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values | fineproofs_reward_mean float64 0.75 0.84 | fineproofs_reward_std float64 0.25 0.32 | fineproofs_num_rewards int64 128 128 | qwen35_9b_full_credit_rate float64 0 0.5 | qwen35_9b_mean_grade float64 0.55 0.77 | full_credit_delta_vs_provided float64 -0.84 -0.25 | problem_id stringclasses 3
values | rollout_index int64 0 7 | gpt_oss_grade float64 0.14 1 | gpt_oss_points float64 1 7 | gpt_oss_max_points float64 7 7 | rollout_status stringclasses 1
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values | proof stringlengths 3.2k 8.92k | grade_feedback stringlengths 109 237 | rubric stringclasses 3
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 0 | 1 | 7 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | Let
\[
A,B,C,D,E
\]
be the vertices of a regular pentagon taken in the order \(A\to B\to C\to D\to E\to A\)
(counter‑clockwise).
All its sides have the same length; denote this common length by
\[
AB=BC=CD=DE=EA=s .
\]
--------------------------------------------------------------------
### 1. Geometry of a ... | The solution correctly sets up coordinates, finds F, computes AF explicitly, and algebraically verifies AE+AF=BE, meeting all Chain C checkpoints. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 1 | 1 | 7 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | Let
\[
A,B,C,D,E
\]
be the vertices of a regular pentagon taken in this order clockwise.
Denote the common side–length by
\[
|AB|=|BC|=|CD|=|DE|=|EA|=:s .
\]
--------------------------------------------------------------------
### 1. Length of a diagonal
The diagonal \(BE\) joins two vertices that are separ... | The solution fully satisfies all checkpoints of Chain C with correct coordinate setup, computation of F, AF, and verification of AE+AF=BE. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 2 | 1 | 7 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | Let
\[
ABCD E
\]
be a regular pentagon taken in the order \(A\!-\!B\!-\!C\!-\!D\!-\!E\) around the figure.
Denote the common side length by
\[
s=|AB|=|BC|=|CD|=|DE|=|EA|.
\]
The diagonal \(BE\) is therefore the longer diagonal of a regular pentagon; it is well‑known (and easy to obtain from the law of sines in ... | The solution correctly identifies all required angles, expresses BE, applies the Sine Rule, simplifies trig values, and verifies the equality, satisfying all Chain B checkpoints. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 3 | 0.142857 | 1 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | \[
\text{In the regular pentagon }A B C D E \;(AB=BC=CD=DE=EA)
\qquad\qquad (\ast )
\]
let
* \(F\) be the point where the line through \(C\) perpendicular to \(CD\) meets the line
\(AB\);
* \(s=AB=BC=CD=DE=EA\) (the side–length);
* \(d=AC=BD=CE= \) the length of a diagonal.
The statement to be proved is
\[
... | The solution correctly states the diagonal length in terms of the side (using the golden ratio) but does not satisfy the other requested checkpoints (no auxiliary point, incomplete angle identification, no sine rule, no coordinate work). | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 4 | 0.428571 | 3 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | \[
\text{Let }ABCD E\text{ be a regular pentagon, }|AB|=|BC|=|CD|=|DE|=|EA|=:s .
\]
Denote by \(F\) the intersection of the line \(AB\) with the line through
\(C\) which is perpendicular to the side \(CD\).
We have to prove
\[
AE+AF=BE .\tag{1}
\]
-------------------------------------------------------------------... | Earned points for stating diagonal length, setting up the final equation and reducing it, but missed angle identification, sine rule, and had an internal inconsistency in the value of AF, resulting in a deduction. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 5 | 0.857143 | 6 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | **Solution**
--------------------------------------------------------------------
### 1. Notation
Let
* \(AB=BC=CD=DE=EA=s\) – the common side of the regular pentagon,
* \(\varphi=\dfrac{1+\sqrt5}{2}\) – the golden ratio.
In a regular pentagon the length of every diagonal equals \(\varphi s\);
for example
... | Earned most points for correct coordinate setup, finding F, calculating AF and verifying the identity, but lost a point for the unjustified step converting the obtained intersection parameter to the correct segment proportion. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 6 | 1 | 7 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | \[
\text{Let }ABCDE\text{ be a regular pentagon, the vertices being taken in this order.}
\]
--------------------------------------------------------------------
### 1. Notation
Denote the common side‑length of the pentagon by
\[
s=AB=BC=CD=DE=EA .
