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Example 3 Let $n$ be a fixed integer, $n \geqslant 2$.
(1) Determine the smallest constant $c$ such that the inequality
$$
\sum_{1 \leqslant i \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c \cdot\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$
holds for all non-negative real numbers $x_{1}, x_{2}, \cdots... | Solution: (1) When non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are not all 0, let $x=\frac{\sum_{1 \leq 1<j \leqslant n} x_{i} x_{j}}{\left(\sum_{i=1}^{n} x_{i}\right)^{2}}$,
$$
\begin{array}{l}
y=\frac{\sum_{1 \leq 1<j<k \leq n} x_{i} x_{j} x_{k}\left(x_{i}+x_{j}+x_{k}\right)}{\left(\sum_{i=1}^{n} x_{1}\ri... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,182 |
II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$. | Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$
Similarly, the common root of the last two equations... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,183 |
Three, (25 points) It is known that the lengths of the three sides of the acute triangle $\wedge A B C$ and the length of one of its altitudes are four consecutive integers, and this altitude divides $\triangle A B C$ into two right triangles with integer side lengths. Try to find the lengths of the three sides of $\tr... | Three, as shown in Figure 7, let the height be $A D$, and the four consecutive integers be $n, n+1, n+2, n+3$. Also, assume $A B < A C$, then $A D < A B < A C$. Therefore, $A D = n$ or $n+1$.
When $A D = n+1$, only $B C = n$. Thus, $A B = n+2, A C = n+3$. So, $B D = \sqrt{2 n+3}, C D = 2 \sqrt{n+2}$, and both are inte... | 13, 14, 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,184 |
1. Given that $x, y$ are two unequal positive numbers, then $A=$ $\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2}, B=\frac{x+y}{2}-\sqrt{x y}, C=\sqrt{x y}-$ $\frac{2}{\frac{1}{x}+\frac{1}{y}}$
(A) $A>B>C$
(B) $A>C>B$
(C) $B>A>C$
(D) $B>C>A$ | $-1 .(\mathrm{C})$
$$
\begin{aligned}
A-B & =\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y}-(x+y) \\
& \leq \sqrt{2\left(\frac{x^{2}+y^{2}}{2}+x y\right)}-(x+y)=0 .
\end{aligned}
$$
$\because x, y$ are not equal, hence the equality cannot be achieved, $\therefore B>A$.
Also, $A-C$
$$
\begin{array}{l}
=\left(\sqrt{\frac{x^{2}+... | B>A>C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 713,185 |
2. The functions $y=f(x)$ and $y=g(x)$ have the same domain. For any $x$ in the domain, $f(x)+f(-x)=0$, $g(x) g(-x)=1$, and when $x \neq 0$, $g(x) \neq 1$. Then $F(x)=\frac{2 f(x)}{g(x)-1}+f(x)$ is $(\quad)$.
(A) odd function
(B) even function
(C) both odd and even function
(D) neither odd nor even function | 2. (B).
From the given, when $x \neq 0$,
$$
\begin{array}{l}
F(-x)=\frac{2 f(-x)}{g(-x)-1}+f(-x) \\
=\frac{-2 g(x) f(x)}{1-g(x)}-f(x) \\
=\frac{2 f(x)+2 f(x)[g(x)-1]}{g(x)-1}-f(x) \\
=\frac{2 f(x)}{g(x)-1}+f(x)=F(x) .
\end{array}
$$
Therefore, $F(x)=\frac{2 f(x)}{g(x)-1}+f(x)$ is an even function. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,186 |
3. Given that $a$ and $b$ are non-zero constants. If $M=a \sin \theta+b \cos \theta$, $N=\sqrt{a^{2}+b^{2}} \sin \left(\theta+\arctan \frac{b}{a}\right)$, then for any $\theta$ ( ).
(A) $M=N$
(B) $M \neq N$
(C) Only when $a>0$, $M=N$
(D) Only when $b>0$, $M=N$ | 3. (C).
It is easy to know that $M=\sqrt{a^{2}+b^{2}} \sin (\theta+\varphi)$, where $\cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}, \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}$.
If $M=N$, then $\varphi=2 k \pi+\arctan \frac{b}{a}$.
Also, $\arctan \frac{b}{a} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, hence $\cos \varph... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,187 |
4. As shown in Figure 1, in a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, points $E$ and $F$ are the centers of faces $B B_{1} C_{1} C$ and $A B C D$, respectively. Then the distance between the skew lines $E F$ and $A_{1} C_{1}$ is ( ).
(A) $\frac{a}{2}$
(B) $\frac{\sqrt{2}}{2} a$
(C) $\frac{\sqrt{... | 4. (C).
As shown in Figure 3, connect $A_{1} C$ and $A B_{1}$.
Since $A_{1} C_{1} \parallel A C$, the distance from line $E F$ to $A_{1} C_{1}$ is equal to the distance from line $A_{1} C_{1}$ to plane $A B_{1} C$, which is also equal to the distance from point $A_{1}$ to plane $A B_{1} C$. Let this distance be $h$. T... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,188 |
5. Given a periodic sequence $\left\{x_{n}\right\}$ satisfying $x_{n}=\left|x_{n-1}-x_{n-2}\right|(n$ $\geqslant 3$ ). If $x_{1}=1, x_{2}=a \geqslant 0$, then when the period of the sequence is the smallest, the sum of the first 2002 terms of the sequence is ( ).
(A) 2002
(B) 1335
(C) 1949
(D) 1428 | 5.(B).
Let the sequence $\left\{x_{n}\right\}$ have a period of $T, T \neq 1, T \geqslant 2$.
If $T=2$, then $x_{3}=|a-1|=1$. This gives $a=0$ or 2.
From $x_{4}=\left|x_{3}-x_{2}\right|=|1-a|=a$, we get $a=\frac{1}{2}$. This contradicts $a=0$ or 2. Therefore, $T \geqslant 3$.
When $T=3$, from $x_{4}=|| a-1|-a|=1$, we ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,189 |
6. Let points $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $l$ be the right directrix. If there exists a point $M$ on the ellipse such that $\left|M F_{1}\right|$, $\left|M F_{2}\right|$, and the distance $d$ from point $M$ to $l$ form a geometric progre... | 6. (A).
From $\left|M F_{2}\right|^{2}=\left|M F_{1}\right| \cdot d$ we get $\frac{\left|M F_{1}\right|}{\left|M F_{2}\right|}=\frac{\left|M F_{2}\right|}{d}=e$, i.e., $\left|M F_{1}\right|=e\left|M F_{2}\right|$.
Also, $\left|M F_{1}\right|+\left|M F_{2}\right|=2 a$.
Solving (1) and (2) we get
$\left|M F_{1}\right|=\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,190 |
1. Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=1,\left|z_{2}\right|=2,3 z_{1}-$ $z_{2}=2+\sqrt{3} \mathrm{i}$. Then $2 z_{1}+z_{2}=$ $\qquad$ . | II. $1.3-\sqrt{3} \mathrm{i}$ or $-\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i}$.
From $3 z_{1}-z_{2}=2+\sqrt{3} \mathrm{i}$, we get
$\left|3-\frac{z_{2}}{z_{1}}\right|=\frac{|2+\sqrt{3} i|}{\left|z_{1}\right|}=\sqrt{7}$.
