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Example 6 On a plane, there are 7 points, and some line segments can be connected between them, so that any three points among the 7 points must have two points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | Proof: First, to meet the requirements of the problem, the number of line segments to be connected must be greater than or equal to 9.
Below, we discuss in 4 cases.
(1) If among the 7 points, there exists 1 point not connected to any other points, by the problem's condition, the remaining 6 points must each be connecte... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,720 |
II. (16 points) As shown in Figure 5, quadrilateral $P Q M N$ is an inscribed quadrilateral of square $A B C D$.
(1) If $M P \parallel B C$ or $N Q \parallel A B$, prove:
$S_{\text {quadrilateral } P Q N}=\frac{1}{2} S_{\square A B C D}$;
(2) If $S_{\text {quadrilateral } P Q M N}=\frac{1}{2} S_{\square A B C D}$, can ... | (1) Suppose $M P / / B C$. Then
$S_{\triangle Q M P}=S_{\triangle M P P}=\frac{1}{2} S_{\square \triangle M P D}$.
Similarly, $S_{\triangle M M P}=\frac{1}{2} S_{\square M B C P}$.
(2) It can always be concluded that $M P / / B C$ or $N Q / / A B$.
If $M P / / B C$, the assertion is already true.
If $M P$ is not paral... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,722 |
Three, (16 points) Let $n$ be a positive integer, and $d_{1}<d_{2}<$ $d_{3}<d_{4}$ be the 4 smallest consecutive positive integer divisors of $n$. If $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$, find the value of $n$. | If $n$ is odd, then $d_{1} 、 d_{2} 、 d_{3} 、 d_{4}$ are all odd. Hence $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \equiv 1+1+1+1 \equiv 0(\bmod 4)$. Contradiction.
If $4 \mid n$, then $d_{1}=1, d_{2}=2$. By $d_{i}^{2} \equiv 0$ or $1(\bmod 4)$, we have $n \equiv 1+0+d_{3}^{2}+d_{4}^{2} \neq 0(\bmod 4)$. Also a contradi... | 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,723 |
Four. (18 points) As shown in Figure 6, given $\triangle A B C$, and $S_{\triangle A B C}=$ 1. $D, E$ are moving points on $A C, A B$ respectively, $B D$ and $C E$ intersect at point $P$, such that $S_{\text {quadrilateral } B C D E}=$ $\frac{16}{9} S_{\triangle B P C}$. Find the maximum value of $S_{\triangle D E P}$. | Let $\frac{A E}{A B}=x, \frac{A D}{A C}=y$, then
$\frac{S_{\triangle A E D}}{S_{\triangle A B C}}=\frac{A E \cdot A D}{A B \cdot A C}=x y$.
Since $S_{\triangle A B C}=1$, we have
$S_{\triangle A D E}=x y, S_{\text {quadrilateral } C D E}=1-x y$.
In $\triangle A B D$, by Menelaus' theorem, we get
$$
\frac{B P}{P D} \cdo... | \frac{1}{18} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,724 |
1. Simplify $\frac{1}{4+\sqrt{59+30 \sqrt{2}}}+\frac{1}{3-\sqrt{66-40 \sqrt{2}}}$, the result is ( ).
(A) irrational number
(B) proper fraction
(C) odd number
(D) even number | \begin{array}{l}\text { - 1.D. } \\ \text { Original expression }=\frac{1}{4+\sqrt{(5 \sqrt{2}+3)^{2}}}+\frac{1}{3-\sqrt{(5 \sqrt{2}-4)^{2}}} \\ =\frac{1}{7+5 \sqrt{2}}+\frac{1}{7-5 \sqrt{2}}=-14 .\end{array} | -14 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,725 |
2. The lengths of the four sides of a cyclic quadrilateral in sequence are 5, 10, 11, and 14. Then the area of this quadrilateral is ( ).
(A) $78 \frac{1}{2}$
(B) $97 \frac{1}{2}$
(C) 90
(D) 102 | 2.C.
