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II. (Total 50 points) For positive integers $a, b$, define the operation $a * b$ as follows:
(1) $a * 1 = 1 * a = a$,
(2) $a * (b+1) = a * b + a$.
Prove that for any positive integer $c$, we have
$$
(a+b) * c = a * c + b * c \text{. }
$$ | For a positive integer $c$, we use mathematical induction.
When $c=1$, by operation (1), we have
$$
(a+b) * 1=a+b=a * 1+b * 1 \text {, }
$$
the proposition holds.
Assume that the proposition holds for $c=k$, i.e.,
$$
(a+b) * k=a * k+b * k .
$$
Then, by operation (2) and the induction hypothesis, we have
$$
\begin{arr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,531 |
Three, (Total 50 points) To welcome the year 2000, a certain place organized a table tennis lucky spring tournament. First, 2000 players were selected through ID numbers, numbered 1, 2, $\cdots, 2000$. Any two of them can form a doubles team, and the sum of the numbers of each team is called their "sum number." It is s... | Three, due to the pairwise subtraction from 1 to 2000 (here it is specified that the larger number minus the smaller number), there are only 1999 numbers from 1 to 1999 (making 1999 drawers). And any 65 numbers $a_{1}, a_{2}, \cdots, a_{65}$ (it is specified that $a_{i}a_{\mathrm{s}}, a_{i}>a_{i}$, such that
$$
\left\{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,532 |
Initial 89. Given that $a, b, c$ are positive numbers. Prove:
$$
\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2 .
$$ | Proof: $\because(b+c-a)^{2} \geqslant 0$,
$$
\begin{array}{l}
\therefore(a+b+c)^{2} \geqslant 4 a(b+c), \\
\therefore \frac{1}{b+c} \geqslant \frac{4 a}{(a+b+c)^{2}}, \\
\therefore \sqrt{\frac{a}{b+c} \geqslant \frac{2 a}{a+b+c} .}
\end{array}
$$
Similarly, we can get $\sqrt{\frac{b}{c+a}} \geqslant \frac{2 b}{a+b+c}$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,533 |
In $\triangle A B C$, $\angle A B C=70^{\circ}, \angle A C B=$ $30^{\circ}, P, Q$ are two points inside the triangle. $\angle Q B C=\angle Q C B=10^{\circ}$, $\angle P B Q=\angle P C B=20^{\circ}$. Prove: $A, P, Q$ are collinear. | Proof: As shown in the figure, on the extension of $BA$: take a point $D$ such that $DC = BC$. Draw a perpendicular from $B$ to $AC$ intersecting $DC$ at $E$. Let $PC$ intersect $BE$ at $F$. Connect $AF$, $AE$, $EP$, and $AP$.
It is easy to see that $CP \perp AB$ and $BF \perp AC$, so $F$ is the orthocenter of $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,534 |
Let $a, b, c$ be the lengths of the three sides of a triangle, $n \in \mathbf{N}, n \geqslant$
2. Prove:
$$
\frac{\sqrt[n]{a^{2}+b^{2}}+\sqrt[n]{b^{2}+c^{2}}+\sqrt[n]{a^{2}+c^{2}}}{\sqrt[n]{(a+b+c)^{2}}}>\frac{2+\sqrt[n]{2}}{\sqrt[n]{4}} .
$$ | Proof: Let \( A = \sqrt[n]{a^2 + b^2} + \sqrt[n]{b^2 + c^2} + \sqrt[n]{c^2 + a^2} \).
Using the arithmetic-geometric mean inequality, we have
\[
\begin{array}{l}
\frac{\sqrt[n]{2}}{2}\left(a^2 + b^2\right) + (n-1) \cdot \sqrt[n]{2}\left(\frac{A}{2 + \sqrt[n]{2}}\right)^n \\
\geqslant n \cdot \sqrt[n]{a^2 + b^2} \cdot \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,535 |
Example 1 Let $\{x\}=x-[x]$.
(1) Find a real number $x$ that satisfies
$$
\{x\}+\left\{\frac{1}{x}\right\}=1 \text{; }
$$
(2) Prove that any $x$ satisfying the above equation is not a rational number.
(1990, National Junior High School Mathematics Competition) | Explanation: How to find a real number $x$, given the hint (2), $x$ cannot be a rational number. Through experimentation, when $x$ is a rational number, $\{x\} + \left\{\frac{1}{x}\right\} = 1$ does not hold. Then, what kind of irrational number should we find? If you have a good understanding of "numbers," you would n... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,537 |
Example 2 Let $n$ be a given positive integer greater than 1. Prove that there exists a unique positive integer $A < n^{2}$ such that
$$
n \left\lvert\,\left[\frac{n^{2}}{A}\right]+1\right.
$$ | Explanation: The expression $\left[\frac{n^{2}}{A}\right]$ in the problem needs to be understood first. In fact, $\left[\frac{n^{2}}{A}\right]$ is the quotient of $n^{2}$ divided by $A$, that is,
$$
n^{2}=\left[\frac{n^{2}}{A}\right] A+r, 0 \leqslant r < A.
$$
If $r \geqslant 2$, then $n^{2} \geqslant \left[\frac{n^{2}... | A=n+1 | Number Theory | proof | Yes | Yes | cn_contest | false | 711,538 |
Example 3: If $a^{2}+b^{2}+c^{2}+3 \leqslant a b+3 b+2 c$, find the values of $a$, $b$, and $c$.
(1992, Hungarian Mathematical Olympiad) | Solution: The original inequality is equivalent to
$$
\left(a-\frac{b}{2}\right)^{2}+\frac{3}{4}(b-2)^{2}+(c-1)^{2} \leqslant 0 .
$$
By the property of non-negative numbers, we have
$$
a-\frac{b}{2}=b-2=c-1=0 \text {. }
$$
Solving, we get $a=1, b=2, c=1$. | a=1, b=2, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,539 |
Prove: If and only if there exists a positive integer $k$, such that $n=2^{k-1}$, then $2^i$ can divide $n!$.
(1985, Canadian Mathematical Olympiad) | Explanation: Solving this problem hinges on two key points: First, the issue of $n=2^{k-1}$. For any given positive integer $n$, there always exists a $k$ such that $2^{k-1} \leqslant n < 2^{k}$. In this problem, $n=2^{k-1}$, so $2^{k-1}$ is precisely an integer that satisfies this inequality. Therefore, the inequality... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,540 |
Example 5 Proof: For any natural number $k(k \geqslant 2)$, there exists an irrational number $r$, such that for every natural number $m$,
$$
\left[r^{m}\right] \equiv-1(\bmod k),
$$
(1987, IMO Shortlist) | Consider the conclusion $\left[r^{m}\right] \equiv-1(\bmod k)$. This is equivalent to $\left[r^{m}\right]+1 \equiv 0(\bmod r)$. The floor function makes it inconvenient to think about. Note the following fact: for any natural number $k(k \geqslant 2)$, such integers $p, q$ exist:
\[
\left\{\begin{array}{l}
r=\frac{p+\s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,542 |
2. Prove: If the positive integer $A$ is not a perfect square, then there exists a natural number $n$, such that $A=\left[n+\sqrt{n}+\frac{1}{2}\right]$. | Let $n=A-[\sqrt{A}]$, prove $A=\left[n+\sqrt{n}+\frac{1}{2}\right]$. In fact, let $[\sqrt{A}]=x$, then $x^{2}<A<(x+1)^{2}, x^{2}-x+1 \leqslant A-x \leqslant x^{2}+x$, that is
$$
\begin{array}{l}
\sqrt{x^{2}-x+1}+\frac{1}{2} \leqslant \sqrt{A-x}+\frac{1}{2} \\
\leqslant \sqrt{x^{2}+x}+\frac{1}{2} .
