question_id int64 1 50 | prompt stringlengths 530 1.97k | solution stringlengths 568 7.39k | answer stringlengths 9 121 | answer_type stringclasses 3
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1 | Divide each side of an equilateral triangle \(ABC\) into four equal parts. Through each division point, draw lines parallel to the other two sides. The sides of triangle\(ABC\) together with these parallel lines define \(15\) lattice points. Find the smallest integer \(n\) such that, no matter which \(n\) points are ch... | Let the three trisection points on side \({AB}\) from \(A\) to \(B\) be \(L,F,W\); the three trisection points on side \({BC}\) from \(B\) to \(C\) be \(V,D,U\); and the three trisection points on side \({CA}\) from \(C\) to \(A\) be \(N,E,M\). The intersection point of \({MV},{LU},{EF}\) is \(P\); the intersection poi... | \boxed{6} | number |
3 | Find all positive integers \(\mathbf{n}\) such that for any: \({a}_{n} \geq {a}_{n - 1} \geq {a}_{n - 2} \geq \cdots \cdots {a}_{2} \geq {a}_{1} > 0,\) satisfying \(\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k} = \mathop{\sum }\limits_{{k = 1}}^{n}\frac{1}{{a}_{k}}\), the inequality \(\mathop{\prod }\limits_{{k = 1}}^{n}{... | For \(n=1\), it clearly holds.
For \(n=2\), we have \(a_2a_1=1\), so \(a_1\cdot a_2^2=a_2\geq\sqrt{a_2a_1}=1\), which also holds.
For \(n=3\), we have \(a_1a_2+a_2a_3+a_1a_3=a_1a_2a_3(a_1+a_2+a_3)\leq\frac{1}{3}(a_1a_2+a_2a_3+a_1a_3)^2\). Therefore, \(a_1a_2+a_2a_3+a_1a_3\geq3\), and \(a_1\cdot(a_2)^2\cdot(a_3)^3=a_1... | \boxed{\{1, 2, 3\}} | set |
4 | Let $x_1,x_2,\ldots,x_{2025}$ be real numbers in the interval $[0,1]$, and let $f(x)$ be a real-valued function defined on $[0,1]$. Find the minimum possible value of
\[
\max_{x_1,x_2,\ldots,x_{2025}\in[0,1]}
\left|
\sum_{k=1}^{2025} f(x_k)
-
\left(\sum_{k=1}^{2025} x_k\right)^2
\right|.
\]
After solving the above pro... | Let
$$
m=\max_{x_1,x_2,\dots,x_{2025}}\left|\sum_{k=1}^{2025}f(x_k)-\left(\sum_{k=1}^{2025}x_k\right)^2\right|.
$$
We have
$$
\begin{aligned}
4m\geq&\left|\sum_{k=1}^{2025}f(1)-2025^2\right|+\left|\sum_{k=1}^{2025}f(0)\right|\\
&+\left|\sum_{k=1}^{1012}f(1)+\sum_{k=1013}^{2025}f(0)-1012^2\right|+\left|\sum_{k=1}^{1012}... | \boxed{512578}
| number |
5 | The sequence $\{a_n\}_{n=1}^{\infty}$ consists of positive terms, with $a_1=7$, $a_2=2$, and satisfies the recurrence relation
$$8a_{n+2}^4=3+4a_{n+1}+a_n\quad(n\in\mathbb{N}^*).$$
Find the general term formula for this sequence.
After solving the above problem, please output your final answer in the following format:
... | The following can be proved by mathematical induction:
\[
a_{n+1}=\sqrt{\frac{1+a_n}{2}}\quad(n\in\mathbb{N}^*).
\]
For \(n=1\), the statement can be verified directly.
Assume that for \(n=k\) we have \(a_{k+1}=\sqrt{\dfrac{1+a_k}{2}}\), i.e. \(a_k=2a_{k+1}^2-1\). Then
\[
8a_{k+2}^4=3+4a_{k+1}+a_k=2+4a_{k+1}+2a_{k+... | $$\boxed{\dfrac{(2+\sqrt{3})^{2^{2-n}}+(2-\sqrt{3})^{2^{2-n}}}{2}}$$ | variable |
6 | Let \({x}_{1},{x}_{2},\cdots ,{x}_{2024}\) be positive real numbers such that \({x}_{k} + {x}_{m} \geq {km}\) for any \(1 \leq k < m \leq {2024}\). Find the minimum value of \({x}_{1} + {x}_{2} + \cdots + {x}_{2024}\).