As shown in Figure 3, let $AB=5$, $BC=10$, $CD=11$, $DA=14$. Since $5^{2}+14^{2}=221=10^{2}+11^{2}$, that is, $AB^{2}+AD^{2}=BC^{2}+CD^{2}$, hence $BD$ is the diameter, thus, $\angle A$ and $\angle C$ are both right angles. Therefore, the area $S$ of quadrilateral $ABCD$ is $S=\frac{1}{2} \times 5 \times 14+\frac... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,726 |
3. Let $r \geqslant 4, a=\frac{1}{r}-\frac{1}{r+1}$,
$$
b=\frac{1}{\sqrt{r}}-\frac{1}{\sqrt{r+1}}, c=\frac{1}{r(\sqrt{r}+\sqrt{r+1})} \text {. }
$$
Then among the following options, the one that must be true is ( ).
(A) $a>b>c$
(B) $b>c>a$
(C) $c>a>b$
(D) $c>b>a$ | 3. D.
Since $r \geqslant 4$, we have $\frac{1}{\sqrt{r}}+\frac{1}{\sqrt{r+1}}\frac{\sqrt{r+1}-\sqrt{r}}{\sqrt{r} \cdot \sqrt{r+1}}=\frac{1}{\sqrt{r}}-\frac{1}{\sqrt{r+1}}=b .
\end{array}
$
Therefore, $c>b>a$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,727 |
4. In Figure 1, the three shaded areas are formed by two circles with a radius of 1 and their external common tangents. If the area of the middle shaded part is equal to the sum of the areas of the upper and lower shaded parts, then the length of the common chord of the two circles is ( ).
(A) $\frac{\sqrt{5}}{2}$
(B) ... | 4.D.
Let the distance between the centers be $d$. From the condition, it is easy to know that the area of rectangle $A B C D$ equals the area of one circle, i.e., $2 d=\pi$, so,
$$
\begin{array}{l}
O_{1} P=\frac{d}{2}=\frac{\pi}{4}, \\
M N=2 M P=2 \sqrt{1-\left(\frac{\pi}{4}\right)^{2}}=\frac{1}{2} \sqrt{16-\pi^{2}} .... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,728 |
5. Given the quadratic function $f(x)=a x^{2}+b x+c$ whose graph is shown in Figure 2, let
$$
\begin{array}{l}
p=|a-b+c|+|2 a+b|, \\
q=|a+b+c|+|2 a-b| .
\end{array}
$$
Then ().
(A) $p>q$
(B) $p=q$
(C) $p<q$
(D) The relationship between $p$ and $q$ cannot be determined | 5.C.
From the graph, we know $f(0)=0=c, f(1)=a+b+c>0$. Thus, $a+b>0, f(-1)=a-b+c=a-b$. Therefore,
$$
\left.\frac{b}{2 a}0\right\}.
$$
And $2 a-b=a+(a-b)<0$, so
$$
\begin{array}{l}
p=-(a-b)+2 a+b=a+2 b, \\
q=a+b-(2 a-b)=2 b-a, \\
p-q=(a+2 b)-(2 b-a)=2 a<0 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,729 |
6. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are distinct positive odd numbers, satisfying
$$
\begin{array}{l}
\left(2005-x_{1}\right)\left(2005-x_{2}\right)\left(2005-x_{3}\right) . \\
\left(2005-x_{4}\right)\left(2005-x_{5}\right)=24^{2},
\end{array}
$$
then, the last digit of $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x... | 6. A.
It is evident that the five factors on the left side of the equation are different even numbers. Decomposing \(24^2\) into the product of five distinct even numbers, there is only one unique form:
$$
24^{2}=2 \times(-2) \times 4 \times 6 \times(-6) .
$$
Therefore, the five factors \(2005-x_{i} (i=1,2,3,4,5)\) s... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 715,730 |
1. Among the natural numbers not exceeding 100, add up all the numbers that are multiples of 3 or 5, the sum is $\qquad$ . | 1.2418.