Also, $\left|\frac{z_{2}}{z_{1}}\right|=\frac{\left|z_{2}\right|}{\left|z_{1}\right|}=2$, so $\fra... | 3-\sqrt{3} \mathrm{i} \text{ or } -\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,191 |
2. Given $x \geqslant 0, x^{2}+(y-4)^{2} \leqslant 4$, let $u=$ $\frac{x^{2}+\sqrt{3} x y+2 y^{2}}{x^{2}+y^{2}}$. Then the range of $u$ is $\qquad$ . | $$
\begin{array}{l}
2.2 \leqslant w \leqslant \frac{5}{2} \\
w-\frac{1}{2}=\frac{\frac{1}{2}(x+\sqrt{3} y)^{2}}{x^{2}+y^{2}}=\frac{2\left(\frac{x+\sqrt{3} y}{2}\right)^{2}}{\left(\sqrt{x^{2}+y^{2}}\right)^{2}}
\end{array}
$$
As shown in Figure 4, let $P(x, y)$, the line $l: x+\sqrt{3} y=0$,
$$
PA \perp l \text{ at } A... | 2 \leqslant w \leqslant \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,192 |
Example 4 Let the constant $a$ be such that the inequality concerning $x$
$$
\frac{1}{1+\sqrt{x}} \geqslant a \sqrt{\frac{x}{x-1}}
$$
has non-zero real solutions. Find the maximum value of $a$. | Let $f(x)=\frac{1}{1+\sqrt{x}} \cdot \sqrt{\frac{x-1}{x}}=\frac{\sqrt{x-1}}{x+\sqrt{x}}$, then $a \leqslant \max _{x>1}\{f(x)\}$. Below we find $\max _{x>1}\{f(x)\}$.
$$
\begin{array}{l}
\because f^{\prime}(x) \\
=\frac{1}{2 \sqrt{x-1}(x+\sqrt{x})}-\frac{\sqrt{x-1}}{(x+\sqrt{x})^{2}}\left(1+\frac{1}{2 \sqrt{x}}\right) ... | \sqrt{\frac{5 \sqrt{5}-11}{2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,193 |
3. Given in the tetrahedron $S-ABC$, the sum of the three dihedral angles at each vertex of the base triangle is $180^{\circ}$, and the sides of the base triangle are $\sqrt{3}$, $2$, and $\sqrt{5}$. Then the volume of the tetrahedron is $\qquad$. | 3. $\frac{\sqrt{6}}{3}$.
As shown in Figure 5, the three lateral faces of the tetrahedron $S-ABC$ are unfolded on the base plane.
$$
\begin{array}{r}
\because S_{1} A=S_{2} A \\
=S A, \text { and } \angle S_{1} A B+ \\
\angle B A C+\angle S_{2} A C=
\end{array}
$$
$180^{\circ}$,
$$
\therefore S_{1} 、 A 、 S_{2} \text {... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,194 |
4. Let $f_{1}(x)=\frac{2}{1+x}$. Define $f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. And let $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}$. Then $a_{100}=$ $\qquad$ . | 4. $-\frac{1}{2^{101}}$.
Since $f_{1}(0)=2$, we have $a_{1}=\frac{1}{4}$.
From $f_{1}(x)=\frac{2}{1+x}$, we get $f_{n}(0)=\frac{2}{1+f_{n-1}(0)}$.
Thus, $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}=-\frac{1}{2} \frac{f_{n-1}(0)-1}{f_{n-1}(0)+2}$,
which means $a_{n}=-\frac{1}{2} a_{n-1}$.
Therefore, $a_{n}=\frac{1}{4}\left(-\f... | -\frac{1}{2^{101}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,195 |
5. Given an ellipse $\frac{x^{2}}{2}+k y^{2}=1$ with its foci on the $x$-axis, points $A$ and $B$ are the two intersection points of a line passing through the origin with the ellipse. If the number $k$ allows for another point $C$ on the ellipse such that $\triangle A B C$ is an equilateral triangle, then for all such... | 5. $\frac{2 \sqrt{3}}{3}$.
Let $|O A|=r_{1},|O C|=r_{2}$. Then, since $\triangle A B C$ is an equilateral triangle, we know that $O C \perp A B$, and $r_{2}=\sqrt{3} r_{1}$.
Let $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right)$. Then it is easy to know that the coordinates of point $C$ are $\left(r_{2} \cos \left(... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,196 |
6. Given the system of equations $\left\{\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1, \\ x^{2}+y^{2}=50\end{array}\right.$ has only integer solutions. Then the number of real pairs $(a, b)$ that satisfy the condition is $\qquad$ . | 6.60.
It is easy to know that the equation $x^{2}+y^{2}=50$ has 12 sets of integer solutions: $(\pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$, which correspond to 12 integer points on the circle. For each pair of real numbers that satisfy the condition, there corresponds a line in the coordinate system. Therefore, th... | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,197 |
Three. (20 points) Given $a_{i} \in \mathbf{R}^{+}$, and $a_{i} \geqslant a_{i+1}, i=1,2$, $\cdots, n-1$. Prove:
$$
\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{n}+a_{1}} \geqslant \frac{n}{2} .
$$ | Three, let $A=\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{n}+a_{1}}$,
$$
B=\frac{a_{2}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{2}+a_{3}}+\cdots+\frac{a_{1}}{a_{n}+a_{1}} .
$$
Then $A+B=n$,
$$
\begin{aligned}
A-B & =\frac{a_{1}-a_{2}}{a_{1}+a_{2}}+\frac{a_{2}-a_{3}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,198 |
Four. (20 points) Given $n$ points in space that are not coplanar $(n \geqslant 4)$. Is there definitely a plane that passes through only three of these $n$ points? Please prove your conclusion. | When $4 \leqslant n \leqslant 7$, such a plane definitely exists, while when $n \geqslant 8$, such a plane may not exist.
Proof: When $4 \leqslant n \leqslant 7$,
If there are three points collinear, then such a plane obviously exists;
If there are four points collinear among these $n$ points, then $n \geqslant 6$. Whe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,199 |
Five. (20 points) If there exist four points on a plane curve such that the figure formed by these four points is a rhombus, then the curve is said to have an inscribed rhombus. Given the hyperbola $c_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, and the hyperbola $c_{2}$: $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$... | Lemma 1 If the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an inscribed rhombus, then the center of this rhombus must be the origin.
Assume the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ has an inscribed rhombus $A B C D$, with coordinates $A\left(x_{A}, y_{A}\right)$, $B\left(x_{B},... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,200 |
One, (50 points) As shown in Figure 2, points $P$ and $Q$ are on the circumcircle of $\triangle ABC$ (excluding $A$, $B$, and $C$). The reflections of point $P$ over the lines $BC$, $CA$, and $AB$ are points $U$, $V$, and $W$, respectively. The lines $QU$, $QV$, and $QW$ intersect the lines $BC$, $CA$, and $AB$ at poin... | (1) As shown in Figure 8, let perpendiculars be drawn from point $P$ to $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $X$, $Y$, and $Z$ respectively.
By symmetry, $XY$ is the midline of $\triangle PUV$.
Thus, $UV \parallel XY$.
Similarly, $VW \parallel YZ$, $WU \parallel XZ$.
By Simson's theorem, $X$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,201 |
Two. (50 points) Given $x_{i} \geqslant 0(i=1,2, \cdots, n), n \geqslant 2$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \frac{k}{j} x_{k} x_{j}=1$. Try to find the maximum and minimum values of $\sum_{j=1}^{n} x_{i}$. | $$
\begin{array}{l}
\because\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{i \leqslant i<i \leqslant n} \frac{k}{j} x_{i} x_{j}=1, \\
\therefore \sum_{i=1}^{n} x_{i} \geqslant 1 .
\end{array}
$$
Therefore, the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1, and the equality holds if and... | \sqrt{\sum_{k=1}^{n} \frac{1}{2 k-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,202 |
Three. (50 points) Given that $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is a permutation of the natural numbers 1, 2, $\cdots, n$, and satisfies: for any $1 \leqslant i \leqslant n-1$, we have $a_{1}+i \leqslant a_{i+1}+i+1$.
(1) If we denote $x_{i}$ as the position index of the number $i(1 \leqslant i \leqslant n)$ i... | Three, (1) Proof 1: Assume the conclusion does not hold, then there must exist $j, 1 \leqslant j \leqslant n-1$, such that $x_{j} \geqslant x_{j+1}+2$.
Without loss of generality, let $x_{j}=x_{j+1}+m, m \in \mathbf{N}, m \geqslant 2$.
First, we prove: For any integer $t \in [0, m]$, we have
$a_{x+1}+1 \geqslant j$.
If... | 2^{n-1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,203 |
Example 5 Find natural numbers $a$, $b$, $c$, such that for any $n \in$ N, $n>2$, we have
$$
\begin{array}{l}
b-\frac{c}{(n-2)!} \\
<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<b .
\end{array}
$$
(1996, World Inter-Cities Mathematical Competition) | Solution: Let $b_{n}=b-\frac{2^{3}-a}{2!}-\frac{3^{3}-a}{3!}-\cdots$ $-\frac{n^{3}-a}{n!}=\frac{P(n)}{n!}$
where $P(n)$ is a polynomial in $n$.