After solving the above problem, please output your final answer in the following format:
### The fin... | From the assumption: for any positive integer \(1 \leq k \leq 1012\), we have
\[
x_{2k-1}+x_{2k} \geq (2k)(2k-1),
\]
thus
\[
x_{1}+x_{2}+\cdots+x_{2024} \geq \sum_{k=1}^{1012} (2k)(2k-1). \tag{①}
\]
When \(x_{2k-1}=2k^{2}-2k+1,\;x_{2k}=2k^{2}-1\;(1 \leq k \leq 1012)\), equality in (①) holds, hence the desired minimum i... | \boxed{1382935444} | number |
7 | \[
\left\lfloor \frac{n!}{(n+1)(n+2)} \right\rfloor
\]
represents the greatest integer less than or equal to
\(\frac{n!}{(n+1)(n+2)}\). Find all possible values of
\[
\left\lfloor \frac{n!}{(n+1)(n+2)} \right\rfloor
- \left\lfloor \frac{1}{32} \cdot \frac{n!}{(n+1)(n+2)} \right\rfloor \times 32.
\]
After solving the... | For any positive integer \(n>10\), if a prime \(p\leq n\), an integer \(k\geq 1\), and
\[
p \mid \bigl(n^{2}+3n+2\bigr),
\]
then
\[
p,2p,\dots,(p^{k-1}-1)p \in \{2,3,\dots,n\}\;\;\Rightarrow\;\;p^{\,p^{k-1}-1}\mid n!.
\]
Moreover, when \(p=2\) or \(p=3\), by choosing \(k\leq 3\), we can obtain
\[
p^{k+1}\mid n!.
\]
In ... | \boxed{\{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30\}} | set |
8 | Given that \(a\) is a positive integer and \(b, c\) are positive real numbers, find the maximum value of
\[
\frac{16(\cos a)^5 - \cos(5a)}{2\cos a + \cos(3a)} \cdot \frac{a^2 b + b^2 c}{a^3 + b^3 + c^3}.
\]
After solving the above problem, please output your final answer in the following format:
### The final answer i... | Since \(a\) is a positive integer, we have: \(2\cos a + \cos(3a) = 4(\cos a)^{3} - \cos a \neq 0\).\(\cos(5a) = 16(\cos a)^{5} - 20(\cos a)^{3} + 5\cos a \Rightarrow \frac{16(\cos a)^{5} - \cos(5a)}{2\cos a + \cos(3a)} = 5\).\(3a^{2}b \leq \frac{a^{3}}{\sqrt[3]{3 - \sqrt{5}}} + \frac{a^{3}}{\sqrt[3]{3 - \sqrt{5}}} + \b... | \boxed{\frac{{15} + 5\sqrt{5}}{6} \cdot \sqrt[3]{{14} - 6\sqrt{5}}} | number |
9 | Given that \(x, y, z\) are all positive real numbers, consider the equation in \(w\):
\[
1121610\sqrt{3270}\, w^{9} - 407425607\, w^{6} + 10360232 = 0
\]
which has \(k\) real roots \(w_{1}, w_{2}, \ldots, w_{k}\), satisfying
\[
w_{1} < w_{2} < \cdots < w_{k}.
\]
If
\[
m = (w_{1} + w_{k}) \cdot w_{\left\lfloor \tf... | Let \(t = \tfrac{5145\sqrt{3270}}{47524}\). Then the equation in \(w\) simplifies as
\[
t w^{9} - (1+t^{2})w^{6} + 1 = 0 \;\;\Rightarrow\;\; (t w^{3}-1)(w^{6}-t w^{3}-1)=0.
\]
Hence, there are three real roots:
\[
w_{1} = \sqrt[3]{\tfrac{t - \sqrt{4+t^{2}}}{2}},\quad
w_{2} = \sqrt[3]{\tfrac{1}{t}},\quad
w_{3} = \sqrt[... | \boxed{6} | number |
10 | Let \(x,y,z\) be positive real numbers such that \({xy} + {xz} + {yz} \neq 1\), \(\frac{\left( {{x}^{2} - 1}\right) \left( {{y}^{2} - 1}\right) }{xy} + \frac{\left( {{x}^{2} - 1}\right) \left( {{z}^{2} - 1}\right) }{xz} + \frac{\left( {{z}^{2} - 1}\right) \left( {{y}^{2} - 1}\right) }{zy} = 4\). Find the minimum value ... | \[
\frac{(x^{2}-1)(y^{2}-1)}{xy}+\frac{(x^{2}-1)(z^{2}-1)}{xz}+\frac{(z^{2}-1)(y^{2}-1)}{yz}=4
\;\;\Rightarrow\;\;
\Bigl(\tfrac{1}{xy}+\tfrac{1}{yz}+\tfrac{1}{xz}-1\Bigr)\Bigl(xy+xz+yz-1\Bigr)=0.
\]
Hence: \(xyz = x+y+z \;\;\Rightarrow\;\; x=\tfrac{y+z}{yz-1}\).
\[
\frac{x^{2}}{169}+9(yz-1)^{2}+\frac{169}{9(y+z)^{2}}... | \boxed{-9126} | number |
11 | Let \(x_{1}, x_{2}, \ldots, x_{n}\) be positive numbers satisfying
\[
x_{1} + x_{2} + \cdots + x_{n} = n.
