problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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4. Let $f(x)=(x+a)(x+b)$ (where $a, b$ are given positive real numbers), and $n \geqslant 2$ be a given integer. For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1}+x_{2}+\cdots+x_{n}=1
$$
find the maximum value of
$$
F=\sum_{1 \leq i<j \leq n} \min \left\{f\left(x_{i}\right), f\left(x_{j}\... | 4. Solution 1
$$
\begin{array}{l}
\min \left\{f\left(x_{i}\right), f\left(x_{j}\right)\right\} \\
=\min \left\{\left(x_{i}+a\right)\left(x_{i}+b\right),\left(x_{j}+a\right)\left(x_{j}+b\right)\right\} \\
\leqslant \sqrt{\left(x_{i}+a\right)\left(x_{i}+b\right)\left(x_{j}+a\right)\left(x_{j}+b\right)} \\
\leqslant \frac... | \frac{n-1}{2}\left(\frac{1}{n}+a+b+n a b\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,911 |
5. Let $n$ be a positive even number without square factors, $k$ be an integer, and $p$ be a prime number, satisfying
$$
p<2 \sqrt{n}, p \nmid_{n, p} \mid\left(n+k^{2}\right) \text {. }
$$
Prove: $n$ can be expressed as $a b+b c+c a$, where $a, b, c$ are distinct positive integers. | 5. Since $n$ is even, hence $p \neq 2$.
Also, $p \nmid$, so $p \nmid k$.
Assume without loss of generality that $00$, and $c \neq a, b$.
By the AM-GM inequality, we have $\frac{n}{k}+k \geqslant 2 \sqrt{n}>p$, thus $n+k^{2}>p k$.
This implies $c>0$.
If $c=a$, then $\frac{n+k^{2}}{p}-k=k$, which means
$$
n=k(2 p-k) \te... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,912 |
6. Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a$, $b$, and $c$ in $S$ such that $a+b$, $b+c$, and $c+a$ are all in the set $A$. | 6. Let $a<b<c$. Let
$x=a+b, y=a+c, z=b+c$.
Then $x\langle y\langle z, x+y\rangle z$, and $x+y+z$ is even. (1)
Conversely, if there exist $x, y, z \in A$ satisfying property (1), then take
$$
a=\frac{x+y-z}{2}, b=\frac{x+z-y}{2}, c=\frac{y+z-x}{2},
$$
we have $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,913 |
1. Each interior angle of a convex polygon is $150^{\circ}$, then this convex polygon has ( ) diagonals.
(A) 42
(B) 54
(C) 66
(D) 78 | -、1. B.
Since each exterior angle is $30^{\circ}$, the number of sides of this convex polygon is 12. Therefore, there are $\frac{12 \times 9}{2}=54$ diagonals. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,914 |
2. As shown in Figure 1, the diagonals of
rectangle $A B C D$ intersect at point $O, A E$ bisects
$\angle B A D$ and intersects $B C$ at
point $E$. If $\angle C A E=$
$15^{\circ}$, then $\angle B O E=$ ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 2. D.
From the given conditions, we know $\angle B A E=45^{\circ}$. Thus,
$$
\angle B A O=60^{\circ}, \angle B E A=45^{\circ} \text {. }
$$
Therefore, $\triangle A B O$ is an equilateral triangle, and $\triangle A B E$ is an isosceles right triangle.
Hence, $B E=A B=B O, \angle A B O=60^{\circ}$.
Thus, $\angle O B E=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,915 |
3. Let the roots of the equation $(x-a)(x-b)-x=0$ be $c$ and $d$. Then the roots of the equation $(x-c)(x-d)+x=0$ are ( ).
(A) $a, b$
(B) $-a,-b$
(C) $c, d$
(D) $-c,-d$ | 3. A.
From the condition, we have
$$
\begin{array}{l}
(x-a)(x-\dot{b})-x=(x-c)(x-d) \\
\Rightarrow(x-c)(x-d)+x=(x-a)(x-b) .
\end{array}
$$
Therefore, the roots of $(x-c)(x-d)+x=0$ are $a$ and $b$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,916 |
4. If the inequality $2|x-1|+3|x-3| \leqslant a$ has a solution, then the minimum value of the real number $a$ is ( ).
(A) 1
(B) 2
(C) 4
(D) 6 | 4. C.
By the geometric meaning of absolute value,
$$
\begin{array}{l}
2|x-1|+3|x-3| \\
=2(|x-1|+|x-3|)+|x-3| \\
\geqslant 2 \times 2+0=4,
\end{array}
$$
When $x=3$, the equality holds. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,917 |
Example 6 Given three quadratic equations in $x$
$$
\begin{array}{l}
a x^{2}+b x+c=0, \\
b x^{2}+c x+a=0, \\
c x^{2}+a x+b=0
\end{array}
$$
have exactly one common real root. Then the value of $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) $0(\mathrm{~B}) 1$
(C) 2
(D) 3 | Let $x_{0}$ be their common real root. Then
$$
\begin{array}{l}
a x_{0}^{2}+b x_{0}+c=0, \\
b x_{0}^{2}+c x_{0}+a=0, \\
c x_{0}^{2}+a x_{0}+b=0 .
\end{array}
$$
Adding and rearranging the above three equations, we get
$$
(a+b+c)\left(x_{0}^{2}+x_{0}+1\right)=0 \text {. }
$$
Since $x_{0}^{2}+x_{0}+1=\left(x_{0}+\frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,918 |
5. If any two sides of a triangle are not equal, it is called an "irregular triangle". Among all triangles with vertices on a cube, the number of irregular triangles is ( ).
(A) 18
(B) 24
(C) 30
(D) 36 | 5. B.
As shown in Figure 3, let's assume the edge length of the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ is 1. Then the side length of the triangle can only be $1, \sqrt{2}, \sqrt{3}$, where the edge length is 1, the face diagonal length is $\sqrt{2}$, and the body diagonal length is $\sqrt{3}$.
Fix one body diagonal... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,919 |
6. The number of positive integer solutions $(x, y)$ for the indeterminate equation $x^{2}-2 y^{2}=5$ is ().
(A) 0
(B) 2
(C) 4
(D) infinitely many | 6. A.
If the equation has a positive integer solution $(x, y)$, note that a perfect square is congruent to 0 or 1 modulo 4, thus, $x$ is odd, $y$ is even.
Let $x=2 m+1, y=2 n$. Substituting in, we get
$$
m(m+1)-2 n^{2}=1 \text {. }
$$
Since $m(m+1)$ is even, $2 n^{2}$ is even, this leads to a contradiction.
Therefore... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,920 |
1. The graph of the quadratic function $y=x^{2}-a x+2$ is symmetric about the line $x=1$. Then the minimum value of $y$ is $\qquad$ . | $$
=, 1.1 .
$$
From the condition, we know that $a=2$. Then $y=(x-1)^{2}+1$ has a minimum value of 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,921 |
2. Given $a=\sqrt{3}-1$. Then the value of $a^{2012}+2 a^{2011}-$ $2 a^{2010}$ is $\qquad$. | 2. 0 .
From the condition we know
$$
\begin{array}{l}
(a+1)^{2}=3 \Rightarrow a^{2}+2 a-2=0 \text {. } \\
\text { Then } a^{2012}+2 a^{2011}-2 a^{2010} \\
=a^{2010}\left(a^{2}+2 a-2\right)=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,922 |
3. In $\triangle A B C$, it is known that $A B=\sqrt{39}, B C=6$, $C A=\sqrt{3}$, $M$ is the midpoint of side $B C$, and a perpendicular line is drawn from point $B$ to the extension of $A M$, with the foot of the perpendicular being $D$. Then the length of segment $B D$ is $\qquad$ | 3. $\frac{3}{2}$.
From the given condition, we know $A B^{2}=B C^{2}+C A^{2}$. By the converse of the Pythagorean theorem, $\angle A C B=90^{\circ}$.