Obviously, the degree of $P(n)$ cannot be greater than 2.
Let $P(n)=k n^{2}+l n+m$.
From $b_{n+1}-b_{n}=-\frac{(n+1)^{3}-a}{(n+1)!}$
$$
=\frac{P(n+1)}{(n+1)!}-\frac{P(n)}{n!}
$... | a=5, b=9, c_{\text{min}}=4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,204 |
117. Find all integers $k$ such that the quadratic equation $k x^{2}-2 \times(3 k-1) x+9 k-1=0$ has at least one integer root. | Solution: From the original equation, we get
$$
k(x-3)^{2}=1-2 x \text {. }
$$
$x$ is an integer root, clearly $x \neq 3, k \neq 0$. Thus, $k=$ $\frac{1-2 x}{(x-3)^{2}}$. Furthermore, since $k$ is an integer, we have
$$
|1-2 x| \geqslant(x-3)^{2} \text {. }
$$
If $x \leqslant 0$, then $1-2 x \geqslant x^{2}-6 x+9$, wh... | -3, -7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,205 |
In $\triangle A B C$, $A P$ bisects $\angle A, B Q$ bisects $\angle B$, $P, Q$ are on $B C, C A$ respectively.
Prove: $A B+B P=A Q+B Q$ if and only if $\angle A B C=120^{\circ}$ or $\angle A B C=2 \angle C$. | Proof: Necessity.
As shown in Figure 1, it is known that $A B + B P = A Q + B Q$. Extend $A B$ to $E$ such that $B E = B P$.
On $A C$, take $Q F = B Q$, and connect $P E, P F, B F$.
(i) If $F$ coincides with $C$, then $\angle Q B C = \angle Q C B \Rightarrow \angle A B C = 2 \angle C$.
(ii) If $F$ does not coincide wit... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,206 |
117. Point $P(a, b)$ is in the first quadrant. A line $l$ is drawn through point $P$, intersecting the positive $x$-axis and $y$-axis at points $A$ and $B$, respectively. $O$ is the origin, and $m$ is a positive integer. Find the slope $k_{1}$ of line $l$ that minimizes $P A^{m} + P B^{m}$. | Solution: Draw $PC \perp x$-axis at $C$, $PD \perp y$-axis at $D$, then we have $PD = a$, $PC = b$. Let $\frac{PB}{PA} = \lambda$, from Figure 3 we know $\triangle PAC \sim \triangle BPD$.
Thus, $\frac{PB}{PA} = \frac{BD}{PC} = \frac{PD}{AC} = \lambda$,
$BD = \lambda PC = \lambda b$.
By the Pythagorean theorem,
$$
\beg... | k_l = -\left(\frac{b}{a}\right)^{\frac{m}{m+2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,207 |
118. Let there be $n$ points $(n \geq 4)$ on a sphere with radius $R$, and these points do not all lie within a single hemisphere. Prove that there exists a plane passing through three of these $n$ points such that the distance from the center $O$ of the sphere to this plane is no greater than $\frac{R}{3}$. | Proof: If the plane determined by any three of these $n$ points passes through the center of the sphere, then the proposition holds. Therefore, we can assume that the center of the sphere is not in the plane determined by any three of these $n$ points.
Among all these planes, there must be one that is closest to the c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,208 |
Example 6 Given a natural number $n \geqslant 2$, find the smallest positive number $\lambda$, such that for any $a_{i} \geqslant 0,0 \leqslant b_{i} \leqslant \frac{1}{2}(i=1,2, \cdots$, $n)$, and $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, we have $a_{1} a_{2} \cdots a_{n} \leqslant \lambda \sum_{i=1}^{n} a_{i} b_... | Solution: If there exists some $j(j=1,2, \cdots, n)$ such that $a_{j}=0$, then the original inequality holds for any $\lambda \geqslant 0$. Therefore, we can assume that for all $i$, $a_{i}>0$.
$\because f(x)=\frac{1}{x}$ is a convex function on $(0,+\infty)$ and $\sum_{i=1}^{n} b_{i}=1$, hence by Jensen's inequality w... | \frac{1}{2(n-1)^{n-1}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,209 |
1. Let $\lambda$ be a given positive number. Find the largest constant $c(\lambda)$ such that for any $x_{1}, x_{2} \geqslant 0$, we have $x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \geqslant c\left(x_{1}+x_{2}\right)^{2}$. | (Tip: In the original inequality, let $x_{1}=x_{2}=1$. Get $c \leqslant$ $\frac{\lambda+2}{4}$. Then consider whether $c=\frac{\lambda+2}{4}$ holds for the original inequality. Consider two cases: $1^{\circ}$ When $\lambda<2$. Use the comparison method to prove $x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2}$ $\geqslant \frac... | c_{\text {max }}=\frac{\lambda+2}{4} \text{ for } \lambda < 2; \, c_{\text {max }}=1 \text{ for } \lambda \geq 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,210 |
2. Let $x+y=k, x, y \in \mathbf{R}^{+}$. Try to find the maximum value of $k$ such that the inequality $\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right) \geqslant\left(\frac{k}{2}+\frac{2}{k}\right)^{2}$ always holds. | (Notice that when $x=y$, the above equation obviously holds. Therefore, without loss of generality, assume $x>y$, and use the mean substitution: let $m=\frac{k}{2}, x=m+t, y=m-t, 0<t<m$, rewrite the original inequality as
$$
\left(m+t+\frac{1}{m+t}\right)\left(m-t+\frac{1}{m-t}\right) \geqslant\left(m+\frac{1}{m}\right... | 2 \sqrt{2+\sqrt{5}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,211 |
3. For any natural numbers $m, n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Find the maximum value of $\lambda$.
| (Let $G=|(m, n)| m<\sqrt{7} n, m, n \in \mathbf{N}$. $\lambda_{\text {max }}=\min _{(m, n \in 6} 17 n^{2}-m^{2}$, then perform $\bmod 7$ analysis on $7 n^{2}-m^{2}$, obtaining $\lambda_{\text {max }}=3$. ) | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,212 |
5. Determine the smallest real number $c$ such that for any sequence satisfying the condition:
$$
x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}(n=1,2, \cdots)
$$ | (Let $x_{1}=2^{-1}$, satisfying the problem's conditions. We can get $c \geqslant \frac{1}{\sqrt{2}-1}$. $\frac{1-\sqrt{\frac{1}{2^{n}}}}{\sqrt{1-\frac{1}{2^{n}}}}$. Let $n \rightarrow+\infty$. We get $c \geqslant \sqrt{2}+1$. Below, we use mathematical induction to prove $\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant(\sqrt{2}... | \sqrt{2}+1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,214 |
Example $\mathbf{3}$ A factory's production on the first day does not exceed 20 units, and thereafter, the daily production increases every day, but the amount of increase each time does not exceed 20 units. Prove: When the daily production reaches 1995 units, the total number of products produced by the factory will n... | Explanation: Suppose the daily output on the $n$-th day after the factory starts operation reaches 1995 pieces. If the daily output on the first day is $a_{1}$ pieces, and the increase in output on the $i$-th day is $a_{i}$ pieces, then the daily output on the $n$-th day should be
$$
a_{1}+a_{2}+\cdots+a_{n}=1995 .\lef... | 100500 | Algebra | proof | Yes | Yes | cn_contest | false | 713,215 |
Example 6 There are 1994 matches on the table, two children, A and B, take turns to take 1, 2 or 3 matches each time, the one who can take the last match wins. Now A takes first, which child will win? How should he play to win this game? | Solution: When it's A's turn, if the number of matches is $4k+1$, $4k+2$, or $4k+3$, then A can win. The specific strategy is: for $4k+1$, $4k+2$, or $4k+3$ matches (favorable state), A takes $1$, $2$, or $3$ matches respectively, leaving $4k$ matches (unfavorable state) for B to take. When it's A's turn, it is always ... | A | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,216 |
Example 7 A group of children sit in a circle to share candies. The teacher asks them to each take an even number of candies first, and then adjust according to the following rules: all children simultaneously give half of their candies to the child on their right, and those whose number of candies becomes odd ask the ... | Solution: Let the child with the most candies have $2M$ pieces before a certain adjustment. The child with the least has $2m$ pieces, and $M > m$. Observe the changes in the number of candies each child has after one adjustment:
(1) Each child's number of candies remains between $2m$ and $2M$. Fact: If a child original... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,217 |
Example 8 Given any positive integer, add its digits, the sum can be a one-digit or multi-digit number. If it is not a one-digit number, add the digits of the sum again, and continue this process until a one-digit number is obtained. If this one-digit number is one of $2, 3, 5, 6$, prove that the original given integer... | Prove: Since any positive integer can be expressed in one of the following five forms:
$$
9 n, 9 n \pm 1, 9 n \pm 2, 9 n \pm 3, 9 n \pm 4 \text {. }
$$
Using proof by contradiction. Suppose the given positive integer is of one of the following forms:
$$
\begin{array}{l}
(9 n)^{2}=9 k, (9 n \pm 1)^{2}=9 k+1, \\
(9 n \p... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,218 |
Example 9 The integers corresponding to each vertex of a regular pentagon sum to a positive number. If three consecutive vertices correspond to integers $x, y, z$, and the middle $y < 0$, then the following operation is performed: the integers $x, y, z$ are replaced by $x+y, -y, z+y$. As long as at least one of the fiv... | Solution: For convenience in calculation, write the five numbers in a column: $v$, $w, x, y, z$, noting that $z$ and $v$ are adjacent. Suppose $y0$, so the difference above is negative, meaning that the sum of the squares of the pairs of the five numbers decreases after the operation. However, the sum of the squares of... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,219 |
Example 10: There are 6 apples in each of three plates, Jia, Yi, and Bing. The apples are moved as follows: Plate Jia remains unchanged, 1 apple is moved from one plate to another; Plate Yi remains unchanged, 2 apples are moved from one plate to another; Plate Bing remains unchanged, 3 apples are moved from one plate t... | Solution: In the following tables, one, two, three, four, and five represent the first, second, third, fourth, and fifth moves, respectively. A, B, and C represent the three plates A, B, and C. According to the problem, the numbers in Table 1 indicate the number of apples moved into or out of each plate during each mov... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,220 |
4. Every positive integer greater than 2 can be expressed as the sum of several distinct natural numbers. For any positive integer $n(n \geqslant 3)$, find the maximum number of these addends $A(n)$.