\end{array}
$$
From ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,544 |
3. (1) Prove: There exists a positive real number $\lambda$, such that for any positive integer $n, \left[\lambda^{n}\right]$ has the same parity as $n$;
(2) Find a positive real number $\lambda$ that satisfies (1). | (提示: If $\lambda$ is the positive root of the equation $x^{2}-\alpha x-b=0$, where $\alpha$ is a positive odd number, $b$ is a positive even number, and $b<\alpha$, then this problem can be proven. In fact, let $S_{n}=\lambda^{n}+\mu^{n},-1<\mu<0$, then we have $S_{n+2}-$ $\alpha S_{n+1}-b S_{n}=0$. Thus, $S_{n+2}$ has... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,545 |
Example 1 As shown in Figure 1, two equilateral triangles $\triangle ABC$ and $\triangle A_{1} B_{1} C_{1}$ (vertices arranged clockwise) are placed on a plane, and the midpoints $D$ of sides $BC$ and $B_{1} C_{1}$ coincide. Try to find:
(1) the angle between $A A_{1}$ and $B B_{1}$; (2) the ratio $A A_{1}: B B_{1}$. | \begin{array}{l}\text { Sol: } \because B B_{1} \xrightarrow{S\left(D, 90^{\circ}, \sqrt{3}\right)} A A_{1}, \\ \therefore A A_{1} \perp B B_{1} \text { and } A A_{1}: B B_{1}=\sqrt{3} .\end{array} | A A_{1} \perp B B_{1} \text{ and } A A_{1}: B B_{1}=\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,546 |
Example 2 As shown in Figure $2, M$ is the midpoint of side $BC$ of $\triangle ABC$, and $\triangle ABD \backsim \triangle ACM$. Prove: $DM \parallel AC$.
保留源文本的换行和格式，直接输出翻译结果。 | Proof: Let $E$ be the symmetric point of $M$ with respect to $AC$, then $\triangle A B D \backsim \triangle A C E$.
$$
\text{Let } \angle B A D = \angle C A E = \theta, \frac{A D}{A B} = \frac{A E}{A C} = k \text{.}
$$
We have
$$
B C \xrightarrow{\mathrm{S}(\mathrm{A}, \theta, k)} D E,
$$
the midpoint $M$ of $B C$ bec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,547 |
Example 3 If $\triangle XYZ$ is similar to $\triangle ABC$, and $X$ is on $BC$, $Y$ is on $AC$, $Z$ is on $AB$. Prove that the orthocenter of $\triangle XYZ$ is the circumcenter of $\triangle ABC$.
(31st IMO Shortlist) | Proof: As shown in Figure 3, let \( X', Y', Z' \) be the midpoints of the three sides of \( \triangle ABC \). It is easy to prove that \( \triangle X'Y'Z' \sim \triangle ABC \). Let \( O \) be the orthocenter of \( \triangle X'Y'Z' \). Since \( OX' \perp Y'Z' \) and \( Y'Z' \parallel BC \), then \( OX' \perp BC \). Als... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,548 |
Example 4 As shown in Figure 4, let $\triangle ABC$ be an acute triangle. Construct isosceles right triangles $\triangle BCD$, $\triangle ABE$, and $\triangle CAF$ outside $\triangle ABC$, where $\angle BDC$, $\angle BAE$, and $\angle CFA$ are right angles. Construct an isosceles right triangle $\triangle EFG$ outside ... | Proof: Let the midpoint of $BE$ be $H$, then $AD$ after a spiral similarity transformation $S\left(B, 45^{\circ}, \sqrt{2}\right)$ becomes $EC$. $EC$ after a spiral similarity transformation $S\left(A, 45^{\circ}, \frac{1}{2}\right)$ becomes $HF$. $HF$ after a spiral similarity transformation $S\left(E, 45^{\circ}, \sq... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,549 |
Example 4 Given $a+x^{2}=1991, b+x^{2}=$ $1992, c+x^{2}=1993$, and $a b c=24$. Find the value of the algebraic expression $\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$. | ```
Solution: Eliminating $x^{2}$ from the known conditions, we get $a-b=$
$$
\begin{array}{l}
-1, b-c=-1, a-c=-2 . \text { Then } \\
\text { the original expression }=\frac{a^{2}+b^{2}+c^{2}-a b-b c-a c}{a b c} \\
= \frac{(a-b)^{2}+(b-c)^{2}+(a-c)^{2}}{2 a b c} \\
= \frac{(-1)^{2}+(-1)^{2}+(-2)^{2}}{2 \times 24}=\frac... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,550 |
Example 6 As shown in Figure 6, on the sides of any $\triangle ABC$, construct $\triangle BPC$, $\triangle QAC$, and $\triangle ARB$ outward such that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, and $\angle ABR = \angle BAR = 15^{\circ}$. Prove:
(1) $\angle QRP = 90^{\circ}$;
(2) $QR... | Proof: Let the other vertex of the triangle constructed outside $\triangle ABC$ with $AB$ as a side be $D$, and connect $RD$. Then
$\triangle BCP \backsim \triangle BDR \cong \triangle ADR \backsim \triangle ACQ$.
Thus, $RP$ undergoes a homothetic rotation transformation $S\left(B, 45^{\circ}, \frac{\sin 105^{\circ}}{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,552 |
Example 7 As shown in Figure 7, given $\angle P A D=\angle E A B$,
$$
\begin{array}{l}
\angle A B E=\angle C B Q, \\
\angle Q C B=\angle E C D, \\
\angle C D E=\angle A D P, \\
\angle A E B=\angle C E D . \text { Find }
\end{array}
$$
Proof:
(1) $P 、 E 、 Q$ are collinear;
$$
\text { (2) } P E: E Q=\sin Q: \sin P \text... | Proof: Let $\angle P A D=\alpha, \angle A B E=\beta, \angle Q C B$ $=\gamma, \angle C D E=\delta, \angle A E B=\theta$, then $P E$ undergoes a rotational transformation $S\left(A,-\alpha, \frac{\sin P}{\sin 0}\right.$; it becomes $D F$, such that $\angle E F A=\delta$. Here, $F$ lies on the line $A B$, and $D F$ underg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,553 |
Example 8 On the plane: there is an $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$. The area is $S$, and there is another point $M$. $M$ rotates around $A$ by an angle $\alpha$ to get points $M_{i}(i=1,2, \cdots, n)$. Find the area of the $n$-sided polygon $M_{1} M_{2} \cdots M_{n}$.
(30th IMO Preliminary Problem) | Prove: As shown in Figure 8, in the isosceles $\triangle A_{i} M M_{i}$, since
$$
\angle A_{i} M M_{i}=\frac{180^{\circ}-\alpha}{2} \text {, }
$$
$\frac{M M_{i}}{M A_{i}}=2 \sin \frac{\alpha}{2}$, it can be seen that point $A_{i}$ undergoes a homothetic rotation transformation $S\left(M,-\frac{180^{\circ}-\alpha}{2}, 2... | 4 S \sin ^{2} \frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,554 |
Example 9 (1) Let $\triangle A B C$ be rotated by a small angle of $180^{\circ}$ around the center of its circumcircle to obtain $\triangle A_{1} B_{1} C_{1}$. The corresponding sides $A B$ and $A_{1} B_{1}$ intersect at point $C_{2}$, $B C$ and $B_{1} C_{1}$ intersect at point $A_{2}$, and $C A$ and $C_{1} A_{1}$ inte... | Proof: As shown in Figure 9, if the chord $Q_{1} Q_{2}$ of circle $\odot O$ is obtained by rotating the chord $P_{1} P_{2}$ around point $O$ by an angle $\alpha$, then the intersection point $K$ of the lines $P_{1} P_{2}$ and $Q_{1} Q_{2}$ can be obtained by the spiral similarity transformation $S\left(O, \frac{\alpha}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,555 |
Example 1 Let $a_{i}$ and $b_{i} (i=1,2, \cdots, n)$ be positive real numbers, and $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$. Prove that:
$$
\sum_{i=1}^{n}\left(\frac{a_{i}^{2}}{a_{i}+b_{i}}\right) \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i} \text {. }
$$
(1991, Asia Pacific Mathematical Olympiad) | Prove: When $a_{i}=b_{i}$, the inequality holds as an equality, and $\frac{a_{i}}{\sqrt{a_{i}+b_{i}}}=\frac{\sqrt{a_{i}+b_{i}}}{2}$. Construct the inequality
$$
\left[\frac{a_{i}}{\sqrt{a_{i}+b_{i}}}-\frac{\sqrt{a_{i}+b_{i}}}{2}\right]^{2} \geqslant 0,
$$
i.e., $\frac{a_{i}^{2}}{a_{i}+b_{i}}+\frac{b_{i}-3 a_{i}}{4} \g... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,556 |
Example 2 Proof: For all positive real numbers $a, b, c$,
$$
\begin{array}{l}
\left(a^{3}+b^{3}+a b c\right)^{-1}+\left(b^{3}+c^{3}+a b c\right)^{-1} \\
+\left(c^{3}+a^{3}+a b c\right)^{-1} \leqslant(a b c)^{-1} .
