ProofRank release

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EXAMPLES.md

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+# Ex. 1: Conciseness disagreement
+
+
+**Problem** `imo-bench_number_theory-042`
+> **Problem statement:** Find all positive integers $k$ such that the number of positive divisors of $2k^3$ is $2k$.
+
+**Solution 1 (GLM-5)**
+
+
+We are to find all positive integers \(k\) such that the number of positive divisors of \(2k^3\) equals \(2k\).  
+Let \(d(n)\) denote the number of positive divisors of \(n\). If \(n=\prod p_i^{e_i}\) then \(d(n)=\prod (e_i+1)\).  
+
+---
+
+### Case I: \(k\) odd  
+
+If \(k=1\), then \(2k^3=2\) and \(d(2)=2=2k\), so \(k=1\) works.  
+
+Assume \(k>1\) odd. Write its prime factorization  
+\[
+k=\prod_{i=1}^{r} p_i^{a_i},\qquad p_i\;\text{odd primes},\;a_i\ge 1.
+\]  
+Then  
+\[
+k^3=\prod p_i^{3a_i},\qquad 2k^3=2\prod p_i^{3a_i},
+\]  
+so  
+\[
+d(2k^3)=(1+1)\prod (3a_i+1)=2\prod (3a_i+1).
+\]  
+Setting \(d(2k^3)=2k\) gives  
+\[
+2\prod (3a_i+1)=2k \quad\Longrightarrow\quad \prod (3a_i+1)=k=\prod p_i^{a_i}. \tag{1}
+\]  
+
+We show that (1) has no solution with \(k>1\).  
+
+For each \(i\) define \(r_i=(3a_i+1)/p_i^{a_i}\). Their product is \(1\), so at least one \(r_i\ge 1\) (otherwise product \(<1\)). Moreover \(r_i=1\) would imply \(3a_i+1=p_i^{a_i}\), which is impossible: for \(a_i=1\) we get \(4=p_i\) (contradiction because \(p_i\) is odd \(\ge 3\)); for \(a_i\ge 2\) the left side is at most \(3a_i+1\) while the right side is at least \(p_i^2\ge 9\), again impossible. Hence some \(r_i>1\).  
+
+Now we prove a useful inequality.  
+
+**Lemma.** For any odd prime \(p\) and integer \(a\ge 1\),  
+\[
+3a+1 \le \frac{4}{3}\,p^a,
+\]  
+with equality iff \(p=3\) and \(a=1\).  
+
+*Proof.* If \(p=3\) the claim is \(3(3a+1)\le 4\cdot 3^a\). For \(a=1\) we have equality. For \(a\ge 2\) the right side grows exponentially while the left side grows linearly, so the inequality holds. If \(p\ge 5\) then \(p^a\ge 5^a>3^a\), so  
+\[
+\frac{3a+1}{p^a}<\frac{3a+1}{3^a}\le\frac{4}{3},
+\]  
+where the last inequality follows from the \(p=3\) case. ∎  
+
+From the lemma, the only way to have \(r_i>1\) is \(p_i=3\) and \(a_i=1\), giving \(r_i=4/3\). Therefore any solution of (1) with \(k>1\) must contain the prime \(3\) with exponent exactly \(1\). Then the factor \((3\cdot1+1)=4\) appears in the left product, making \(\prod(3a_i+1)\) even. But the right‑hand side \(k\) is odd (product of odd primes). This contradiction shows that no odd \(k>1\) satisfies (1). Hence the only odd solution is \(k=1\).  
+
+---
+
+### Case II: \(k\) even  
+
+Write \(k=2^a\cdot m\) with \(a\ge 1\) and \(m\) odd. Let  
+\[
+m=\prod_{i=1}^{s} q_i^{b_i},\qquad q_i\;\text{odd primes},\;b_i\ge 1.
+\]  
+Then  
+\[
+k^3=2^{3a}\prod q_i^{3b_i},\qquad 2k^3=2^{3a+1}\prod q_i^{3b_i},
+\]  
+so  
+\[
+d(2k^3)=(3a+2)\prod (3b_i+1).
+\]  
+Setting equal to \(2k=2^{a+1}\prod q_i^{b_i}\) yields  
+\[
+(3a+2)\prod (3b_i+1)=2^{a+1}\prod q_i^{b_i}. \tag{2}
+\]  
+
+Rewrite (2) as  
+\[
+\prod_{i=1}^{s}\frac{3b_i+1}{q_i^{b_i}}=\frac{2^{a+1}}{3a+2}. \tag{3}
+\]  
+
+Denote the left side by \(P\). By the lemma, each factor satisfies  
+\[
+\frac{3b_i+1}{q_i^{b_i}}\le\frac{4}{3},
+\]  
+with equality only for \((q_i,b_i)=(3,1)\). Moreover, the only factor that can exceed \(1\) is the one for \((3,1)\) (which equals \(4/3\)); all other factors are \(<1\). Consequently  
+\[
+P\le\frac{4}{3},
+\]  
+with equality possible only if the set of odd primes consists exactly of a single factor \(3\) with exponent \(1\). If there are no odd primes (\(m=1\)), then \(P=1\).  
+
+Now examine the right side of (3).  
+
+* If \(a\ge 3\), then  
+\[
+\frac{2^{a+1}}{3a+2}\ge\frac{2^{4}}{11}=\frac{16}{11}>\frac{4}{3}.
+\]  
+(For larger \(a\) the fraction increases because the numerator grows exponentially while the denominator grows linearly.) Hence we would need \(P>4/3\), impossible. Therefore \(a\) cannot be \(\ge 3\).  
+
+Thus only \(a=1\) and \(a=2\) remain.  
+
+---
+
+#### Subcase \(a=2\)  
+
+Then \(3a+2=8\) and \(2^{a+1}=8\), so (2) simplifies to  
+\[
+\prod (3b_i+1)=\prod q_i^{b_i}=m.
+\]  
+This is exactly equation (1) for the odd integer \(m\). By the same reasoning as in Case I (or by Lemma 1 applied to the product), the only possibility is \(m=1\). Hence \(k=2^2\cdot1=4\).  
+Check: \(2\cdot4^3=128\), \(d(128)=8=2\cdot4\).  
+
+---
+
+#### Subcase \(a=1\)  
+
+Now \(3a+2=5\) and \(2^{a+1}=4\). Equation (2) becomes  
+\[
+5\prod (3b_i+1)=4\prod q_i^{b_i}. \tag{4}
+\]  
+Since the left side is divisible by \(5\), the right side must be divisible by \(5\). Because \(\gcd(5,4)=1\), the product \(\prod q_i^{b_i}\) must contain the prime \(5\). Let the exponent of \(5\) in \(m\) be \(e\ge 1\). Write \(m=5^e\cdot m'\) with \(5\nmid m'\). Then (4) rewrites as  
+\[
+5\cdot(3e+1)\prod_{q_i\neq5}(3b_i+1)=4\cdot5^e\prod_{q_i\neq5}q_i^{b_i}.
+\]  
+Dividing by \(5\) gives  
+\[
+(3e+1)\prod_{q_i\neq5}(3b_i+1)=4\cdot5^{e-1}\prod_{q_i\neq5}q_i^{b_i}. \tag{5}
+\]  
+Take the product of ratios for the primes different from \(5\):  
+\[
+Q:=\prod_{q_i\neq5}\frac{3b_i+1}{q_i^{b_i}}=\frac{4\cdot5^{e-1}}{3e+1}. \tag{6}
+\]  
+
+Again, by the lemma each factor in \(Q\) is at most \(4/3\), and as argued earlier, \(Q\le 4/3\). Thus  
+\[
+\frac{4\cdot5^{e-1}}{3e+1}\le\frac{4}{3}\quad\Longrightarrow\quad 5^{e-1}\le\frac{3e+1}{3}. \tag{7}
+\]  
+
+For \(e=1\): \(5^{0}=1\le\frac{4}{3}\) holds.  
+For \(e=2\): \(5\le\frac{7}{3}\approx2.33\) false.  
+For \(e\ge3\) the left side grows exponentially while the right side grows linearly, so (7) fails. Hence \(e=1\) is forced.  
+
+With \(e=1\), (6) gives  
+\[
+Q=\frac{4\cdot5^{0}}{3\cdot1+1}=1.
+\]  
+Thus  
+\[
+\prod_{q_i\neq5}\frac{3b_i+1}{q_i^{b_i}}=1\quad\Longrightarrow\quad \prod_{q_i\neq5}(3b_i+1)=\prod_{q_i\neq5}q_i^{b_i}.
+\]  
+This is again an equation of type (1) for the odd integer \(m'\) (the part of \(m\) not containing \(5\)). As before, the only possibility is that there are no such primes, i.e. \(m'=1\). Therefore \(m=5^1=5\) and \(k=2^1\cdot5=10\).  
+Check: \(2\cdot10^3=2000=2^4\cdot5^3\), \(d(2000)=5\cdot4=20=2\cdot10\).  
+
+---
+
+All possibilities have been examined. The positive integers \(k\) satisfying \(d(2k^3)=2k\) are exactly  
+
+\[
+\boxed{1,\ 4,\ 10}.
+\]
+
+**Solution 2 (Gemini-3.1-Pro)**
+
+
+To find all positive integers $k$ such that the number of positive divisors of $2k^3$ is $2k$, we start by using the Fundamental Theorem of Arithmetic. We can express the prime factorization of $k$ as
+\[ k = 2^a p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m} \]
+where $a \ge 0$ is a non-negative integer, $m \ge 0$ is the number of distinct odd prime factors, $p_1, p_2, \dots, p_m$ are distinct odd primes, and $a_1, a_2, \dots, a_m \ge 1$ are positive integers.
+
+Using this representation, we can write $2k^3$ as:
+\[ 2k^3 = 2 \left( 2^a p_1^{a_1} \cdots p_m^{a_m} \right)^3 = 2^{3a+1} p_1^{3a_1} \cdots p_m^{3a_m} \]
+The number of positive divisors of a positive integer $N = p_1^{e_1} \cdots p_k^{e_k}$ is given by the well-known divisor function formula $d(N) = (e_1+1) \cdots (e_k+1)$. Applying this to $2k^3$, we get:
+\[ d(2k^3) = (3a+2)(3a_1+1)(3a_2+1) \cdots (3a_m+1) \]
+We are given that $d(2k^3) = 2k$. Substituting $k$ and $d(2k^3)$ into this equation gives:
+\[ (3a+2) \prod_{i=1}^m (3a_i+1) = 2^{a+1} \prod_{i=1}^m p_i^{a_i} \quad \text{--- (Equation 1)} \]
+
+**Step 1: Exclusion of the prime factor 3**
+Assume for the sake of contradiction that one of the odd prime factors of $k$ is $3$ (say, $p_1 = 3$). 
+Because $a_1 \ge 1$, the right-hand side of Equation 1 is divisible by $3^{a_1}$, making it a multiple of $3$.
+This implies that the left-hand side, $(3a+2) \prod_{i=1}^m (3a_i+1)$, must also be a multiple of $3$. 
+However, let's examine the factors on the left-hand side modulo $3$:
+- $3a+2 \equiv 2 \pmod 3$, so it is not divisible by $3$.
+- For each $i$ such that $1 \le i \le m$, $3a_i+1 \equiv 1 \pmod 3$, so none of these terms are divisible by $3$.
+
+Since $3$ is a prime number, a product of integers that are not divisible by $3$ cannot be divisible by $3$. Therefore, the left-hand side is not divisible by $3$, establishing a direct contradiction. 
+Our assumption must be false, meaning $k$ cannot have $3$ as a prime factor. Thus, all odd prime factors of $k$ must be strictly greater than $3$ (i.e., $p_i \ge 5$ for all $1 \le i \le m$).
+
+**Step 2: Bounding the Components**
+We can rearrange Equation 1 by grouping the factors corresponding to the powers of $2$ and the odd primes:
+\[ \frac{3a+2}{2^{a+1}} \prod_{i=1}^m \frac{3a_i+1}{p_i^{a_i}} = 1 \]
+Let us define $g(a) = \frac{3a+2}{2^{a+1}}$ for $a \ge 0$, and $f(p, x) = \frac{3x+1}{p^x}$ for primes $p \ge 5$ and integers $x \ge 1$. The equation becomes:
+\[ g(a) \prod_{i=1}^m f(p_i, a_i) = 1 \quad \text{--- (Equation 2)} \]
+
+Let's find the absolute maximum value of $f(p, x)$ for $p \ge 5$ and $x \ge 1$.
+Since $p \ge 5$, we know $p^x \ge 5^x$, which gives $f(p, x) \le \frac{3x+1}{5^x}$.
+- For $x=1$, $\frac{3(1)+1}{5^1} = \frac{4}{5} = 0.8$.
+- To see that this is the maximum, we check the ratio of consecutive terms for $x \ge 1$:
+\[ \frac{\frac{3(x+1)+1}{5^{x+1}}}{\frac{3x+1}{5^x}} = \frac{3x+4}{5(3x+1)} = \frac{3x+4}{15x+5} \]
+For $x \ge 1$, it is evident that $15x+5 > 3x+4$, meaning the ratio is strictly less than $1$. Thus, the sequence strictly decreases, and the maximum is securely achieved at $x=1$.
