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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 13 Given $a+b+c=0, a^{3}+b^{3}+c^{3}$ $=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | Solution: From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$ - $\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)$, we have $a b c=0$. Therefore, at least one of $a, b, c$ is 0.
Assume $c=0$, then $a, b$ are opposites,
$$
\therefore a^{15}+b^{15}+c^{15}=0 \text {. }
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,271 |
Example 14 If $x^{3}+x^{2}+x+1=0$, then the value of $y=$ $x^{97}+x^{98}+\cdots+x^{103}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2 | Given: $\because x^{3}+x^{2}+x+1=0$, then from
$$
x^{4}-1=(x-1)\left(1+x+x^{2}+x^{3}\right)
$$
we get $x^{4}=1$.
$$
\begin{aligned}
\therefore \text { the original expression }= & x^{97}\left(1+x+x^{2}+x^{3}\right) \\
& +x^{100}\left(x+x^{2}+x^{3}\right) \\
= & x^{97} \times 0+\left(x^{4}\right)^{25} \times(-1) \\
= &... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,272 |
Example: 15 Given positive integers $x, y, z$ satisfy $x^{3}-y^{3}-z^{3}=3 x y z, x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original text's line breaks and format. | $$
\begin{array}{l}
\text { Solution: } x^{3}-y^{3}-z^{3}-3 x y z \\
=x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
=(x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z\right. \\
\left.+z^{2}\right)-3 y z(x-y-z) \\
=(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
=\frac{1}{2}(x-y-z)\left[(x+y)^{2}+(y-z)^{2}\right. \\
\left.+(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,273 |
Example 1 Let $A_{1} A_{2} A_{3} A_{4}$ be a cyclic quadrilateral inscribed in $\odot O$, and let $H_{1}, H_{2}, H_{3}, H_{4}$ be the orthocenters of $\triangle A_{2} A_{3} A_{4}$, $\triangle A_{3} A_{4} A_{1}$, $\triangle A_{4} A_{1} A_{2}$, and $\triangle A_{1} A_{2} A_{3}$, respectively. Prove that $H_{1}, H_{2}, H_... | Solution: Establish a complex plane with the center $O$ as the origin. Let the complex numbers corresponding to points $A_{1}, A_{2}, A_{3}, A_{4}$ be $z_{1}, z_{2}, z_{3}, z_{4}$, respectively, then
$$
\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right| .
$$
By the formula for the orthocenter ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,274 |
Example 2 Find the maximum value of $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$ $-\sqrt{x^{4}-x^{2}+1}$.
(1992, National High School Competition)
Analysis: From $f(x)=\sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}}$ $-\sqrt{x^{2}+\left(x^{2}-1\right)^{2}}$, let $Z_{1}\left(x-3, x^{2}-2\right)$, $Z_{2}\left(x, x^{2}-1\right)$, then $f... | Let $z_{1}=(x-3)+\left(x^{2}-2\right) \mathrm{i}, z_{2}=x+\left(x^{2}-1\right) \mathrm{i}$. From $\left|z_{1}\right|-\left|z_{2}\right| \leqslant\left|z_{1}-z_{2}\right|$, we have
$$
\begin{aligned}
f(x)= & \left|(x-3)+\left(x^{2}-2\right) \mathrm{i}\right| \\
& -\left|x+\left(x^{2}-1\right) \mathrm{i}\right| \\
& \leq... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,275 |
Example $6 \triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ are two isosceles right triangles with legs both equal to $2 a$. As shown in Figure 6, they are stacked together, with $\triangle A B C$ fixed in position, and the midpoints of the legs $A C$ and $B C$ being $M$ and $N$, respectively. Keeping ... | Solution: In figure $\mathrm{o}$, it is easy to prove that quadrilateral $A E B^{\prime} M$ is a parallelogram.
Then $B^{\prime} E=A M=\frac{1}{2} A C=a$.
$$
\therefore C^{\prime} E=2 a-a=a \text {. }
$$
Therefore, $E$ is the midpoint of $B^{\prime} C^{\prime}$.
Similarly, $D$ is the midpoint of $A^{\prime} C$.
$$
\th... | \frac{5}{4} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,276 |
Example 4 Proof: An octagon with all sides of rational length and all interior angles equal is centrally symmetric.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: Let the 8 vertices of the octagon be $A_{1}$, $A_{2}, \cdots, A_{8}$ in sequence, and the lengths of the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $\cdots$, $A_{7} A_{8}$, $A_{8} A_{1}$ be denoted as $a_{1}$, $a_{2}$, $\cdots$, $a_{7}$, $a_{8}$ respectively. Establish a complex plane with $A_{1}$ as the origin and the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,278 |
Let 5 be a point inside an acute triangle $\triangle ABC$ such that $\angle ADB = \angle ACB + 90^{\circ}$, and $AC \cdot BD = AD \cdot BC$. Find the value of $\frac{AB \cdot CD}{AC \cdot BD}$. | Solution: Taking vertex $C$ as the origin and $CA$ as the real axis to establish a complex plane, let $|\overrightarrow{C A}|=r$. According to the geometric meaning of complex number multiplication and division, vector $\overrightarrow{D B}$ can be seen as $\overrightarrow{D A}$ after rotation and scaling. From the giv... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,279 |
Question: Find the number of five-digit numbers formed by the digits $1, 2, 3, 4, 5, 6$ such that at least three digits are different, and $1, 6$ are not adjacent. | Solution 1: Let $A$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$ with at least three different digits, $B$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$ with no $1, 6$ adjacent, and $R$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$. By the set theory formula, we have
$$
... | 5880 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,280 |
3. In $\angle A$, there is a fixed point $P$, draw a line through $P$ intersecting the two sides at $B$ and $C$. When does $\frac{1}{P B}+\frac{1}{P C}$ reach its maximum value? | $$
\begin{array}{l}
\frac{1}{P B}+\frac{1}{P C}=\frac{1}{\rho_{B}}+\frac{1}{\rho_{C}} \\
=\frac{\sin (\theta+\alpha)}{\rho_{0} \sin \alpha} \\
+\frac{\sin (\theta-\beta)}{\rho_{0} \sin \beta} \\
=\frac{\sin (\theta+\alpha) \sin \beta+\sin (\theta-\alpha) \sin \alpha}{\rho_{0} \sin \alpha \cdot \sin \beta} \\
=\frac{\si... | \frac{1}{P B}+\frac{1}{P C} \text{ reaches its maximum value when } \theta=90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,281 |
4. In the right triangle $\triangle A B C$, $A D$ is the altitude on the hypotenuse $B C$. Connecting the incenter of $\triangle A B D$ with the incenter of $\triangle A C D$, the line intersects sides $A B$ and $A C$ at points $K$ and $L$ respectively. The areas of $\triangle A B C$ and $\triangle A K L$ are $S$ and $... | - Prompt: Let the inradii of $\triangle A B D$ and $\triangle A C D$ be $r_{1}$ and $r_{2}$, $A D = h$, and $\angle A B C = \alpha$. Then the coordinates of $M$ and $N$ are
$$
\begin{array}{l}
\left(h-r_{1}, r_{1}\right),\left(r_{2}, h-r_{2}\right) . \\
\therefore k_{M N}=\frac{h-r_{2}-r_{1}}{r_{2}-h+r_{1}}=-1 .
\end{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,282 |
$5 . D$ is the midpoint of $A G$, and on the same side of $A G$, construct congruent quadrilaterals $A B C D$ and $D E F G$, both having incircles with centers $O$ and $I$ respectively. Prove: $A O$, $C E$, and $G I$ are concurrent. | (Given: The equation of $\odot O$ is: $x^{2}+y^{2}=1$. Since $G I / / O D$, therefore $O$ is the midpoint of $A K$. Hence
$$
K\left(-x_{A},-y_{A}\right) \text {. }
$$
Also, since $A B E D$ is a parallelogram,
$$
\therefore x_{E}=x_{B}+x_{D}-x_{A}, y_{E}=y_{B}+y_{D}-y_{A} \text {. }
$$
Therefore, the equation of $C E$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,283 |
Let $a, b, c$ be the three sides of $\triangle ABC$, with $a \leqslant b \leqslant c$, and $R$ and $r$ be the circumradius and inradius of $\triangle ABC$, respectively. Let $f=a+b-2R-2r$. Try to determine the sign of $f$ using the size of angle $C$.
(Xu Yichao, problem contributor) | Let $A, B, C$ represent the three interior angles of $\triangle ABC$:
$$
\begin{array}{l}
a=2 R \sin A, b=2 R \sin B, \\
r=4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} .