By the Pythagorean theorem, we have
$$
A M=\sqrt{A C^{2}+C M^{2}}=2 \sqrt{3}=2 A C \text {. }
$$
Therefore, $\angle D M B=\angle C M A=30^{\circ}$. Hence, $B D=\frac{1}... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,923 |
4. In a chess tournament, there are $n$ female players and $9 n$ male players. Each player plays one game against each of the other $10 n-1$ players. The scoring system is as follows: the winner gets 2 points, the loser gets 0 points, and in the case of a draw, each player gets 1 point. After the tournament, it was fou... | 4.1.
Let the total score of the girls be $m$. Then the total score of the boys is $4 m$. According to the problem,
$$
\begin{array}{l}
\frac{10 n(10 n-1)}{2} \times 2=5 m \\
\Rightarrow m=2 n(10 n-1) .
\end{array}
$$
Since each player competes in $10 n-1$ matches, the maximum score is $2(10 n-1)$, so the total score ... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,924 |
1. (20 points) Given that $x_{1}$ and $x_{2}$ are the two real roots of the quadratic equation in $x$
$$
x^{2}+(3 a-1) x+2 a^{2}-1=0
$$
such that
$$
\left(3 x_{1}-x_{2}\right)\left(x_{1}-3 x_{2}\right)=-80
$$
holds. Find all possible values of the real number $a$. | Three, 1. From the given conditions,
$$
\begin{array}{l}
\Delta=(3 a-1)^{2}-4\left(2 a^{2}-1\right) \\
=a^{2}-6 a+5 \geqslant 0 .
\end{array}
$$
Solving this, we get $a \geqslant 5$ or $a \leqslant 1$.
From the relationship between roots and coefficients, we know
$$
\begin{array}{l}
x_{1}+x_{2}=-(3 a-1), x_{1} x_{2}=2... | -\frac{33}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,925 |
2. (25 points) Given the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $M\left(x_{1}, 0\right), N\left(x_{2}, 0\right)$ $\left(0<x_{1}<x_{2}\right)$, and passes through the point $A(0,1)$. A line $l$ passing through point $A$ intersects the $x$-axis at point $C$ and the parabola at point $B$ (differe... | 2. From the conditions, we know that the parabola opens upwards, and its two intersections with the $x$-axis are to the right of the $y$-axis.
Since $\triangle C A N$ is an isosceles right triangle, point $C$ is on the left side of the $x$-axis, and
$\angle C A N=90^{\circ}, \angle A C N=45^{\circ}$.
Thus, $C(-1,0), N... | y=4 x^{2}-5 x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,926 |
3. (25 points) As shown in Figure 2, $A D$ and $A H$ are the angle bisector and altitude of $\triangle A B C$ $(A B > A C)$, respectively, and $M$ is the midpoint of side $A D$. The circumcircle of $\triangle M D H$ intersects $C M$ at point $E$. Prove: $\angle A E B = 90^{\circ}$. | 3. Connect $M H$ and $E H$.
From $M$ being the midpoint of the hypotenuse $A D$ of the right triangle $\triangle A H D$, we know $M A = M H = M D$.
Thus, $\angle M H D = \angle M D H$.
Since $M, D, H, E$ are concyclic,
$\angle M E H = \angle B D M = \angle C H M$.
Also, $\angle C M H = \angle H M E$, so
$\triangle C M... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,927 |
Example 7 Two quadratic equations with unequal leading coefficients $(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0$, (1) and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$ ( $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$. | Given the conditions $a>1, b>1, a \neq b$.
Assume the common root of equations (1) and (2) is $x_{0}$. Then
$$
\begin{array}{l}
(a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)$ - (4) $\times(a-1... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,929 |
2. According to the main data bulletin of the 6th National Population Census in 2010, the total population within China's territory was 1,339,724,852 people. Compared to the 5th National Population Census in 2000, it increased by 73.9 million people over 10 years. The average annual growth rate over these 10 years is (... | 2. $0.569 \%$ | 0.569 \% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,930 |
3. Let $AB$ be a chord of the unit circle $\odot O$. If the area of $\odot O$ is exactly equal to the area of the square with side $AB$, then $\angle AOB=$ $\qquad$ (to 0.001 degree). | 3. $124.806^{\circ}$ | 124.806^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,931 |
4. In the region
$$
\left\{\begin{array}{l}
\frac{x^{2}}{9}+y^{2} \leqslant 1 \\
x \geqslant 0
\end{array}\right.
$$
the maximum radius of the circle that can be contained is $\qquad$ . | 4. $\frac{2 \sqrt{2}}{3}$ | \frac{2 \sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,932 |
6. When $s$ and $t$ take all real values,
$$
(s+7-|\cos t|)^{2}+(s-2|\sin t|)^{2}
$$
the minimum value is $\qquad$ | 6. 18
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "6. 18" is already in a numerical form that is universal, it does not require translation. Here is the retained format:
6. 18 | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,934 |
7. Let the right vertex of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $A$. If there exists a point $P$ on the ellipse such that $\angle O P A$ $=\frac{\pi}{2}$ ( $O$ is the origin), then the range of $\frac{b}{a}$ is | 7. $\left(0, \frac{\sqrt{2}}{2}\right)$ | \left(0, \frac{\sqrt{2}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,935 |
8. A six-digit number $\overline{6 x y z x 6}(x 、 y 、 z$ can be the same $)$ is a perfect square. Then all such six-digit numbers are | 8. $630436,678976,689896$ | 630436,678976,689896 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,936 |
II. (20 points) As shown in Figure 1, in the isosceles right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, and $D$ is a moving point on the hypotenuse $AB$. Connect $CD$, and draw $BE \perp CD$, with the foot of the perpendicular being $E$. Find the minimum value of $\frac{S_{\triangle ACD}}{S_{\triangle BCE}}$. | II. Solution 1 Let the length of the legs of the isosceles right triangle be $1$, and $\angle A C D=\alpha$. Then
$$
\angle C B E=\alpha, \angle A D C=\frac{3 \pi}{4}-\alpha, B E=\cos \alpha \text {. }
$$
In $\triangle A C D$, by the Law of Sines, we have
$$
\begin{array}{l}
\frac{C D}{\sin A}=\frac{A C}{\sin \angle A... | 2 \sqrt{2}-2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,937 |
Three. (20 points) It is known that $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 9 and common difference 7.
(1) Prove: The sequence $\left\{a_{n}\right\}$ contains infinitely many perfect squares;
(2) The 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the nth term? | (1) The general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=9+7(n-1) \text {. }
$$
When $n=7 k^{2} \pm 6 k+1(k=0,1, \cdots)$, we have
$$
\begin{array}{l}
a_{n}=9+7(n-1) \\
=9+7\left(7 k^{2} \pm 6 k\right)=(7 k \pm 3)^{2} .
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has infinitel... | 17201 | Number Theory | proof | Yes | Yes | cn_contest | false | 724,938 |
Four. (20 points) Let the sets
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}
$$
be two sets composed of positive integers, satisfying:
(1) $A \cap B=\varnothing$;
(2) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}$;
(3) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}... | When $n=3$, because
$$
\begin{array}{l}
1+5+6=12=2+3+7, \\
1^{2}+5^{2}+6^{2}=62=2^{2}+3^{2}+7^{2},
\end{array}
$$
so, $A_{3}=\{1,5,6\}, B_{3}=\{2,3,7\}$ satisfy the conditions of the problem.
When $n=4$, because
$$
\begin{array}{l}
1+4+6+7=18=2+3+5+8, \\
1^{2}+4^{2}+6^{2}+7^{2}=102=2^{2}+3^{2}+5^{2}+8^{2},
\end{array}... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,939 |
1. If the equations
$$
x^{2}+b x+1=0 \text { and } x^{2}-x-b=0
$$
have a common root, find the value of $b$. | Prompt: Example 1. Use the substitution method to find the common root \( x_{0}=-1 \).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,940 |
(20 points) (1) Construct a $2 \times 2$ matrix with integer elements such that the sums of each row and each column are four distinct perfect squares;
(2) Construct a $3 \times 3$ matrix with integer elements such that the sums of each row and each column are six distinct perfect squares;
(3) Construct a $4 \times 4$ ... | $$
\begin{array}{rr}
-、(1) & \text { Row sum } \\
\left(\begin{array}{ll}
1 & 15 \\
0 & 49
\end{array}\right) & 16 \\
49
\end{array}
$$
Column sum $\quad 1 \quad 64$
$$
\begin{array}{ccc}
(2) & & \\
\text { Row sum } \\
\left(\begin{array}{ccc}
1 & 15 & 0 \\
0 & 49 & 51 \\
0 & 0 & 625
\end{array}\right) & 16 \\
100 \\... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,941 |
(20 points) In the Cartesian coordinate system, given points $A(-2,0), B(2,0), P$ is a moving point on the plane, let $|P A|=a,|P B|=b$, satisfying $a b=2$.
