Update Automated MNLP evaluation report (2026-05-18)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
 - **Owner(s):** group **kth**
-- **Generated at:** 2026-05-17T04:52:19+00:00 (UTC)
+- **Generated at:** 2026-05-18T02:30:25+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 | 0.3100 | ok |
+| Math | 0.3200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,66 +24,70 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `4`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `4`
+- **reference**: `$\frac{a}{2}$`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `\frac{a}{2}`
 - **completion**:
 
     ```text
     <think>
-    We can rewrite the equation as:
-    \[ 10^x = 10 + 9990 \Rightarrow 10^x = 10000 \Rightarrow 10^x = 10^4 \]
+    Since \(P\) is the circumcenter of triangle \(XYZ\), \(P\) is equidistant from the vertices \(X\), \(Y\), and \(Z\). Therefore, \(PX = PY = PZ\).
     
-    Since the bases are the same, we can equate the exponents:
-    \[ x = 4 \]
+    Since \(Q\) is the midpoint of \(YZ\), \(Q\) is also the midpoint of \(XY\), because \(XY = XZ\).
     
-    The answer is:
-    \[
-    </think>
-    
-    Therefore, the answer is \boxed{4}.
-    ```
-
-**Incorrect** (1 shown)
-
-- **reference**: `2 - \frac{\pi}{2}`
-- **overall** (0/8 completions correct)
-- **extracted** (✗): `\frac{1}{2}`
-- **completion**:
-
-    ```text
-    <think>
-    Let $P$ be the probability that Derek eventually becomes debt-free. We can derive a recursive equation for $P$.
+    When \(R\) is the image of \(Q\) after reflection across the side \(XZ\), \(R\) is on the line \(XY\), and \(QR = XZ = XY\).
     
-    When Derek tosses the coin, there are two possible outcomes:
+    Since \(XQYS\) is a rectangle, \(XQ = QS = YS\).
     
-    1. Heads: He reduces his debt by 1 MIT. This means his debt becomes $\pi - 1$ MITs.
-    2. Tails: He reduces his debt by 2 MITs. This means his debt becomes $\pi - 2$ MITs.
+    Since \(Q\) is the midpoint of \(YZ\), \(QY = QZ\).
     
-    However, if his debt becomes nonpositive, he is
+    Since \(R\) is on the line \(XY\), \(QR\) is perpendicular to \(XZ\).
     
-    ... [truncated 540 chars] ...
+    Since \(XQYS\) is a rectangle, \(XQ\) is perpendicular to \(YS\).
     
-    ly becomes debt-free given a debt of $\pi - 1$ MITs, and $P_{\pi - 2}$ is the probability that Derek eventually becomes debt-free given a debt of $\pi - 2$ MITs.
+    Since \(QY = QZ\), and \(XQ = QS\), we have \(XQ = QY\).
     
-    We can continue this process recursively, but we notice a pattern. The recursive equation can be rewritten as:
+    Therefore, triangle \(XQY\) is a right triangle.
     
-    \[ P = \frac{1}{2} \left( P_{\pi - 1} + P_{\pi - 2} \right) \]
+    Since \(\angle YXZ = a\), we have \(\angle QXY = 90^\circ - a\).
     
-    Using the same logic, we can derive the following recursive equation:
+    Since \(XQYS\) is a rectangle, \(\angle QYS = 90^\circ\).
     
-    \[ P = \frac{1}{2} \left( \frac{1}{2} \left( P_{\pi - 2} + P_{\pi - 3} \right) + \frac{1}{2} \left( P_{\pi - 3} + P_{\pi - 4} \right) \right) \]
+    Since \(QR\) is perpendicular to \(XZ\), \(\angle QRS = 90^\circ\).
     
-    And so on.
+    Since \(XQ = QS\), triangle \(QXS\) is isosceles.
     
-    Notice that the recursive equation can be simplified to:
+    Therefore, \(\angle QXS = \angle QSX = 90^\circ - \frac{a}{2}\).
     
-    \[ P = \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \cdots \right) + \frac{1}{2} \left( \frac{1}{2} \left( \cdots \right) \right) \right) + \frac{1}{2} \left( \frac{1}{2} \left( \cdots \right) + \frac{1}{2} \left( \frac{1}{2} \left( \cdots \right) \right) \right) \right) + \cdots \right) \]
+    Since \(\angle QRS = 90^\circ\), we have \(\angle PRS = 180^\circ - \angle QRS - \angle QXS = 180^\circ - 90^\circ - (90^\circ - \frac{a}{2}) = \frac{a}{2}\).
     
-    This is a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$. The sum of the series is:
+    Therefore, \(\angle PRS =
+    </think>
     
-    \[ P = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{1}{2} \div \frac{1}{2} = \frac{1}{2} \cdot 2 =
+    Therefore, the answer is \boxed{\frac{a}{2}}.
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `104`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `51`
+- **completion**:
+
+    ```text
+    <think>
+    ## Step 1:  To solve this problem, we first need to understand the given tetrahedron $WXYZ$ and the conditions for point $P$ being equidistant from all four faces.
+    ## Step 2:  The tetrahedron $WXYZ$ has edge lengths $WX = YZ = \sqrt{41}$, $WY = XZ = \sqrt{80}$, and $WZ = XY = \sqrt{89}$.
+    ## Step 3:  We are given that point $P$ lies inside the tetrahedron and is equidistant from all four of its faces. This common distance is denoted as $d$.
+    ## Step 4:  To find the value of $d$, we can use the properties of the tetrahedron and the given edge lengths.
+    ## Step 5:  We can start by finding the area of the triangle $XYZ$. Using Heron's formula, we can calculate the area as $\frac{1}{2}\sqrt{41\cdot89\cdot80}$.
+    ## Step 6:  Then, we can find the volume of the tetrahedron $WXYZ$ using the formula $\frac{1}{3}\cdot(\text{area of base})\cdot\text{height}$. The height is the distance from point $W$ to the plane containing the base $XYZ$.
+    ## Step 7:  We can then use the formula for the volume of a tetrahedron to express the volume in terms of the distance $d$ and the areas of the faces.
+    ## Step 8:  By equating the two expressions for the volume, we can solve for the value of $d$.
+    ## Step 9:  Finally, we can express the common distance $d$ in the form $\frac{a\sqrt{b}}{c}$ and calculate the sum $a+b+c$.
+    
+    The final answer is: $
     </think>
     
-    Therefore, the answer is \boxed{\frac{1}{2}}.
+    Therefore, the answer is \boxed{51}.
     ```