\end{array}
$$
Thus,
$$
\begin{aligned}
f= & 2 R\left(\sin A+\sin B-1-4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\righ... | f>0 \Leftrightarrow C<\frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,284 |
II. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=\hat{j}, a_{2}=1, a_{n}=$ $\frac{1}{2} n a_{n-1}+\frac{1}{2} n(n-1) a_{n-2}+(-1)^{n}\left(1-\frac{n}{2}\right), n \geqslant$ 3. Try to find the simplest expression for $i_{n}=a_{n}+2 C_{n}^{1} a_{n-1}+3 C_{n}^{2} a_{n-2}+\cdots+(n-1)$. $\mathrm{C}_{n... | Solution 1: Let $b_{n}=\frac{a_{n}}{n!}$, then
$b_{1}=0, b_{2}=\frac{1}{2}, b_{3}=\frac{1}{3}$,
$b_{n}=\frac{1}{2}\left(b_{n-1}+b_{n-2}\right)+(-1) \cdot \frac{1-\frac{n}{2}}{n!}$
Let $g_{n}=\frac{1}{n} f_{n}=\frac{1}{n} \sum_{k=1}^{n}(n-k+1) G_{4}^{-a_{t}}$
$=\sum_{k=1}^{n} \frac{n-k+1}{(n-k)!} \cdot b_{k}$.
$\therefo... | 2 \cdot n! - (n + 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,285 |
Three, a table tennis club organizes an exchange activity, arranging a doubles schedule that meets the following rules. The rules are:
(i) each participant belongs to at most two pairs;
(ii) any two different pairs play against each other at most once;
(iii) any two people in the same pair do not meet as opponents in a... | Solution: (I) Let $a_{1}<a_{2}<\cdots<a_{k}$, and let the number of matches a certain participant plays in a scheduled tournament be $a_{k}$.
If this participant only pairs with one partner, then there are $a_{k}$ pairs playing doubles with this participant, thus there are at least $a_{k}$ other people. The total numb... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,286 |
Let $n \geqslant 2$. For an $n$-tuple of ordered real numbers
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), \\
\text { let } \quad b_{k}=\max _{1 \leqslant i \leqslant k}, k=1,2, \cdots, n .
\end{array}
$$
The array $B=\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ is called the "innovation array" of $A$;... | Solution 1: Let $A=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be a permutation of $1,2, \cdots, n$. Then the innovation order of $A$ is 2 if and only if its innovation array contains exactly two different elements $a_{1}$ and $n$. It is easy to see that $a_{1}$ can be any one of $1,2, \cdots, n-1$.
When $a_{1}=m$, the ... | n-\frac{n-1}{1+\frac{1}{2}+\cdots+\frac{1}{n-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,288 |
Six, in a certain exam, there are 5 multiple-choice questions, each with 4 different answers to choose from. Each person selects exactly 1 answer for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where each pair of sheets has at most 3... | Solution: The minimum possible value of $n$ is 25.
Let the 4 possible answers to each question be denoted as $1,2,3,4$. The answers on each test paper are denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let
$$
\begin{array}{l}
\{(1, h, i, j, k),(2, h, i, j, k),(3, h, i, j, k), \\
(4, h, i, j, k)\},... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,290 |
For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$. Find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$. | Solution: Use the adjustment method to explore the maximum value of the sum in the problem.
(i) First, for $x, y > 0$, we compare $(x+y)^{4} - (x+y)^{5} + 0^{4} - 0^{5}$ with $x^{4} - x^{5} + y^{4} - y^{5}$:
$$
\begin{array}{l}
(x+y)^{4} - (x+y)^{5} + 0^{4} - 0^{5} - \left(x^{4} - x^{5} + y^{4} - y^{5}\right) \\
= x y ... | \frac{1}{12} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,291 |
Sure, here is the translated text:
```
II. Try to find all prime numbers $p$ that satisfy the following conditions:
For any prime number $q<p$, if $p=b_{q}+r, 0 \leqslant r<\dot{q}$, then there does not exist an integer $a$ greater than 1 such that $a^{i}$ divides $r$.
``` | Solution: It is easy to verify that $p=2,3,5,7$ all satisfy the conditions. Now let's discuss the prime $p \geqslant 11$. If $p$ satisfies the conditions, then we have:
(i) $p-4$ has no prime factors greater than 4;
(ii) $p-8$ has no prime factors greater than 8;
(iii) $p-9$ has no prime factors greater than 9.
From (i... | 2,3,5,7,13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,292 |
Three, let $S=\{1,2, \cdots, 15\}$. From $S$, take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ that satisfy the following conditions:
(i) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(ii) $\left|A_{i} \cap A_{j}\right| \leqslant 3,1 \leqslant i<j \leqslant n$;
(iii) For any three-element subset $M$ of $S$, there exists some... | Let $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be any family of sets that satisfies the conditions of the problem. For any $a \in S$, let the number of sets in the family $\mathscr{A}$ that contain $a$ be denoted by $r(a)$. These $r(a)$ sets each contain $C_{6}^{2}:=15$ three-element subsets that include ... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,293 |
Four, a circle is tangent to sides $AB$ and $BC$ of a convex quadrilateral $ABCD$ at points $G$ and $H$, respectively, and intersects diagonal $AC$ at points $E$ and $F$. What necessary and sufficient condition must $ABCD$ satisfy so that there exists another circle passing through $E$ and $F$, and tangent to the exten... | The required necessary and sufficient condition is
$$
A B+A D=C B+C D \text {. }
$$
(i) Proof of necessity:
Suppose the circles passing through points $E$ and $F$ are tangent to the extensions of $D A$ and $D C$ at points $J$ and $K$, respectively. Then we have
$$
\begin{aligned}
A B+A D & =B G+G A+A D \\
& =B G+J A+A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,294 |
Given a positive integer $m \geqslant 2$. Prove:
(1) There exist integers $x_{1}, x_{2}, \cdots, x_{2 m}$ such that
$$
x_{i} x_{m+i}=x_{i+1} x_{m+i-1}+1,1 \leqslant i \leqslant m \text {. }
$$
(2) For any integer sequence $x_{1}, x_{2}, \cdots, x_{2 m}$ satisfying condition (*), it is possible to construct an integer s... | Proof: (1) Taking \( x_{1}=\cdots=x_{m}=1 \), from
\[
x_{i} x_{m+i}=x_{i+1} x_{m+i-1}+1, \quad 1 \leqslant i \leqslant m,
\]
we have
\[
\begin{array}{l}
x_{m+i}=x_{m+i-1}+1, \quad 1 \leqslant i \leqslant m-1, \\
x_{2 m}=x_{m+1} x_{2 m-1}+1.
\end{array}
\]
Thus, \( x_{m+1}=x_{m}+1=2 \).
We have
\[
\begin{array}{l}
x_{k+... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,295 |
5. Originally, $m$ students planned to engage in a letter-writing activity during the summer, with each person having to write a letter to the other $(m-1)$ students. Later, $n$ more students became interested in the activity. Given that $n>1$, and due to the addition of $n$ students, 74 more letters were written, then... | 5.
18 | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,302 |
6. Given that $a$ and $b$ are real numbers, one root of the equation $x^{2}=a x+b$ is 6, and the absolute value of the other root is less than 2. Then the locus of the vertex of the parabola $y=-x^{2}+a x+b$ is $\qquad$ | 6. $y=x^{2}-12x+36(2<x<4)$ | y=x^{2}-12x+36(2<x<4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,303 |
II. The side lengths $a, b, c (a \leqslant b \leqslant c)$ of $\triangle A B C$ simultaneously satisfy the following three conditions:
(1) $a, b, c$ are all integers;
(2) $a, b, c$ form a geometric sequence;
(3) At least one of $a$ and $c$ equals 100.
Find all possible solutions for the triplet $(a, b, c)$. | (ii) If $a=100$, then $b^{2}=100 c, \therefore 10 \mid b$.
Also, $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}<0, b^{2}+100 b-100^{2}>0$, thus
$50(\sqrt{5}-1)<b \leqslant 100$.
Noticing that $10 \mid b$, then $b$ can take $70,80,90,100$. The corresponding $c=49,64,81,100$.
In summary, the triplets $(a, b, c)$ have 1... | (49,70,100),(64,80,100),(81,90,100),(100,100,100),(100,110,121),(100,120,144),(100,130,169),(100,140,196),(100,150 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,308 |
Example 9 Does there exist a point $P$ in $\triangle ABC$ such that any straight line passing through $P$ will divide the $\triangle ABC$ into two parts of equal area? Why? | Solution: As shown in Figure 9, assume that point $P$ satisfying the conditions exists. Connect $A P$ and extend it to intersect $B C$ at $D$, and connect $B P$ and extend it to intersect $A C$ at $E$.
Then $S_{\triangle A B P}=S_{\triangle A C D}$.
Therefore, $B D=C D$.
Similarly, $A E=C E$.
Draw $G H / / B C$ through... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,309 |
Three, let $a, b, c, d$ be four distinct real numbers such that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=b d$.
Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$. | Three, let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we know $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$. Thus, the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
... | -12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,310 |
For any set of $n$ points in the plane, if the distance between any two points is determined, then this point set is called "stable". Prove that in a set of $n$ $(n \geqslant 4)$ points in the plane, $P$, with no three points collinear, and where $\frac{1}{2} n(n-3)+4$ distances between pairs of points are determined, ... | Sure, here is the translation:
---
Four, we first agree that two points are connected by an edge if the distance between them is determined. We will use mathematical induction to prove this problem.
When $n=4$, $\frac{1}{2} n(n-3)+4=6$, there are only $\mathrm{C}_{4}^{2}=6$ distances between the four points, all of w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,311 |
1. Let $a, b, c$ be real numbers, $abc \neq 0$ and $a+b=$ $c$. Then the value of $\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+$ $\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ is ( ).
(A) -1
(B) 1
(C) 2
(D) 3 | $-1 .(\mathrm{B})$.
$$
\begin{aligned}
\text { Original expression }= & \frac{(b-c)^{2}+2 b c-a^{2}}{2 b c} \\
& +\frac{(c-a)^{2}+2 c a-b^{2}}{2 c a} \\
& +\frac{(a+b)^{2}-2 a b-c^{2}}{2 a b} \\
= & \frac{2 b c}{2 b c}+\frac{2 c a}{2 c a}-\frac{2 a b}{2 a b}=1 .
\end{aligned}
$$
Another solution: Take $a=b=1, c=2$ and... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,312 |
2. Let $a b \neq 0$, in the same coordinate system, the graphs of the functions $y=a x^{2}+b$ and $y=\frac{a b}{|x|}$ can only be ( ).
| 2. (A).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,313 |
3. In $\triangle A B C$, $B C=3$, the inradius $r$ $=\frac{\sqrt{3}}{2}$. Then the value of $\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{2}{\sqrt{3}}$
(C) $\frac{3 \sqrt{3}}{2}$
(D) $2 \sqrt{3}$ | 3. (D).
As shown in the right figure, the incenter $I$ of the triangle is the intersection of the three angle bisectors, so $\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}$ $=\frac{B D}{I D}+\frac{C D}{I D}=\frac{B C}{r}=2 \sqrt{3}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,314 |
4. Given real numbers $\alpha, \beta$ satisfy $\alpha^{2}+3 \alpha-1=0, \beta^{2}$ $-3 \beta-1=0$, and $\alpha \beta \neq 1$. Then the value of $\alpha^{-2}+3 \beta$ is ( ).
(A) 1
(B) 3
(C) -3
(D) 10 | 4. (D).
From $\alpha^{2}+3 \alpha-1=0$ we know $\alpha \neq 0$ and $\alpha^{-2}-3 \alpha^{-1}-1=0$, meaning $\alpha^{-1}$ and $\beta$ are the two distinct roots of the equation $x^{2}-3 x-1=0$ (since $\alpha \beta \neq 1$), hence
$$
\alpha^{-2}+3 \beta=3 \alpha^{-1}+1+3 \beta=3\left(\alpha^{-1}+\beta\right)+1=10 .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,315 |
5. Given that $M$ and $N$ are two distinct points on a plane, there are $m$ lines passing through $M$ but not through $N$ (called $M$-type lines), and $n$ lines passing through $N$ but not through $M$ (called $N$-type lines). If each $M$-type line intersects with each $N$-type line, and each line is divided into severa... | 5. (C).
On each line of class $M$, there are $n$ intersection points, these $n$ points along with point $M$ divide the line into $n+2$ segments; similarly, each line of class $N$ is divided into $m+2$ segments by its $m$ intersection points along with point $N$, so the total number of segments is $m(n+2)+n(m+2)=2(m n+... | 2(m n+m+n) | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 711,316 |
6. In $\triangle A B C$, $\angle A=30^{\circ}, \angle C=90^{\circ}$, with $C$ as the center and $C B$ as the radius, a circle is drawn intersecting side $A B$ at $M$ and side $A C$ at $N, C M$ intersects $B N$ at point $P$. If $A N=1$, then $S_{\triangle C P N}-S_{\triangle B P M}=(\quad)$.
(A) $\frac{1}{8}$
(B) $\frac... | 6. (A).
As shown in the right figure, it is easy to know that $M$ is the midpoint of $AB$. Let $BC=x$, then we have $(2x)^{2}=x^{2}+(x+1)^{2}$, which simplifies to $x^{2}-x=\frac{1}{2}$. Therefore,
$$
\begin{array}{l}
S_{\triangle X P N}-S_{\triangle B P M}=S_{\triangle C B N}-S_{\triangle C R M} \\
=\frac{1}{2} x^{2}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,317 |
1. Simplify $\frac{a+1}{a+1-\sqrt{1-a^{2}}}+\frac{a-1}{\sqrt{1-a^{2}}+a-1}$ $(0<|a|<1)$ The result is $\qquad$ | $\begin{array}{l}\text { Original expression }=\frac{(\sqrt{a+1})^{2}}{\sqrt{a+1}(\sqrt{a+1}-\sqrt{1-a})} \\ \quad+\frac{-(\sqrt{1-a})^{2}}{\sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,318 |
Example 10 Let $\triangle A B C$ have three inscribed squares on its sides (with two vertices on one side of the triangle, and the other two vertices on the other two sides of the triangle) with equal areas. Prove: $\triangle A B C$ is an equilateral triangle. | Proof: As shown in Figure 10, let
the area of $\triangle ABC$ be $S$,
the altitude from $A$ to $BC$ be $h_{a}$, and the side length of the inscribed square on $BC$ be $x$. Then, from $\triangle A^{\prime} B^{\prime} C^{\prime} \leadsto \triangle ABC$, we get
\[
\frac{x}{a}=\frac{h_{a}-x}{h_{a}}.
\]
Solving for $x$, we ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,320 |
3. Given integers $x, y$ satisfy $15 x y=21 x+$ $20 y-13$. Then $x y=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.6 .
From the original equation, we get
$$
\begin{array}{c}
(3 x-4)(5 y-7)=15 . \\
\because 3 \nmid 3 x-4, \text { and } 5 \times 5 x-7, \text { so it must be that } \\
3 x-4=5 \text { and } 5 y-7=3, \\
\text { or } 3 x-4=-5 \text { and } 5 y-7=-3 .
\end{array}
$$
The former can solve for $x=3$ and $y=2$, while the ... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,321 |
4. In $\triangle A B C$, $A B=A C, D$ is the midpoint of $B C$, $B E \perp A C$ at $E$, intersecting $A D$ at $P$. Given $B P=3$, $P E=1$. Then $P A=$ $\qquad$ . | 4. $\sqrt{3}$.
As shown in the right figure, from $A D \perp B C$ and $B E \perp A C$, we know that $P, D, C, E$ are concyclic, so $B D \cdot B C = B P \cdot B E = 12$.
Solving this, we get $B D = \sqrt{6}$.
Thus, $P D = \sqrt{B P^{2} - B D^{2}} = \sqrt{3}$.
Furthermore, since $A, B, D, E$ are concyclic, we have $P A ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,322 |
One, (Full score 20 points) A newly built oil tank was found to be leaking oil uniformly from the bottom after being filled. To ensure safety and minimize losses, the oil needs to be pumped out before repairs can be made. There are several small oil pumps of the same power available. If 5 pumps work together, it takes ... | Let the oil extraction rate of the oil pump be $x$ units/hour, the internal oil storage be $y$ units, and the oil leakage rate be $z$ units/hour. Suppose $n$ oil pumps are needed to extract all the oil within 3 hours. Then we have
$$
\left\{\begin{array}{l}
50 x = y - 10 z, \\
56 x = y - 8 z, \\
3 n x \geqslant y - 3 z... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,323 |
II. (Full marks 25 points) As shown in the figure, $P$ is the intersection of the circumcircle of $\triangle ABC$ and the perpendicular bisector of side $AB$, with $P$ and $C$ on opposite sides of $AB$, and $M$ is the midpoint of $AB$. Let $\angle PCB = \alpha, \angle PBC = \beta$. Prove:
$$
\frac{AC}{AM} = \frac{\sin ... | II. As shown in the figure, draw $PN \perp AC$, with the foot of the perpendicular at $N$, and connect $PA$.
(1) When $\beta$ is a right angle, $N$ coincides with $A$, and the conclusion $\frac{AM}{AC}=\sin \alpha$ is obviously valid.
(2) When $\beta$ is an acute angle, $N$ is on the extension of $CA$. It is easy to kn... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,324 |
Three. (Full marks 25 points) The product of $n$ consecutive natural numbers from 1 to $n$ is called the factorial of $n$, denoted as $n$! (For example, 5! $=5 \times 4 \times 3 \times 2 \times 1$). How many consecutive zeros are at the end of 1999! ? Explain your reasoning.