(1) Find the range of values for $a$;
(2) Find the locus of point $P$, and draw its graph. | (1) Since $|A B|=4, a b=2$, by the properties of a triangle, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b \geqslant 4, \\
a+4 \geqslant b, \\
b+4 \geqslant a
\end{array}, \Rightarrow\left\{\begin{array}{l}
a+\frac{2}{a} \geqslant 4, \\
a+4 \geqslant \frac{2}{a}, \\
\frac{2}{a}+4 \geqslant a
\end{array}\right... | \sqrt{6}-2 \leqslant a \leqslant 2-\sqrt{2} \text { or } 2+\sqrt{2} \leqslant a \leqslant 2+\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,942 |
Three. (20 points) Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=[n \sqrt{2}]$ ([ $\left.x\right]$ denotes the greatest integer not exceeding the real number $x$).
(1) Prove that $2, 2^{2}, 2^{3}, 2^{4}, 2^{5}$ are all terms of the sequence $\left\{a_{n}\right\}$;
(2) Is $2^{6}$ a term o... | $$
\begin{array}{l}
\text { (3) First, prove: There exist infinitely many positive integers } k \text { such that } \\
\left\{\frac{2^{k}}{\sqrt{2}}\right\}>1-\frac{1}{\sqrt{2}},
\end{array}
$$
where $\{x\}$ denotes the fractional part of $x$.
In fact, if there are only finitely many positive integers $k$ such that
$$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,943 |
2. The foci of the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ are $F_{1}$ and $F_{2}$. If a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$ | 2. 9 .
It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$.
Then $\left(P F_{1}+P F_{2}\right)^{2}=10^{2}$.
In the right triangle $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}^{2}+P F_{2}^{2}=8^{2} \text{. }
$$
From equations (1) and (2), we get
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9 .
$$ | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,945 |
3. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, \frac{a_{2 k}}{a_{2 k-1}}=2, \frac{a_{2 k+1}}{a_{2 k}}=3(k \geqslant 1) \text {. }
$$
Then the sum of its first 100 terms $S_{100}=$ $\qquad$ | 3. $\frac{3}{5}\left(6^{50}-1\right)$.
Notice, $\frac{a_{2 k+1}}{a_{2 k-1}}=\frac{a_{2 k+1}}{a_{2 k}} \cdot \frac{a_{2 k}}{a_{2 k-1}}=6$,
$$
\begin{array}{l}
\frac{a_{2 k+2}}{a_{2 k}}=\frac{a_{2 k+2}}{a_{2 k+1}} \cdot \frac{a_{2 k+1}}{a_{2 k}}=6, \\
a_{1}=1, a_{2}=2 .
\end{array}
$$
Therefore, $a_{2 k-1}=6^{k-1}, a_{... | \frac{3}{5}\left(6^{50}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,946 |
4. If $4n+1$ and $6n+1$ are both perfect squares, then the smallest positive integer $n$ is $\qquad$ | 4. 20 .
Obviously, $4 n+1, 6 n+1$ are both odd square numbers.
Let $6 n+1=(2 m+1)^{2}=4 m(m+1)+1$.
Then $3 n=2 m(m+1)$.
Since $m(m+1)$ is even, $4 \mid n$.
Let $n=4 k$. Then
$4 n+1=16 k+1, 6 n+1=24 k+1$.
When $k=1,2,3,4$, $4 n+1, 6 n+1$ are not both square numbers, but when $k=5$, i.e., $n=20$, $4 n+1=81$, $6 n+1=121$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,947 |
5. The maximum value of the function $y=2 x-5+\sqrt{11-3 x}$ is $\qquad$ . | 5. $\frac{65}{24}$.
Let $\sqrt{11-3 x}=t$. Then
$$
\begin{array}{l}
6 y=12 x-30+6 \sqrt{11-3 x} \\
=-4 t^{2}+6 t+14 \\
=-\left(2 t-\frac{3}{2}\right)^{2}+\frac{65}{4} \leqslant \frac{65}{4} .
\end{array}
$$
Thus, $y \leqslant \frac{65}{24}$.
When $t=\frac{3}{4}$, i.e., $x=\frac{167}{48}$, the equality holds. | \frac{65}{24} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,948 |
6. As shown in Figure 1, in the unit cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $D_{1} A_{1}$, respectively. Then the distance from point $B_{1}$ to the plane containing $\triangle E F G$ is . $\qquad$ | 6. $\frac{\sqrt{3}}{2}$.
Solution 1 As shown in Figure 3, the plane passing through points $E$, $F$, and $G$ intersects the cube to form a regular hexagon. It is easy to see that this plane perpendicularly bisects the diagonal $B_{1} D$ of the cube.
Since $B_{1} D=\sqrt{3}$, the distance from point $B_{1}$ to the pla... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,949 |
7. $\sin ^{2} 130^{\circ}+\sin 70^{\circ} \cdot \cos 80^{\circ}=$ | $\begin{array}{l}7 \cdot \frac{3}{4} \cdot \\ \sin ^{2} 130^{\circ}+\sin 70^{\circ} \cdot \cos 80^{\circ} \\ =\cos ^{2} 40^{\circ}+\sin 70^{\circ} \cdot \sin 10^{\circ} \\ =\frac{1+\cos 80^{\circ}}{2}+\sin 70^{\circ} \cdot \sin 10^{\circ} \\ =\frac{1}{2}+\frac{1}{2}\left(\cos 80^{\circ}+2 \sin 70^{\circ} \cdot \sin 10^... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,950 |
2. Let $a$, $b$, and $c$ be three distinct real numbers such that the equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+b x+c=0
$$
have a common real root, and the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$
also have a common real root. Find $a+b+c$. | Prompt: Example 3. Answer: $a+b+c=-3$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Example 3. Answer: $a+b+c=-3$.
| a+b+c=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,951 |
8. If the four digits of the four-digit number $\overline{a b c d}$ satisfy $a+b=c+d$, then it is called a "good number" (for example, 2011 is a good number). Then, the number of good numbers is $\qquad$ | 8. 615 .
Let $k=a+b=c+d$.
Since $1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant 9$, then
$1 \leqslant k \leqslant 18$.
When $1 \leqslant k \leqslant 9$, in the above equation, $a$ can take any value in $\{1,2$, $\cdots, k\}$, $c$ can take any value in $\{0,1, \cdots, k\}$, and once $a 、 c$ are determined, $... | 615 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,952 |
9. (20 points) In $\triangle A B C$, the measures of the three interior angles satisfy $\frac{\angle A}{\angle B}=\frac{\angle B}{\angle C}=\frac{1}{3}$.
Find the value of $T=\cos A+\cos B+\cos C$. | Given $\angle A=\theta, \angle B=3 \theta, \angle C=9 \theta$.
From $\theta+3 \theta+9 \theta=\pi \Rightarrow \theta=\frac{\pi}{13}$.
