Update Automated MNLP evaluation report (2026-06-11)

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:** 2026-06-11T04:19:46+00:00 (UTC)
+- **Generated at:** 2026-06-11T06:23:10+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._
@@ -17,7 +17,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3000 | ok |
+| Math | 0.3300 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,85 +30,102 @@ _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{1}{21}`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `-\frac{1}{21}`
 - **completion**:
 
     ```text
     <think>
-    To solve for $x$ in the equation $10^x - 10 = 9990$, we can first isolate $10^x$ by adding $10$ to both sides of the equation:
+    The equation of the curve is \(y = x^2 + \frac{2}{x}\). The straight line intersects this curve at three distinct points.
     
-    \[ 10^x - 10 + 10 = 9990 + 10 \Rightarrow 10^x = 10000 \]
+    Let the equation of the line be \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
     
-    Now, we can solve for $x$ by taking the logarithm base $10$ of both sides:
+    Substitute \(y = mx + b\) into the equation of the curve:
+    \[ mx + b = x^2 + \frac{2}{x} \Rightarrow x^2 - mx - \frac{2}{x} - b = 0 \]
     
-    \[ \log_{10}(10^x) = \log_{10}(10000) \Rightarrow x \log_{10}(10) = \log_{10}(10000) \Rightarrow x = \frac{\log_{10}(10000)}{\log_{10}(10)} \]
+    Since two of the intersection
     
-    Since $\log_{10}(10) = 1$, we have:
+    ... [truncated 823 chars] ...
+    
+    }\) into (1) to find \(b\):
+    \[ -6\left(\frac{286}{21}\right) - b = -\frac{107}{3} \Rightarrow -\frac{1716}{21} - b = -\frac{107}{3} \Rightarrow b = -\frac{107}{3} + \frac{1716}{21} \Rightarrow b = \frac{-107 \cdot 7 + 1716}{21} \Rightarrow b = \frac{-749 + 1716}{21} \Rightarrow b = \frac{967}{21} \]
     
-    \[ x = \frac{\log_{10}(10000)}{1} = \log_{10}(10000) \]
+    Now that we have \(m = \frac{286}{21}\) and \(b = \frac{967}{21}\), we can substitute these values into the equation of the line:
+    \[ y = \frac{286}{21}x + \frac{967}{21} \]
     
-    We can simplify $\log_{10}(10000)$ as follows:
+    To find the \(x\)-coordinate of the remaining point, substitute \(y = mx + b\) into the equation of the curve:
+    \[ \frac{286}{21}x + \frac{967}{21} = x^2 + \frac{2}{x} \Rightarrow x^2 - \frac{286}{21}x - \frac{2}{x} - \frac{967}{21} = 0 \]
     
-    \[ \log_{10}(10000) = \log_{10}(10^4) = 4 \log_{10}(10) = 4 \]
+    Multiply through by \(21x\) to clear the fractions:
+    \[ 21x^3 - 286x^2 - 42 - 967x = 0 \Rightarrow 21x^3 - 286x^2 - 967x - 42 = 0 \]
     
-    Therefore, the solution to the equation is:
+    Since \(x = 6\) and \(x = 7\) are solutions, we can factor out \((x - 6)\) and \((x - 7)\) from the equation:
+    \[ (x - 6)(x - 7)(21x + 1) = 0 \]
     
-    \[ x =
+    This gives us the possible values of \(x\): \(x = 6\), \(x = 7\), and \(x = -\frac{1}{21}\).
+    
+    So, the \(x\)-coordinate of the remaining point is:
+    \[
     </think>
     
-    Therefore, the answer is \boxed{4}.
+    Therefore, the answer is \boxed{-\frac{1}{21}}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `504`
+- **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `720`
+- **extracted** (✗): `100`
 - **completion**:
 
     ```text
     <think>
-    The area of the trapezoid is given by the formula:
-    \[ A = \frac{1}{2}(r + s)h \]
-    where \(h\) is the height of the trapezoid.
+    We know that the distance between the centers of the two balloons is equal to the sum of their radii. Let's denote the centers of the balloons with radii $11$, $13$, and $19$ as $A$, $B$, and $C$ respectively. Then, we have:
     
-    We are given that the area of the trapezoid is \(72\), so:
-    \[ 72 = \frac{1}{2}(r + s)h \Rightarrow 144 = (r + s)h \]
+    \[ AB = 11 + 13 = 24 \]
+    \[ BC = 13 + 19 = 32 \]
+    \[ AC = 11 + 19 = 30 \]
     
-    Since the circle is inscribed in the trapezoid, the diameter of the circle is equal to the height of the trapezoid:
-    \[ 2r = h \Rightarrow h = 2r \]
+    Since the centers of the balloons lie on one side of the plane, the triangle $ABC$ is a trian
     
-    ... [truncated 823 chars] ...
+    ... [truncated 34 chars] ...
     