+Therefore, $f(p, x) \le \frac{4}{5}$, with equality if and only if $p=5$ and $x=1$.
+
+Next, let's analyze $g(a)$ for integers $a \ge 0$:
+- $g(0) = \frac{2}{2} = 1$
+- $g(1) = \frac{5}{4} = 1.25$
+- $g(2) = \frac{8}{8} = 1$
+For $a \ge 1$, let's look at the difference between consecutive terms:
+\[ g(a+1) - g(a) = \frac{3(a+1)+2}{2^{a+2}} - \frac{3a+2}{2^{a+1}} = \frac{3a+5 - 2(3a+2)}{2^{a+2}} = \frac{1 - 3a}{2^{a+2}} \]
+Because $1 - 3a < 0$ for all $a \ge 1$, $g(a)$ strictly decreases for $a \ge 1$. Thus, the absolute maximum of $g(a)$ is $g(1) = \frac{5}{4}$.
+
+**Step 3: Case Analysis to Find All Solutions**
+We evaluate Equation 2 based on $m$, the number of distinct odd prime factors of $k$.
+
+- **Case 1: $m \ge 2$**
+Using our maximum bounds, the product of the sequence terms is at most:
+\[ g(a) \prod_{i=1}^m f(p_i, a_i) \le \frac{5}{4} \left( \frac{4}{5} \right)^m = \frac{5}{4} \cdot \frac{16}{25} \left( \frac{4}{5} \right)^{m-2} = \frac{4}{5} \left( \frac{4}{5} \right)^{m-2} \]
+Because $m \ge 2$, the maximum possible value is $\frac{4}{5} < 1$. So no solution can exist for $m \ge 2$.
+
+- **Case 2: $m = 1$**
+Equation 2 simplifies to $g(a) f(p_1, a_1) = 1$.
+Since $f(p_1, a_1) \le \frac{4}{5}$, we must have $g(a) \ge \frac{5}{4}$. The only evaluated integer $a$ fulfilling this condition is $a=1$, dictating exactly $g(1) = \frac{5}{4}$.
+To perfectly satisfy the equation, we are forced to have $f(p_1, a_1) = \frac{4}{5}$. As established earlier, this equality holds true if and only if $p_1=5$ and $a_1=1$. 
+This provides the solution $k = 2^1 \cdot 5^1 = 10$.
+
+- **Case 3: $m = 0$**
+This reflects $k$ having no odd prime factors, so $k = 2^a$. The empty product defaults to $1$, leaving Equation 2 as $g(a) = 1$.
+Looking at the behavior of $g(a)$:
+$g(0) = 1 \implies a = 0 \implies k = 2^0 = 1$.
+$g(1) = 1.25 \neq 1$.
+$g(2) = 1 \implies a = 2 \implies k = 2^2 = 4$.
+Since $g(a)$ is strictly decreasing for $a \ge 2$, $g(a) < 1$ for all $a \ge 3$, leaving no other solutions.
+This provides the solutions $k = 1$ and $k = 4$.
+
+**Verification:**
+- For $k=1$: $2k^3 = 2$. Divisors of 2 are $1, 2$, giving $d(2) = 2$. And $2k = 2(1) = 2$.
+- For $k=4$: $2k^3 = 2(64) = 128 = 2^7$. Divisors of $2^7$ is $7+1 = 8$. And $2k = 2(4) = 8$.
+- For $k=10$: $2k^3 = 2(1000) = 2000 = 2^4 \cdot 5^3$. Number of divisors is $(4+1)(3+1) = 20$. And $2k = 2(10) = 20$.
+
+All solutions hold perfectly without error.
+
+\boxed{1, 4, 10}
+
+**Explanation**: We point to the parts of each solution that were perceived as verbose by the annotators. GLM-5’s approach is clear, and overall there are seemingly not many places to cut, but includes some comments on the given step that are not really necessary, i.e. restating the statement, or long explanations such as “This is again an equation of type (1) for the odd integer (m') (the part of (m) not containing (5))”. Some comments can be considered overly long, e.g., "Their product is $1$, so at least one $r_i \ge 1$ (otherwise product $<1$). Moreover $r_i=1$ would imply... Hence some $r_i>1$.", and the quantity $\prod \frac{3\cdot+1}{\text{prime}^{\cdot}}$ is introduced as $r_i$ in Case I, renamed $P$ in Case II, and renamed $Q$ in the subcase $a=1$. 
+
+In Solution 2, the redundancy is concentrated in framing: the exact same argument (FTA plus the number-of-divisors count) appears much more briefly in Solution 1 than in Solution 2, and it is surrounded by commentary sentences ("Let's find the absolute maximum value...", "the maximum is securely achieved", "All solutions hold perfectly without error"). Further, Solution 1 doesn't explain well its linear vs exponential trend inequalities.
+
+Which of these weighs more is a subjective judgment call. This pair is precisely an instance of the failure mode we report in App. B.2, where annotators agreed post hoc that different solutions are verbose in different ways and that local perception of redundancy is unreliable. We selected it for the repository because it illustrates that failure mode, not because it is representative of the full annotation set (see R5 and R6 below).
+
+
+# Ex. 2: Human disagreement on Cognitive Ease
+
+**Problem**: `shortlist_2025_6`
+**Problem Statement**: Jerry places at most one rook in each cell of a $2025 \times 2025$ grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them. Determine the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.
+
+**Solution 1 (GLM-5):**
+
+
+We are given a $2025 \times 2025$ grid. A rook attacks another rook if they are in the same row or column and there is no other rook between them. For a rook, the number of rooks it attacks is the sum of the number of nearest rooks to the left and right in its row, plus the number of nearest rooks above and below in its column. Thus a rook can attack at most four others. The problem asks for the maximum number of rooks that can be placed (at most one per cell) such that no rook attacks four others, i.e., every rook attacks at most three others.
+
+**Reformulation**  
+For a given placement, consider a rook at $(i,j)$.  
+- In its row $i$, let the rooks be at columns $c_1<c_2<\dots<c_m$. The rook at column $c_k$ attacks the rooks at $c_{k-1}$ (if $k>1$) and $c_{k+1}$ (if $k<m$). Thus it attacks two rooks in its row exactly when it is not the leftmost nor the rightmost in that row.  
+- Similarly, in its column $j$, it attacks two rooks exactly when it is not the topmost nor the bottommost in that column.
+
+Consequently, a rook attacks four others precisely when it is **not** the leftmost nor the rightmost in its row **and** not the topmost nor the bottommost in its column. The condition "no rook attacks four others" is therefore equivalent to: *every rook is either a leftmost or rightmost rook in its row, or a topmost or bottommost rook in its column (or both)*.
+
+We call such rooks *protected* in at least one direction.
+
+**Notation**  
+Let $n=2025\ge 2$.  
+Define the set of *border cells* $B$ as those cells with row $=1$ or row $=n$ or column $=1$ or column $=n$. The number of border cells is $|B|=2n+2(n-2)=4n-4$.  
+Define the set of *interior cells* $I$ as the remaining cells: rows $2,\dots,n-1$ and columns $2,\dots,n-1$. $|I|=(n-2)^2$.
+
+For a placement $S$ (set of occupied cells), write $X=S\cap I$ (interior rooks) and $Y=S\cap B$ (border rooks). We aim to prove $|S|\le 4n-4$.
+
+**Key Lemma**  
+In any admissible placement, there is an injective mapping from $X$ to $B\setminus S$ (empty border cells). Consequently $|X|\le |B\setminus S|$.
+
+*Proof of Lemma.*  
+Take any interior rook $r=(i,j)\in X$ ($2\le i\le n-1$, $2\le j\le n-1$). By the condition, $r$ is either a row‑extreme (leftmost or rightmost in its row) or a column‑extreme (topmost or bottommost in its column). We assign an empty border cell $f(r)\in B\setminus S$ according to the following rule:
+
+- If $r$ is a row‑extreme (i.e., leftmost or rightmost in row $i$), then:  
+  * If it is the leftmost rook in row $i$, let $f(r)=(i,1)$.  
+  * If it is the rightmost rook in row $i$, let $f(r)=(i,n)$.  
+  (If it is the only rook in its row, it is both leftmost and rightmost; we may choose e.g. the leftmost assignment.)
+
+  Because $j\ge 2$, $j\neq 1$; if $r$ is leftmost, there is no rook in row $i$ with column $<j$, hence $(i,1)$ is empty. Similarly, if $r$ is rightmost, $(i,n)$ is empty.
+
+- Otherwise, $r$ is not a row‑extreme, so it must be a column‑extreme (topmost or bottommost in column $j$). Then:  
+  * If it is the topmost rook in column $j$, let $f(r)=(1,j)$.  
+  * If it is the bottommost rook in column $j$, let $f(r)=(n,j)$.  
+  (If it is the only rook in its column, we may choose the topmost assignment.)
+
+  Because $i\ge 2$, $i\neq 1$; if $r$ is topmost, there is no rook in column $j$ with row $<i$, so $(1,j)$ is empty. Similarly, bottommost forces $(n,j)$ empty.
+
+Now we show that $f$ is injective.
+
+- If two different interior rooks both fall under the row‑extreme case, their images are of the form $(i,1)$ or $(i,n)$ with $i$ being their respective rows. Since rows are distinct for rooks in different rows, and for two rooks in the same row one would be assigned $(i,1)$ and the other $(i,n)$, the images are distinct.
+- If both fall under the column‑extreme case, their images are $(1,j)$ or $(n,j)$ with $j$ being their columns. Columns are distinct for rooks in different columns, and for two rooks in the same column one gets $(1,j)$ and the other $(n,j)$, so images are distinct.
+- If one rook is assigned via the row‑extreme case and the other via the column‑extreme case, their images belong to disjoint sets: row‑extreme images lie on the left or right border with row $i\in\{2,\dots,n-1\}$ (so column $1$ or $n$), while column‑extreme images lie on the top or bottom border with column $j\in\{2,\dots,n-1\}$ (so row $1$ or $n$). The only possible overlap would be at the four corners, but none of these images is a corner because $i\notin\{1,n\}$ in the first case and $j\notin\{1,n\}$ in the second. Hence the images are distinct.
+
+Thus $f$ is injective, and each $f(r)$ is indeed an empty border cell. Therefore $|X|\le |B\setminus S|$. ∎
+
+**Upper bound**  
+We have  
+\[
+|S| = |Y| + |X| = (|B| - |B\setminus S|) + |X| \le |B| = 4n-4.
+\]
+Hence any admissible placement contains at most $4n-4$ rooks.
+
+**Construction achieving the bound**  
+Place rooks on every border cell. This uses exactly $4n-4$ rooks. We verify that no rook attacks four others:
+
+- A rook on the top row (row $1$):  
+  * Its row is fully occupied, so it has both a left and a right neighbor → $2$ attacks from the row.  
+  * If it is not a corner (i.e., column $j$ with $2\le j\le n-1$), its column contains rooks only at rows $1$ and $n$ (since interior rows have no rook in that column), so it has exactly one neighbor in the column (the rook at $(n,j)$) → $1$ attack. Total $3$.  
+  * If it is a corner, say $(1,1)$, it is leftmost in its row (only right neighbor) and topmost in its column (only down neighbor) → $2$ attacks.
+
+- A rook on the bottom row is symmetric.
+
+- A rook on the left border (column $1$) with $2\le i\le n-1$:  
+  * Its row $i$ contains exactly two rooks: at columns $1$ and $n$ (since the row is not top/bottom). Thus it is leftmost, so it has one neighbor in the row (the rook at $(i,n)$) → $1$ attack.  
+  * Its column $1$ is fully occupied, so it has both an up and a down neighbor → $2$ attacks. Total $3$.
+
+- A rook on the right border is symmetric.
+
+Thus every rook attacks at most three others, satisfying the condition.
+
+Therefore the maximum number of rooks for $n=2025$ is $4\cdot2025-4 = 8096$.
+
+\[
+\boxed{8096}
+\]
+
+**Solution 2 (GPT-5.4):**
+
+ 
+\[
+\textbf{Claim: The maximum number of rooks is }8096.
+\]
+
+Let \(m\) be the number of rooks in a valid arrangement.  
+Also let
+
+- \(r\) = number of rows containing at least one rook,
+- \(c\) = number of columns containing at least one rook.
+
+For each rook \(X\), define:
+
+- \(h(X)\) = number of rooks attacked by \(X\) in its row,
+- \(v(X)\) = number of rooks attacked by \(X\) in its column.
+
+Since a rook can attack at most one rook to the left and at most one rook to the right, we have \(h(X)\in\{0,1,2\}\). Similarly \(v(X)\in\{0,1,2\}\).  
+The condition in the problem says that no rook attacks \(4\) others, so
+\[
+h(X)+v(X)\le 3
+\quad\text{for every rook }X.
+\]
+
+## Step 1: Sum of horizontal and vertical attacks
+
+Consider one occupied row containing \(k\) rooks.
+
+If we list those \(k\) rooks from left to right, then the only attacking pairs in that row are the \(k-1\) consecutive pairs. Each such pair contributes \(1\) to the horizontal attack count of each of its two rooks, so the total horizontal contribution of this row is
+\[
+2(k-1).