\]
Find the largest positive integer \(n\) such that
\[
\frac{1}{x_{1}^{2}} + \frac{1}{x_{2}^{2}} + \cdots + \frac{1}{x_{n}^{2}}
\;\geq\;
x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}.
\]
After solving the above prob... | First, we prove that the original inequality holds for \(n=10\).
Let
\[
x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}=10+90t^{2}, \quad t\in[0,1).
\]
By the Cauchy-Schwarz inequality,
\[
9\bigl(10+90t^{2}-x_{1}^{2}\bigr)
=9(x_{2}^{2}+x_{3}^{2}+\cdots+x_{10}^{2})
\geq (x_{2}+x_{3}+\cdots+x_{10})^{2}
=(10-x_{1})^{2},
\]
which g... | \boxed{10} | number |
12 | Space vectors \(\overrightarrow{A_n}(x_{n}, y_{n}, z_{n}) \; (1 \leq n \leq 2052)\) satisfy that
\(\sqrt{1+x_{n}}, \sqrt{1+y_{n}}, \sqrt{1+z_{n}} \; (1 \leq n \leq 2052)\) are all real numbers, and
\[
\sum_{n=1}^{2052} x_{n}^{3} = \sum_{n=1}^{2052} z_{n}^{3} = \sum_{n=1}^{2052} y_{n} - 2052 = 0,
\]
\[
\sum_{n=1}^{2... | From the problem statement we have:
\[
\sum_{n=1}^{2052} y_{n} = 2052,\qquad
\sum_{n=1}^{2052} y_{n}^{2} = 2052^{2}-1.
\]
By the EV-Theorem, the minimum value of
\[
\sum_{n=1}^{2052} y_{n}^{3}
\]
is attained if and only if exactly \(2050\) of the \(y_{n}\) are zero.
At that point,
\[
\sum_{n=1}^{2052} y_{n}^{3} = 2052... | \boxed{\frac{3}{5} \cdot \sqrt{{4678570} + {2280}\sqrt{4210702}}} | number |
13 | Let $\{a_k\}$ be a sequence of non-negative integers, and $\{b_k\}$ be a sequence of positive real numbers. Given that:
1.$$\left(\sum_{n=1}^{2023} a_n b_n\right)\left( \sum_{k=0}^{673} \frac{1}{(k+675)(1348-k)} \right)= \sum_{k=1}^{674}\frac{1}{k(2k-1)}$$
2.$$\frac{\sum_{k=1}^{1364176}\frac{1}{k(2k-1)}}{\sum_{k=0}^{13... | Solution:
\[
\frac{\sum_{k=1}^{674} \frac{1}{k(2k-1)}}{\sum_{k=0}^{673} \frac{1}{(k+675)(1348-k)}}
= \frac{2\left(\sum_{k=1}^{1348}\frac{1}{k} - \sum_{k=1}^{674}\frac{1}{k}\right)}
{\tfrac{2}{2023}\left(\sum_{k=1}^{1348}\frac{1}{k} - \sum_{k=1}^{674}\frac{1}{k}\right)}
= 2023 \;\;\Rightarrow\;\; \sum_{n=1}^{2023} a_n b... | \boxed{\frac{33497570861567}{2}} | number |
14 | Let \(\{a_n\}\) be a sequence of positive integers, and let \([x]\) denote the greatest integer less than or equal to \(x\). Consider the system of equations in \(x,y\):
\[
\big([y] \cdot x^2 + 2n + 2\big)\left(\sum_{k=0}^{n-1} \frac{1}{(k+n+1)(2n-k)}\right)
= \sum_{k=1}^{n} \frac{1}{k(2k-1)},
\]
\[
4\big([x]y^2\big... | Solution:
$$\frac{\sum_{k=1}^{n} \frac{1}{k(2k-1)}}{\sum_{k=0}^{n-1} \frac{1}{(k+n+1)(2n-k)}}= \frac{2\left(\sum_{k=1}^{2n}\frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right)}{\frac{2}{3n+1}\left(\sum_{k=1}^{2n}\frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right)}
= 3n+1 \;\;\Rightarrow\;\; [y]\cdot x^2 = n-1 $$
$4([x]\cdot y^2)... | \boxed{59} | number |
15 | Let \( a, b, c, d \) be real numbers such that \[4a^4 + \frac{c^4}{16} + b^4 + b^3d^2 + 27 + \frac{d^2}{4} + a^2c^2 = 3c^2 + 24a^2 + 8b^2 + 4d^2.\] Find the minimum value of \[\frac{4a^4 + \frac{c^4}{16} + a^2c^2 + 12a^2 + \frac{3}{2}c^2 + b^4 + \frac{d^4}{4} + 10 + b^2d^2 + 2b^2 + d^2 }{\sqrt{2a^2b^2 + a^2d^2 + 2a^2 +... | Solution: Let
$$x = 2a^2 + 3 + \frac{c^2}{4}, \quad y = b^2 + \frac{d^2}{2} + 1.$$
We have
\[
4a^4 + \frac{c^4}{16} + b^4 + b^2d^2 + 27 + \frac{d^2}{4} + a^2c^2
= 3c^2 + 24a^2 + 8b^2 + 4d^2
\;\;\Rightarrow\;\; (x-9)^2 + (y-5)^2 = 25.
