problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 3 If the digits of a four-digit number are reversed to form a new four-digit number, the new number is exactly four times the original number. Find the original number.
(1988, Nanjing Mathematical Olympiad Selection Contest) | Solution: Let the required four-digit number be $\overline{a b c d}$, then the new four-digit number is $\overline{d c b a}$. According to the problem, we have
$\overline{a b c d}+\overline{d c b a}=5 \overline{a b c d}$.
Obviously, the last digit of $a+d$ can only be 0 or 5.
$\because 4 a \leqslant d \leqslant 9$, and... | 2178 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,551 |
Example 3 Given $a>b>0, \sin \theta=\frac{2 a b}{a^{2}+b^{2}}$ (where $\left.0<\theta<\frac{\pi}{2}\right), A_{n}=\left(a^{2}+b^{2}\right)^{n} \sin n \theta$. Prove: For all natural numbers $n, A_{n}$ is an integer.
(1990, National High School Mathematics Competition) | Prove, construct the conjugate of $A_{n}$ as $B_{n}=\left(a^{2}+\right.$ $\left.b^{2}\right)^{n} \cos n \theta$. We prove a stronger statement: $A_{n}$ and $B_{n}$ are integers.
(1) For $n=1$, $\sin \theta=\frac{2 a b}{a^{2}+b^{2}}, \cos \theta=$ $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$. Thus, $A_{1}=2 a b, B_{1}=a^{2}-b^{2}$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,552 |
Example 4 Given $a, b, c, d \in R, a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$. Prove: $(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant 6$.
(USA Mathematical Competition) | Proof: Let the left side of the inequality be $A$, and construct the conjugate of $A$ as $B=(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b-$
$$
c)^{4}+(b-d)^{4}+(c-d)^{4} \text {. }
$$
Then, we have
$$
\begin{array}{l}
A+B=6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}\right. \\
\left.+2 a^{2} d^{2}+2 b^{2} c^{2}+2 b^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,553 |
Example 5 Let $k$ be a given positive integer, and $A=k+\frac{1}{2}+\sqrt{k^{2}+\frac{1}{4}}$. Prove: For all natural numbers $n, A^{n}$'s integer part $\left[A^{n}\right]$ is divisible by $k$.
(Hubei Province Mathematics Competition Training Question) | \[
\sqrt{k^{2}+\frac{2}{4}},
\]
and let \( U_{n}=A^{n}+B^{n}-1 \).
We will prove that \( U_{n} \) is divisible by \( k \).
Since \( A+B=2 k+1 \) and \( A B=k \), \( A \) and \( B \) are the roots of the equation \( x^{2}-(2 k+1) x+k=0 \). Therefore,
\[
\begin{array}{l}
A^{n}-(2 k+1) A^{n-1}+k A^{n-2}=0, \\
B^{n}-(2 k+1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,554 |
Example 6 Prove: $\prod_{k=1}^{134}\left(1+\operatorname{ctg} k^{\circ}\right)=2^{57}$. | Proof: Let \( A = \left(1 + \operatorname{ctg} 1^{\circ}\right)\left(1 + \operatorname{ctg} 2^{\circ}\right) \cdots \left(1 + \operatorname{ctg} 134^{\circ}\right) \). Construct its conjugate:
\[
\begin{array}{l}
A = \left(1 + \operatorname{ctg} 134^{\circ}\right)\left(1 + \operatorname{ctg} 133^{\circ}\right) \ldots \... | 2^{57} | Algebra | proof | Yes | Yes | cn_contest | false | 709,555 |
Column 7 Given that $a$ and $b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$. Prove: for every natural number $n$, we have $(a+b)^{n}-a^{n}-$ $b^{n} \geqslant 2^{2 n}-2^{n+1}$.
(National High School Mathematics Competition). | Proof: Let the left side of the inequality be $A$, then we have
$$
\begin{aligned}
A= & C_{n}^{1} a^{n-1} b+C_{n}^{2} a^{n-2} b^{2}+\cdots \\
& +C_{n}^{n-1} a b^{n-1} .
\end{aligned}
$$
Reversing the order, we get
$$
\begin{aligned}
A= & C_{n}^{n-1} a^{n-1} b+C_{n}^{n-2} a^{n-2} b^{2}+\cdots \\
& +C_{n}^{1} a b^{n-1} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,556 |
Example 8 When $x$ equals $\frac{1}{1993}, \frac{1}{1992}, \ldots, \frac{1}{2}, 1, 2$, $\cdots, 1992, 1993$, calculate the value of the algebraic expression $\frac{x^{2}}{1+x^{2}}$, then add up all the results obtained. What is the sum?
(1993-1994 Wuhan and Four Cities Mathematics Competition) | Solution: Let $A(x)=\frac{x^{2}}{1+x_{1}^{2}}$, by substituting $x$ with $\frac{1}{x}$, we get the conjugate of $A(x)$, denoted as $B(x)$, which is $B(x)=A\left(\frac{1}{x}\right)=$ $\frac{1}{1+x^{2}}$. Then,
$$
A(x)+B(x)=1 .
$$
Thus, the desired sum
$$
\begin{aligned}
S= & A\left(\frac{1}{1993}\right)+\cdots+A\left(\... | 1992.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,557 |
Let $n=1990$. Find $\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}\right.$ $-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}$ ).
(1990, National High School Mathematics Competition) | Solution: Let the required expression be $A$. After transformation, we get
$$
\begin{array}{l}
A=\frac{1}{2^{n}}\left(\sqrt{3^{0}}-\sqrt{3^{2}} C_{n}^{2}+\sqrt{3^{4}} C_{n}^{4}\right. \\
-\sqrt{3^{6}} C_{n}^{6}+\cdots+\sqrt{3^{1988}} C_{n}^{1988} \\
\left.-\sqrt{3^{1990}} C_{n}^{1990} \cdot\right) \text {. } \\
\end{ar... | -\frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,558 |
Example 1 There are $n$ stones, and two people take turns to take them. It is stipulated that the number of stones taken each time is a positive divisor of the current number of stones, and the one who takes the last stone loses. Question: Who has a winning strategy? | Solution: The final losing position is a single stone, and the player who creates this state wins. Note that 1 is an odd number, which is a significant characteristic of a losing position. Therefore, we call a state with an odd number of stones an odd state. If a player can make a clever move to always reach a losing s... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,559 |
Example 2: Two players, A and B, take turns removing $p^{k}$ pieces from $n$ pieces (where $p$ is any prime number, $k=0$ or 1, meaning they can take 1 or a prime number of pieces). The player who takes the last piece wins. Who has a winning strategy? | Through $n=1,2,3,4$ for trials, it is found that:
When $n=1,2,3$, the first player has a winning strategy; when $n=4$, the second player has a winning strategy. Thus, we conjecture:
When $4 \mid n$, the second player has a winning strategy; when $4 \nmid n$, the first player has a winning strategy.
We only need to pro... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,560 |
Example 3 There are two piles of matches, with 100 and 252 matches respectively. Two people take turns to take matches, each time they can take matches from any one pile, and the number of matches taken must be a positive divisor of the number of matches in the other pile. The one who takes the last match wins. Questio... | Solution: Consider the winner's last move to take matches, it must be that one pile has already been emptied, and the other pile still has some left, and he takes all the remaining matches at once, indicating that $(r, 0)$ is a winning state, and $(0,0)$ is a losing state.
Now consider the initial state $(100,252)$, n... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,561 |
Example 4 A four-digit number has the sum of the squares of its first and last digits equal to 13, and the sum of the squares of its middle two digits equal to 85. If 1089 is subtracted from this number, the result is the same four digits written in reverse order. What is the original four-digit number? $\qquad$
(1994,... | Solution: Let the original four-digit number be $\overline{a b c d}$. According to the problem, we have
$$
\begin{array}{l}
a^{2}+d^{2}=13, \\
b^{2}+c^{2}=85, \\
\overline{a b c d}-1089=\overline{d c b a} .
\end{array}
$$
Analyzing from the first digit, it is easy to know
$$
a-d=1 \text {. }
$$
From (1) and (4), we g... | 3762 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,562 |
Example 4 There are two piles of matches, with $m$, $n$, and $t$ matches respectively. Two people take turns to take matches, with the rule that each time they can only take from one pile, and must take at least one. If the one who takes the last match wins, who has a sure-win strategy? If the one who takes the last ma... | Solution: First, consider some special cases to obtain the following winning and losing positions:
Winning positions: $(0,0,3),(3,3,4),(0,3,4)$,
$$
(1,4,6),
$$
Losing positions: $(0,3,3),(1,4,5),(1,8,9)$.
