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
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3. If $[x]$ represents the greatest integer not greater than $x$, $\{x\}=x-[x]$, then the solution to the equation $x+2\{x\}=3[x]$ is $\qquad$ . | 3. $\frac{5}{3}$.
Substitute $x=[x]+\{x\}$ into the original equation to get $3\{x\}=2[x]$. From $0<\{x\}=\frac{2}{3}[x]<1$, we have $[x]=1,\{x\}=\frac{2}{3}$, thus $x=\frac{5}{3}$. | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,076 |
4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD=4$, $AC$ and $BD$ intersect at $E$, $AE=6$, and the lengths of $BE$ and $DE$ are both integers. Then the length of $BD$ is $\qquad$ | 4.7.
From $\overparen{B C}=\overparen{C D}, \angle B D C=\angle C A D, \angle A C D=\angle A C D$, we know $\triangle D C E \backsim \triangle A C D$, so $\frac{C E}{C D}=\frac{C D}{A C}$, which means $C E \cdot A C=C D^{2}$. Therefore, $C E(C E+6)=16$, solving this gives $C E=2$. Also, from $B E \cdot D E=A E \cdot C... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,077 |
For any natural number $n$, there exists a unique pair of $k$ and $t$ such that
$$
n=\frac{k(k-1)}{2}+t, 0 \leqslant t<k .
$$ | $$
\begin{array}{c}
\text { I. There exists } k>0 \text { such that } \\
\frac{k(k-1)}{2} \leqslant nk_{1}, \text { hence we get } \\
\frac{k(k-1)}{2}-\frac{k_{1}\left(k_{1}-1\right)}{2}=t_{1}-t .
\end{array}
$$
The right side of equation (2) $t_{1}-t<k_{1}$, and since $k \geqslant k_{1}+1$, the left side of equation ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,078 |
II. (Full marks 25 points) As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $M$, and the angle between their two external common tangents is $60^{\circ}$. The line connecting the centers intersects $\odot O_{1}$ and $\odot O_{2}$ at $A$ and $B$ (different from $M$), respectively. A line... | From the given conditions, we have $O_{1} E \perp P E, O_{1} F \perp P F, O_{1} E= O_{1} F$, thus $O_{1}$ lies on the bisector of $\angle E P F$. Similarly, $O_{2}$ lies on the bisector of $\angle E P F$. Therefore, $P A$ is the bisector of $\angle E P F$. Since $O_{2} Q / / P E$, we have $\angle Q O_{2} O_{1}=\angle E... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,079 |
iii. (Full marks 25 points) A mall installs an escalator between the first and second floors, which moves upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor at the same time (the escalator itself is also moving). If the boy and the girl are both considered to be moving at a ... | (i) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, and the stairs have $s$ levels. Then the boy's speed is $2x$ levels/min. According to the problem, we have:
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} ;
\end{array}\right.
$$
Solving, we ge... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,080 |
1. The solution set of the equation $\sqrt{x^{2}+1}-\sqrt{1 x^{2}+2}=1-$ $\frac{1}{\sqrt[3]{x^{2}+1}}$ is ( ).
(A) $\{1.5,2 \sqrt{2}\}$
(B) $\left\{\frac{1}{10}, \frac{1}{\sqrt{2}}\right\}$
(C) $\left\{\frac{1}{\pi}, \sqrt{\pi}\right\}$
(D) $\varnothing$ | -,1.D.
For any $x \in R, \sqrt{x^{2}+1}-\sqrt{x^{2}+2}<0$.
And $1-\frac{1}{\sqrt[3]{x^{2}+1}} \geqslant 0$, left $\neq$ right.
That is, for any $x \in R$, the equation is not satisfied, the solution set of the equation is the empty set: $\varnothing$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,081 |
Example 7 If at least one of A's height or weight is greater than B's, then A is said to be no less than B. Among 100 young men, if someone is no less than the other 99, he is called a good young man. Then, the maximum number of good young men among 100 young men is ( ).
(A) 1
(B) 2
(C) 50
(D) 100
(1995, National High ... | Obvi
ously, the $k$th
person $a_{k}$, his
height is not
less than the previous $(k-$
Analysis: We can construct 100 people $a_{i}(i=1,2,3, \cdots, 100)$ with heights and weights that are oppositely symmetric.
1) person, whose weight is not less than the next $(100-k)$ people, together not less than $(k-1)+(100-k)=99$ p... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 710,084 |
5. $\operatorname{tg} x=1$ is a ( ) of $x=\frac{5 \pi}{4}$.
(A) necessary condition, but not a sufficient condition
(B) sufficient condition, but not a necessary condition
(C) sufficient condition, and also a necessary condition
(D) neither a sufficient condition, nor a necessary condition | 5. A.
$\operatorname{tg} \frac{5 \pi}{4}=1$. Conversely, if $\operatorname{tg} x=1$, it does not necessarily mean that $x=\frac{5 \pi}{4}$. Therefore, $\operatorname{tg} x=1$ is a necessary but not sufficient condition for $x=\frac{5 \pi}{4}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,085 |
6. A square piece of paper $A B C D$, folded along the diagonal $A C$, so that point $D$ is outside the plane $A B C$. At this time, the angle formed by $D B$ and the plane $A B C$ is definitely not equal to ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$ | 6.D.
As shown in the figure, if $D B \perp$ plane $B A C$, then $\angle D B A=90^{\circ}$. In $\triangle A D B$, $A D=A B$, $\angle A D B=\angle A B D=90^{\circ}$. The sum of the interior angles of $\triangle A B D$ will be greater than $180^{\circ}$, which contradicts the triangle angle sum theorem. Therefore, the an... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,086 |
1. Let $f(x)=x^{10}+2 x^{9}-2 x^{8}-2 x^{7}+x^{6}$ $+3 x^{2}+6 x+1$, then $f(\sqrt{2}-1)=$ | $=1.4$.
Let $x=\sqrt{2}-1$, then $x+1=\sqrt{2} \Rightarrow (x+1)^{2}=2 \Rightarrow x^{2}+2x-1=0$. That is, $x=\sqrt{2}-1$ is a root of $x^{2}+2x-1=0$. But $f(x)=(x^{8}-x^{6}+3)(x^{2}+2x-1)+4$, so $f(\sqrt{2}-1)=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,087 |
$\begin{array}{l}\text { 2. } n \in N \text {, then } 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+ \\ \frac{1}{1+2+3+\cdots+n}=\end{array}$ | $\begin{array}{l}\text { 2. } \frac{2 n}{n+1} . \\ 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n} \\ =\frac{1}{\frac{1 \cdot 2}{2}}+\frac{1}{\frac{2 \cdot 3}{2}}+\frac{1}{\frac{3 \cdot 4}{2}}+\cdots+\frac{1}{\frac{n(n+1)}{2}} \\ =\frac{2}{1 \cdot 2}+\frac{2}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\cdots+\frac{... | \frac{2 n}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,088 |
3. If
$$
\dot{z}=\frac{(1+i)^{2000}(6+2 i)-(1-i)^{1998}(3-i)}{(1+i)^{1996}(23-7 i)+(1-i)^{1994}(10+2 i)} \text {, }
$$
then $|z|=$ . $\qquad$ | 3.1.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,089 |
4. The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-\right.$ $3)^{2001}$ is expanded and like terms are combined. The sum of the coefficients of the odd powers of $x$ in the resulting expression is $\qquad$. | 4. -1 .
$$
\text { Let } \begin{aligned}
f(x)= & \left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-3\right)^{2001} \\
= & A_{0}+A_{1} x+A_{2} x^{2}+\cdots+A_{4001} x^{4001} \\
& +A_{4002} x^{4002} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Then } A_{0}+A_{1}+A_{2}+\cdots+A_{4001}+A_{4002} \\
\quad=f(1)=0, \\
A_{0}-A_... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,090 |
6. On the coordinate plane, the area of the plane region bounded by the conditions $\left\{\begin{array}{l}y \geqslant-|x|-1, \\ y \leqslant-2|x|+3\end{array}\right.$ is | 6. 16 .