\]
Since $x>0,\; y>0$, set $y = kx$. Then
\[
(x-9)^2 + (y-5)^2 = 25
\;\;\Rightar... | \boxed{\frac{9\sqrt{3}}{2}} | number |
16 | Let $n$ be a positive integer such that $ 2 \leq n \leq 5 $. For each such $n$, let $A_nB_nC_nD_n$ be a rectangle with vertex $A_n$ at $(-n, 1 + 6 \sqrt{2} - \sqrt{2}n)$, and vertex $B_n$ the reflection of $A_n$ through the origin. The line segments $A_nC_n$ and $B_nD_n$ are parallel to the $x$-axis, and the line segme... | Solution: By case analysis to remove the absolute value and using derivatives to test monotonicity, we obtain the points on the curve \((y-|x|)^2+x^2=1\).
The maximum \(y\)-coordinate is \(\sqrt{2}\), attained at \(\left(\pm\frac{\sqrt{2}}{2},\;\sqrt{2}\right)\);
the minimum \(y\)-coordinate is \(-1\), attained at \((... | \boxed{68} | number |
17 | Let \(\{x_1, x_2, \ldots, x_{2025}\} = \{1, 2, \ldots, 2025\}\).
Find the greatest positive integer less than
\[
\sum_{i=1}^{2024} \frac{x_i}{x_{i+1}}.
\]
After solving the above problem, please output your final answer in the following format:
### The final answer is: $\boxed{<your answer>}$
Example:
### The final an... | Solution: Let $y_i =
\begin{cases}
x_i & \text{if } x_i < x_{2025} \\
x_i - 1 & \text{if } x_i > x_{2025}
\end{cases},
\quad
z_i =
\begin{cases}
\dfrac{1}{x_{1+i}} & \text{if } x_{1+i} > x_1 \\[6pt]
\dfrac{1}{1+x_{1+i}} & \text{if } x_{1+i} < x_1
\end{cases}$
By the rearrangement (sorting) inequality:
... | \boxed{2016} | number |
18 | Let complex numbers \(z_1, z_2, \ldots, z_{2025}\) satisfy
\[
\sum_{k=1}^{2025} z_k = 0,\qquad \sum_{k=1}^{2025} |z_k|^2 = 1.
\]
Find the minimum possible value of
\[
\max_{1 \le i < j \le 2025} |z_i - z_j|.
\]
After solving the above problem, please output your final answer in the following format:
### The final answ... | Solution:On one hand, equality holds when we take 675 copies of each of $\dfrac{1}{2025}, \dfrac{1}{2025}\omega, \dfrac{1}{2025}\omega^2$, where $\omega = e^{\frac{2\pi i}{3}}$.
First, we prove a lemma.
Lemma: If the distance between any two of three points in a plane is at most $1$, then these three points can be co... | \boxed{\frac{\sqrt{3}}{45}} | number |
19 | Let \(a_1 < a_2 < \cdots < a_{38}\) be positive integers with
\[
a_1 + a_2 + \cdots + a_{38} = 1997.
\]
Define
\[
b_k = \min\!\left(\sum_{i=1}^{k} a_i,\; \sum_{i=k+1}^{38} a_i\right),
\]
and let
\[
S = \max(b_1, b_2, \ldots, b_{38}),
\]
where \(b_{38} = 0\).
Find the minimum possible value of \(S\).
After solving the ... | Solution:On one hand, let
$$a_i = i \;(1 \leq i \leq 36), \quad a_{37} = 665, \quad a_{38} = 666.$$
We have
$$\sum_{i=1}^{38} a_i = 1997, \quad S = 666.$$
On the other hand, we prove that $S \geq 666$.
Let $k \;(1 \leq k \leq 37)$ be the index such that
$$a_1+a_2+\cdots+a_k \leq a_{k+1}+\cdots+a_{38} \quad \text{and} ... | \boxed{666} | number |
20 | A school has newly formed \(3\) classes. It is known that for any two students belonging to different classes, in the third class there are exactly \(10\) students who know both of them and \(10\) students who know neither of them. Find the total number of students across the three classes.
