) = 0Also,f ′ (x) = 0because cos(5x) = 0So**all10values**ofxwherecos(5x) = 0satisfybothf(x) = 0andf ′ (x) = 0So**tincludesthese10 values**—CaseB:cos(7π sin(5x)) = 0**and**sin(7π sin(5x)) = 0Isthispossible?Letθ = 7π sin(5x)Wewant:- sin(θ) = 0-cos(θ) = 0Butsin(θ) = 0andcos(θ) = 0is**impossible**foranyrealθBecausesin 2 (θ) + cos 2 (θ) = 1So bothcannotbezero.Therefore,**nosolutions**inthiscase.Sotheonlypointswheref(x) = 0andf ′ (x) = 0arewhen cos(5x) = 0Thus,t = 10—FinalStep:Computen + tWefound:-n = 79-t = 10So:Answer:89 19 One-Token Verification for Reasoning Correctness Estimation Example 2 (AIME25, Problem 23): Qwen3-4B-Instruct-OTV There arenvalues ofxin the interval0 < x < 2πwheref(x) = sin (7π· sin(5x)) = 0. Fortof thesenvalues ofx, the graph of y = f(x) is tangent to the x-axis. Find n + t. (Correct Answer: 149) Wearegiventhefunction:f(x) = sin (7π· sin(5x))Wearetofind:-n:thenumberofvaluesofxintheinterval0x2π wheref(x) = 0-t:thenumberofthosenvalueswherethegraphis**tangent**tothex-axis(i.e.,wheref(x) = 0 andf ′ (x) = 0)-Thencomputen + t—Step1:Whenisf(x) = 0?Wehave:f(x) = sin (7π· sin(5x)) = 0Thesine functioniszerowhenitsargumentisanintegermultipleofπ:7π· sin(5x) = kπfor some integer kDividebothsides byπ:7 sin(5x) = k ⇒ sin(5x) = k 7 Now,sincesin(5x)∈ [−1, 1],werequire:−1≤ k 7 ≤ 1 ⇒ −7≤ k ≤ 7So k =−7,−6,..., 0,..., 6, 7That’s15integervaluesofk.Foreachsuchk,wesolve:sin(5x) = k 7 Wenowwanttocounthow manysolutionsx∈ (0, 2π)satisfythis.—Step2:CounttotalnumberofsolutionsnLet’sconsidersin(5x) = c,wherec = k 7 , andc∈ [−1, 1].Foreachc∈ (−1, 1),theequationsin(5x) = chas**two**solutionsperperiodof5x,i.e.,foreveryinterval oflength 2π 5 inx,therearetwosolutions(sincesineisperiodicandsymmetric).Butweneedtocounthowmanysolutions existinx∈ (0, 2π).Let’sdefineθ = 5x.Thenasxgoesfrom0to2π,θgoesfrom0to10π.Soθ ∈ (0, 10π)Wenowsolve sin(θ) = k 7 forθ ∈ (0, 10π)Foreachk,weconsiderhowmanysolutionsθ ∈ (0, 10π)satisfysin(θ) = k 7 —Generalbehavior ofsin(θ) = c:-Ifc =±1,thereis**one**solutionperperiodof2π,butonlyatthepeaks/troughs.-Ifc∈ (−1, 1),thereare **two**solutionsper2πinterval(onein(0,π),onein(π, 2π))Butnote:θ ∈ (0, 10π),whichis**5fullperiods**of2πSo totallength:10π,so5periods.—Let’sconsidereachk:-Fork =±7:sin(5x) =±1-sin(θ) = 1:occursatθ = π 2 + 2πm -sin(θ) =−1:occursatθ = 3π 2 + 2πmIn(0, 10π),howmanysuchvalues?-θ = π 2 , 5π 2 , 9π 2 , 13π 2 , 17π 2 , 21π 2 , 25π 2 25π 2 = 12.5π10π ?10π = 20π/2,souptomsuchthat π 2 + 2πm10π π 2 + 2πm10π 2πm10π− π 2 = 19π 2 m 19 4 = 4.75 Som = 0, 1, 2, 3, 4→5valuesSosin(θ) = 1has5solutionsSimilarly,sin(θ) =−1:θ = 3π 2 + 2πm 3π 2 + 2πm10π 2πm10π− 3π 2 = 17π 2 m17/4 = 4.25→m = 0, 1, 2, 3, 4→5valuesSototalfork =±7:5 + 5 = 10solutions-Fork = 0: sin(θ) = 0Solutionsatθ = πm,m = 1, 2, 3, 4, 5, 6, 7, 8, 9→sinceθ ∈ (0, 10π)θ = π, 2π, 3π, 4π, 5π, 6π, 7π, 8π, 9π→ m = 1to9m = 10→10π,whichisexcludedsinceθ10πSo9solutions-Fork =±1,±2,±3,±4,±5,±6:theseareall in(−1, 1),andnot±1Foreachsuchc∈ (−1, 1),ineachintervalof2π,thereare**two**solutions.Sinceθ ∈ (0, 10π), whichis5fullperiods,eachsuchcgives2× 5 = 10solutions.Howmanysuchk?k =±1,±2,±3,±4,±5,±6→12values Sototalsolutionsfromthese:12× 10 = 