\end{array}
$$
(1997, 26th United States of America Mathematical Olympiad) | Proof: Left side of the original equation
$$
\begin{aligned}
= & {\left[(a-b)\left(a^{2}-b^{2}\right)+a^{2} b+a b^{2}+a b c\right]^{-1} } \\
& +\left[(b-c)\left(b^{2}-c^{2}\right)+b^{2} c+c b^{2}+a b c\right]^{-1} \\
& +\left[(c-a)\left(c^{2}-a^{2}\right)+c^{2} a+c a^{2}+a b c\right]^{-1} \\
\leqslant & \left(a^{2} b+a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,557 |
Example $\mathbf{4}$ Let $a, b, c \in \mathbf{R}^{+}$. Prove that:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} \text {. }
$$
(1963, Moscow Mathematical Olympiad) | Proof: Assuming $a \geqslant b \geqslant c>0$, then
$\frac{1}{b+c} \geqslant -\frac{1}{c+}-\geqslant \frac{1}{a+b}$.
Thus, by the rearrangement inequality, we have
(ordered sum) $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\geqslant \frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$ (disordered sum),
$\frac{a}{b+c}+\frac{b}{c+... | \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,559 |
Example 5 (W $\mathrm{W}$ Janous Conjecture) Let $x, y, z \in \mathbf{R}^{+}$. Prove that: $\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \geqslant 0$. | Prove: Let the left side of the original inequality be \( M \), \( z+x=a \), \( x+y=b \), \( y+z=c \), then \( y-x=c-a \), \( z-y=a-b \), \( x-z=b-c \), \( a, b, c \in \mathbf{R}^{+} \). Therefore,
\[
\begin{aligned}
& M=\frac{b(c-a)}{a}+\frac{c(a-b)}{b}+\frac{a(b-c)}{c} \\
= & \frac{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,560 |
Example 7 Let $\alpha, \beta$ both be acute angles. Prove:
$$
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \geqslant 9 \text {. }
$$
For what values of $\alpha, \beta$ does equality hold?
(1979, National High School Mathematics Contest) | Proof: From $\sin ^{2} \beta+\cos ^{2} \beta=1$, by splitting the left side of the inequality, we get
$$
\begin{array}{l}
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \\
=\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \cos ^{2} \beta}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \b... | \alpha=\arctan \sqrt{2}, \beta=\frac{\pi}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,562 |
Example 9 Let $\alpha, \beta, \gamma$ be acute angles, and $\sin ^{2} \alpha+$
$$
\begin{array}{l}
\sin ^{2} \beta+\sin ^{2} \gamma=1 \text{. Then } \\
\quad \frac{\sin ^{3} \alpha}{\sin \beta}+\frac{\sin ^{3} \beta}{\sin \gamma}+\frac{\sin ^{3} \gamma}{\sin \alpha} \geqslant 1 .
\end{array}
$$ | Proof: The left side of the inequality is
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\frac{\sin ^{4} \alpha}{\sin \alpha \cdot \sin \beta}+\frac{\sin ^{4} \beta}{\sin \beta \cdot \sin \gamma}+\frac{\sin ^{4} \gamma}{\sin \gamma \cdot \sin \alpha}\right]} \\
\cdot(\sin \alpha \cdot \sin \beta+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,564 |
Example 11 Given any 13 real numbers. Prove that there exist at least two real numbers (denoted as $x, y$) satisfying
$$
0 \leqslant \frac{x-y}{1+x y} \leqslant 2-\sqrt{3} \text {. }
$$ | Proof: Let any 13 real numbers be denoted as $\tan \theta_{i}$, $\theta_{i} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(i=1,2, \cdots, 13)$. If we let these two angles be $\alpha, \beta(\alpha=\beta)$, then $0<\alpha-\beta \leqslant \frac{\pi}{12}$. Let $x=\tan \alpha, v=\tan \beta$, then $\tan (\alpha-\beta)=\frac{x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,566 |
Example 14 Let $x, y \in \mathbf{R}$, H satisfies $|x|<1,|y|$ $<1$. Prove: $\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geqslant \frac{2}{1-x y}$. | Proof: Since $|x|<1,|y|<1$. Therefore,
$$
0<x^{2}<1,0<y^{2}<1,|x y|<1 \text {. }
$$
By the sum formula of an infinite geometric series $S = \frac{a_{1}}{1-q}$, we derive the series. Then, summing up, we have
$$
\begin{array}{l}
\frac{1}{1-x}+\frac{1}{1-y^{2}} \\
=\left(1+x^{2}+x^{4}+\cdots\right)+\left(1+y^{2}+y^{4}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,569 |
Example 15 Given $a_{1} , a_{2} \in \mathbf{R}^{+}$. Prove:
$$
\sqrt{1+\left(\frac{a_{1}+a_{2}}{2}\right)^{2}} \leqslant \frac{\sqrt{1+a_{1}^{2}}+\sqrt{1+a_{2}^{2}}}{2} .
$$ | Prove: The original inequality is equivalent to the inequality
$$
\begin{array}{l}
\sqrt{1+\left(\frac{a_{1}+a_{2}}{2}\right)^{2}}+\sqrt{1+\left(\frac{a_{1}+a_{2}}{2}\right)^{2}} \\
\leqslant \sqrt{1+a_{1}^{2}}+\sqrt{1+a_{2}^{2}} .
\end{array}
$$
When $a_{1}=a_{2}$, the equality holds.
Thus, the left side is the minim... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,570 |
Example 6 If $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=$ $\frac{c+a}{b}$, then $\frac{(a+b)(b+c)(c+a)}{a b c}=$ $\qquad$
(1985, Hangzhou City Junior High School Mathematics Competition) | Let $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=k$. Then $a+b=c k, b+c=a k, c+a=b k$.
We have $2(a+b+c)=k(a+b+c)$.
If $a+b+c \neq 0$, then $k=2$, the original expression $=8$;
If $a+b+c=0$, then $a+b=-c, k=$ -1. The original expression $=-1$. | 8 \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,571 |
Example 16 For any positive numbers $x, y$, the inequality holds
$$
\frac{1}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}} \geqslant \frac{2^{\frac{5}{4}}}{\sqrt[4]{x+y}}.
$$
(18th All-Russian High School Mathematics Competition, Grade 10) | Prove: The original inequality is equivalent to the inequality
$$
\sqrt[4]{\left(\frac{\sqrt{x+y}}{\sqrt{x}}\right)^{2}}+\sqrt[4]{\left(\frac{\sqrt{x+y}}{\sqrt{y}}\right)^{2}} \geqslant \sqrt[4]{2^{5}} \text {. }
$$
Construct a right triangle with $\sqrt{x}$ and $\sqrt{y}$ as the legs, then the hypotenuse is $\sqrt{x+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,572 |
Example $\mathbf{1 7}$ If $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, and
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i}=s, m \in \mathbf{N} \text {, then } \\
\quad \sum_{i=1}^{n}\left(a_{i}+a_{i}^{-1}\right)^{m} \geqslant n\left(\frac{s}{n}+\frac{n}{s}\right)^{m} .
\end{array}
$$ | Proof: (i) When $m=1$,
$$
\because \sum_{i=1}^{n} a_{i} \cdot \sum_{i=1}^{n} a_{i}^{-1} \geqslant n^{2}, \therefore \sum_{i=1}^{n} a_{i}^{-1} \geqslant \frac{n^{2}}{s} \text {. }
$$
Hence
$$
\begin{array}{l}
\sum_{i=1}^{n}\left(a_{i}+a_{i}^{-1}\right)=\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} a_{i}^{-1} \\
\geqslant s+\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,573 |
Zhou Da's Theorem: If in the gap between two mutually internally tangent circles, seven others are successively inscribed, then
$$
\frac{1}{r_{1}}-\frac{3}{r_{2}}+\frac{3}{r_{3}}-\frac{1}{r_{4}}=0 .
$$
Here, $r_{i}$ successively represents the radii of the four circles.
To prove Zhou Da's Theorem, we first prove a lem... | Proof: As shown in Figure 1, in $\triangle O_{1} O O_{2}$, by the cosine rule, we have
$$
\begin{array}{c}
\cos \angle O_{1} O O_{2} \\
=\frac{O O_{1}^{2}+O O_{2}^{2}-O_{1} O_{2}^{2}}{2 O O_{1} \cdot O O_{2}} \\
=\frac{\left(r-r_{1}\right)^{2}+\left(r-r_{2}\right)^{2}-\left(r_{1}+r_{2}\right)^{2}}{2\left(r-r_{1}\right)... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,574 |
The fifth problem of the 1999 Canadian Mathematical Olympiad is as follows ${ }^{[1]}$ :
Let $x, y, z$ be non-negative real numbers satisfying $x+y+z=1$. Prove:
$$
x^{2} y+y^{2} z+z^{2} x \leqslant \frac{4}{27}.
$$
And find the conditions under which equality holds.