After solving the above prob... | Solution: Let $A,B,C$ be three classes with $a,b,c$ students, respectively. We represent each student by a point in the plane such that no three points are collinear. The points corresponding to the students in classes $A,B,C$ form sets $A,B,C$ containing $a,b,c$ points, respectively. No edge is drawn between points co... | \boxed{120} | number |
21 | Given that \(x, y, z\) are integers satisfying the system
\[
\begin{cases}
x^2 - 7y^4 = z,\\[6pt]
7z^4+\displaystyle\frac{\sum_{k=1}^{z} \frac{1}{k(2k-1)}}{\sum_{k=0}^{\,z-1} \frac{1}{(k+z+1)(2z-k)}}
= \displaystyle\frac{\sum_{i=0}^{174545562} \frac{1}{i(2i-1)}}{\sum_{i=0}^{174545561} \frac{1}{(i+174545563)(349091124-i... | Solution:
\[\frac{\sum_{k=1}^{z} \frac{1}{k(2k-1)}}{\sum_{k=0}^{z-1} \frac{1}{(k+z+1)(2z-k)}}= \frac{2\left(\sum_{k=1}^{2z}\frac{1}{k} - \sum_{k=1}^{z}\frac{1}{k}\right)}{\frac{2}{3z+1}\left(\sum_{k=1}^{2z}\frac{1}{k} - \sum_{k=1}^{z}\frac{1}{k}\right)}
= 3z+1\]
Similarly, we get
\[
\frac{\sum_{i=1}^{174545562} \frac{1... | \boxed{230} | number |
22 | Let variables \(x, y, c\) satisfy \(x+y+c > \sqrt{13}\) with \(x, y \ge 0\).
Find the minimum value of
\[
\frac{(2+x^2)\,(4+y^2)\,\big(49+6c^2+4c\sqrt{3+3c^2}\big)}
{\big(7\sqrt{3}\,c+14\sqrt{1+c^2}\big)\,(x+y)}.
\]
After solving the above problem, please output your final answer in the following format:
### The final... | Solution:
Let
\[
f(c) = \frac{49+6c^2+4c\cdot\sqrt{3+3c^2}}{7\sqrt{3}c+14\sqrt{1+c^2}},
\]
then
\[
f'(c) = \frac{2\sqrt{3}}{7} - \frac{7\left(\sqrt{3}+\dfrac{2c}{\sqrt{1+c^2}}\right)}{(\sqrt{3}c+2\sqrt{1+c^2})^2}.
\]
When $c=\sqrt{3}$, $f'(c)=0$; when $c>\sqrt{3}$, $f'(c)>0$; when $c<\sqrt{3}$, $f'(c)<0$.
Therefore, ... | \boxed{\frac{208\cdot 3^{1/4}}{21}} | number |
23 | It is known that the equation in \(u, v\), \(u^2 - 3v^4 = 1\), has at most four positive integer solutions; and the equation in \(\lambda, \iota\), \(4\lambda^2 - 3\iota^2 = 1\), has at most four positive integer solutions. If a prime \(p\) satisfies \(p \equiv 5 \pmod{8}\), and integers \(x, y\) satisfy
\[
\left(\frac... | Solution:
Let
\[
u = \sqrt[3]{\frac{1+x^2}{2} + \frac{1}{6}\sqrt{9x^4+\frac{4x^6}{3}+9+18x^2}},
\quad
w = \sqrt[3]{\frac{1+x^2}{2} - \frac{1}{6}\sqrt{9x^4+9+\frac{4x^6}{3}+18x^2}}.
\]
Then
\[
u^3+w^3 = 1+x^2,
\quad
u^3w^3 = -\frac{x^6}{27} \Rightarrow uw = -\frac{x^2}{3},
\quad
u+w=1.
\]
Because $(x+1, \; x^2-x+1... | \boxed{-1} | number |
28 | For each positive integer \(k \ge 2\), define \(f(k)\) to be the product of the distinct prime factors of \(k\). All the coefficients of the polynomial function \(9P(x)\) are rational numbers. Suppose \(45 \le P(2025) \le 81\), and there are infinitely many positive integers \(n\) for which \(P(n) = f(n)\).
How many po... | Solution: From the definition of $f$, for any positive integer $k\geq 2$, we have $2\leq f(k)\leq k$. Therefore, there exist infinitely many positive integers $n$ such that $2\leq P(n)\leq n$. Thus, $P(x)$ must be a constant function or a linear function, and the coefficient of the leading term must be positive.
(1) $... | \boxed{42} | number |
29 | Let \(a\) be a prime greater than \(100\) with \(a \equiv 3 \pmod{5}\).
For complex numbers \(z_n\) \((1 \le n \le a)\), both the real and imaginary parts are nonzero.
For each \(n\), let \(d_n\) be the distance from the point \((n,\lvert z_n\rvert)\) to the line
\[
y=\frac{202}{5}x+2025.