\]
The diagonal \(BE\) is well known to satisfy
\[
\frac{B... | The solution correctly follows Chain B, identifying angles, using the sine rule, computing lengths, and establishing the final equality with proper justification. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | representative | 0.75 | 0.316152 | 128 | 0.5 | 0.767857 | -0.25 | 0fc95550361f0675 | 7 | 0.714286 | 5 | 7 | judged | In the regular pentagon $ABCDE$ , the perpendicular at $C$ to $CD$ meets $AB$ at $F$ . Prove that $AE+AF=BE$ .
*Proposed by Alireza Cheraghi* | \[
\text{In a regular pentagon }ABCDE\text{ let }AB=BC=CD=DE=EA=s .
\]
Denote by \(F\) the point where the line through \(C\) which is
perpendicular to \(CD\) meets the line \(AB\).
We shall prove
\[
AE+AF=BE .
\tag{1}
\]
--------------------------------------------------------------------
### 1. Angles in \(\tr... | Correctly applied sine rule and algebraic reduction, but failed to properly identify angle CFB=54° and introduced an incorrect right‑angle claim, losing points on the angle identification checkpoint. | Checkpoints (7 pts total)
**Chain A: Auxiliary Point / Synthetic (Official Solution)**
* **(1 pt)** Correctly states the length of diagonal $BE$ in terms of side $x$ (e.g., $BE = 2x \sin 54^{\circ}$ or $BE = \phi x$) **OR** identifies relevant angles (e.g., $\angle CFB = 54^{\circ}$).
* **(1 pt)** Defines auxiliar... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 0 | 1 | 7 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
To prove that the line $CI$ is tangent to the circumcircle of triangle $BMN$, we use coordinate geometry.
### 1. Coordinate System Setup
Let us set up a Cartesian coordinate system with the right angle $A$ at the origin $(0,0)$.
Let the vertices of the triangle $ABC$ be:
$$ A = (0, 0) $$
$$ B = (c, 0) $$
$$ C = (0,... | Student gave a correct coordinate setup, found the circumcircle of BMN analytically, and proved tangency via the distance‑to‑line equals radius condition, satisfying all checkpoints. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 1 | 0.285714 | 2 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
Let us set up a Cartesian coordinate system. Since $\triangle ABC$ is a right triangle with the right angle at $A$, we place $A$ at the origin $(0,0)$. Let the vertices $B$ and $C$ lie on the positive $x$ and $y$ axes, respectively.
Thus, we can denote the coordinates as follows:
$$A = (0,0), \quad B = (c, 0), \quad ... | The solution provides a correct analytic verification of tangency (2 pts) but omits the required point K and cyclicity arguments (0 pts for the other checkpoints). | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 2 | 0.571429 | 4 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
Let $A$ be the vertex of the right angle at the origin $(0,0)$ of a Cartesian coordinate system. Let the vertices of the right triangle be $A(0,0)$, $B(b,0)$, and $C(0,c)$ with $b, c > 0$.
The hypotenuse $BC$ has length $a = \sqrt{b^2+c^2}$.
The incenter $I$ has coordinates $(r,r)$ where $r$ is the inradius. For a ri... | Analytic setup and tangency verification are correct (2 pts each), but the student never introduced point K nor proved BNMK cyclic, losing the 3‑point section. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 3 | 1 | 7 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
**Proof:**
Let us place the right triangle $ABC$ in the Cartesian coordinate system. Since $\angle A = 90^\circ$, we can set the coordinates of the vertices as $A(0, 0)$, $B(c, 0)$, and $C(0, b)$, where $b = AC$ and $c = AB$.
The hypotenuse $BC$ has length $a = \sqrt{b^2 + c^2}$.
The incenter $I$ of a right triangl... | The solution provides a correct analytic setup, finds the circumcenter and radius of BMN, and proves CI tangent by equating distance to radius, satisfying all checkpoints. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 4 | 0.142857 | 1 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
We provide a complete proof using coordinate geometry and geometric properties.
### 1. Coordinate System Setup
Let us place the triangle $ABC$ in the Cartesian plane with vertex $A$ at the origin $(0,0)$. Since $\angle A = 90^\circ$, we can align the legs of the triangle with the axes.