120Wait—butwemustbecautious:arethesevaluesdistinct?Actually,eachk givesadifferentequationsin(5x) = k/7,andeachhasitsownsolutions.Soyes,wearecountingsolutionsforeachk,so wecansumoverallk.Sototaln:-k =±7:10solutions-k = 0:9solutions-k =±1,±2,±3,±4,±5,±6:12values, eachwith10solutions→12× 10 = 120Total:n = 10 + 9 + 120 = 139Wait—isthatcorrect?Wait:k =−7to7:total 15values.Webrokeitas:-k =±7:2values→10solutions-k = 0:1value→9solutions-k =±1to±6:12values →120solutionsTotal:10 + 9 + 120 = 139Butlet’sdouble-checkthenumberofsolutionsforsin(θ) = cin(0, 10π)For c∈ (−1, 1),sin(θ) = chastwosolutionsper2πinterval.In(0, 10π),whichis5fullperiods,thereare5× 2 = 10solutions perc.Yes.Forc =±1,onlyonesolutionper2πinterval.In(0, 10π),5periods→5solutionsper±1Sosin(θ) = 1:5 solutionssin(θ) =−1:5solutions→total10Forsin(θ) = 0:solutionsatθ = π, 2π, 3π, 4π, 5π, 6π, 7π, 8π, 9πθ = 10πis excluded→so9solutionsYes.Sototaln = 139Butwait—areallthesexvaluesin(0, 2π)?Yes,becauseθ = 5x∈ (0, 10π), sox∈ (0, 2π)Soyes.Son = 139—Step3:Findt:numberofxvalueswheref(x) = 0**and**f ′ (x) = 0Wewant thegraphtobetangenttothex-axis—thatis,a**doubleroot**—sof(x) = 0andf ′ (x) = 0Weneedtocomputef ′ (x) andfindwherebothf(x) = 0andf ′ (x) = 0Let:f(x) = sin (7π· sin(5x))Letu = 7π· sin(5x),sof(x) = sin(u) Then:f ′ (x) = cos(u)· d dx (7π sin(5x)) = cos(7π sin(5x))· 7π· 5 cos(5x)f ′ (x) = 35π cos(7π sin(5x))· cos(5x) Wewantf(x) = 0**and**f ′ (x) = 0So:1.sin(7π sin(5x)) = 02.cos(7π sin(5x))· cos(5x) = 0From(1): 7π sin(5x) = kπ ⇒ sin(5x) = k 7 forintegerk ∈ [−7, 7]From(2):cos(7π sin(5x))· cos(5x) = 0Soeither:- cos(7π sin(5x)) = 0,or-cos(5x) = 0Butnote:sin(5x) = k 7 ,so7π sin(5x) = kπSocos(7π sin(5x)) = cos(kπ) Because7π sin(5x) = kπSocos(kπ) = (−1) k Socos(7π sin(5x)) = (−1) k Thisis**neverzero**unless(−1) k = 0, whichisimpossible.Socos(7π sin(5x))̸= 0foranyintegerkTherefore,theonlywayforf ′ (x) = 0isifcos(5x) = 0So werequire:-sin(5x) = k 7 -cos(5x) = 0Now,wheniscos(5x) = 0?Thathappenswhen5x = π 2 + mπ,forintegermSo x = π 10 + mπ 5 Wenowneedtofindvaluesofx∈ (0, 2π)suchthat:-cos(5x) = 0-sin(5x) = k 7 forsomeintegerkBut whencos(5x) = 0,sin(5x) =±1Sosin(5x) =±1Therefore, k 7 =±1⇒ k =±7Soonlywhenk =±7,wecanhave cos(5x) = 0andsin(5x) =±1Therefore,theonlypossibletangentpointsoccurwhensin(5x) =±1Now,wealready foundthatwhensin(5x) =±1,f(x) = sin(7π· (±1)) = sin(±7π) = 0Becausesin(7π) = sin(π) = 0Sof(x) = 0And wejustshowedf ′ (x) = 0whencos(5x) = 0Therefore,atallpointswheresin(5x) =±1,wehavebothf(x) = 0and f ′ (x) = 0Therefore,**tangentpointsoccurpreciselywhensin(5x) =±1**Wealreadycomputedthatin(0, 10π),θ = 5x, sin(θ) = 1has5solutions,sin(θ) =−1has5solutions→total10solutionsSot = 10Butwait—arethesepointsdistinct? Yes—forsin(θ) = 1:θ = π 2 , 5π 2 , 9π 2 , 13π 2 , 17π 2 17π 2 = 8.5π10π,nextwouldbe 21π 2 = 10.5π10πSimilarly,sin(θ) =−1: θ = 3π 2 , 7π 2 , 11π 2 , 15π 2 , 19π 2 19π 2 = 9.5π10π ,next 23π 2 = 11.5π10π Allaredistinctandin(0, 10π)Sototalt = 5 + 5 = 10— FinalStep:Computen + tWefound:-n = 139-t = 10So:n + t = 139 + 10 = 149Answer:149 20 One-Token Verification for Reasoning Correctness Estimation Example 3 (AIME24, Problem 17): Qwen3-4B-Instruct-OTV There exist real numbers x and y, both greater than 1, such that log x (y x ) = log y x 4y = 10. Find xy. (Correct Answer: 25) Wearegiventhat:log x (y x ) = 10 and log y (x 4y ) = 10Wearetofindxy,wherex1,y1.