For $\left\{\begin{array}{l}y \geqslant-|x|-1, \\ y \leqslant-2|x|+3\end{array}\right.$, discuss for $x \geqslant 0, x<0$.
When $x \geqslant 0$, it is $\left\{\begin{array}{l}x \geqslant 0, \\ y \geqslant-x-1, \\ y \leqslant-2 x+3 .\end{array}\right.$
When $x<0$, it is $\left\{\begin{array}{l}x<0, \\ y \geqsla... | 16 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,091 |
Three. (Full marks 20 points) Is $\operatorname{tg} 5^{\circ}$ a rational number or an irrational number? Please prove your conclusion. | Three, $\operatorname{tg} 5^{\circ}$ is an irrational number. The reason is as follows.
If $\operatorname{tg} 5^{\circ}$ is a rational number, then by $\operatorname{tg} 10^{\circ}=\frac{2 \operatorname{tg} 5^{\circ}}{1-\operatorname{tg}^{2} 5^{\circ}}$ we know $\operatorname{tg} 10^{\circ}$ is a rational number. Again... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 710,092 |
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
保留源文本的换行和格式，所以翻译结果如下：
Four, (Full marks 20 po... | Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{a... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,093 |
Five. (Full marks 20 points) $a, b, c$ are all real numbers, $a \neq b, b \neq c, c \neq a$. Prove:
$$
\frac{3}{2} \leqslant \frac{|a+b-2 c|+|b+c-2 a|+|c+a-2 b|}{|a-b|+|b-c|+|c-a|}<2 .
$$ | $$
\begin{array}{l}
\text { 5. Let } f(a, b, c) \\
=\frac{|a+b-2 c|+|b+c-2 a|+|c+a-2 b|}{|a-b|+|b-c|+|c-a|},
\end{array}
$$
where $a \neq b, b \neq c, c \neq a$.
Since $|a+b-2 c|=|(a-c)+(b-c)|$
$$
\leqslant|a-c|+|b-c|,
$$
the equality holds if and only if $a-c$ and $b-c$ have the same sign.
Similarly, $|b+c-2 a| \leq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,094 |
Example 8 Let $x \in\left(-\frac{1}{2}, 0\right)$. Then
$$
\begin{array}{l}
a_{1}=\cos (\sin x \pi), a_{2}=\sin (\cos x \pi), \\
a_{3}=\cos (x+1) \pi
\end{array}
$$
the size relationship is ( ).
(A) $a_{3}<a_{2}<a_{1}$
(B) $a_{1}<a_{3}<a_{2}$
(C) $a_{3}<a_{1}<a_{2}$
(D) $a_{2}<a_{3}<a_{1}$
(1996, National High School ... | Analysis: By the uniqueness of the answer, the size relationship must satisfy the special value $x_{0}=-\frac{1}{4} \in\left(-\frac{1}{2}, 0\right)$. At this time,
$$
\begin{aligned}
a_{1} & =\cos \left[\sin \left(-\frac{\pi}{4}\right)\right]=\cos \left(-\frac{\sqrt{2}}{2}\right) \\
& =\cos \frac{\sqrt{2}}{2}, \\
a_{2}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,095 |
4. The solution set $(x, y)$ that satisfies the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-14 x-10 y+58=0, \\
\sqrt{x^{2}+y^{2}-16 x-12 y+100}+\sqrt{x^{2}+y^{2}+4 x-20 y+104}=2 \sqrt{29}
\end{array}\right.
$$
is ( ).
(A) $\left(\frac{217+5 \sqrt{415}}{29}, \frac{180-2 \sqrt{415}}{29}\right)$
(B) $\left... | 4.C.
The system of equations becomes
$$
\left\{\begin{array}{l}
(x-7)^{2}+(y-5)^{2}=4^{2}, \\
\sqrt{(x-8)^{2}+(y-6)^{2}}+\sqrt{(x+2)^{2}+(y-10)^{2}} \\
=\sqrt{10^{2}+4^{2}} .
\end{array}\right.
$$
From (1), we know that the solution $(x, y)$ lies on a circle with center $(7,5)$ and radius 4; from (2), we know that th... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,096 |
5. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $E$ is the midpoint of $D C$, and $F$ is the midpoint of $B B_{1}$. Then the volume of the tetrahedron $A D_{1} E F$ is $\qquad$ | 5. $\frac{5}{24}$.
Draw a line through $D_{1}$ parallel to $\mathrm{AF}$ intersecting the extension of $\mathrm{CD}$ at $G$. Then $G D_{1} / /$ plane $F A E$. Connect $F G, A G$, we get
$$
\begin{array}{l}
V_{A D_{1} E F} \\
=V_{D_{1}-A E F} \\
=V_{G-A E F} \\
=V_{F-A G E} .
\end{array}
$$
The base of the tetrahedron... | \frac{5}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,097 |
一、(满分 50 分) $O$ and $H$ are the circumcenter and orthocenter of an acute-angled $\triangle ABC$, respectively. Point $D$ is on $AB$, with $AD = AH$, and point $E$ is on $AC$, with $AE = AO$. Prove: $DE = AE$.
---
Translation:
I. (Full marks 50 points) $O$ and $H$ are the circumcenter and orthocenter of an acute-angl... | $-、$ Construct the circumcircle of $\triangle A B C$, connect $O$ to the circle at $F$, and connect $B F, A F$.
Since $F B \perp B C, A H \perp B C$,
$$
\therefore F B / / A H \text {. }
$$
Also, since $F A \perp A C$,
$B H \perp A C$
$$
\therefore F A / / B H \text {, }
$$
then $A F B H$ is a parallelogram.
$$
B F=A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,098 |
II. (Full marks 50 points) In a factory, $m$ workers together proposed $n$ different ($n>1$) reasonable suggestions. It was found that any two workers have at least one suggestion in common, but no two workers proposed exactly the same set of suggestions. Prove: $m \leqslant 2^{n-1}$.
untranslated part:
将上面的文本翻译成英文，请... | Let $A$ denote the set of all distinct suggestions, and $A_{i}$ represent the set of suggestions made by the $i$th worker $(i=1,2, \cdots, m)$, then
$$
\begin{array}{l}
|A|=n, A_{i} \cap A_{j} \neq \varnothing, A_{i} \neq A_{j}(1 \leqslant i<j \leqslant n), \\
A=A_{1} \cup A_{2} \cup \cdots \cup A_{m} . \\
\because A_{... | m \leqslant 2^{n-1} | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,099 |
Three. (Full marks 50 points) There are 21 points on a circle. Prove: Among all the arcs formed with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$. | Three points on a circle divide the circle into three arcs, at least one of which is greater than $120^{\circ}$. Connecting the endpoints of the arc that does not exceed $120^{\circ}$ with a chord, we can see that among any three points on the circle, at least two points are connected by a chord (referred to as an "edg... | 100 | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,100 |
Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$. | Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,101 |
66. Let the 6-digit number $n=\overline{a b c d e f}$, $n^{\prime}=\overline{c d e f a b}$, $n^{\prime \prime}=\overline{\text { efabcd. }}$.
(1) Find $n$ such that $2 n-111111=n^{\prime}$;
(2) Find $n$ such that $3 n-222222=n^{\prime \prime}$. | Solution: (1) Let $\overline{a d}=x, \overline{c d e f}=y$. Then $2(10000 x+y)-111111=x+100 y$. Simplifying, we get $2875 x=14 y+15873$. Therefore, (1) $x$ is an odd number;
(2) $2875 x \leqslant 14 \times 9999+15873 \Rightarrow x \leqslant 54$;
(3) $7(408 x-2 y-2267)=4-x \Rightarrow 7 \mid 4-x$.
Thus, $x \in\{11,25,3... | n=111111,253968,396825,539682 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,102 |
65. In $\triangle ABC$, prove that:
$$
\sin A + 2 \sin \frac{B}{2} + 3 \sin \frac{C}{3} \leqslant 3,
$$
and determine the conditions under which equality holds. | Proof: As shown in the figure, construct a unit
circle $O, P_{1}, P_{2}, P_{3}$ as three points on the
circumference, and $\angle P_{1} O P_{2}$
$=2 \angle A, \angle P_{2} O P_{3}=2$
$\angle B, \angle P_{3} O P_{1}=2 \angle C$.