+\]
+Summing over all occupied rows,
+\[
+\sum_X h(X)=2(m-r).
+\]
+Exactly the same argument for columns gives
+\[
+\sum_X v(X)=2(m-c).
+\]
+
+## Step 2: Define a “defect”
+
+For each rook \(X\), define
+\[
+\delta(X)=3-h(X)-v(X).
+\]
+Because \(h(X)+v(X)\le 3\), we have \(\delta(X)\ge 0\).
+
+Now sum over all rooks:
+\[
+\sum_X \delta(X)
+= \sum_X \bigl(3-h(X)-v(X)\bigr)
+= 3m-\sum_X h(X)-\sum_X v(X).
+\]
+Substituting the formulas from Step 1,
+\[
+\sum_X \delta(X)
+=3m-2(m-r)-2(m-c)
+=2r+2c-m.
+\]
+So
+\[
+\sum_X \delta(X)=2r+2c-m. \tag{1}
+\]
+
+## Step 3: The top and bottom occupied rows force at least \(4\) total defect
+
+If all rooks lie in a single row, then
+\[
+m\le 2025<8096,
+\]
+so such an arrangement cannot beat \(8096\). Thus, for the upper bound, we may assume there are at least two occupied rows.
+
+Let \(T\) be the top occupied row and \(B\) the bottom occupied row. These are distinct.
+
+Take a rook \(X\) in the top occupied row \(T\). There is no rook above it, so in its column it can attack at most one rook. Hence
+\[
+v(X)\le 1.
+\]
+Therefore
+\[
+\delta(X)=3-h(X)-v(X)\ge 2-h(X).
+\]
+Summing over all rooks \(X\) in row \(T\),
+\[
+\sum_{X\in T}\delta(X)\ge \sum_{X\in T}(2-h(X)).
+\]
+Suppose row \(T\) contains \(k\) rooks. As above, the total horizontal attack count in that row is \(2(k-1)\). Hence
+\[
+\sum_{X\in T}(2-h(X))
+=2k-2(k-1)=2.
+\]
+So
+\[
+\sum_{X\in T}\delta(X)\ge 2.
+\]
+
+By exactly the same argument, the bottom occupied row \(B\) also contributes at least \(2\):
+\[
+\sum_{X\in B}\delta(X)\ge 2.
+\]
+
+Since \(T\) and \(B\) are different rows, these contributions come from different rooks, so altogether
+\[
+\sum_X \delta(X)\ge 4.
+\]
+Using (1),
+\[
+2r+2c-m\ge 4,
+\]
+hence
+\[
+m\le 2r+2c-4.
+\]
+Finally, since \(r\le 2025\) and \(c\le 2025\),
+\[
+m\le 2\cdot 2025+2\cdot 2025-4=8096.
+\]
+
+So no valid arrangement can have more than \(8096\) rooks.
+
+## Step 4: Construction with \(8096\) rooks
+
+Place rooks in every boundary cell of the \(2025\times 2025\) grid.
+
+The number of boundary cells is
+\[
+4\cdot 2025-4=8096.
+\]
+
+Now check the condition:
+
+- A corner rook attacks exactly \(2\) rooks: one along its row and one along its column.
+- A non-corner rook on the top or bottom edge attacks \(3\) rooks: the nearest rook to its left, the nearest rook to its right, and the rook on the opposite edge in its column.
+- A non-corner rook on the left or right edge also attacks \(3\) rooks: the nearest rook above, the nearest rook below, and the rook on the opposite edge in its row.
+
+Thus no rook attacks \(4\) others.
+
+So \(8096\) rooks are achievable.
+
+Therefore the maximum possible number is
+\[
+\boxed{8096}.
+\]
+
+**Explanation:** The first solution uses a more rigorous and formal notation, especially in the proof of its lemma, and the definition of border and interior cells, while the second solution presents an overall more abstract idea that requires some internal visualization. Both cases can be deemed more complex, depending on the reader’s background, or their experience.
+
+# Ex. 3: Human disagreement on Computational Complexity
+
+**Problem:** `imo-bench_combinatorics-011`
+**Problem statement:** A soccer player named Ronaldo stands on a point on a circle with circumference $1$. Given an infinite sequence of positive real numbers $c_1, c_2, c_3, \dots$, Ronaldo successively runs distances $c_1, c_2, c_3, \dots$ around the circle, each time choosing to runs either clockwise or counterclockwise.
+
+Determine the largest constant $C > 0$ with the following property: for every sequence of positive real numbers $c_1, c_2, c_3, \dots$ with $c_i < C$ for all $i$, Ronaldo can (after studying the sequence) ensure that there is some point on the circle that it will never visit or run across.
+
+**Solution 1 (Gemini-3-Flash):**
+
+
+To determine the largest constant $C > 0$ such that for every sequence of positive real numbers $c_1, c_2, c_3, \dots$ with $c_i < C$, Ronaldo can choose directions $s_i \in \{-1, 1\}$ such that he never visits or runs across some point $p$ on the circle, we analyze the conditions under which he can avoid covering the entire circle.
+
+Let the circle be represented by the interval $[0, 1)$ with $0 \equiv 1$. Ronaldo starts at $x_0 = 0$. In each step $i$, he moves a distance $c_i$ to $x_i = (x_{i-1} + s_i c_i) \pmod 1$, covering the arc $L_i$ of length $c_i$ between $x_{i-1}$ and $x_i$.
+
+1. Suppose $C > 1/2$. We show that Ronaldo can be forced to cover the circle. Let $c_1 = c_2 = 1/2 + \epsilon$ for some $0 < \epsilon < C - 1/2$. 
+   - In step 1, Ronaldo moves distance $c_1$, covering the arc $L_1$ of length $1/2 + \epsilon$. Assume without loss of generality that he moves clockwise, so $x_1 = 1/2 + \epsilon$ and $L_1 = [0, 1/2 + \epsilon]$.
+   - In step 2, Ronaldo moves distance $c_2 = 1/2 + \epsilon$.
+     - If he moves clockwise ($s_2 = 1$), the arc $L_2$ is $[1/2 + \epsilon, (1/2 + \epsilon) + (1/2 + \epsilon)] = [1/2 + \epsilon, 1 + 2\epsilon]$. In $\mathbb{R}/\mathbb{Z}$, this covers $[1/2 + \epsilon, 1] \cup [0, 2\epsilon]$. Combined with $L_1$, the covered set is $[0, 1/2 + \epsilon] \cup [1/2 + \epsilon, 1] = [0, 1]$, the entire circle.
+     - If he moves counter-clockwise ($s_2 = -1$), the arc $L_2$ is the set of points obtained by starting at $1/2 + \epsilon$ and moving $1/2 + \epsilon$ backwards. The complement of $L_1$ is the arc $(1/2 + \epsilon, 1)$ of length $1 - (1/2 + \epsilon) = 1/2 - \epsilon$. Since $c_2 = 1/2 + \epsilon > 1/2 - \epsilon$, the arc $L_2$ covers the entire complement of $L_1$.
+   In either case, Ronaldo covers the entire circle after two steps. Thus, $C$ cannot be greater than $1/2$.
+
+2. Suppose $c_i < 1/2$ for all $i$. We show that Ronaldo can always avoid some point. He can use the greedy strategy: choose $s_i = -\text{sgn}(x_{i-1})$, where $x_{i-1}$ is represented in the range $(-1/2, 1/2]$.
+   - Starting with $x_0 = 0$, we have $x_1 = s_1 c_1 \in (-1/2, 1/2)$ since $c_1 < 1/2$.
+   - For each step $n$, if $x_{n-1} \in [0, 1/2)$, then $s_n = -1$ gives $x_n = x_{n-1} - c_n$. Since $0 < c_n < 1/2$, $x_n \in (0-1/2, 1/2-0) = (-1/2, 1/2)$.
+   - Similarly, if $x_{n-1} \in (-1/2, 0)$, then $s_n = 1$ gives $x_n = x_{n-1} + c_n \in (-1/2, 1/2)$.
+   - This strategy ensures that $x_n \in (-1/2, 1/2)$ for all $n$. Each arc $L_n$ has length $c_n < 1/2$ and connects $x_{n-1}$ and $x_n$, both of which are in the open interval $(-1/2, 1/2)$. Because $c_n$ is the shorter arc distance (since $c_n < 1/2$), the arc $L_n$ is contained entirely within the interval $(-1/2, 1/2)$.
+   - Consequently, the union of all arcs $\bigcup_{i=1}^\infty L_i$ is contained in $(-1/2, 1/2)$, meaning the point $1/2$ is never visited or run across.
+
+Thus, the largest such constant $C$ is $1/2$.
+
+\boxed{1/2}
+
+**Solution 2 (GPT-5.4):**
+
+ 
+We claim that the largest possible constant is \( \frac12 \).
+
+First, we use a convenient reformulation.
+
+If there is a point \(P\) on the circle that Ronaldo never visits and never runs across, then we may **cut** the circle at \(P\). What remains is an open interval of length \(1\), which we identify with \((0,1)\). Ronaldo’s starting point becomes some \(x_0\in(0,1)\), and after each move his new position is
+
+\[
+x_n=x_{n-1}\pm c_n,
+\]
+
+because in the cut interval, “clockwise” and “counterclockwise” simply mean moving right or left by distance \(c_n\). Also, every segment he runs is contained in \((0,1)\), since touching an endpoint would mean touching \(P\).
+
+Conversely, if we can choose signs so that a walk with steps \(c_1,c_2,\dots\) stays entirely inside \((0,1)\), then after gluing the endpoints back together we get a strategy on the circle that avoids the cut point \(P\).
+
+So the problem is equivalent to this:
+
+> Find the largest \(C\) such that for every sequence \(0<c_n<C\), one can choose left/right moves so that the whole walk stays inside an open interval of length \(1\).
+
+---
+
+## 1. Lower bound: \(C=\frac12\) works
+
+Assume \(0<c_n<\frac12\) for all \(n\).
+
+Choose \(P\) to be the point opposite Ronaldo’s starting point on the circle. After cutting at \(P\), the start is at the midpoint of the interval, so
+
+\[
+x_0=\frac12.
+\]
+
+Now use the following rule:
+
+- if \(x_{n-1}<\frac12\), move to the right, so \(x_n=x_{n-1}+c_n\);
+- if \(x_{n-1}>\frac12\), move to the left, so \(x_n=x_{n-1}-c_n\);
+- if \(x_{n-1}=\frac12\), either choice is fine.
+
+We check that this always keeps \(x_n\) inside \((0,1)\).
+
+### Case 1: \(x_{n-1}\le \frac12\)
+
+Then we move right:
+
+\[
+x_n=x_{n-1}+c_n.
+\]
+
+Since \(x_{n-1}>0\) and \(c_n>0\), we have \(x_n>0\). Also,
+
+\[
+x_n=x_{n-1}+c_n<\frac12+\frac12=1.
+\]
+
+So \(x_n\in(0,1)\).
+
+### Case 2: \(x_{n-1}\ge \frac12\)
+
+Then we move left:
+
+\[
+x_n=x_{n-1}-c_n.
+\]
+
+Because \(x_{n-1}\ge \frac12\) and \(c_n<\frac12\),
+
+\[
+x_n=x_{n-1}-c_n>\frac12-\frac12=0.
+\]
+
+Also \(x_n<1\) because \(x_{n-1}<1\). Thus again \(x_n\in(0,1)\).
+
+So by induction, every position \(x_n\) stays in \((0,1)\). Hence the cut point \(P\) is never visited and never crossed.
+
+Therefore every sequence with all terms \(<\frac12\) is avoidable, so
+
+\[
+C\ge \frac12.
+\]
+
+---
+
+## 2. Upper bound: no \(C>\frac12\) can work
+
+Now suppose \(C>\frac12\). We will construct a sequence with all terms \(<C\) for which avoiding a point is impossible.
+
+Choose numbers \(a,b\) such that
+
+\[
+\frac12<b<a<\min(C,1).
+\]
+
+Consider the infinite sequence
+
+\[
+c_1=a,\quad c_2=b,\quad c_3=a,\quad c_4=b,\quad \dots
+\]
+
+Assume for contradiction that Ronaldo can avoid some point \(P\). Cut the circle at \(P\). Then his positions satisfy
+
+\[
+x_n=x_{n-1}+\varepsilon_n c_n,
+\qquad \varepsilon_n\in\{+1,-1\},
+\]
+
+with all \(x_n\in(0,1)\).
+
+We now show this is impossible.
+
+### Claim: consecutive signs must be opposite
+
+Suppose for some \(n\) we had \(\varepsilon_{n+1}=\varepsilon_n\). Then
+
+\[
+x_{n+1}-x_{n-1}
+=(x_n-x_{n-1})+(x_{n+1}-x_n)
+=\varepsilon_n c_n+\varepsilon_{n+1} c_{n+1}.
+\]
+
+Since the signs are equal, this becomes
+
+\[
+x_{n+1}-x_{n-1}=\pm(c_n+c_{n+1}).
+\]
+
+But \(c_n,c_{n+1}\) are one \(a\) and one \(b\), so
+
+\[
+|x_{n+1}-x_{n-1}|=a+b.
