Automated MNLP evaluation report (2026-06-06)

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-emainelpe/math_model`](https://huggingface.co/cs-552-2026-emainelpe/math_model)
 - **Owner(s):** group **emainelpe**
-- **Generated at:** 2026-06-03T08:04:32+00:00 (UTC)
+- **Generated at:** 2026-06-06T17:20:53+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._
 
+## Evaluated checkpoint
+
+- **Commit:** [`6db58e7`](https://huggingface.co/cs-552-2026-emainelpe/math_model/commit/6db58e7736e81594484aa94c9ee1bd2b1991f810)
+- **Message:** trigger 16k CI re-evaluation
+- **Committed:** 2026-06-03T09:40:03+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3100 | ok |
+| Math | 0.3400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,76 +30,94 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `4`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `4`
+- **reference**: `90`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `90`
 - **completion**:
 
     ```text
     <think>
-    We have the equation $10^x - 10 = 9990$. To solve for $x$, we first add 10 to both sides to isolate the term with $x$:
-    \[ 10^x = 9990 + 10 \Rightarrow 10^x = 10000 \]
-    
-    Now, we recognize that $10000 = 10^4$, so we can rewrite the equation as:
-    \[ 10^x = 10^4 \]
-    
-    Since the bases are the same, we can equate the exponents:
-    \[ x = 4 \]
-    
-    Therefore, the solution for $x$ is:
-    \[ \boxed{4} \]
+    ## Step 1:  To solve this problem, we first need to understand the given information and the relationships between the various points and lines involved.
+    ## Step 2:  We are given a triangle $XYZ$ with $\angle X = 120^\circ$. The incenter is denoted as $J$, and $N$ is the midpoint of $YZ$.
+    ## Step 3:  We draw a line through $N$ parallel to $XJ$, intersecting the circle with diameter $YZ$ at $U$ and $V$, with $U$ chosen so that $X$ and $U$ are on the same side of line $YZ$.
+    ## Step 4:  Through $U$, we draw the line perpendicular to $VJ$, intersecting $XY$ at $R$ and $XZ$ at $S$.
+    ## Step 5:  We need to determine the measure of $\angle RJS$ in degrees.
+    ## Step 6:  To approach this problem, we can start by analyzing the properties of the incenter and the relationships between the various points and lines involved.
+    ## Step 7:  We can also use the properties of parallel lines and perpendicular lines to establish relationships between the various angles and segments involved.
+    ## Step 8:  By carefully analyzing the given information and applying the properties of triangles and circles, we can determine the measure of $\angle RJS$.
+    ## Step 9:  We can use the fact that the incenter is the intersection of the angle bisectors of the triangle, and the properties of parallel and perpendicular lines to establish the necessary relationships.
+    ## Step 10:  By following these steps, we can determine the measure of $\angle RJS$ and provide the final answer.
+    
+    The final answer is: $\boxed{90}
     </think>
     
-    \boxed{4}
+    \boxed{90}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `2 - \frac{\pi}{2}`
+- **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `1`
+- **extracted** (✗): `970`
 - **completion**:
 
     ```text
     <think>
-    Let $p$ be the probability that Derek eventually becomes debt-free.
+    Let the centers of the balloons be $A$, $B$, and $C$, with radii $r_A = 11$, $r_B = 13$, and $r_C = 19$, respectively.
+    
+    Since the centers of the balloons all lie on one side of the plane, we can assume that the plane is parallel to the $xy$-plane.
+    
+    The distance between the centers of the balloons is equal to the sum of their radii:
+    \[ AB = r_A + r_B = 11 + 13 = 24 \]
+    \[ BC = r_B + r_C = 13
+    
+    ... [truncated 453 chars] ...
+    
+    )^2 : 2 - XY^2 - YZ^2 - XZ^2 \]
     
-    Derek starts with a debt of $\pi$ MIT.
+    Substituting the given values, we get:
+    \[ XZ^2 = (24)^2 + (32)^2 + (30)^2 : 2 - 560 - YZ^2 - XZ^2 \]
     
-    On each day, he has a $\frac{1}{2}$ chance of reducing his debt by 1 MIT and a $\frac{1}{2}$ chance of reducing it by 2 MITs.
+    Simplifying, we get:
+    \[ XZ^2 = 576 + 1024 + 900 : 2 - 560 - YZ^2 - XZ^2 \]
     
-    If he reduces his debt by 1 MIT, his new debt is $\pi - 1$ MIT. If he reduces it by 2 MITs, his new debt is $\pi - 2$ MIT.
+    \[ XZ^2 = 2500 - 560 - YZ^2 - XZ^2 \]
     
-    If his debt is nonpositive, he i
+    \[ 2XZ^2 = 1940 - YZ^2 \]
     
-    ... [truncated 88 chars] ...
+    \[ XZ^2 = 970 - YZ^2 : 2 \]
     
-    onsider the cases:
+    We can also use the fact that $XY^2 + YZ^2 + XZ^2 = (XY + YZ)^2 + (YZ + XZ)^2 + (XZ + XY)^2 : 2$ to find $YZ^2$:
+    \[ YZ^2 = (XY + YZ)^2 + (YZ + XZ)^2 + (XZ + XY)^2 : 2 - XY^2 - XZ^2 - YZ^2 \]
     
-    Case 1: Derek reduces his debt by 1 MIT.
+    Substituting the given values, we get:
+    \[ YZ^2 = (24)^2 + (32)^2 + (30)^2 : 2 - 560 - XZ^2 - YZ^2 \]
     
-    The probability of this happening is $\frac{1}{2}$. In this case, his new debt is $\pi - 1$ MIT.
+    Simplifying, we get:
+    \[ YZ^2 = 576 + 1024 + 900 : 2 - 560 - XZ^2 - YZ^2 \]
     
-    Case 2: Derek reduces his debt by 2 MITs.
+    \[ YZ^2 = 2500 - 560 - XZ^2 - YZ^2 \]
     
-    The probability of this happening is $\frac{1}{2}$. In this case, his new debt is $\pi - 2$ MIT.
+    \[ 2YZ^2 = 1940 - XZ^2 \]
     
-    Now, let's consider the probability of Derek becoming debt-free.
+    \[ YZ^2 = 970 - XZ^2 : 2 \]
     
-    The probability of Derek becoming debt-free in the first day is $\frac{1}{2}$ if his debt is reduced by 1 MIT, or 0 if his debt is reduced by 2 MITs.
+    Substituting this expression for $YZ^2$ into the expression for $XZ^2$, we get:
+    \[ XZ^2 = 970 - (970 - XZ^2 : 2) : 2 \]
     
-    The probability of Derek becoming debt-free in the second day is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ if his debt is reduced by 1 MIT in the first day, or 0 if his debt is reduced by 2 MITs in the first day.
+    \[ XZ^2 = 970 - 485 + XZ^2 : 4 \]
     
-    In general, the probability of Derek becoming debt-free on the $n$-th day is $\frac{1}{2^n}$.
+    \[ 4XZ^2 = 485 + 4XZ^2 \]
     
-    The probability of Derek becoming debt-free is the sum of the probabilities of him becoming debt-free on each day:
+    \[ 4XZ^2 - 4XZ^2 = 485 \]
     
-    \[ p = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]
+    \[ 0 = 485 \]
     
-    This is a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$. The sum of the series is:
+    This is a contradiction, so there must be an error in the problem statement.
     
-    \[ p = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \boxed{1} \]
+    However, if we ignore the contradiction and continue with the solution, we get:
+    \[ XZ^2 = \boxed{970} \]
     </think>
     
-    \boxed{1}
+    \boxed{970}
     ```