The sum of multiples of 3 from $1 \sim 100$ is $S_{3}=3(1+2+\cdots+33)=1683$,
The sum of multiples of 5 from $1 \sim 100$ is $S_{5}=5(1+2+\cdots+20)=1050$, the sum of multiples of 15 from $1 \sim 100$ is $S_{15}=15(1+2+\cdots+6)=315$. Therefore, $S=S_{3}+S_{5}-S_{15}=2418$. | 2418 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,732 |
2. If $\sqrt{7 x^{2}+9 x+13}+\sqrt{7 x^{2}-5 x+13}=$ $7 x$, then $x=$ $\qquad$ | 2. $\frac{12}{7}$.
Obviously, $x>0$. From the condition, we have
$$
\frac{14 x}{\sqrt{7 x^{2}+9 x+13}-\sqrt{7 x^{2}-5 x+13}}=7 x \text {. }
$$
Then $\sqrt{7 x^{2}+9 x+13}-\sqrt{7 x^{2}-5 x+13}=2$.
Thus, $2 \sqrt{7 x^{2}+9 x+13}=7 x+2$.
Squaring and simplifying, we get $21 x^{2}-8 x-48=0$.
Therefore, the positive root... | \frac{12}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,733 |
3. If real numbers $x, y$ satisfy
$$
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \frac{x}{5^{3}+4^{2}}+\frac{y}{5^{3}+6^{3}}=1,
$$
then $x+y=$ $\qquad$ . | 3.432 .
It is known that $3^{3}$ and $5^{3}$ are the two roots of the equation $\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1$, which means they are the roots of the equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0
$$
By the relationship between roots and coefficients, we hav... | 432 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,734 |
4. Given an acute triangle $\triangle A B C$ with three interior angles satisfying $\angle A>\angle B>\angle C$, let $\alpha$ represent the minimum of $\angle A-\angle B$, $\angle B-\angle C$, and $90^{\circ}-\angle A$. Then the maximum value of $\alpha$ is $\qquad$ | $$
\begin{array}{l}
\alpha .15^{\circ} . \\
\alpha=\frac{3 \alpha+2 \alpha+\alpha}{6} \\
\leqslant \frac{3\left(90^{\circ}-\angle A\right)+2(\angle A-\angle B)+(\angle B-\angle C)}{6} \\
=\frac{270^{\circ}-(\angle A+\angle B+\angle C)}{6}=\frac{90^{\circ}}{6}=15^{\circ} .
\end{array}
$$
On the other hand, when
$$
\ang... | 15^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,735 |
1. (20 points) Given that $a$, $b$, and $c$ are real numbers, $a c < 0$, and $\sqrt{2} a + \sqrt{3} b + \sqrt{5} c = 0$.
Prove: The quadratic equation $a x^{2} + b x + c = 0$ has a root greater than $\frac{3}{4}$ and less than 1. | 1. From the condition, we have $\frac{\sqrt{3}}{\sqrt{5}} b+c=-\frac{\sqrt{2}}{\sqrt{5}} a$. Let
$$
y=a x^{2}+b x+c .
$$
When $x=\frac{\sqrt{3}}{\sqrt{5}}$, we have
$$
y_{1}=\frac{3}{5} a+\frac{\sqrt{3}}{\sqrt{5}} b+c=\frac{3-\sqrt{10}}{5} a \text {. }
$$
When $x=1$, we have
$$
\begin{array}{l}
y_{2}=a+b+c=a+b+c-\fra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,736 |
2. (25 points) In acute $\triangle ABC$, $AB > AC$, $CD$ and $BE$ are altitudes to sides $AB$ and $AC$ respectively. Through point $D$, draw a perpendicular to $BC$ intersecting $BE$ at point $F$ and the extension of $CA$ at point $P$. Through point $E$, draw a perpendicular to $BC$ intersecting $CD$ at point $G$ and t... | 2. As shown in Figure 4, from $AB > AC$, we know that $DE$ must intersect the extension of $BC$, and let the intersection point be $T$. Let the perpendiculars through $D$ and $E$ intersect $BC$ at $M$ and $N$, respectively, then
$$
\begin{array}{l}
CN \cdot CB \\
= CE^2, \\
BM \cdot BC = BD^2. \\
\text{Therefore, } \fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,737 |
3. (25 points) $a$, $b$, $c$ are positive integers, and $a^{2}+b^{3}=$ $c^{4}$. Find the minimum value of $c$.