\]
Choose one complex number a... | Solution: Let $x_i = i^{d_i}-1$. The given conditions are equivalent to:
\[
\sum_{i=1}^a x_i = a, \quad \sum_{i=1}^a x_i^2 = a^2-1, \quad \sum_{i=1}^a x_i^3 = a^3 - \tfrac{3}{2}a
\]
According to the EV-Theorem, under the conditions $\sum_{i=1}^a x_i = a$ and $\sum_{i=1}^a x_i^2=a^2-1$, the sum $\sum_{i=1}^a x_i^3$ atta... | \boxed{\frac{21}{103}} | number |
30 | Let \( p, q \) be positive integers satisfying the system of congruences:
\[
\begin{cases}
p \equiv 3 \pmod{911}, \\
q \equiv 2 \pmod{997}, \\
p \equiv 738 \pmod{q};
\end{cases}
\]
Let \( \{ a_n \}, \{ b_n \}, \{ c_n \} \) be periodic sequences of rational numbers satisfying the equations:
\[
\left[ 4b_n + 1 - 3\sin(\p... | Solution:
\[
0 = \sum_{i=1}^p b_i^3, \quad 4b_n+1-3\sin(\pi \cdot a_n)=0 \Rightarrow b_n = \frac{3\sin(\pi \cdot a_n)-1}{4}\geq -1 \Rightarrow (b_n+1)\left(b_n-\tfrac{1}{2}\right)^2 \geq 0 \Rightarrow \frac{p}{3} \geq \sum_{i=1}^p b_i
\]
\[
0 = \sum_{i=1}^q c_i^3, \quad 4c_n-1+3\sin(\pi \cdot a_n)=0 \Rightarrow c_n = \... | \boxed{3736} | number |
31 | Let \((a_i, b_i)\), for \(1 \leq i \leq 2025\), be 2025 distinct points on the curve
\[3x^2 + 3y^2 = 8 + 4xy + 2(x + y)|x - y|\]
Let \(d_i\) be the distance from the point \((a_i, b_i)\) to the line \(x + y = 6\), and let \(\left\lfloor d_i \right\rfloor\) denote the greatest integer not exceeding \(d_i\).
It is given... | Solution:
Rotating by $45^\circ$, we get: the equation of the curve is $x^2 + 5y^2 - 4x|y| = 8$; the equation of the line is $y=3\sqrt{2}$.
By case analysis, the range of y is found to be $[-2\sqrt{2},2\sqrt{2}]$, therefore $d_i \in [\sqrt{2},5\sqrt{2}]$.
Let the function be
\[
f(m)=\frac{21-15\sqrt{2}}{2}\left(m^3-3... | \boxed{2295} | number |
32 | Let \(x\) and \(y\) be positive rational numbers that satisfy:
\[
1 + \sqrt{1 - 2y} = y + \frac{4x - 2 + 2\sqrt{1 - 2x}}{3 - 5x - \sqrt{1 - 2x}} + \sqrt[3]{\frac{1 + x^2}{2} + \frac{1}{6} \cdot \sqrt{9x^4 + \frac{4x^6}{3} + 9 + 18x^2}}+ \sqrt[3]{\frac{1 + x^2}{2} - \frac{1}{6} \cdot \sqrt{9x^4 + \frac{4x^6}{3} + 9 + 18... | Solution:
Let
\[
a = \sqrt[3]{\frac{1 + x^2}{2}+ \frac{\sqrt{9x^4+\frac{4x^6}{3}+9+18x^2}}{6}}, \quad b = \sqrt[3]{\frac{1 + x^2}{2}- \frac{\sqrt{9x^4+\frac{4x^6}{3}+9+18x^2}}{6}}
\]
We have:
\[
a^3 + b^3 = 1 + x^2, \quad a^3b^3=-\frac{x^6}{27} \Rightarrow a+b = 1
\]
The given condition simplifies to
\[
\sqrt{1 - 2y}=y... | \boxed{225} | number |
33 | An album contains 25 photos, each showing three people, subject to the following conditions: (1) In each photo, the person on the left is a full brother of the person in the middle, and the person on the right is the daughter of the person in the middle. (2) The person in the middle is different in each of the 25 photo... | Solution:
At least \( 34 \) different people appear in this photo album.
Construction: Let \( x \to y \) denote that \( x \) is a parent of \( y \), and let \( x - y \) denote that \( x \) and \( y \) are siblings.