Let the coordinates of the ve... | Only earned the single point for correctly noting that MN is parallel to AI; all other required elements (defining K, proving cyclicity, and establishing tangency rigorously) are missing or incorrect. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 5 | 0.2857 | 2 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
**Proof:**
We will use coordinate geometry to prove that the line $CI$ is tangent to the circumcircle of triangle $BMN$.
Let the vertices of the triangle be placed in the Cartesian coordinate plane as follows:
* Let $A$ be the origin $(0,0)$.
* Since $\angle A = 90^\circ$, we place $AB$ along the $x$-axis and $... | Student omitted any discussion of point K and cyclicity, but correctly used analytic geometry to show CI is tangent to the circumcircle of BMN. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 6 | 0.4286 | 3 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
Let us introduce a Cartesian coordinate system to solve the problem rigorously.
**1. Setting up the Coordinate System**
Let the vertex $A$ be the origin $(0,0)$. Since $\angle A = 90^\circ$, we can place the legs $AB$ and $AC$ along the positive $x$-axis and $y$-axis respectively.
Let the side lengths of the triangl... | Earned full credit for coordinate setup but gave no definition or proof of point K and cyclicity, and only partially justified the analytic tangency condition. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | rollout_contrast | 0.816964 | 0.293996 | 128 | 0.25 | 0.553573 | -0.566964 | 2afbd059c1887c52 | 7 | 0.714286 | 5 | 7 | judged | Let $I$ denote the center of the circle inscribed in the right triangle $ABC$ with right angle at the vertex $A$ . Next, denote by $M$ and $N$ the midpoints of the lines $AB$ and $BI$ . Prove that the line $CI$ is tangent to the circumscribed circle of triangle $BMN$ .
(Patrik Bak, Josef Tkadlec) |
Let us set up a Cartesian coordinate system. Let vertex $A$ be at the origin $(0,0)$. Since $\angle A = 90^\circ$, we align the legs of the triangle along the coordinate axes.
Let $AB$ lie on the $x$-axis and $AC$ lie on the $y$-axis.
The coordinates of the vertices are $A(0,0)$, $B(c,0)$, and $C(0,b)$, where $c = |A... | The solution correctly derived the circumcircle of BMN and verified tangency via distance = radius, earning full analytic points, but omitted the required identification of point K, losing the initial 2 points. | Checkpoints (7 pts total)**
* **2 pts**: Identify the point $K$ (intersection of $PM$ and $CI$, or projection of $B$ onto $CI$) and establish its key properties ($A,B,K,C$ concyclic or $BK \perp CI$).
* Award full points for citing "Iran Lemma" or "Fact 5" to state that the projection of $B$ onto $CI$ lies on ... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 0 | 0.857143 | 6 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer.**
All the required sets are the two–element sets
\[
X=\{a , a^{3}\},\qquad a\in \mathbb Z_{\ge 2}.
\]
No other set of positive integers (with at least two elements) enjoys the
property
\[
\forall\,m,n\in X,\;n>m\;\exists k\in X\;:\;n=m k^{2}. \tag{1}
\]
------------------------------------------------... | Correctly proved 1∉X, derived a₂=a₁³, and gave a convincing (though slightly informal) contradiction for a third element, earning most points. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 1 | 0.857143 | 6 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer**
\[
\boxed{\;X=\{a,\;a^{3}\}\qquad (a\in \mathbb N ,\;a\ge 1)\;}
\]
These are the only sets of positive integers containing at least two elements that satisfy
\[
\forall\,m,n\in X,\;m<n\;\Longrightarrow\;\exists k\in X\;:\; n=m\,k^{2}. \tag{1}
\]
---------------------------------------------------------... | Student derived a2 = a^3 and proved no third element, but omitted a rigorous proof that 1 cannot belong to X. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 2 | 0.857143 | 6 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer.**
All the required sets are exactly the two–element sets
\[
X=\{a,\;a^{3}\}\qquad (a\in \mathbb {N},\;a\ge 2).
\]
No set containing three or more different positive integers satisfies the
condition.