Every positive integer greater than 2 can be expressed as the sum of several distinct natural numbers. For any positive... | (For any positive integer $n(n \geqslant 3)$, there must exist a positive integer $m$ such that
$$
1+2+\cdots+m \leqslant n<1+2+\cdots+m+(m+1),
$$
i.e., $\frac{m(m+1)}{2} \leqslant n<\frac{(m+1)(m+2)}{2}$.
At this time, $n$ can be expressed as the sum of at most $m$ different positive integers $\left(n=1+2+\cdots+(m-1... | \left\lfloor \frac{-1+\sqrt{8 n+1}}{2} \right\rfloor | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,221 |
5. Given $n$ points $A_{1}, A_{2}, \cdots, A_{n}(n \geqslant 3)$ in the plane, no three of which are collinear, determine $k$ lines from $k$ point pairs (i.e., draw a line through each point pair of the $k$ point pairs), such that these $k$ lines do not form a triangle with all three vertices being the given points. Fi... | (提示: 设 $A_{1} 、 A_{2}$ 连接一条直线. 由于 $A_{1}$ 至多连 $n-1$ 条直线, 而 $A_{3}$ 不能与 $A_{1} 、 A_{2}$ 相连, 所以 $A_{3}$ 至多连 $n-3$ 条直线. 设 $A_{3}$ 与 $A_{4}$ 相连, 则 $A_{5}$ 不能与 $A_{1}$ 、 $A_{2} 、 A_{3} 、 A_{4}$ 相连, $A_{5}$ 至多连 $n-5$ 条直线. 如此继续.
$$
\begin{aligned}
k & \leqslant(n-1)+(n-3)+(n-5)+\cdots \\
& =\left\{\begin{array}{l}
(n-1)+(n-3)... | \frac{n^2}{4} \text{ for even } n, \frac{n^2-1}{4} \text{ for odd } n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,222 |
In 1996, the second trial of the National High School Mathematics Competition, the plane geometry problem: As shown in Figure 1, circles $O_{1}$ and $O_{2}$ are tangent to the lines containing the sides of $\triangle A B C$, with $E$, $F$, $G$, and $H$ being the points of tangency. The extensions of $E G$ and $F H$ int... | This problem has many proof methods, which have been widely explored by everyone. We will not repeat them here. By utilizing the conclusion of this competition problem and with the help of some well-known theorems, we obtain two new conclusions related to the excircles of a triangle as follows:
Proposition: Let $\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,223 |
1. The monotonic increasing interval of the function $f(x)=\log _{\frac{1}{2}}\left(x^{2}-2 x-3\right)$ is ( ).
(A) $(-\infty,-1)$
(B) $(-\infty, 1)$
(C) $(1,+\infty)$
(D) $(3,+\infty)$ | -1. (A).
From $x^{2}-2 x-3>0$ we have $x<-1$ or $x>3$, so the domain of the function $\log _{\frac{1}{2}}\left(x^{2}-2 x-3\right)$ is $x<-1$ or $x>3$. The quadratic function $u=x^{2}-2 x-3$ is monotonically decreasing in $(-\infty,-1)$ and monotonically increasing in $(3,+\infty)$. Since $\log _{\frac{1}{2}} u$ is mono... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,224 |
2. If real numbers $x, y$ satisfy $(x+5)^{2}+(y-12)^{2}=14^{2}$, then the minimum value of $x^{2}+y^{2}$ is ( ).
(A) 2
(B) 1
(C) $\sqrt{3}$
(D) $\sqrt{2}$ | 2. (B).
As shown in Figure 6, $(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center at point $C(-5, 12)$ and radius 14. Let $P$ be any point on the circle, then $|O P| \geqslant |C P|-|O C|=14-13=1$.
When points $C$, $O$, and $P$ are collinear, the equality holds. Therefore, the minimum value of $P$ to point $O$ is 1... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,225 |
Example 4 Given $n$ $(n \geqslant 2)$ positive integers $x_{1}, x_{2}$, $\cdots, x_{n}$, arrange them in non-decreasing order as $x_{1} \leqslant x_{2} \leqslant \cdots$ $\leqslant x_{n}$. If the sum of these $n$ positive integers equals their product, find the maximum value of $x_{n}$. | Explanation: An integer array that meets the requirements of the problem exists, for example, $2+2=2 \times 2, 1+2+3=1 \times 2 \times 3, 1+1+2+4=1 \times 1 \times 2 \times 4$, etc. From these examples, we can see that the maximum value of $x_{n}$ is exactly $n$.
We estimate the maximum value of $x_{n}$ as a whole.
Fir... | n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,226 |
3. The function $f(x)=\frac{x}{1-2^{x}}-\frac{x}{2}(\quad)$.
(A) is an even function but not an odd function
(B) is an odd function but not an even function
(C) is both an even function and an odd function
(D) is neither an even function nor an odd function | 3. (A).
The domain of the function $f(x)$ is $(-\infty, 0) \cup(0,+\infty)$. When $x \neq 0$, since
$$
\begin{array}{l}
f(-x)=\frac{-x}{1-2^{-x}}-\frac{-x}{2}=\frac{-x 2^{x}}{2^{x}-1}+\frac{x}{2} \\
=\frac{x+x\left(2^{x}-1\right)}{1-2^{x}}+\frac{x}{2}=\frac{x}{1-2^{x}}-\frac{x}{2}=f(x),
\end{array}
$$
therefore, $f(x... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,227 |
4. The line $\frac{x}{4}+\frac{y}{3}=1$ intersects the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at points $A$ and $B$. There is a point $P$ on the ellipse such that the area of $\triangle P A B$ is 3. How many such points $P$ are there?
(A) 1
(B) 2
(C) 3
(D) 4 | 4.(B).