---
Note: The original text uses "畍居" whic... | Three, consider the prime factorization of 1999!
$$
1999!=2^{\alpha_{1}} \cdot 3^{\sigma_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \ldots
$$
where the various $\alpha_{i}$ are non-negative integers. The number of trailing zeros in 1999! depends on $\alpha_{1}$ and $\alpha_{3}$, and is equal to the smaller ... | 496 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,325 |
1. Given $\sin \alpha \cdot \cos \beta=-\frac{1}{2}$. Then the range of $\cos \alpha \cdot \sin \beta$ is ( ).
(A) $\left[-1, \frac{1}{2}\right]$
(B) $\left[-\frac{1}{2}, 1\right]$
(C) $\left[-\frac{3}{4}, \frac{3}{4}\right]$
(D) $\left[-\frac{1}{2}, \frac{1}{2}\right]$ | $-1 .(D)$
$$
\text { Method i: } \begin{aligned}
& \cos ^{2} \alpha \cdot \sin ^{2} \beta=\left(1-\sin ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right) \\
& =1-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right)+\sin ^{2} \alpha \cdot \cos ^{2} \beta \\
& =\frac{5}{4}-\left(\sin ^{2} \alpha+\cos ^{2} \beta\right) \\
& =\frac... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,326 |
2. A person is chasing a car stopped at a traffic light at a uniform speed of $6 \mathrm{~m}$ /s. When he is $25 \mathrm{~m}$ away from the car, the traffic light changes from red to green, and the car starts to move with a uniform acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$. Then ( ).
(A) the person can catch up ... | 2. (D).
As shown in Figure 3, suppose the car starts moving at point $C$, while the person is at point A at this moment. After $t$ seconds, the car reaches point $D$, with the distance $C D=\frac{1}{2} a t^{2}$; the person meanwhile reaches point $B$, with the distance $A B=v t$. According to the problem, the distance... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,327 |
3. Given that $a, b$ are unequal positive numbers, insert two sets of numbers
$$
\begin{array}{l}
x_{1}, x_{2}, \cdots, x_{n}, \\
y_{1}, y_{2}, \cdots, y_{n},
\end{array}
$$
such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence. The... | 3. (B).
Solution 1: From the arithmetic sequence, we have $x_{1}+x_{n}=x_{2}+x_{n-1}=\cdots=a+b$, thus
$$
\begin{array}{l}
\frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \\
=\frac{\left(x_{1}+x_{n}\right)+\left(x_{2}+x_{n-1}\right)+\cdots+\left(x_{n}+x_{1}\right)}{2 n} \\
=\frac{n(a+b)}{2 n}=\frac{a+b}{2}=\sqrt{a b}+2\left(\frac{... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 711,328 |
4. Given the lengths of the three face diagonals of a rectangular prism are 5, $4, x$. Then the range of values for $x$ is ( ).
(A) $(2, \sqrt{41})$
(B) $(3,9)$
(C) $(3, \sqrt{41})$
(i)) $(2,9)$ | 4. (C).
Solution 1: As shown in Figure 1, let $A_{1} D_{1}=a, A_{1} B_{1}=b$, $A A_{1}=c$, then we have
$$
\left\{\begin{array}{l}
a^{2}+c^{2}=4^{2} \\
b^{2}+c^{2}=5^{2}
\end{array}\right.
$$
Thus, $0<c<4$,
and $x=\sqrt{a^{2}+b^{2}}=\sqrt{4^{2}+5^{2}-2 c^{2}}$.
Substituting (3) into (2), we get
$$
3<x<\sqrt{41} \text... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,329 |
5. Given the line $l_{1}: y=a x+3 a+2$ and $l_{2}: y$ $=-3 x+3$ intersect at a point in the first quadrant. Then the range of values for $a$ is $(\quad)$.
(A) $\left(-\frac{1}{2}, \frac{1}{3}\right)$
(B) $\left(-\infty, \frac{1}{3}\right)$
(C) $\left(-3, \frac{1}{3}\right)$
(D) $\left(-\frac{1}{2},+\infty\right)$ | 5. (A).
As shown in Figure 6, $l_{1}$ is a family of lines passing through the fixed point $M(-3,2)$; $l_{2}$ intersects the $x$-axis at $A(1,0)$ and the $y$-axis at $B(0,3)$. For the intersection point of the two lines to be in the first quadrant, it must and only must lie on the line segment $AB$. First, the points ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,330 |
Example 1 Find the real solutions of the system of equations
$$
\left\{\begin{array}{l}
x^{3}+x^{3} y^{3}+y^{3}=17, \\
x+x y+y=5
\end{array}\right.
$$
(1990, Shaoxing City Junior High School Mathematics Competition) | Solution: Let $x+y=u, xy=v$, the original system of equations is transformed into $\left\{\begin{array}{l}u^{3}+v^{3}-3uv=17, \\ u+v=5 .\end{array}\right.$
$(2)^{3}-(1)$ gives $uv=6$.
From (2) and (3), we get
$\left\{\begin{array}{l}u=2, \\ v=3\end{array}\right.$ or $\left\{\begin{array}{l}u=3, \\ v=2 .\end{array}\righ... | (x, y) = (1, 2) \text{ or } (2, 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,331 |
Example 2 The system of equations $\left\{\begin{array}{l}x+y+xy=19, \\ x^{2}+y^{2}=25\end{array}\right.$ has the solution
(First Songliao Cup Junior High School Mathematics Competition) | Solution: Let $x+y=u, xy=v$, the original system of equations can be transformed into
$$
\left\{\begin{array}{l}
u+v=19 \\
u^{2}-2 v=25 .
\end{array}\right.
$$
(1) $\times 2+$ (2) gives $u^{2}-2 u-63=0$.
Solving, we get $u=-9$ or $u=7$. Substituting into (1), we get $v=28$ or $v=12$.
Therefore, the original system of ... | x+y=7, xy=12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,332 |
2. Solve the system of equations $\left\{\begin{array}{l}x^{2} y^{2}+x^{2}+y^{2}=9, \\ x^{2} y+x y^{2}-x-y=3 .\end{array}\right.$ | (Solution: $(x, y)=(1,2),(2,1),(-2,1),(1$, $-2),(0,-3),(-3,0)$. | (x, y)=(1,2),(2,1),(-2,1),(1,-2),(0,-3),(-3,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,333 |
1. Given that $(x-1)^{2}$ divides the polynomial $x^{4}+a x^{3}-$ $3 x^{2}+b x+3$ with a remainder of $x+1$. Then $a b=$ $\qquad$ . | II, 1.0.
From the given, we have
$$
\begin{array}{l}
x^{4}+a x^{3}-3 x^{2}+b x+3 \\
=(x-1)^{2}\left(x^{2}+\alpha x+\beta\right)+x+1 \\
=x^{4}+(\alpha-2) x^{3}+(\beta+1-2 \alpha) x^{2}+(1+\alpha- \\
2 \beta) x+1+\beta .
\end{array}
$$
Then $a=\alpha-2$,
$$
\begin{array}{l}
-3=\beta+1-2 \alpha, \\
b=1+\alpha-2 \beta, \\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,334 |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and suppose they satisfy $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=$ 999, then $m=$ . $\qquad$ | 2.1999.
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\sin B \cdot \cos A+\cos B \cdot \sin A)} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin C \cdot \sin (A+B)} \\
=\cos C \cdot \frac{... | 1999 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,335 |
3. If $a, b, c, x, y, z$ are real numbers, and $a^{2}+b^{2}+c^{2}=25, x^{2}+y^{2}+z^{2}=36, a x+b y+c z=$ 30, then $\frac{a+b+c}{x+y+z}=$ $\qquad$ | 3. $\frac{5}{6}$.
From the given conditions, we have
$$
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
=(a x+b y+c z)^{2} .
\end{array}
$$
By the condition for equality in the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\frac{a}{x}=\frac{b}{y}=\frac{c}{z} . \\
\text { Let ... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,336 |
4. Let $\mathrm{i}=\sqrt{-1}$, define a complex number sequence as follows:
$$
z_{1}=0, z_{n+1}=z_{n}^{2}+\mathrm{i}, n \geqslant 1 \text {. }
$$
Then, in the complex plane, the distance from $z_{111}$ to the origin is
$\qquad$ | $4 \cdot \sqrt{2}$.
