Update Automated MNLP evaluation report (2026-06-01)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-bilko/math_model`](https://huggingface.co/cs-552-2026-bilko/math_model)
 - **Owner(s):** group **bilko**
-- **Generated at:** 2026-05-29T06:54:15+00:00 (UTC)
+- **Generated at:** 2026-06-01T06:13:57+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 | 0.3700 | ok |
+| Math | 0.3400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,76 +24,97 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1600`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `1600`
+- **reference**: `\frac{7}{4}`
+- **overall** (6/8 completions correct)
+- **extracted** (✓): `\frac{7}{4}`
 - **completion**:
 
     ```text
-    First, we will simplify the expression inside the given equation:
+    To find the value of $b$ in the direction vector of the red line given as $\begin{pmatrix} -7 \\ b \end{pmatrix}$, we need to determine the slope of the red line and compare it with the slope of the line segment $AB$ on the graph.
+    
+    First, we calculate the slope of line segment $AB$. The coordinates of points $A$ and $B$ are given as $(-5,4)$ and $(-1,3)$, respectively. The slope $m_{AB}$ is calculated using the formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\frac{y_2 - y_1}{x_2 - x_1}$. Applying this formula to points $A$ and $B$ gives:
+    
     \[
-    441 + 2 \times 21 \times 19 + 361
+    m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{-1 - (-5)} = \frac{-1}{4}.
     \]
     
-    1. Calculate \(2 \times 21 \times 19\):
-       \[
-       2 \times 21 \times 19 = 42 \times 19
-       \]
-       Compute \(42 \times 19\):
-       \[
-       42 \times 19 = 798
-       \]
+    Next, we compare this slope with the slope of the direction vector $\begin{pmatrix} -7 \\ b \end{pmatrix}$, which can be calculated similarly. The slope $m$ is given by $\frac{b - 0}{-7 - 0} = \frac{b}{-7}$. To find the value of $b$ that makes the slope of the direction vector equal to the slope of line segment $AB$, we set the two slopes equal to each other and solve for $b$:
     
-    2. Substitute \(798\) back into the expression:
-       \[
-       441 + 798 + 361
-       \]
+    \[
+    \frac{b}{-7} = \frac{-1}{4}.
+    \]
     
-    3. Add \(441\) and \(361\) first:
-       \[
-       441 + 361 = 802
-       \]
+    Solving for $b$, we multiply both sides of the equation by $-7$:
     
-    4. Add the remaining \(798\):
-       \[
-       802 + 798 = 1600
-       \]
+    \[
+    b = \frac{-1}{4} \cdot -7 = \frac{7}{4}.
+    \]
     
-    Thus, \(x =
+    Therefore, the value of $b$ that makes the slope of the direction vector equal to the slope of line segment $AB$ is
     </think>
     
-    \boxed{1600}\).
+    $\boxed{\frac{7}{4}}$.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `125`
+- **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `64`
+- **extracted** (✗): `0.1064349083`
 - **completion**:
 
     ```text
-    To solve the given problem, we start by considering the rhombus \(WXYZ\) with an acute angle \(\angle ZWX\) and an incircle. Point \(Q\) is on the incircle, and the perpendicular distances from \(Q\) to lines \(ZW\), \(WX\), and \(XY\) are \(9\), \(5\), and \(16\), respectively. We need to determine the perimeter of rhombus \(WXYZ\).
+    Given three spherical balloons with radii \(11\), \(13\), and \(19\), and each pair externally tangent, we need to find \(XZ^2\) where \(XY^2 = 560\).
     
-    1. **Understanding the Geometry:**
-       - Since \(Q\) is on the i
+    1. **Identify the distances between centers**:
+       - Since the balloons are externally tangent, the distance between the centers of any two balloons is the sum of their radii:
+         \[
+         XY = 11 + 13 = 24 \quad \text{and} \quad XZ = 11 + 19 = 30
     
-    ... [truncated 526 chars] ...
+    ... [truncated 655 chars] ...
     
-    } = \text{base} \times \text{height}\), where the base is the length of one side of the rhombus and the height is the inradius.
+    XY^2 = 560 \Rightarrow \cos \angle XYZ = \frac{560 + XZ^2 - 576}{2 \cdot 4\sqrt{35} \cdot XZ}
+         \]
+         Substituting \(XZ^2 = 290 + 304\cos \angle XYZ\):
+         \[
+         \cos \angle XYZ = \frac{560 + 290 + 304\cos \angle XYZ - 576}{2 \cdot 4\sqrt{35} \cdot XZ}
+         \]
+         \[
+         \cos \angle XYZ = \frac{294 + 304\cos \angle XYZ - 576}{80\sqrt{35} \cdot XZ}
+         \]
+         Simplifying:
+         \[
+         \cos \angle XYZ = \frac{-282 + 304\cos \angle XYZ}{80\sqrt{35} \cdot XZ}
+         \]
+         \[
+         80\sqrt{35} \cdot XZ (\cos \angle XYZ) = -282 + 304\cos \angle XYZ
+         \]
+         Solving for \(XZ\):
+         \[
+         80\sqrt{35} XZ + 304XZ = 282
+         \]
+         \[
+         XZ (80\sqrt{35} + 304) = 282
+         \]
+         \[
+         XZ = \frac{282}{80\sqrt{35} + 304}
+         \]
+         Approximating:
+         \[
+         XZ^2 \approx \frac{282^2}{(80\sqrt{35} + 304)^2}
+         \]
     
-    3. **Using the Given Distances:**
-       - The perpendicular distances from \(Q\) to \(ZW\), \(WX\), and \(XY\) being \(9\), \(5\), and \(16\) respectively imply that these are the sides of the rhombus.
-       - The inradius is the minimum distance from the center to any side, which is \(5\).
+    3. **Calculation**:
+       - Calculate \(XZ^2\):
+         \[
+         XZ^2 = \left( \frac{282}{80\sqrt{35} + 304} \right)^2 \approx \frac{79524}{(80 \cdot 5.916 + 304)^2} \approx \frac{79524}{(473.28 + 304)^2} \approx \frac{79524}{777.28^2} \approx 0.1064349083
+         \]
+       - Therefore,
+         \[
+         XZ^2 = \boxed{0.1064349083}
+         \]
     
-    4. **Calculating the Side Length of the Rhombus:**
-       - Since the inradius is \(5\), the side length \(a\) of the rhombus can be calculated using the formula for the area of a rhombus: \(\text{Area} = a \cdot \text{inradius}\).
-       - We also know that the area can be expressed in terms of the diagonals \(d_1\) and \(d_2\) as \(\text{Area} = \frac{d_1 \cdot d_2}{2}\).
-       - However, without knowing the lengths of the diagonals, we use the side lengths \(a\), \(b\), and \(c\) where \(a = b = c = 16\).
-    
-    5. **Finding the Perimeter:**
-       - Since all sides of the rhombus are equal, the perimeter \(P\) of the rhombus is \(4 \times a\).
-       - Substituting \(a = 16\) into the perimeter formula, we get \(P = 4 \times 16 = 64\).
+    </think>
     
-    Thus, the perimeter of rhombus \(WXYZ\) is \(\boxed{64}\).
+    $\boxed{0.1064349083}$
     ```