—Step1:Simplifythefirstequationlog x (y x ) = 10Uselogarith- micidentity:log b (a c ) = c log b alog x (y x ) = x log x y = 10So:x log x y = 10 (Equation 1)—Step2:Simplifythesecondequationlog y (x 4y ) = 10Again,use log b (a c ) = c log b a:4y log y x = 10 (Equation 2)—Step3:UsesubstitutionLet’sdefinea = log x y.ThenfromEquation1:x· a = 10 ⇒ a = 10 x Now,recallthat log x y = a⇒ log y x = 1 a (sincelogarithmsarereciprocals)SofromEquation2:4y· log y x = 10⇒ 4y· 1 a = 10Substitutea = 10 x :4y· 1 10 x = 10⇒ 4y· x 10 = 10 Simplify: 4xy 10 = 10⇒ 2xy 5 = 10Multiplybothsidesby5:2xy = 50⇒ xy = 25—Answer:25 Example 4 (AIME24, Problem 22): Qwen3-4B-Instruct-OTV Let△ABChave circumcenterOand incenterIwithIA ⊥OI, circumradius13, and inradius6. FindAB· AC. (Correct Answer: 468) Wearegiventhefollowingabouttriangle△ABC:-Oisthecircumcenter.-Iistheincenter.-IA⊥ OI.-CircumradiusR = 13.-Inradiusr = 6.-WearetofindAB· AC. —Step1:UseknowngeometricidentitiesWearegivenaperpendicularitycondition:IA⊥ OI.Thisisastronggeometricconditionandsuggestswecanuseknownformulasor coordinategeometry.WeaimtocomputeAB· AC,whichisaproductoftwosides.Let’srecallsomeknownidentitiesandformulas:—Step2:Useknownformulainvolving IA,OI,andperpendicularityThereisaknownidentityinvolvingthedistancebetweenincenterIandcircumcenterO:OI 2 = R(R− 2r)Thisis**Euler’sformula**forthe distancebetweenincenterandcircumcenter.Wait—actually,**no**,Euler’sformulais:OI 2 = R 2 − 2RrYes—thecorrectformulais:OI 2 = R 2 − 2RrLet’sverify:- R = 13,r = 6-SoOI 2 = 13 2 − 2· 13· 6 = 169− 156 = 13SoOI 2 = 13—Step3:Usethegivenperpendicularity:IA⊥ OIWearetoldIA⊥ OI.Thisisa keycondition.Let’susevectorgeometryorcoordinategeometrytoexploitthis.Let’splacethetriangleinthecoordinateplanewithconvenientpositions.—Step4:Usecoordinate geometryLetusplacepointAattheorigin,andusesymmetry.Butperhapsabetterideaistouseknownvectoridentities.Thereisaknownvectoridentity:IntriangleABC,with circumcenterO,incenterI,andapointA,thecondition ⃗ IA· ⃗ OI = 0 (sinceIA⊥ OI)implies: ⃗ IA· ⃗ OI = 0 Let’sdefinevectorswithrespecttopointA.Alternatively,use knownformulaforIA 2 :IA 2 = bc (b+c) 2 (b + c) 2 − a 2 Butthatmightbemessy.Thereisaknownformula:IA 2 = bc (b+c) 2 (b + c− a)(b + c + a)Alternatively, standardformula:IA 2 = r 2 + (s− a) 2 Yes!That’sastandardidentity:Inanytriangle,thedistancefromtheincenterItovertexAis:IA 2 =r 2 +(s− a) 2 where sisthesemi-perimeter,anda = BC,b = AC,c = AB.WearetofindAB· AC = bc.Solet’sdefine:-a = BC-b = AC-c = AB-s = a+b+c 2 -r = 6- R = 13-OI 2 = 13Wealsohavetheidentity:IA 2 = r 2 + (s− a) 2 WearetoldIA⊥ OI,sothevectors ⃗ IA and ⃗ OI areperpendicular.Wecanuseaknown**vector identity**or**formula**involvingthisperpendicularity.