Then bisect $\angle P_{1} O P_{2}$, divide $\angle P_{2} O P_{3}$ into four equal parts, and... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,103 |
66. Given that $a$, $b$, $c$ are positive numbers, and $abc \leqslant 1$.
Prove: $\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geqslant a+b+c$. | Proof: First, when $abc=1$, without loss of generality, assume $a \geqslant 1$, then we can get
$$
\begin{array}{l}
(a-1)(1-b)(1-c) \\
=a+b+c-(ab+bc+ac).
\end{array}
$$
(i) When $b \geqslant 1, c \leqslant 1$ or $b \leqslant 1, c \geqslant 1$,
we have $a+b+c \leqslant ab+bc+ac$,
then
$$
\begin{array}{l}
a^{2}b+b^{2}c+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 710,104 |
Example 9 In a regular $n$-sided pyramid, the range of the dihedral angle formed by two adjacent side faces is ( ).
(A) $\left(\frac{n-2}{n} \pi, \pi\right)$
(B) $\left(\frac{n-1}{n} \pi, \pi\right)$
(C) $\left(0, \frac{\pi}{2}\right)$
(D) $\left(\frac{n-2}{n} \pi, \frac{n-1}{n} \pi\right)$
(1994, National High School ... | Analysis: It is easy to know that when the plane angle approaches infinitely close to the vertex of the base regular $n$-sided polygon, its value tends to $\frac{(n-2) \pi}{n}$. When $n \rightarrow +\infty$, the plane angle $\rightarrow \pi$. Therefore, the correct choice is (A). | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,105 |
Example 10 There exists a positive integer $n$, such that $\sqrt{p+n}+\sqrt{n}$ is an integer for the prime number $p(\quad)$.
(A) does not exist
(B) there is only one
(C) there are more than one but only a finite number
(D) there are infinitely many
(1996, National High School Mathematics Competition) | Analysis: Let $\sqrt{p+n}+\sqrt{n}=k(k$ be a positive integer), making the problem a prime solution equation for $p$. Rearranging and squaring gives $p=k(k-2 \sqrt{n})$.
Since $p$ is a prime, we can let $k-2 \sqrt{n}=1$, thus,
$$
k=2 \sqrt{n}+1=p \text {. }
$$
Obviously, only the square of a composite number for $n$ w... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,106 |
Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competi... | Analysis: The inequality can be transformed into
$$
\begin{array}{l}
k \leqslant-\frac{\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993}{\log _{\frac{x_{0}}{x_{3}}} 1993} \\
=f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \text {. } \\
\end{array}
$$
Thus,
$$
\{k\}_{\text {m... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 710,107 |
Example 3 As shown in the figure, $A D$, $A E$, and $A M$ are the altitude, angle bisector, and median of $\triangle A B C$, respectively, $\angle 1=\angle 2$. Prove: $\angle B A C=90^{\circ}$. | Prove: Draw the circumcircle $\odot O$ of $\triangle ABC$ intersecting the extension of $AE$ at $N$, and connect $MN$.
$\because \angle BAE = \angle CAE, \therefore \overparen{BN} = \overparen{CN}$.
$\because MB = MC, \therefore MN \perp BC$.
$\because AD \perp BC, \therefore MN \parallel AD$.
Thus, $\angle N = \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,108 |
Example 12 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=$ 0. Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Competition) | Analysis: Since $2 a=x^{3}+\sin x=(-2 y)^{3}+$ $\sin (-2 y)$, if we let $f(t)=t^{3}+\sin t$, then we have $f(x)$ $=f(-2 y)$.
And $f(t)$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so by monotonicity we have
$$
x=-2 y \text {, hence } x+2 y=0 \text {. }
$$
Thus $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,109 |
Example 1 Given $f(x)=\frac{x^{2}}{1+x^{2}}$. Then the sum
$$
\begin{array}{l}
f\left(\frac{1}{1}\right)+f\left(\frac{2}{1}\right)+\cdots+f\left(\frac{100}{1}\right)+f\left(\frac{1}{2}\right) \\
+f\left(\frac{2}{2}\right)+\cdots+f\left(\frac{100}{2}\right)+\cdots f\left(\frac{1}{100}\right)+ \\
f\left(\frac{2}{100}\rig... | List the square number table as shown in the figure, and add the two number tables after rotating around the main diagonal by $180^{\circ}$,
$$
\begin{array}{ccccc}
f\left(\frac{1}{1}\right) & f\left(\frac{2}{1}\right) & f\left(\frac{3}{1}\right) & \cdots & f\left(\frac{100}{1}\right) \\
f\left(\frac{1}{2}\right) & f\l... | 5000 | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,110 |
Example 2 Given the sequence $a_{k}=2^{k}(1 \leqslant k \leqslant n)$. Then the sum of all possible products $a_{i} a_{j}(1 \leqslant i \leqslant j \leqslant n)$ is
(1997, Shanghai High School Mathematics Competition) | Solution: Given $a_{i} a_{j}=2^{i+j}(1 \leqslant i \leqslant j \leqslant n)$, the table is as follows: $\square$
$$
\begin{array}{ccccc}
2^{1+1} & 2^{1+2} & 2^{1+3} & \cdots & 2^{1+n} \\
& 2^{2+2} & 2^{2+3} & \cdots & 2^{2+n} \\
& & 2^{3+3} & \cdots & 2^{3+n} \\
& & & \cdots & \cdots \\
& & & & 2^{n+n}
\end{array}
$$
... | \frac{4}{3}\left(2^{n}-1\right)\left(2^{n+1}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,111 |
Example 3 Find the sum $S_{n}=1^{2}+2^{2}+\cdots+n^{2}$. | The geometric meaning of the sum $S_{n}$ is the $n$-layer stacking of a unit cube, which, when mapped onto a plane, results in a table of $n$ square numbers represented by 1, arranged in a stepped shape:
\begin{tabular}{ccccccc}
1 & & & & & \\
1 & 1 & & & & \\
1 & 1 & &. & & \\
1 & 1 & 1 & & & \\
1 & 1 & 1 & & & \\
1 &... | S_{n}=\frac{1}{6} n(n+1)(2 n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,112 |
Example 4 Find the sum $S_{n}=n \cdot 1+(n-1) \cdot 3+(n$ $-2) \cdot 5+\cdots+1 \cdot(2 n-1)$. | Solution: Each term in $S_{n}$ can be viewed as the sum of 1 $n$, 3 $n-1$s, ..., and $2n-1$ 1s. Thus, we can construct the following triangular number table (for convenience, only the case for $n=5$ is shown):
$$
\begin{array}{lllllllll}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
& 2 & 2 & 2 & 2 & 2 & 2 & 2 & \\
& & 3 & 3 & ... | \frac{1}{6} n(n+1)(2 n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,113 |
Example 5 Given an arithmetic sequence $\left\{a_{n}\right\}$. Try to find the sum of the first $n$ terms of the sequence: $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, a_{1}+a_{2}+\cdots$ $+a_{n}$. | Solve: When the equilateral triangular number table as shown in the figure is rotated $60^{\circ}$ clockwise and counterclockwise around the center, and the three number tables are superimposed, the sum of the three numbers in each position is $n+2$.
1
12
$\begin{array}{lll}1 & 2 & 3\end{array}$
```
... ... ...