Then $T=\cos \theta+\cos 3 \theta+\cos 9 \theta$
$$
\begin{aligned}
= & \cos \theta+\cos 3 \theta-\cos 4 \theta \\
= & 2 \cos \theta \cdot \cos 2 \theta-2 \cos ^{2} 2 \theta+1>1, \\
T^{2... | \frac{1+\sqrt{13}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,953 |
10. (25 points) As shown in Figure 2, $D$, $E$, and $F$ are points on the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, and $DE$ intersects $AB$ at point $F_{0}$, $EF$ intersects $BC$ at point $D_{0}$, and $FD$ intersects $CA$ at point $E_{0}$. Prove that $AD$, $BE$, and $CF$ are concurrent if and only i... | 10. By Menelaus' theorem, we know that points $D_{0}, E_{0}, F_{0}$ are collinear if and only if
$$
\frac{A E_{0}}{E_{0} C} \cdot \frac{C D_{0}}{D_{0} B} \cdot \frac{B F_{0}}{F_{0} A}=1 .
$$
By Ceva's theorem, we know that lines $A D, B E, C F$ are concurrent if and only if
$$
\frac{B D}{D C} \cdot \frac{C E}{E A} \cd... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,954 |
11. (25 points) 20 wizards gather on an isolated island. During this period, any three wizards have cursed some other wizards together. Prove: there must exist a wizard who has been cursed by at least nine other wizards. | 11.20 wizards can form $\mathrm{C}_{20}^{3}$ "trio groups". Each group has cursed at least one person, so the number of wizards who have been cursed is at least $\mathrm{C}_{20}^{3}$ times.
Let W be the wizard who has been cursed the most, and he has been cursed by $m$ trio groups. Then,
$$
m \geqslant \frac{\mathrm{C... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,955 |
2. If the modulus of the complex number $(a+2 \mathrm{i})(1+\mathrm{i})$ is 4, then the value of the real number $a$ is ( )
(A) 2
(B) $2 \sqrt{2}$
(C) $\pm 2$
(D) $\pm 2 \sqrt{2}$ | 2. C.
From the problem, we have
$$
\sqrt{2} \cdot \sqrt{a^{2}+4}=4 \Rightarrow a= \pm 2 .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,957 |
4. A chord $A B$ is drawn through the right focus $F_{2}$ of the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with an inclination angle of $45^{\circ}$. Then $|A B|$ is ( ).
(A) $\frac{2 \sqrt{6}}{3}$
(B) $\frac{4 \sqrt{6}}{3}$
(C) $\frac{4 \sqrt{2}}{3}$
(D) $\frac{4 \sqrt{3}}{3}$ | 4. C.
It is known that the right focus of the ellipse is $(1,0)$. Then $l_{A B}: y=x-1$.
Substituting the above equation into the ellipse equation, we get
$$
\begin{array}{l}
3 x^{2}-4 x=0 \Rightarrow x_{1}=0, x_{2}=\frac{4}{3} \\
\Rightarrow|A B|=\sqrt{2\left(x_{1}-x_{2}\right)^{2}}=\frac{4 \sqrt{2}}{3} .
\end{array}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,959 |
5. Function
$$
f(x)=\left\{\begin{array}{cc}
1-5^{-x} & x \geqslant 0, \\
5^{x}-1 & x<0 .
\end{array}\right.
$$
Then the function is ( ).
(A) Monotonically increasing function, odd function
(B) Monotonically decreasing function, even function
(C) Monotonically increasing function, even function
(D) Monotonically decre... | 5. A.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,960 |
8. In the plane region $\{(x, y)|| x| \leqslant 1| y \mid, \leqslant 1\}$, it is always true that $a x - 2 b y \leqslant 2$. Then the area of the plane region formed by the moving point $P(a, b)$ is ( ).
(A) 4
(B) 8
(C) 16
(D) 32 | 8. A.
The plane region $\{(x, y)|| x| \leqslant 1, | y | \leqslant 1\}$ has four boundary points $(-1,-1),(-1,1),(1,-1),(1,1)$ that satisfy $a x - 2 b y \leqslant 2$, which gives us
$$
\begin{array}{l}
a + 2 b \leqslant 2, a - 2 b \leqslant 2, \\
-a - 2 b \leqslant 2, -a + 2 b \leqslant 2 .
\end{array}
$$
From this, ... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,964 |
9. Given the function
$$
f(x)=\sin \left(2 x-\frac{\pi}{6}\right)-m
$$
has two zeros in $\left[0, \frac{\pi}{2}\right]$. Then the range of $m$ is
(A) $\left(\frac{1}{2}, 1\right)$
(B) $\left[\frac{1}{2}, 1\right]$
(C) $\left[\frac{1}{2}, 1\right)$
(D) $\left(\frac{1}{2}, 1\right]$ | 9. C.
The problem is equivalent to the function $f(x)=\sin \left(2 x-\frac{\pi}{6}\right)$ intersecting the line $y=m$ at two points in the interval $\left[0, \frac{\pi}{2}\right]$. Therefore, the range of values for $m$ is $\left[\frac{1}{2}, 1\right)$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,965 |
12. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first 15 terms $S_{15}=30$. Then $a_{1}+a_{8}+a_{15}=$ $\qquad$ | 12.6.
From $S_{15}=30 \Rightarrow a_{1}+7 d=2$.
Therefore, $a_{1}+a_{8}+a_{15}=3\left(a_{1}+7 d\right)=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,968 |
13. Vectors
$$
a=(1, \sin \theta), b=(\cos \theta, \sqrt{3})(\theta \in \mathbf{R}) \text {. }
$$
Then the range of values for $|a-b|$ is $\qquad$ . | 13. $[1,3]$.
$$
\begin{array}{l}
|a-b|=\sqrt{(1-\cos \theta)^{2}+(\sin \theta-\sqrt{3})^{2}} \\
=\sqrt{5-2(\cos \theta-\sqrt{3} \sin \theta)} \\
=\sqrt{5-4 \sin \left(\frac{\pi}{6}-\theta\right)} \in[1,3] .
\end{array}
$$ | [1,3] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,969 |
14. Right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base $\triangle A B C$ is an equilateral triangle, $P$ and $E$ are moving points on $B B_{1}$ and $C C_{1}$ (including endpoints), $D$ is the midpoint of side $B C$, and $P D \perp P E$. Then the angle between line $A P$ and $P E$ is $\qquad$ | 14. $90^{\circ}$.
Since plane $ABC \perp$ plane $BCC_{1}B_{1}, AD \perp BC$, therefore, $AD \perp$ plane $BCC_{1}B_{1}$. Hence $AD \perp PE$. Also, $PE \perp PD, PE \perp$ plane $APD$, then $PE \perp PA$, which means the angle is $90^{\circ}$. | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,970 |
15. Let $x, y$ be real numbers. Then
$$
\max _{S x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=
$$
$\qquad$ | 15. 4 .
$$
\begin{array}{l}
\text { Given } 5 x^{2}+4 y^{2}=10 x \\
\Rightarrow 4 y^{2}=10 x-5 x^{2} \geqslant 0 \\
\Rightarrow 0 \leqslant x \leqslant 2 .
\end{array}
$$
Then $4\left(x^{2}+y^{2}\right)=10 x-x^{2}$
$$
\begin{array}{l}
=25-(5-x)^{2} \leqslant 25-3^{2} \\
\Rightarrow x^{2}+y^{2} \leqslant 4 .
\end{array... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,971 |
16. There are 2011 street lamps numbered $1,2, \cdots, 2011$ on the road. To save electricity, it is required to turn off 300 of these lamps, but no two adjacent lamps can be turned off at the same time, and the lamps at both ends cannot be turned off. The number of ways to satisfy these conditions is $\qquad$ (express... | 16. $C_{1710}^{300}$.
The problem is equivalent to inserting 300 dark lamps into 1711 paths. Therefore, there are $\mathrm{C}_{1710}^{300}$ ways to turn off the lamps. | C_{1710}^{300} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,972 |
Example 1: There are $n$ people registered to participate in four sports events: A, B, C, and D. It is stipulated that each person must participate in at least one event and at most two events, but events B and C cannot be registered for simultaneously. If in all different registration methods, there must be at least o... | Let the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{T}}\right)$ represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is 1 (for example, if participating in event 甲, then $a_{\text {甲 }}=1... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,973 |
Example 2 Select 2012 numbers arbitrarily from $1,2, \cdots, 4000$. Prove: there must exist two numbers whose difference is exactly 100. | Prove that by the remainder of dividing by 100, $0,1, \cdots, 99$, the set of positive integers can be divided into 100 drawers.