Let's examine the characteristics of several winning and losing positions. On the surface, there seems to be no pa... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,563 |
Example 5, Two people take turns picking up matches from $n(n>1)$ matches, each time picking up at least one match, and the first person cannot take all the matches at once. And for $r>1$, the number of matches taken on the $r$-th turn cannot exceed twice the number of matches taken on the $(r-1)$-th turn. The rule is ... | Solution: First, consider some special cases to obtain several winning and losing positions as follows:
Winning positions: $4,6,7,9, \cdots$,
Losing positions $\mathrm{a} 2,3,5,8, \cdots$.
From this, we conjecture: when $n$ is a Fibonacci number, the initial state is a losing position; when $n$ is not a Fibonacci numbe... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,564 |
Example 1 Prove that $f(x)=\frac{1}{\sqrt{1-x}}$ is strictly convex on $(0,1)$. | Proof: Let any $a, b, c, d \in (0,1), a < b < c < d$, and $b - a = d - c$. Then we have
$$
0 < 1 - d < 1 - c < 1 - b < 1 - a.
$$
Thus,
$$
\begin{array}{l}
\frac{b-a}{(\sqrt{1-b}+\sqrt{1-a}) \sqrt{(1-b)(1-a)}} \\
<\frac{d-c}{(\sqrt{1-d}+\sqrt{1-c}) \sqrt{(1-d)(1-c)}},
\end{array}
$$
which means $\frac{\sqrt{1-a}-\sqrt... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 709,565 |
Example 2 In $\triangle A B C$, prove that:
$$
\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2} \text {. }
$$
Equality holds if and only if $A=B=C$. | Prove: When $A=B=C$, $\sin A+\sin B+\sin C = \frac{3 \sqrt{3}}{2}$.
When $A, B, C$ are not all equal, by symmetry, we can assume $A \leqslant B \leqslant C$ (equality cannot hold simultaneously).
From $A+B+C=\pi$, we get $A<\frac{\pi}{3}<C$.
Let $A^{\prime}=\frac{\pi}{3}, B^{\prime}=B, C^{\prime}=A+C-\frac{\pi}{3}$. T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,566 |
Example 3 (Jensen's Inequality) Let $f$ be a strictly convex function on the interval $I$. Then for any $x_{i} \in I (i=1,2, \cdots, n)$, we have
$$
f\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f\left(x_{i}\right) .
$$
Equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$. | Prove: When $x_{1}=x_{2}=\cdots=x_{n}$, the inequality holds with equality.
When $x_{i}(i=1,2, \cdots, n)$ are not all equal, since the positions of $x_{1}, x_{2}, \cdots, x_{n}$ are symmetric, we can assume without loss of generality that the smallest among $x_{i}(i=1,2, \cdots, n)$ is $x_{1}$, and the largest is $x_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,567 |
There are four points $A$, $B$, $C$, $D$ on a straight line, $AB: BC: CD=2: 1: 3$. Circles $\odot O_{1}$ and $\odot O_{2}$ are constructed with $AC$ and $BD$ as diameters, respectively. The two circles intersect at $E$ and $F$. Find $ED: EA$.
(1996, Huanggang Region, Hubei Province, Junior High School Mathematics Compe... | Connect $E B$ 、
$$
\begin{array}{l}
E C, \text { let } \angle E C B \\
=\angle 1, \angle E B C \\
=\angle 2, B C=
\end{array}
$$
a, connect $E O_{2}$. Then
$$
\begin{array}{c}
A C=3 a ; \\
A E=3 a \cdot \sin \angle 1, E C=3 a \cdot \cos \angle 1 .
\end{array}
$$
Given $B D=4 a$, we have
$$
\begin{array}{l}
E D=4 a \cd... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,568 |
Question: Let $P$ be a point inside $\triangle ABC$, $\angle APB - \angle ACB = \angle APC - \angle ABC$, and let $D$ and $E$ be the incenters of $\triangle APB$ and $\triangle APC$, respectively. Prove that $AP$, $BD$, and $CE$ intersect at one point. | Proof: Draw perpendiculars from $P$ to the three sides, as shown in the figure, with the feet of the perpendiculars being $R$, $S$, and $T$.
Connecting $RS$, $ST$, and $TR$, it is easy to see that:
$$
P, R, A, S; \\
P, T, B, R; P, S, C, T
$$
are all sets of four concyclic points. Then:
$$
\begin{array}{l}
\angle APB -... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,569 |
1. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2), x_{1}$ $=a, x_{2}=b$, and let $S_{n}=x_{1}+x_{2}+\cdots+x_{n}$. Then which of the following conclusions is correct? ( ).
(A) $x_{100}=-a, S_{100}=2 b-a$
(B) $x_{100}=-b, S_{100}=2 b-a$
(C) $x_{100}=-b, S_{100}=b-a$
(D) $x_{100... | ,- 1. A.
After calculation, the first few terms of the sequence $\left\{x_{n}\right\}$ are
$$
\begin{array}{l}
a, b, b-a,-a,-b, a \sigma-b, a, b, b-a,-a,-b, \\
a-b, \cdots,
\end{array}
$$
From this, we can see that $x_{n+6}=x_{n}$, i.e., $\left\{x_{n}\right\}$ is a sequence with a period of 6.
$$
\begin{array}{l}
\the... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,570 |
2. As the pen, correctly illuminated
Body $A B C D$, $E$ is on
edge $A B$, $F$ is on the
edge $C D$, such that
$$
\begin{array}{l}
\therefore \frac{A E}{E B}=\frac{C F}{F D}=\lambda \\
(0<\lambda<+\infty) .
\end{array}
$$
Let $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}$, where
$\alpha_{\lambda}$ represents the angl... | 2. D.
Since $A B C D$ is a regular tetrahedron, $A C \perp B D$. Draw $E G / / A C$ intersecting $B C$ at $G$, and connect $G F$. Then $\alpha_{\lambda}=\angle G E F$, and $\frac{C G}{G B}=\frac{A E}{E B}=$ $\frac{C F}{F D}$. Therefore, $G F / / B D$. Hence $G F \perp E G$, and $\beta_{\lambda}=\angle E F G$. So,
$$
\... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,571 |
3. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$. Then the number of such sequences is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
(Proposed by the Problem Committee) | 3. C.
Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. Then, according to the problem, we have
$$
n a+\frac{n(n-1)}{2} d=97^{2} \text {, }
$$
which simplifies to $[2 a+(n-1) d] n=2 \times 97^{2}$.
Since $n$ is a natural number not less than 3, and 97 is a prime number, the possi... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,572 |
Example 5 Try to find the largest and smallest seven-digit numbers, without repeated digits, that can be formed from the 7 digits $0,1,2 ; 3,4,5,6$ and are divisible by 165 (require a brief reasoning process).
(199d, Huanggang Region, Hubei Province Junior High School Mathematics Competition) | Solution: From $165=3 \times 5 \times 11$, we know that the required number must be divisible by $3, 5, 11$.
Since $0+1+2+3+4+5+6=21$,
the seven-digit number must be divisible by 3.
According to the characteristics of numbers divisible by 11, let the sum of the four digits in the odd positions of the required seven-dig... | 6431205, 1042635 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,573 |
4. In the Cartesian coordinate system, the curve represented by the equation $m\left(x^{2}+y^{2}+2 y\right.$ $+1)=(x-2 y+3)^{2}$ is an ellipse, then the range of $m$ is .
(A) $(0,1)$
(B) $(1,+\infty)$
(C) $(0,5)$
(D) $(5,+\infty)$
(Hunan contribution) | 4. D.
From the given, $\frac{\sqrt{x^{2}+(y+1)^{2}}}{\frac{x-2 y+3}{\sqrt{1^{2}+(-2)^{2}}}}=\sqrt{\frac{5}{m}}$. This indicates that the ratio of the distance from $(x, y)$ to the fixed point $(0, -1)$ to the distance from $(x, y)$ to the fixed line $x-2 y+3=0$ is a constant $\sqrt{\frac{5}{m}}$. By the definition of ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,574 |
5. Let $f(x)=x^{2}-\pi x, \alpha=\arcsin \frac{1}{3}, \beta=\operatorname{arctg} \frac{5}{4}$, $\gamma=\arccos \left(-\frac{1}{3}\right), \delta=\operatorname{arcctg}\left(-\frac{5}{4}\right)$. Then ( ).
(A) $f(\alpha)>f(\beta)>f(\delta)>f(\gamma)$
(B) $f(\alpha)>f(\delta)>f(\beta)>f(\gamma)$
(C) $f(\delta)>f(\alpha)>f... | 5. B.