保留了源文本的换行和格式。 | 3. Clearly, $c>1$. From $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$, if we take $c^{2}-a=b, c^{2}+a=b^{2}$, then at this time
$$
c^{2}=\frac{b(b+1)}{2} \text {. }
$$
Examining $b$ from small to large to make the right side a perfect square, we find that when $b=8$, $c=6$, and thus, $a=28$.
Next, we show that ther... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,738 |
2. (25 points) In acute $\triangle ABC$, $AB > AC$, $CD$ and $BE$ are altitudes to sides $AB$ and $AC$ respectively, $DE$ intersects the extension of $BC$ at point $T$, a perpendicular from $D$ to $BC$ intersects $BE$ at point $F$, and a perpendicular from $E$ to $BC$ intersects $CD$ at point $G$. Prove that $F$, $G$, ... | 2. As shown in Figure 5, let the perpendiculars through $D$ and $E$ intersect $BC$ at $M$ and $N$ respectively. Then,
$$
\begin{array}{l}
C N \cdot C B \\
=C E^{2}, \\
B M \cdot B C \\
=B D^{2} . \\
\text { Hence } \frac{C N}{B M}=\frac{C E^{2}}{B D^{2}} .
\end{array}
$$
And $G N=\frac{B D}{C D} \cdot C N, F M=\frac{C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,739 |
3. (25 points) In the sum $0^{2}+1^{2}+2^{2}+3^{2}+\cdots+$ $2005^{2}$, some of the “+” signs can be changed to “-” signs. If the resulting algebraic sum is $n$, the number $n$ is called “representable”. Which of the first 10 positive integers 1, 2, $3, \cdots, 10$ are representable? Explain your reasoning. | 3. In $0^{2}+1^{2}+2^{2}+3^{2}+\cdots+2005^{2}$, there are 1003 (an odd number) odd numbers, so no matter how " + " or " - " signs are added, the algebraic sum must be odd. Therefore, among the first 10 positive integers, the numbers that can be expressed must be among $1,3,5,7,9$.
Below, we show that these 5 numbers c... | 1,3,5,7,9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,740 |
1. The range of the function $y=f(x)$ defined on $\mathbf{R}$ is $[m, n]$. Then the range of $y=f(x-1)$ is $(\quad)$.
(A) $[m, n]$
(B) $[m-1, n-1]$
(C) $[f(m-1), f(n-1)]$
(D) Cannot be determined | 1. A.
When the graph of a function is shifted left or right, the range of the function does not change. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,741 |
2. Let the arithmetic sequence $\left\{a_{n}\right\}$ satisfy $3 a_{8}=5 a_{13}$, and $a_{1}>0, S_{n}$ be the sum of its first $n$ terms. Then the largest $S_{n}\left(n \in \mathbf{N}_{+}\right)$ is ( ).
(A) $S_{10}$
(B) $S_{11}$
(C) $S_{20}$
(D) $S_{21}$ | 2.C.
Let the common difference of the arithmetic sequence be $d$. According to the problem, we have $3\left(a_{1}+7 d\right)=5\left(a_{1}+12 d\right)$, solving this gives $d=-\frac{2}{39} a_{1}$. Thus, $a_{n}=a_{1}+(n-1) d=a_{1}-\frac{2}{39} a_{1}(n-1)$ $=a_{1}\left(\frac{41}{39}-\frac{2}{39} n\right)$.
To maximize $S... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,743 |
3. The equation $\log _{2} x=3 \cos x$ has $(\quad)$ solutions.
(A) 1
(B) 2
(C) 3
(D) 4 | 3.C.