Since $2012=20 \times 100+12$, by the pigeonhole principle, among the 2012 selected numbers, there must be 21 numbers belonging to the same drawer, meaning the difference between any two of ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,974 |
Example 4 Let the function be
$$
\begin{array}{l}
f(x)=\sqrt{10-6 \cos x}+\sqrt{\frac{17}{8}-\frac{3 \sqrt{2}}{2} \sin x}+ \\
\sqrt{\frac{17}{8}-\frac{\sqrt{2}}{2} \cos x-2 \sin x} \text {. } \\
\end{array}
$$
Then the minimum value of $f(x)$ is $\qquad$ | Notice,
$$
\begin{aligned}
f(x)= & \sqrt{(\cos x-3)^{2}+\sin ^{2} x}+ \\
& \sqrt{\cos ^{2} x+\left(\sin x-\frac{3 \sqrt{2}}{4}\right)^{2}}+ \\
& \sqrt{\left(\cos x-\frac{\sqrt{2}}{4}\right)^{2}+(\sin x-1)^{2}} .
\end{aligned}
$$
As shown in Figure 4, let $A(3,0), B\left(0, \frac{3 \sqrt{2}}{4}\right), C\left(\frac{\sq... | 3 \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,975 |
6. Given three points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$ on the unit circle $x^{2}+y^{2}=1$ that satisfy
$$
x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3}=0 \text {. }
$$
then $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=$ | 6. $\frac{3}{2}$.
Let $x_{1}=\cos \alpha, x_{2}=\cos \beta, x_{3}=\cos \gamma$,
$$
y_{1}=\sin \alpha, y_{2}=\sin \beta, y_{3}=\sin \gamma .
$$
From the given conditions, the circumcenter, centroid, and orthocenter of $\triangle A B C$ coincide, indicating that it is an equilateral triangle.
$$
\begin{array}{l}
\text ... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,976 |
7. Let $M=\{1,2,3,4,5\}$. Then the number of mappings $f: M \rightarrow M$ such that
$$
f(f(x))=f(x)
$$
is $\qquad$ | 7. 196.
For the case where $M$ contains $n$ elements, from
$$
f(f(x))=f(x) \text {, }
$$
we know that for any $a \in M$ (let $f(c)=a$), we have
$$
f(f(c))=f(a)=f(c)=a .
$$
If the range of $f$ contains $k$ elements, then the $n-k$ elements of set $M$ that are not in the range have $k^{n-k}$ possible mappings. Therefo... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,977 |
8. Given a positive integer $n$. Then the positive integers $m$ for which there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$ satisfying
$$
\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m
$$
are $\qquad$ | 8. $1,2, \cdots n$.
First, we have $m \leqslant \frac{1}{1}+\frac{2}{2}+\cdots+\frac{n}{n}=n$.
Take $x_{1}=n, x_{2}=2 n, \cdots, x_{n}=n^{2}$. Then $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=1$.
Take $x_{1}=1, \cdots, x_{k}=k, x_{k+1}=(n-k)(k+1)$, $\cdots, x_{n}=n(n-k)$. Then
$$
\begin{array}{l}
\frac{1}{... | 1,2, \cdots n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,978 |
9. (16 points) Let $P$ be any point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The tangent line to the hyperbola at $P$ intersects the $x$-axis at point $Q$, and $F_{1}$ and $F_{2}$ are the foci. Prove that $PQ$ bisects $\angle F_{1} P F_{2}$. | Let $P\left(x_{0}, y_{0}\right)$ be any point on the left branch of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
$$
Then,
$$
l_{P Q}: \frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1 \text {. }
$$
Thus, the point $Q\left(\frac{a^{2}}{x_{0}}, 0\right)$.
Therefore, $\frac{\left|F_{1} Q\right|}{\left|Q F_{2}\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,979 |
10. (20 points) Find
$$
f(x)=|x-1|+2|x-2|+\cdots+2011|x-2011|
$$
the minimum value. | 10. Obviously, when $x=2011$, $f(x)$ has no minimum value.
The following assumes $x \in[1,2011]$.
When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \time... | 794598996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,980 |
11. (20 points) The volume of a regular $n$-sided pyramid is $V$. Find a necessary and sufficient condition independent of $n$ for the surface area of the regular $n$-sided pyramid to be minimized. | 11. Let the radius of the base of a regular $n$-sided pyramid be $r$, and the angle between the side and the base be $\theta\left(0<\theta<\frac{\pi}{2}\right)$.
Then the height of the pyramid is
$$
h=r \cos \frac{\pi}{n} \cdot \tan \theta,
$$
The area of the base, a regular $n$-sided polygon, is
$$
\begin{array}{l}
n... | \cos \theta = \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,981 |
One, (40 points) As shown in Figure $1, \odot O_{1}$ is internally tangent to $\odot O_{2}$ at point $P, \odot O_{1}$'s chord $A B$ is tangent to $\odot O_{2}$ at point $C$.
(1) If the radii of $\odot O_{1}$ and $\odot O_{2}$ are $R$ and $r$ respectively, find $\frac{A C^{2}}{A P^{2}}$;
(2) If $P C$ intersects $\odot O... | (1) Since $A C^{2}=A E \cdot A P$, we have $\frac{A C^{2}}{A P^{2}}=\frac{A E}{A P}=\frac{O_{1} O_{2}}{O_{1} P}=\frac{R-r}{R}$.
(2) As shown in Figure 2, extend $P A$ to point $M$ such that $A M = F B$, and connect $F M$ and $G M$.
Since $\angle M A G = \angle F B G$ and $A G = B G$, we have $\triangle M A G \cong \tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,982 |
Given a positive integer $n(n>1)$, and $x_{1}$, $x_{2}, \cdots, x_{n} \in[a, a+1](a \in \mathbf{R})$. Prove:
$$
\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)^{2} \leqslant \frac{1}{4} .
$$ | II. Fix $x_{2}, x_{3}, \cdots, x_{n}$. Then
$$
\begin{array}{c}
f=\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)^{2} \\
=\left(\frac{1}{n}-\frac{1}{n^{2}}\right) x_{1}^{2}-\frac{2 x_{1}}{n^{2}} \sum_{j=2}^{n} x_{j}+ \\
\frac{1}{n} \sum_{i=2}^{n} x_{i}^{2}-\left(\frac{1}{n} \sum_{j=2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,983 |
$$
\begin{array}{l}
\text { Three. (50 points) Let the set } \\
P=\{1,2, \cdots, 2012\} .
\end{array}
$$
Prove: For any $n \in \mathbf{N}_{+}$, there exist $k \in P$ and a positive integer $m$ such that $n=\sum_{i=1}^{2012}\left[m \mathrm{e}^{k-i}\right]$, where $[x]$ denotes the greatest integer not exceeding the rea... | Consider the set
$$
A=\left\{m \mathrm{e}^{k} \mid k \in P, m \in \mathbf{N}_{+}\right\} \text {. }
$$
For any $m_{1} \mathrm{e}^{k_{1}}, m_{2} \mathrm{e}^{k_{2}} \in A$, and
$$
\begin{array}{l}
m_{1} \mathrm{e}^{k_{1}}=m_{2} \mathrm{e}^{k_{2}} \\
\Leftrightarrow \mathrm{e}^{k_{1}-k_{2}}=\frac{m_{2}}{m_{1}} \\
\Leftri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,984 |
Four. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\},
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedron. Question: What is the maximum number of triangles in graph $G$? | Four, first prove a lemma.
Lemma If in a space graph with $n$ points there is no triangle, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$ ( $[x]$ represents the largest integer not exceeding the real number $x$).
Proof Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, among which, the numbe... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,985 |
Example 5 Let real numbers $x, y$ satisfy $3 x^{2}+4 y^{2}=48$. Then
$$
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}
$$
The maximum value is $\qquad$ [3] | Notice that,
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5} \\
=\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x-1)^{2}+(y+2)^{2}},
\end{array}
$$
and the point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$.
As shown in Figure 5, let the left and right foci of the ellipse be $F_{1}(-2, ... | 8+\sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,986 |
In $\triangle A B C$, let the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively. $A D$ is the angle bisector of $\angle A$, intersecting $B C$ at point $D$, and $A D = B C$. Prove:
$$
2 \leqslant \frac{b}{c}+\frac{c}{b}<2 \sqrt{2} .
$$ | Solve: The left inequality is obviously true.