```
$\b... | S_{n}=\frac{1}{6} n(n+1)\left(2 a_{1}+a_{n}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,114 |
Example 6 Find the value of $1^{2}+3^{2}+\cdots+(2 n-1)^2$ | To construct a triangular number table, add 1 to the left: (1, 3, ..., 2n-1) as the first row. Then
$$
\begin{aligned}
S_{n}= & 2[(2 n-1)+((2 n-1)+(2 n \\
& -3))+\cdots+((2 n-1)+\cdots+3+
\end{aligned}
$$
$$
\begin{aligned}
& 1)]-[1+3+\cdots+(2 n-1)] \\
= & 2 \cdot \frac{1}{6} n(n+1)[2(2 n-1)+1] \\
& -n^{2} \\
= & \fra... | \frac{1}{3} n\left(4 n^{2}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,115 |
Example 1 Let $\triangle ABC$ be an acute-angled triangle, with circumcenter $O$ and circumradius $R$. Let $AO$ intersect the circumcircle of $\triangle BOC$ at another point $A'$, $BO$ intersect the circumcircle of $\triangle COA$ at another point $B'$, and $CO$ intersect the circumcircle of $\triangle AOB$ at another... | Proof: As shown in the figure, let the intersection points of $AO$ with $BC$, $BO$ with $CA$, and $CO$ with $AB$ be $D$, $E$, and $F$ respectively. The areas of $\triangle AOB$, $\triangle DOC$, and $\triangle COA$ are denoted as $S_{1}$, $S_{2}$, and $S_{3}$ respectively. From the fact that points $B$, $O$, $C$, and $... | OA' \cdot OB' \cdot OC' \geq 8R^3 | Geometry | proof | Yes | Yes | cn_contest | false | 710,116 |
Example 2 The center of a semicircle is $O$, the diameter is $A B$, a line intersects the semicircle at $C$ and $D$, and intersects $A B$ at $M (M B < M A, M C < M D)$. Let $K$ be the other intersection point of the circumcircles of $\triangle A O C$ and $\triangle D O B$ except for point $O$. Prove that $\angle M O$ i... | Prove: Connect $K B, K C$. From the fact that $A, O, C, K$ are concyclic, we know
$$
\angle O A C=\angle O K C \text{. }
$$
From the fact that $B, O, D, K$ are concyclic, we know
$$
\angle B D O=\angle B K O \text{. }
$$
Therefore, $\angle A M D=\angle A B D-\angle B D C=$ $\angle B D O-\angle O A C=\angle B K O-\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,117 |
Proposition 1 Let the finite set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, $f$ be a mapping from $A$ to $A$. Denote $f_{1}(x)=f(x)$, $f_{r+1}(x)=f\left[f_{r}(x)\right](x \in A, r \in N)$. Then $f$ is a one-to-one mapping if and only if: for any $a_{i} \in A$, there exists $m_{i} \in N, 1 \leqslant m_{i} \leqslant... | Proof: Necessity. If $f$ is a one-to-one mapping, then $f_{1}\left(a_{i}\right)=a_{1}$ (at this time $\left.m_{i}=1\right)$ or else $f_{1}\left(a_{i}\right)=a_{i_{1}}$ $f_{1}\left(a_{i_{1}}\right)=a_{i_{1}}$. Otherwise, $a_{i_{1}}$ would have two pre-images $a_{i}$ and $a_{i_{1}}$ in $A$, which contradicts the fact tha... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,118 |
Example 4 In the concave quadrilateral $ABCD$, $\angle BAD=$ $\angle 1=\angle 2=45^{\circ}$. Prove: $AC=BD$ | Prove: Construct the circumcircle $\odot O$ of $\triangle A B D$ intersecting the extension of $D C$ at $E$, and connect $A E, B E$. Then $\angle 3=\angle 1=\angle B A D=\angle 4=\angle 2=45^{\circ}$, so $\angle 5=45^{\circ}$. Knowing that $A B$ perpendicularly bisects $C E$, then $A C=A E=B D$. | AC=BD | Geometry | proof | Yes | Yes | cn_contest | false | 710,119 |
Example 1 Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a finite set with $n$ elements, and $f$ be a one-to-one mapping from $A$ to $A$. Denote $f_{1}(x)=f(x), f_{k+1}(x)=f\left[f_{k}(x)\right]$. Prove that there exists $m \in \mathbb{N}$, such that for any $x \in A$, we always have $f_{m}(x)=$ (1993, Jiangsu P... | Proof: According to the known conditions, from Proposition 1, for any $-a_{i} \in A$, there exists $1 \leqslant m_{i} \leqslant n$, such that $f_{m_{i}}\left(a_{i}\right)=$ $a_{i}(i=1,2, \cdots, n)$.
Taking the least common multiple $m$ of all $m_{i}$, then from Proposition 2, for any $x \in A$, it must be that $f_{m}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,120 |
Example 2. Let $D=\{1,2, \cdots, 10\}, f$ be a one-to-one mapping from $D$ to $D$. Let $f_{1}(x)=f(x), f_{n+1}(x)=$ $f\left[f_{n}(x)\right]$. Prove: There exists a permutation $x_{1}, x_{2}$, $\cdots, x_{n}$ of $D$, such that the following equation holds:
$$
\sum_{i=1}^{10} x_{i} f_{2520}\left(x_{i}\right)=220 \text {.... | Proof: Since $f$ is a one-to-one mapping from $D$ to $D$, by Proposition 1, for any $i \in D$, there exists $m_{i}\left(1 \leqslant m_{i} \leqslant 10\right)$, such that $f_{m_{i}}(i)=i$.
Considering $2520=2^{3} \cdot 3^{2} \cdot 5 \cdot 7$, which is the least common multiple of $1,2, \cdots, 10$, it follows from the ... | 220 | Combinatorics | proof | Yes | Yes | cn_contest | false | 710,121 |
Example 3 Let the set $A=\{1,2, \cdots, 10\},$ and the mapping $f$ from $A$ to $A$ satisfies the following two conditions:
(1) For any $x \in A, f_{30}(x)=x$;
(2) For each $k \in \mathbb{Z}^{+}, 1 \leqslant k \leqslant 29$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.
Find the total number of such m... | Solution: Notice that $10=5+3+2,30=5 \times 3 \times 2$. This suggests dividing $A$ into three disjoint subsets
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\} \cup\left\{b_{1}, b_{2}, b_{3}\right\}
$$
$\cup\left\{c_{1}, c_{2}\right\}$.
Since $f$ satisfies conditions (1) and (2), $f$ is a one-to-one mapping from ... | 120960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,122 |
Example 4 Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A \rightarrow A$, such that its third composite mapping $f \cdot f \cdot f$ is the identity mapping. How many such $f$ are there?
(1996. Japan Mathematical Olympiad Preliminary) | Solution: Since the threefold composite mapping on set A is the identity mapping, it follows from Proposition 1 and Proposition 2 that there are three types of mappings $f$ that meet the conditions:
(1) $f$ is the identity mapping;
(2) There exists a three-element mapping cycle $a \rightarrow b \rightarrow c \rightarro... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,123 |
Example 1 Constructing equations to solve problems:
Let $x=\frac{\sqrt{5}+1}{2}$, then $\frac{x^{3}+x+1}{x^{5}}=$ $\qquad$ $(1988$, Shanghai Junior High School Mathematics Competition) | If we substitute $x=\frac{\sqrt{5}+1}{2}$ directly, it is a clumsy method, leading to complicated calculations. If we regard $\frac{\sqrt{5}+1}{2}$ as a root of a quadratic equation, then the other root is $\frac{1-\sqrt{5}}{2}$. Thus, we can construct the equation
$$
x^{2}-x-1=0 \text {. }
$$
Let one root of this equ... | \frac{1}{2}(\sqrt{5}-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,124 |
For example, if $2 \alpha$ and $\beta$ are the two roots of equation (1), let
$$
u_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}}(n=1,2, \cdots) \text {. }
$$
Then $u_{n}$ is a positive integer. | Proof: $u_{1}=\frac{\alpha-\beta}{\sqrt{5}}$
$$
\begin{aligned}
& =\frac{\frac{1}{2}(1+\sqrt{5})-\frac{1}{2}(1-\sqrt{5})}{\sqrt{5}}=1 . \\
u_{2}= & \frac{\alpha^{2}-\beta^{2}}{\sqrt{5}}=\frac{(\alpha-\beta)(\alpha+\beta)}{\sqrt{5}} . \\
\because \alpha & +\beta=1, \\
\therefore u_{2} & =\frac{\alpha-\beta}{\sqrt{5}}=1 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,125 |
Example 3 Given the equation.
$$
x^{2}+a x+1=b
$$
has two roots that are natural numbers. Prove: $a^{2}+b^{2}$ is a composite number. (Hua Luogeng Mathematical School Question) | To prove: To show that $a^{2}+b^{2}$ is a composite number, it is necessary to show that there exist two natural numbers $p$ and $q$ greater than 1, such that
$$
a^{2}+b^{2}=p q \text {. }
$$
Let $x_{1}$ and $x_{2}$ be the two roots of equation (1). By the relationship between roots and coefficients, we have
$$
\left\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,126 |
$$
\sum \frac{r_{a}}{h_{a}} \geqslant 3
$$
Strengthened to
$$
\sum \frac{r_{a}}{h_{a}} \geqslant \frac{3 R}{2 r},
$$
where $R, r, h_{a}, h_{b}, h_{c}, r_{a}, r_{b}, r_{c}$ are the circumradius, inradius, altitudes, and exradii of $\triangle ABC$, respectively. $\Sigma$ denotes the cyclic sum.