Draw the graphs of the functions $y=\log _{2} x$ and $y=3 \cos x$, and study their intersection points. It can be observed that there are 3 solutions in total. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,744 |
4. Given a quadratic equation in $x$
$$
x^{2}+\left(a^{2}-1\right) x+a-2=0
$$
one root is greater than 1, and the other root is less than 1. Then ( ).
(A) $-11$
(C) $-21$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
The tr... | 4.C.
Let $f(x)=x^{2}+\left(a^{2}-1\right) x+a-2$, its graph opens upwards. According to the problem, $f(1)<0$, that is,
$$
1^{2}+\left(a^{2}-1\right) \times 1+a-2<0 \text {. }
$$
Simplifying, we get $a^{2}+a-2<0$, solving this yields $-2<a<1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,745 |
5. Given that $\alpha, \beta$ are acute angles, $\sin \alpha=x, \cos \beta=y$, $\cos (\alpha+\beta)=-\frac{3}{5}$. Then the functional relationship between $y$ and $x$ is $(\quad)$.
(A) $y=-\frac{3}{5} \sqrt{1-x^{2}}+\frac{4}{5} x\left(\frac{3}{5}<x<1\right)$
(B) $y=-\frac{3}{5} \sqrt{1-x^{2}}+\frac{4}{5} x(0<x<1)$
(C)... | 5. A.
$$
\begin{array}{l}
y=\cos \beta=\cos [(\alpha+\beta)-\alpha] \\
=\cos (\alpha+\beta) \cdot \cos \alpha+\sin (\alpha+\beta) \cdot \sin \alpha \\
=-\frac{3}{5} \sqrt{1-x^{2}}+\frac{4}{5} x .
\end{array}
$$
Since $y \in(0,1)$, then $0<-\frac{3}{5} \sqrt{1-x^{2}}+\frac{4}{5} x<1$.
Thus, $x \in\left(\frac{3}{5}, 1\r... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,746 |
6. The domain of the function $y=\sin x$ is $[a, b]$, and the range is $\left[-1, \frac{1}{2}\right]$. Then the maximum value of $b-a$ is ( ).
(A) $\pi$
(B) $2 \pi$
(C) $\frac{4 \pi}{3}$
(D) $\frac{5 \pi}{3}$ | 6.C.
As shown in Figure 2.
To make the function $y=\sin x$ have a range of $\left[-1, \frac{1}{2}\right]$ on the domain $[a, b]$, the maximum value of $b-a$ is $\frac{\pi}{6}-\left(-\frac{7 \pi}{6}\right)=\frac{4 \pi}{3}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,747 |
1. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and its smallest positive period is 2. Then the value of $f(-1)$ is $\qquad$ . | 1.0.
Since $f(x+2)=f(x)$, let $x=-1$, then we have $f(-1+2)=f(-1)$,
which means $f(1)=f(-1)=-f(1)$.
Therefore, $f(-1)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,748 |
2. Let the set $M=\left\{x \mid x=2^{n}-2^{k}\right.$, where $n, k \in$ $\mathbf{N}$, and $n>k\}, P=\{x \mid 1912 \leqslant x \leqslant 2004$, and $x$ $\in \mathbf{N}\}$. Then, the sum of all elements in the set $M \cap P$ is $\qquad$. | 2.3904.
Since $2^{10}=1024,2^{11}=2048$, all elements in the set $M \cap P$ must be of the form $2^{11}-2^{k}$ (where $k \in \mathbf{N}$).
Also, $1912 \leqslant 2^{11}-2^{k} \leqslant 2004$, so $44 \leqslant 2^{k} \leqslant 136$.
The only values of $k$ that satisfy this condition are 6 and 7, so the set $M \cap$ $P$ c... | 3904 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,749 |
3. Given $f(x)=\frac{1}{2^{x}+\sqrt{2}}$. Then
$$
\begin{array}{l}
f(-2003)+f(-2002)+\cdots+f(-1)+ \\
f(0)+f(1)+\cdots+f(2003)+f(2004) \\
=
\end{array}
$$ | $3.1002 \sqrt{2}$.