After considering inequality (1), the author derive... | Proof: Let the median of $x, y, z$ be $y$, and define
$$
f(x, y, z)=x^{m} y^{n}+y^{m} z^{n}+z^{m} x^{n} \text {. }
$$
We can prove the following inequality chain:
$$
\begin{array}{l}
f(x, y, z) \stackrel{\mathbb{D}}{\leqslant} f(x+z, y, 0) \\
\leqslant \frac{m^{m} n^{n}}{(m+n)^{m+n}} .
\end{array}
$$
First, we prove ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,575 |
1. Let the average of $a, b, c$ be $M$, the average of $a, b$ be $N$, and the average of $N, c$ be $P$. If $a>b>c$, then the relationship between $M$ and $P$ is ( ).
(A) $M=P$
(B) $M>P$
(C) $M<P$
(D) Uncertain | 1. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,576 |
Example 7 Given that $a, b, c$ are real numbers, and $\frac{ab}{a+b} = \frac{1}{3}, \frac{bc}{b+c} = \frac{1}{4}, \frac{ac}{c+a} = \frac{1}{5}$. Then, the value of $\frac{abc}{ab+bc+ca}$ is $\qquad$
(1997, Hope Cup Mathematics Invitational for Junior High School Second Round) | Solving: Transform the given equations into
$$
\frac{a+b}{a b}=3, \frac{b+c}{b c}=4, \frac{a+c}{a c}=5 \text {. }
$$
Then $\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{a}+\frac{1}{c}=5$.
Adding the three equations yields $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$.
Thus, $\frac{a b+b c+a c}{a b c}=\frac{... | \frac{1}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,582 |
10. As shown in Figure 2, there are two utility poles $A B$ and $(I)$ on the construction site. They are $15 \mathrm{~m}$ apart, and points $A$ and $(I)$ are $4 \mathrm{~m}$ and $6 \mathrm{~m}$ above the ground, respectively. Steel cables are stretched from these points to points $E$ and $D$, and from $B$ and $F$ on th... | $10 . \frac{12}{5}$ (or 2.4 ) | 2.4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,586 |
12. A mall
sells a product, and due to
the purchase price being reduced by
$6.4 \%$ compared to the original
purchase price, the profit margin increased by 8 percentage points. Therefore, the original profit margin for selling this product is $\qquad$ . (Profit margin $=$ $\frac{\text { selling price - purchase price ... | $12.17 \%$ | 12.17\% | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,588 |
13. Let $m$ be a real number not less than -1, such that the equation $x^{2}+2(m-2) x+m^{2}-3 m+3=0$ has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$. Find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$. | $$
\begin{aligned}
\Delta & =4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
& =-4 m+4>0 .
\end{aligned}
$$
Then $m<1$.
Combining the given conditions, we have $-1 \leqslant m<1$.
$$
\begin{array}{l}
\text { (1) } \because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=4(m-2)^{2}-2\left(m^{2}-3 m+3\right) ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,589 |
14. As shown in Figure 4, given that the radius of the circumcircle $\odot O$ of quadrilateral $ABCD$ is 2, the intersection point of diagonals $AC$ and $BD$ is $E, AE=$ $EC, AB=\sqrt{2} AE$, and $BD=$ $2 \sqrt{3}$. Find the area of quadrilateral $ABCD$. | 14. From the given, we have $A B^{2}=2 A E^{2}=A E \cdot A C$. Therefore,
$$
\frac{A B}{A C}=\frac{A E}{A B} \text {. }
$$
Also, since $\angle E A B=\angle B A C$, it follows that $\triangle A B E \sim \triangle A C B$.
Thus, $\angle A B E=\angle A C B$, which implies $A B=A D$.
As shown in Figure 5, connect $A O$, in... | 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,590 |
15. A 33-story building has an elevator that starts on the first floor. It can accommodate a maximum of 32 people and can only stop once at one of the floors from the 2nd to the 33rd. For each person, walking down one floor of stairs results in 1 point of dissatisfaction, and walking up one floor of stairs results in 3... | 15. It is easy to see that these 32 people live on each of the 2nd to 33rd floors, one person per floor. For each person who takes the elevator to go up or down, the floor they live on must be no less than the floor of the person who walks up directly. In fact, if a person living on the $s$-th floor takes the elevator,... | 316 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,591 |
1. The result of $|-2| \div\left(-\frac{1}{2}\right)^{2}+(-2)^{3}$ is ( ).
(A) 2
(B) 1
(C) -1
(D) 0 | 1. 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 | 711,592 |
2. Given $-a<b<a$. Simplify the result of $|a+b|+|a-b|$ is ( ).
(A) $2 a$
(B) $2 b$
(C) $-2 b$
(1) $2(a+b)$ | 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 | MCQ | Yes | Yes | cn_contest | false | 711,594 |
3. Due to 1, in $\triangle A . B C$, $A B=A C, B E \perp A C, \emptyset$ is the midpoint of $A B$, and $D E=B E$. Then the measure of $\angle C$ is ( ).
(A) $65^{\circ}$
(B) $70^{\circ}$
(C) $75^{\circ}$
(D) $80^{\circ}$ | 3. 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 | 711,595 |
Example 9 Five numbers $a, b, c, d, e$, their pairwise sums are $183, 186, 187, 190, 191, 192, 193, 194, 196, 200$. If $a<b<c<d<e$, then the value of $a$ is $\qquad$ | Analysis: From the problem, we know that $a+b=183, a+c=$ $186, \cdots, d+e=200$. If we can find the value of $a+b+c+$ $d+e$, and note that these ten numbers are the sums of $a$, $b$, $c$, $d$, and $e$ taken two at a time, and that $a$, $b$, $c$, $d$, and $e$ each appear 4 times, then the sum of these ten numbers, 1912,... | 91 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,604 |
Three, (16 points) An engineering
team contracted two projects, Project A and Project B, with the workload of Project A being twice that of Project B. In the first half of the month, all workers worked on Project A, and in the second half of the month, the workers were divided into two equal groups, one group continued... | Three, suppose this construction team has $x$ people, and the monthly work volume per person is 1. Then the work volume for site A is $\frac{\bar{x}}{2}+\frac{1}{2}$ $\frac{x}{2}$, and the work volume for site B is $\frac{x}{2} \times \frac{1}{2}+1$. According to the problem, we get $\frac{x}{2}+\frac{x}{4}=2\left(\fra... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,610 |
Four, (20 points) A four-digit number, the sum of this four-digit number and the sum of its digits is 1999. Find this four-digit number and explain your reasoning. | Let this number be abcd. According to the problem, we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d+a+b+c+d \\
=1999,
\end{array}
$$
which simplifies to $1001 a+101 b+11 c+2 d=1999$.
(1) Clearly, $a=1$. Otherwise, $1001 a>2000$. Subtracting $1001$ from both sides, we get
$$
101 b+11 c+2 d=998 \text{. }
$$
(2) The maxim... | 1976 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,611 |
Five. (20 points) As shown in Figure 6, in the isosceles right $\triangle ABC$, $\angle BAC = 90^{\circ}, AD = AE$, $AF \perp BE$ intersects $BC$ at point $F$, and through $F$ a line $FG \perp CD$ intersects the extension of $BE$ at point $G$. Prove: $BG = AF + FG$.
