Datasets:
problem_id large_stringlengths 24 24 | source large_stringclasses 41
values | year int64 1.96k 2.02k | problem large_stringlengths 36 2.65k | reference_solution large_stringlengths 60 11.9k | is_aime bool 2
classes | trace large_stringlengths 2.37k 158k | solution large_stringlengths 1.09k 6.61k | accepted_at_iteration int64 1 25 | final_hint_level int64 0 4 | n_interventions float64 2 29 ⌀ | inserted_texts listlengths 2 29 ⌀ | hint_span_starts listlengths 0 60 | hint_span_ends listlengths 0 60 | flag_shortened bool 1
class | flag_cot_leak bool 1
class | flag_wrong_problem bool 1
class | flag_excessive_hinting bool 2
classes |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
6941c012838f2bbb09132888 | AIME | 1,983 | Let $x$, $y$, and $z$ all exceed 1 and let $w$ be a positive number such that \[\log_x w = 24,\quad \log_y w = 40 \quad\text{and}\quad \log_{xyz} w = 12.\]Find $\log_z w$. | sol: $$\log_x{w}=24\Longrightarrow \log_w{x}=\frac{1}{24}$$ $$\log_y{w}=40\Longrightarrow \log_w{y}=\frac{1}{40}$$ $$\log_w{x}+\log_w{y}=\log_w{xy}=\frac{1}{15}$$ $$\log_{xyz}{w}=12\Longrightarrow \log_{w}{xyz}=\frac{1}{12}$$ $$\log_{w}{xyz}-\log_{w}{xy}=\log_{w}{z}=\frac{1}{60}$$ $$\log_{w}{z}=\frac{1}{60}\Longrightar... | true | Okay, let's try to solve this problem. So we have three variables x, y, z all greater than 1, and a positive number w. We're given the logarithms of w with bases x, y, and xyz, and we need to find log base z of w. Hmm, logarithms with different bases. I remember that logarithms can be converted using the change of base... | To solve the problem, we are given the following logarithmic relationships:
$$
\log_x w = 24, \quad \log_y w = 40, \quad \log_{xyz} w = 12,
$$
and we are to find $\log_z w$.
---
### Step 1: Use the Change of Base Formula
We recall that for any positive real numbers $a$, $b$, and $c$ (with $b \neq 1$ and $c \neq 1$... | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb09132889 | AIME | 1,983 | Let $f(x) = |x - p| + |x - 15| + |x - p - 15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \le x \le 15$. | Solution It is best to get rid of the absolute value first. Under the given circumstances, we notice that $|x-p|=x-p$ , $|x-15|=15-x$ , and $|x-p-15|=15+p-x$ . Adding these together, we find that the sum is equal to $30-x$ , of which the minimum value is attained when $x=15$ . The answer is thus $15$ . | true | Okay, let's try to solve this problem step by step. The function given is \( f(x) = |x - p| + |x - 15| + |x - p - 15| \), where \( 0 < p < 15 \), and we need to find its minimum value on the interval \( p \leq x \leq 15 \).
First, I remember that absolute value functions are piecewise linear, and their minima often oc... | To determine the minimum value of the function
$$
f(x) = |x - p| + |x - 15| + |x - p - 15|
$$
on the interval $ p \leq x \leq 15 $, where $ 0 < p < 15 $, we begin by analyzing the behavior of each absolute value term within this interval.
---
### Step 1: Identify Critical Points and Their Behavior
The critical poi... | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb0913288a | AIME | 1,983 | What is the product of the real roots of the equation \[x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}\,\,?\] | Let $y=x^2+18x+30$ , so that the given equation rewrites as $y=2\sqrt{y+15}$ . Squaring, $$y^2=2^2(y+15)=4y+60\implies y^2-4y-60=(y+6)(y-10)=0,$$ which has solutions $y=-6$ and $y=10$ . However, we require that $y\geq0$ in order for $y=2\sqrt{y+15}$ , so we eliminate $y=-6$ to obtain $y=x^2+18x+30=10$ , or $x^2+18x+20=... | true | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the product of the real roots of this equation: x squared plus 18x plus 30 equals 2 times the square root of (x squared plus 18x plus 45). Hmm, let's see. It looks li... | To solve the equation
$$
x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45},
$$
we begin by recognizing that both sides involve the same quadratic expression. Let us define:
$$
t = x^2 + 18x + 45.
$$
This substitution simplifies the right-hand side (RHS), which becomes $2\sqrt{t}$, and the left-hand side (LHS) becomes:
$$
x... | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb0913288b | AIME | 1,983 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy]
size(150); defaultp... | <quote> A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy]
size(150); ... | true | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have a machine-shop cutting tool that's a notched circle, meaning it's a circle with some notches, but in this case, the shape is given by points A, B, C on or related to the cir... | To solve the problem, we are given a notched circle with radius $\sqrt{50}$ cm, and a right triangle $ABC$ where:
- $AB = 6$ cm,
- $BC = 2$ cm,
- $\angle ABC = 90^\circ$.