Let $P_{1}(4 \cos \alpha, 3 \sin \alpha)$ $\left(0<\alpha<\frac{\pi}{2}\right)$, i.e., point $P_{1}$ is on the ellipse in the first quadrant. As shown in Figure 7, consider the area $S$ of quadrilateral $P_{1} A O B$:
$$
\begin{aligned}
S & =S_{\text {Cax }_{1}}+S_{\text {COses }_{1}} \\
& =-\frac{1}{2} \times ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,228 |
6. The volume of the solid of revolution obtained by rotating the figure bounded by the curves $x^{2}=4 y, x^{2}=-4 y, x=4, x=-4$ around the $y$-axis is $V_{1}$; the volume of the solid of revolution obtained by rotating the figure composed of points $(x, y)$ that satisfy $x^{2}+y^{2} \leqslant 16, x^{2}+(y-2)^{2} \geq... | 6. (C).
As shown in Figure 1, the two figures rotate around the $y$-axis to form solids of revolution that are confined between two parallel planes 8 units apart. If any plane perpendicular to the $y$-axis is used to intersect these solids of revolution, and the distance of the intersection from the origin is $|y|$, t... | C | Calculus | MCQ | Yes | Yes | cn_contest | false | 713,230 |
7. Given complex numbers $z_{1}$ and $z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3$. If the angle between the vectors they correspond to is $60^{\circ}$, then $\left|\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right|=$ $\qquad$ . | $=.7 . \frac{\sqrt{133}}{7}$.
As shown in Figure 8, by the cosine rule we get
$$
\begin{array}{l}
\left|z_{1}+z_{2}\right|=\sqrt{19}, \\
\left|z_{1}-z_{2}\right|=\sqrt{7} .
\end{array}
$$
Therefore, $\left|\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right|=\frac{\sqrt{133}}{7}$. | \frac{\sqrt{133}}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,231 |
8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ . | 8.3.
It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).
When $n=8$, $T_{r+1}=\mathrm{... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,232 |
10. Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$ | 10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,234 |
11. Given $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, the minimum value of $|x|-$ $|y|$ is $\qquad$ . | $11 \cdot \sqrt{3}$.
$$
\left\{\begin{array} { l }
{ x + 2 y > 0 , } \\
{ x - 2 y > 0 , } \\
{ ( x + 2 y ) ( x - 2 y ) = 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x>2|y| \geqslant 0, \\
x^{2}-4 y^{2}=4 .
\end{array}\right.\right.
$$
By symmetry, we only need to consider $y \geqslant 0$. Since $x>0$, we only... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,235 |
Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team.... | Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,236 |
13. As shown in Figure 3. Given point $A(0,2)$ and two points $B$、$C$ on the parabola $y^{2}=x+4$. Such that $A B \perp B C$. Find the range of the y-coordinate of point $C$. | 13. As shown in Figure 3, let $B\left(y_{1}^{2}-4, y_{1}\right), C\left(y^{2}-4, y\right)$.
Obviously, $y_{1}^{2}-4 \neq 0$. Therefore, $k_{A B}=\frac{y_{1}-2}{y_{1}^{2}-4}=\frac{1}{y_{1}+2}$.
Since $A B \perp B C$, we have $k_{B C}=-\left(y_{1}+2\right)$.
Thus, $\left\{\begin{array}{l}y-y_{1}=-\left(y_{1}+2\right)\lef... | y \leqslant 0 \text{ or } y \geqslant 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,237 |
14. As shown in Figure 4, there is a sequence of curves $P_{0}, P_{1}, P_{2}, \cdots$, where it is known that the area of the equilateral triangle enclosed by $P_{0}$ is 1. $P_{\lambda+1}$ is obtained by performing the following operation on $P_{A}$: divide each side of $P_{k}$ into three equal parts, construct an equi... | 14. (1) As shown in Figure 4, when operating on $P_{0}$, it is easy to see that each side of $P_{0}$ becomes 4 sides of $P_{1}$, so the number of sides of $P_{1}$ is $3 \times 4$; similarly, when operating on $P_{1}$, each side of $P_{1}$ becomes 4 sides of $P_{2}$, so the number of sides of $P_{2}$ is $3 \times 4^{2}$... | \frac{8}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,238 |
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$, $a \neq 0$ ) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find... | 15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$. Therefore, $-\frac{b}{2a}=-1, b=2a$.
From (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From (1), $f(1) \geqslant 1$, and from (2), $f(1) \leqslant 1$,
thus $f(1)=1$, i.e., $a+b+c=1$.
Also, $a-b+c=0$, so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,239 |
一, (50 points) As shown in Figure 5, in $\triangle A B C$, $\angle A=$ $60^{\circ}, A B>A C$, point $O$ is the circumcenter, altitudes $B E$ and $C F$ intersect at point $H$. Points $M$ and $N$ lie on segments $B H$ and $H F$ respectively, and satisfy $B M = C N$. Find the value of $\frac{M H + N H}{O H}$. | As shown in Figure 9, take $B K = C H$ on $B E$, and connect $O B, O C, O K$. By the properties of the circumcenter of a triangle, we know $\angle B O C = 2 \angle A = 120^{\circ}$. By the properties of the orthocenter of a triangle, we know $\angle B H C = 180^{\circ} - \angle A = 120^{\circ}$.
Thus, $\angle B O C = \... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,240 |
II. (50 points) Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x) = x^{3} + a x^{2} + b x + c$ have three real roots $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2} - x_{1} = \lambda$;
(2) $x_{3} > \frac{1}{2}\left(x_{1} + x_{2}\right)$.
Find the maximum value of $\frac{2 a^{3} + 27 c - 9 a b}{\lambda^{3}... | Due to $f(x)=f(x)-f\left(x_{3}\right)$
$$
=\left(x-x_{3}\right)\left[x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b\right] \text {, }
$$
Therefore, $x_{1} 、 x_{2}$ are the roots of the equation
$$
x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b=0
$$
From (1) we get
$$
\left(a+x_{3}\right)^{2}-4\left(x_{3}^{2}+a x_... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,241 |
Three, (50 points) Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (... | Three, let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2$, $\cdots, 7)$, the problem is to find the number of positive integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+\cdots+x_{7}=270
$$
under the conditions $71 x_{i}(1 \leqslant i \leqslant 4)$ and $131 x_{j}(5 \leqslant j \leqslant ... | 42244 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,242 |
1. Let the sequence $\left\{x_{n} \mid\right.$ satisfy $x_{1}=\frac{1}{2}, x_{n+1}=x_{n}+\frac{x_{n}^{2}}{n^{2}}$. Prove: $x_{2001}<1001$.
(Li Weiguo) | 1. Prove by mathematical induction: For any $n \in \mathbf{N}$, we have $x_{n} \leqslant \frac{n}{2}$.
When $n=1$, it is known from the condition that (1) holds.
Assume that (1) holds for $n=k$, i.e., $x_{k} \leqslant \frac{k}{2}$. When $n=k+1$, we have
$$
\begin{array}{l}
x_{k+1}=x_{k}+\frac{x_{k}^{2}}{k^{2}} \leqslan... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,243 |
2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant... | Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,244 |
3. Let $n$ and $m$ be positive integers of different parity, and $n > m$. Find all integers $x$ such that $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}$ is a perfect square.
(Pan Chengdu) | 3. Let $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}=A^{2}$, where $A \in \mathbf{N}$, then
$$
A^{2}=\prod_{i=m}^{n-1}\left(x^{2^{4}}+1\right) \text {. }
$$
Notice that for $i \neq j,\left(x^{2^{2}}+1, x^{2^{2}}+1\right)=\left\{\begin{array}{ll}1, & 2 \mid x, \\ 2, & 2 \nmid x .\end{array}\right.$
Case 1: If $2 \mid x$, then $\for... | x=0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,245 |
4. Let $x$, $y$, $z$ be positive real numbers, and $x+y+z \geqslant xyz$. Find the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{xyz}$.
(Feng Zhigang) | 4. Notice that, when $x=y=z=\sqrt{3}$, $x+y+z=x y z$, and $\frac{x^{2}+y^{2}+z^{2}}{x y z}=\sqrt{3}$.
Below we prove that the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.
In fact, we have
$x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}(x+y+z)^{2}$
$\geqslant\left\{\begin{array}{l}\frac{1}{3}(x y z)^{2} \g... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,246 |
Example 6 Given an integer array consisting of 121 integers, each integer in this array takes a value between 1 and 1000 (inclusive of 1 and 1000), and repeated values are allowed. The arithmetic mean of these numbers is $m$, and there is a unique "mode" (the number that appears most frequently) $M$ in this set of numb... | Obviously, to make $D$ as large as possible, the mode $M$ should be as large as possible. For this purpose, let $M=1000$, and the mean $m$ should be as small as possible, so the other numbers should be as small as possible, i.e., $1,2,3, \cdots$.