+\]
+
+Because \(a>\frac12\) and \(b>\frac12\),
+
+\[
+a+b>1.
+\]
+
+On the other hand, both \(x_{n-1}\) and \(x_{n+1}\) lie in the interval \((0,1)\), so their distance must be strictly less than \(1\). Contradiction.
+
+Therefore for every \(n\),
+
+\[
+\varepsilon_{n+1}=-\varepsilon_n.
+\]
+
+So the signs must alternate forever.
+
+---
+
+### What alternating signs imply
+
+There are only two possibilities.
+
+#### Possibility 1: \(+,-,+,-,\dots\)
+
+Then every pair of moves changes the position by
+
+\[
+a-b>0.
+\]
+
+Indeed,
+
+\[
+x_2=x_0+a-b,\qquad
+x_4=x_2+a-b=x_0+2(a-b),
+\]
+
+and in general
+
+\[
+x_{2k}=x_0+k(a-b).
+\]
+
+Since \(a-b>0\), choose \(k\) so large that
+
+\[
+k(a-b)>1-x_0.
+\]
+
+Then \(x_{2k}>1\), impossible because \(x_{2k}\in(0,1)\).
+
+#### Possibility 2: \(-,+,-,+,\dots\)
+
+Now every pair of moves changes the position by
+
+\[
+-(a-b)<0,
+\]
+
+so similarly
+
+\[
+x_{2k}=x_0-k(a-b).
+\]
+
+Choose \(k\) so large that
+
+\[
+k(a-b)>x_0.
+\]
+
+Then \(x_{2k}<0\), again impossible.
+
+Both possibilities lead to contradictions.
+
+So for this sequence \(a,b,a,b,\dots\), no point can be left untouched. Hence the required property fails for every \(C>\frac12\). Therefore
+
+\[
+C\le \frac12.
+\]
+
+---
+
+Since we proved both \(C\ge \frac12\) and \(C\le \frac12\), the largest constant is
+
+\[
+\boxed{\frac12}.
+\]
+
+**Explanation:** In Solution 1, while the notation is much sparser, and clearer, verifying each claim about the inequalities may take the reader significantly longer. On the other hand, even though Solution 2 introduces more symbols and performs more algebraic steps as a whole, they are overall simpler to parse.
+
+# Ex. 4: GPT-OSS-120B disagrees with human judges on pairwise comparisons
+
+**Problem:** `aime_2025_27`
+
+**Problem statement:** Let $A_1 A_2 A_3 \ldots A_{11}$ be an $11$-sided non-convex simple polygon with the following properties:
+
+\begin{itemize}
+\item For every integer $2 \le i \le 10$,  the area of $\triangle A_i A_{1} A_{i+1}$  is equal to $1$.
+\item For every integer $2 \le i \le 10$, $\cos(\angle A_i A_{1} A_{i+1}) = \frac{12}{13}$.
+\item The perimeter of the $11$-gon $A_1 A_2 A_3 \dots A_{11}$  is equal to $20$.
+\end{itemize}
+
+Then $A_1 A_2 + A_1 A_{11} = \frac{m \sqrt{n} - p}{q}$ where $m, n, p$, and $q$ are positive integers, $n$ is not divisible by the square of any prime, and no prime divides all of $m, p$, and $q$. Find $m + n + p + q$.
+
+**Solution 1 (Qwen-3.5-397B):**
+
+Let $x_i$ denote the length of the segment $A_1 A_i$ for $i = 1, 2, \dots, 11$. Since $A_1$ is a vertex, $x_1 = 0$. We are interested in finding the sum $x_2 + x_{11}$.
+
+**Step 1: Analyze the properties of the triangles**
+For every integer $2 \le i \le 10$, consider the triangle $\triangle A_i A_1 A_{i+1}$. Let $\alpha_i = \angle A_i A_1 A_{i+1}$. We are given that $\cos(\alpha_i) = \frac{12}{13}$. Since $\alpha_i$ is an angle in a triangle, $\sin(\alpha_i) > 0$, so:
+$$ \sin(\alpha_i) = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13} $$
+The area of $\triangle A_i A_1 A_{i+1}$ is given as $1$. Using the sine area formula:
+$$ \text{Area} = \frac{1}{2} x_i x_{i+1} \sin(\alpha_i) = 1 $$
+Substituting $\sin(\alpha_i) = \frac{5}{13}$:
+$$ \frac{1}{2} x_i x_{i+1} \left(\frac{5}{13}\right) = 1 \implies x_i x_{i+1} = \frac{26}{5} $$
+This relation holds for $i = 2, 3, \dots, 10$.
+
+**Step 2: Determine the sequence of lengths $x_i$**
+From the relation $x_i x_{i+1} = \frac{26}{5}$, we can see that $x_{i+2} = \frac{26/5}{x_{i+1}} = x_i$. Thus, the sequence of lengths $x_2, x_3, \dots, x_{11}$ is periodic with period 2.
+Let $x_2 = a$ and $x_3 = b$. Then $x_4 = a, x_5 = b$, and so on.
+The sequence is $a, b, a, b, a, b, a, b, a, b$.
+Specifically, $x_2 = a$ and $x_{11} = b$.
+We have the product $ab = \frac{26}{5}$. We need to find the sum $S = a + b$.
+
+**Step 3: Express the perimeter in terms of $a$ and $b$**
+The perimeter of the polygon is the sum of the lengths of its edges:
+$$ P = A_1 A_2 + \sum_{i=2}^{10} A_i A_{i+1} + A_{11} A_1 = a + \sum_{i=2}^{10} A_i A_{i+1} + b $$
+There are 9 terms in the summation. Let $L_i = A_i A_{i+1}$. By the Law of Cosines on $\triangle A_i A_1 A_{i+1}$:
+$$ L_i^2 = x_i^2 + x_{i+1}^2 - 2 x_i x_{i+1} \cos(\alpha_i) $$
+Since $x_i x_{i+1} = \frac{26}{5}$ and $\cos(\alpha_i) = \frac{12}{13}$, the term $2 x_i x_{i+1} \cos(\alpha_i)$ is constant:
+$$ 2 \left(\frac{26}{5}\right) \left(\frac{12}{13}\right) = \frac{48}{5} $$
+Thus, $L_i^2 = x_i^2 + x_{i+1}^2 - \frac{48}{5}$.
+For any $i$, the pair $(x_i, x_{i+1})$ is either $(a, b)$ or $(b, a)$. In both cases, $x_i^2 + x_{i+1}^2 = a^2 + b^2$. Therefore, all edge lengths $A_i A_{i+1}$ are equal. Let $L = \sqrt{a^2 + b^2 - \frac{48}{5}}$.
+The perimeter equation becomes:
+$$ 20 = a + b + 9L = S + 9\sqrt{a^2 + b^2 - \frac{48}{5}} $$
+Substituting $a^2 + b^2 = (a+b)^2 - 2ab = S^2 - \frac{52}{5}$:
+$$ 20 = S + 9\sqrt{S^2 - \frac{52}{5} - \frac{48}{5}} = S + 9\sqrt{S^2 - \frac{100}{5}} = S + 9\sqrt{S^2 - 20} $$
+
+**Step 4: Solve for $S$**
+Rearranging the equation:
+$$ 9\sqrt{S^2 - 20} = 20 - S $$
+Squaring both sides (noting that $20-S \ge 0$):
+$$ 81(S^2 - 20) = (20 - S)^2 $$
+$$ 81S^2 - 1620 = 400 - 40S + S^2 $$
+$$ 80S^2 + 40S - 2020 = 0 $$
+Dividing by 20:
+$$ 4S^2 + 2S - 101 = 0 $$
+Using the quadratic formula for the positive root (since $S > 0$):
+$$ S = \frac{-2 + \sqrt{2^2 - 4(4)(-101)}}{2(4)} = \frac{-2 + \sqrt{4 + 1616}}{8} = \frac{-2 + \sqrt{1620}}{8} $$
+Simplifying $\sqrt{1620} = \sqrt{324 \times 5} = 18\sqrt{5}$:
+$$ S = \frac{-2 + 18\sqrt{5}}{8} = \frac{9\sqrt{5} - 1}{4} $$
+
+**Step 5: Final Calculation**
+We have $A_1 A_2 + A_1 A_{11} = S = \frac{9\sqrt{5} - 1}{4}$.
+Comparing this to the form $\frac{m \sqrt{n} - p}{q}$, we identify:
+$m = 9$, $n = 5$, $p = 1$, $q = 4$.
+These satisfy the conditions: $n=5$ is square-free, and $\gcd(9, 1, 4) = 1$.
+The required sum is:
+$$ m + n + p + q = 9 + 5 + 1 + 4 = 19 $$
+
+\boxed{19}
+
+**Solution 2 (GPT-5.4):**
+
+
+\[
+\text{Let }x_i=A_1A_i \qquad (2\le i\le 11).
+\]
+
+For each \(i=2,3,\dots,10\), consider \(\triangle A_iA_1A_{i+1}\).
+
+Since
+\[
+\cos\angle A_iA_1A_{i+1}=\frac{12}{13},
+\]
+and this is an angle of a triangle, its sine is positive, so
+\[
+\sin\angle A_iA_1A_{i+1}
+=\sqrt{1-\left(\frac{12}{13}\right)^2}
+=\frac{5}{13}.
+\]
+
+Now use the area formula:
+\[
+1=\text{Area}(\triangle A_iA_1A_{i+1})
+=\frac12\,x_i x_{i+1}\sin\angle A_iA_1A_{i+1}
+=\frac12\,x_i x_{i+1}\cdot \frac{5}{13}.
+\]
+Hence
+\[
+x_i x_{i+1}=\frac{26}{5}\qquad (2\le i\le 10).
+\]
+
+So all consecutive products are equal:
+\[
+x_2x_3=x_3x_4=\cdots=x_{10}x_{11}=\frac{26}{5}.
+\]
+Because every \(x_i>0\), from
+\[
+x_i x_{i+1}=x_{i+1}x_{i+2}
+\]
+we may divide by \(x_{i+1}\) to get
+\[
+x_i=x_{i+2}.
+\]
+Therefore the distances from \(A_1\) alternate between two values. Set
+\[
+u=x_2=x_4=x_6=x_8=x_{10},\qquad
+v=x_3=x_5=x_7=x_9=x_{11}.
+\]
+Then
+\[
+uv=\frac{26}{5}.
+\]
+
+Now compute the side lengths \(A_iA_{i+1}\) for \(i=2,\dots,10\).  
+By the Law of Cosines in \(\triangle A_iA_1A_{i+1}\),
+\[
+A_iA_{i+1}^2=x_i^2+x_{i+1}^2-2x_ix_{i+1}\cdot \frac{12}{13}.
+\]
+Since \((x_i,x_{i+1})\) is always \((u,v)\) or \((v,u)\), all these nine side lengths are equal. Call their common length \(s\). Then
+\[
+s^2=u^2+v^2-2uv\cdot \frac{12}{13}.
+\]
+
+Let
+\[
+t=u+v=A_1A_2+A_1A_{11}.
+\]
+Because \(u^2+v^2=t^2-2uv\), we get
+\[
+s^2=t^2-2uv-\frac{24}{13}uv
+=t^2-\left(2+\frac{24}{13}\right)uv
+=t^2-\frac{50}{13}\cdot \frac{26}{5}.
+\]
+Now
+\[
+\frac{50}{13}\cdot \frac{26}{5}=20,
+\]
+so
+\[
+s^2=t^2-20.
+\]
+
+Next use the perimeter condition. The perimeter is
+\[
+A_1A_2+(A_2A_3+\cdots +A_{10}A_{11})+A_{11}A_1
+= u+9s+v=t+9s.
+\]
+Since the perimeter is \(20\),
+\[
+t+9s=20,
+\]
+so
+\[
+s=\frac{20-t}{9}.
+\]
+
+Substitute this into \(s^2=t^2-20\):
+\[
+\left(\frac{20-t}{9}\right)^2=t^2-20.
+\]
+Multiply both sides by \(81\):
+\[
+(20-t)^2=81(t^2-20).
+\]
+Expand:
+\[
+400-40t+t^2=81t^2-1620.
+\]
+So
+\[
+80t^2+40t-2020=0.
+\]
+Divide by \(20\):
+\[
+4t^2+2t-101=0.
+\]
+Now solve:
+\[
+t=\frac{-2\pm\sqrt{2^2-4\cdot 4\cdot (-101)}}{8}
+=\frac{-2\pm\sqrt{4+1616}}{8}
+=\frac{-2\pm\sqrt{1620}}{8}.
+\]
+Since
+\[
+\sqrt{1620}=\sqrt{324\cdot 5}=18\sqrt5,
+\]
+we get
+\[
+t=\frac{-2\pm 18\sqrt5}{8}
+=\frac{-1\pm 9\sqrt5}{4}.
+\]
+Because \(t=u+v>0\), we must take
+\[
+t=\frac{9\sqrt5-1}{4}.
+\]
+
+Thus
+\[
+A_1A_2+A_1A_{11}=\frac{9\sqrt5-1}{4}.
+\]
+So
+\[
+m=9,\quad n=5,\quad p=1,\quad q=4,
+\]
+and
+\[
+m+n+p+q=9+5+1+4=19.