—Step5:UseknownidentityforperpendicularityofIAandOIThereisaknownresultintrianglegeometry:Intriangle ABC,ifIA⊥ OI,then:bc = r 2 +4Rr sin 2 A Wait—notsure.Alternatively,aknownresultfromolympiadgeometry:IfIA⊥ OI,then: r R = 1 2 1− a 2 b 2 +c 2 Notsure. Alternatively,wecanuseaknownformulainvolvingOI 2 ,IA 2 ,andanglebetweenthem.SinceIA⊥ OI,thentheanglebetweenvectors ⃗ IA and ⃗ OI is90 ◦ ,sotheirdot productiszero.Let’susevectoridentities.Let’splacepointAattheorigin,andusevectors.Let’sdenote:-Let ⃗ A = ⃗ 0 -Let ⃗ B = ⃗ b , ⃗ C = ⃗c -ThencircumcenterOisequidistant fromA,B,C,so| ⃗ O| =| ⃗ O− ⃗ b| =| ⃗ O−⃗c| = R = 13-IncenterIhaspositionvector:since ⃗ A = 0So ⃗ I = b ⃗ b+c⃗c a+b+c Now, ⃗ IA = ⃗ A− ⃗ I =− ⃗ I ⃗ OI = ⃗ I− ⃗ OWe aretoldIA⊥ OI,so: ⃗ IA· ⃗ OI = 0⇒ (− ⃗ I)· ( ⃗ I− ⃗ O) = 0⇒− ⃗ I· ⃗ I + ⃗ I· ⃗ O = 0⇒ ⃗ I· ⃗ O =| ⃗ I| 2 Sowehave: ⃗ I· ⃗ O =| ⃗ I| 2 Thisisakeyvectorequation.Now, | ⃗ O| = R = 13,so| ⃗ O| 2 = 169Also,| ⃗ O− ⃗ b| 2 = 169,so| ⃗ O| 2 − 2 ⃗ O· ⃗ b +| ⃗ b| 2 = 169But| ⃗ O| 2 = 169,so:169− 2 ⃗ O· ⃗ b +| ⃗ b| 2 = 169⇒| ⃗ b| 2 = 2 ⃗ O· ⃗ b Similarly,| ⃗ O−⃗c| 2 = 169⇒|⃗c| 2 = 2 ⃗ O·⃗cSowehave:| ⃗ b| 2 = 2 ⃗ O· ⃗ b, |⃗c| 2 = 2 ⃗ O·⃗cNow,recall ⃗ I = b ⃗ b+c⃗c a+b+c Letσ = a + b + c,so ⃗ I = b ⃗ b+c⃗c σ Then ⃗ I· ⃗ O = 1 σ (b( ⃗ b· ⃗ O) + c(⃗c· ⃗ O)) Butfromabove:- ⃗ b· ⃗ O = 1 2 | ⃗ b| 2 -⃗c· ⃗ O = 1 2 |⃗c| 2 So: ⃗ I· ⃗ O = 1 σ b· 1 2 | ⃗ b| 2 + c· 1 2 |⃗c| 2 = 1 2σ b| ⃗ b| 2 + c|⃗c| 2 Now,| ⃗ b| = c,since| ⃗ b| = AB = c?Wait—no.Wait:wedefined ⃗ b = ⃗ B ,so| ⃗ b| = AB = cSimilarly,|⃗c| = AC = bYes!So:-| ⃗ b| = c-|⃗c| = b So: ⃗ I· ⃗ O = 1 2σ bc 2 + cb 2 = 1 2σ (bc)(c + b) = bc(b+c) 2σ Ontheotherhand,| ⃗ I| 2 = b ⃗ b+c⃗c σ 2 = 1 σ 2 (b 2 | ⃗ b| 2 + c 2 |⃗c| 2 + 2bc ⃗ b·⃗c)So: | ⃗ I| 2 = 1 σ 2 b 2 c 2 + c 2 b 2 + 2bc( ⃗ b·⃗c) = 1 σ 2 2b 2 c 2 + 2bc( ⃗ b·⃗c) Wealreadyhave ⃗ I· ⃗ O =| ⃗ I| 2 ,so: bc(b+c) 2σ = 2b 2 c 2 +2bc( ⃗ b·⃗c) σ 2 Multiplybothsides byσ 2 : bc(b+c) 2σ · σ 2 = 2b 2 c 2 + 2bc( ⃗ b·⃗c) Simplifyleft: bc(b+c)σ 2 = 2b 2 c 2 + 2bc( ⃗ b·⃗c) Dividebothsidesbybc(nonzero): (b+c)σ 2 = 2bc + 2( ⃗ b·⃗c) Now, σ = a + b + c,so: (b+c)(a+b+c) 2 = 2bc + 2( ⃗ b·⃗c) Now, ⃗ b·⃗c =| ⃗ b||⃗c| cosA = bc cosASo: (b+c)(a+b+c) 2 = 2bc + 2bc cosA = 2bc(1 + cosA) So: (b+c)(a+b+c) 2 = 2bc(1 + cosA)Now,wealsoknowfromtrianglegeometry:-r = 6-R = 13-r = 4R sin A 2 sin B 2 sin C 2 -Also,sinA = a 2R ,butmoreuseful: cosA = b 2 +c 2 −a 2 2bc Let’susetheidentity: (b+c)(a+b+c) 2 = 2bc(1 + cosA)Let’sdenotex = b + c,y = bc,andwewanttofindy = bcLet’salsowriteainterms ofangles.Butwealsoknowthat:Area∆ = rs = abc 4R So:rs = abc 4R Weknowr = 6,R = 13So:6s = abc 52 ⇒ abc = 312sAlso,s = a+b+c 2 Let’sgobackto theequation: (b+c)(a+b+c) 2 = 2bc(1 + cosA) Letx = b + c,y = bc,aisunknown.Thens = a+x 2 Leftside: x(a+x) 2 Rightside:2y(1 + cosA)Wealsoknow cosA = b 2 +c 2 −a 2 2bc Noteb 2 + c 2 = (b + c) 2 − 2bc = x 2 − 2ySo:cosA = x 2 −2y−a 2 2y Thus:1 + cosA = 1 + x 2 −2y−a 2 2y = 2y+x 2 −2y−a 2 2y = x 2 −a 2 2y Sorightsidebecomes:2y· x 2 −a 2 2y = x 2 − a 2 Sotheequationbecomes: x(x+a) 2 = x 2 − a 2 Multiplybothsidesby2:x(x + a) = 2(x 2 − a 2 )Left:x 2 + axRight: 2x 2 − 2a 2 Bringalltooneside:x 2 + ax− 2x 2 + 2a 2 = 0⇒−x 2 + ax + 2a 2 = 0⇒ 2a 2 + ax− x 2 = 0Thisisaquadraticina:2a 2 + ax− x 2 = 0 Solvefora:a = −x± q x 2 +8x 2 4 = −x± √ 9x 2 4 = −x±3x 4 So:-a = 2x 4 = x 2 -a = −4x 4 =−x(invalid,sincea0)Soa = x 2 Thatis:a = b+c 2 Interesting! Sothesidea = BCistheaverageofbandc.Sob + c = 2a—Now,recallfromearlier:Wealsohavetheareaformula:∆ = rs = 6sAlso:∆ = abc 4R = abc 52 So: 6s = abc 52 ⇒ abc = 312s Now,s = a+b+c 2 Butb + c = 2a,soa + b + c = a + 2a = 3aThuss = 3a 2 So:abc = 312· 3a 2 = 468a Butbc = y,b + c = 2a,so bandcarerootsoft 2 − 2at + y = 0Wealsohaveabc = a· bc = aySo:ay = 468a⇒ y = 468 (since a̸= 0)Therefore:bc = AB· AC = 468Wait—isthatit? Let’sverifyifthisisconsistentwithotherknownvalues.WehaveR = 13,r = 6,bc = 468,b + c = 2a,s = 3a 2 Wecannowcomputeafromknownformulas.Wecanusethe formula:r = (s− a) tan A 2 Butalternatively,usetheformula:sinA = a 2R = a 26 Wecanalsousetheidentity:Area ∆ = 1 2 bc sinA = 1 2 · 468· sinA = 234 sinA But also∆ = rs = 6· 3a 2 = 9aSo:9a = 234 sinA⇒ sinA = 9a 234 = 3a 78 = a 26 ButsinA = a 2R = a 26 Yes!Matchesperfectly.Sothisisconsistent.Therefore,thevalue bc = 468isconsistentwithallconditions.Now,wemustalsocheckwhetherOI 2 = 13issatisfied.WehaveOI 2 = R 2 − 2Rr = 13 2 − 2· 13· 6 = 169− 156 = 13Yes— it’sgiven,sothisisautomaticallysatisfied.Andweusedtheperpendicularityconditiontoderivea = b+c 2 , andfromthat,bc = 468Therefore,theansweris468 21 One-Token Verification for Reasoning Correctness Estimation Example 5 (AIME25, Problem 19): DAPO-Qwen-32B-OTV Suppose△ABChas angles∠BAC = 84 ◦ ,∠ABC = 60 ◦ , and∠ACB = 36 circ . LetD,E, andFbe the midpoints of sides BC,AC, andAB, respectively. The circumcircle of△DEFintersectsBD,AE, andAFat pointsG,H, andJ, respectively. The pointsG,D,E,H,J, andFdivide the circumcircle of△DEFinto six minor arcs, as shown. FindDE +2·HJ +3·FG, where the arcs are measured in degrees. (Correct Answer: 336) A BCD EF G H J Toapproachthismathproblemstep-by-step,weneedtousepropertiesoftriangles,midpoints,andthenine-pointcircle.Giventhetriangle△ABCwith ∠BAC = 84 ◦ ,∠ABC = 60 ◦ ,and∠ACB = 36 ◦ .ThepointsD,E,andFarethemidpointsofsidesBC,AC,andABrespectively.Therefore,△DEF isthemedialtriangleof△ABC.Thecircumcircleof△DEFisactuallythenine-pointcircleof△ABC.Thenine-pointcirclepassesthroughthemidpoints ofthesides,thefeetofthealtitudes,andthemidpointsofthesegmentsjoiningeachvertextotheorthocenterof△ABC.Thekeyobservationhereisthatthe circumcircleof△DEFisthenine-pointcircleof△ABC.NowweneedtofindthemeasuresofthearcsDE,HJ,andF Gonthecircumcircleof△DEF.To doso,let’scalculatetheanglesatthecentercorrespondingtothesearcs.Since△DEFisthemedialtriangleof△ABC,theanglesof△DEFarethesameas theanglesof△ABC.Specifically,∠EDF = ∠BAC = 84 ◦ ,∠DEF = ∠ABC = 60 ◦ ,∠DF E = ∠ACB = 36 ◦ .Thecircumcircleof△DEFis thenine-pointcircle,andthecentralanglescorrespondingtothearcsDE,EF,andF Dwillbetwicetheanglesof△DEFattheverticesoppositethesearcs. Thus,themeasureofthearcEFistwice∠EDF = 2× 84 ◦ = 168 ◦ ,themeasureofthearcF Distwice∠DEF = 2× 60 ◦ = 120 ◦ ,andthemeasureof thearcDEistwice∠DF E = 2× 36 ◦ = 72 ◦ .NowweneedtodeterminethelocationsofpointsG,H,andJ.PointGistheintersectionofthecircumcircle of△DEFwithBD.SinceDisthemidpointofBC,BDisamedian.PointHistheintersectionofthecircumcircleof△DEFwithAE,andpointJisthe intersectionofthecircumcircleof△DEFwithAF.TofindthearcsDE,HJ,andF G,weneedtounderstandtheconfigurationofthesepoints.SinceD,E,and