Thus, the key lies in the frequency of the mode $M$, i.e., how many time... | 947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,247 |
1. Find all real numbers $x$ such that $\left[x^{3}\right]=4 x+3$. Here $[y]$ denotes the greatest integer not exceeding the real number $y$.
（Yang Wenpeng） | 1. Let $x$ be a real number that satisfies the condition, then we can set $x=\frac{k}{4}$ $(k \in \mathbf{Z})$, and $\left[\frac{k^{3}}{64}\right]=k+3$. Therefore, $3 \leqslant \frac{k^{3}}{64}-k256$, which is a contradiction.
Hence, $k \in\{-7,-6, \cdots,-1,9\}$.
By calculating each, we find that the $k$ satisfying (1... | x=-\frac{5}{4} \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,248 |
2. $P$ is a point outside $\odot O$, and two tangents from $P$ to $\odot O$ touch the circle at points $A$ and $B$. Let $Q$ be the intersection of $PO$ and $AB$, and let $CD$ be any chord of $\odot O$ passing through $Q$. Prove: $\triangle PAB$ and $\triangle PCD$ have the same incenter.
(Liu Kangning) | 2. As shown in Figure 1, let $R$ be the intersection point of line segment $O P$ and $\odot O$, and $E$ be the intersection point of $P D$ and $\odot O$ (different from $D$).
$$
\begin{array}{l}
\because C Q \cdot Q D \\
=A Q \cdot Q B=A Q^{2}, \\
P Q \cdot Q O=A Q^{2}, \\
\therefore C Q \cdot Q D=P Q \cdot Q O .
\end{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,249 |
3. Find all real numbers $x \in\left[0, \frac{\pi}{2}\right]$, such that
$$
(2-\sin 2 x) \sin \left(x+\frac{\pi}{4}\right)=1 \text {, }
$$
and prove your conclusion.
(Li Shenghong) | 3. Let $\sin \left(x+\frac{\pi}{4}\right)=t$, i.e., $\sin x+\cos x=\sqrt{2} t$. Then,
$1+\sin 2 x=2 t^{2}$, i.e., $\sin 2 x=2 t^{2}-1$. Therefore, $t\left(3-2 t^{2}\right)=1$,
which simplifies to $2 t^{3}-3 t+1=0$.
Notice that $t=1$ is a solution to the above equation, hence $(t-1)\left(2 t^{2}+2 t-1\right)=0$.
Since... | x=\frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,250 |
4. We call $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j} \neq \varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 14-partition $A_{1}, A_{2}, \cdots, A_{14}$ of $A=\{1,2, \cdots, m\}$, ... | 4. (i) If $ma > b$, and $a - b \geqslant 14$. Therefore, $b \leqslant a - 141 + \frac{14}{42} = \frac{4}{3}$.
Hence, the positive integer $m \geqslant 56$.
(ii) If $m = 56$, then for any partition of $A$ into $A_{1}, A_{2}, \cdots, A_{14}$, among the numbers $42, 43, \cdots, 56$, there must be two numbers belonging to ... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,251 |
1. Let set $A=\left\{x^{2}, x+1,-3\right\}, B=\{x-5,2 x$ $\left.-1, x^{2}+1\right\}$, satisfying $A \cap B=\{-3\}$. Then the value of $x$ is ( ).
(A) 2
(B) 1
(C) 0
(D) -1 | - 1.(D).
Substitute $x=2,1,0,-1$ into sets $A$ and $B$, considering the uniqueness of set elements and check if $A \cap B=\{-3\}$, we find $x=-1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,252 |
4. The necessary and sufficient condition for the line $a x+b y+c=0(a, b, c \neq 0)$ to be symmetric to the line $p x+q y+m=0(p, q, m \neq 0)$ with respect to the $y$-axis is ( ).
(A) $\frac{b}{q}=\frac{c}{m}$
(B) $-\frac{a}{p}=\frac{b}{q}$
(C) $\frac{a}{p}=\frac{b}{q} \neq \frac{c}{m}$
(D) $-\frac{a}{p}=\frac{b}{q}=\f... | 4. (D).
The line symmetric to the line $a x+b y+c=0(a, b, c \neq 0)$ with respect to the $y$-axis is obtained by replacing $x$ with $-x$, resulting in $-a x+b y+c=0$. Note that the line $-a x+b y+c=0$ coincides with the line $p x+q y+m=0$ if and only if the necessary and sufficient condition is met. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,255 |
5. Given that $\left\{a_{n}\right\}$ is a geometric sequence, and $a_{n}>0, a_{2} a_{4}+2 a_{3} a_{5}$ $+a_{4} a_{6}=2025$. Then, the value of $a_{3}+a_{5}$ is ( ).
(A) 15
(B) 25
(C) 35
(D) 45 | 5.(D).
From $2025=a_{2} a_{4}+2 a_{3} a_{5}+a_{4} a_{6}=a_{3}^{2}+2 a_{3} a_{5}+a_{5}^{2}$ $=\left(a_{3}+a_{5}\right)^{2}$, we know the answer is (D). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,256 |
7. Using the digits $0,1,2,3,4$ to form a five-digit number without repeating digits, the number of five-digit numbers where exactly one even digit is sandwiched between two odd digits is ( ).
(A) 48
(B) 36
(C) 28
(D) 12 | 7. (C). From $\mathrm{C}_{3}^{1} \mathrm{P}_{2}^{2} \mathrm{P}_{3}^{3}-\mathrm{C}_{2}^{1} \mathrm{P}_{2}^{2} \mathrm{P}_{2}^{2}=28$ we get. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,259 |
8. The circle $\rho=D \cos \theta+E \sin \theta$ is tangent to the line of the polar axis if and only if ( ).
(A) $D \cdot E=0$
(B) $D \cdot E \neq 0$
(C) $D=0, E \neq 0$
(D) $D \neq 0, E=0$ | 8.(C).
From $\rho=D \cos \theta+E \sin \theta$, the rectangular coordinate equation is $\left(x-\frac{D}{2}\right)^{2}+\left(y-\frac{E}{2}\right)^{2}=\frac{1}{4}\left(D^{2}+E^{2}\right)$, or from $\rho=E \sin \theta(E \neq 0)$, it represents a circle tangent to the line of the polar axis. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,260 |
10. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the trisection point of $A_{1} A$, and $F$ is the trisection point of $C_{1} C$, with $A E=2 A_{1} E$ and $C F=2 C_{1} F$. A section is made through $B, E, F$ in the cube. Among the following projection diagrams of the section on the respective faces, which one ... | 10.(D).
When projected onto the $D_{1} D B B_{1}$ face, it should form a line segment. | D | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,261 |
11. The solution set of the equation $x^{2}|x|+|x|^{2}-x^{2}-|x|=0$ in the complex number set corresponds to the figure in the complex plane is ( ).
(A) Several points and a line
(B) Unit circle and a line
(C) Several lines
(D) Origin and unit circle | 11. (D).
The original equation can be transformed into $\left(x^{2}+|x|\right)(|x|-1)=0$, from which we know that (D) is correct. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,262 |
12 . Arrange the odd positive integers $1,3,5$, $7, \cdots$ into five columns, as shown in the table on the right. If the arrangement continues in the format of the table, the column in which 2001 is located, counting from the left, is ( ).
$\begin{array}{rrrrr} & 1 & 3 & 5 & 7 \\ 15 & 13 & 11 & 9 & \\ & 17 & 19 & 21 &... | 12.(B).