After calculation, we get
$$
z_{1}=0, z_{2}=\mathrm{i}, z_{3}=\mathrm{i}-1, z_{4}=-\mathrm{i}, z_{5}=\mathrm{i}-1 \text {. }
$$
Since $z_{3}=z_{5}$, then
$$
z_{111}=z_{109}=z_{107}=\cdots=z_{5}=z_{3}=\mathrm{i}-1 .
$$
Therefore, the distance from $\approx_{111}$ to the origin is $\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,337 |
5. Four lines $l_{1}: x+3 y-15=0, l_{2}: k x$ $-y-6=0, l_{3}: x+5 y=0, l_{4}: y=0$ form a quadrilateral. To make this quadrilateral have a circumscribed circle, the value of $k$ is $\qquad$ . | 5. $-\frac{4}{7}$.
Let the equation of the linear system passing through the four vertices of the quadrilateral be $(x+3 y-15)(x+5 y)+\lambda(k x-y-6) y=0$. From $\left\{\begin{array}{l}15-\lambda=1, \\ 8+k \lambda=0\end{array}\right.$, we get $k=-\frac{4}{7}$. | -\frac{4}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,338 |
6. In a regular tetrahedron $ABCD$, the line segment connecting the midpoint of edge $AD$ and the centroid of face $BCD$ is $MN$, and the line segment connecting the midpoint of edge $CD$ and the centroid of face $ABC$ is $PQ$. Then the angle formed by $MN$ and $PQ$ is $\qquad$ | 6. $\arccos \frac{1}{18}$.
Let the edge length of a regular tetrahedron be $a$. On the line segment $AM$, intercept the line segment $ME = \frac{a}{3}$, and construct the height $DF$ from plane $BCD$ and the height $AF$ from plane $ABC$, as shown in Figure 1. Then, $FN = \frac{1}{3} FD$, $FQ = \frac{1}{3} FA$, $NQ \pa... | \arccos \frac{1}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,339 |
Three. (20 points) Given the function $f(x)=\frac{\sin x}{x}$. Please prove using elementary methods that $f(x)$ is a decreasing function on $x \in\left(0, \frac{\pi}{2}\right)$. | $$
\begin{array}{r}
\text { Three, let } \\
\forall x_{1}, x_{2} \in(0, \\
\left.\frac{\pi}{2}\right), x_{1}0 .
\end{array}
$$
Thus, $x_{2} \sin x_{1}>x_{1} \sin x_{2}$.
$$
\begin{array}{l}
\because x_{1} x_{2}>0, \\
\therefore \frac{\sin x_{1}}{x_{1}}>\frac{\sin x_{2}}{x_{2}} .
\end{array}
$$
Therefore, $f(x)$ is a ... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 711,340 |
Four, (20 points) Prove that if the complex number $z$ satisfies $11 z^{10}+$ $10 \mathrm{i} z^{9}+10 \mathrm{i} z-11=0$, then $|z|=1$.
| $$
\begin{array}{l}
z^{9}(11 z+10 \mathrm{i})=11-10 \mathrm{i} z . \\
\therefore|z|^{9} \cdot|11 z+10 \mathrm{i}|=|11-10 \mathrm{i} z| .
\end{array}
$$
Let $z=x+y \mathrm{i}, x, y \in \mathbf{R}^{+}$, then
$$
\begin{array}{l}
|z|^{2}=x^{2}+y^{2}, \\
|11 z+10 i|^{2}=121 x^{2}+(11 y+10)^{2} \\
=121\left(x^{2}+y^{2}\righ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,341 |
Five. (20 points) For what value of $a$ does the graph of the function $f(x)=(x+a)(|x+1-a|+|x-3|)-2x+4a$ have a center of symmetry?
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | For the function $f(x)=(x+a)(|x+1-a|+|x-3|)-2x+4a$, on the right interval $x \geqslant \max \{3, a-1\}$, it takes the form
$$
\begin{array}{l}
f(x)=(x+a)(2x-a-2)-2x+4a \\
=2x^{2}+(a-4)x+q_{1} ;
\end{array}
$$
On the left interval $x \leqslant \min \{3, a-1\}$, it takes the form
$$
\begin{aligned}
f(x) & =(x+a)(-2x+a+2... | a=-\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,342 |
一、(50 points) Given $\triangle A B C$. Construct $-\triangle A^{\prime} B^{\prime} C^{\prime}$ with $\sin A$, $\sin B$, and $\sin C$ as its sides. With $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ as centers, construct circles with radii $\cos A$, $\cos B$, and $\cos C$ respectively. Then these three circles must inter... | $\triangle A^{\prime} B^{\prime} C^{\prime} \backsim \triangle A I C, \angle A^{\prime}=\angle A, \angle B^{\prime}=$ $\angle B, \angle C^{\prime}=\angle C$.
As shown in Figure 3, in $\triangle A^{\prime} B^{\prime} C^{\prime}$,
$A^{\prime} B^{\prime}=\sin \angle C, B^{\prime} C^{\prime}=$
$$
\sin \angle A, C^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,343 |
3. Solve the system of equations $\left\{\begin{array}{l}y^{2}=x^{3}-3 x^{2}+2 x, \\ x^{2}=y^{3}-3 y^{2}+2 y .\end{array}\right.$ | (Solution: $(x, y)=(0,0),(2+\sqrt{2}, 2-\sqrt{2})$, $(2-\sqrt{2}, 2+\sqrt{2})$. | (x, y)=(0,0),(2+\sqrt{2}, 2-\sqrt{2}), (2-\sqrt{2}, 2+\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,344 |
$$
\left\{\begin{array}{l}
\sqrt{x(1-y)}+\sqrt{y(1-x)}=\frac{1}{2}, \\
\sqrt{x(1-x)}+\sqrt{y(1-y)}=\frac{\sqrt{3}}{4} .
\end{array}\right.
$$
Solve the system of equations in the set of real numbers: | From (2), we know that $x(1-x) \geqslant 0, y(1-y) \geqslant 0$. Therefore, $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$.
When $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$, (1) also makes sense, so we can let
$$
x=\sin ^{2} \alpha, y=\sin ^{2} \beta,
$$
where $\alpha, \beta$ satisfy $0 \leqslant \a... | \left(\frac{1}{4}, 0\right),\left(0, \frac{1}{4}\right),\left(1, \frac{3}{4}\right),\left(\frac{3}{4}, 1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,345 |
Three, (50 points) There are 12 balls, all identical in color and size, but one of them is defective in weight, though it is unknown whether this ball is heavier or lighter than the standard. Can the defective ball be identified using a balance scale in just three weighings (without using weights)?
Can the defective b... | Three, number 12 balls as $1,2, \cdots, 12$, then the following weighing method can be designed:
\begin{tabular}{ccc}
& Left Pan & Right Pan \\
First Time & $1,5,6,12$ & $2,3,7,11$ \\
Second Time & $2,4,6,10$ & $1,3,8,12$ \\
Third Time & $3,4,5,11$ & $1,2,9,10$
\end{tabular}
Each weighing can result in three outcomes... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,346 |
Let $n$ be a natural number, $\alpha_{n} \backslash \beta_{n}\left(\alpha_{n}>\beta_{n}\right)$ are the integer parts of the roots of the quadratic equation $x^{2}-2(n+2) x+3(n+1)=0$. Find the value of $\frac{\alpha_{1}}{\beta_{1}}+\frac{\alpha_{2}}{\beta_{2}}+\cdots+\frac{\alpha_{99}}{\beta_{99}}$. | Solution: Let the two roots of the original equation be $x_{n}, x_{n}^{\prime}$, then by the root formula we get
$$
\begin{array}{l}
x_{n}=n+2+\sqrt{n^{2}+n+1}, \\
x_{n}^{\prime}=n+2-\sqrt{n^{2}+n+1} . \\
\because n<\sqrt{n^{2}+n+1}<n+1, \\
2 n+2<n+2+\sqrt{n^{2}} \overline{n+1}<2 n+3,
\end{array}
$$
E. $\quad<n+2-\sqrt... | 10098 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,347 |
Initial 88. As shown in the figure, the side length of the square $ABCD$ is 1. With $A$ as the center and 1 as the radius, the arc intersects the semicircle with $BC$ as the diameter at $P$. Connect and extend $BP$ to intersect $CD$ at $E$, and connect and extend $DP$ to intersect $BC$ at $F$. Prove that the area of qu... | $$
\begin{array}{l}
E C^{2}=E P \cdot E B, E D^{2}=E P \cdot E B, \\
\therefore E C=E D .