+\]
+
+\[
+\boxed{19}
+\]
+
+**Explanation:** Both annotators gave the comparison a tying grade, given that the methodology is more or less the same, and the computations barely differ. On the other hand, the LLM judge put the Qwen-3.5 solution as the more computationally complex one, likely due to being more verbose in its computations. This is the main failure case of our judges, where they exhibit some slight preferences, and are more likely to notice subtle differences that they latch on to.

LICENSE

@@ -0,0 +1,454 @@
+ProofRank dataset license
+=========================
+
+This license covers the DATA in this project: the problem statements, the human-written reference solutions, the model-generated solutions, and the judgments and fields derived from them. The code in this repository is licensed separately under the MIT License; see LICENSE.
+
+The problems are drawn from publicly available mathematics competitions and previously released benchmarks, and the human-written reference solutions are collected from public Art of Problem Solving posts and official competition materials. We abide by the sources' terms and conditions and redistribute this material under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International license reproduced below, for non-commercial research use and with proper credit to the original competitions.
+
+The terms of the original sources continue to apply to the redistributed text, independently of the license below:
+
+  * Art of Problem Solving: the forum content is NOT AVAILABLE FOR MODEL TRAINING and remains subject to the applicable AoPS restrictions, which require attribution when the material is used as input to an AI system.
+
+  * Official competition problems and previously released benchmarks: typically in the public domain or released under comparable conditions, which likewise forbid use of the material for training.
+
+Accordingly, this data MUST NOT be used to train any AI model. Users are responsible for complying with the original terms that apply to each entry.
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README.md

@@ -0,0 +1,44 @@
+---
+license: cc-by-nc-sa-4.0
+language:
+- en
+pretty_name: ProofRank Qualitative Disagreement Examples
+tags:
+- mathematics
+- competition-math
+- proof-evaluation
+- llm-as-a-judge
+- pairwise-comparison
+- qualitative-analysis
+size_categories:
+- n<1K
+viewer: false
+---
+
+# ProofRank: Qualitative Examples of Grading Disagreements
+
+Companion artifact to the ICLR 2027 submission **"Not All Proofs Are Equal: Evaluating LLM Proof Quality Beyond Correctness"**.
+
+We include supplementary examples in [`examples.md`](https://huggingface.co/datasets/Anon539823983/ProofRank-Examples/blob/main/EXAMPLES.md). Each one contains the problem ID and statement, two complete model-written solutions, and a short commentary explaining the nature of the disagreement.
+
+## Contents
+
+| # | Dimension | Problem | Solutions compared | What it illustrates |
+|---|-----------|---------|--------------------|---------------------|
+| 1 | Conciseness (human-human disagreement) | `imo-bench_number_theory-042` | GLM-5 vs. Gemini-3.1-Pro | Two solutions that are verbose in *different* ways, one padded with local restatements throughout, the other prefaced by a long recap of elementary facts, making the "more concise" call ambiguous. |
+| 2 | Cognitive ease (human–human disagreement) | `shortlist_2025_6` | GLM-5 vs. GPT-5.4 | A rigorous, notation-heavy combinatorial argument vs. a more abstract counting argument requiring internal visualizatio, which reads as "easier" depends on the annotator's background. |
+| 3 | Computational complexity (human–human disagreement) | `imo-bench_combinatorics-011` | Gemini-3-Flash vs. GPT-5.4 | Sparser notation whose individual claims are slower to verify vs. more symbols and algebraic steps that are each simple to parse. |
+| 4 | LLM judge vs. human consensus | `aime_2025_27` | Qwen-3.5-397B vs. GPT-5.4 | Two near-identical solutions that both human annotators graded as a tie, while the LLM judge (GPT-OSS-120B) declared one more computationally complex. |
+
+## Licensing
+
+The problems are drawn from publicly available mathematics competitions and previously released benchmarks, and the human-written reference solutions are collected from public Art of Problem Solving posts and official competition materials. We abide by the sources' terms and conditions and redistribute this material under [**CC BY-NC-SA 4.0**](https://creativecommons.org/licenses/by-nc-sa/4.0/) - see [`LICENSE`](LICENSE) - for non-commercial research use and with proper credit to the original competitions.
+
+The terms of the original sources continue to apply to the redistributed text, independently of that license:
+
+- **Art of Problem Solving**: the forum content is **not available for model training** and remains subject to the applicable AoPS restrictions, which require attribution when the material is used as input to an AI system.
+- **Official competition problems and previously released benchmarks**: typically in the public domain or released under comparable conditions, which likewise forbid use of the material for training.
+
+Accordingly, this data **must not be used to train any AI model**. Users are responsible for complying with the original terms that apply to each entry.
+
+Work produced by us - code, results, and evaluation scripts - is published under the more permissive **MIT** license.

examples_agreement.md

@@ -0,0 +1,709 @@
+# High-Agreement Annotation Examples (Supplement for R5)
+
+## Index
+
+| Example | Solutions | Agreements | Key feature |
+|---|---|---|---|
+| E1. `aime_2025_27` | Gemini-3-Flash vs. Gemini-3.1-Pro | 5/5 match, 1 decisive + 4 agreed ties | Verbosity separates the solutions, while other constructs untouched by length |
+| E2. `shortlist_2025_6` | Gemini-3-Flash vs. StepFun-3.5-Flash | 5/5 match, all 5 decisive | Unanimous decisions in both directions |
+| E3. `imo-bench_geometry-013` | GLM-5 vs. Qwen-3.5 | 5/5 match, 3 decisive + 2 agreed ties | Mix between 3 decisive qualities, while finding equality in others |
+| E4. `aime_2026_9` | Qwen-3.5 vs. GPT-5.4 | 4/5 match, 4 decisive | The solution with fewer words is unanimously judged more verbose |
+
+---
+
+# E1. `aime_2025_27`: verbosity separates, all other constructs untouched
+
+**Problem:**
+
+> Let $A_1 A_2 A_3 \ldots A_{11}$ be an $11$-sided non-convex simple polygon with the following properties:
+>
+> - For every integer $2 \le i \le 10$, the area of $\triangle A_i A_{1} A_{i+1}$ is equal to $1$.
+> - For every integer $2 \le i \le 10$, $\cos(\angle A_i A_{1} A_{i+1}) = \frac{12}{13}$.
+> - The perimeter of the $11$-gon $A_1 A_2 A_3 \dots A_{11}$ is equal to $20$.
+>
+> Then $A_1 A_2 + A_1 A_{11} = \frac{m \sqrt{n} - p}{q}$ where $m, n, p$, and $q$ are positive integers, $n$ is not divisible by the square of any prime, and no prime divides all of $m, p$, and $q$. Find $m + n + p + q$.
+
+**Annotator verdicts** (which solution is more verbose / more elegant / ...):
+
+| dimension | annotator 1 | annotator 2 |
+|---|---|---|
+| Verbosity | Solution 2 | Solution 2 |
+| Elegance | tie | tie |
+| Insight | tie | tie |
+| Complexity | tie | tie |
+| Computational Challenge | tie | tie |
+
+Solution 1: Gemini-3-Flash. Solution 2: Gemini-3.1-Pro.
+
+**Discussion.** The two solutions follow the same computational route (area formula to a constant product $r_i r_{i+1}$, alternating lengths, Law of Cosines, quadratic equation in $S$) and differ almost entirely in how much text carries it. Both annotators judge the longer one more verbose, as it also give verbatim commentary about the performed algebra, and give unanimous ties on all four remaining dimension, which shows the annotators did not let raw length impose a difference in the other measurements, as only the wording mattered.
+## Solution 1 - gemini/gemini-3-flash
+
+To solve the problem, let the lengths of the segments $A_1 A_i$ be $r_i$ for $i=2, 3, \dots, 11$. For each $i \in \{2, 3, \dots, 10\}$, we are given the area of $\triangle A_i A_1 A_{i+1}$ is $1$ and $\cos(\angle A_i A_1 A_{i+1}) = 12/13$. Let $\theta_i = \angle A_i A_1 A_{i+1}$. Since $0 < \theta_i < \pi$ and $\cos \theta_i = 12/13$, we have:
+$$\sin \theta_i = \sqrt{1 - \cos^2 \theta_i} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{169-144}{169}} = \frac{5}{13}.$$
+The area of $\triangle A_i A_1 A_{i+1}$ is given by $\frac{1}{2} r_i r_{i+1} \sin \theta_i = 1$. Substituting $\sin \theta_i = 5/13$, we get:
+$$\frac{1}{2} r_i r_{i+1} \cdot \frac{5}{13} = 1 \implies r_i r_{i+1} = \frac{26}{5}.$$
+Let $k = 26/5$. For $i=2, 3, \dots, 10$, we have $r_i r_{i+1} = k$. This implies:
+$r_2 r_3 = k$, $r_3 r_4 = k \implies r_4 = r_2$, $r_4 r_5 = k \implies r_5 = r_3$, and so on.
+Thus, $r_{2j} = r_2$ and $r_{2j+1} = r_3 = k/r_2$. Let $x = r_2$ and $y = r_{11} = k/r_2$.
+
+The sides of the $11$-gon are $s_1, s_2, \dots, s_{11}$, where $s_1 = A_1 A_2 = r_2 = x$ and $s_{11} = A_{11} A_1 = r_{11} = y$. For $i \in \{2, 3, \dots, 10\}$, the length $s_i = A_i A_{i+1}$ is found using the law of cosines in $\triangle A_i A_1 A_{i+1}$:
+$$s_i^2 = r_i^2 + r_{i+1}^2 - 2 r_i r_{i+1} \cos \theta_i = x^2 + y^2 - 2k \cdot \frac{12}{13}.$$
+Substituting $k = 26/5$ and $\cos \theta_i = 12/13$:
+$$s_i^2 = x^2 + y^2 - 2 \cdot \frac{26}{5} \cdot \frac{12}{13} = x^2 + y^2 - \frac{48}{5}.$$
+Let $s = s_i$ for $i=2, \dots, 10$. The perimeter is $x + 9s + y = 20$. Let $S = x + y$. Then $x^2 + y^2 = S^2 - 2xy = S^2 - 52/5$. Substituting this into the equation for $s^2$:
+$$s^2 = S^2 - \frac{52}{5} - \frac{48}{5} = S^2 - 20.$$
+The perimeter equation becomes $S + 9\sqrt{S^2 - 20} = 20$. Squaring both sides:
+$$81(S^2 - 20) = (20 - S)^2 \implies 81S^2 - 1620 = 400 - 40S + S^2 \implies 80S^2 + 40S - 2020 = 0.$$
+Dividing by $20$, we get $4S^2 + 2S - 101 = 0$. Solving for $S > 0$ using the quadratic formula:
+$$S = \frac{-2 + \sqrt{4 - 4(4)(-101)}}{8} = \frac{-2 + \sqrt{1620}}{8} = \frac{-2 + 18\sqrt{5}}{8} = \frac{9\sqrt{5} - 1}{4}.$$
+The expression for $A_1 A_2 + A_1 A_{11}$ is $x + y = S = \frac{9\sqrt{5} - 1}{4}$.
+The values of $m, n, p, q$ are $m=9, n=5, p=1, q=4$. We check: $n=5$ is square-free, $m, n, p, q$ are positive integers, and $\gcd(9, 1, 4) = 1$. The sum $m + n + p + q = 9 + 5 + 1 + 4 = 19$.
+
+Final Answer: \boxed{19}
+
+
+## Solution 2 - gemini/gemini-31-pro
+
+To find the sum $A_1 A_2 + A_1 A_{11}$, we will first use the given information about the triangles formed by the vertices $A_1, A_i,$ and $A_{i+1}$ to find the lengths of the edges of the polygon.
+
+**Step 1: Finding the lengths of the segments from $A_1$**
+Let $r_i = A_1 A_i$ denote the distance from $A_1$ to the vertex $A_i$ for $2 \le i \le 11$.
+For each integer $i$ from 2 to 10, consider the triangle $\triangle A_i A_1 A_{i+1}$. The area of a triangle can be expressed as:
+$$\text{Area} = \frac{1}{2} A_1 A_i \cdot A_1 A_{i+1} \sin(\angle A_i A_1 A_{i+1})$$
+We are given that $\cos(\angle A_i A_1 A_{i+1}) = \frac{12}{13}$. Because the internal angle of a triangle must be strictly between $0^\circ$ and $180^\circ$, its sine must be positive. Thus, we can find the sine using the Pythagorean identity:
+$$\sin(\angle A_i A_1 A_{i+1}) = \sqrt{1 - \cos^2(\angle A_i A_1 A_{i+1})} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{169 - 144}{169}} = \frac{5}{13}$$
+We are also given that the area of each of these 9 triangles is $1$. Substituting the known values into the area formula gives:
+$$1 = \frac{1}{2} r_i r_{i+1} \left(\frac{5}{13}\right)$$
+Multiplying both sides by $\frac{26}{5}$ yields:
+$$r_i r_{i+1} = \frac{26}{5} \quad \text{for } i = 2, 3, \ldots, 10$$
+This equation gives us a system:
+$$r_2 r_3 = \frac{26}{5}, \quad r_3 r_4 = \frac{26}{5}, \quad r_4 r_5 = \frac{26}{5}, \dots$$
+From $r_2 r_3 = r_3 r_4$, and knowing $r_3 > 0$ (since lengths are positive), we can divide by $r_3$ to conclude that $r_2 = r_4$.