The numbers in the first column leave a remainder of 15 when divided by 16, the numbers in the second column leave a remainder of 1 or 13 when divided by 16, the numbers in the third column leave a remainder of 3 or 11 when divided by 16, the numbers in the fourth column leave a remainder of 5 or 9 when divide... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,263 |
13. The hyperbola that has the same asymptotes as $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ and passes through the point $(-3,2 \sqrt{3})$ is $\qquad$. | $$
=.13 .1 .
$$
Let $\frac{x^{2}}{9}-\frac{y^{2}}{16}=\lambda$, substituting $(-3,2 \sqrt{3})$ yields $\lambda=\frac{1}{4}$. | \frac{x^{2}}{9}-\frac{y^{2}}{16}=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,264 |
14. Given that $A B C-A_{1} B_{1} C_{1}$ is a regular triangular prism, $A B=B C$ $=C A=2, A A_{1}=\sqrt{2}, D$ and $E$ are the midpoints of $A C$ and $B C$ respectively. Then the angle formed by $A_{1} D$ and $C_{1} E$ is $\qquad$ . | $14.60^{\circ}$.
Take the midpoint $F$ of $A_{1} B_{1}$, then $D E \Perp \frac{1}{2} A B \Perp A_{1} F$. Therefore, $E F \mathbb{\mathbb { I }} D A_{1}$.
$\because E F=A_{1} D=C_{1} E=\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}$,
and $C_{1} F=\frac{\sqrt{3}}{2} A_{1} B_{1}=\sqrt{3}$,
$\therefore \triangle C_{1} E F_{1}$ is an... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,265 |
15. Let $x, y \in \mathbf{R}$ and satisfy $x^{2}+4 y^{2}=4$. Then the maximum and minimum values of $x^{2}+2 x y$ $+4 y^{2}$ are $\qquad$. | 15.6,2.
Solution 1: Let $x=2 \cos \theta, y=\sin \theta, 0 \leqslant \theta<2 \pi$.
By $x^{2}+4 y^{2}=4$, then $x^{2}+2 x y+4 y^{2}=4+4 \sin \theta \cdot \cos \theta=4+2 \sin 2 \theta$. When $\sin 2 \theta=1$, $x^{2}+2 x y+4 y^{2}$ reaches its maximum value 6; when $\sin 2 \theta=-1$, $x^{2}+2 x y+4 y^{2}$ reaches its... | 6,2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,266 |
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero. Then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to | 16. -3 .
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right) \\
& +\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c} \\
= & \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3=-3 .
\end{aligned}
$$ | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,267 |
2. Divide 1996 into the sum of 19 positive integers, prove that the maximum value of their product is $105^{18} \times 106$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (Hint: Let $a_{1}+a_{2}+\cdots+a_{14}=1996, a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{19}$.
If $a_{19}>a_{1}+1$, then let $a_{19}^{\prime}=a_{19}-1, a_{1}^{\prime}=a_{1}+1$. Such an adjustment keeps the sum of these 19 numbers unchanged. However, because
$$
\begin{array}{l}
a_{1}^{\prime} a_{19}^{\prime}=\le... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 713,268 |
17. Calculate $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\cdots$ $+\frac{2001}{1999!+2000!+2001!}$
The value is $\qquad$ | 17. $\frac{1}{2}-\frac{1}{2001!}$.
From $k!+(k+1)!+(k+2)!=k!(k+2)^{2}$, we have
$$
\begin{array}{l}
\frac{k+2}{k!+(k+1)!+(k+2)!} \\
=\frac{1}{k!(k+2)}=\frac{1}{(k+1)!}-\frac{1}{(k+2)!} \text {. } \\
\end{array}
$$
Therefore, the original expression $=\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{... | \frac{1}{2}-\frac{1}{2001!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,269 |
18. Given the quadratic function $f(x)=4 x^{2}-4 a x+\left(a^{2}-2 a\right.$ +2 ) has a minimum value of 2 on $0 \leqslant x \leqslant 1$. Find the value of $a$. | $$
\text { Three, 18. } f(x)=4\left(x-\frac{a}{2}\right)^{2}-2 a+2 \text {. }
$$
When $0 \leqslant \frac{a}{2} \leqslant 1$ i.e., $0 \leqslant a \leqslant 2$, $f_{\text {min}}(x)=-2 a+2=2$. This gives $a=0$. When $\frac{a}{2}>1$, i.e., $a>2$,
$$
f_{12}(x)=f(1)=4-4 a+a^{2}-2 a+2=2 \text {. }
$$
We can find $a=3-\sqrt{... | a=0 \text{ or } a=3+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,270 |
19. Let the sequence $\left\{a_{n}\right\}$ with at least four terms have the sum of the first $n$ terms $S_{n}=n p a_{n}\left(n \in \mathbf{N}^{+}, p\right.$ is a constant $)$, and $a_{1} \neq a_{2}$. What kind of sequence is $\left\{a_{n}\right\}$? Explain your reasoning. | 19. When $n=1$, $a_{1}=p a_{1} \Rightarrow a_{1}=0$ or $p=1$.
When $p=1$, we have $a_{1}+a_{2}=2 a_{2} \Rightarrow a_{1}=a_{2}$. This contradicts the given $a_{1} \neq a_{2}$, hence $p \neq 1$.
When $a_{1}=0$, then $a_{2} \neq 0$, from $a_{2}=2 p a_{2}$ we get $p=\frac{1}{2}$.
From $a_{1}+a_{2}+a_{3}=\frac{3}{2} a_{3}... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,271 |
20. As shown in Figure 1, given that the base of the pyramid $P-ABCD$ is a square with side length 4, and $PD \perp$ the base $ABCD$. Let $PD=6$, and $M$, $N$ be the midpoints of $PB$ and $AB$ respectively.
(1) Find the volume of the tetrahedron $P-DMN$;
(2) Find the tangent value of the dihedral angle $M-DN-C$. | 20. As shown in Figure 2, let $A C$ and $B D$ intersect at $O$, and connect $M O$ and $P N$.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
M O \perp \text { base } A B C D, \\
M O=\frac{1}{2} P D=3 .
\end{array}\right. \\
\end{array}
$$
(1) Since $N$ is the midpoint of $A B$, $\therefore S_{\triangle O V B}=\f... | \frac{3 \sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,272 |
21. Now buy soda in batches for $a$ people to drink. According to the merchant's rules, every $b\left(a>b>1, a, b \in \mathbf{N}^{+}\right)$ empty bottles can be exchanged for one bottle of soda, so it is not necessary to buy $a$ bottles of soda. How many bottles of soda must be bought at least to ensure that each of t... | 21. Assume the minimum number of bottles to buy is $x$.
After drinking, the number of bottles exchanged for the first time does not exceed $\frac{x}{b}$. After drinking again, the number of bottles exchanged for the second time does not exceed $\frac{x}{b^{2}}$. This continues, and when $k$ is sufficiently large, $\fr... | a-\left[\frac{a}{b}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,273 |
22. For a rhombus $A_{1} B_{1} C D$ with side length 1, the diagonals intersect at $A_{2}$. Draw $A_{2} B_{2} / / A_{1} B_{1}$ intersecting $B_{1} C$ at $B_{2}$, connect $B_{2} D$ intersecting $A_{1} C$ at $A_{3}$, and draw $A_{3} B_{3} / / A_{1} B_{1}$ intersecting $B_{1} C$ at $B_{3}, \cdots \cdots$, continuing this ... | 22. As shown in Figure 3, from the given conditions we know $A_{1} B_{1}=1, A_{2} B_{2}=\frac{1}{2}$.
Guess: $A_{n} B_{n}=\frac{1}{n}$.
Prove by mathematical induction.
When $n=1$, it is obviously true.
Assume that when $n=k$, we have $A_{k} B_{k}=\frac{1}{k}$. By the similarity of triangles, we have $\frac{A_{k+1} B_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,274 |
1. Question: How many real roots does the equation $x^{2}|x|-5 x|x|+2 x=0$ have (where $|x|$ represents the absolute value of $x$)? | 1.4 .
The original equation simplifies to $x(x|x|-5|x|+2)=0$. After removing the absolute value signs and discussing, we get
$$
x_{1}=0, x_{2}=\frac{5+\sqrt{17}}{2}, x_{3}=\frac{5-\sqrt{17}}{2}, x_{4}=\frac{5-\sqrt{33}}{2}
$$
as the four real roots of the original equation. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,275 |
3. Team A and Team B each send out 5 players to participate in a chess broadcast tournament according to a pre-arranged order. The two teams first have their No. 1 players compete; the loser is eliminated, and the winner then competes with the No. 2 player of the losing team, …, until all players of one side are elimin... | 3.252 .