保留源文本的换行和格式，直接输出翻译结果。 | $$
\begin{array}{l}
\because A P \perp B E, \\
\therefore \angle A B E=\angle C A P. \\
\text { Also, } \because C P \perp A C, \\
A B=A C, \\
\therefore \mathrm{Rt} \triangle A B E \cong \mathrm{Rt} \triangle A C P. \text { Thus, } \angle 1=\angle P. \\
\end{array}
$$
Then
$$
B E=A P.
$$
In $\triangle A B E$ and $\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,612 |
1. Given that the perimeter of an isosceles triangle is 12. Then the range of the length of the legs $a$ is ( ).
(A) $a>6$
(B) $a<3$
(C) $4<a<7$
(I)) $3<a<6$ | 1.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text "1.D" seems to be a label or a choice in a list, and it doesn't require translation as it is already in a form that is commonly used in both Chinese... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,613 |
Example 10 Given $\left(x^{2}-x+1\right)^{6}=a_{12} x^{12}+$
$$
\begin{array}{l}
a_{11} x^{11}+\cdots+a_{2} x^{2}+a_{1} x+a_{0} \text {. Then } a_{12}+a_{10} \\
+\cdots+a_{2}+a_{0}=\ldots
\end{array}
$$
(Fifth National Partial Provinces Junior High School Mathematics Correspondence Competition) | Solution: In the given equation, let $x=1$, we get
$$
a_{12}+a_{11}+a_{10}+\cdots+a_{2}+a_{1}+a_{0}=1 \text {; }
$$
Let $x=-1$, we get
$$
a_{12}-a_{11}+a_{10}-\cdots+a_{2}-a_{1}+a_{0}=3^{6} \text {. }
$$
Adding the above two equations, we get
$$
\begin{array}{l}
2\left(a_{12}+a_{10}+\cdots+a_{2}+a_{0}\right)=1+3^{6} ... | 365 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,615 |
4. Given that a trapezoid is divided into two similar triangles by one of its diagonals. If the ratio of the two non-parallel sides is $\frac{1}{4}$, then the ratio of the two bases is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{4}$
(C) $\frac{1}{8}$
(D) $\frac{1}{16}$ | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,617 |
5. In the Cartesian coordinate system, given points $A(0,0), B(0,1)$, $C(0,2), D(1,0), E(2,0), F(1,1), G(1,2)$, $H(2,1), K(2,2)$: then the number of squares formed by any four of the points $A, B, C, D, E, F, G, H, K$ is $\qquad$; the number of right-angled triangles formed by any three of the points $A, B, C, D, E, F,... | $5.6,44$ | 5, 44 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,624 |
Example 1 Prove that the equation
$$
\begin{array}{l}
{[x]+[2 x]+[4 x]+[8 x]+[16 x]+[32 x]} \\
=12345
\end{array}
$$
has no real solution.
(1981, Canadian Mathematical Olympiad) | Proof: Assume the equation has a real solution $x$.
$$
\begin{array}{l}
\text { Let } x=[x]+r, 0 \leqslant r<1 . \text { Then, } \\
n x=n[x]+n r, \\
{[n x]=n[x]+[n r] .}
\end{array}
$$
Thus, the original equation can be transformed into
$$
\begin{array}{l}
63[x]+\sum_{k=0}^{5}\left[2^{k} r\right]=12345 . \\
\because 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,626 |
Example 2 Find the sum of all solutions to the equation $[3 x+1]=2 x-\frac{1}{2}$.
(1987, All-China Junior High School Mathematics Competition) | Solution: We can consider $[3 x+1]$ as a whole, for example, an integer $t$, thus transforming the given equation into an inequality related to $t$.
Let $[3 x+1]=t$, then $t$ is an integer, and
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$
Thus, the original equation becomes
$$
t=2 x-\frac{1}{2} \text {, }
$$
which means $... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,627 |
5. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.
(1989, Soviet Union Mathematical Olympiad) | (Hint: Using the inequality $\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x}$, solve for $x=5026$, at this point $n=7$.) | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,628 |
Example 1. Let $a$ and $b$ be two given positive integers, and satisfy $\frac{a+1}{b}+\frac{b+1}{a}$ is an integer. Prove: the greatest common divisor of $a$ and $b$ does not exceed $\sqrt{a+b}$.
(20th Russian Mathematical Competition) | Analysis and Solution: Let the greatest common divisor of $a$ and $b$ be $d$, then there is the standard decomposition
$$
a=a_{1} d, b=b_{1} d,
$$
where $a_{1}$ and $b_{1}$ are coprime integers.
Substitute the above decomposition of $a$ and $b$ into the given fraction and simplify, we get
$$
\frac{d^{2}\left(a_{1}^{2}... | d \leqslant \sqrt{a+b} | Number Theory | proof | Yes | Yes | cn_contest | false | 711,629 |
Example 2 Given natural numbers $a, b, p, q, r, s$ satisfying $qr - ps = 1$ and $\frac{p}{q} < \frac{a}{b} < \frac{r}{s}$. Prove: $b \geqslant q + s$.
(1968-1969 Polish Mathematical Olympiad) | Analysis and solution: The problem gives the size relationship between three fractions. To facilitate comparison, we perform a common denominator operation, obtaining
$$
\frac{p b s}{q b s}<\frac{a q s}{b q s}<\frac{r q b}{s q b}.
$$
These three numbers have the same denominator, so we have
$$
p b s<a q s<r q b \text{... | b \geqslant q + s | Number Theory | proof | Yes | Yes | cn_contest | false | 711,630 |
Example 3 Given natural numbers $a, b, c, d, n$ satisfy
$$
n^{2}<a<b<c<d<(n+1)^{2} .
$$
Prove: $a d$ is not equal to $b c$. | Analysis and Proof: The conclusion to be proved suggests the use of proof by contradiction, i.e., assuming \( a d = b c \) and deriving a contradiction. The problem states that \( a, b, c, d \) are all between two consecutive perfect squares, so the differences between them are small, which should be the reason why the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,631 |
Example 4 Try to determine all positive integer pairs $(a, b)$ such that $a b^{2}+b+7$ divides $a^{2} b+$ $a+b$.
(IMO 38th problem) | Analysis and Solution: According to the problem, the quotient of $a^{2} b+a+b$ divided by $a b^{2} + b + 7$ is an integer. If the division operation's quotient and remainder can both be expressed in terms of $a$ and $b$, the problem would be easier to solve. However, this is unlikely because both the divisor and the di... | (11,1),(49,1),\left(7 k^{2}, 7 k\right)(k \in \mathbf{N}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,632 |
Example 5 Prove that there is only one array composed of three natural numbers greater than 1 with the following property: the product of any two numbers plus 1 is divisible by the third number.
(1965, Bulgarian Mathematical Competition) | Analysis and Solution: Let these three natural numbers be $a, b, c$. Then according to the problem, we have
$$
c \mid ab + 1, \quad a \mid bc + 1, \quad b \mid ca + 1.
$$
This problem is somewhat similar to Example 4, where we changed the dividend. Now, we similarly consider a transformation, but the goal is not to re... | 2, 3, 7 | Number Theory | proof | Yes | Yes | cn_contest | false | 711,633 |
Example 6 Find the largest positive integer $n$ that satisfies the following condition: $n$ is divisible by all positive integers less than $\sqrt[3]{n}$. (1998, Asia Pacific Mathematical Olympiad) | Analysis and Solution: The problem involves radicals, which are inconvenient to handle. If we set the largest integer less than $\sqrt[3]{n}$ as $k$, then the original problem can be transformed into the following equivalent form:
Let $k$ and $n$ be positive integers, satisfying
$$
k^{3} < n < (k+1)^{3},
$$
i.e., it h... | 420 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,634 |
Try to prove: For any natural number $n$, among the binomial coefficients $\mathrm{C}_{n}^{k}(0 \leqslant k \leqslant n)$, the number of odd ones is a power of 2. | Proof: Represent $n$ in binary form, we have
$$
n=2^{\beta_{1}}+2^{\beta_{2}}+\cdots+2^{\beta_{k}}\left(\beta_{1}>\beta_{2}>\cdots>\right.
$$
$\beta_{k}$ ).
We have
$$
\begin{aligned}
(1+x)^{n}= & (1+x)^{2^{\beta_{1}}}(1+x)^{2^{\beta_{2}}} \cdots \\
& \cdot(1+x)^{2^{\beta_{k}}} .
\end{aligned}
$$
Therefore,
$$
\begin{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,635 |
Question $B$ Proof: When and only when $n$ is of the form $s p^{k}-1$ $(1 \leqslant s<p)$, $\mathrm{C}_{n}^{m}(0 \leqslant m \leqslant n)$ are all not divisible by the prime number $p$. | To prove this problem, we first prove two lemmas.
Lemma 1: Let the $p$-ary representation of $n$ be
$$
n=a_{k} p^{k}+a_{k-1} p^{k-1}+\cdots+a_{1} p+a_{0}.
$$
Then the exponent of $p$ in $n!$ is $\frac{n-s(n)}{p-1}$, where $s(n) = a_{k}+a_{k-1}+\cdots+a_{1}+a_{0}$ is the sum of the digits of $n$.