According to the problem, the graph of $f(x)$ is symmetric about the line $x=\frac{\pi}{2}$, and it is monotonically decreasing in $\left(-\infty, \frac{\pi}{2}\right)$, and monotonically increasing in $\left(\frac{\pi}{2},+\infty\right)$. Therefore, when $\left|x_{1}-\frac{\pi}{2}\right|>\left|x_{2}-\frac{\pi}{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,575 |
6. If three lines $a$, $b$, and $c$ in space are pairwise skew lines, then the number of lines that intersect with $a$, $b$, and $c$ is ( ).
(A) 0 lines
(B) 1 line
(C) more than 1 but a finite number of lines
(D) infinitely many lines
（Shaanxi contribution） | 6. D.
First, regardless of the positional relationship between $a$, $b$, and $c$, we can always construct a parallelepiped $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D_{1} D$ lies on line $c$. Then, take any point $M$ on the extension of $D D_{1}$. Draw a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,576 |
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ (Proposed by the Problem Group) | $=、 1.2$.
The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1)=f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,577 |
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 lines $l$, then $\lambda=$
(Proposed by the Problem Committee) | 2. 4 .
First, note the following conclusion: For a chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points, the chord attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,578 |
3. Given the complex number $z$ satisfies $\left|2 z+\frac{1}{z}\right|=1$. Then the principal value range of the argument of $z$ is $\qquad$
(Proposed by the Problem Committee) | 3. $k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta$
$$
\leqslant k \pi+\frac{\pi}{2}+\frac{1}{2} \arccos \frac{3}{4}(k=0, 1) \text {. }
$$
Let $z=r(\cos \theta+\sin \theta)$. Then
$$
\begin{array}{l}
\left|2 z+\frac{1}{z}\right|^{2}=\left(4 r^{2}+\frac{1}{r^{2}}\right)+4 \cos 2 \theta=1, \\
4 r^{... | k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta \leqslant k \pi+\frac{\pi}{2}+\frac{1}{2} \arccos \frac{3}{4}(k=0, 1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,579 |
4. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2, AB=2$, and points $S$, $A$, $B$, $C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$ (Proposed by the Problem Group) | 4. $\frac{\sqrt{3}}{3}$.
In the isosceles right triangle $ABC$ with hypotenuse $AB$ and midpoint $D$, $D$ is the circumcenter of $\triangle ABC$. Given $SA=SB=SC=2$, it follows that the projection of $S$ onto the base $ABC$ is $D$. Therefore, the center of the sphere $O$ lies on $SD$ and $OA=OB=OS$. Thus, $O$ is the c... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,580 |
5. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can randomly jump to one of the two adjacent vertices each time. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences can the frog h... | 5. 26 .
As shown in the figure, it is clear that the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only of the following two scenarios:
(1) The frog reaches point $D$ after 3 jumps, with 2 ways to do so;
(2) The frog stops after a total of 5 jumps. In thi... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,581 |
6. Let $a=\lg x+\lg \left[(y z)^{-1}-1\right], b=\lg x^{-1}+$ $\lg (x y z+1), c=\lg y+\lg \left[(x y z)^{-1}+1\right]$, and let the maximum of $a, b, c$ be $M$. Then the minimum value of $M$ is $\qquad$
(Fujian contribution) | 6. Ig2.
From the known conditions, we have $a=\lg \left(x y^{-1}+z\right), b=\lg (y z+$ $\left.x^{-1}\right), c=\operatorname{ig}\left[(x z)^{-1}+y\right]$. Let the maximum number among $x y^{-1}+z, y z+x^{-1}$, $(x z)^{-1}+y$ be $u$, then $M=\lg u$.
From the known conditions, $x, y, z$ are all positive numbers, thus
... | \lg 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,582 |
Three. (This question is worth 20 points) Let $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, and $x+y+z = \frac{\pi}{2}$. Find the maximum and minimum values of the product $\cos x \sin y \cos z$.
(Beijing contribution)
Translate the above text into English, please retain the original text's line breaks and for... | Three, from the known conditions we get
$$
\begin{array}{l}
x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-\left(\frac{\pi}{12}+\frac{\pi}{12}\right)=\frac{\pi}{3}, \\
\sin (x-y) \geqslant 0, \sin (y-z) \geqslant 0 .
\end{array}
$$
Thus, $\cos x \sin y \cos z=\frac{1}{2} \cos x[\sin (y+z)+\sin (y-z)]$
$$
\begin{array}{... | \frac{1}{8} \text{ and } \frac{2+\sqrt{3}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,583 |
Example 6 What is the smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers?
(11th American Invitational Mathematics Examination (AIME))
| Solution: Let the required positive integer be $A$. According to the problem, $A$ can be expressed as (where $p$, $n$, $k$ are all integers)
$$
\begin{aligned}
A & =(p+1)+(p+2)+\cdots+(p+9) \\
& =9 p+45, \\
A & =(n+1)+(n+2)+\cdots+(n+10) \\
& =10 n+55, \\
A & =(k+1)+(k+2)+\cdots+(k+11) \\
& =11 k+66 .
\end{aligned}
$$
... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,584 |
Four. (This question is worth 20 points) Let the two branches of the hyperbola $x y=1$ be $C_{1}$ and $C_{2}$ (as shown in the figure), and the three vertices of the equilateral triangle $P Q R$ are located on this hyperbola:
(1) Prove that $P, Q$, and $R$ cannot all be on the same branch of the hyperbola;
(2) Suppose ... | (1) By contradiction.
Assume the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ lie on a branch of a hyperbola $C_1$, with coordinates $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively. Without loss of generality, assume $0 < y_{1} < y_{2} < y_{3}$. The... | (2 - \sqrt{3}, 2 + \sqrt{3}) \text{ and } (2 + \sqrt{3}, 2 - \sqrt{3}) | Geometry | proof | Yes | Yes | cn_contest | false | 709,585 |
Five. (This question is worth 20 points) Let non-zero complex numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ satisfy
$$
\left\{\begin{array}{l}
\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{a_{5}}{a_{4}}, \\
a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \\
\quad=4\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\... | $$
\text { V. Let } \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{a_{5}}{a_{4}}=q \text {. }
$$
Then $a_{2}=a_{1} q, a_{3}=a_{1} q^{2}, a_{4}=a_{1} q^{3}, a_{5}=a_{1} q^{4}$. From the given conditions, we have
$$
\begin{array}{l}
a_{1}\left(1+q+q^{2}+q^{3}+q^{4}\right) \\
=\frac{4}{a_{1} q^{4}}\lef... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,586 |
一、(This problem
50 points) As shown in the figure, given two circles
$\odot O_{1}$ and $\odot O_{2}$ with unequal radii
intersect at points $M$ and $N$, and
$\odot O_{1}$ and $\odot O_{2}$ are internally tangent to
$\odot O$ at points $S$ and $T$,
respectively. Prove that $O M \perp$
$M N$ if and only if $S, N, T$ are ... | Given the circles $\odot O_{1}$, $\odot O_{2}$, and $\odot O$ with radii $r_{1}$, $r_{2}$, and $r$ respectively. From the conditions, we know that points $O$, $O_{1}$, and $S$ are collinear, and points $O$, $O_{2}$, and $T$ are collinear, with $OS = OT = r$. Connect $OS$, $OT$, $SN$, $NT$, $O_{1}M$, $O_{1}N$, $O_{2}M$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,587 |
2. (This question is worth 50 points) Question: When and only when do real numbers $x_{0}, x_{1}, \cdots, x_{n} \left(n \geqslant\right.$ ? satisfy what conditions so that there exist real numbers $y_{0}, y_{1}, \cdots, y_{n}$ such that
$$
z_{0}=z_{1}^{2}+z_{2}^{2}+\cdots+z_{n}^{2}
$$
- holds, where $z_{k}=x_{k}+i y_{k... | Sure, here is the translation:
---
II. It is easy to see that $(1) \Leftrightarrow$
$$
\left\{\begin{array}{l}
\sum_{k=1}^{n} x_{k}^{2}-x_{0}^{2}=\sum_{k=1}^{n} y_{k}^{2}-y_{0}^{2}, \\
\sum_{k=1}^{n} x_{k} y_{k}=x_{0} y_{0} .
\end{array}\right.
$$
If there exist real numbers $y_{0}, y_{1}, \cdots, y_{n}$ such that (... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,588 |
Three. (This question is worth 50 points) In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number. The number in the $i$-th row and $j$-th column is denoted as $x_{i, j} (i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are r... | Three, the minimum value of $k$ is 97.