Due to the tournament rules, the first team to achieve 5 victories is declared the winner, even if some players have not participated. However, we can consider the players who did not participate as having lost. Thus, we get a complete ten-game result for one side: five wins and five losses. A match process, i... | 252 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,277 |
4. Let $x_{1}=\sqrt[3]{3}, x_{2}=\left(x_{1}\right)^{\sqrt[3]{3}}$, for $n>1$ define $x_{n+1}$ $=\left(x_{n}\right)^{\sqrt[3]{3}}$. Find the smallest positive integer $n$ such that $x_{n}=27$. | 4.7.
It is known that $x_{n}=x_{1}^{x_{1}^{n-1}}(n>1)$, and $27=(\sqrt[3]{3})^{(\sqrt[3]{3})^{6}}$. Therefore, $n-1=6$. Hence, $n=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,278 |
3. A math competition has a total of 15 questions. The table below shows the number of people who got $n$ $(n=0,1,2, \cdots, 15)$ questions correct:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people who got $n$ questions correct & 7 & 8 & 10 & 21 & $\cdot... | (Tip: The number of people who got at most 3 questions correct is $7+8+10+21$ $=46$ people.
The number of people who got at least 12 questions correct is $15+6+3+1=25$ people.
The total number of questions correctly answered by those who got at most 3 questions correct is 91 questions,
The total number of questions cor... | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,279 |
5. Find all positive integers $n$ such that the sum $1+2+3+\cdots+n$ is a three-digit number composed of the same digit. | 5. $n=36$.
$1+2+\cdots+n=\frac{n(n+1)}{2}$, which means solving the equation $\frac{n(n+1)}{2} = 111,222, \cdots, 999$, or equivalently $n(n+1)=2 \times 3 \times 37 \times k, k = 1,2, \cdots, 9$. It can be seen that $n=36, k=6$ is the only integer solution. | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,280 |
$6 . A 、 B 、 C 、 D$ Four football teams are playing a round-robin tournament (i.e., any two teams will play against each other). After several matches, the match situations of teams $A 、 B 、 C$ are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & Matches & Wins & Losses & Draws & Goals Scored & Goals Conceded \\
\... | 6. Since Team C has played 3 matches, we start with Team C. Team C won 2 matches and drew 1, so the match results of Team C against other teams can only be 1:0 (win), 1:0 (win), and 0:0 (draw). Since Team B drew 1 match, the match result between Team C and Team B is 0:0, and the match results between Team C and Team A,... | 3, 1, 2, 8, 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,281 |
8 . Write $1,2, 3, 4, 5, 6,7,8,9, 10, 11, 12, 13, 14, 15, 16$ in a $4 \times 4$ table, one number per cell. Add up the four numbers in each (row) and each (column), requiring each sum to be the same. Try to write the numbers in each cell: | 8. The sum of four rows $=1+2+\cdots+16$
$=136$, so the sum of each row $=\frac{136}{4}=34$. Each row of 4 numbers can be divided into two pairs, making the sum of each pair equal to 17, and ensuring that the sum of each column of 4 numbers also equals 34, then we have $17=16+1=15+2$
$$
\begin{array}{l}
=14+3=13+4=12+5... | \begin{array}{cccc}
1 & 12 & 8 & 13 \\
14 & 7 & 11 & 2 \\
15 & 6 & 9 & 4 \\
16 & 5 & 3 & 10 \\
\end{array} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,283 |
9. In $\triangle A B C$, it is known that $\tan A=\frac{5}{12}$, and the foot of the perpendicular from $A$ to $B C$ divides the segment $B C$ into two parts, with lengths 3 and 17, respectively. Try to find the perimeter of $\triangle A B C$. | $$
9.20+\sqrt{1212+48 \sqrt{627}}+\sqrt{1492+48 \sqrt{627}} .
$$
As shown in Figure 4, let $O$ be the circumcenter of $\triangle ABC$. Draw $AK \perp BC, OH \perp BC$, with the feet of the perpendiculars being $K$ and $H$ respectively. Then draw $OM \perp AK$, with $M$ being the foot of the perpendicular, and connect ... | 20+\sqrt{1212+48 \sqrt{627}}+\sqrt{1492+48 \sqrt{627}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,284 |
10. Question: Among $1,2,3, \cdots, 1999,2000,2001$, what is the maximum number of numbers that can be chosen such that the sum of any three chosen numbers is divisible by 21? | 10. Take at most 95.
Let $a, b, c, d$ be 4 of the numbers taken. According to the problem, $a+b+c$ and $b+c+d$ are both multiples of 21. Therefore, the difference $a - d$ is also a multiple of 21. By the arbitrariness of $a$ and $d$, it can be concluded that the difference between any two of the numbers taken is a mul... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,285 |
11. To place three squares with side lengths of $1 \mathrm{~cm}$ inside a circular dish. The squares must not have any part outside the dish and must not overlap: what is the minimum radius of the circular dish?
To place three squares with side lengths of $1 \mathrm{~cm}$ inside a circular dish. The squares must not h... | 11. $\frac{5 \sqrt{17}}{16}$.
The placement of three squares can be considered in the following scenarios:
First scenario: Placed side by side to form a $3 \mathrm{~cm} \times 1 \mathrm{~cm}$ rectangle, as shown in Figure 5. The diameter of the circular disc must be at least the diagonal of the rectangle, which is $\s... | \frac{5 \sqrt{17}}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,286 |
12. Given that the radius of the large circle in Figure 1 is $R$, and the three smaller circles inside the large circle are pairwise tangent to each other and to the large circle. Their radii are $2r$, $r$, and $r$. Try to find $\frac{r}{R}$. | 12. $\frac{4 \sqrt{2}-5}{2}$.
It is easy to prove that $O$ and $B$ lie on the common tangent of $\odot A$ and $\odot A^{\prime}$, so we only need to consider half of the figure. (See Figure 9 and Figure 10)
Let $x=M O, y=M B$. Then we have
$$
M N=x+R=y+2 r,
$$
which gives $x=y-R+2 r$.
Also, $A O=O P-A P=R-r, A B=A K+... | \frac{4 \sqrt{2}-5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,287 |
1. Let $A_{1}$ be the center of the inscribed square of the acute triangle $\triangle ABC$, where two vertices of the inscribed square lie on side $BC$, one vertex on side $AB$, and one vertex on side $AC$. Similarly, define the centers of the inscribed squares with two vertices on sides $AC$ and $AB$ as $B_{1}$ and $C... | Proof: As shown in Figure 5, let the line $A A_{1}$ intersect $B C$ at $X$, the line $B B_{1}$ intersect $C A$ at $Y$, and $C C_{1}$ intersect $A B$ at $Z$. By the converse of Ceva's theorem, it suffices to prove that
$$
\frac{B X}{X C} \cdot \frac{C Y}{Y A} \cdot \frac{A Z}{Z B}=1.
$$
Let the side length of the squar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,288 |
3. Let $G$ be the centroid of $\triangle ABC$. Determine the position of point $P$ in the plane of $\triangle ABC$ such that $AP \cdot AG + BP \cdot BG + CP \cdot CG$ is minimized, and express this minimum value in terms of the side lengths of $\triangle ABC$.
Translate the above text into English, please retain the o... | Let $a, b, c$ be the lengths of the sides opposite to vertices $A, B, C$ of $\triangle ABC$, respectively. We will prove that
$$
A P \cdot A G + B P \cdot B G + C P \cdot C G
$$
attains its minimum value when $P$ is the centroid $G$, and the minimum value is
$$
\begin{array}{l}
A G^{2} + B G^{2} + C G^{2} \\
=\frac{1}... | \frac{1}{3}(a^2 + b^2 + c^2) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,289 |
Example 1 Find the smallest positive number $\alpha$, such that there exists a positive number $\beta$, for which the inequality
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{\alpha}}{\beta}
$$
holds for $0 \leqslant x \leqslant 1$. | For any $x \in [0,1]$, there is an identity
$$
\begin{array}{l}
(\sqrt{1+x}+\sqrt{1-x}-2)(\sqrt{1+x}+ \\
\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right)=-2 x^{2}
\end{array}
$$
Since on the closed interval
$$
\begin{aligned}
0 & 0$, the following similar inequality holds:
$$
\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{\alph... | \alpha=2, \beta=4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,290 |
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