\end{array}
$$
Draw $E G / / B C$ through point $E$ intersecting $D F$ at $G$, and connect $P C$. It is easy to see that
$$
\begin{array}{l}
\frac{P E}{P C}=\frac{P C}{P B}=\frac{C E}{B C}=\frac{1}{2} \text {. } \\
\therefore \fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,348 |
87. Let $H$ be the orthocenter of a non-obtuse $\triangle ABC$ with side lengths $a, b, c$. Prove:
$$
a^{2}+b^{2}+c^{2} \geqslant\left(A I+B H+C H^{2}\right)^{2} \text {. }
$$
and specify the condition for equality. | Proof: Let the circumradius of $\triangle ABC$ be $R$. From $AH = \frac{c \cos A}{\sin C} = 2R \cos A$ and similar expressions, and using the Law of Sines, the inequality to be proven is equivalent to
$$
\begin{array}{l}
\sum \sin^2 A \geqslant \left(\sum \cos A\right)^2 \\
\Leftrightarrow \sum (\cos B + \cos C)^2 \leq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,349 |
4. Solve the system of equations $\left\{\begin{array}{l}x^{3}-7 x-3 y=0, \\ y^{3}-7 y-3 x=6 .\end{array}\right.$ | (From (1)-(2), after factorization, we can get $x-y=0$, (3) or $x^{2}+x y+y^{2}-4=0$. (4) (4) $\cdot x-(1)$, after factorization, we can get $(x+y)(x y+3)=0 .(x, y)=(0,0)$, $(-\sqrt{10}, \sqrt{10}),(-\sqrt{10},-\sqrt{10}),(2,-2),(-2,2)$, $\left(\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}\right), \quad\left(\frac{1-\sq... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,351 |
6. The solution to the equation $x=\left(x^{2}-2\right)^{2}-2$ is $\qquad$ _. $(1985$, Shanghai Junior High School Mathematics Competition) | (Let $y=x^{2}-2 . x_{1}=2, x_{2}=-1, x_{3,4}=$
$$
\frac{-1 \pm \sqrt{5}}{2} \text {. })
$$ | \frac{-1 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,353 |
7. Solve the equation $\frac{11 x^{2}-6}{7-12 x^{2}}=\sqrt{\frac{7 x+6}{12 x+11}}$. | (Prompt: Let $\frac{11 x^{2}-6}{7-12 x^{2}}=\sqrt{\frac{7 x+6}{12 x+11}}=y$. Transform the equation into a Type II cyclic symmetric system of equations. $x_{1}=\frac{3}{4}, x_{2}=$ $\left.\frac{-5-11 \sqrt{21}}{74}.\right)$ | x_{1}=\frac{3}{4}, x_{2}=\frac{-5-11 \sqrt{21}}{74} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,354 |
Example 1 Function
$$
\begin{aligned}
F(x)= & \mid \cos ^{2} x+2 \sin x \cdot \cos x-\sin ^{2} x \\
& +A x+B \mid
\end{aligned}
$$
The maximum value $M$ of the function on $0 \leqslant x \leqslant \frac{3 \pi}{2}$ is related to the parameters $A$ and $B$. For what values of $A$ and $B$ is $M$ minimized? Prove your con... | Analysis: First, estimate the possible conclusion through specialization. Let $A=0, B=0$. At this point, $F(x)$ becomes
$$
f(x)=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right| .
$$
Within $0 \leqslant x \leqslant \frac{3 \pi}{2}$, there are three points $\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}$ where ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,355 |
Example 2 There is a pile of goods stacked in a regular manner, with each layer arranged in a rectangle. The bottom layer has 30 items on one side and 15 items on the other, and each subsequent layer has one less item on each side, until the top layer is a straight line. If this pile of goods is restacked into a square... | Solution: The total number of items in this pile of goods is
$$
\begin{array}{l}
30 \times 15 + 29 \times 14 + 28 \times 13 + \cdots + 16 \times 1 \\
= (15 + 15) \times 15 + (15 + 14) \times 14 + \cdots \\
+ (15 + 1) \times 1 \\
= 15 \times (15 + 14 + \cdots + 1) + 15^2 + 14^2 \\
+ \cdots + 1^2 \\
= 3040 .
\end{array}... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,356 |
Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which t... | Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \t... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,357 |
Example 4 Let $D$ be a point on the side $AB$ of $\triangle ABC$, point $D$ moves along a direction parallel to $BC$ to point $E$ on side $AC$; then from point $E$ along a direction parallel to $AB$ to point $F$ on side $BC$; then from point $F$ along a direction parallel to $CA$ to point $G$ on side $AB$, $\cdots \cdo... | Analysis: We can estimate through graphical experiments that point $D$ can return to the original starting point after 6 times.
In fact, as shown in
Figure 1, by the intercept theorem of parallel lines,
we get
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,358 |
Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
\dot{x}^{2}+y^{2}+z^{2}=\frac{7}{2}, \\
x^{3}+y^{3}+z^{3}=\frac{9}{2} .
\end{array}\right.
$$ | Given: $\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2} +2(x y+y z+z x)$,
Substituting (1) and (2) into the above equation, we get
$$
\begin{aligned}
x y+ & y z+z x=\frac{11}{4} . \\
\text { Also, } \because & x^{3}+y^{3}+z^{3}-3 x y z \\
& =(x+y+z)\left[x^{2}+y^{2}+z^{2}\right. \\
& -(x y+y z+z x)],
\end{aligned}
$$
Substitut... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,359 |
Example 5 Find all positive integers $a, b, c$ such that $1 < a < b < c$ and $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.
(33rd IMO) | - Analysis: First, estimate
$$
s=\frac{a b c-1}{(a-1)(b-1)(c-1)}(s \in \mathbf{N})
$$
the range. For this, let $c=a-1, y=b-1, z=c-1$, then $1 \leqslant x1$;
$$
\text { Also, } \begin{aligned}
s & =\frac{(x+1)(y+1)(z+1)-1}{x y z} \\
& <\frac{(x+1)(y+1)(z+1)}{x y z} \\
& =\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\ri... | (3,5,15) \text{ and } (2,4,8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,360 |
Example 6 Let the equation $x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x$ $+a_{0}=0$ have real coefficients, and satisfy the condition:
$$
0<a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n-1} \leqslant 1 .
$$
Given that $\lambda$ is a complex root of this equation, and satisfies the condition $|\lambda|$ $\geqslant 1$. ... | Proof: $\because 0=(\lambda-1)\left(\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots\right.$
$$
\begin{aligned}
& \left.+a_{1} \lambda+a_{0}\right) \\
= & \lambda^{n+1}+\left(a_{n-1}-1\right) \lambda^{n} \\
& +\left(a_{n-2}-a_{n-1}\right) \lambda^{n-1} \\
& +\cdots+\left(a_{0}-a_{1}\right) \lambda-a_{0},
\end{aligned}
$$
$$
\b... | \lambda^{n+1}=1 | Algebra | proof | Yes | Yes | cn_contest | false | 711,361 |
Example 7 The function $f(n)$ is defined on the set of positive integers and takes non-negative integer values, and for all $m, n$ we have
$$
\begin{aligned}
f(m+n)-f(m)-f(n) & =0 \text{ or } 1, \\
\text{and } f(2)=0, f(3)>0, f(9999) & =3333 .
\end{aligned}
$$
Find $f(1982)$. | Solution: Let $m=n=1$, then
$f(2)=2 f(1)+(0$ or 1$)$.
Since $f(2)=0, f(1)$ is a non-negative integer, hence $f(1)$ $=0$.
Let $m=2, n=1$, then
$f(3)=f(2)+f(1)+(0$ or 1$)$.
By $f(3)>0, f(2)=0, f(1)=0$ we get $f(3)$ $=1$.
Next, we prove: For $kk$, then $f(3 k) \geqslant k+1$,
$$
\begin{aligned}
f(9999) & \geqslant f(9999-... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,362 |
For positive integers $a, n$, define $F_{n}(a)=q+r$, where $q, r$ are non-negative integers, $a=q n+r$, and 0 $\leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$
F_{n_{6}}\left(F_{... | To find $A$, we first prove a lemma.
Lemma: If there exists a positive integer $n$, such that for all positive integers $a \leqslant B$ we have $F_{n}(a) \leqslant 2 k$ (where $k$ is a given positive integer), then the largest positive integer $13=k^{2}+3 k$.
Proof: Since $a=q n+r, F_{n}(a)=q+r \leqslant 2 k$, we fir... | 53590 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,363 |
Example 1 Let any real numbers $x$, $y$ satisfy $|x|<1$, $|y|<1$. Prove:
$$
\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geqslant \frac{2}{1-x y} \text {. }
$$
(19th Moscow Mathematical Competition) | Proof: $\because|x|<1,|y|<1$,
$$
\therefore x^{2}<1, y^{2}<1 \text {. }
$$
Using the formula for the sum of an infinite geometric series in reverse, we have
$$
\begin{array}{l}
\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \\
=\left(1+x^{2}+x^{4}+\cdots\right) \\
+\left(1+y^{2}+y^{4}+\cdots\right) \\
=2+\left(x^{2}+y^{2}\right)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,364 |
Example 2 Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be an arithmetic sequence and a geometric sequence, respectively, and $a_{1}=b_{1}>0, a_{2}=b_{2}>0$. Try to compare the size of $a_{n}$ and $b_{n}$. | Let $\{a_{n}\}$ have a common difference of $d$, and $\{b_{n}\}$ have a common ratio of $q$. Clearly, $q>0$.
$$
\begin{aligned}
& \because a_{2}=b_{2}, \\
& \therefore a_{1}+d=a_{1} q, \text { i.e., } d=a_{1}(q-1) . \\
& \therefore a_{n}-b_{n}=a_{1}+(n-1) d-a_{1} q^{n-1} \\
= & a_{1}+(n-1) a_{1}(q-1)-a_{1} q^{n-1} \\
=... | a_{n} \leqslant b_{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,365 |
Example 3 Find the positive integer solutions of the equation $7^{x}-3 \times 2^{y}=1$.