(1) Take $x_{i, j}=\left\{\begin{array}{ll}0, & 4(j-1)+1 \leqslant i \leqslant 4 j \\ \frac{1}{24}, & \text { otherwise } i .\end{array}\right.$ $(j=1,2, \cdots, 25)$
At this time, $\sum_{j=1}^{25} x_{i, j}=0+24 \times \frac{1}{24} .(i=1,2, \cdots, 100)$
It satisfies the conditio... | 97 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,589 |
1. Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove
$$
\frac{ab}{a^{5}+b^{5}+ab}+\frac{bc}{b^{5}+c^{5}+bc}+\frac{ca}{c^{5}+a^{5}+ca} \leqslant 1,
$$
and specify the conditions under which equality holds. | Proof: From
$$
a^{5}+b^{2}-a^{2} b^{2}(a+b)=\left(a^{2}-b^{2}\right)\left(a^{2}-b^{3}\right) \geqslant 0,
$$
we have $a^{5}-i-\dot{u}^{3} \geqslant a^{2}(a+b)$.
$$
\begin{array}{l}
\therefore \frac{a b}{a^{5}+b^{5}+a b}=\frac{a^{2} b^{2} c}{a^{5}+b^{5}+a^{2} b^{2} c} \\
\leqslant \frac{a^{2} b^{2} c}{a^{2} b^{2}(a+b)+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,590 |
3. Given $a>2,\left\{a_{n}\right\}$ is recursively defined as follows:
$$
a_{0}=1, a_{1}=a, a_{n+1}=\left(\frac{a_{n}^{2}}{a_{n-1}^{2}}-2\right) a_{n} \text {. }
$$
Prove: For any $k \in N$, we have
$$
\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}<\frac{1}{2}\left(2+a-\sqrt{a^{2}-4}\right) .
$$ | Proof: Let $f(x)=x^{2}-2$. Then
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=f^{(1)}\left(\frac{a_{n}}{a_{n-1}}\right)=f^{(2)}\left(\frac{a_{n-1}}{a_{n-2}}\right)=\cdots \\
=f^{(n)}\left(\frac{a_{1}}{a_{0}}\right)=f^{(n)}(a) .
\end{array}
$$
Thus, $a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \cdots \c... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,592 |
4. Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers, and not all zero.
(a) Prove: The equation $x^{n}-a_{1} x^{n-1}-\cdots-a_{n-1} x-a_{n}=0$ has exactly one positive real root;
(b) Let $A=\sum_{j=1}^{n} a_{j}, B=\sum_{j=1}^{n} j a_{j}$, and let $R$ be the positive real root of the above equation. Prove: ... | Prove: (a) When $x>0$,
$$
\begin{array}{l}
x^{n}-a_{1} x^{n-1}-\cdots-a_{n-1} x-a_{n}=0 \\
\Leftrightarrow \frac{a_{n}}{x^{n}}+\frac{a_{n-1}}{x^{n-1}}+\cdots+\frac{a_{1}}{x}=1 .
\end{array}
$$
Let $f(y)=a_{n} y^{n}+a_{n-1} y^{n-1}+\cdots+a_{1} y$.
Since $f(y)$ is continuous and strictly increasing when $y \geqslant 0$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,593 |
5. Let $P(x)$ be a polynomial function with real coefficients $P(x)=a x^{3}+$ $b x^{2}+c x+d$. Prove: If for any $|x|<1$, we have $|P(x)|$ $\leqslant 1$, then
$$
|a|+|b|+|c|+|d| \leqslant 7 .
$$ | Prove: Given that $P(x)$ is a continuous function and when $|x|<1$, $|P(x)| \leqslant 1$. Therefore, when $|x| \leqslant 1$, $|P(x)| \leqslant 1$. Let $x=\lambda$ and $\frac{\lambda}{2}$ (where $\lambda= \pm 1$), we get
$$
\begin{aligned}
& |\lambda a+b+\lambda c+d| \leqslant 1, \\
& \left|\frac{\lambda}{8} a+\frac{1}{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,594 |
Example 7 A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2=2401=49^{2}=$ $(48... | Solution: Let the first two digits of the four-digit number be $c_{1}$, and the last two digits be $c_{2}$. Then $10 \leqslant c_{1} \leqslant 99,1 \leqslant c_{2} \leqslant 99$. The four-digit number can be represented as $100 c_{1}+c_{2}$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_... | 1805 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,595 |
6. Let be an even integer. Prove: There exists a positive integer $k$, satisfying $k=f(x)(x+1)^{n}+g(x)\left(x^{n}+1\right)$, where $f(x), g(x)$ are some integer-coefficient polynomials. If $k$ represents the smallest $k$ satisfying the above equation, express $k_{0}$ as a function of $n$. | Prove: (i) When $n$ is even, $\left((x+1)^{n}, x^{n}+1\right)=1$, so there exist rational coefficient polynomials $f^{*}(x), g^{*}(x)$, such that
$$
1=f^{*}(x)(x+1)^{n}+g^{*}(x)\left(x^{n}+1\right) \text {. }
$$
Let $k$ be a common multiple of the denominators of all coefficients of $f^{*}(x), g^{*}(x)$, and set $f(x)... | k_{0}=2^{t} | Algebra | proof | Yes | Yes | cn_contest | false | 709,596 |
7. Let $f$ be a function mapping the set of real numbers $\mathbb{R}$ to itself, and for any $x \in \mathbb{R}$, we have $|f(x)| \leqslant 1$, and
$$
f\left(x+\frac{13}{42}\right)+f(x)=f\left(x+\frac{1}{6}\right)+f\left(x+\frac{1}{7}\right).
$$
Prove: $f$ is a periodic function, i.e., there exists a non-zero real numb... | Proof: For any $x \in \mathbb{R}$, we have
$$
\begin{array}{l}
f\left(x+\frac{13}{42}\right)+f(x) \\
=f\left(x+\frac{7}{42}\right)+f\left(x+\frac{6}{42}\right),
\end{array}
$$
Thus, $f\left(x+\frac{7}{42}\right)-f(x)$
$$
\begin{array}{l}
=f\left(x+\frac{13}{42}\right)-f\left(x+\frac{6}{42}\right) \\
=f\left(x+\frac{19... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,597 |
10. Let $H$ be the orthocenter of $\triangle ABC$, $P$ a point on the circumcircle of the triangle, $E$ the foot of the altitude $BH$, and suppose that $PAQB$ and $PARC$ are parallelograms, with $AQ$ and $HR$ intersecting at $X$. Prove: $EX \parallel AP$. | Prove: Connect $P R$ intersecting $A C$ at $M$, then $M$ is the midpoint of $A C$, and also the midpoint of $P R$. Draw the diameter $B D$ of the circumcircle of $\triangle A B C$, and connect $D A$, $D C$, $H A$, $H C$.
$$
\begin{array}{l}
\because D A \perp A B, H C \perp A B, \\
\therefore D A / / H C .
\end{array}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,599 |
13. Let $\triangle A B C$ be an equilateral triangle, and $P$ be a point inside it. The segments $A P$, $B P$, and $C P$ intersect the sides $B C$, $C A$, and $A B$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Prove:
$$
A_{1} B_{1} \cdot B_{1} C_{1} \cdot C_{1} A_{1} \geqslant A_{1} B \cdot B_{1} C \cdot C_{1... | Proof: By the cosine theorem,
$$
\begin{aligned}
A_{1} B_{1}^{2}= & A_{1} C^{2}+B_{1} C^{2} \\
& -A_{1} C \cdot B_{1} C \\
\geqslant & 2 A_{1} C \cdot B_{1} C \\
& -A_{1} C \cdot B_{1} C \\
= & A_{1} C \cdot B_{1} C .
\end{aligned}
$$
Similarly, \( B_{1} C_{1}^{2} \geqslant B_{1} A \cdot C_{1} A \), \( C_{1} A_{1}^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,601 |
1. In $\odot O$, there are two inscribed angles $\alpha_{1}, \alpha_{2}$, and the lengths of the chords they intercept are $a_{1}, a_{2}$, respectively. Then, the true proposition among the following is ( , ).
(A) If $\alpha_{1}>\alpha_{2}$, then $a_{1}>a_{2}$
(B) If $\alpha_{1}=\alpha_{2}$, then $a_{1}=a_{2}$
(C) If $... | - 1. B.
In figure $a, \alpha_{1}>90^{\circ}>\alpha_{2}$, but $a_{1}<a_{2}$, which can negate (A) and (C). In figure $b, a_{1}=a_{2}$, but $\alpha_{1}<90^{\circ}<\alpha_{2}$, which can negate (D). Therefore, the answer is (B). In fact, in the same circle, equal inscribed angles intercept equal arcs, and equal arcs corr... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,602 |
3. As shown in the figure, in quadrilateral $A B C D$, $B C>$
$C D>D A, O$ is the midpoint of $A B$, and connect $O C, O D$.
If $\angle A O D=\angle D O C$
$=\angle C O B=60^{\circ}$, then the relationship between $B C$ and $C D+D A$ is ( ).