Take \( 34 \) people as follows, \( 25 \) of whom are female:
$$ g_i \to g_{i+5} \to g_{i+10} \to g_{i+... | \boxed{34} | number |
34 | Find $C_{\max}$ such that $\forall x, y, z \in \mathbb{R}$,$$f = 1 + |x+y+z| + |xy+yz+zx| + |xyz| \geq C(|x| + |y| + |z|)$$
After solving the above problem, please output your final answer in the following format:
### The final answer is: $\boxed{<your answer>}$
Example:
### The final answer is: $\boxed{123}$
The final... | Solution:
Let \( f(t) = \frac{1+3t^2+2t^3}{4t} \quad (0 < t \leq \frac{1}{2}) \). Then \( f'(t) = \frac{4t^3 + 3t^2 - 1}{4t^2} \) has a unique zero in \( (0, \frac{1}{2}) \) at \( t_0= \frac{\sqrt[3]{7 + 4\sqrt{3}} + \sqrt[3]{7 - 4\sqrt{3}} - 1}{4}\). Thus, \( f(t) \) has a minimum value \( f(t_0)=\frac{3(16+(\sqrt[3]{... | \boxed{\frac{3(16+(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}-1)^2)}{32(\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}-1)}} | number |
35 | In a game, a positive integer $n$ not exceeding 10000 is initially written on a blackboard. Two players, Player 1 and Player 2, take turns making moves, with Player 1 going first. A move consists of erasing a positive integer $k$ greater than 1 from the blackboard and replacing it with five non-negative integers that s... | For a given positive integer \(n\), let \(f\left( n\right)\) denote the number of operations that can be performed on this number. It is not difficult to see that if \(f\left( n\right)\) is odd, the first player has a winning strategy, and if \(f\left( n\right)\) is even, the second player has a winning strategy.
For ... | \boxed{7657} | number |
36 | On a 19*19 board, a black or a white stone is placed on each of the 361 squares. An operation consists of choosing one row and one column and reversing the color of all 37 stones in that row and column. Find the largest integer n such that, regardless of the initial arrangement of stones, it is always possible to perfo... | First, consider the $18\times18$ space in the upper-left corner. Select a point and perform one operation on all 35 pieces of its row and column, then this point is affected by 35 operations, its color flipped; other points in the same row or column with it are operated on 18 times, and points not in the same row and c... | \boxed{343} | number |
37 | In $\triangle ABC$ with $AB < AC < BC$, let $O$ and $I$ be its circumcenter and incenter, respectively. Let $M_b$ and $M_c$ be the midpoints of the arc $AC$ (not containing $B$) and the arc $AB$ (not containing $C$), respectively, on the circumcircle $\odot O$. Let $S_b$ be a point on $\odot O$ such that $IS_b \perp BS... | Solution:
Lemma 1. In \(\bigtriangleup {ABC}\), let \(I\) be the incenter and let \(S\) be a point on the circumcircle such that \({IS} \bot {AS}\). Let \(M\) be the midpoint of the arc \(\overset{⏜}{BC}\) containing point \(A\), and let \(D\) be the point of tangency of the incircle \(\odot I\) with side \({BC}\).... | \boxed{\frac{(a-b)(a-c)(b-c)}{8}\cdot \left( \frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc-(a+b-c)(a-b+c)(-a+b+c)} \right)^3} | variable |
43 | Let \(f(x),g(x),h(x)\) be functions defined on the set of positive integers.
Let \(\left\lfloor \tfrac{x}{2}\right\rfloor\) denote the greatest integer less than or equal to \(\tfrac{x}{2}\).
Define \(f(x)\) to be the number of integers in the list \(1,2,3,\dots,\left\lfloor \tfrac{x}{2}\right\rfloor\) that are coprim... | Solution:
When \(b=2^{\beta}\) with \(\beta\ge 7\):
\[
f(b)=2^{\beta-2},\qquad g(b^{2})=2^{2\beta},\qquad h\bigl(g(b^{2})\bigr)\le 2\beta\cdot 2,
\]
hence \(2^{\beta-1}\le 4\beta\), a contradiction.
When \(b=2^{\beta}\) with \(\beta\le 6\): checking case by case shows that only \(\beta=6\) works, i.e. \(b=64\) is a... | \boxed{\{40, 64, 75, 100, 150\}} | set |
44 | Let $a=\log_3\bigg(\frac{1}{2}\sum\limits_{n=0}^{676}(-1)^n{2028 \choose 3n}\bigg)$. Calculate the sum of distinct prime factors of $a^2+(a-1)^4$.
After solving the above problem, please output your final answer in the following format:
### The final answer is: $\boxed{<your answer>}$
Example:
### The final answer is: ... | Solution:
If let \(\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2}i\), \({\omega }^{3} = 1\); thus
\({1}^{{2028} - n} + {\omega }^{{2028} - n} + {\left( {\omega }^{2}\right) }^{{2028} - n} = \left\{ \begin{array}{ll} 0 & \text{ if }3 \nmid n \\ 3 & \text{ if }3 \mid n \end{array}\right.\), then
\[
\mathop{\sum }\... | \boxed{8281323}
| number |
45 | Find the largest positive integer \(n\) that satisfies the following property: there exists a sequence of \(n\) positive integers \({a}_{1},\cdots ,{a}_{n}\) such that \({a}_{i} \leq {2}^{2025}\left( {\forall 1 \leq i \leq n}\right)\), and for any \(1 \leq i < j \leq n\), there is no non-empty proper subset \(A \s... | First, we show that \(n = {2}^{2026}\) does not satisfy the condition. Take any \({a}_{1},\cdots ,{a}_{n} \in \left\{ {1,2,\cdots ,{2}^{2025}}\right\}\), we define a sequence recursively as follows \({s}_{1},\cdots ,{s}_{n} \in \{ 1, - 1\}\):
1. \({s}_{1} = 1\).
