Update Automated MNLP evaluation report (2026-06-03)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-TopHaylin/math_model`](https://huggingface.co/cs-552-2026-TopHaylin/math_model)
 - **Owner(s):** group **TopHaylin**
-- **Generated at:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-03T08:04:32+00:00 (UTC)
 - **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
 
 _This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3300 | ok |
+| Math | 0.3200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,9 +24,9 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{7}{4}`
-- **overall** (5/8 completions correct)
-- **extracted** (✓): `\frac{7}{4}`
+- **reference**: `$\frac{a}{2}$`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `\frac{a}{2}`
 - **completion**:
 
     ```text
@@ -34,31 +34,39 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     </think>
     
-    The slope of the line is the ratio of the $y$-component to the $x$-component of the direction vector, which is $\frac{b}{-7}$. This slope is also equal to the slope of the line between points $A$ and $B$. 
+    1. **Identify Key Points and Properties**:
+        - Given: Isosceles triangle $XYZ$ with $XY = XZ$, $P$ is the circumcenter, $Q$ is the midpoint of $YZ$, and $R$ is the reflection of $Q$ across $XZ$.
+        - $S$ is such that $XQYS$ forms a rectangle.
+        - $\angle YXZ = a$.
     
-    The slope of the line between $A$ and $B$ is given by:
-    \[
-    \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{-1 - (-5)} = \frac{-1}{4}
-    \]
+    2. **Properties of the Circumcenter $P$**:
+        - Since $P$ is the circumcenter of $\triangle XYZ$, it is equidistant from all vertices of the triangle.
+        - Therefore, $PX = PY = PZ$.
     
-    Setting the slope from the direction vector equal to the slope from the line between $A$ and $B$ gives:
-    \[
-    \frac{b}{-7} = \frac{-1}{4}
-    \]
+    3. **Reflection and Symmetry**:
+        - Reflecting $Q$ across $XZ$ to get $R$ implies that $Q$ and $R$ are symmetric with respect to $XZ$.
+        - Since $Q$ is the midpoint of $YZ$, and $R$ is the reflection of $Q$ across $XZ$, $R$ is symmetric to $Q$ about $XZ$.
+    
+    4. **Rectangle $XQYS$**:
+        - $XQYS$ forms a rectangle, implying that $QY = XS$ and $XY = QS$.
+        - Given $XY = XZ$, this symmetry will also hold for $QS$ and $XY$.
+    
+    5. **Calculate $\angle PRS$**:
+        - Since $R$ is symmetric to $Q$ about $XZ$, and $XZ$ is one of the equal sides of the isosceles triangle, $\angle PRS$ will be equal to $\angle QYS$.
+        - $\angle QYS$ is half of $\angle YXZ$ because $Q$ and $S$ are symmetric about $XY$.
+        - Therefore, $\angle PRS = \frac{a}{2}$.
     
-    Solving for $b$, we multiply both sides by $-7$:
+    ### Conclusion:
     \[
-    b = \frac{-1}{4} \times -7 = \frac{7}{4}
+    \boxed{\frac{a}{2}}
     \]
-    
-    Thus, the value of $b$ is $\boxed{\frac{7}{4}}$.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `756`
+- **reference**: `104`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `6864`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
@@ -66,42 +74,34 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     </think>
     
-    1. **Identify the centers of the spheres**:
-       Let the centers of the spheres with radii $11$, $13$, and $19$ be denoted by $A$, $B$, and $C$ respectively. 
-    
-    2. **Calculate the distances between the centers**:
-       - The distance $AB$ can be found using the Pythagorean theorem, since the spheres are externally tangent. Thus, 
-         \[
-         AB = \sqrt{11^2 + 13^2} = \sqrt{121 + 169}
-    
-    ... [truncated 495 chars] ...
+    To find the distance from point $P$ to the faces of tetrahedron $WXYZ$, we use the formula for the volume of a tetrahedron:
+    \[
+    V = \frac{1}{3} \times \text{Area of one face} \times \text{Height}.
+    \]
+    Let the common distance from $P$ to each face be $d$. Since $P$ is equidistant from all four faces, we can express the volume $V$ as:
+    \[
+    V = \frac{1}{3} \times \text{Total area of al
     
-    $ are the centers of the circular cross-sections. This relationship is derived from the fact that the plane cuts each sphere into two equal parts, hence the distances from the centers to the plane are equal.
+    ... [truncated 823 chars] ...
     
-    4. **Use the given distance $XY^2 = 560$ to find $XZ^2$**:
-       - By the symmetry and the fact that the centers $A$, $B$, and $C$ are on one side of the plane, the distances $XY$ and $XZ$ can be related through the distances $AB$, $BC$, and $CA$.
-       - We know from the problem statement that the distances $AB$, $BC$, and $CA$ are $\sqrt{290}$, $\sqrt{530}$, and $\sqrt{482}$ respectively.
-       - Using the given $XY^2 = 560$, we calculate $XZ^2$ as follows:
-         \[
-         XZ^2 = XY^2 + 2 \cdot AB \cdot BC \cdot \cos(\angle ABC) + BC^2.
-         \]
-       - Since $\cos(\angle ABC)$ is the cosine of the angle between $AB$ and $BC$, and given the symmetry of the problem, we can simplify this to:
-         \[
-         XZ^2 = 560 + 2 \cdot 11 \cdot 13 \cdot 19 + 530.
-         \]
+    t{41}+\sqrt{80}+\sqrt{89}}{2}-\sqrt{41}\right)\left(\frac{\sqrt{41}+\sqrt{80}+\sqrt{89}}{2}-\sqrt{80}\right)\left(\frac{\sqrt{41}+\sqrt{80}+\sqrt{89}}{2}-\sqrt{89}\right)}.
+    \]
+    - For face $XYZ$ with sides $\sqrt{80}$, $\sqrt{89}$, and $\sqrt{89}$, the area is:
+    \[
+    A_4 = \sqrt{\left(\frac{\sqrt{80}+\sqrt{89}+\sqrt{89}}{2}\right)\left(\frac{\sqrt{80}+\sqrt{89}+\sqrt{89}}{2}-\sqrt{80}\right)\left(\frac{\sqrt{80}+\sqrt{89}+\sqrt{89}}{2}-\sqrt{89}\right)\left(\frac{\sqrt{80}+\sqrt{89}+\sqrt{89}}{2}-\sqrt{89}\right)}.
+    \]
     
-    5. **Simplify the expression**:
-       - Calculate the product:
-         \[
-         2 \cdot 11 \cdot 13 \cdot 19 = 2 \cdot 2887 = 5774.
-         \]
-       - Add the values:
-         \[
-         560 + 5774 + 530 = 6864.
-         \]
+    Assuming all faces are congruent, we can simplify the calculation:
+    \[
+    \text{Total area} = 4 \times A_1.
+    \]
     
-    ### Conclusion:
+    Using the volume formula, solve for $d$:
     \[
-    \boxed{6864}
+    d = \frac{3V}{4A_1}.
     \]
+    
+    Finally, express $d$ in the form $\frac{a\sqrt{b}}{c}$ and calculate $a+b+c$. This requires detailed computation of the areas and volumes of the faces, as well as the common distance $d$.
+    
+    Conclusion: The sum $a+b+c$ is found by calculating the common distance $d$ and simplifying it to the form $\frac{a\sqrt{b}}{c}$, where $a$, $b$, and $c$ are integers, and $a$ and $c$ are relatively prime. The specific values depend on the detailed calculations of the areas and volumes of the tetrahedron's faces.
     ```