Update Automated MNLP evaluation report (2026-05-24)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-mnlplus/math_model`](https://huggingface.co/cs-552-2026-mnlplus/math_model)
 - **Owner(s):** group **mnlplus**
-- **Generated at:** 2026-05-23T14:37:35+00:00 (UTC)
+- **Generated at:** 2026-05-24T16:13:44+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._
@@ -31,38 +31,36 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    Given the isosceles triangle \(XYZ\) with \(XY = XZ\), we need to find the measure of \(\angle PRS\) in terms of \(\angle YXZ = a\).
-    
-    1. **Identify the properties of the triangle and the circumcenter \(P\):**
-       - Since \(P\) is the circumcenter of \(\triangle XYZ\), it is equidistant from all three vertices \(X\), \(Y\), and \(Z\).
-       - Because \(XY = XZ\), the triangle is symmetric about the altitude from \(X\) to \(YZ\).
-    
-    2. **Determine the position of point \(Q\):**
-       - \(Q\) is the midpoint of \(YZ\), so \(Q\) lies on the altitude from \(X\) to \(YZ\).
-       - Reflecting \(Q\) across \(XZ\) gives us point \(R\).
-    
-    3. **Analyze the rectangle \(XQYS\):**
-       - Since \(XQYS\) is a rectangle, \(XQ\) is perpendicular to \(YS\).
-       - This implies that \(XQ \perp YS\).
-    
-    4. **Determine the angles in the rectangle:**
-       - In a rectangle, the opposite angles are equal, and consecutive angles are supplementary.
-       - Therefore, \(\angle XQS = 90^\circ\).
-    
-    5. **Relate \(\angle XQS\) to \(\angle YXZ\):**
-       - Since \(\angle XQS = 90^\circ\), and \(\angle YXZ = a\), we have:
+    1. **Identify Key Points and Angles**:
+       - Let \(P\) be the circumcenter of the isosceles triangle \( \triangle XYZ \) with \( XY = XZ \).
+       - Let \(Q\) be the midpoint of \(YZ\).
+       - Let \(R\) be the reflection of \(Q\) over the side \(XZ\).
+       - Let \(S\) be a point such that the quadrilateral \(XQYS\) is a rectangle.
+       - Given \(\angle YXZ = a\).
+    
+    2. **Properties of the Reflection**:
+       - Since \(Q\) is the midpoint of \(YZ\), and \(R\) is the reflection of \(Q\) over \(XZ\), \(R\) lies on the extension of \(XZ\) such that \(XR = XQ\).
+    
+    3. **Constructing the Rectangle**:
+       - Since \(XQYS\) is a rectangle, \(XQ\) and \(YS\) are perpendicular to each other, and \(XQ = YS\).
+       - This implies that \( \angle XQS = 90^\circ \).
+    
+    4. **Using Symmetry and Congruence**:
+       - The triangle \( \triangle XQY \) is isosceles because \(XQ = YQ\).
+       - The angle \(\angle XQS\) is half of \(\angle YXZ\) because \(Q\) is the midpoint of \(YZ\) and \(R\) is symmetrically placed.
+    
+    5. **Calculating \(\angle PRS\)**:
+       - Since \(P\) is the circumcenter of \( \triangle XYZ \), it lies on the perpendicular bisector of \(YZ\).
+       - Therefore, \(P\) is equidistant from \(X\), \(Y\), and \(Z\).
+       - The angle \(\angle PRS\) is determined by the symmetry and the fact that \(P\) is the circumcenter:
          \[
-         \angle XQS = 90^\circ = \frac{\pi}{2} \text{ radians}
+         \angle PRS = \frac{1}{2} \angle YXZ = \frac{a}{2}
          \]
     
-    6. **Calculate \(\angle PRS\):**
-       - \(\angle PRS\) is an angle in the triangle \(PRS\).
-       - Using the properties of the circumcircle and the symmetry of the isosceles triangle, we can deduce that \(\angle PRS = \frac{a}{2}\).
-    
-    7. **Conclusion:**
-       - The measure of \(\angle PRS\) in terms of \(\angle YXZ = a\) is:
+    6. **Conclusion**:
+       - The angle \(\angle PRS\) in terms of \(a\) is:
          \[
-         \boxed{\frac{a}{2}}
+         
          \]
     </think>
     
@@ -73,102 +71,46 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 - **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `350`
+- **extracted** (✗): `1120`
 - **completion**:
 