(A) $B C>C D+D A$
(B) $B C=C D+D A$
(C) $B C<C D+D A$
(D) Cannot be determined | 3. C.
Solution 1: As shown in the figure, draw a line $C D$ through $O M \therefore A B$ at: $M$. Then $A, B$ are symmetric about $C M$. Take the symmetric point $E$ of $D$ about $O M$. Since $O D$ and $O C$ are symmetric about $O M$, $E$ lies on $O C$. Connect $M E, E B$, then we have
$$
M D=M E, A D=B E \text {. }
$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,604 |
For real numbers $a, b, c, d$, let
$$
\begin{aligned}
M= & \left(a^{2}+b^{2}-1\right)^{2}+\left(c^{2}+d^{2}-1\right)^{2} \\
& +2(a c+b d)^{2}, \\
N= & \left(a^{2}+c^{2}-1\right)^{2}+\left(b^{2}+d^{2}-1\right)^{2} \\
& +2(a b+c d)^{2} .
\end{aligned}
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M... | 4. B.
$$
\begin{aligned}
M= & \left(a^{4}+b^{4}+1-2 a^{2}-2 b^{2}+2 a^{2} b^{2}\right)+\left(c^{4}+d^{4}\right. \\
+ & \left.1-2 c^{2}-2 d^{2}+2 c^{2} d^{2}\right)+2\left(a^{2} c^{2}+b^{2} d^{2}\right. \\
+ & 2 a b c d) \\
= & \left(a^{4}+c^{4}+1-2 a^{2}-2 c^{2}+2 a^{2} c^{2}\right)+\left(b^{4}+d^{4}\right. \\
& \left.... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,605 |
Example 8 Try to find all natural number triples $(A, B, C)$, such that
$$
A^{2}+B-C=100, A+B^{2}-C=124 .
$$
(1990, Leningrad (now St. Petersburg) Mathematical Olympiad Third Round, Grade 8) | Solution: From the problem,
$A^{2}+B-C=100$,
$A+B^{2}-C=124$.
(2) - (1), $A+B^{2}-A^{2}-B=24$,
$\quad(B-A)(A+B-1)=24$.
Since $A, B, C$ are natural numbers,
we can obtain the following system of equations
(1) $\left\{\begin{array}{l}B-A=1, \\ A+B-1=24\end{array}\right.$
(2) $\left\{\begin{array}{l}B-A=24, \\ A+B-1=1\en... | A=12, B=13, C=57 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,606 |
5. As shown in the figure, take any point $E$ on the side $B C$ inside the circumcircle of the equilateral $\triangle A B C$, and connect $A E$ to intersect $B C$ at $F$. Take a point $P$ on the segment $E F$, and connect $P B, P C$.
Then the relationship between $P A$ and $P B+P C$ is ( ).
(A) $P A>P B+P C$
(B) $P A=... | 5. C.
Obviously, $\angle B C P < \angle B A P$, as shown in the figure. Now, rotate $\triangle B P C$ $60^{\circ}$ inward around point $B$ to get $\triangle B Q A$, such that $Q A = P C$, and $Q A$ is inside $\triangle B A P$. Connect $Q P$, then $\triangle B Q P$ is an equilateral triangle, $Q P = P B$.
In $\triangle... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,607 |
6. There are four racing boats numbered (1), (2), (3), and (4), with speeds of $v_{1}$, $v_{2}$, $v_{3}$, and $v_{4}$ kilometers per hour, respectively, and they satisfy $v_{1}>v_{2}>v_{3}>v_{4}>v_{\text {water }}>0$, where $v_{\text {water }}$ is the speed of the river current (kilometers/hour). They are participating... | 6. A.
1 hour after departure, the distances between boats (1), (2), (3) and boat (4) are
$$
S_{i}=\left[\left(v_{i}-v_{\text {t }}\right)-\left(v_{k}+v_{i}\right)\right] \times 1=v_{i}+v_{i} .
$$
The time it takes for each boat to catch up with boat (4) is
$$
\begin{array}{l}
t_{i}=\frac{v_{i}+v_{4}}{\left(v_{i}+v_{\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,608 |
1. Given the numbers $a, b, c$ satisfy
$$
\frac{c-a}{2(a-b)}=\frac{2(b-c)}{c-a} \text {. }
$$
then the value of $a+b-2c$ is $\qquad$ | Solution 1: From the given, we have
$$
\frac{(c-a)+2(a-b)}{2(a-b)}=\frac{2(b-c)+(c-a)}{c-a},
$$
which simplifies to $\frac{a+c-2 b}{2(a-b)}=-\frac{a+c-2 b}{c-a}$.
Thus, $0=(a+c-2 b)\left[\frac{1}{2(a-b)}+\frac{1}{c-a}\right]$
$$
=\frac{(a+c-2 b)^{2}}{2(a-b)(c-a)} \text {. }
$$
This gives $a+c-2 b=0$.
Solution 2: From... | a+c-2b=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,609 |
2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$ | 2. 0 .
From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,610 |
3. In $\triangle A B C$, $A B$ is the longest side, $\sin A \sin B=$ $\frac{2-\sqrt{3}}{4}$. Then the maximum value of $\cos A \cos B$ is $\qquad$ . | 3. $\frac{2-\sqrt{3}}{4}$.
Given that $AB$ is the longest side, $A$ and $B$ are both acute angles, so we have
$$
\cos A=\sqrt{1-\sin ^{2} A}, \cos B=\sqrt{1-\sin ^{2} B} .
$$
Thus, $\cos A \cos B=\sqrt{1-\left(\sin ^{2} A+\sin ^{2} B\right)+\sin ^{2} A \sin ^{2} B}$
$$
\begin{array}{l}
=\sqrt{1-2 \sin A \sin B-(\sin ... | \frac{2+\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,611 |
4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ . | 4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$
and $n$ is not a multiple of 5, so among $n+2,... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,612 |
One, (25 points) As shown in the figure, take point $D$ on side $AB$ of $\triangle ABC$, connect $CD$, draw $DE$ $/ / BC$ intersecting $AC$ at $E$, and draw $EF / / CD$ intersecting $AB$ at $F$. Prove: $AB \geqslant 4 DF$.
保留源文本的换行和格式，直接输出翻译结果如下：
```
One, (25 points) As shown in the figure, take point $D$ on side $AB... | 1. From $D E / / B C$, we get
$$
\frac{D E}{B C}=\frac{A D}{A B} \text {, }
$$
and $\angle E D F=\angle C B D$.
From $E F / / C D$, we get
$$
\angle E F D=\angle C D B \text {. }
$$
From (2) and (3), $\triangle D E F \sim \triangle B C D$. Thus,
$$
\frac{D F}{D B}=\frac{D E}{B C} \stackrel{\text { by (1) }}{A B},
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,613 |
II. (25 points) Given the parabola $y=x^{2}+p x+q$ has a point $M\left(x_{0}, y_{0}\right)$ located below the $x$-axis.
(1) Prove: The parabola must intersect the $x$-axis at two points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}<x_{2}$.
(2) Prove: $x_{1}<x_{0}<x_{2}$.
(3) When point $M$ is $(1,... | (1) Given that point $M\left(x_{0}, y_{0}\right)$ is on the parabola and below the $x$-axis, we have
$$
\left\{\begin{array}{l}
y_{0} < 0, \\
x_{0}^{2} + p x_{0} + q = y_{0} < 0.
\end{array}
\right.
$$
This indicates that the discriminant of the quadratic equation
$$
x^{2} + p x + q = 0
$$
is greater than 0, thus the... | x_{1} = 0, x_{2} = 1998 \text{ or } x_{1} = -1996, x_{2} = 2 | Algebra | proof | Yes | Yes | cn_contest | false | 709,614 |
1. $\sqrt{1-\sin ^{2} 1997}-\sqrt{1-\cos ^{2} 1997}$ equals ).
(A) $\cos 1997-\sin 1997$
(B) $-\cos 1997-\sin 1997$
(C) $-\cos 1997+\sin 1997$
(D) $\cos 1997+\sin 1997$ | - 1. (D).
It is easy to calculate that the terminal side of a 1997 radian angle is in the $\mathrm{N}$-th quadrant. Therefore,
$$
\begin{aligned}
\text { original expression } & =|\cos 1997|-|\sin 1997| \\
& =\cos 1997+\sin 1997 .
\end{aligned}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,616 |
Example 9 The decimal representation of a natural number $x$ does not
contain the digit 0, and it satisfies $x \cdot \bar{x}=1000+\Pi(x)$, where $\bar{x}$ is the natural number obtained by writing the digits of $x$ in reverse order, and $\Pi(x)$ is the product of the digits of $x$. Find all such natural numbers $x$.