Using $f(x)+f(1-x)=\frac{\sqrt{2}}{2}$, we get the result | 3.1002 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,750 |
4. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=99^{\frac{1}{99}}, a_{n}=$ $\left(a_{n-1}\right)^{a_{1}}$. When $a_{n}$ is an integer, the smallest positive integer $n$ is $\qquad$ | 4. 100 .
Obviously, for any positive integer $n$, we have $a_{n}>0$.
From $a_{n}=\left(a_{n-1}\right)^{a_{1}}$, taking the common logarithm on both sides, we get
$\lg a_{n}=a_{1} \lg a_{n-1}$.
Therefore, $\left\{\lg a_{n}\right\}$ is a geometric sequence with the first term $\lg a_{1}$ and the common ratio $a_{1}$. So... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,751 |
5. Let the function $f(x)=\frac{x+3}{\sqrt{x-a}}(a \in \mathbf{R})$. If $f(x)$ is an increasing function on $(1,+\infty)$, then the range of values for $a$ is $\qquad$ . | 5. $(-\infty,-1]$.
Let $t=\sqrt{x-a}$, then the original function becomes
$$
f(t)=\frac{t^{2}+3+a}{t}=t+\frac{3+a}{t} \text {. }
$$
Since $f(x)$ is an increasing function on $(1,+\infty)$, $f(t)$ is an increasing function on $(\sqrt{1-a},+\infty)$.
When $3+a \leqslant 0$, i.e., $a \leqslant-3$, it clearly meets the r... | (-\infty,-1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,752 |
3. For $i=1,2, \cdots, n$, we have $\left|x_{i}\right|<1$, and
$$
\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right| \text {. }
$$
Find the minimum value of the positive integer $n$. | Given $19=\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|-\mid x_{1}+x_{2}+\cdots+x_{n}|\leqslant| x_{1}|+| x_{2}|+\cdots+| x_{n} \mid<n$, it follows that $n \geqslant 20$. When $n=20$, take
$$
x_{i}=\left\{\begin{array}{cc}
0.95, & \text { when } i \text { is odd: } \\
-0.95, & \text { when } i \text {... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,753 |
Three. (20 points) Let $a_{n}>0\left(n \in \mathbf{N}_{+}\right), a_{1}=5$, when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{7}{a_{n}-a_{n-1}}+6$. Find the general term formula of the sequence $\left\{a_{n}\right\}$. | The given condition can be transformed into
$$
a_{n}^{2}-a_{n-1}^{2}=6 a_{n}-6 a_{n-1}+7 \text {. }
$$
Completing the square yields $\left(a_{n}-3\right)^{2}-\left(a_{n-1}-3\right)^{2}=7$.
Therefore, $\left(a_{n}-3\right)^{2}=7 n-3$.
Hence, $a_{n}=\sqrt{7 n-3}+3$. | a_{n}=\sqrt{7 n-3}+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,755 |
Four. (20 points) Given the function
$$
f(x)=4 \sin x \cdot \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+\cos 2 x \text {. }
$$
(1) Let the constant $\omega>0$, if $y=f(\omega x)$ is an increasing function on the interval $\left[-\frac{\pi}{2}, \frac{2 \pi}{3}\right]$, find the range of $\omega$;
(2) Let the set $A=... | (1) $f(x)=4 \sin x \cdot \frac{1-\cos \left(\frac{\pi}{2}+x\right)}{2}+\cos 2 x$ $=2 \sin x \cdot(1+\sin x)+1-2 \sin ^{2} x=2 \sin x+1$.
Since $f(\omega x)=2 \sin \omega x+1$ is an increasing function on $\left[-\frac{\pi}{2}, \frac{2 \pi}{3}\right]$, we have
$\left[-\frac{\pi}{2}, \frac{2 \pi}{3}\right] \subseteq\left... | (1, 4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,756 |
Five. (20 points) As shown in Figure 1, given that the circumcenter of acute triangle $\triangle ABC$ is $O$, the midpoints of line segments $OA$ and $BC$ are points $M$ and $N$ respectively, $\angle ABC = 4 \angle OMN$, and $\angle ACB = 6 \angle OMN$. Find the degree measure of $\angle OMN$.