Proof: Among the num... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,636 |
Test $C \quad m \in \mathbf{N}$, define $f(m)$ as the number of factor 2's in $m!$ (i.e., the largest integer $k$ such that $2^{k} \mid m!$). Prove: there are infinitely many $m \in \mathbf{N}$, satisfying
$$
m-f(m)=1989 \text {. }
$$ | Proof: It is only necessary in Lemma 1 to let $n=m, s(n) = 1989, p=2$, then
$$
f(m)=\frac{m-1989}{2-1} .
$$
Thus, $m-f(m)=1989$.
Since 1989 is the number of non-zero digits of $m$ in binary, there are infinitely many $m$ whose binary representation has exactly 1989 non-zero digits, hence the conclusion holds.
Using L... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,637 |
Example 3 Solve the equation
$$
x^{2}-2 x-3=12 \cdot\left[\frac{x-1}{2}\right] \text {. }
$$
(1990, Sichuan Province Junior High School Mathematics Competition) | Solution: The original equation can be rewritten as
$$
\left(\frac{x-1}{2}\right)^{2}-1=3 \cdot\left[\frac{x-1}{2}\right] \text {. }
$$
From $\left[\frac{x-1}{2}\right]$ being an integer and equation (1), we know that $\left(\frac{x-1}{2}\right)^{2}$ is also an integer.
Let $\left[\frac{x-1}{2}\right]=m$, then
$$
\fra... | x=1+2 \sqrt{7} \text { or } x=1+2 \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,638 |
In 1935, the famous mathematicians P. Erdos and G. Szekeres proved a well-known proposition:
For any positive integer $n \geqslant 3$, there exists a number $f(n)$, such that if and only if $m \geqslant f(n)$, any set of $m$ points in the plane, with no three points collinear, contains $n$ points that form a convex $n... | Lemma 1: Let the vertex set of an arbitrary convex hull $K$ be denoted as $S$. If there are $m$ points inside the convex hull forming a convex $m$-gon, and this convex $m$-gon has a side $A_{2} A_{3}$, with its adjacent sides $A_{1} A_{2}$ and $A_{3} A_{4}$ (where $A_{1}, A_{2}, A_{3}, A_{4}$ are all vertices), and if ... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,640 |
- Example 1 Sheet Layout Problem. There is a square cardboard with a side length of $a$, to make a regular triangular pyramid model, the base and sides cannot be spliced, and the volume should be maximized. How should the lines be drawn, and what are the volume and material utilization rate? | Solution: In a square with side length $a$, two isosceles triangles with base length $a$ and base angles of $15^{\circ}$ are removed. The figure 8 line is used as the unfolded diagram of the required regular triangular pyramid, which can be folded and assembled into the regular triangular pyramid $A-B C D$.
By the sine... | 0.04 a^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,641 |
Example 3 In the tetrahedron $A-B C D$, $\angle D A B+$ $\angle B A C+\angle D A C=90^{\circ}$, and $\angle A D B=\angle B D C$ $=\angle A D C=90^{\circ}$. If $D B=a, D C=b$, try to calculate the volume of the tetrahedron $A-B C D$. | Solution: Using Figure 12, the lateral surface of the triangular pyramid unfolds to $A D_{1} B C D_{2}$. Extend $D_{1} B$ and $D_{2} C$ to intersect at $P$, then $A D_{1} P D_{2}$ is a square. We have
Rt $\triangle B P C \cong \mathrm{Rt} \triangle C D B$.
Let the side length of the square be $x$, i.e., $A D=x$.
From $... | \frac{1}{6} a b(a+b) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,643 |
Example 5 The edge lengths of the tetrahedron $S-A B C$ are $S C=A B=$ $a, S A=B C=b, S B=C A=c$. Try to calculate the volume of this tetrahedron $S-A B C$. | Solution: As shown in Figure 15 of Example 4, we can deduce:
$$
\begin{array}{l}
x^{2}+z^{2}=4 c^{2}, \\
x^{2}+y^{2}=4 b^{2}, \\
y^{2}+z^{2}=4 a^{2} .
\end{array}
$$
Then, by adding and simplifying, we get
$$
x^{2}+y^{2}+z^{2}=2\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$
(4) - (3) yields
$$
x^{2}=2\left(b^{2}+c^{2}-a... | \frac{1}{12} \sqrt{2\left(a^{2}+b^{2}-c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\left(c^{2}+a^{2}-b^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,645 |
Proposition Let $a, b \in \mathbf{R}^{+}$, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
\left\{\begin{array}{l}
x_{1}=a+\frac{b}{x_{n}}, \\
x_{2}=a+\frac{b}{x_{1}}, \\
\cdots \cdots \\
x_{n}=a+\frac{b}{x_{n-1}} .
\end{array}\right.
$$
Then when $n$ is odd,
$$
x_{1}=x_{2}=\cdots=x_{n} \text {; }
$$
When $... | Proof: First, $x_{1}, x_{2}, \cdots, x_{n}$ have the same sign (if one of them, for example $x_{1}x_{2}$, by equations (2) and (3) of (4), we know $x_{2}x_{4}, \cdots, x_{n}>x_{1}$, and by equations (1) and (2), we know $x_{1}x_{2}$, which is also impossible. Therefore,
$$
x_{1}=x_{2}=\cdots=x_{n} \text {. }
$$
When $... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,647 |
Proposition Let $m_{a}, m_{b}, m_{c}, w_{a}, w_{b}, w_{c}, r_{a}, r_{b}, r_{c}$ denote the medians, angle bisectors, and exradii of $\triangle A B C$, respectively. Then
$$
m_{a} m_{b} m_{c} w_{a} w_{b} w_{c} \geqslant\left(r_{a} r_{b} r_{c}\right)^{2} .
$$
Equality holds if and only if $\triangle A B C$ is equilatera... | Proof: From the formula for the radius of the excircle, we get
$$
\begin{aligned}
\left(r_{a} r_{b} r_{c}\right)^{2} & =s^{3}(s-a)(s-b)(s-c) \\
& =s(s-a) s(s-b) s(s-c) .
\end{aligned}
$$
Therefore, to prove inequality (2), it suffices to prove the following three inequalities:
$$
\begin{array}{l}
m_{a} w_{a} \geqslant... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,648 |
Example 4 Solve the equation $r^{3}-[x]=3$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Let $x=[x]+r, 0 \leqslant r < 1$,
i.e., $-1 < r < 0$.
From the given equation, it is clear that $x > 0$, thus $x > 1$.
If $x \geqslant 2$, then
$$
x^{3}-x=x\left(x^{2}-1\right) \geqslant 2 \times 3=6 \text {. }
$$
Equation (2) contradicts Equation (1), hence $x < 2$.
$$
\therefore 1 < x < 2, [x]=1 \text {. }... | x=\sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,649 |
In $\triangle A B C$, let $a, b, c$ be the lengths of the three sides, $s$ be the semi-perimeter, $\triangle$ be the area, $R, r$ be the radii of the circumcircle and incircle, respectively; $h_{a}, h_{b}, h_{c}, r_{a}, r_{b}, r_{c}$ be the altitudes and the radii of the excircles of $\triangle A B C$. $\sum$ denotes t... | Proof: From the trigonometric identities
$$
\begin{array}{l}
\triangle=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c} \\
=r_{a}(s-a)=r_{b}(s-b)=r_{c}(s-c) \\
=r s, \\
a b c=4 R \triangle, \\
\sum b c=s^{2}+4 R r+r^{2[2]} \\
\end{array}
$$
we get $\sum \frac{h_{a}}{r_{a}}=\sum \frac{\frac{2 \triangle}{a}}{... | 4-\frac{2 r}{R} \leqslant \sum \frac{h_{a}}{r_{a}} \leqslant \frac{2 R}{r}+\frac{2 r}{R}-2 | Inequalities | proof | Yes | Yes | cn_contest | false | 711,650 |
Let $F$ be the Fermat point inside $\triangle ABC$, and extend $AF, BF, CF$ to intersect the opposite sides at $A', B', C'$, respectively. Let $AA' = x, BB' = y, CC' = z$. Paper [1] conjectures
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant \frac{2}{3}\left(\frac{1}{R}+\frac{1}{r}\right).
$$
where $R, r$ are the cir... | Lemma 1 Let $F$ be the Fermat point inside $\triangle ABC$, $l$ be the sum of the distances from $F$ to the three vertices, and $r, s$ be the inradius and semiperimeter, respectively. Then
$$
l \geqslant \sqrt{s^{2}+9 r^{2}}.
$$
Lemma 2 Let $\triangle$ denote the area of $\triangle ABC$, and the other conditions are t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,651 |
1. Circle $\Gamma_{1}$ and circle $\Gamma_{2}$ intersect at points $M$ and $N$. Let $l$ be the common tangent of circles $\Gamma_{1}$ and $\Gamma_{2}$ that is closer to $M$. $l$ is tangent to circle $\Gamma_{1}$ at point $A$ and to circle $\Gamma_{2}$ at point $B$. The line through point $M$ parallel to $l$ intersects ... | Proof: Let $K$ be the intersection of $A N$. According to the figure theorem, $A K^{2}=K T \cdot K M=B K^{2}$. In other words, $K$ is the midpoint of $A B$. Since $P Q / / A B$, $M$ is the midpoint of $P Q$. Therefore, we only need to prove that $E M \perp P Q$.