(1... | Solution: From the problem, we know that $x$ cannot be a one-digit number.
If $x$ is a two-digit number, let $x=10a+b$.
Then $\bar{x}=10b+a$
(where $a, b$ are natural numbers less than 10),
and it satisfies
$$
(10a+b)(10b+a)=1000+ab, \text{ (1) }
$$
which simplifies to $10ab+a^2+b^2=100$.
$$
\begin{array}{l}
\text { A... | 24 \text{ or } 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,617 |
2. The complex number $z$ satisfies $z+\frac{1}{z} \in R$ and $|z-2|=$ $\sqrt{2}$. Then the number of such complex numbers is ( ).
(A) 1
(B) 2
(C) 3 $^{-}$
(D) 4 | 2. (D).
The complex number $z$ that satisfies $z+\frac{1}{z} \in R$ corresponds to points on the $x$-axis (excluding the origin) or on the unit circle.
The points that satisfy $|z-2|=\sqrt{2}$ lie on a circle with center at $(2,0)$ and radius $\sqrt{2}$.
The latter clearly intersects the former at 4 points (intersec... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,618 |
3 Given that $a$ and $b$ are real numbers. Then $x+y>a+b$ and $x y>a b$ implies $x>a$ and $y>b$ ( ).
(A) Sufficient but not necessary condition
(B) Necessary but not sufficient condition
(C) Sufficient and necessary condition
(D) Neither sufficient nor necessary condition | 3. (B).
When $a>0, b>0$, from $x>a, y>b$ we can deduce that $x+y > a+b$, and $xy > ab$.
However, the converse is not necessarily true. From $a>0, b>0, x+y>a+b, xy>ab$, it does not necessarily follow that $x>a, y>b$. For example, $x=1, y=\frac{1}{3}, a=\frac{1}{3}, b=\frac{3}{5}$ satisfies $1+\frac{1}{3}>\frac{1}{3}+\... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 709,619 |
4. $a, b$ are two positive integers, their least common multiple is 465696. Then the number of such ordered pairs of positive integers $(a, b)$ is ( ) .
(A) 144
(B) 724
(C) 1008
(D) 1155 | 4. (D).
From $[a, b]=465696=2^{5} \cdot 3^{3} \cdot 7^{2} \cdot 11$, we know that
$$
a=2^{a_{1}} \cdot 3^{a_{2}} \cdot 7^{a_{3}} \cdot 11^{\alpha_{4}}, b=2^{\beta_{1}} \cdot 3^{\beta_{2}} \cdot 7^{\beta_{3}} \cdot 11^{\beta_{4}},
$$
where $\alpha_{i}$ and $\beta_{i}$ are non-negative integers, and
$$
\begin{array}{l}... | 1155 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,620 |
5. The roots of the equation $2 x^{2}+p x+q=0$ are $\sin \alpha$ and $\cos \alpha$. Then, in the $p o q$ coordinate plane, the figure of the point $(p, q)$ is ( ).
| 5. (B).
From the relationship between roots and coefficients, we have
$$
\left\{\begin{array}{l}
-\frac{p}{2}=\sin \alpha+\cos \alpha, \\
\frac{q}{2}=\sin 2 \alpha .
\end{array}\right.
$$
Substituting (2) into (1) gives $\frac{p^{2}}{4}=1+q$. Therefore, $q=\frac{1}{4} p^{2}-1$.
Since $|\sin \alpha+\cos \alpha| \leqsl... | B | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,621 |
6. For a cube wooden block $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, take points $P, Q, R$ on the three edges passing through vertex $A_{1}$, such that $A_{1} P=A_{1} Q=A_{1} R$. After cutting off the tetrahedron $A_{1}-P Q R$, use the section $\triangle P Q R$ as the base to drill a triangular pri... | 6. (B).
Through point $R$, a line parallel to $A_{1} C$ intersects $A C$ at a point $R^{\prime}$. Through point $Q$, a line parallel to $A_{1} C$ intersects $B_{1} C$ at a point $Q^{\prime}$. A plane through $R Q$ and parallel to $A_{1} C$ intersects the side face of the triangular prism at a point $C_{0}$ on the edge... | 6 | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,622 |
1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is | Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,623 |
3. The base of the pyramid $S-ABC$ is an equilateral triangle $ABC$, the side face $SAC$ is perpendicular to the base, and the dihedral angles formed by the other two side faces with the base are both $45^{\circ}$. Then the value of the dihedral angle $A$ $SC-B$ is $\qquad$ . (Express using inverse trigonometric functi... | 3. $\operatorname{arctg} \sqrt{7}$.
Let $SO \perp AC$ at $O$, so $SO \perp$ plane $ABC$. Draw $SD \perp BC$ at $D$, $SE \perp AB$ at $E$, and connect $OD$, $OE$. Then $OD \perp BC$, $OE \perp AB$. According to the conditions, $\angle SDO = \angle SEO = 45^{\circ}$.
Thus, $SO - OE = OD$.
From the congruence of $\triang... | \operatorname{arctg} \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,625 |
4. If $2 x+y \geqslant 1$, then the minimum value of the function $u=y^{2}-2 y+x^{2}$ $+4 x$ is $\qquad$ | 4. $u_{\min }=-\frac{9}{5}$.
By completing the square from $u=y^{2}-2 y+x^{2}+4 x$, we get
$$
(x+2)^{2}+(y-1)^{2}=u+5 \text {. }
$$
The left side $(x+2)^{2}+(y-1)^{2}$ can be interpreted as the square of the distance between point $P(x, y)$ and the fixed point $A(-2,1)$. The constraint $2 x+y \geqslant 1$ indicates t... | -\frac{9}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,626 |
6. Given that $a$, $b$, and $c$ are the lengths of the three sides of a right triangle, and for a natural number $n$ greater than 2, the following holds:
$$
\left(a^{n}+b^{n}+c^{n}\right)^{2}=2\left(a^{2 n}+b^{2 n}+c^{2 n}\right) .
$$
Then $n=$ | 6. 4 .
Let $x=a^{\frac{n}{2}}, y=b^{\frac{n}{2}}, z=c^{\frac{\pi}{2}}$, then
$$
\begin{aligned}
0= & 2\left(a^{2 n}+b^{2 n}+c^{2 n}\right)-\left(a^{n}+b^{n}+c^{n}\right)^{2} \\
= & 2\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
= & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} \\
=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,629 |
Three. (20 points) In the pyramid $S-ABC$, $SA=4, SB \geqslant 7, SC \geqslant 9, AB=5, BC \leqslant 6, AC \leqslant 8$. Try to find the maximum volume of the pyramid $S-ABC$. | Three, let the pyramid $S-ABC$ satisfy the given conditions:
$$
\begin{array}{c}
S A \leqslant 4, S B \geqslant 7, S C \geqslant 9, \\
A B=5, B C \leqslant 6, A C \leqslant 8 .
\end{array}
$$
We estimate the volume of the pyramid using $\triangle A B S$ as the base.
According to the cosine rule, we have
$$
\cos \angle... | 8 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,630 |
Four. (20 points) The sequence $\left\{a_{n}\right\}$ satisfies the conditions
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4, a_{5}=5, \\
a_{6}=119,
\end{array}
$$
When $n \geqslant 5$, $a_{n+1}=a_{1} a_{2} \cdots a_{n}-1$.
Prove: $a_{1}^{2}+a_{2}^{2}+\cdots+a_{70}^{2}=a_{1} a_{2} \cdots a_{70}$. | Let $b_{n}=a_{1} a_{2} \cdots a_{n}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}$.
We prove: When $n \geqslant 5$, we have $b_{n+1}=b_{n}-1$.
In fact,
$$
\begin{aligned}
b_{n+1}= & a_{1} a_{2} \cdots a_{n} a_{n+1}-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}-a_{n+1}^{2} \\
= & a_{1} a_{2} \cdots a_{n}\left(a_{1} a_{2} \cdots a_{n}-1\r... | a_{1}^{2}+a_{2}^{2}+\cdots+a_{70}^{2}=a_{1} a_{2} \cdots a_{70} | Number Theory | proof | Yes | Yes | cn_contest | false | 709,631 |
Five. (20 points) Given that $f(x)$, $g(x)$, and $h(x)$ are all quadratic trinomials in $x$. Prove that the equation
$$
f(g(h(x)))=0
$$
cannot have roots $1,2,3,4,5,6,7,8$. | Five, using proof by contradiction. Assume that the composite equation of the three quadratic trinomials $f(x), g(x), h(x)$, $f(g(h(x)))=0$, has 8 roots which are exactly $1,2,3,4,5,6,7,8$.