In fact, the optimal for... | Proof: Given the familiar formulas $r_{a}=\frac{\Delta}{s-a}, h_{a}=\frac{2 \Delta}{a}$, and $(s-a)(s-b)(s-c)=\frac{\Delta^{2}}{s}=s r^{2}, \sum b c=s^{2} + 4 R r + r^{2}$, and noting
$$
\begin{array}{l}
a(s-b)(s-c) \\
=-(s-a)(s-b)(s-c) \\
\quad+s(s-b)(s-c) \\
=-(s-a)(s-b)(s-c) \\
\quad+s\left[-s^{2}+a s+b c\right],
\e... | \sum \frac{r_{a}}{h_{a}} = \frac{2 R}{r} - 1 | Inequalities | proof | Yes | Yes | cn_contest | false | 710,127 |
Example 5. Let $C$ and $D$ be two fixed points on the chord $AE$. Find a point $P$ on the arc of the segment such that $\angle CPD$ is maximized.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Draw auxiliary circle $\odot O'$ through $C$ and $D$, intersecting $\overparen{A P B}$ at $M$ and $N$. Connect $M$ and $N$ and extend to intersect the extension of $A B$ at $E$. Draw the tangent to $\widehat{A P B}$ from $E$, with the point of tangency at $P$. Then, by $E P^{2}=E N \cdot E M, E M \cdot E N=E ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,130 |
6. As shown in the figure, in right $\triangle A C B$, $C D$ is the altitude on the hypotenuse $A B$, and $D E$ is the altitude on the hypotenuse $B C$ of $\mathrm{Rt} \triangle C D B$. If $B E=$ $6, C E=4$, then the length of $A D$ is $\qquad$ | 6. $\frac{4}{3} \sqrt{15}$ | \frac{4}{3} \sqrt{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,134 |
7. Given $x-y-2=0, 2 y^{2}+y-4=0$. Then the value of $\frac{x}{y}-y$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7.
$$
\frac{3}{2}
$$ | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 710,135 |
Example $6 P$ is any point on the side $A B$ of $\triangle A B C$, draw $P Q / / A C$ intersecting $B C$ at $Q$, draw $P R / / B C$ intersecting $A C$ at $R$. Is there a fixed point $M$ other than $C$, such that $C, Q, R, M$ are concyclic? Prove it. | Given: Through $A$ draw $\odot O$ tangent to $B C$ at $C$, through $B$ draw $\odot O^{\prime}$ tangent to $A C$ at $C$, $\odot O$ and $\odot O^{\prime}$ intersect at a fixed point $M$ other than $C$.
$$
\begin{array}{l}
\because \angle 3=\angle 4, \angle 5=\angle 6, \\
\therefore \triangle M A C \sim \triangle M C B, \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,141 |
13. If the equation $\frac{a}{1997}|x|-x-1997=0$ has only negative roots, then the range of the real number $a$ is $\qquad$ ـ. | 13.
$$
-1997 < a \leqslant 1997
$$ | -1997 < a \leqslant 1997 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,142 |
$$
\begin{array}{r}
\text { 2. } \triangle A B C \text { has side lengths of } \\
\text { 4, 5, and 6. } \triangle A^{\prime} B^{\prime} C^{\prime} \sim \triangle A B C,
\end{array}
$$
$$
\begin{array}{l}
\text { If the difference between the longest and shortest } \\
\text { sides of } \triangle A^{\prime} B^{\prime} ... | 2. $\frac{15}{16} \sqrt{7}$ | \frac{15}{16} \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,146 |
Example 7 As shown in the figure, $CD$ is the altitude on the hypotenuse of Rt $\triangle ABC$, $O$ is a point on $AC$, $OA$ $=OB=a$. Find: $OD^{2}+$ $CD^{2}=$ ? | Analysis: $C D^{2}=D A \cdot D B$, and the positions of $D A, D B, D O$ resemble the form of intersecting chords. Since $O A=O B$, we construct $\odot O$ with $O$ as the center and $O A$ as the radius, intersecting the extension of $D O$ at $E$ and $F$. We have
$$
\begin{array}{l}
D A \cdot D B=D E \cdot D F=(O E-O D)(... | a^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,152 |
Three, let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
Let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit di... | $$
\text { Three, } \because 1^{3}=1,2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125 \text {. }
$$
$\therefore(x, y, z)$ has the following 10 possibilities:
(1) $(1,8,7) ;(2)(1,8,4) ;(3)(1,8,5)$;
(4) $(1,7,4) ;(5)(1,7,5) ;(6)(1,4,5)$;
(7) $(8,7,4) ;(8)(8,7,5) ;(9)(8,4,5)$;
$(10)(7,4,5)$.
Then the values of $(x-y)(y-z)(z-x)$ are
$$... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,157 |
Four, on a circular road, there are 4 middle schools arranged clockwise: $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferred ... | Let $A_{1}$ High School transfer $x_{1}$ color TVs to $A_{2}$ High School (if $x_{1}$ is negative, it means $A_{2}$ High School transfers $|x_{1}|$ color TVs to $A_{1}$ High School. The same applies below), $A_{2}$ High School transfer $x_{2}$ color TVs to $A_{3}$ High School, $A_{3}$ High School transfer $x_{3}$ color... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,158 |
3. Given that the perimeter of a triangle is less than 15, the lengths of its three sides are all prime numbers, and one of the sides has a length of 3. How many such triangles are there?
(A) 4
(B) 5
(C) 6
(D) 7 | $3 . B$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 710,161 |
4. In $\triangle A B C$, it is known that $\angle B=60^{\circ}, \angle C=$ $75^{\circ}$, and the area of $\triangle A B C$ is $\frac{1}{2}(3+\sqrt{3})$. If $B C=$ $a$, then $a$ equals $(\quad)$.
(A) 1
(B) $\sqrt{2}$
(C) 2
(D) $\sqrt{3}$ | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,162 |
Example $8 . \triangle A B C$'s external angle bisector $A D$ intersects the extension of $B C$ at D. Prove:
$$
\begin{array}{c}
D B \cdot D C \\
D A^{2}=A B \cdot A C
\end{array}
$$ | Prove: Construct the circumcircle $\odot O$ of $\triangle ABC$ intersecting the extension of $DA$ at $E$, and connect $BE$.
$$
\begin{array}{l}
\because \angle 1=\angle 2=\angle 3, \angle 4=\angle E, \\
\therefore \triangle A B E \sim \triangle A D C, \\
\text { so } \frac{A B}{A D}=\frac{A E}{A C} \text {, thus } A B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,163 |
8. In an acute $\triangle A B C$, $2 \angle B=\angle C$. Then the relationship between $A B$ and $2 A C$ is ( ).
(A) $A B=2 A C$
(B) $A B2 A C$
(D) Cannot be determined | 8. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,167 |
2. If regardless of the value of $k$, $x=-1$ is always a solution to the equation $\frac{k x+a}{2}-\frac{2 x-b k}{3}=1$, then $a=$ $\qquad$ ,$b=$ $\qquad$ . | 2. $a=\frac{2}{3}, b=\frac{3}{2}$ | a=\frac{2}{3}, b=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,169 |
4. In Rt $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the altitude on the hypotenuse $A B$, and $C E$ is the angle bisector of $\angle C$. If $\frac{A E}{E B}=\frac{2}{3}$, then $\frac{A D}{D B}=$ $\qquad$ . | 4. $\frac{4}{9}$ | \frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,171 |
Four. (This question is worth 20 points) Let $a, b, c$ be distinct non-zero real numbers. Prove that the three equations
\[
\begin{array}{l}
a x^{2}+2 b x+c=0, \\
b x^{2}+2 c x+a=0, \\
c x^{2}+2 a x+b=0
\end{array}
\]
cannot all have two equal real roots. | Four, using proof by contradiction.