---
Translation:
Five.... | Let $\angle O M N=\theta$, then $\angle A B C=4 \theta, \angle A C B=6 \theta$.
Thus, $\angle B A C=180^{\circ}-10 \theta$.
Also, $\angle N O C=\frac{1}{2} \angle B O C=\angle B A C=180^{\circ}-10 \theta$.
$\angle M O C=\angle A O C=2 \angle A B C=8 \theta$,
Therefore, $\angle M O N=8 \theta+\left(180^{\circ}-10 \theta... | 12^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,757 |
Six. (20 points) Does there exist a function $f(x)$ defined on the set of real numbers $\mathbf{R}$, such that for any $x \in \mathbf{R}$, we have
$$
f(f(x))=x,
$$
and $f(f(x)+1)=1-x$?
If it exists, write down a function that satisfies the conditions; if it does not exist, please explain the reason. | Six, such a function does not exist.
Below is a proof by contradiction.
If there exists $f(x)$ such that all conditions hold, we first prove that $f(x)$ is a one-to-one mapping.
For any $a, b \in \mathbf{R}$, if $f(a)=f(b)$, by equation (1) we have
$$
a=f(f(a))=f(f(b))=b \text{, }
$$
i.e., $f(x)$ is a one-to-one mappi... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,758 |
2. For the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{3}=1$, the foci are $F_{1}$ and $F_{2}$, and point $P$ is on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then, $\left|P F_{1}\right|$ is ( times $\left|P F_{2}\right|$.
(A) 7
(B) 5
(C) 4
(D) 3 | 2. A.
The x-coordinate of point $P$ is 3, so the absolute value of the y-coordinate of point $P$ is $\sqrt{\frac{3}{2}}$. Meanwhile, $P F_{2}$ is perpendicular to the $x$-axis, hence
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=4 \sqrt{3} \text{. }
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,760 |
3. For the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, the sums of the first $n$ terms are $S_{n}$ and $T_{n}$, respectively. For all positive integers $n$, $\frac{S_{n}}{T_{n}}=\frac{2 n}{3 n+1}$. Then $\frac{a_{7}}{b_{9}}$ equals ( ).
(A) $\frac{7}{9}$
(B) $\frac{14}{27}$
(C) $\frac{2}{9}$... | 3. D.
Using the property of the sum of the first $n$ terms of an arithmetic sequence, we can set
$$
S_{n}=2 k n^{2}, T_{n}=k(3 n+1) n .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,761 |
5. The function $f(x)=a x^{4}+b \sin ^{3} x+c x^{2}+d x+$ 2 satisfies $f(1)=7, f(-1)=9$ and $f(2)+f(-2)$ $=124$. Then $f(\sqrt{2})+f(-\sqrt{2})=(\quad)$.
(A) 34
(B) 36
(C) 38
(D) 40 | 5.D.
Using $f(1)=7, f(-1)=9$ we get $a+c=6$;
Using $f(2)+f(-2)=124$ we get $32a+8c=120$.
Thus, $a=3, c=3$. We have $f(\sqrt{2})+f(-\sqrt{2})=40$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,763 |
4. Given a sequence of cards numbered from 1 to $2 n+1$, what is the maximum number of cards that can be selected so that in the selected cards, no card's number is equal to the sum of the numbers of any other two selected cards?