     ```text
     <think>
-    Let the centers of the balloons be \(A\), \(B\), and \(C\) with radii \(11\), \(13\), and \(19\) respectively. The distance between \(A\) and \(B\) is \(11 + 13 = 24\), between \(B\) and \(C\) is \(13 + 19 = 32\), and between \(A\) and \(C\) is \(11 + 19 = 30\). The plane cuts the balloons such that the centers \(X\), \(Y\), and \(Z\) are equidistant from the respective centers of the balloons.
-    
-    Let \(XY = YZ = ZX = d\). Then, \(X\) is the circumcenter of triangle \(ABC\). Using the Law of Cosines in triangle \(ABC\):
-    
-    \[
-    AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC) = AC^2
-    \]
-    
-    \[
-    24^2 + 32^2 - 2 \cdot 24 \cdot 32 \cdot \cos(\angle ABC) = 30^2
-    \]
-    
-    \[
-    576 + 1024 - 1536 \cos(\angle ABC) = 900
-    \]
-    
-    \[
-    1600 - 1536 \cos(\angle ABC) = 900
-    \]
-    
-    \[
-    700 = 1536 \cos(\angle ABC)
-    \]
-    
-    \[
-    \cos(\angle ABC) = \frac{700}{1536} = \frac{175}{384}
-    \]
-    
-    Thus,
-    
-    \[
-    \cos(\angle ABC) = \frac{175}{384}
-    \]
-    
-    Then,
-    
-    \[
-    \cos(\angle BAX) = \cos(\angle BAY) = \cos(\angle CAX) = \frac{175}{384}
-    \]
-    
-    Using the Law of Cosines in triangle \(AXY\):
-    
-    \[
-    AX^2 = AY^2 + XY^2 - 2 \cdot AY \cdot XY \cdot \cos(\angle AYX)
-    \]
-    
-    \[
-    d^2 = d^2 + d^2 - 2 \cdot d^2 \cdot \frac{175}{384}
-    \]
-    
-    \[
-    d^2 = 2d^2 - \frac{350}{384}d^2
-    \]
-    
-    \[
-    d^2 = 2d^2 - \frac{350}{384}d^2
-    \]
-    
-    \[
-    \frac{350}{384}d^2 = d^2
-    \]
-    
-    \[
-    \frac{350}{384} = 1
-    \]
-    
-    \[
-    \frac{350}{384} = \frac{175}{192}
-    \]
-    
-    Thus,
-    
-    \[
-    d^2 = \frac{175}{192}
-    \]
-    
-    Finally,
+    We start with the given radii of the balloons: $r_1 = 11$, $r_2 = 13$, and $r_3 = 19$. The centers of the balloons are $X$, $Y$, and $Z$, respectively, and the distances between the centers are $XY$, $YZ$, and $XZ$. We are given that $XY^2 = 560$.
+    
+    1. **Calculate $YZ^2$**:
+       Using the Pythagorean theorem in triangle $XYZ$:
+       \[
+       YZ^2 = XY^2 + XZ^2 - 2 \cdot XY \cdot XZ \cdot \cos(\theta)
+       \]
+       Since $X$, $Y$, and $Z$ are the centers of the spheres and the triangles formed are right-angled, we have:
+       \[
+       YZ^2 = XY^2 + XZ^2 - 2 \cdot XY \cdot XZ \cdot \cos(90^\circ) = XY^2 + XZ^2
+       \]
+       Given $XY = \sqrt{560}$, we substitute and solve:
+       \[
+       YZ^2 = 560 + XZ^2
+       \]
+    
+    2. **Calculate $XZ^2$**:
+       Using the Pythagorean theorem in triangle $XZP$ where $P$ is the point of tangency on the third sphere:
+       \[
+       XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(90^\circ) = XY^2 + YZ^2
+       \]
+       Substituting $XY^2 = 560$ and $YZ^2 = 560 + XZ^2$:
+       \[
+       XZ^2 = 560 + (560 + XZ^2) = 1120 + XZ^2
+       \]
+       Solving for $XZ^2$:
+       \[
+       XZ^2 = 1120
+       \]
     
+    Conclusion:
     \[
-    XZ^2 = \frac{175}{192} \cdot 2 = \frac{350}{192} = \frac{175}{96}
-    \]
     
-    Thus,
-    
-    \[
-    XZ^2 = 175 \times 2 = 350
     \]
-    
-    Conclusion:
-    \(\)
     </think>
     
-    \boxed{350}
+    \boxed{1120}
     ```