We are to find the **square of the distance from point $B$ to the center of the circle**, denoted as $BO^2$, where $O$ is the center of the circle.... | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb0913288c | AIME | 1,983 | Suppose that the sum of the squares of two complex numbers $x$ and $y$ is 7 and the sum of the cubes is 10. What is the largest real value that $x + y$ can have? | Let $s=x+y$ and $p=xy$ , so that $x^2+y^2=s^2-2p=7$ and $x^3+y^3=s(s^2-3p)=10$ , or, equivalently, $s^2-3p=\tfrac{10}{s}$ . Seeking to create elimination in $p$ , we multiply the two equations by $3$ and $2$ in that order to obtain $3s^2-6p=21$ and $2s^2-6p=\tfrac{20}{s}$ , which after subtracting yields $s^2=21-\tfrac... | true | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so the problem says we have two complex numbers x and y, sum of their squares is 7, sum of their cubes is 10, and we need to find the largest real value of x + y. Hmm, complex numbers,... | To solve the problem, we are given two complex numbers $ x $ and $ y $ such that:
$$
x^2 + y^2 = 7 \quad \text{and} \quad x^3 + y^3 = 10,
$$
and we are to find the **largest real value** of $ x + y $.
---
### Step 1: Use Symmetric Sums
Let us define:
- $ s = x + y $
- $ p = xy $
Using standard algebraic identitie... | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb0913288d | AIME | 1,983 | Let $a_n = 6^n + 8^n$. Determine the remainder on dividing $a_{83}$ by 49. | Click to reveal hidden text The divisor can be rewritten as $7^2$ . Rewrite the dividend as: $(7-1)^{83}+(7+1)^{83}$ When we expand this out using the Binomial Theorem, we'll get the sum of a bunch of different powers of $7$ multiplied by the corresponding coefficient. We are dividing by $7^2$ , so any term with $7$ ra... | true | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so I need to find the remainder when a₈₃ = 6⁸³ + 8⁸³ is divided by 49. That means compute (6⁸³ + 8⁸³) mod 49. Hmm, modular arithmetic with exponents... I remember something about Euler... | To determine the remainder when $ a_{83} = 6^{83} + 8^{83} $ is divided by 49, we approach the problem using modular arithmetic and binomial expansion. The key insight lies in recognizing that both 6 and 8 can be expressed as $ 7 - 1 $ and $ 7 + 1 $, respectively. This allows us to simplify the expression using the bin... | 2 | 0 | 3 | [
"I need to keep thinking and dive deeper into the logic of the binomial expansion. I am NOT ready to conclude or write the final answer yet. Wait, I should re-derive the exact terms that contribute to the sum modulo $49$ from the binomial expansion of $(7-1)^n + (7+1)^n$. I need to pay close attention to the consta... | [
18296,
18669,
19131
] | [
18667,
19129,
19527
] | false | false | false | false |
6941c012838f2bbb0913288e | AIME | 1,983 | Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices of three being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a f... | Solution: In total there will be $\dbinom{25}{3} = \dfrac{25\cdot24\cdot23}{3\cdot2\cdot1}=25\cdot23\cdot4$ ways to choose $3$ knights of course. Next, lets do casework on the amount of knights next to each other. First case, we have $2$ knights next to each other. We have $25$ ways to choose the $2$ adjacent knights a... | true | This is a complex or challenging question, and it is difficult to provide a direct and correct answer. I need to think about it.
Well, so we have 25 knights sitting around a round table, and we need to find the probability that when we pick three, at least two are adjacent. Probability is usually favorable outcomes ov... | To solve the problem, we are given 25 knights seated around a **circular** table. We are to compute the probability $ P $ that **at least two** of three randomly chosen knights are sitting **next to each other**, and then express this probability as a reduced fraction to find the sum of its numerator and denominator.