(16th All-Russian Mathematics Competition) | When $1 \leqslant y \leqslant 4$, the equation has two sets of solutions: $x=y=1$ and $x=2, y=4$.
When $y \geqslant 5$, by reversing the geometric series sum formula, we have
$$
\begin{aligned}
\frac{7^{x}-1}{7-1} & =7^{x-1}+7^{x-2}+\cdots+1 \\
& =\frac{3 \times 2^{y}}{7-1}=2^{y-1} .
\end{aligned}
$$
Therefore, $x$ mu... | x=y=1 \text{ and } x=2, y=4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,366 |
Example 6 Prove: If $n$ is a positive odd number, then $1+x^{n}$ can be divided by $1+x$, but cannot be divided by $1-x$.
| Prove: $\frac{1+x^{n}}{1+x}=\frac{1-(-x)^{n}}{1-(-x)}(n$ is an odd number $)$.
From the formula $a_{1}+a_{1} q+a_{1} q^{2}+\cdots+a_{1} q^{n-1}$ $=\frac{a_{1}\left(1-q^{n}\right)}{1-q}(q \neq 1)$, it shows that $\frac{a_{1}\left(1-q^{n}\right)}{1-q}$ is a
polynomial in $q$, so $\frac{1-(-x)^{n}}{1-(-x)}$ is a polynomia... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,368 |
Example 5 Let $R_{x}$ denote the positive integer in decimal notation consisting of $x$ ones. Determine whether $Q=\frac{R_{24}}{R_{4}}$ is a decimal number consisting of several 1s and 0s, and find the number of 0s in $Q$.
(44th American High School AHSME) | Solution: By reversing the formula for the sum of a geometric series, we have
$$
\begin{aligned}
Q & =\frac{R_{24}}{R_{4}}=\frac{9 R_{24}}{9 R_{4}} \\
& =\frac{10^{24}-1}{10^{4}-1}=\frac{\left(10^{4}\right)^{6}-1}{10^{4}-1} \\
& =10^{20}+10^{16}+10^{12}+10^{8}+10^{4}+1 \\
& =100010001000100010001 .
\end{aligned}
$$
Th... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,369 |
Example 4 Given that $a$ and $b$ are the two real roots of the quadratic equation $t^{2}-t-1=0$. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1+x, \\
\frac{x}{b}+\frac{y}{a}=1+\hat{y} .
\end{array}\right.
$$
(1000th Zu Chongzhi Cup Junior High School Mathematics Invitational Competit... | Solving (1)-(2), factorizing, we get
$$
\left(\frac{1}{a}-\frac{1}{b}-1\right)(x-y)=0 \text {. }
$$
From the given conditions, we have
$$
\begin{array}{l}
a+b=1, a b=-1 . \\
\therefore \frac{1}{a}-\frac{1}{b}-1=-b+a-(a+b) . \\
\quad=-2 b \neq 0 .
\end{array}
$$
Thus, from (3) we get $x=y$. Substituting into (1), simp... | x=-\frac{1}{2}, y=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,370 |
Example 7 Prove: $4^{n}+1$ cannot be divisible by 7 $(n \in$ N). | Prove: When $n=3 k(k \in \mathbf{N})$,
$$
\begin{aligned}
\frac{4^{3 k}+1}{7} & =-\frac{\left(1-64^{k}\right)+2}{1-64} \times(-9) \\
& =9 t+\frac{2}{7}(t \in \mathbf{N}) .
\end{aligned}
$$
This expression leaves a remainder of 2 when divided by 7.
When $n=3 k+1(k \in \mathbf{N})$,
$$
\begin{aligned}
\frac{4^{3 k+1}+1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,371 |
Example 8 Prove that $N=\frac{5^{125}-1}{5^{25}-1}$ is not a prime number.
(33rd IMO Preliminary Question) | Prove: Let $5^{25}=x$, then
$$
\begin{aligned}
N= & \frac{5^{125}-1}{5^{25}-1}=\frac{x^{5}}{x-1} \\
= & x^{4}+x^{3}+x^{2}+x+1 \\
= & \left(x^{2}+3 x+1\right)^{2}-5 x(x+1)^{2} \\
= & {\left[\left(x^{2}+3 x+1\right)+5^{13}(x+1)\right] } \\
& \cdot\left[\left(x^{2}+3 x+1\right)-5^{13}(x+1)\right] .
\end{aligned}
$$
Obvio... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,372 |
Example 9 Find all prime numbers $p, q, r$ that satisfy $p^{q}+q^{p}=r$.
| Solution: Clearly, $p \neq q$. Without loss of generality, assume $p<q$.
$\because r$ is a prime number,
$\therefore p$ and $q$ cannot both be odd.
Thus, $p=2$.
When $q$ is a prime number not less than 5, we have
$$
\begin{array}{l}
p^{q}+q^{p}=2^{q}+q^{2} \\
=\left(2^{q}+1\right)+\left(q^{2}-1\right) \\
=3 \times \fra... | p=2, q=3, r=17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,373 |
Example 10 Given a natural number $a \geqslant 2$, the sets are
$$
\begin{array}{l}
A=\left\{y \mid y=a^{x}, x \in \mathbf{N}\right\}, \\
B=\{y \mid y=(a+1) x+b, x \in \mathbf{N}\} .
\end{array}
$$
Does there exist a $b$ in the interval $[1, a]$ such that $C=A \cap B$ $\neq \varnothing$? If it exists, find all possibl... | $$
\begin{array}{l}
\text { Solution: } \because a \geqslant 2, a \in \mathbf{N}, x \leqslant \mathbf{N}, \\
\therefore a \in \mathbf{N}, \text { only if } (a+1) x \in \mathbf{N} . \\
\text { Since } A \cap B \cdot 7^{\prime}-6, \\
\therefore z \in \mathbf{N} .
\end{array}
$$
According to the problem, we only need to ... | b=1 \text{ or } b=a | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,374 |
Example 1 Given $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}$, and $x_{1}+$
$$
\begin{array}{l}
x_{2}+\cdots+x_{n}=A(A>0), x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \\
=\frac{A^{2}}{n-1}(n \in \mathbf{N} \text { and } n \geqslant 2) .
\end{array}
$$
Prove: $0 \leqslant x_{i} \leqslant \frac{2 A}{n}(i=1,2, \cdots, n)$.
(1988... | Proof: Given $x_{1}+x_{2}+\cdots+x_{n}-A=0$, introduce an undetermined parameter $t$, then
$$
\begin{array}{l}
\frac{A^{2}}{n-1}-x_{1}^{2}=x_{2}^{2}+x_{3}^{2}+\cdots+x_{n}^{2} \\
=x_{2}^{2}+x_{3}^{2}+\cdots+x_{n}^{2}+t\left(x_{1}+x_{2}+\right. \\
+x_{n}-A \text { ) } \\
=\left(x_{2}+\frac{t}{2}\right)^{2}+\left(x_{3}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,375 |
Example 2 Let $a, b, c \in \mathbf{R}^{+}$, and $abc=1$. Prove that:
$$
\begin{array}{l}
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \\
\geqslant \frac{3}{2} .
\end{array}
$$
(36th IMO) | Prove: Introducing a positive parameter $t$, noting that $a b c=1$, we get
$$
\begin{array}{l}
\frac{1}{a^{3}(b+c)}+t a(b+c) \\
=\frac{a b c}{a^{3}(b+c)}+t \cdot \frac{b+c}{b c} \\
\geqslant 2 \sqrt{\frac{a b c}{a^{3}(b+c)} \cdot t \cdot \frac{b+c}{b c}} \\
=\frac{2 \sqrt{t}}{a} .
\end{array}
$$
Similarly, we have
$$
... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,376 |
Example 3 Let $n \in \mathbf{N}$. Does there exist $2 n+1$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n+1}$, satisfying the conditions:
(i) $\left|a_{k}\right|<1, k=1,2, \cdots, 2 n+1$;
(ii)
$$
\begin{array}{l}
\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{2 n+1}\right| \\
-\left|a_{1}+a_{2}+\cdots+a_{2 n+1}\righ... | Solution: There do not exist such $2n+1$ numbers.