Suppose the three equations in the problem each have two equal real roots. Then we have
$$
\left\{\begin{array}{l}
\Delta_{1}=4 b^{2}-4 a c=0, \\
\Delta_{2}=4 c^{2}-4 a b=0, \\
\Delta_{3}=4 a^{2}-4 b c=0 .
\end{array}\right.
$$
Adding the three equations, we get
$$
a^{2}+b^{2}+c^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,173 |
Example 9 Given that $B I$ and $C E$ are the angle bisectors of $\triangle A B C$, $B D = C E$. Prove: $A B = A C$.
---
The translation maintains the original text's line breaks and format. | Prove: Construct the circumcircles $\odot O_{1}$ and $\odot O_{2}$ of $\triangle A B D$ and $\triangle A C E$, intersecting at $F$. Connect $F D$ and extend it to intersect $\odot O_{2}$ at $S$. Connect $F E$ and extend it to intersect $\odot O_{1}$ at $T$. Connect $A T$, $A S$, and $A F$. Let $\angle A=2 \alpha, \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,174 |
3. Given that $a, b, c$ are three real numbers not all zero. Then, the nature of the roots of the equation $x^{2}+(a+b+c) x+$ $\left(a^{2}+b^{2}+c^{2}\right)=0$ with respect to $x$ is ( ).
(A) has two negative roots
(B) has two positive roots
(C) has two real roots of opposite signs
(I) has no real roots | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,178 |
5. Given that $a$ is any real number. In the following questions, the number of correct conclusions is ( ).
(1) The solution to the equation $a x=0$ is $x=0$;
(2) The solution to the equation $a x=a$ is $x=1$;
(3) The solution to the equation $a x=1$ is $x=\frac{1}{a}$;
(4) The solution to the equation $|a| x=a$ is $x=... | $5 . A$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,180 |
6. Let $x$ be an acute angle, and satisfy $\sin x=3 \cos x$. Then $\sin x \cos x$ equals ( ).
(A) $\frac{1}{6}$
(B) $\frac{1}{5}$
(C) $\frac{2}{9}$
(i) $\frac{3}{10}$ | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 710,181 |
7. As shown in the figure, the diameter of circle ○A is equal to the side length of equilateral $\triangle A B C$, the perimeter of equilateral $\triangle A B^{\prime} C^{\prime}$ is the same as that of $\triangle A B C$, and $B^{\prime} C^{\prime}$ is tangent to $\odot A$. Then ( ).
(A) $\angle B^{\prime} A C^{\prime}... | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 710,182 |
Example 10 As shown in the figure, $B D$ is the angle bisector of $\triangle A B C$, $A B=A C$, $A D+B D>B C$. Prove: $\angle A<100^{\circ}$. | Proof: Construct the circumcircle $\odot O$ of $\triangle ABD$ intersecting $BC$ at $E$, and connect $IE$. Let $\angle 1 = \angle 2 = \alpha$, then $\angle 4 = 2\alpha$.
$$
\begin{array}{c}
\because AB = AC, \\
\therefore \angle C = \angle ABC = 2\alpha, \\
\text{ then } \angle 3 = 3\alpha, \angle 5 = 4\alpha, \angle 6... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,185 |
4. As shown in the figure, in the right triangle $\triangle A B C$, $A B$ is the hypotenuse, point $P$ is on the median $C D$, $A C=3 \mathrm{~cm}, B C=$ $4 \mathrm{~cm}$, let the distance between point $P$ and $C$ be $x \mathrm{~cm}$, and the area of $\triangle A P B$ be $y \mathrm{~cm}^{2}$. Then, the functional rela... | $\begin{array}{l}\text { 4. } y=-2.4 x \\ +6 \quad 0 \leqslant x<2.5\end{array}$
The translation is:
$\begin{array}{l}\text { 4. } y=-2.4 x \\ +6 \quad 0 \leqslant x<2.5\end{array}$
Note: The text is already in English, so no translation was necessary. The format and content are preserved as requested. | y=-2.4x + 6 \quad 0 \leqslant x < 2.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,188 |
8. As shown in the figure, the perimeter of square $A B C D$ is $40 \mathrm{~m}$. Two people, Jia and Yi, start from $A$ and $B$ respectively at the same time and walk along the edges of the square. Jia walks $35 \mathrm{~m}$ per minute in a counterclockwise direction, and Yi walks $30 \mathrm{~m}$ per minute in a cloc... | $\begin{array}{l}8 . \\ (6,10)\end{array}$ | (6,10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,192 |
Three, (Full marks 12 points) A, B, and C together solved 100 math problems. Each of them solved 60 of these problems. Problems solved by only one person are called difficult problems, and problems solved by all three are called easy problems. Try to determine: Are there more difficult problems or easy problems? By how... | Three, let the number of "easy questions" solved by all three people be $x$, and the number of "difficult questions" solved only by one person be $y_{1}, y_{2}, y_{3}$, respectively. Then the total number of "difficult questions" is $y=y_{1}+y_{2}+y_{3}$. From the problem and the diagram, we get the system of equations... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 710,193 |
$$
\begin{array}{l}
\text { Four, (Full marks } 12 \text { points) Given quadrilateral } A B C D \text {, from } \\
\text { (1) } A B / / D C \text {; (2) } A B=D C \text {; (3) } A D / / \\
B C \text {; (4) } A D=B C \text {; (5) } \angle A=\angle C \text {; (6) } \angle B= \\
\angle D \text {, choose two conditions t... | From the 6 conditions given in the question, any 2 conditions can be chosen, resulting in only $5+4+3+2+1=\frac{1}{2} \times 5(5+1)=15$ combinations. Among these, there are 9 scenarios that can deduce that quadrilateral $ABCD$ is a parallelogram:
(1) From (1) and (3): A quadrilateral with two pairs of opposite sides pa... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,194 |
Five. (Full marks 12 points) Prove that regardless of the value of $k$, the graph of the linear function
$$
(2 k-1) x-(k+3) y-(k-11)=0
$$
always passes through a fixed point. | Five, this problem is about a "pencil of lines," which can be solved by first finding the intersection coordinates of two special lines. Then, prove that all other lines must pass through this point. Since the given function is a linear function, we have $k+3 \neq 0$. By setting $k=1$ and $k=2$ respectively,
$$
\left\{... | (2,3) | Algebra | proof | Yes | Yes | cn_contest | false | 710,195 |
Example 1: Xiao Zhang is riding a bicycle on a road next to a double-track railway. He notices that every 12 minutes, a train catches up with him from behind, and every 4 minutes, a train comes towards him from the opposite direction. If the intervals between each train are constant, the speeds are the same, and both t... | Solution: Let the trains depart from the station ahead and behind Xiao Zhang every $x$ minutes, Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$, then
$$
\left\{\begin{array}{l}
4\left(v_{1}+v_{2}\right)=x v_{2}, \\
12\left(v_{2}-v_{1}\right)=x v_{2} .
\end{array}\right.
$$
Solving, we get $x=6$ (... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,196 |
Example 2: Person A and Person B start from points $A$ and $B$ respectively at the same time and walk towards each other. They meet at point $C$, which is 10 kilometers away from $A$. After meeting, both continue at the same speed, reach $B$ and $A$ respectively, and immediately return. They meet again at point $D$, wh... | Solution: Let the distance between $A$ and $B$ be $x$ kilometers, the speed of person A be $v_{1}$ kilometers/hour, and the speed of person B be $v_{2}$ kilometers/hour, then
$$
\left\{\begin{array}{l}
\frac{10}{v_{1}}=\frac{x-10}{v_{2}}, \\
\frac{x+3}{v_{1}}=\frac{2 x-3}{v_{2}} .