2. If \({s}_{1},\cdots ,{s}_{m - 1}\) have been chosen,... | \boxed{{2}^{2026}-1} | number |
47 | Assume that graph $G$ has $50$ vertices, and for any two non-adjacent vertices, they have exactly two common neighbors. Vertices $v_1$ and $v_2$ are adjacent and have no common neighbors, the degree of $v_1$ is $d(v_1)=11$. The parameter $M$ is defined as: if such a graph $G$ exists, then $M=max(d(v_2)-d(v_1))$; otherw... | Assume the graph $G$ described in the problem exists. Let $A$ be the set of neighbors of $v_1$ (excluding $v_2$); then $|A| = 10$. Let $B$ be the set of neighbors of $v_2$ (excluding $v_1$); then $|B| = d - 1$, where $d$ is the degree of $v_2$. Since $v_1$ and $v_2$ have no common neighbors, $A$ and $B$ are disjoint. C... | \boxed{72} | number |
48 | A city is going to be built as a $10\times 10$ square grid of blocks. Now, the city is planning to set up security booths in some blocks. To ensure the safety of the entire city, city safety regulations require that for any block (regardless of whether it has a security booth itself), at least one of its four adjacent ... | A minimum of 30 security booths is needed. The proof: We use $(i, j)$ to denote the block in row $i$ and column $j$, where $1 \le i, j \le 10$.
1. Prove the minimum required number (Lower Bound). First, we prove that at least 30 security booths are needed. We define the distance $d(i, j)$ from a block $(i, j)$ to the ... | \boxed{30} | number |
49 | In the Cartesian plane, there is a lamp at every integer lattice point, each with two states: ``on'' and ``off''. The lamp board features a special switching system: pressing any lamp toggles the states of all lamps in its entire row and entire column (on to off, and off to on). Define $f(x)=x^2-18x+23$ and $g(x)=37x+1... | First, let's solve $f(f(x))=g(x)$. Assume there exist some solutions $x$ such that $f(x) = -x - a$, where $a$ is a constant. Substitute the expression for $f(x)$: $x^2 - 18x + 23 = -x - a \implies x^2 - 17x + (23 + a) = 0$ Now, substitute the relation $f(x) = -x - a$ into the left side of the original equation $f(f(x))... | $$
\boxed{16}
$$ | number |
50 | Let the polynomial $f(x)=x^6 - 11x^5 + 28x^4 + 2x^3 + 57x^2 + 13x + 30$, and let $\alpha$ be the smallest positive real root of the equation $f(x)=0$. Set $N$ as the remainder of the closest integer to $\alpha^{35}$ divided by 1498.
A logistics company warehouses pallets with serial numbers $m=1,2,\dots, M$. To simpli... | The entire solution process is divided into two parts. First, we calculate the value of $N$; then, using the value of $N$, we determine the counting range $M$ for the second part and solve the counting problem. Part One: Calculate the value of $N$. The given polynomial equation is $f(x)=x^6 - 11x^5 + 28x^4 + 2x^3 + 57x... | \boxed{180} | number |
AMO-Bench-P
This is the 39-problem, parser-graded subset of
meituan-longcat/AMO-Bench.
It keeps every row whose answer_type is not description:
subset = dataset.filter(lambda row: row["answer_type"] != "description")
The retained answer types are:
answer_type |
Rows |
|---|---|
number |
34 |
set |
3 |
variable |
2 |
| Total | 39 |
The upstream evaluator verifies these three types programmatically. It sends
the 11 excluded description rows to an LLM judge. The upstream project calls
this parser-only selection AMO-Bench-P and implements --only_parser by
skipping rows where answer_type == "description".
Provenance
- Source dataset revision:
2f422616c25d862984408fbbfaed63a961e8e025 - Source evaluator revision:
52e1f378e4bcb0c593e860be38f9251b1a192571 - Source Parquet SHA-256:
eb08f5239c0cc092e13552c31e505a7fdb8e8140aa8e227beedd62fc442cde58 - Filtered Parquet SHA-256:
e282346a1c721284c4dd8bc7527000b46489acf104a6c3b0d81085060c02ff01
The row schema, values, order, and original question_id values are preserved.
Only the 11 description rows are removed.
Use
After uploading this directory as the root of a Hugging Face dataset repository:
from datasets import load_dataset
dataset = load_dataset("YOUR_USERNAME/YOUR_DATASET_NAME", split="test")
Use the official AMO-Bench evaluator for answer extraction and verification.
License and citation
AMO-Bench is released under the MIT License. See LICENSE and the
upstream dataset card
for its citation.
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