-    xr \Rightarrow 72 = r^2 \]
+    $. We can use Heron's formula to find the area of the triangle $ABC$:
     
-    Now we can solve for \(r\):
-    \[ r^2 = 72 \Rightarrow r = \sqrt{72} \Rightarrow r = 6\sqrt{2} \]
+    \[ s = \frac{24 + 32 + 30}{2} = 43 \]
+    \[ \text{Area} = \sqrt{s(s - 24)(s - 32)(s - 30)} = \sqrt{43 \cdot 19 \cdot 11 \cdot 13} = 43 \cdot 11 \cdot \sqrt{13} \]
     
-    Now we can find \(s\):
-    \[ s = 3r + 2x \Rightarrow s = 3(6\sqrt{2}) + 2x \Rightarrow s = 18\sqrt{2} + 2x \]
+    Now, we need to find the area of the triangle $XYZ$. Since the plane cuts the balloons in three congruent circular cross-sections, the triangle $XYZ$ is similar to the triangle $ABC$. Let's denote the distance between the centers $X$ and $Z$ as $d$. Then, we have:
     
-    We can also use the fact that the area of the trapezoid is equal to the sum of the areas of the two triangles formed by the circle and the trapezoid. The area of each triangle is given by the formula:
-    \[ A = \frac{1}{2}bh \]
-    where \(b\) is the base of the triangle and \(h\) is the height of the triangle.
+    \[ XY = \frac{24}{3} = 8 \]
+    \[ YZ = \frac{32}{3} \]
+    \[ XZ = d \]
     
-    The area of the trapezoid is given by:
-    \[ 72 = 2 \left( \frac{1}{2} \cdot x \cdot r \right) + 2 \left( \frac{1}{2} \cdot (r - x) \cdot r \right) \Rightarrow 72 = xr + r^2 - xr \Rightarrow 72 = r^2 \]
+    Using the similarity of the triangles, we can write:
     
-    Now we can solve for \(x\):
-    \[ r^2 = 72 \Rightarrow x = \frac{r^2 - 72}{2r} \Rightarrow x = \frac{72 - 72}{12\sqrt{2}} \Rightarrow x = 0 \]
+    \[ \frac{XY}{AB} = \frac{YZ}{BC} = \frac{XZ}{AC} \]
     
-    Now we can find \(s\):
-    \[ s = 18\sqrt{2} + 2x \Rightarrow s = 18\sqrt{2} + 2(0) \Rightarrow s = 18\sqrt{2} \]
+    Substituting the values, we get:
     
-    Finally, we can find \(r^2+s^2\):
-    \[ r^2+s^2 = (6\sqrt{2})^2 + (18\sqrt{2})^2 \Rightarrow r^2+s^2 = 72 + 648 \Rightarrow r^2+s^2 = 720 \]
+    \[ \frac{8}{24} = \frac{\frac{32}{3}}{32} = \frac{d}{30} \]
     
-    So, the answer is:
-    \[
+    Simplifying, we get:
+    
+    \[ \frac{1}{3} = \frac{1}{3} = \frac{d}{30} \]
+    
+    Solving for $d$, we get:
+    
+    \[ d = 30 \cdot \frac{1}{3} = 10 \]
+    
+    Now, we can find the area of the triangle $XYZ$:
+    
+    \[ \text{Area}_{XYZ} = \frac{1}{3} \cdot \text{Area}_{ABC} = \frac{1}{3} \cdot 43 \cdot 11 \cdot \sqrt{13} = \frac{43 \cdot 11}{3} \cdot \sqrt{13} \]
+    
+    Finally, we can find $XZ^2$:
+    
+    \[ XZ^2 = d^2 = 10^2 =
     </think>
     
-    Therefore, the answer is \boxed{720}.
+    Therefore, the answer is \boxed{100}.
     ```