Otherwise, suppose there exist numbers $a_{1}, a_{2}, \cdots, a_{2 n+1}$ that satisfy the conditions. Introduce the parameter $t$, and set
$$
\left|a_{1}+a_{2}+\cdots+a_{2 n+1}\right|=t,
$$
then $t \geqslant 0$.
Remove the absolute value, and separate the positive and ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,377 |
Example 4 Let $S$ denote the set of all real numbers greater than -1, determine all functions $f: S \rightarrow S$, satisfying the following two conditions:
(i) For all $x$ and $y$ in $S$, we have
\[
\begin{array}{l}
f(x+f(y)+x f(y)) \\
=y+f(x)+y f(x) ;
\end{array}
\]
(ii) In every interval within $-1 < x < 0$ and $0 <... | Let $x=y$, then
$$
\begin{array}{l}
f(x+f(x)+x f(x)) \\
=x+f(x)+x f(x) .
\end{array}
$$
Fix $x$, let $x+f(x)+x f(x)=t$, then
$$
f(t)=t \text{. }
$$
In (1), let $x=t$, then
$$
f\left(t^{2}+2 t\right)=f(t+f(t)+t f(t)) \text{. }
$$
If $t>0$, then $(t+2) t>t>0$.
But $\frac{f(t)}{t}=\frac{f((t+2) t)}{(t+2) t}=1$ contradi... | f(x)=-\frac{x}{x+1} | Algebra | proof | Yes | Yes | cn_contest | false | 711,378 |
Example 5 Given any 13 different real numbers. Prove that there exist at least two, say $x$ and $y$ satisfying the inequality:
$$
0<\frac{x-y}{1+x y}<\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}} \text {. }
$$
(1989, Singapore Mathematical Competition) | Proof: Let 13 different numbers be $a_{1}, a_{2}, \cdots, a_{13}$. Since $\tan x$ maps $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ to $(-\infty, +\infty)$, we can find different numbers $\theta_{i} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that
$$
\tan \theta_{i}=a_{i} \quad (i=1,2, \cdots, 13).
$$
Divide ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,379 |
Example 6 Given that $x_{1}, x_{2}, \cdots, x_{n}$ are pairwise distinct positive integers. Prove that for any positive integer $n$, the inequality
$$
\frac{x_{1}}{1^{2}}+\frac{x_{2}}{2^{2}}+\cdots+\frac{x_{n}}{n^{2}} \geqslant \frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n}
$$
holds.
(20th IMO) | Proof: Introduce parameters $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$.
Let $\frac{1}{1}=\frac{1}{x_{1}}+\alpha_{1}, \frac{1}{2}=\frac{1}{x_{2}}+\alpha_{2}, \cdots$,
$$
\frac{1}{n}=\frac{1}{x_{n}}+\alpha_{n} \text {. }
$$
Since $x_{1}, x_{2}, \cdots, x_{n}$ are the same positive integers, adding the above $n$ equati... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,380 |
Example 5 Find all real solutions $x, y$ of the system of equations
$$
\left\{\begin{array}{l}
x^{4}+y^{2}-x y^{3}-\frac{9}{8} x=0, \\
y^{4}+x^{2}-x^{3} y-\frac{9}{8} y=0
\end{array}\right.
$$
(1985, AIME Mathematics Competition) | Solving (1)-(2), factorizing, we get
$$
\begin{array}{l}
(x-y)\left(x^{3}+y^{3}+2 x^{2} y+2 x y^{2}-\frac{9}{8}\right) \\
=0 . \\
\therefore x-y=0
\end{array}
$$
or \(x^{3}+y^{3}+2 x^{2} y+2 x y^{2}-\frac{9}{8}=0\).
From (3), we get \(y=x\). Substituting into (1), and simplifying, we get \(x^{2}-\frac{9}{8} x=0\).
Sol... | (x_1, y_1) = (0, 0), (x_2, y_2) = \left(\frac{9}{8}, \frac{9}{8}\right), (x_3, y_3) = (1, \frac{1}{2}), (x_4, y_4) = \left(\frac{1}{2}, 1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,381 |
For example, $7 f$ is a function defined on $(1,+\infty)$ and takes values in $(1,+\infty)$, satisfying the condition: for any $x > 1, y > 1$ and $u > 0, v > 0$, we have
$$
f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}}
$$
Determine all such functions $f$.
(Fourth National High School Ma... | Solution 1: Take $x=y=a, u=v=b$, then for any $a>1, b>0$, we have $f\left(a^{2b}\right) \leqslant f(a)^{\frac{1}{2b}}$.
Let $a=10, 2b=\lg x$, then for any $x>1$ we have $f(x) \leqslant f(10)^{\frac{1}{8x}}$.
Again, let $a=x, 2b=\frac{1}{\lg x}$, then for any $x>1$, we have
$f(x) \geqslant f(10)^{\frac{1}{8x}}$.
Thus, t... | f(x)=c^{\frac{1}{8x}} (c>1) | Algebra | proof | Yes | Yes | cn_contest | false | 711,382 |
4. Given that $a, b, c$ are integers satisfying $1<a<b<c$, and $a b c \mid(a b-1)(b c-1)(c a-1)$. Find $a, b, c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | (It is easy to know that $a b c \mid(a b+b c+c a-1)$. Let $k=$ $\frac{a b+b c+c a-1}{a b c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c}<\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ $=\frac{13}{12}$. Therefore, $k=1$. Consequently, we get $(a, b, c)=(2,3,5)$. ) | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 711,383 |
5. If $s=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10^{6}}}$. Try to find the integer part of $s$. | $\begin{array}{l}\text { (Hint: } s=1+\frac{2}{2 \sqrt{2}}+\frac{2}{2 \sqrt{3}}+\cdots+\frac{2}{2 \sqrt{10^{6}}} \\ \\ 2\left(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\cdots+\frac{1}{\sqrt{10^{6}+1}+\sqrt{10^{6}}}\right)= \\ \left.2\left(-1+\sqrt{10^{6}+1}\right)>2\left(-1+10^{3}\right)=1998 .\right)\end{array... | 1998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,384 |
In a circle with center $C$, chord $CD$ is perpendicular to diameter $AB$, and chord $AE$ bisects radius $OC$. Prove that chord $DE$ bisects chord $BC$.
(21st Russian Mathematical Olympiad). | $$
\begin{array}{l}
\because AB \text { is the diameter, } \\
AB \perp CD, \\
\therefore \overparen{AC}=\overparen{AI} \\
\Rightarrow \angle CBA=\angle DBA \\
\Rightarrow \angleDBG=2 \angle CBA . \\
\because \angle AOF=\angle CBA+\angle OCB, \\
\angle CBA=\angle OCB, \\
\therefore \angle AOF=2 \angle CBA . \\
\text { T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,385 |
Let $x_{0}, x_{1}, \cdots, x_{n}$ be positive integers, and $x_{0}<x_{1}$ $<\cdots<x_{n}, n \geqslant 1,[x, y]$ be the least common multiple of $x$ and $y$, and let
$$
S_{n}=\frac{1}{\left[x_{0}, x_{1}\right]}+\frac{1}{\left[x_{1}, x_{2}\right]}+\cdots+\frac{1}{\left[x_{n-1}, x_{n}\right]} \text {. }
$$
Try to find th... | When $n=1$, it is obvious that when $x_{0}=1, x_{1}=2$, the maximum value of $S_{1}$ is $\frac{1}{2}=1-\frac{1}{2^{2}}$;
When $n=2$, then when $x_{0}=1, x_{1}=2, x_{2}=2^{2}$, the maximum value of $S_{2}$ is $\frac{3}{4}=1-\frac{1}{2^{2}}$;
Continuing this approximation, we conjecture:
The maximum value of $S_{n}$ is ... | 1-\frac{1}{2^{n}} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,388 |
In 1971, JU. I. Gerasimov gave the following triangle inequality:
Let $P$ be any point inside $\triangle ABC$, and the distances from $P$ to the three sides $BC$, $CA$, and $AB$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively, with $BC = a$, $CA = b$, and $AB = c$. Then
$$
\frac{r_{1} r_{2}}{a b} + \frac{r_{2} r_{3}}{b... | Proof: Let $\angle B P C=\alpha, \angle C P A=\beta, \angle A P B=\gamma$. In $\triangle P B C$, by the Law of Cosines, we have
$$
\begin{aligned}
a & =\sqrt{P B^{2}+P C^{2}-2 P B \cdot P C \cdot \cos \alpha} \\
& \geqslant \sqrt{2 P B \cdot P C-2 P B \cdot P C \cdot \cos \alpha} \\
& =\sqrt{2 P B \cdot P C(1-\cos \alp... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,389 |
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