+By continuing this reasoning ($r_{i} r_{i+1} = r_{i+1} r_{i+2} \implies r_i = r_{i+2}$), the lengths alternate. Therefore, we have:
+$$r_2 = r_4 = r_6 = r_8 = r_{10} = x$$
+$$r_3 = r_5 = r_7 = r_9 = r_{11} = y$$
+for some positive real numbers $x$ and $y$ such that $xy = \frac{26}{5}$. Notice that $A_1 A_2 = x$ and $A_1 A_{11} = y$. The quantity we want to find is $A_1 A_2 + A_1 A_{11} = x + y$. Let $S = x + y$.
+
+**Step 2: Finding the lengths of the edges $A_i A_{i+1}$**
+For $2 \le i \le 10$, we can find the side length $A_i A_{i+1}$ using the Law of Cosines on $\triangle A_i A_1 A_{i+1}$:
+$$A_i A_{i+1}^2 = A_1 A_i^2 + A_1 A_{i+1}^2 - 2(A_1 A_i)(A_1 A_{i+1}) \cos(\angle A_i A_1 A_{i+1})$$
+Since one of $\{A_1 A_i, A_1 A_{i+1}\}$ is $x$ and the other is $y$, we have $A_1 A_i^2 + A_1 A_{i+1}^2 = x^2 + y^2$. Furthermore, their product is $xy = \frac{26}{5}$, and we are given the cosine of the angle is $\frac{12}{13}$. Thus:
+$$A_i A_{i+1}^2 = x^2 + y^2 - 2 \left(\frac{26}{5}\right) \left(\frac{12}{13}\right) = x^2 + y^2 - \frac{48}{5}$$
+We can express $x^2 + y^2$ in terms of $S = x + y$:
+$$x^2 + y^2 = (x+y)^2 - 2xy = S^2 - 2\left(\frac{26}{5}\right) = S^2 - \frac{52}{5}$$
+Substituting this back into our expression for the squared side length:
+$$A_i A_{i+1}^2 = \left(S^2 - \frac{52}{5}\right) - \frac{48}{5} = S^2 - \frac{100}{5} = S^2 - 20$$
+So, the length of each of these 9 edges is $d = \sqrt{S^2 - 20}$.
+
+**Step 3: Utilizing the perimeter to solve for $S$**
+The perimeter of the 11-gon $A_1 A_2 \ldots A_{11}$ is the sum of the lengths of all 11 of its edges:
+$$\text{Perimeter} = A_1 A_2 + \left( \sum_{i=2}^{10} A_i A_{i+1} \right) + A_{11} A_1$$
+Substituting our defined variables ($A_1 A_2 = x$, $A_{11} A_1 = y$, and $A_i A_{i+1} = \sqrt{S^2 - 20}$):
+$$\text{Perimeter} = x + 9\sqrt{S^2 - 20} + y = (x + y) + 9\sqrt{S^2 - 20} = S + 9\sqrt{S^2 - 20}$$
+We are given that the perimeter is equal to 20. Thus, we establish the equation:
+$$S + 9\sqrt{S^2 - 20} = 20$$
+Isolating the square root gives:
+$$9\sqrt{S^2 - 20} = 20 - S$$
+Squaring both sides (which requires that $20 - S \ge 0$ for a valid solution):
+$$81(S^2 - 20) = (20 - S)^2$$
+$$81S^2 - 1620 = 400 - 40S + S^2$$
+Subtracting the right side from the left to set the equation to zero:
+$$80S^2 + 40S - 2020 = 0$$
+Dividing by 20 leaves a simpler quadratic equation:
+$$4S^2 + 2S - 101 = 0$$
+Using the quadratic formula to solve for $S$:
+$$S = \frac{-2 \pm \sqrt{2^2 - 4(4)(-101)}}{2(4)} = \frac{-2 \pm \sqrt{4 + 1616}}{8} = \frac{-2 \pm \sqrt{1620}}{8}$$
+We simplify the radical, noting that $1620 = 324 \times 5 = 18^2 \times 5$, so $\sqrt{1620} = 18\sqrt{5}$:
+$$S = \frac{-2 \pm 18\sqrt{5}}{8} = \frac{-1 \pm 9\sqrt{5}}{4}$$
+Because $S = x + y$ must represent the sum of two positive lengths, $S$ must be strictly positive. Since $9\sqrt{5} > 1$, we take the positive root:
+$$S = \frac{9\sqrt{5} - 1}{4}$$
+*(Note: $\sqrt{5} \approx 2.236 \implies S \approx 4.781$, confirming that $20 - S \ge 0$ is indeed satisfied, ensuring no extraneous solutions were generated by squaring).*
+
+**Step 4: Formatting the solution**
+We have found that $A_1 A_2 + A_1 A_{11} = S = \frac{9\sqrt{5} - 1}{4}$.
+This is exactly in the required format $\frac{m\sqrt{n} - p}{q}$, giving:
+$$m = 9, \quad n = 5, \quad p = 1, \quad q = 4$$
+We verify the given conditions: $m, n, p, q$ are positive integers, $n=5$ is square-free, and no prime divides all of $m, p,$ and $q$ (since $p=1$).
+
+Finally, finding the required sum:
+$$m + n + p + q = 9 + 5 + 1 + 4 = 19$$
+
+\boxed{19}
+
+
+---
+
+# E2. `shortlist_2025_6`: full agreement, all verdicts decisive
+
+**Problem:**
+
+> Jerry places at most one rook in each cell of a $2025 \times 2025$ grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them. Determine the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 other rooks.
+
+**Annotator verdicts** (which solution is more verbose / more elegant / ...):
+
+| dimension | annotator 1 | annotator 2 |
+|---|---|---|
+| Verbosity | Solution 2 | Solution 2 |
+| Elegance | Solution 2 | Solution 2 |
+| Insight | Solution 2 | Solution 2 |
+| Complexity | Solution 1 | Solution 1 |
+| Computational Challenge | Solution 1 | Solution 1 |
+
+Solution 1: Gemini-3-Flash. Solution 2: StepFun-3.5-Flash.
+
+**Discussion.** The two annotators agree exactly on all five dimensions, with no ties, on a pair of comparable length (526 vs 653 words). The longer solution (Solution 2) is unanimously more verbose, yet also more elegant and more insightful (its two claims, one border cell forced empty per interior rook and distinct forced cells, carry the whole bound), while the shorter Solution 1 is unanimously more conceptually complex and more computationally challenging (edge and interior rook set definitions, inclusion-exclusion, extremal-row bookkeeping). Thus, we can see that judging verbosity does not require an extreme length gap, and raw length does not bleed into the other four verdicts.
+
+## Solution 1 - gemini/gemini-3-flash
+
+To find the maximum number of rooks Jerry can place on a $2025 \times 2025$ grid such that no rook attacks 4 other rooks, let $n = 2025$. A rook $r$ at cell $(i, j)$ attacks another rook if they are in the same row or column and no other rooks are between them. Each rook can attack at most 2 rooks in its row (one to its left and one to its right) and at most 2 rooks in its column (one above and one below). Let $a_H(r)$ and $a_V(r)$ be the number of rooks $r$ attacks in its row and column, respectively. Then $a(r) = a_H(r) + a_V(r)$, with $a_H(r), a_V(r) \in \{0, 1, 2\}$.
+
+The condition $a(r) \le 3$ for all rooks $r$ means that no rook can simultaneously attack 2 rooks in its row and 2 rooks in its column. That is, $a_H(r) = 2$ and $a_V(r) = 2$ is impossible. Let $I_H$ be the set of "interior" rooks in their rows ($a_H(r) = 2$) and $I_V$ be the set of "interior" rooks in their columns ($a_V(r) = 2$). The condition is $I_H \cap I_V = \emptyset$.
+
+Let $R$ be the set of all rooks placed, so $N = |R|$. Let $E_H$ be the set of "edge" rooks in their rows, which are rooks that are not interior rooks ($E_H = R \setminus I_H$). Similarly, $E_V = R \setminus I_V$. The condition $I_H \cap I_V = \emptyset$ is equivalent to $R = E_H \cup E_V$. Thus, by the principle of inclusion-exclusion:
+$$N = |E_H \cup E_V| = |E_H| + |E_V| - |E_H \cap E_V|$$
+Let $r_i$ be the number of rooks in row $i$ and $c_j$ be the number of rooks in column $j$. The number of horizontal edge rooks in row $i$ is $\min(r_i, 2)$. Thus, $|E_H| = \sum_{i=1}^n \min(r_i, 2) \le 2n$. Similarly, $|E_V| = \sum_{j=1}^n \min(c_j, 2) \le 2n$.
+
+Let $i_{\min}$ and $i_{\max}$ be the first and last non-empty rows. Any rook $(i_{\min}, j)$ is the first rook in column $j$ because there are no rooks in rows $i < i_{\min}$. Thus, all $r_{i_{\min}}$ rooks in the first row are vertical edge rooks. The horizontal edge rooks in this row (of which there are $\min(r_{i_{\min}}, 2)$) are therefore in $E_H \cap E_V$. Similarly, the $\min(r_{i_{\max}}, 2)$ horizontal edge rooks in the last row are in $E_H \cap E_V$. If $i_{\min} < i_{\max}$, these two sets are disjoint, so $|E_H \cap E_V| \ge \min(r_{i_{\min}}, 2) + \min(r_{i_{\max}}, 2)$.
+
+Substituting this into our equation for $N$:
+$$N \le \sum_{i=1}^n \min(r_i, 2) + 2n - (\min(r_{i_{\min}}, 2) + \min(r_{i_{\max}}, 2)) = \sum_{i \neq i_{\min}, i_{\max}} \min(r_i, 2) + 2n$$
+There are at most $n-2$ terms in the sum, each at most 2. Thus:
+$$N \le 2(n-2) + 2n = 4n - 4$$
+For $n = 2025$, this gives $N \le 4(2025) - 4 = 8100 - 4 = 8096$. Jerry can achieve this by placing rooks along the entire boundary of the grid. For such a configuration, each corner rook attacks 2 rooks, and each other boundary rook attacks 3 rooks.
+
+The maximum number of rooks Jerry can place is 8096.
+
+\boxed{8096}
+
+
+## Solution 2 - stepfun/3.5-flash
+
+The maximum number of rooks Jerry can place on a \(2025 \times 2025\) grid such that no rook attacks four other rooks is \(4 \cdot 2025 - 4 = 8096\).
+
+**Proof:**
+A rook attacks another if they share a row or column with no rook between them. Thus a rook attacks a neighbor in each of the four directions (left, right, up, down) if such a neighbor exists. Hence a rook attacks four others exactly when there is at least one rook to its left, right, above, and below. Equivalently, the rook is **not** the leftmost or rightmost in its row, and **not** the topmost or bottommost in its column. Therefore, the condition "no rook attacks 4 other rooks" means every rook is either leftmost or rightmost in its row, or topmost or bottommost in its column.
+
+Define **border cells** as those with row index \(1\) or \(n\) or column index \(1\) or \(n\); there are \(4n-4\) such cells. All other cells are **interior**.
+
+**Claim 1.** If a rook is placed in an interior cell \((i,j)\) (with \(2 \le i \le n-1\) and \(2 \le j \le n-1\)), then at least one of the border cells \((i,1)\), \((i,n)\), \((1,j)\), \((n,j)\) must be empty.
+*Proof.* The rook must be extreme in its row or column.
+- If it is leftmost in row \(i\), then no rook exists in row \(i\) with column \(< j\); in particular \((i,1)\) is empty.
+- If it is rightmost, then \((i,n)\) is empty.
+- If it is topmost in column \(j\), then \((1,j)\) is empty.
+- If it is bottommost, then \((n,j)\) is empty. ∎
+
+**Claim 2.** Two different interior rooks cannot force the same border cell to be empty.
+*Proof.* Consider a border cell \(B\).
+- If \(B = (i,1)\), it can be forced empty only by an interior rook that is leftmost in row \(i\); there is at most one such rook.
+- If \(B = (i,n)\), only a rightmost in row \(i\).
+- If \(B = (1,j)\), only a topmost in column \(j\).
+- If \(B = (n,j)\), only a bottommost in column \(j\).
+Thus at most one interior rook can be responsible for a given border cell. ∎
+
+Let \(I\) be the number of interior rooks and \(B\) the number of rooks on border cells. By Claim 1, each interior rook forces at least one border cell to be empty. By Claim 2, the forced border cells are distinct, so the number of empty border cells is at least \(I\). The total number of border cells is \(4n-4\), so the number of empty border cells equals \((4n-4) - B\). Hence \((4n-4) - B \ge I\), i.e., \(B + I \le 4n-4\). Since the total number of rooks is \(N = B + I\), we have \(N \le 4n-4\).
+
+**Construction achieving \(4n-4\):** Place a rook on every border cell. This gives exactly \(4n-4\) rooks.