By the angle bisector formula, we get
$$
\begin{array}{l}
A D=\frac{2 \sqrt{b c p(p-a)}}{b+c}\left(p=\frac{a+b+c}{2}\right) . \\
\text { Then } A D^{2}=\frac{b c\left[(b+c)^{2}-a^{2}\right]}{(b+c)^{2}}=a^{2} \\
\Rightarrow\left[(b+c)^{2}+b c\right] a^{2}=b c(b+c)^{2} \\
\Ri... | 2 \leqslant \frac{b}{c}+\frac{c}{b}<2 \sqrt{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 724,987 |
As shown in Figure 1, points $D$ and $E$ are on the sides $AB$ and $AC$ of equilateral $\triangle ABC$, respectively, satisfying $BD = AE$. Connecting $CD$ and $BE$ intersect at point $O$. Given that $BO = 2$ and $CO = 5$, find the length of $AO$.
---
The above text translated into English, preserving the original te... | Solving, we know
$$
\begin{array}{l}
\triangle D B C \cong \triangle E A B \\
\Rightarrow \angle B C D=\angle A B E . \\
\text { Then } \angle B O C=180^{\circ}-(\angle B C D+\angle O B C) \\
=180^{\circ}-(\angle A B E+\angle O B C) \\
=180^{\circ}-60^{\circ}=120^{\circ} .
\end{array}
$$
As shown in Figure 2, rotate
$... | \sqrt{19} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,988 |
317 Given $h_{a} 、 h_{b} 、 h_{c}$ and $m_{a} 、 m_{b} 、 m_{c}$ are the lengths of the altitudes and medians to the sides $a 、 b 、 c$ of $\triangle A B C$, respectively. Prove:
$$
\frac{m_{a}}{h_{b}+h_{c}} \cdot \frac{m_{b}}{h_{c}+h_{a}} \cdot \frac{m_{c}}{h_{a}+h_{b}} \leqslant \frac{1}{8} .
$$ | Proof As shown in Figure 3, let $G$ be the centroid of $\triangle ABC$.
It is easy to know that $m_{a}=\frac{3}{2} GA, S_{\triangle CAC}=\frac{1}{3} S_{\triangle ABC}$.
Let $\angle BAE=\angle A_{1}, \angle CAE=\angle A_{2}$. Then
$$
\begin{array}{l}
\frac{h_{b}}{m_{a}}=\frac{4 S_{\triangle ABC}}{3 b GA}=\frac{4 S_{\tri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,989 |
Given $x_{i}(i=1,2, \cdots, n) \in \mathbf{R}_{+}$, $n \geqslant 4$, and $\prod_{i=1}^{n} x_{i}=1$. Prove:
$$
\sum_{i=1}^{n} \frac{1}{x_{i}}+\frac{2 n+1}{\sum_{i=1}^{n} x_{i}} \geqslant \frac{(n+1)^{2}}{n} .
$$ | Proof Note
$$
A=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}+\frac{2 n+1}{x_{1}+x_{2}+\cdots+x_{n}} .
$$
By symmetry, we can assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. Let $f\left(x_{1}\right)=A$.
Considering $x_{2}, x_{3}, \cdots, x_{n-1}$ as constants, $x_{n}=\frac{1}{x_{1} x_{2} \cdo... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,990 |
Example 6 Given the sets
$$
\begin{array}{l}
M=\left\{(x, y) \left\lvert\, y \geqslant \frac{1}{4} x^{2}\right.\right\}, \\
N=\left\{(x, y) \left\lvert\, y \leqslant-\frac{1}{4} x^{2}+x+7\right.\right\}, \\
D_{r}\left(x_{0}, y_{0}\right)=\left\{(x, y)!\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\ri... | Solution: Clearly, $M \cap N \neq \varnothing$.
As shown in Figure 6, in the Cartesian coordinate system,
$$
y=\frac{1}{4} x^{2}, y=-\frac{1}{4} x^{2}+x+7
$$
have vertices at $O(0,0)$ and $A(2,8)$, respectively, and the two parabolas are symmetric about point $B(1,4)$.
Therefore, the figure $T$ corresponding to $M \c... | \sqrt{\frac{25-5 \sqrt{5}}{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,991 |
Example 7 Given a set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both intege... | Solve: Establish a complex plane, let
$$
z=x+y \mathrm{i}(x, y \in \mathbf{R}) \text {. }
$$
From $\left|z_{1}\right|=1$, we know $z_{1}^{2}$ is also on the unit circle $\odot 0$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \\
=\left|z_{1}^{2}-1\right|^{2}
\end{array}
$$
This represents the... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,992 |
Example 8 If for all $\theta \in \mathbf{R}$, the complex number
$$
z=(a+\cos \theta)+(2 a-\sin \theta) \mathrm{i}
$$
has a modulus not exceeding 2, then the range of the real number $a$ is
$\qquad$ [6] | Let $\left\{\begin{array}{l}x=a+\cos \theta, \\ y=2 a-\sin \theta .\end{array}\right.$ Then $\left\{\begin{array}{l}x^{2}+y^{2} \leqslant 2^{2}, \\ (x-a)^{2}+(y-2 a)^{2}=1 .\end{array}\right.$ As shown in Figure 8, establish a complex plane, and take $Z(x, y), P(a, 2 a)$.
Then point $P$ lies on the line $y=2 x$, and $... | \left(-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,993 |
Example 9 If three positive real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
x^{2}+x y+y^{2}=\frac{25}{4}, \\
y^{2}+y z+z^{2}=36, \\
z^{2}+z x+x^{2}=\frac{169}{4},
\end{array}
$$
find the value of $x y+y z+z x$.
[7] | It is known that the three equations can be transformed into
$$
\begin{array}{l}
x^{2}+y^{2}-2 x y \cos 120^{\circ}=\left(\frac{5}{2}\right)^{2}, \\
y^{2}+z^{2}-2 y z \cos 120^{\circ}=\left(\frac{12}{2}\right)^{2}, \\
z^{2}+x^{2}-2 z x \cos 120^{\circ}=\left(\frac{13}{2}\right)^{2} .
\end{array}
$$
As shown in Figure ... | 10 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,994 |
Example 10 Given positive real numbers $\alpha \sqrt{\beta} \gamma$ satisfying $\alpha+\beta+\gamma=1$. Prove: For any positive real numbers $x, y, z$, we have
$$
\begin{array}{c}
\alpha(y+z) \sqrt{\frac{x(7 x+4 y+10 z)}{7}+\frac{(2 y-5 z)^{2}}{49}}+ \\
\beta(z+x) \sqrt{4 x(x+y+z)+(y-z)^{2}}+ \\
\gamma(x+y) \sqrt{\frac... | Prove as shown in Figure 10, construct $\triangle ABC$, such that
$$
\begin{array}{l}
AB = x + y, \\
BC = y + z, \\
CA = z + x.
\end{array}
$$
On the lines $BC$, $CA$, and $AB$, take points $D$, $E$, and $F$ respectively, satisfying
$$
\begin{array}{l}
2 \overrightarrow{BD} = 5 \overrightarrow{DC} \\
\overrightarrow{E... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,995 |
Example 11 Given that $x, y, z$ are real numbers, satisfying
$$
\begin{array}{c}
x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}, \\
y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}, \\
z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}},
\end{array}
$$
and $x+y+z=\frac{m}{\sqrt{n}}\left(m, n \in \mathbf{... | Solve as shown in Figure 11, construct an acute triangle $\triangle ABC$, such that
$$
\begin{array}{l}
BC=x, \\
CA=y, \\
AB=z .
\end{array}
$$
Draw perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $AD=u$, $BE=v$, and $CF=w$. T... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,996 |
1. Given $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+6 \geqslant 0, \\
x+y \geqslant 0, \\
x \leqslant 3 .
\end{array}\right.
$$
If the maximum value of $z=a x+y$ is $3 a+9$, and the minimum value is $3 a-3$, then the range of $a$ is ( ).
(A) $0 \leqslant a \leqslant 1$
(B) $-1 \leqslant a \leqslant 0$
(C) $-1 \leqs... | Prompt: Imitate Example 1, Signature: C.