... | 2 | 0 | 4 | [
"I need to keep thinking and dive deeper into the logic of counting favorable outcomes. I am NOT ready to conclude or write the final answer yet. So, the immediate goal is to make sure my counting methods are absolutely rigorous and fully justified. I should really re-derive the combinatorial formula for counting n... | [
28458,
29018,
29394,
29588
] | [
29016,
29392,
29586,
30012
] | false | false | false | false |
6941c012838f2bbb0913288f | AIME | 1,983 | What is the largest 2-digit prime factor of the integer $n = \binom{200}{100}$? | "<quote> What is the largest 2-digit prime factor of the integer $n = \\binom{200}{100}$ ? </quote> (...TRUNCATED) | true | "This is a complex or challenging question, and it is difficult to provide a direct and correct answ(...TRUNCATED) | "To determine the **largest 2-digit prime factor** of the binomial coefficient $ \\binom{200}{100} $(...TRUNCATED) | 1 | 0 | null | null | [] | [] | false | false | false | false |
6941c012838f2bbb09132890 | AIME | 1,983 | Find the minimum value of
\[\frac{9x^2 \sin^2 x + 4}{x \sin x}\]
for $0 < x < \pi$. | "$\\frac{9x^2\\sin^2x+4}{x\\sin x}=9x\\sin x+\\frac4{x\\sin x}\\ge12$ by AM-GM. Equality is at $x\\s(...TRUNCATED) | true | "This is a complex or challenging question, and it is difficult to provide a direct and correct answ(...TRUNCATED) | "To find the minimum value of the expression \n$$\n\\frac{9x^2 \\sin^2 x + 4}{x \\sin x}\n$$\nfor $(...TRUNCATED) | 2 | 0 | null | null | [
9746
] | [
10426
] | false | false | false | false |
6941c012838f2bbb09132891 | AIME | 1,983 | "The numbers 1447, 1005, and 1231 have something in common: each is a four-digit number beginning wi(...TRUNCATED) | "Solution Suppose the two identical digits are both one. Since the thousands digits must be one, the(...TRUNCATED) | true | "Okay, let's try to solve this problem. The question is asking for the number of four-digit numbers (...TRUNCATED) | "To determine how many four-digit numbers beginning with 1 have **exactly two identical digits**, we(...TRUNCATED) | 1 | 0 | null | null | [] | [] | false | false | false | false |
SPLICE traces
Training data for the paper Splicing Hints into Reasoning Traces: Learning from Demonstrator Failures (arXiv:2610.13996). SPLICE is a data-generation workflow in which a tutor model, holding a reference solution, watches a reasoning model (the demonstrator) write a trace, cuts the trace back to its last verified step when progress stalls, splices in a hint, and lets generation resume. The five configs are the five finetuning datasets of the paper, one per trained student.
| Config | Traces | What it is |
|---|---|---|
splice-nf |
2,130 | every trace the SPLICE loop accepted (no filter) |
splice |
2,009 | splice-nf minus the traces flagged as compression artifacts, reasoning leaked into the solution, or answering a different problem (121 traces) |
splice-hc |
1,790 | splice minus 219 traces with maximum-level hinting and no verified progress |
untutored-correct |
768 | the problems the demonstrator solved unaided (no hint was ever spliced) |
untutored |
2,130 | the demonstrator's own traces with no tutor, the distillation baseline |
Demonstrator: Qwen3-30B-A3B-Thinking. Tutor: Gemini 2.5 Pro, with cheaper family members for lighter stages. Student trained on each config: Qwen3-4B-Instruct-2507.
Problems
2,130 problems with reference solutions from the corresponding Art of Problem Solving threads: AIME 1983–2023 (778) and proof problems (1,352) from the IMO 1959–2024, the IMO Shortlist 1966–2023 and Longlists, team selection tests, Baltic Way, Tuymaada, USAMO and the Balkan MO Shortlist, none later than 2024. The set is disjoint from the paper's evaluation exams (AIME 2024–2026; Canadian, Iranian and IMO olympiads 2024–2026).
Schema
All configs share the first block. The SPLICE configs add the generation record, which is what the paper's failure-mode filters are computed from.
| Column | Type | Description |
|---|---|---|
problem_id |
string | stable id, shared with the other SPLICE datasets |
source |
string | contest name, e.g. IMO Shortlist |
year |
int | contest year |
problem |
string | problem statement (LaTeX) |
reference_solution |
string | the AoPS solution the tutor was given |
is_aime |
bool | answer-based problem |
trace |
string | reasoning trace, the finetuning target before the end-of-reasoning marker |
solution |
string | final written solution after the marker |
| SPLICE configs only | ||
accepted_at_iteration |
int | loop iteration at which the trace was accepted |
final_hint_level |
int | hint level (0–4) when the trace was accepted |
n_interventions |
int | number of hints spliced into the trace; null where the per-hint record was not stored |
inserted_texts |
list[string] | each hint as restyled and inserted, in order; null where not stored |
hint_span_starts |
list[int] | character offsets into trace where detected hint passages begin |
hint_span_ends |
list[int] | matching end offsets |
flag_shortened |
bool | compression artifact from the Shorten stage (97 traces) |
flag_cot_leak |
bool | reasoning leaked into the final solution (16) |
flag_wrong_problem |
bool | write-up answers a different problem (10) |
flag_excessive_hinting |
bool | maximum-level hinting without verified progress (219) |
splice is splice-nf with the first three flags false; splice-hc adds
the fourth. The filters are reproducible from splice-nf alone. The
per-hint texts exist for 1,277 of the 1,362 hinted traces; the hint spans
come from the paper's hint-passage detector and cover 1,428 traces.
Training format
Each student was finetuned on one conversation per trace:
system: training_system_prompt.md (this repo)
user: problem
assistant: <think>\n{trace}\n</think>\n\n{solution}
with loss on the assistant turn, under the recipe in Appendix J of the paper.
Licensing
Traces, hints, and generation records are released under CC BY 4.0.
Problem statements and reference solutions are reproduced from Art of
Problem Solving forum threads for research use; each row's source and
year identify the contest.
Citation
@article{mahdavi2026splice,
title = {Splicing Hints into Reasoning Traces: Learning from Demonstrator Failures},
author = {Mahdavi, Hamed and Mahdavinia, Pouria and Farhadi, Majid and Honavar, Vasant},
journal = {arXiv preprint arXiv:2610.13996},
year = {2026}
}
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