\end{array}\right.
$$
Eliminating the... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,197 |
Example 3 Let $H$ be the orthocenter of acute $\triangle A B C$, and let $A$ draw tangents $A P, A Q$ to the circle with $B C$ as its diameter, with points of tangency being $P, Q$ respectively. Prove that $P, H, Q$ are collinear.
(1996. China Mathematical Olympiad) | Proof: Let the midpoint of $BC$ be $O$. Connect $AO$ and $PQ$ intersecting at point $G$, then $AO \perp PQ$.
Let $AD, BE$ be the two altitudes of $\triangle ABC$.
Then $E$ lies on $\odot O$.
$$
\begin{array}{l}
\because \angle HEC=90^{\circ} \\
=\angle HDC,
\end{array}
$$
Therefore, $H, D, C, E$ are concyclic.
Thus, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,198 |
Example 4 A line $DE$ bisects the perimeter of $\triangle ABC$, and at the same time, line $DE$ bisects the area of $\triangle ABC$. Prove: line $DE$ passes through the incenter $O$ of $\triangle ABC$.
(3rd Henan Province Junior High School Mathematics Competition) | Proof: Let the inradius of $\triangle ABC$ be $r$. Therefore, the area of $\triangle ABC$ is
$$
\frac{1}{2}(AB + AC +
$$
$BC) \cdot r$.
From the given conditions, we have
$$
\begin{aligned}
BD + BE & =\frac{1}{2}(AB + AC + BC) . \\
\therefore S_{\triangle ABC} & =(BD + BE) \cdot r \\
& =2\left(S_{\triangle BOD} + S_{\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,199 |
Example 5 (Menelaus' Theorem Converse) On the sides $AB$, $BC$, $CA$ of $\triangle ABC$ (or their extensions), take points $X$, $Y$, $Z$ respectively, and if $\frac{AX}{XB} \cdot \frac{BY}{YC} \cdot \frac{CZ}{ZA}=1$ holds. Then $X$, $Y$, $Z$ are collinear. | Proof: Let line $A B$ intersect $Y Z$ at $X'$. By Menelaus' theorem, we have
$$
\begin{array}{l}
\overline{A X^{\prime}} \overline{E^{\prime}} \cdot \frac{B Y}{Y C} \cdot \frac{C Z}{Z A} \\
=1 \\
\because \frac{A X}{X B} \cdot \frac{E Y}{Y C} \cdot \frac{C Z}{Z A} \\
=1, \\
\therefore \frac{A X^{\prime}}{X^{\prime} B}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,200 |
Example 6 In trapezoid $ABCD$, the legs $AB = CD$. Triangle $\triangle ABC$ is rotated around point $C$ to obtain $\triangle A'B'C$. Prove: The midpoints of segments $A'D$, $BC$, and $B'C$ are collinear.
(1989, All-Soviet Union Mathematical Olympiad) | Proof: Let $K$ and $M$ be the midpoints of $B'C$ and $BC$ respectively. From
$$
\begin{array}{l}
B'K = \frac{1}{2} BC \\
= CM, \\
B'A' = BA \\
= CD, \\
\angle B' = \angle B = \angle DCM,
\end{array}
$$
we get $\triangle CMD \cong \triangle B'KA'$.
$$
\therefore A'K = DM, \angle CMD = \angle B'KA'.
$$
In the isosceles... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,201 |
Example 7 As shown in the figure, in $\triangle ABC$, $X$, $Y$, and $Z$ are points on the extensions of sides $BC$, $CA$, and $AB$, respectively, and $XA$, $YB$, and $ZC$ are tangents to the circumcircle of $\triangle ABC$. Prove that $X$, $Y$, and $Z$ are collinear. | Given:
$$
\begin{array}{l}
\angle X A B=\angle X C A, \angle X=\angle X, \\
\therefore \triangle A X B \backsim \triangle C X A .
\end{array}
$$
Therefore, $\frac{B X}{X C}=\frac{S_{\triangle A X B}}{S_{\triangle C X A}}=\left(\frac{A B}{A C}\right)^{2}$.
Similarly, $\frac{C Y}{Y A}=\left(\frac{B C}{B A}\right)^{2}, \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,202 |
Example 8 Three
congruent circles have a common point $O$, and are all inside a given triangle, with each circle being tangent to two sides of the triangle. Prove: The incenter, circumcenter, and point $O$ of this triangle are collinear. | Proof: Let the incenter of $\triangle ABC$ be $I$, and the circumcenter be $K$. The centers of the three circles are $A^{\prime}, B^{\prime}, C^{\prime}$. Since $\odot A^{\prime}$ is tangent to $AB$ and $AC$, $A^{\prime}$ lies on the angle bisector $IA$ of $\angle BAC$. Similarly, $B^{\prime}$ lies on $IB$, and $C^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,203 |
Example 9 Given that $P, R$ are two points on ray $AX$, $Q, S$ are two points on ray $BY$, and it satisfies $\frac{AP}{BQ}=\frac{AR}{RS}=\lambda, M, N, T$ are points on segments $AB, PQ, RS$ respectively, and $\frac{AM}{MB}=\frac{PN}{NQ}=\frac{RT}{TS}=k$. What is the positional relationship of points $M, N, T$?
(1994, ... | Solution: As shown in the figure, establish a rectangular coordinate system, and let $\dot{x}_{D}$ and $y_{D}$ represent the x-coordinate and y-coordinate of point $D$, respectively.
According to the problem, let $\frac{A R}{A P}=\frac{B S}{B Q}=u$. Also, let $x_{P}=x_{A} +a_{1}, y_{P}=y_{A}+b_{1}, x_{Q}=x_{B}+a_{2}, ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 710,204 |
Example 10 (Euler's Theorem) Prove: In any triangle, the circumcenter, centroid, and orthocenter are collinear, and the distance from the circumcenter to the centroid is half the distance from the centroid to the orthocenter. | Proof: As shown in the figure, let the circumcenter $O$ of $\triangle ABC$ be the starting point, then the position vector of the centroid $G$ is $\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+$ $\overrightarrow{O B}+\overrightarrow{O C})$.
Extend $BO$ to intersect the circumcircle of $\triangle ABC$ at point $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 710,205 |
Example 1 Given $\sin x + \cos x = \frac{1}{5}$, and $0 \leqslant x < \pi$. Then the value of $\tan x$ is $\qquad$
(1993, Sichuan Province High School Mathematics Competition) | Solution: Since $\sin x + \cos x = 2 \cdot \frac{1}{10}$, then $\sin x$, $\frac{1}{10}$, and $\cos x$ form an arithmetic sequence. Therefore, we set
$$
\left\{\begin{array}{l}
\sin x = \frac{1}{10} - d, \\
\cos x = \frac{1}{10} + d.
\end{array}\right.
$$
where $-\frac{7}{10} \leqslant d \leqslant \frac{1}{10}$.
Substi... | -\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,206 |
Example 2 Find three real numbers $x, y, z$ such that they simultaneously satisfy the following equations:
$$
\begin{array}{l}
2 x+3 y+z=13, \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82 .
\end{array}
$$
(1992, Friendship Cup International Mathematics Competition) | Solution: From $(1)$, we get $2 x+3 y=2 \cdot \frac{13-z}{2}$. Then $2 x, \frac{13-z}{2}, 3 y$ form an arithmetic sequence. Therefore, let
$$
2 x=\frac{13-z}{2}+d, 3 y=\frac{13-z}{2}-d .
$$
Substitute into (2), and simplify and complete the square, to get
$3(z-4)^{2}+4\left(d-\frac{3}{2}\right)^{2}=0$.
$\therefore z=4... | x=3, y=1, z=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,207 |
Example 3 A person walks from place A to place B, and there are regular buses running between A and B, with equal intervals for departures from both places. He notices that a bus going to A passes by every 6 minutes, and a bus going to B passes by every 12 minutes. How often do the buses depart from their respective st... | Analysis: Let the distance between two consecutive buses in the same direction be $s$, and the time be $t$ minutes, then the speed of the bus is $\frac{s}{t}$.