--------------------------------------------------------------------
### 1. Reformulation of the cond... | Derived a2 = a^3 and proved no third element, but failed to justify that 1 cannot belong to X, incurring a one‑point penalty. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 3 | 0.857143 | 6 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer**
\[
\boxed{\;X=\{a,\;a^{3}\}\qquad (a\in \mathbb Z_{>0})\;}
\]
These and only these sets satisfy
\[
\forall \,m,n\in X,\;n>m\;\Longrightarrow\;\exists k\in X:\;n=m\,k^{2}.
\]
--------------------------------------------------------------------
### 1. Preliminary observations
*All numbers are positi... | Derived the needed a₂ = a₁³ and proved no third element can exist, but omitted the required justification that 1 cannot belong to X. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 4 | 0.857143 | 6 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer.**
The only subsets \(X\subset \mathbb N\) with \(|X|\ge 2\) that satisfy
\[
\forall\,m,n\in X,\; n>m\;\Longrightarrow\;\exists k\in X\text{ such that }n=m k^{2}
\tag{1}
\]
are the two‑element sets
\[
\boxed{X=\{a,\;a^{3}\}}\qquad (a\in\mathbb N,\;a\ge 2).
\]
-----------------------------------------... | Correctly derived the only possible set {a,a^3} and proved no third element can exist, but omitted a justified proof that 1 cannot belong to X. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
easy | Easy | baseline_disagreement | 0.839286 | 0.24646 | 128 | 0 | 0.714232 | -0.839286 | 51f417eeaa1c4735 | 5 | 0.571 | 4 | 7 | judged | Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$ , where $n>m$ , there exists an element $k\in X$ such that $n=mk^2$ . | **Answer.**
The required sets are exactly the two‑element sets
\[
X=\{a,\;a^{3}\},
\qquad a\in \mathbb Z_{>0},\;a\ge 2 .
\]
--------------------------------------------------------------------
### 1. What the condition means
Let
\[
X\subseteq \mathbb Z_{>0},\qquad |X|\ge 2,
\]
and suppose that
> (∗) f... | Correctly obtained X={a,a^3} and proved no third element, but failed to justify that 1 cannot belong to X and used a>1 without proof, losing points on the first checkpoint and the multiplier justification. | Checkpoints (7 pts total)
* **1 pt: Proving $1 \notin X$ (or $a_1 > 1$).**
* Showing that if $1 \in X$, the condition leads to a contradiction (e.g., finding an element strictly between the first and second smallest elements).
* **2 pts: Deriving the relation $a_2 = a_1^3$.**
* **1 pt:** Establishing t... | lm-provers/FineProofs-RL | fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722 | Qwen/Qwen3.5-9B | openai/gpt-oss-120b |
End of preview. Expand in Data Studio
FineProofs Qwen3.5-9B Rollout View
A curated visualization sample from fineproofs_qwen35_9b_resp81920_gptoss120b_m32_20260722. Each selected problem contributes all eight
Qwen3.5-9B proofs and their GPT-OSS-120B rubric grades. This is a viewing aid, not an evaluation set.
Difficulty strata
FineProofs has no populated categorical difficulty field. difficulty_band is defined here from
the provided extra.reward_mean, an independent empirical success rate over 128 prior evaluations:
| Split | Provided success rate |
|---|---|
very_hard |
[0.00, 0.10) |
hard |
[0.10, 0.30) |
medium |
[0.30, 0.60) |
easy |
[0.60, 0.90) |
very_easy |
[0.90, 1.00] |
There are 15 selected problems and 120 rows. Every problem retains all eight rollouts. Selection covers a representative problem, a high within-problem grade-contrast problem, and a large baseline-vs-9B disagreement in each stratum.
Aggregate comparison
Across all 5,227 problems:
- Provided success rate: 0.399
- 9B full-credit success rate: 0.387
- Difference (9B - provided): -0.012
- Pearson correlation: 0.806
- 9B mean normalized rubric grade: 0.557
The binary comparison uses gpt_oss_grade == 1. qwen35_9b_mean_grade is reported separately because
partial-credit rubric quality is not the same quantity as success probability.
Interactive view: https://huggingface.co/spaces/asingh15/fineproofs-rollout-explorer
Source problems: lm-provers/FineProofs-RL. Generated proofs: Qwen/Qwen3.5-9B.
Grades: openai/gpt-oss-120b.
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