+- For a rook on row 1 at column \(j\): if \(j=1\) or \(j=n\), it is leftmost/rightmost in its row and also topmost/bottommost in its column. If \(2 \le j \le n-1\), row 1 is fully filled so the rook is not leftmost/rightmost, but column \(j\) contains only rooks at rows 1 and \(n\) (since interior rows have no rook in column \(j\)), so it is topmost.
+- For a rook on row \(n\): analogous.
+- For a rook on column 1 at row \(i\) with \(2 \le i \le n-1\): row \(i\) contains only rooks at columns 1 and \(n\) (since interior columns are empty), so the rook is leftmost. Column 1 is fully filled, so it is not topmost/bottommost, but row-extreme suffices.
+- For a rook on column \(n\): analogous.
+Thus every rook is extreme in its row or column, so the condition holds.
+
+Therefore the maximum is exactly \(4n-4\). For \(n = 2025\), this is \(4 \cdot 2025 - 4 = 8100 - 4 = 8096\).
+
+\[
+\boxed{8096}
+\]
+
+
+---
+
+# E3. `imo-bench_geometry-013`: full agreement, decisive verdicts plus agreed ties
+
+**Problem:**
+
+> In triangle $XYZ$, we have $XY = YZ$, $\angle Y = 20^\circ$. Point $P$ on $XZ$ is such that $XP : PZ = 1 : 2$, point $Q$ is the projection of $Z$ to $YP$. Find angle $\angle XQY$ in terms of degree.
+
+**Annotator verdicts** (which solution is more verbose / more elegant / ...):
+
+| dimension | annotator 1 | annotator 2 |
+|---|---|---|
+| Verbosity | Solution 1 | Solution 1 |
+| Elegance | Solution 2 | Solution 2 |
+| Insight | tie | tie |
+| Complexity | tie | tie |
+| Computational Challenge | Solution 1 | Solution 1 |
+
+Solution 1: GLM-5. Solution 2: Qwen-3.5.
+
+**Discussion.** We observe 5/5 agreement with 3 decisive verdicts and 2 agreed ties, on two solutions of the same method family (coordinate bashing). Both annotators independently record that the two solutions are comparable on insight and conceptual complexity (or rather lack thereof due to the heavily computational approaches). They also agree that Solution 1 is more verbose and computationally heavier and the Qwen solution is more elegant as it spares some heavy computations. All 3 aspects are largely due to the choice of coordinate system in Solution 1 requiring further computations and explanations down the line.
+
+## Solution 1 - glm/glm-5
+
+**Proof.**
+We place the triangle in a coordinate system. Because angles are invariant under scaling, rotation and translation, we may set
+
+\[
+Y = (0,0),\qquad X = (1,0),\qquad Z = (\cos 20^\circ,\; \sin 20^\circ),
+\]
+
+so that \(XY = YZ = 1\) and \(\angle XYZ = 20^\circ\).
+
+---
+
+### 1. Point \(P\) on \(XZ\)
+
+Given \(XP : PZ = 1 : 2\), the section formula gives
+
+\[
+P = \frac{2X + Z}{3}
+   = \left( \frac{2 + \cos 20^\circ}{3},\; \frac{\sin 20^\circ}{3} \right).
+\]
+
+---
+
+### 2. Point \(Q\) - foot from \(Z\) to \(YP\)
+
+Since \(Y\) is the origin, line \(YP\) is the line through the origin with direction vector \(P\). The orthogonal projection of \(Z\) onto this line is
+
+\[
+Q = \frac{Z \cdot P}{P \cdot P}\, P.
+\]
+
+Compute the dot products:
+
+\[
+\begin{aligned}
+Z \cdot P &= \cos 20^\circ\cdot\frac{2+\cos 20^\circ}{3} + \sin 20^\circ\cdot\frac{\sin 20^\circ}{3} \\
+          &= \frac{1}{3}\bigl(2\cos 20^\circ + \cos^2 20^\circ + \sin^2 20^\circ\bigr)
+           = \frac{1 + 2\cos 20^\circ}{3},
+\end{aligned}
+\]
+
+\[
+\begin{aligned}
+P \cdot P &= \left(\frac{2+\cos 20^\circ}{3}\right)^2 + \left(\frac{\sin 20^\circ}{3}\right)^2 \\
+          &= \frac{1}{9}\bigl((2+\cos 20^\circ)^2 + \sin^2 20^\circ\bigr)
+           = \frac{5 + 4\cos 20^\circ}{9}.
+\end{aligned}
+\]
+
+Hence
+
+\[
+t = \frac{Z \cdot P}{P \cdot P}
+   = \frac{(1+2\cos 20^\circ)/3}{(5+4\cos 20^\circ)/9}
+   = \frac{3(1+2\cos 20^\circ)}{5+4\cos 20^\circ}.
+\]
+
+Thus
+
+\[
+Q = tP = \left(
+          \frac{(1+2\cos 20^\circ)(2+\cos 20^\circ)}{5+4\cos 20^\circ},\;
+          \frac{(1+2\cos 20^\circ)\sin 20^\circ}{5+4\cos 20^\circ}
+        \right).
+\]
+
+Denote \(D = 5 + 4\cos 20^\circ\) for brevity.
+
+---
+
+### 3. Cosine of \(\angle XQY\)
+
+Vectors from \(Q\):
+
+\[
+\overrightarrow{QX} = X - Q = (1 - Q_x,\; -Q_y), \qquad
+\overrightarrow{QY} = Y - Q = (-Q_x,\; -Q_y).
+\]
+
+Their dot product and lengths:
+
+\[
+\overrightarrow{QX} \cdot \overrightarrow{QY}
+= (1 - Q_x)(-Q_x) + (-Q_y)(-Q_y) = -Q_x + Q_x^2 + Q_y^2 = |Q|^2 - Q_x,
+\]
+
+\[
+|\overrightarrow{QY}| = |Q|.
+\]
+
+Therefore
+
+\[
+\cos \angle XQY = \frac{|Q|^2 - Q_x}{|Q| \cdot |\overrightarrow{QX}|}.
+\]
+
+We compute \(|Q|^2\) and \(|\overrightarrow{QX}|^2\).
+
+- **\(|Q|^2\)**
+
+  \[
+  |Q|^2 = Q_x^2 + Q_y^2
+        = \frac{(1+2\cos 20^\circ)^2}{D^2}\bigl((2+\cos 20^\circ)^2 + \sin^2 20^\circ\bigr).
+  \]
+
+  But \((2+\cos 20^\circ)^2 + \sin^2 20^\circ = 5 + 4\cos 20^\circ = D\), so
+
+  \[
+  |Q|^2 = \frac{(1+2\cos 20^\circ)^2}{D}.
+  \]
+
+- **\(|\overrightarrow{QX}|^2\)**
+
+  \[
+  |\overrightarrow{QX}|^2 = (1 - Q_x)^2 + Q_y^2 = 1 - 2Q_x + |Q|^2.
+  \]
+
+  Substituting the expressions for \(Q_x\) and \(|Q|^2\):
+
+  \[
+  \begin{aligned}
+  1 - 2Q_x + |Q|^2
+  &= 1 - \frac{2(1+2\cos 20^\circ)(2+\cos 20^\circ)}{D} + \frac{(1+2\cos 20^\circ)^2}{D} \\
+  &= 1 + \frac{ (1+2\cos 20^\circ)^2 - 2(1+2\cos 20^\circ)(2+\cos 20^\circ) }{D}.
+  \end{aligned}
+  \]
+
+  The numerator simplifies:
+
+  \[
+  (1+2\cos 20^\circ)^2 - 2(1+2\cos 20^\circ)(2+\cos 20^\circ)
+  = (1+2\cos 20^\circ)\bigl[(1+2\cos 20^\circ) - 2(2+\cos 20^\circ)\bigr]
+  = (1+2\cos 20^\circ)(-3).
+  \]
+
+  Hence
+
+  \[
+  1 - 2Q_x + |Q|^2 = 1 - \frac{3(1+2\cos 20^\circ)}{D}
+                   = \frac{D - 3(1+2\cos 20^\circ)}{D}.
+  \]
+
+  With \(D = 5 + 4\cos 20^\circ\),
+
+  \[
+  D - 3(1+2\cos 20^\circ) = (5+4\cos 20^\circ) - 3 - 6\cos 20^\circ = 2 - 2\cos 20^\circ = 2(1-\cos 20^\circ).
+  \]
+
+  Therefore
+
+  \[
+  |\overrightarrow{QX}|^2 = \frac{2(1-\cos 20^\circ)}{D}.
+  \]
+
+---
+
+### 4. Simplification of \(\cos \angle XQY\)
+
+Now compute the numerator and denominator of the cosine.
+
+**Numerator**
+
+\[
+\begin{aligned}
+|Q|^2 - Q_x
+&= \frac{(1+2\cos 20^\circ)^2}{D} - \frac{(1+2\cos 20^\circ)(2+\cos 20^\circ)}{D} \\
+&= \frac{1+2\cos 20^\circ}{D}\bigl[(1+2\cos 20^\circ) - (2+\cos 20^\circ)\bigr] \\
+&= \frac{1+2\cos 20^\circ}{D}\,(\cos 20^\circ - 1).
+\end{aligned}
+\]
+
+**Denominator**
+
+\[
+|Q| \cdot |\overrightarrow{QX}|
+= \frac{1+2\cos 20^\circ}{\sqrt{D}} \cdot \sqrt{\frac{2(1-\cos 20^\circ)}{D}}
+= \frac{(1+2\cos 20^\circ)\sqrt{2(1-\cos 20^\circ)}}{D}.
+\]
+
+Thus
+
+\[
+\cos \angle XQY
+= \frac{ \dfrac{(1+2\cos 20^\circ)(\cos 20^\circ - 1)}{D} }
+       { \dfrac{(1+2\cos 20^\circ)\sqrt{2(1-\cos 20^\circ)}}{D} }
+= \frac{\cos 20^\circ - 1}{\sqrt{2(1-\cos 20^\circ)}}.
+\]
+
+Because \(\cos 20^\circ - 1 = -(1-\cos 20^\circ)\),
+
+\[
+\cos \angle XQY = -\frac{1-\cos 20^\circ}{\sqrt{2(1-\cos 20^\circ)}}
+                = -\sqrt{\frac{1-\cos 20^\circ}{2}}.
+\]
+
+---
+
+### 5. Trigonometric identity
+
+Using \(1 - \cos \theta = 2\sin^2(\theta/2)\) with \(\theta = 20^\circ\):
+
+\[
+\sqrt{\frac{1-\cos 20^\circ}{2}} = \sqrt{\frac{2\sin^2 10^\circ}{2}} = \sin 10^\circ.
+\]
+
+Hence
+
+\[
+\cos \angle XQY = -\sin 10^\circ.
+\]
+
+---
+
+### 6. Determination of the angle
+
+Since \(\sin 10^\circ = \cos 80^\circ\), we have
+
+\[
+\cos \angle XQY = -\cos 80^\circ = \cos(180^\circ - 80^\circ) = \cos 100^\circ.
+\]
+
+The angle \(\angle XQY\) is a geometric angle formed by two segments, so it lies in \((0^\circ, 180^\circ)\). Its cosine is negative, therefore the angle must be obtuse and uniquely
+
+\[
+\angle XQY = 100^\circ.
+\]
+
+\[
+\boxed{100^\circ}
+\]
+
+
+## Solution 2 - qwen/qwen35_397b_a17b_high
+
+To find the measure of $\angle XQY$, we use a Cartesian coordinate system.
+
+### 1. Coordinate Setup
+Let $M$ be the midpoint of $XZ$. Since $XY = YZ$, $YM$ is the perpendicular bisector of $XZ$. We place $M$ at the origin $(0,0)$ and $XZ$ along the x-axis.
+*   Let $X = (-a, 0)$ and $Z = (a, 0)$.
+*   Since $\triangle XYZ$ is isosceles with $\angle Y = 20^\circ$, the base angles are $\angle X = \angle Z = \frac{180^\circ - 20^\circ}{2} = 80^\circ$.
+*   The height $YM = a \tan 80^\circ = a \cot 10^\circ$. Thus, $Y = (0, a \cot 10^\circ)$.
+*   Point $P$ lies on $XZ$ such that $XP : PZ = 1 : 2$. Since $XZ = 2a$, $XP = \frac{2}{3}a$.
+*   The coordinate of $P$ is $x_P = -a + \frac{2}{3}a = -\frac{a}{3}$. So, $P = (-\frac{a}{3}, 0)$.