Since $C D / / A B$, point $A$ is the midpoint of the ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,653 |
2. Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove:
$$
\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 .
$$ | Prove: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$,
where $x, y, z$ are positive real numbers. Then the original inequality becomes
$$
(x-y+z)(y-z+x)(z-x+y) \leqslant x y z \text {. }
$$
Let $u=x-y+z, v=y-z+x, w=z-x+y$.
Since the sum of any two of these three numbers is positive, at most one of them can be negat... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,654 |
3. Let $n \geqslant 2$ be a positive integer. Initially, there are $n$ grasshoppers on a straight line, and they are not all at the same point. For any given positive real number $\lambda$, the following "move" can be defined:
(1) Select any three grasshoppers, let them be at points $A$ and $B$, with $A$ to the left of... | To make the flea jump as far to the right as possible, a reasonable strategy is to choose the position of the leftmost flea as point $A$ and the position of the rightmost flea as point $B$ at each move. According to this strategy, assume that after $k$ moves, the maximum distance between the fleas is $d_{k}$, and the m... | \lambda \geqslant \frac{1}{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,655 |
4. A magician has one hundred cards, each with a number from 1 to 100. He places these one hundred cards into three boxes, one red, one white, and one blue. Each box must contain at least one card.
A participant selects two of the three boxes and then picks one card from each of the selected boxes, announcing the sum ... | Solution: There are 12 different methods. Consider the integers from 1 to 100. For simplicity, define the color of the box in which integer $i$ is placed as the color of that integer. Use $\mathrm{r}$ to represent red, $\mathrm{w}$ to represent white, and $\mathrm{b}$ to represent blue.
(1) There exists some $i$ such t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,656 |
5. Determine whether there exists a positive integer $n$ such that: $n$ is divisible by exactly 2000 distinct prime numbers, and $2^{n}+1$ is divisible by $n$. | Proof: Existence. We use induction to prove a more general proposition:
For every natural number $k$, there exists a natural number $n=n(k)$, such that $n|2^n+1$, $3|n$, and $n$ is exactly divisible by $k$ distinct prime numbers.
When $k=1$, $n(1)=3$ satisfies the proposition.
Assume for $k \geqslant 1$ there exists a... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,657 |
6. Let $\mathrm{AH}_{1}$, $B\mathrm{H}_{2}$, and $\mathrm{CH}_{3}$ be the altitudes of the acute triangle $\triangle \mathrm{ABC}$. The incircle of $\triangle A B C$ touches the sides $B C$, $C A$, and $A B$ at points $T_{1}$, $T_{2}$, and $T_{3}$, respectively. Let the lines $l_{1}$, $l_{2}$, and $l_{3}$ be the reflec... | Proof: Let $M_{1}$ be the symmetric point of $T_{1}$ with respect to the angle bisector of $\angle A$, and $M_{2}$ and $M_{3}$ be the symmetric points of $T_{2}$ and $T_{3}$ with respect to the angle bisectors of $\angle B$ and $\angle C$, respectively. Clearly, $M_{1}$, $M_{2}$, and $M_{3}$ lie on the incircle of $\tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,658 |
Example 5 Let $n$ be a positive integer. How many solutions does $x^{2}-\left[x^{2}\right]=$ $(x-[x])^{2}$ have in $1 \leqslant x \leqslant n$?
$(1982$, Swedish Mathematical Olympiad) | Solution: Let $x=n$ be a solution of the equation.
Suppose $1 \leqslant x<n$, then $m=[x], x=m+r$.
Substituting into the given equation, we get
$$
\begin{array}{l}
m^{2}+2 m r+r^{2}-\left[m^{2}+2 m r+r^{2}\right]=r^{2}, \\
2 m r=\left[2 m r+r^{2}\right] .
\end{array}
$$
Thus, $2 m r$ is an integer. Considering $0 \leq... | n^2 - n + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,660 |
3. Prove: There exist two strictly increasing sequences of integers $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, such that for any natural number $n$, $a_{n}\left(a_{n}+1\right)$ divides $b_{n}^{2}+1$. | Proof: First, we prove a lemma.
Lemma If $c, d \in \mathbf{N}$, and $d^{2} \mid c^{2}+1$, then there exists $b \in \mathbf{N}$, such that
$$
d^{2}\left(d^{2}+1\right) \mid b^{2}+1 \text {. }
$$
Proof of the lemma: Since $d^{2} \mid \left(c+d^{2} c-d^{3}\right)^{2}+1$,
$$
d^{2}+1 \mid \left(c+d^{2} c-d^{3}\right)^{2}+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,661 |
4. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that
$$
\frac{1}{p}=0 . a_{1} a_{2} \cdot... | (1) Proof: The length of the smallest repeating cycle of $\frac{1}{p}$ is the $d$ that satisfies $10^{d}-1$.
Let $\varphi$ be a prime, $N_{q}=10^{2 q}+10^{q}+1$, then $N_{q} \equiv 3 \pmod{9}$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$, then $p_{q}$ is not divisible by 3. Since $N_{q}$ is a factor of $10^{3 q... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false | 711,662 |
5. Let $n, k$ be positive integers, and $n$ is not divisible by 3, $k \geqslant$ $n$. Prove: There exists a positive integer $m$, such that $m$ is divisible by $n$, and the sum of its digits is $k$.
| Proof: Let $n=2^{a} 5^{b} p$, where $a, b$ are non-negative integers, and $(p, 10)=1$. It is sufficient to prove that there exists a positive integer $M$, such that $p \mid M$, and the sum of the digits of $M$ equals $k$. In fact, it is enough to set $m=$ $M \cdot 10^{c}$, where $c=\max \{a, b\}$.
Since $(p, 10)=1$, b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,663 |
6. Prove: For each real number $M$, there exists an arithmetic sequence with infinitely many terms, such that
(1) each term is a positive integer, the common difference is not divisible by 10;
(2) the sum of the digits of each term exceeds $M$.
| Proof: We prove that the common difference of this arithmetic sequence is of the form $10^{m}+1$, where $m$ is a positive integer.
Let $a_{0}$ be a positive integer,
$$
a_{n}=a_{0}+n\left(10^{m}+1\right)=\overline{b_{s} b_{s-1} \cdots b_{0}},
$$
where $s$ and the digits $b_{0}, b_{1}, \cdots, b_{s}$ depend on $n$.
If ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,664 |
1. Given that $M$ is any point inside $\triangle A B C$. Prove:
$$
\begin{array}{l}
\min \{M A, M B, M C\}+M A+M B+M C \\
<A B+A C+B C .
\end{array}
$$ | Proof: First, we prove a lemma.