If the line $x=a$ is the axis of symmetry of the parabola $y=h(x)$, then $h(x_1)=h(x_2)$ if and only if $x_1 + x_2 = 2a$. The pol... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,632 |
One, (50 points) The product of all elements in a finite set $S$ is called the "product number" of the set $S$. Given the set $M=$ $\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots, \frac{1}{99}, \frac{1}{100}\right\}$, determine the sum of the "product numbers" of all even-numbered (2, 4, $\cdots$, 98) element sub... | Let the set $M=\left\{\frac{1}{2}, \frac{1}{3},-\frac{1}{4}, \cdots, \frac{i}{99}, \frac{1}{150}\right\}, M$ contains 99 different elements. Let $G$ be the sum of the "product numbers" of all even-numbered element subsets of $M$, and $H$ be the sum of the "product numbers" of all odd-numbered element subsets of $M$. It... | 24.255 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,633 |
II. (50 points) The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$. Prove: The sufficient and necessary condition for $ABCD$ to be a tangential quadrilateral is that the inradii $r_{1}, r_{2}, r_{3}, r_{4}$ of $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$ satisfy the relation
$$
\frac{... | In the convex quadrilateral $ABCD$, $AC$ and $BD$ intersect at $O$. Let $OA=x, OB=y, OC=z, OD=t, \angle AOB=\alpha$. By the cosine rule, we have
$$
\begin{array}{l}
a^{2}=x^{2}+y^{2}-2xy \cos \alpha, \\
b^{2}=y^{2}+z^{2}+2yz \cos \alpha, \\
c^{2}=z^{2}+t^{2}-2zt \cos \alpha, \\
d^{2}=t^{2}+x^{2}+2tx \cos \alpha.
\end{a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,634 |
Three, (50 points) $a_{1}, a_{2}, \cdots, a_{1:} ; b_{1}, b_{2}, \cdots, b_{11}$ are different permutations of $1,2,3,4,5,6,7,8,9,10,11$.
Prove: $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{11} b_{11}$ contain at least two numbers that have the same remainder when divided by 11. | Three, assuming the remainders of $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{11} b_{11}$ when divided by 11 are all different, then these remainders equal the 11 values $0,1,2,3,4,5,6,7,8,9,10$. For convenience, without loss of generality, we can assume that the remainder of $a_{1} b_{1}$ when divided by 11 is 0.
Let $k=a_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,635 |
Given $\alpha, \beta, \gamma$ are the three interior angles of $\triangle A B C$, and $\frac{\sin \beta \cos \gamma}{\sin \alpha}=\frac{\sin \gamma \cos \alpha}{\sin \beta}=\frac{\sin \alpha \cos \beta}{\sin \gamma}$. Prove: $\triangle A B C$ is an equilateral triangle. | Proof: Let $\alpha, \beta, \gamma$ be the angles opposite to sides $a, b, c$. Then, according to the Law of Sines, the Law of Cosines, and the Proportionality Theorem, from the given equation we have
$$
\begin{array}{l}
\frac{a^{2}+b^{2}-c^{2}}{a^{2}}=\frac{b^{2}+c^{2}-a^{2}}{b^{2}}=\frac{c^{2}+a^{2}-b^{2}}{c^{2}} \\
=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,637 |
Given 57 Let $a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{2} x^{2}-x+1=0$ have $n$ positive real roots. Prove:
$$
0<2^{2} a_{2}+\cdots+2^{n} a_{n} \leqslant\left(\frac{n-2}{n}\right)^{2}+1 .
$$ | Proof: Let $x_{1}, x_{2}, \cdots, x_{n}$ be the $n$ positive real roots of a polynomial. By Vieta's formulas, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=-\frac{a_{n-1}}{a_{n}}, \\
\sum_{i \neq j} x_{i} x_{j}=\frac{a_{n-2}}{a_{n}}, \cdots, \\
\prod_{i=1}^{n} x_{i}+\cdots+\prod_{i=2}^{n} x_{i}=\frac{-(-1)^{n-1}}{a_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,638 |
2. In the parallelogram $ABCD$, $P$ is the midpoint of $BC$. A line parallel to $BD$ through $P$ intersects $CD$ at $Q$. Connect $PA, PD, QA, QB$. Then, in the figure, the number of triangles with the same area as $\triangle ABP$ is ( ) besides $\triangle ABP$.
(A) 3
(B) 4
(C) 5
(D) 6 | Sure, here is the translation:
```
Bamboo:
```
Please note that "筞" is a traditional Chinese character which means "bamboo" (竹 in simplified Chinese). If you have more context or a different meaning for "筞," please let me know! | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,641 |
9. On a circle with radius 1, two chords $A B=1, A C=$ $\sqrt{2}$ are drawn. Then the degree measure of $\angle B A C$ is
保留源文本的换行和格式，直接输出翻译结果。 | 0. $105^{\circ}$ or $15^{\circ}$ | 105^{\circ} \text{ or } 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,642 |
Three. (9 points) As shown in the figure, $A$, $B$, $C$, $D$ are four points on $\odot O$, and $D$ is the midpoint of $\overparen{A C}$. The extensions of $A D$ and $B C$ intersect at $E$. Compare the sizes of $A D \cdot B E$ and $A B \cdot D E$. (The proof must not directly use the Angle Bisector Theorem)
---
Transl... | $\frac{ED}{EB}=\frac{DC}{BA}$.
But $DC=AD$, so $\frac{ED}{BE}=\frac{AD}{AB}$,
which means $AD \cdot BE=AB \cdot DE$. | AD \cdot BE = AB \cdot DE | Geometry | proof | Yes | Yes | cn_contest | false | 709,648 |
Four. (18 points) Given the parabola $y=x^{2}+p x+q$ with a point $M\left(x_{0}, y_{0}\right)$ located below the $x$-axis.
(1) Prove: The given parabola must intersect the $x$-axis at two points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, where $x_{1}<x_{2}$;
(2) Prove: $x_{1}<x_{0}<x_{2}$;
(3) When point $M... | (1) Given point $M\left(x_{0}, y_{0}\right)$ is on the parabola and below the $x$-axis, we have
$$
\left\{\begin{array}{l}
y_{0}0$.
This indicates that the discriminant of the quadratic equation
$$
x^{2}+p x+q=0
$$
is greater than 0, thus the equation (1) has two real roots, denoted as $x_{1}$ and $x_{2}\left(x_{1}<x_... | x_{1}=0, x_{2}=3 \text{ or } x_{1}=-1, x_{2}=2 | Algebra | proof | Yes | Yes | cn_contest | false | 709,649 |
Five. (9 points) The first 24 digits of $\pi$ are 3. 14159265358979323846264. Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Prove that $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots$ $\left(a_{21}-a_{22}\right)\left(a_{25}-a_{24}\right)$ must be even. | Five, by observing the approximate value of $\pi$, we can see that among 24 digits, 13 are odd numbers, and 11 are even numbers. When the 13 odd numbers are placed into 12 parentheses for subtraction, there must be one parenthesis where two odd numbers are subtracted, making this parenthesis necessarily even. This even... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,650 |
Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse? | Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the rig... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,651 |
Seven, (15 points) There are $n(n \geqslant 6)$ table tennis players participating in a round-robin tournament. The results of the matches show that: among any 5 players, there is 1 player who wins against the other 4, and there is 1 player who loses to the other 4. Prove that there must be 1 player who wins against th... | Seven, assuming no one wins all, the number of wins for $n$ people can only take $0,1,2, \cdots, n-2$, a total of $n-1$ values. Among these, there must be two people with the same number of wins.
Let $A$ and $B$ have the same number of wins, and $A$ beats $B$. Then, among those who lose to $B$, there must be a $C$ who... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,653 |
15. Let $\{a, h\},\{c, d\}$ be the lengths and widths of two rectangles, respectively, and $a<c<d<b, a b<c d$. Prove: The necessary and sufficient condition for the first rectangle to fit inside the second rectangle is
$$
\begin{array}{l}
\left(b^{2}-a^{2}\right)^{2} \leqslant(b d-a c)^{2}+ \\
(b c-a d)^{2}
\end{array}... | Proof: Suppose the first rectangle is placed inside the second rectangle, and the side of length $b$ of the $\cdots$th rectangle forms an angle $\theta, 00$ with the side of length $c$ of the second rectangle, then the following system holds:
$$
\left\{\begin{array}{l}
b x+a y \leqslant c, \\
a x+b y \leqslant d, \\
x^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,654 |
16. Let $\triangle ABC$ be an acute triangle, with the circumcenter $O$ and circumradius $R$. $AO$ intersects the circumcircle of $\triangle BOC$ at another point $A'$, $BO$ intersects the circumcircle of $\triangle COA$ at another point $B'$, and $CO$ intersects the circumcircle of $\triangle AOB$ at another point $C'... | Prove: Draw the diameter $OD$ of the circle containing $BOC$, and connect $A'D$. Then
$$
\begin{array}{l}
\angle OA'D = \angle OCD = 90^{\circ}. \\
\therefore OA' = OD \cos \angle A'DO \\
= R \cdot \frac{\cos \angle A'DO}{\cos \angle COD}. \\
\end{array}
$$
It is easy to see that $OD \perp BC$.
