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

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:** 2026-06-01T06:13:57+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._
@@ -31,37 +31,34 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    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 \(\angle YXZ = a\).
+    1. **Identify Key Points and Geometry:**
+       - Given: Isosceles triangle \(XYZ\) with \(XY = XZ\).
+       - \(P\) is the circumcenter of \(XYZ\).
+       - \(Q\) is the midpoint of \(YZ\).
+       - \(R\) is the reflection of \(Q\) over \(XZ\).
+       - \(S\) is such that \(XQYS\) is a rectangle.
+       - Given angle \(\angle YXZ = a\).
     
-    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 \(XR = XQ\).
+    2. **Identify the Symmetry and Midpoint Properties:**
+       - Since \(Q\) is the midpoint of \(YZ\), \(XQ\) is perpendicular to \(YZ\).
+       - Because \(XQYS\) is a rectangle, \(XQ\) is also perpendicular to \(YS\).
     
-    3. **Constructing the Rectangle**:
-       - Since \(XQYS\) is a rectangle, \(XQ\) and \(YS\) are perpendicular to each other, and \(XQ = YS\).
-       - This implies that \( \angle XQS = 90^\circ \).
+    3. **Analyze the Reflection and Rectangle Properties:**
+       - \(R\) is the reflection of \(Q\) over \(XZ\), which implies that \(XZ\) is the perpendicular bisector of \(QR\).
+       - Since \(XQYS\) is a rectangle, \(XQ\) is parallel to \(YS\) and equal in length.
     
-    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.
+    4. **Determine the Angle \(\angle PRS\):**
+       - \(P\) is the circumcenter of \(\triangle XYZ\), which means it lies on the perpendicular bisectors of the sides of the triangle.
+       - Since \(XQ\) is perpendicular to \(YZ\) and \(XQYS\) is a rectangle, \(XQ\) is also perpendicular to \(YS\).
+       - Because \(R\) is the reflection of \(Q\) over \(XZ\), \(XZ\) bisects \(QR\).
+       - Therefore, \(\angle PRS\) is half of \(\angle YXZ\) because \(P\) is the circumcenter.
     
-    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:
-         \[
-         \angle PRS = \frac{1}{2} \angle YXZ = \frac{a}{2}
-         \]
+    5. **Final Calculation:**
+       - Given \(\angle YXZ = a\), then \(\angle PRS = \frac{a}{2}\).
     
-    6. **Conclusion**:
-       - The angle \(\angle PRS\) in terms of \(a\) is:
-         \[
-         
-         \]
+    \[
+    
+    \]
     </think>
     
     \boxed{\frac{a}{2}}
@@ -71,7 +68,7 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 - **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `1120`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
@@ -81,36 +78,57 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     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:
-       \[
-       YZ^2 = 560 + XZ^2
-       \]
+       YZ^2 = XY^2 + XZ^2 - 2 \cdot XY \cdot XZ \cdot \cos(\thet
     
-    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 $XZ^2$:
-       \[
-       XZ^2 = 1120
-       \]
+    ... [truncated 819 chars] ...
     
-    Conclusion:
-    \[
+    uld use the spheres' radii and the distances between their centers. Using the Pythagorean theorem in triangle $XYZ$:
+        \[
+        XZ^2 = XY^2 + YZ^2
+        \]
+        Given $XY^2 = 560$, and $YZ^2 = 560 + XZ^2$:
+        \[
+        XZ^2 = 560 + (560 + XZ^2)
+        \]
+        \[
+        XZ^2 = 1120 + XZ^2
+        \]
+        \[
+        0 = 1120
+        \]
+        This is still a contradiction. Let's re-evaluate the problem setup.
     
-    \]
-    </think>
+    94. **Correct the setup**:
+        The correct approach should use the spheres' radii and the distances between their centers. Using the Pythagorean theorem in triangle $XYZ$:
+        \[
+        XZ^2 = XY^2 + YZ^2
+        \]
+        Given $XY^2 = 560$, and $YZ^2 = 560 + XZ^2$:
+        \[
+        XZ^2 = 560 + (560 + XZ^2)
+        \]
+        \[
+        XZ^2 = 1120 + XZ^2
+        \]
+        \[
+        0 = 1120
+        \]
+        This is still a contradiction. Let's re-evaluate the problem setup.
     
-    \boxed{1120}
+    95. **Correct the setup**:
+        The correct approach should use the spheres' radii and the distances between their centers. Using the Pythagorean theorem in triangle $XYZ$:
+        \[
+        XZ^2 = XY^2 + YZ^2
+        \]
+        Given $XY^2 = 560$, and $YZ^2 = 560 + XZ^2$:
+        \[
+        XZ^2 = 560 + (560 + XZ^2)
+        \]
+        \[
+        XZ^2 = 1120 + XZ^2
+        \]
+        \[
+        0 = 1120
+        \]
+        This is
     ```