Translate the text above into English, preserving the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,997 |
2. Given that $x, y$ are real numbers, satisfying $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $|1 x^{2}+2 x y-y^{2}|$ is $\qquad$ | Notice that, $x^{2}+2 x y-y^{2}=k$ is a hyperbola, with asymptotes
$$
x+(1+\sqrt{2}) y=0, x+(1-\sqrt{2}) y=0 .
$$
Then its conjugate hyperbola is
$$
\begin{array}{l}
y=(\sqrt{2}-1) x, \\
y=-(\sqrt{2}+1) x .
\end{array}
$$
As shown in Figure 12, when the hyperbola is tangent to the circle $x^{2}+y^{2}=1$, $k$ takes th... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,998 |
For example, 32006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find
$$
a_{1}+a_{2}+\cdots+a_{2000}
$$
the minimum value. ${ }^{[1]}$ | First, we prove: For any 119 positive integers \( b_{1}, b_{2}, \cdots, b_{119} \), there must exist a subset (at least one, or all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
\[ b_{1}, b_{1}+b_{2}, b_{1}+b_{2}+b_{3}, \cdots, b_{1}+b_{2}+\cdots+b_{119}. \]
If one of them is a... | 3910 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,999 |
3. Know
$$
\begin{array}{l}
\left\{(x,-y) \mid(x+\sqrt{2})^{2}+y^{2}=a^{2}-3 a+2\right\} \cap \\
\left\{(x, y) \mid x^{2}+(y+\sqrt{2})^{2}<a\right\} \neq \varnothing .
\end{array}
$$
Then the range of real number $a$ is $\qquad$ | $$
\begin{array}{l}
\sqrt{a^{2}-3 a+2}+\sqrt{a}>2>\sqrt{a^{2}-3 a+2}-\sqrt{a} \geqslant 0 \\
\text { or } \sqrt{a^{2}-3 a+2}+\sqrt{a}>2 \geqslant \sqrt{a}-\sqrt{a^{2}}-3 a+2>0 \text {. } \\
\text { Answer: (2.215 3,5.9717). } \\
\end{array}
$$ | (2.2153, 5.9717) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,000 |
4. E. $\operatorname{E}\{\mathrm{A} a=(\cos \alpha, \sin \alpha), b=(\cos \beta, \sin \beta)$, $|a-b|=\frac{2 \cdot 5}{5}$. Given $0<\alpha<\frac{\pi}{2},-\frac{\pi}{2}<\beta<0, \mathrm{H}$. si: $\beta=-\frac{5}{12}$. Then $\sin \alpha=(\quad$.
(A) $\frac{31}{65}$
(B) $\frac{32}{65}$
(C) $\frac{33}{65}$
(D) $\frac{34}{... | Prompt: As shown in the figure
13, take points $A(\cos \alpha, \sin \alpha), B(\cos \beta, \sin \beta)$.
Given $|A B| = \frac{2 \sqrt{5}}{5}$, then $\angle A O B = \alpha - \beta$, and
$$
\begin{array}{l}
\cos (\alpha - \beta) = \frac{1^{2} + 1^{2} - \left(\frac{2 \sqrt{5}}{5}\right)^{2}}{2 \times 1 \times 1} = \frac{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,001 |
6. Let $a, b, c, d, e, f$ be real numbers, and
$$
a x^{2}+b x+c \geqslant\left|d x^{2}+e x+f\right|
$$
for all real numbers $x$. Prove:
$$
4 a c-b^{2} \geqslant\left|4 d f-e^{2}\right| \text {. }
$$ | Consider
$$
\begin{array}{l}
p(x)=a x^{2}+b x+c, \\
q(x)=d x^{2}+e x+f
\end{array}
$$
with the graphs (assuming $d>0$). Then $a>d$.
If the graph of $y=q(x)$ intersects the $x$-axis, then $e^{2}-4 d f \geqslant 0$.
As shown in Figure 14, we know that $p(x) \pm q(x)$ are both non-negative.
Thus, the discriminants of $y=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,002 |
Example 1 Let $a^{2}+1=3 a, b^{2}+1=3 b, \mathrm{H} a \neq b$. Then the value of the algebraic expression $\frac{1}{a^{2}}+\frac{1}{b^{2}}$ is ( ).
(A) 5
(B) 7
(C) 9
(D) 11 | From the given conditions, we have
$$
a^{2}-3 a+1=0, b^{2}-3 b+1=0 \text {, and } a \neq b \text {. }
$$
Then \(a\) and \(b\) are the roots of the quadratic equation
$$
x^{2}-3 x+1=0
$$
Therefore, by Vieta's formulas, we get
$$
\begin{array}{l}
a+b=3, a b=1 . \\
\text { Hence } \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,003 |
Example 2 Given $m=1+\sqrt{2}, n=1-\sqrt{2}$, and $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$. Then the value of $a$ is ( ).
$(\mathrm{A})-5$
(B) 5
(C) -9
(D) 9 | Given that $m+n=2, m n=-1$.
Therefore, $m, n$ can be seen as the roots of the quadratic equation $x^{2}-2 x-1=0$.
Thus, $m^{2}-2 m=1, n^{2}-2 n=1$.
Since $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$,
So, $(7+a)(3-7)=8$.
Solving for $a$ gives $a=-9$.
Therefore, the answer is C. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,004 |
Example 3 Given $a=\frac{1}{2} \sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8}$. Try to find the value of $a^{2}+\sqrt{a^{4}+a+1}$. | Solving, we have
$$
\begin{aligned}
a & =\frac{1}{2} \sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8} \\
& =\frac{-\sqrt{2}+\sqrt{(\sqrt{2})^{2}-4 \times 4 \times(-\sqrt{2})}}{2 \times 4},
\end{aligned}
$$
Therefore, $u$ is a real root of the quadratic equation
$$
4 x^{2}+\sqrt{2} x-\sqrt{2}=0
$$
Thus, $a$ satisfies $a... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,005 |
Example 4 Let $a=\sqrt{7}-1$. Then $3 a^{3}+12 a^{2}-6 a-12=(\quad)$.
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$ | Given $a=\sqrt{7}-1$, we get $a+1=\sqrt{7}$.
Thus, $a^{2}+2 a-6=0$.
Therefore, the expression to be found is
$$
\begin{array}{l}
=3 a^{3}+6 a^{2}+6 a^{2}-6 a-12 \\
=3 a\left(a^{2}+2 a\right)+6 a^{2}-6 a-12 \\
=18 a+6 a^{2}-6 a-12 \\
=6\left(a^{2}+2 a\right)-12 \\
=6 \times 6-12=24 .
\end{array}
$$
Therefore, the answe... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,006 |
Question 1 Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$, line $AD$ and $BC$ intersecting at point $F$, and $G$, $H$ being the midpoints of sides $AB$ and $CD$ respectively. Prove: $EF$ is tangent to the circle passing through points $E$, $G$, and $H$ at point $E. ^{[11[2]}$ | Proof As shown in Figure 1, let $K$ be the midpoint of side $EF$. For the complete quadrilateral FDAEBC, by Newton's theorem, we know that points $K$, $H$, and $G$ are collinear.
Extend $HK$ to point $N$ such that $KN = HK$. Then quadrilateral $EHFN$ is a parallelogram.
Since $\triangle DEC \sim \triangle AEB$, and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,007 |
Question 2 Let quadrilateral $ABCD$ have an inscribed circle, and let line $l$ through point $A$ intersect segment $BC$ and line $DC$ at points $M$ and $N$, respectively. Let $I_{1}$, $I_{2}$, and $I_{3}$ be the incenters of $\triangle ABM$, $\triangle MNC$, and $\triangle NDA$, respectively. Prove that the orthocenter... | To prove, as shown in Figure 2, construct \( I_{3} E \perp I_{1} I_{2} \) at point \( E \), intersecting line \( l \) at point \( H_{1} \), and \( I_{1} F \perp I_{2} I_{3} \) at point \( F \), intersecting line \( l \) at point \( H_{2} \). It suffices to prove that points \( H_{1} \) and \( H_{2} \) coincide.