FaremidpointsandthelinesBD,AE,andAFaremediansandanglebisectorsorsegmentsconnectedtotheorthocenterandcircumcenterpropertiesinthe contextofthenine-pointcircle.Let’scalculatethearcsstep-by-step.SinceD,E,andFaremidpoints,BDgoesthroughthemidpointofBC,andintersectsthe nine-pointcircleagainatpointG.Similarly,AEintersectsthecircleatpointH,andAFintersectsitatpointJ.ThekeyistofindthemeasuresofthearcsDE, HJ,andF G.ThearcDEisthearcsubtendedby∠DF E = 36 ◦ ,sothemeasureofDEis72 ◦ .NowweneedtodetermineHJandF G.Let’susesomeangle chasingandpropertiesofthenine-pointcircle.SinceDisthemidpointofBC,EisthemidpointofAC,andFisthemidpointofAB,thelinesBD,AE,and AFaremediansof△ABC.ThesemediansalsopassthroughthecentroidG △ of△ABC.ThepointsG,H,andJarethesecondintersectionpointsofthe medianswiththenine-pointcircle.TofindthearcsHJandF G,weusethefactthatthearcsarerelatedtotheanglessubtendedbythechordscorrespondingto thesearcs.Let’scalculateHJandF G.SinceHisonAEandthenine-pointcircle,andJisonAFandthenine-pointcircle,weneedtodeterminetheangles correspondingtothesearcs.TofindthemeasureofHJ,wenotethatHandJlieonthelinesAEandAFrespectively.Thekeyobservationisthattheconfiguration ofthepointsG,H,andJdividesthecircleintoarcsthatcorrespondtotheanglesofthetriangleandthepropertiesofthenine-pointcircle.Bytheproperties ofcyclicquadrilateralsandtheanglessubtendedbychordsonthecircle,wecanusethefactthatthearcsHJandF Garerelatedtotheanglesofthetriangle △ABC.Now,let’susethefactthatthenine-pointcircleisthecirclepassingthroughthemidpointsofthesides,thefeetoftheperpendicularsfromthevertices totheoppositesides,andthemidpointsofthesegmentsjoiningtheverticestotheorthocenter.TheconfigurationimpliesthatthearcsHJandF Gcorrespond totheanglessubtendedbythechordsconnectingthesepoints.Aftersomeanglechasing,wecanfindthatthearcsHJandF Garerelatedtotheanglesofthe triangle△DEFandthepropertiesofthenine-pointcircle.Bysymmetryandpropertiesofthenine-pointcircle,wehave:HJcorrespondstothearcbetween theintersectionsofAEandAFwiththenine-pointcircle.TosimplifythefinalexpressionDE + 2· HJ + 3· F G,let’susetheknownmeasures.Wehave alreadyestablishedthat:DE = 72 ◦ .NowweneedtodetermineHJandF G.Bysymmetryandthepropertiesofthenine-pointcircleandtheconfiguration ofthetriangle,wecanusethefactthatthearcsHJandF Garesegmentsofthecirclecorrespondingtotheanglesattheverticesof△ABC.Afteranalyzing theconfigurationandusingthefactthatthesumofthearcsaroundthecircleis360 ◦ ,wecandeterminethat:HJcorrespondstothearcsubtendedbyanangle relatedto∠BAC = 84 ◦ and∠ACB = 36 ◦ ,andsimilarlyforF G.Bythepropertiesofthecircleandtheconfiguration,wefindthat:HJ = DE = 72 ◦ ,and F G = EF = 168 ◦ /2 = a segment related to the configuration.Aftersimplifyingtheconfiguration,wecanusethefactthatthearcsHJandF Gareactually segmentsthatadduptotheremainingpartsofthecircle.Let’scalculatetheexpressionDE + 2· HJ + 3· F Gstepbystep.Sincetheprobleminvolvesthesum ofthearcsonthenine-pointcircleandthegivenconfiguration,let’susethesymmetryandpropertiesoftheangles.Bythepropertiesofthenine-pointcircleand theconfigurationofthetriangle:DE = 72 ◦ ,andIfwedenotethearcscorrespondingtotheconfiguration:LetHJ = x,andF G = y.Bythepropertiesof thecircle,thesumofthearcsaroundthecircleis360 ◦ .Therefore,thesumofthearcsbetweenthepointsD, E, F, G, H,andJshouldaddupto360 ◦ .After calculatingandusingthepropertiesoftheanglesandthecyclicnatureofthecircle,wecandeterminethatthecorrectmeasuresaresuchthattheexpressionsimplifies toaknownconfiguration.Bythegivenconfigurationandtheproblem’ssetup,wehave:DE = 72 ◦ ,HJ = 60 ◦ ,andF G = 36 ◦ .Nowweplugthesevalues intotheexpression:DE + 2· HJ + 3· F G.Usingthevalues:DE = 72 ◦ ,HJ = 36 ◦ ,andF G = 24 ◦ .However,let’susethecorrectconfigurationthat fitsthemathproblem.Let’sdenotethearcssuchthattheyadduptothetotalconfiguration.Let’susetheknownconfigurationthatthesumofthearcsaroundthe circleis360 ◦ andthepropertiesoftheanglesofthetriangle.Aftersimplifyingtheconfigurationandusingthecorrectvalues:DE = 72 ◦ ,HJ = 60 ◦ ,and F G = 48 ◦ .Nowcalculate:DE + 2· HJ + 3· F G = 72 + 2· 60 + 3· 48.Let’ssimplifythisstep-by-step:72 + 2· 60 = 72 + 120 = 192.Thenadd theterm3· F G = 3· 48 = 144.Now,192 + 144 = 336.Thustheexpressionevaluatesto:DE + 2· HJ + 3· F G = 336.Answer:336 22 One-Token Verification for Reasoning Correctness Estimation Example 6 (AIME25, Problem 19): DAPO-Qwen-32B-OTV Suppose△ABChas angles∠BAC = 84 ◦ ,∠ABC = 60 ◦ , and∠ACB = 36 circ . LetD,E, andFbe the midpoints of sides BC,AC, andAB, respectively. The circumcircle of△DEFintersectsBD,AE, andAFat pointsG,H, andJ, respectively. The pointsG,D,E,H,J, andFdivide the circumcircle of△DEFinto six minor arcs, as shown. FindDE +2·HJ +3·FG, where the arcs are measured in degrees. (Correct Answer: 336) A BCD EF G H J Toapproachthismathproblem,let’sbreakitdownstep-by-step.Step1:Identifythekeypropertiesandrelationshipswithin△ABC.Giventheanglesof△ABC:- ∠BAC = 84 ◦ ,-∠ABC = 60 ◦ ,and-∠ACB = 36 ◦ .SinceD,E,andFarethemidpointsofsidesBC,AC,andABrespectively,△DEFisthemedial triangleof△ABC.Therefore,△DEFissimilarto△ABCandeachsideof△DEFisparalleltothecorrespondingsideof△ABC.Thetriangle△DEFis similarto△ABC,andthustheanglesof△DEFarethesameasthoseof△ABC.Specifically:-∠EDF = ∠BAC = 84 ◦ ,-∠DEF = ∠ABC = 60 ◦ , and-∠DF E = ∠ACB = 36 ◦ .Thecircumcircleof△DEFisthenine-pointcircleof△ABC,whichpassesthroughthemidpointsofthesides,thefeet ofthealtitudes,andthemidpointsofthesegmentsfromtheverticestotheorthocenter.Now,weneedtofindthearcsDE,HJ,andF Gonthecircumcircleof △DEF.Since△DEFissimilarto△ABC,theanglessubtendedbythearcsonthecircumcircleof△DEFcorrespondtotheanglesof△DEF.Thecentral anglescorrespondingtothearcsDE,EF,andF Daretwicetheangles∠DF E,∠DEF,and∠EDF,respectively.Therefore,themeasuresofthearcsare: -EF = 2×∠EDF = 2× 84 ◦ = 168 ◦ ,-F D = 2×∠DEF = 2× 60 ◦ = 120 ◦ ,and-DE = 2×∠DF E = 2× 36 ◦ = 72 ◦ .Nowweneedto determinethepositionsofpointsG,H,andJ.PointGistheintersectionofthecircumcircleof△DEFwithBD.SinceDisthemidpointofBC,BDisa medianof△ABC.Similarly,HistheintersectionofthecircumcirclewithAE,andJistheintersectionofthecircumcirclewithAF.SinceDisthemidpoint ofBC,BD =DC.ThelineBDpassesthroughthemidpointDandpointB.BecauseDisthemidpointofBC,BDisamedian.Similarly,AEandAF aremedians.ThekeyobservationhereisthatthepointsG,H,andJdividethecircumcircleof△DEFalongthelinesBD,AE,andAFrespectively.Let’s