+
+### 2. Equations of Lines $YP$ and $ZQ$
+The slope of line $YP$ is:
+$$m_{YP} = \frac{a \cot 10^\circ - 0}{0 - (-a/3)} = 3 \cot 10^\circ$$
+Point $Q$ is the projection of $Z$ onto $YP$, meaning $ZQ \perp YP$. The slope of $ZQ$ is:
+$$m_{ZQ} = -\frac{1}{m_{YP}} = -\frac{1}{3 \cot 10^\circ} = -\frac{\tan 10^\circ}{3}$$
+The equation of line $ZQ$ passing through $Z(a, 0)$ is:
+$$y = -\frac{\tan 10^\circ}{3}(x - a)$$
+The equation of line $YP$ is:
+$$y = 3 \cot 10^\circ(x + \frac{a}{3}) = 3 \cot 10^\circ x + a \cot 10^\circ$$
+
+### 3. Coordinates of $Q$
+To find $Q(x_Q, y_Q)$, we solve the system:
+$$3 \cot 10^\circ x + a \cot 10^\circ = -\frac{\tan 10^\circ}{3}(x - a)$$
+Multiplying by $3 \tan 10^\circ$:
+$$9x + 3a = -x \tan^2 10^\circ + a \tan^2 10^\circ \implies x_Q(9 + \tan^2 10^\circ) = a(\tan^2 10^\circ - 3)$$
+$$x_Q = a \frac{\tan^2 10^\circ - 3}{9 + \tan^2 10^\circ}, \quad y_Q = -\frac{\tan 10^\circ}{3}(x_Q - a) = \frac{4a \tan 10^\circ}{9 + \tan^2 10^\circ}$$
+
+### 4. Slopes and Angle Calculation
+We calculate the slopes of vectors $\vec{QX}$ and $\vec{QY}$:
+*   **Slope of $QX$ ($m_1$):**
+    $$x_X - x_Q = -a - a \frac{\tan^2 10^\circ - 3}{9 + \tan^2 10^\circ} = a \frac{-9 - \tan^2 10^\circ - \tan^2 10^\circ + 3}{9 + \tan^2 10^\circ} = \frac{-2a(\tan^2 10^\circ + 3)}{9 + \tan^2 10^\circ}$$
+    $$m_1 = \frac{0 - y_Q}{x_X - x_Q} = \frac{-4a \tan 10^\circ}{-2a(\tan^2 10^\circ + 3)} = \frac{2 \tan 10^\circ}{\tan^2 10^\circ + 3}$$
+*   **Slope of $QY$ ($m_2$):** Since $Q$ lies on $YP$, $m_2 = m_{YP} = 3 \cot 10^\circ$.
+
+The tangent of the angle $\theta$ between the lines is:
+$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_2 m_1} \right|$$
+Substituting the values:
+$$m_2 - m_1 = \frac{3}{\tan 10^\circ} - \frac{2 \tan 10^\circ}{\tan^2 10^\circ + 3} = \frac{3(\tan^2 10^\circ + 3) - 2 \tan^2 10^\circ}{\tan 10^\circ(\tan^2 10^\circ + 3)} = \frac{\tan^2 10^\circ + 9}{\tan 10^\circ(\tan^2 10^\circ + 3)}$$
+$$1 + m_2 m_1 = 1 + \frac{3}{\tan 10^\circ} \cdot \frac{2 \tan 10^\circ}{\tan^2 10^\circ + 3} = \frac{\tan^2 10^\circ + 9}{\tan^2 10^\circ + 3}$$
+$$\tan \theta = \frac{1}{\tan 10^\circ} = \cot 10^\circ = \tan 80^\circ$$
+The acute angle between the lines is $80^\circ$. To determine the interior angle $\angle XQY$, we check the vector directions. $\vec{QY}$ points in the first quadrant (up-right), while $\vec{QX}$ points in the third quadrant (down-left). The angle between these directions is $180^\circ - 80^\circ = 100^\circ$.
+
+\boxed{100^\circ}
+
+
+---
+
+# E4. `aime_2026_9`: fewer words, unanimously more verbose
+
+**Problem** (middle portion truncated in the source dossier; paste the full statement before sharing):
+
+> Joanne has a blank fair six-sided die and six stickers each displaying a different integer from 1 to 6. Joanne rolls the die and then places the sticker labeled 1 on the top face of the die. She then rolls the die again, places the sticker labeled 2 on the top face, and continues this process to  place the rest of the stickers in order. If the die ever lands with a sticker already on its top face, the new sticker is placed to cover the old sticker. Let $p$ be the conditional probability that at the end of the process exactly one face has been left blank, given that all the even-numbered stickers are visible on faces of the die. Then $p$ can be written as $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
+
+**Annotator verdicts** (which solution is more verbose / more elegant / ...):
+
+| dimension | annotator 1 | annotator 2 |
+|---|---|---|
+| Verbosity | Solution 2 | Solution 2 |
+| Elegance | Solution 2 | Solution 2 |
+| Insight | Solution 2 | Solution 2 |
+| Complexity | Solution 1 | tie |
+| Computational Challenge | Solution 2 | Solution 2 |
+
+Solution 1: Qwen-3.5. Solution 2: GPT-5.4.
+
+**Discussion.** Solution 2 is unanimously judged more verbose than Solution 1. This is direct evidence that the annotators define verbosity by how much text is spent relative to content, and not by word count (539 vs 479). as the solution expresses the counting in details that some might find unnecessary
+
+In parituclar, Solution 2 spells out routine counting in itemized sub-steps and display equations for trivial arithmetic, as well as including unused commentary such as:
+
+> Let \(X_i\) be the face on top on the \(i\)-th roll, for \(i=1,2,\dots,6\). Since the die is fair and the rolls are independent, all \(6^6\) sequences \((X_1,X_2,\dots,X_6)\) are equally likely.
+
+On the other hand, Solution 1 carries denser content per sentence despite being longer. All in all, four of the five dimensions receive identical decisive verdicts with the single non-identical outcome is a preference vs. tie on Complexity.
+
+## Solution 1 - qwen/qwen35_397b_a17b_high
+
+To find the conditional probability $p$, we model the process using random variables $X_k \in \{1, 2, \dots, 6}$ representing the face of the die that lands on top during the $k$-th roll, where sticker $k$ is placed. Since the die is fair, each $X_k$ is independently and uniformly distributed over the set of faces.
+
+### 1. Analyzing the Condition "Even-numbered stickers are visible"
+A sticker $k$ is visible if it is not covered by any sticker $j > k$. This occurs if and only if the face $X_k$ is not rolled again in any subsequent step.
+*   **Sticker 6:** Always visible as it is the last sticker placed.
+*   **Sticker 4:** Visible if $X_4 \neq X_5$ and $X_4 \neq X_6$.
+*   **Sticker 2:** Visible if $X_2 \neq X_3$, $X_2 \neq X_4$, $X_2 \neq X_5$, and $X_2 \neq X_6$.
+
+Let $E$ be the event that stickers 2, 4, and 6 are visible. The constraints on the sequence $(X_1, \dots, X_6)$ are:
+$$X_4 \neq X_5, \quad X_4 \neq X_6, \quad X_2 \neq X_3, \quad X_2 \neq X_4, \quad X_2 \neq X_5, \quad X_2 \neq X_6$$
+
+### 2. Calculating the Size of Event $E$
+We count the number of sequences satisfying these constraints:
+*   $X_1$: 6 choices.
+*   $X_2$: 6 choices.
+*   $X_3$: 5 choices ($X_3 \neq X_2$).
+*   $X_4$: 5 choices ($X_4 \neq X_2$).
+*   $X_5$: 4 choices ($X_5 \neq X_2$ and $X_5 \neq X_4$).
+*   $X_6$: 4 choices ($X_6 \neq X_2$ and $X_6 \neq X_4$).
+
+Total sequences in $E$:
+$$|E| = 6 \times 6 \times 5 \times 5 \times 4 \times 4 = 14400$$
+
+### 3. Analyzing the Condition "Exactly one face is blank"
+Let $B$ be the event that exactly one face is left blank. A face is blank if it never lands on top. For exactly one face to be blank, the set of values $\{X_1, \dots, X_6}$ must have cardinality 5. This implies the sequence consists of 5 distinct values, where one value $R$ appears twice and the other four values $y_1, y_2, y_3, y_4$ appear once.
+
+### 4. Calculating the Size of $B \cap E$
+To satisfy both $B$ and $E$, we choose a missing value $M$ (6 choices) and a repeated value $R$ (5 choices). For each pair $(M, R)$, we must place the two occurrences of $R$ at indices $(i, j)$ such that no constraint $x_i \neq x_j$ is violated. Since the other values are distinct, violations only occur if both occurrences of $R$ are placed at indices that are constrained to be different.
+
+The forbidden pairs of indices for $R$ are:
+$$\{(2, 3), (2, 4), (2, 5), (2, 6), (4, 5), (4, 6)}$$
+There are 6 forbidden pairs out of $\binom{6}{2} = 15$ total pairs. Thus, there are $15 - 6 = 9$ allowed pairs of positions for $R$. For each allowed pair, the remaining 4 distinct values can be arranged in $4! = 24$ ways.
+$$|B \cap E| = (6 \times 5) \times 9 \times 24 = 30 \times 216 = 6480$$
+
+### 5. Final Probability
+The conditional probability $p$ is:
+$$p = \frac{|B \cap E|}{|E|} = \frac{6480}{14400} = \frac{648}{1440} = \frac{9}{20}$$
+Since $m=9$ and $n=20$ are relatively prime, we find:
+$$m + n = 9 + 20 = 29$$
+
+\boxed{29}
+
+
+## Solution 2 - openai/gpt-54
+
+\[
+\boxed{29}
+\]
+
+Let \(X_i\) be the face on top on the \(i\)-th roll, for \(i=1,2,\dots,6\).
+Since the die is fair and the rolls are independent, all \(6^6\) sequences
+\((X_1,X_2,\dots,X_6)\) are equally likely.
+
+A face is blank at the end exactly when it was never on top in any of the six rolls.
+So "exactly one face has been left blank" means that exactly \(5\) distinct faces appear among
+\[
+X_1,X_2,X_3,X_4,X_5,X_6.
+\]
+
+Now let \(E\) be the event that all even-numbered stickers \(2,4,6\) are visible at the end.
+
+- Sticker \(2\) is visible iff no later sticker covers it, so none of rolls \(3,4,5,6\) may land on face \(X_2\).
+- Sticker \(4\) is visible iff rolls \(5,6\) do not land on face \(X_4\).
+- Sticker \(6\) is always visible, since no sticker comes after it.
+
+Thus \(E\) is equivalent to:
+- \(X_2,X_4,X_6\) are pairwise distinct,
+- \(X_3\neq X_2\),
+- \(X_5\neq X_2\) and \(X_5\neq X_4\).
+
+No other restrictions are needed.
+
+So the number of sequences satisfying \(E\) is
+\[
+6\cdot 5\cdot 4\cdot 5\cdot 4\cdot 6.
+\]
+Indeed:
+- \(X_2\): \(6\) choices,
+- \(X_4\): \(5\) choices (not \(X_2\)),
+- \(X_6\): \(4\) choices (not \(X_2,X_4\)),
+- \(X_3\): \(5\) choices (anything but \(X_2\)),
+- \(X_5\): \(4\) choices (anything but \(X_2,X_4\)),
+- \(X_1\): \(6\) choices.
+
+Now we count the sequences satisfying both \(E\) and the condition "exactly one face blank."
+
+First choose \(X_2,X_4,X_6\). There are
+\[
+6\cdot 5\cdot 4
+\]
+ways to do this, and these are three distinct faces. Call them \(a,b,c\). Let the remaining three faces of the die be \(d,e,f\).
+
+Under \(E\),
+- \(X_3\) can be any face except \(a\), so \(X_3\in\{b,c,d,e,f\}\),
+- \(X_5\) can be any face except \(a,b\), so \(X_5\in\{c,d,e,f\}\),
+- \(X_1\) can be any face.
+
+Since \(a,b,c\) are already used by rolls \(2,4,6\), having exactly one blank face overall means:
+
+> among the remaining faces \(d,e,f\), exactly two are used by \(X_1,X_3,X_5\), and one is never used.
+
+Choose which one of \(d,e,f\) stays blank: \(3\) choices.
+
+Suppose \(f\) is the blank one. Then \(X_1,X_3,X_5\) may only use faces from \(\{a,b,c,d,e\}\), and both \(d\) and \(e\) must appear at least once.
+
+Count these:
+
+- Total choices avoiding \(f\):
+  \[
+  4\cdot 3\cdot 5=60,
+  \]
+  since \(X_3\in\{b,c,d,e\}\), \(X_5\in\{c,d,e\}\), \(X_1\in\{a,b,c,d,e\}\).
+
+- Choices with no \(d\):
+  \[
+  3\cdot 2\cdot 4=24.
+  \]
+
+- Choices with no \(e\):
+  \[
+  3\cdot 2\cdot 4=24.
+  \]
+
+- Choices with neither \(d\) nor \(e\):
+  \[
+  2\cdot 1\cdot 3=6.
+  \]
+
+So the number using both \(d\) and \(e\) is
+\[
+60-24-24+6=18.
+\]
+
+Since there are \(3\) choices for which of \(d,e,f\) is blank, for each fixed choice of \(a,b,c\) we get
+\[
+3\cdot 18=54
+\]
+valid triples \((X_1,X_3,X_5)\).
+
+Therefore the number of sequences satisfying both conditions is
+\[
+6\cdot 5\cdot 4\cdot 54.
+\]
+
+Hence
+\[
+p=\frac{6\cdot 5\cdot 4\cdot 54}{6\cdot 5\cdot 4\cdot 5\cdot 4\cdot 6}
+=\frac{54}{120}
+=\frac{9}{20}.
+\]
+
+Thus \(p=\frac{m}{n}\) with \(m=9\) and \(n=20\), so
+\[
+m+n=9+20=29.
+\]
+
+\[
+\boxed{29}
+\]