Clearl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,008 |
Question 3 Given that the three interior angles of $\triangle A B C$, $\angle A$, $\angle B$, and $\angle C$, are $\frac{\pi}{7}$, $\frac{2 \pi}{7}$, and $\frac{4 \pi}{7}$ respectively, and the three angle bisectors intersect the opposite sides at points $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$. Prove: $\triangle A... | Prove as shown in Figure 3, on $BA$ intercept $BD=BC$, connect $B'D$.
Then $\triangle BDB' \cong \triangle BCB'$.
Thus, we can calculate
$\angle ADB'=\frac{3\pi}{7}=\angle AB'D$.
Therefore, $AB'=AD=AB-BC$,
$CB'=AC-AB'=AC-AB+BC$.
Extend $AC$ to point $E$, such that $CE=CA'$, connect $A'E$.
Since $\angle E=\frac{1}{2} \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,009 |
Example 4 In the positive integers not exceeding 2012, take $n$ numbers arbitrarily, so that there are definitely two numbers whose ratio is within the interval $\left[\frac{2}{3} ; \frac{3}{2}\right]$. Try to find the minimum value of $n$.
| Solve for
$$
\begin{array}{l}
\{1,2,4,7,11,17,26,40,61,92,139, \\
209,314,472,709,1064,1597\}
\end{array}
$$
When any two numbers are taken, their ratio is either greater than $\frac{3}{2}$ or less than $\frac{2}{3}$.
Therefore, $n \geqslant 18$.
Next, we prove: When 18 numbers are taken, there exist two numbers whose... | 18 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,010 |
As shown in Figure 1, given that $\odot O$ is the excircle of $\triangle ABC$ on side $BC$, points $D$ and $E$ are on line segments $AB$ and $AC$ respectively, such that $DE \parallel BC$. $\odot O_{1}$ is the incircle of $\triangle ADE$. $O, B$ intersects with $DO, O, C$ intersects with $EO$ at points $F$ and $G$ resp... | Proof As shown in Figure 2, let $\odot O_{1}$ and $\odot O$ touch $AB$ at points $H_{1}$ and $H$, and touch $AC$ at points $K_{1}$ and $K$. $O_{1}D$ intersects $OB$ and $BC$ at points $R$ and $L$, respectively. $OE$ intersects $OC$ and $BC$ at points $S$ and $T$, respectively. $OO_{1}$ intersects $MN$ at point $P$. $RS... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,011 |
1. Given two equations about $x$
$$
x^{2}-x+3 m=0, x^{2}+x+m=0(m \neq 0) \text {. }
$$
If one root of the first equation is three times a root of the second equation, then the value of the real number $m$ is $\qquad$ | - 1. -2 .
Let one root of the latter equation be $\alpha$. And the former equation has a root $3 \alpha$, then
$$
\alpha^{2}+\alpha+m=0,
$$
and
$$
\begin{array}{l}
9 \alpha^{2}-3 \alpha+3 m=0 \\
\Rightarrow 3 \alpha^{2}-\alpha+m=0 .
\end{array}
$$
(2) - (1) gives
$$
2 \alpha^{2}-2 \alpha=0 \Rightarrow \alpha=0 \text ... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,012 |
2. In trapezoid $A B C D$, it is known that $A B / / C D$, $\angle A B C=90^{\circ}, B D \perp A D, B C=5, B D=13$. Then the area of trapezoid $A B C D$ is $\qquad$ . | 2. $\frac{1565}{24}$.
As shown in Figure 2, from the given conditions we know
$$
C D=\sqrt{13^{2}-5^{2}}=12, \angle B D C=\angle A B D \text {. }
$$
Therefore, Rt $\triangle B C D \backsim$ Rt $\triangle A D B$
$$
\Rightarrow \frac{A B}{13}=\frac{13}{12} \Rightarrow A B=\frac{169}{12} \text {. }
$$
Thus, $S_{\text {... | \frac{1565}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,013 |
3. From six cards numbered $1, 2, \cdots, 6$, any three cards are drawn. Then the probability that the numbers on the drawn cards are all greater than or equal to 2 is $\qquad$. | 3. $\frac{1}{2}$.
There are 20 possible ways to draw three cards from six cards. There are 10 possible ways where the card numbers are all greater than or equal to 2, so the required probability is $p=\frac{10}{20}=\frac{1}{2}$. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,014 |
4. Arrange the eight numbers $-7, -5, -3, -2, 2, 4, 6, 13$ as $a, b, c, d, e, f, g, h$, such that
$$
(a+b+c+d)^{2}+(e+f+g+h)^{2}
$$
is minimized. Then this minimum value is $\qquad$ | 4.34.
Let $x=a+b+c+d$. Then
$$
\begin{array}{l}
e+f+g+h=8-x, \\
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
=x^{2}+(8-x)^{2}=2(x-4)^{2}+32 .
\end{array}
$$
From the known eight numbers, the sum of any four numbers $x$ is an integer and cannot be 4, so we get
$$
\begin{array}{l}
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
\geqslant 2+32=34,
\en... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,015 |
5. Given a square $A B C D$ with side length $4, E$ and $F$ are points on sides $A B$ and $B C$ respectively, such that $A E=3, B F=2$. Line segments $A F$ and $D E$ intersect at point $G$. Then the area of quadrilateral $D G F C$ is $\qquad$. | 5. $\frac{84}{11}$.
As shown in Figure 3, extend $C B$ and the extension of $D E$ to intersect at point $K$. Then
$$
\begin{array}{l}
\frac{K B}{K B+4}=\frac{B E}{C D}=\frac{1}{4} \\
\Rightarrow K B=\frac{4}{3} \text {. } \\
\text { Therefore, } \frac{K G}{G D}=\frac{K F}{A D} \\
=\frac{\frac{4}{3}+2}{4}=\frac{5}{6} \... | \frac{84}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,016 |
6. In an isosceles right $\triangle ABC$, it is known that $\angle ACB=$ $90^{\circ}, P$ is a point inside $\triangle ABC$, such that $PA=11, PB=7$, $PC=6$. Then the length of side $AC$ is $\qquad$ | $$
\text { 6. } 85+42 \sqrt{2} \text {. }
$$
Rotate $\triangle P B C$ 90 degrees clockwise around point $C$, so that point $B$ coincides with $A$, and point $P$ moves to position $P^{\prime}$ (as shown in Figure 4), and connect $P P^{\prime}$. Then $\triangle C P P^{\prime}$ is an isosceles right triangle, and $P P^{\... | 85+42 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,017 |
7. There are 10 chess players participating in a round-robin tournament (i.e., each pair of players competes in one match). The rules state that a win earns 2 points, a draw earns 1 point for each player, and a loss earns 0 points. After the tournament, it is found that each player's score is unique, and the second-pla... | 7. 16 .
The last five contestants have to play 10 matches, so the total score of the last five contestants is greater than or equal to $2 \times 10=20$.
Thus, the score of the second-place contestant is greater than or equal to
$$
\frac{4}{5} \times 20=16 \text {. }
$$
The score of the first-place contestant is less ... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,018 |
8. Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $a b c d$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ . | 8. 22 .
$$
\begin{array}{l}
\text { Let } a b c d=k+(k+1)+\cdots+(k+34) \\
=35(k+17),
\end{array}
$$
where $k$ is a positive integer. Then the prime numbers $a, b, c, d$ must include one 5 and one 7.
Assume $a=5, b=7$. Then $c d=k+17$.
When $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,019 |
9. (15 points) As shown in Figure 1, given that the intersection point of the diagonals of rectangle $ABCD$ is $O, \angle DAC=60^{\circ},$ and the angle bisector of $\angle DAC$ intersects side $DC$ at point $S$. Line $OS$ intersects $AD$ and $BL$ with $AC$ at points $L$ and $M$ respectively. Prove: $SM \parallel LC$. | Let the intersection point of line $DC$ and $LB$ be $Q$.
From $\angle SAC = \angle SCA = 30^{\circ}$, we know that $\triangle SAC$ is an isosceles triangle.
Since $O$ is the midpoint of side $AC$, $SO \perp AC$.
Thus, $LO$ is both the altitude and the median of $\triangle LAC$ on side $AC$.
Therefore, $\triangle LAC$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,020 |
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