Lemma $M$ is a point inside the convex quadrilateral $ABCD$. Then
$$
M A+M B<A D+D C+C B \text {. }
$$
Lemma Proof: As shown in Figure 1, let $AM$ intersect the quadrilateral $ABCD$ at $N$, and assume without loss of generality that $N$ is on $CD$. Then
$$
\begin{array}{l}
M A+M B<M A+M... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,665 |
2. For a set of five points in the plane, no three points are collinear, and no four points are concyclic. If a circle is drawn through three points such that there is one known point inside the circle and one known point outside the circle, then the circle is called a "separator." Prove: The number of "separators" is ... | Proof: Let one point in the point set be the inversion center, transforming the other four points into $A, B, C, D$. Then, a "separator" passing through the inversion center is transformed into a line passing through two of the points $A, B, C, D$, and the other two points lie on opposite sides of this line; a "separat... | 4 | Geometry | proof | Yes | Yes | cn_contest | false | 711,666 |
3. A set of points $S$ in space is called "fully symmetric" if $S$ contains at least three points, and satisfies the condition: for any two distinct points $A$ and $B$ in $S$, the perpendicular bisector plane of segment $AB$ is a symmetry plane of the point set $S$. Prove: If a "fully symmetric" point set is a finite p... | Proof: Let $r_{P Q}$ be the symmetry transformation with the perpendicular bisector plane of line segment $P Q$ as the symmetry plane, and $G$ be the centroid of $S$. For any $A, B \in S$, by $r_{A B}(S)=S$, we have $r_{A B}(G)=G$. Therefore, all points in $S$ are equidistant from point $G$, indicating that $S$ lies on... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,667 |
4. For $\triangle A B C$, let point $X$ be on side $A B$ such that $\frac{A X}{X B}=\frac{4}{5}$; point $Y$ be on segment $C X$ such that $C Y=2 Y X$; and point $Z$ be on the extension of $C A$ such that $\angle C X Z=180^{\circ}-\angle A B C$. Let $\sum$ denote the set of all $\triangle A B C$ that satisfy $\angle X Y... | Proof: First, prove a lemma.
Lemma In $\triangle ABC$, $X$ is a point on $AB$, and $XA: XB = m: n, \angle CXB = \alpha, \angle ACX = \beta$. Then
$$
\begin{array}{l}
(m+n) \cot \alpha = n \cot A - m \cot B, \\
m \cot \beta = (m+n) \cot C + n \cot A .
\end{array}
$$
Lemma Proof: Let $CF = h$ be the altitude from $C$ to... | 45^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 711,668 |
5. Given that the incenter of $\triangle A B C$ is $I$, and $\odot O_{1}, \odot O_{2}, \odot O_{3}$ pass through $B, C$, $A, C$, and $A, B$ respectively, and are orthogonal to $\odot I$. $\odot O_{1}$ intersects $\odot O_{2}$ at another point $C^{\prime}$, and similarly, we get points $B^{\prime}$ and $A^{\prime}$. Pro... | Proof: Let the inradius of $\triangle ABC$ be $r$, and the points of tangency with $BC$, $CA$, and $AB$ be $D$, $E$, and $F$, respectively. Let $P$, $Q$, and $R$ be the midpoints of segments $EF$, $FD$, and $DE$ (as shown in Figure 9).
Since $\triangle IBD$ and $\triangle IDQ$ are both right triangles, we have
$$
\beg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,669 |
7. Let $M$ be a point inside a convex quadrilateral $ABCD$ such that $MA = MC$, $\angle AMB = \angle MAD + \angle MCD$, and $\angle CMD = \angle MCB + \angle MAB$. Prove:
$$
AB \cdot CM = BC \cdot MD, \quad BM \cdot AD = MA \cdot CD \text{. }
$$ | Prove: Construct a convex quadrilateral $P Q R S$ and a point $T$ inside it, such that $\triangle P T Q \cong \triangle A M B, \triangle Q T R \sim \triangle A M D, \triangle P T S \sim$ $\triangle C M D$ (as shown in Figure 10, Figure 11).
$$
\begin{array}{l}
\because P T=M A=M C, \\
\therefore T S=\frac{M D \cdot P T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,670 |
8. Points $A, B, C$ divide the circumcircle $\odot O$ of $\triangle ABC$ into arcs. Let $X$ be a moving point on the arc $AB$. $I_{1}, I_{2}$ are the incenters of $\triangle CAX$ and $\triangle CBX$. Prove that the circumcircle of $\triangle XI_{1}I_{2}$ intersects $\odot O$ at a fixed point other than $X$. | Proof: As shown in Figure 12, let $M$ be the midpoint of arc $BC$, then $M I_{2}=M B$.
Let $I$ be the incenter of $\triangle ABC$, then $M I_{2}=M B =M C=M I$. Therefore, $I_{2}$, $B$, $C$, and $I$ lie on the circumference of a circle with $M$ as the center.
Similarly, let $N$ be the midpoint of arc $AC$, then $A$, $I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,672 |
1. Consider the permutations $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$ formed by $\{1,2,3,4,5,6\}$, which can be transformed into 123456 with no more than 100 operations. Find the number of permutations that satisfy this condition.
(12th Korean Mathematical Olympiad) | Solution: For any permutation $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$, take $a_{1}$. If $a_{1}$ is not in position $a_{1}$, take the number $a_{1}^{\prime}$ in position $a_{1}$; if $a_{1}^{\prime}$ is not in position $a_{1}^{\prime}$, take the number $a_{1}^{\prime \prime}$ in position $a_{1}^{\prime}$, and continue unti... | 600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,673 |
2. There are some black dots on a rectangular piece of paper. Now, the paper is to be folded several times along straight lines, with the fold lines not passing through any black dots. Then, a needle is to be inserted into the folded paper, passing through all the black dots without passing through any other points. Pr... | Prove: (i) Suppose there are $n$ collinear points, sequentially denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Fold the paper once along the perpendicular bisector of $A_{1} A_{2}$, then $A_{1}$ and $A_{2}$ will coincide, and $\Lambda_{2}, A_{3}, \cdots, A_{n}$ will still be collinear. Repeating this process, the folded pap... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,674 |
3. In a chessboard with $2 n$ red squares, any two can be reached from one to another, moving each step horizontally or vertically to an adjacent red square. Prove: all the red squares can be divided into $n$ rectangles.
(1999, World City Mathematics Competition) | Proof: When $n=1$, there are 2 adjacent red cells, which obviously form a rectangle.
Assume that when $n \leqslant k$, it is possible to divide $2 k$ connected cells into $k$ rectangles.
When $n=k+1$,
(i) For $2 k+2$ cells, if removing any pair of adjacent red cells still leaves a connected graph, then the conclusion ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,675 |
4. Place some chess pieces on an $n \times n$ chessboard, satisfying the following conditions:
(i) Each empty small square shares a common edge with a small square containing a chess piece;
(ii) For any pair of small squares containing chess pieces, there is a series of small squares containing chess pieces, starting a... | Proof: For convenience, use points on a plane to represent the squares with chess pieces. If two squares with chess pieces are adjacent, connect the points representing them with a straight line, thus forming a graph. By (ii), this graph is a connected graph.
If a square with a chess piece is adjacent to an empty squa... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,676 |
5. There are 2000 nodes, each connected to every other node with a wire. Vasia and Petia take turns cutting these wires. Vasia starts first, and he can only cut one wire at a time, while Petia can cut two or win.
```plaintext
5. There are 2000 nodes, each connected to every other node with a wire. Vasia and Petia take... | Petia can win.
Because there are $C_{2000}^{2}=1999000$ wires, and since 1999000 is a multiple of 4, Vasia cuts one, and Petia cuts three, until four wires are left, at which point it is Vasia's turn to cut. Vasia cuts one, Petia cuts the next two, and the last one is cut by Vasia, so Petia can win.
(Tianjin Normal Uni... | Petia can win | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,677 |
6. The $\mathrm{Y} 2 \mathrm{~K}$ game is played on a $1 \times 2000$ grid, with the following rules: Two players take turns writing $\mathrm{S}$ or $\mathrm{O}$ in an empty cell. The first player to form the sequence SOS in three consecutive cells wins. If the grid is filled without forming SOS, the game is a draw. Pr... | In the game, the player who fills a number in a trap must lose. A grid in the form of $\mathrm{S}|\mathrm{S}$, $\mathrm{~S}| \mathrm{O}$, or $\square \mathrm{O} \mid \mathrm{S}$ is called a "semi-finished product". It is easy to see that the player who first creates a semi-finished product must lose. Suppose the first ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,678 |
1. Given $\sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a}$. Then the value of $\sqrt{4 x+x^{2}}$ is ( ).
(A) $a-\frac{1}{a}$
(B) $\frac{1}{a}-a$
(C) $a+\frac{1}{a}$
(D) cannot be determined | \begin{array}{l}-1 .(\mathrm{B}) . \\ \sqrt{x}=\frac{1}{\sqrt{a}}-\sqrt{a} \geqslant 0, x=\frac{1}{a}+a-2, \frac{1}{a} \geqslant a, \\ 4 x+x^{2}=(x+2)^{2}-4 \\ =\left(\frac{1}{a}+a\right)^{2}-4=\left(\frac{1}{a}-a\right)^{2}, \\ \therefore \sqrt{4 x+x^{2}}=\frac{1}{a}-a .\end{array} | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,679 |
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