Thus, $\angle COD = \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,655 |
17. Let $ABCD$ be a convex quadrilateral, and let $R_{A}, R_{B}, R_{C}, R_{D}$ denote the circumradii of $\triangle DAB, \triangle ABC, \triangle BCD, \triangle CDA$ respectively. Prove that $R_{A} + R_{C} > R_{B} + R_{D}$ if and only if $\angle A + \angle C > \angle B + \angle D$. | Proof: (i) Sufficiency. As shown in the figure, if $\angle A + \angle C > \angle B + \angle D$, then $\alpha > \alpha^{\prime}, \beta > \beta^{\prime}, \theta > \theta^{\prime}, \varphi > \varphi^{\prime}$.
$$
\begin{array}{l}
R_{A} + R_{C} = \frac{1}{2}\left(\frac{A B}{2 \sin \beta^{\prime}} + \frac{A D}{2 \sin \varph... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,656 |
18. In the plane, given a point $O$ and a polygon $F$, $F$ is not necessarily convex. $p$ is the perimeter of $F$, $D$ is the sum of the distances from point $O$ to the vertices of $F$, and $H$ is the sum of the perpendicular distances from point $O$ to the sides of $F$. Prove: $D^{2}-H^{2} \geqslant \frac{p^{2}}{4}$. | Proof: Let the vertices of $F$ be $A_{1}, A_{2}, \cdots, A_{n}; O$ is the point whose perpendiculars to the lines $A_{k} A_{k+1}$ have feet $H_{k}, k=1,2, \cdots, n$ (consider $A_{n+1}=A_{1}$). By the Pythagorean theorem, we have
$$
\begin{array}{l}
O A_{k}^{2}-O H_{k}^{2}=A_{k} H_{k}^{2}, O A_{k+1}^{2}-O H_{k}^{2}=A_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,657 |
19. Four integers are marked on a circle, with a specified direction so that each integer has a unique neighboring number. Each operation involves replacing each number with the difference between it and its neighbor, i.e., for $a, b, c, d$, they are replaced with $a-b, b-c, c-d, d-a$. After 1996 such replacements, can... | Solution: The answer is negative. The proof is as follows.
Let the numbers obtained after $k$ operations be $a_{k}, b_{k}, c_{k}, d_{k}$. Let $n=1996$, then
$$
\begin{array}{l}
b_{n} c_{n}-a_{n} d_{n} \\
=\left(b_{n-1}-c_{n-1}\right)\left(c_{n-1}-d_{n-1}\right) \\
-\left(a_{n-1}-b_{n-1}\right)\left(d_{n-1}-c_{n-1}\rig... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,658 |
21. A sequence of integers $a_{0}, a_{1}, \cdots, a_{n}$ is called a 二次 sequence if for each $i \in\{1,2, \cdots, n\}$, the equation
$$
\left|a_{i}-a_{i-1}\right|=i^{2} \text {. }
$$
holds.
(a) Prove that for any integers $b$ and $c$, there exists a natural number $n$ and a 二次 sequence such that $a_{0}=b, a_{n}=c$. A 二... | Solution: (a) Since $a_{i}-a_{i-1}= \pm i^{2}, 1 \leqslant i \leqslant n$,
and $c-b=a_{n}-a_{0}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)=\sum_{i=1}^{n}\left( \pm i^{2}\right)$.
Let $d=c-b$. We only need to prove that for any integer $d$, there exists a natural number $n$, such that we can make:
“ $\square 1^{2} \square... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false | 709,659 |
22. Find all natural numbers $a$ and $b$ that satisfy the following equation:
$$
\left\lceil\frac{a^{2}}{b}\right\rceil+\left\lceil\frac{b^{2}}{a}\right\rceil=\left\lceil\frac{a^{2}+b^{2}}{a b}\right\rceil+a b .
$$ | Solution: By symmetry, without loss of generality, assume $a \leqslant b$.
Given $x-10 .\end{array}\right.$
First, consider (1). If $b \geqslant a^{2}+2$, then
$$
\begin{array}{l}
b^{3}-\left(a^{2}+1\right) b^{2}-2 a b+a^{3}-a^{2} \\
=b\left[b\left(b-a^{2}-1\right)-2 a\right]+a^{2}-a^{2} \\
\geqslant b(b-2)-a^{2}+a^{2}... | b=a^{2}+1, a \in N \text{ or } a=b^{2}+1, b \in N | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,660 |
27. Determine whether there exist two non-intersecting infinite point sets $A$ and $B$ in the plane that satisfy the following conditions:
（1）In $A \cup B$, no three points are collinear, and the distance between any two points is at least 1.
（2）For any triangle with vertices in $B$, there exists a point in $A$ inside ... | Solution: Such a set of points does not exist. We will prove this by contradiction.
Assume there exist such point sets $A$ and $B$, then the following proposition holds:
For any natural number $n \geqslant 3$, there exists a concave polygon whose vertices are marked points (i.e., points in $A \cup B$), and it has $n$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,665 |
1. The number of natural numbers less than 1000 that are neither divisible by 5 nor by 7 is ().
(A) 658
(B) 648
(C) 686
(D) 688 | ,$- 1 . C$.
Among the 999 natural numbers less than 1000, there are $\left[\frac{999}{5}\right]$ numbers divisible by 5, $\left[\frac{999}{7}\right]$ numbers divisible by 7, and among them, $\left[\frac{999}{5 \times 7}\right]$ numbers are divisible by both 5 and 7. Therefore, the number we are looking for is:
$$
999-\... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,668 |
2. In a triangle, the area (unit: $\mathrm{cm}^{2}$) is numerically equal to its perimeter (unit: $\mathrm{cm}$), then the radius of its inscribed circle is ( ).
(A) $2 \mathrm{~cm}$
(B) $3 \mathrm{~cm}$
(C) $4 \mathrm{~cm}$
(i) Sern | 2. A.
Let the semi-perimeter of the triangle be $p$, and the inradius be $r$, then the area of this triangle is $p \cdot r$. According to the problem, we have $p r=2 p$. Therefore, $r$ $=2$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,669 |
5. Given the system of equations in terms of $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=a-1 \\
x y=a-2
\end{array}\right.
$$
has two real solutions. Then which of the following statements is correct? ( )
(A) $a$ can only be 2
(B) $a$ can only be $\frac{5}{3}$
(C) $a$ can only be 3
(D) None of the above | 5. D.
Team Fire Qian Group is symmetric in $x, y$ of the second degree, so if the system of equations has one solution $(s, t)$, there must be three other solutions $(-s,-t), (t, s)$, and $(-t,-s)$.
For the original system of equations to have two solutions, it must be that $s=t$ or $s=-t$, i.e., $x=y$ or $x=-y$.
Whe... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,672 |
6. Two swimmers start swimming simultaneously from opposite ends of a 90-meter-long swimming pool, one at a speed of 3 meters per second and the other at a speed of 2 meters per second. They swim back and forth for 12 minutes. If the time taken to turn around is negligible, then the number of times they pass each other... | 6. B.
The swimming conditions of the two swimmers within 3 minutes are shown in the graph below:
After 3 minutes, both return to the starting point as they were at the beginning. Within 3 minutes, they pass each other 5 times (as can be seen from the graph, i.e., the intersections of the dashed and solid lines, includ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,673 |
1. Given $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{2}{3}$. Then
$$
\frac{a^{-2}+2^{-2} \cdot c^{-2}+3^{-3} \cdot e^{-2}+4^{-4} \cdot 2^{-2}}{b^{-2}+2^{-2} \cdot d^{-2}+3^{-3} \cdot f^{-2}+4^{-4} \cdot 3^{-2}}=
$$ | $$
\text { II. } 1 . \frac{9}{4} \text {. }
$$
First, raise to the power of -2, then according to the basic properties of fractions, we get
$$
\frac{a^{-2}}{b^{-2}}=\frac{2^{-2} \cdot c^{-2}}{2^{-2} \cdot d^{-2}}=\frac{3^{-2} \cdot e^{-2}}{3^{-2} \cdot f^{-2}}=\frac{4^{-4} \cdot 2^{-2}}{4^{-4} \cdot 3^{-2}}=\frac{9}{4... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,675 |
2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points. | 2. 2 .
Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same pa... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,676 |
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