calculatethearcsstepbystep.SinceBDisalinepassingthroughBandD,andDisthemidpointofBC,thelineBDintersectsthecircumcircleof△DEFat pointsDandG.Similarly,AEintersectsthecircumcircleatpointsEandH,andAFintersectsthecircumcircleatpointsFandJ.Now,weneedtocalculatethe arcsDE,HJ,andF G.SinceD,E,Faremidpointsandthetriangle△DEFisthemedialtriangle,thelinesBD,AE,andAFaremediansof△ABCand perpendicularbisectorsrelativetothenine-pointcircleproperties.Thearcscorrespondingtothesegmentsbetweenthepointsofintersectioncanbederivedusingthe propertiesofcyclicquadrilateralsandtheangles.Tofindthearcs:-DEcorrespondstothearcbetweenpointsDandEonthecircumcircleof△DEF.Fromour earliercalculation,wehaveDE = 72 ◦ .Next,weneedtofindthearcHJ.PointsHandJareintersectionsofthecircumcirclewiththemediansAEandAF respectively.SinceAEandAFaremedians,andconsideringthesymmetryandpropertiesofthemedialtriangleandthenine-pointcircle,thearcsHJandF Gcan becalculatedbasedontheanglessubtendedbythesearcs.Tosimplify,weneedtousethefactthatthepointsD,E,F,G,H,andJdividethecircleintosegments thatcorrespondtotheanglesof△DEFandtheintersectionswiththemedians.Let’scalculateHJandF G.SinceHliesonAEandJliesonAF,thearcsHJ andF Gcorrespondtothesegmentscreatedbytheintersectionsofthemedianswiththecircumcircle.Giventheconfiguration,thearcsHJandF Gcanberelated totheanglessubtendedbythesegmentsofthetriangle.TofindHJ,notethatAEandAFintersectthecircleatpointsHandJ,respectively,creatingarcsthat arerelatedtotheangles∠BAC,∠ABC,and∠ACB.SinceAEisthemedianfromAtoE(themidpointofBC),andsimilarlyforAF,thearcsHJand F Gwilleachsubtendanglesrelatedtotheanglesofthetriangle△ABC.Bysymmetryandpropertiesofthenine-pointcircle,wehavethatthearcsHJand F Gareeachhalfthemeasureofthearcscorrespondingtotheanglesofthetriangle△DEF.TofindthearcsHJandF G,notethatthesearcscorrespondtothe segmentscreatedbythemediansintersectingthecircle.Sincetheconfigurationdividesthecircleintosegmentscorrespondingtotheangles:HJandF Gwilleach behalfthedifferencebetweenthearcscorrespondingtothesegments.Bysymmetry,wecanconcludethat:HJ = F G = 1 2 (the corresponding arc measures). TofindthemeasureofHJandF G,weusethefactthatthetotalcirclesumsto360 ◦ ,andthearcsDE,EF,andF Dsumuptothefullcircle.Sincethe pointsG,H,andJdividethecircleintosegmentscorrespondingtotheintersectionswiththemedians,thearcsHJandF Geachcorrespondtohalfthearcs ofthesegments.Let’scalculatethespecificvalues:SinceAEisthemedianfromAtothemidpointEofBCandAFisthemedianfromAtothemidpoint FofBC,thearcsHJandF GcanbederivedfromtheanglesatA.Thus,HJ = F G = 1 2 (180 ◦ −∠BAC) = 1 2 (180 ◦ − 84 ◦ ) = 1 2 × 96 ◦ = 48 ◦ . Therefore,HJ = 48 ◦ andF G = 48 ◦ .NowwecancalculatetherequiredexpressionDE + 2· HJ + 3· F G.Plugginginthevalueswehave:-DE = 72 ◦ , -HJ = 48 ◦ ,and-F G = 48 ◦ .Sotheexpressionbecomes:DE + 2· HJ + 3· F G = 72 ◦ + 2· 48 ◦ + 3· 48 ◦ .Nowlet’ssimplifystepbystep:First, calculate2· HJ = 2· 48 ◦ = 96 ◦ .Next,calculate3· F G = 3· 48 ◦ = 144 ◦ .Nowplugthesevaluesbackintotheexpression:72 ◦ + 96 ◦ + 144 ◦ .Add themtogetherstepbystep:72 + 96 = 168,andthenadd144totheresult:168 + 144 = 312.Therefore,thefinalvalueis312 ◦ .Answer:312 23