(1) A person's speed relative to the bus going to location A is $\frac{s}{6}$, so the person's speed is $\frac{s}{6}-\frac{s}{t}$.
(2) A person's speed relative... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 710,208 |
Example 3 Given $\sin \varphi \cdot \cos \varphi=\frac{60}{169}$, and $\frac{\pi}{4}<\varphi$ $<\frac{\pi}{2}$. Find the values of $\sin \varphi$ and $\cos \varphi$. | Solution: Since $\sin \varphi \cdot \cos \varphi=\left(\frac{2}{13} \sqrt{15}\right)^{2}$, then $\sin \varphi$, $\frac{2}{13} \sqrt{15}$, and $\cos \varphi$ form a geometric sequence. Therefore, let
$$
\sin \varphi=\frac{2}{13} \sqrt{15} q, \cos \varphi=\frac{2}{13 q} \sqrt{15}(1<q
$$
$<2$ ).
Substituting into $\sin ^{... | \sin \varphi=\frac{12}{13}, \cos \varphi=\frac{5}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,209 |
Example 4 Solve the equation
$$
\sqrt{x^{2}+x+1}-\sqrt{x^{2}}+7 x+5=3 x+2
$$ | Given: $\because \sqrt{x^{2}+x} \div 1, \frac{3 x+2}{2}$, $-\sqrt{x^{2}+7 x+5}$ form an arithmetic sequence, we set
$$
\left\{\begin{array}{l}
\sqrt{x^{2}+x+1}=\frac{3 x+2}{2}-d, \\
\sqrt{x^{2}+7 x+5}=-\frac{3 x+2}{2}-d .
\end{array}\right.
$$
$(1)^{2}-(2)^{2}$, simplifying gives
$$
\begin{array}{l}
-2(3 x+2)=-2(3 x+2)... | x=-\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,210 |
Example 5 Solve the equation $\sqrt{x-1}+\sqrt[3]{2-x}=1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text provided is already in English, here is the equivalent text, maintaining the format:
Example 5... | Solution: Since $\sqrt{x-1}, \frac{1}{2}, \sqrt[3]{2-x}$ form an arithmetic sequence, we set
$$
\left\{\begin{array}{l}
\sqrt{x-1}=\frac{1}{2}-d \\
\sqrt[3]{2-x}=\frac{1}{2}+d .
\end{array}\right.
$$
$(1)^{2}+(2)^{3}$, we get
$$
1=\left(\frac{1}{2}-d\right)^{2}+\left(\frac{1}{2}+d\right)^{3} \text {. }
$$
Simplifying ... | x_{1}=1, x_{2}=2, x_{3}=10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,211 |
Example 6 Solve the system of equations
$$
\left\{\begin{array}{l}
2 x y-5 \sqrt{x y+1}=10, \\
x^{2}+y^{2}=34 .
\end{array}\right.
$$
(1984, Nanjing City Mathematics Competition) | $2(x y+1)-5 \sqrt{x y+1-12}=0.$
Let $\sqrt{x y+1}=t(\geqslant 0)$, then $2 t^{2}-5 t-12=0$.
Solving, we get $t_{1}=4, t_{2}=-\frac{3}{2}$ (discard).
From $t_{1}=4$, we get $x y=15=(\sqrt{15})^{2}$, so $x$, $\sqrt{15}$, $y$ form a geometric sequence. Therefore, let $x=\sqrt{15} q, y=\frac{\sqrt{15}}{q}$, substituting in... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 710,212 |
Example 7 Given real numbers $a$, $b$, $c$ satisfy $a=6-b$, $c^{2}=ab-9$. Prove: $a=b$.
(1983, Tianjin City Mathematics Competition) | Proof: Since $a+b=2 \times 3$, then $a, 3, b$ form an arithmetic sequence. Therefore, let $a=3-d, b=3+d$. Substituting into $c^{2}=a b - 9$, we get $c^{2}+d^{2}=0$.
$$
\therefore c=0, d=0 \text {. }
$$
Thus, $a=3, b=3$. Hence $a=b$. | a=b | Algebra | proof | Yes | Yes | cn_contest | false | 710,213 |
Example 8 Given real numbers $a, b$ satisfy $ab=10, a^{2}+$ $b^{2}=c^{2}$. Prove: $|c| \geqslant 2 \sqrt{5}$. | Proof: Given $ab=10$, we have $ab=(\sqrt{10})^{2}$. Therefore, $a$, $\sqrt{10}$, and $b$ form a geometric sequence. Hence, let $a=\sqrt{10} q, b=\frac{1}{q} \sqrt{10}$. Substituting into $a^{2}+b^{2}=c^{2}$ and simplifying, we get
$$
\begin{array}{l}
10 q^{4}-c^{2} q^{2}+10=0 . \\
\because q^{2} \in R^{+}, \\
\therefor... | |c| \geqslant 2 \sqrt{5} | Algebra | proof | Yes | Yes | cn_contest | false | 710,214 |
Example 9 Prove: $\sin ^{10} x+\cos ^{10} x \geqslant \frac{1}{16}$. | Proof: $\because \sin ^{2} x+\cos ^{2} x=2 \times \frac{1}{2}$,
$\therefore \sin ^{2} x, \frac{1}{2}, \cos ^{2} x$ form an arithmetic sequence. Therefore, let
$$
\begin{array}{l}
\sin ^{2} x=\frac{1}{2}-d, \\
\cos ^{2} x=\frac{1}{2}+d\left(-\frac{1}{2} \leqslant d \leqslant \frac{1}{2}\right) \\
\therefore \sin ^{10} x... | \frac{1}{16} | Inequalities | proof | Yes | Yes | cn_contest | false | 710,215 |
Example 10 Assume $x, y, z$ are all real numbers, $a \geqslant 0$, and they satisfy
$$
\begin{array}{l}
x+y+z=a, \\
x^{2}+y^{2}+z^{2}=\frac{1}{2} a^{2} .
\end{array}
$$
Prove: $x, y, z$ are all non-negative and none of them can be greater than $\frac{2}{3} a$.
(1957, Beijing Mathematical Competition) | Proof: From (1), we get $x+y=2 \cdot \frac{a-z}{2}$, then $x$, $\frac{a-z}{2}$, $y$ form an arithmetic sequence, hence
$$
x=\frac{a-z}{2}-d, y=\frac{a-z}{2}+d .
$$
Substituting into (2), simplifying yields
$$
\begin{array}{l}
3 z^{2}-2 a z=-4 d^{2} . \\
\therefore 3 z^{2}-2 a z \leqslant 0 .
\end{array}
$$
Solving gi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 710,216 |
Can 2123456 be split into the sum of
(1) 64; (2) 128
consecutive natural numbers? | Solution: (1) When $r=64$, since $\frac{2 n}{r}=2 \times 123456 \div 64=3858$ has the same parity as $r=64$, 123456 cannot be decomposed into the sum of 64 consecutive natural numbers; $\square$
(2) When $r=128$, since $\frac{2 n}{r}=2 \times 123456 \div 128=1929$ has a different parity from $r=64$, and $2 \times 1234... | 123456=901+902+\cdots+1028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,220 |
Example 3 Does there exist a positive integer $n$ that satisfies the following two conditions simultaneously:
(1) $n$ can be decomposed into the sum of 1990 consecutive positive integers.
(2) $n$ has exactly 1990 ways to be decomposed into the sum of several (at least two) consecutive positive integers.
(31st IMO Short... | Solution: Suppose $n$ is a natural number that meets the conditions. According to Theorem 3, condition (1) is equivalent to:
(1') $1990+\frac{2 n}{1990}$ is odd, i.e., $\frac{n}{5 \times 199}$ is odd, and $2 n \geqslant 1990 \times 1991$.
By Theorem 2, condition (2) is equivalent to:
(2) $n$ has exactly 1990 distinct o... | n=5^{180} \times 199^{10}, n=5^{10} \times 199^{180} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 710,221 |
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