Translate the above text into English, please keep the original text's line breaks and ... | (First, prove that the number of selected cards cannot be more than $n+1$. On the other hand, the cards numbered $1,3, \cdots, 2 n+1$, these $n+1$ cards with odd numbers satisfy the conditions of the problem. Therefore, the maximum number of cards that can be selected is $n+1$.) | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,764 |
1. Given three positive real numbers $a$, $b$, $c$ satisfying $a+b+c=1$. Then the minimum value of $\frac{1-a b-b c-c a}{\sqrt{a b c}}$ is $\qquad$ | $1.2 \sqrt{3}$.
$$
\begin{array}{l}
\text { Since } a b+b c+c a \leqslant \frac{(a+b+c)^{2}}{3}=\frac{1}{3}, \\
\sqrt{a b c} \leqslant \sqrt{\left(\frac{a+b+c}{3}\right)^{3}}=\frac{\sqrt{3}}{9},
\end{array}
$$
Therefore, the minimum value of $\frac{1-a b-b c-c a}{\sqrt{a b c}}$ is $2 \sqrt{3}$. | 2 \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,766 |
2. The largest natural number that cannot be expressed as the sum of four distinct composite numbers is $\qquad$ .
翻译完成，保留了原文的格式和换行。 | 2. 26 .
The sum of the four smallest composite numbers is $4+6+8+9=27$. It is easy to prove that any natural number greater than 27 can be expressed as the sum of four different composite numbers. | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,767 |
3. The orbit of a certain spacecraft is an ellipse with the Earth's center $F$ as the left focus. It is measured that the distance from the perigee $A$ to the ground is $m \mathrm{~km}$, and the distance from the apogee $B$ to the ground is $n \mathrm{~km}$. The radius of the Earth is $r \mathrm{~km}$. There are four s... | 3. (1),(3),(4).
Taking $F$ as the origin and $AB$ as the $x$-axis to establish a rectangular coordinate system, and let the major axis of the ellipse be $2a$, the minor axis be $2b$. The focal distance is $2c$. Using the definition, it is easy to prove that (1), (3), and (4) are correct. | (1),(3),(4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,768 |
4. Given that $\boldsymbol{a}$ and $\boldsymbol{b}$ are non-zero vectors, and $\boldsymbol{a}+3 \boldsymbol{b}$ is perpendicular to $7 \boldsymbol{a}-5 \boldsymbol{b}$, $\boldsymbol{a}-4 \boldsymbol{b}$ is perpendicular to $7 \boldsymbol{a}-2 \boldsymbol{b}$. Then the angle between vectors $\boldsymbol{a}$ and $\boldsy... | $4.60^{\circ}$.
From $(a+3 b)(7 a-5 b)=0$, we get
$$
7 a^{2}+16 a \cdot b-15 b^{2}=0 \text {. }
$$
From $(a-4 b)(7 a-2 b)=0$, we get
$$
7 a^{2}-30 a \cdot b+8 b^{2}=0 \text {. }
$$
Subtracting the two equations yields $2 \boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b}^{2}$.
Substituting into equation (1) or (2) gi... | 60^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,769 |
5. Let the function $f(x)=\lg \left(a^{x}-b^{x}+9\right)(a>1$, $b>0)$, the domain of the function $f(x)$ is $[1,+\infty)$, and the range of the function $f(x)$ is a subset of $[1,+\infty)$. Then the relationship between the real numbers $a$ and $b$ is $\qquad$ . | $\begin{array}{l}\text { 5. } a \geqslant 1+b . \\ \text { When } x=1 \text {, } f(1)=\lg (a-b+9) \geqslant 1 .\end{array}$ | a \geqslant 1+b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,770 |
Three, (20 points) In a non-right $\triangle ABC$, the side lengths $a$, $b$, and $c$ satisfy $a + c = \lambda b (\lambda > 1)$.
(1) Prove: $\tan \frac{A}{2} \cdot \tan \frac{C}{2} = \frac{\lambda - 1}{\lambda + 1}$.
(Possible formulas to use: $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alph... | (1) Since $a+c=\lambda b$, we have $\sin A+\sin C=\lambda \sin B$,
which means $2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=2 \lambda \sin \frac{B}{2} \cdot \cos \frac{B}{2}$.
Thus, $\cos \frac{A-C}{2}=\lambda \cos \frac{A+C}{2}$.
Therefore, $\cos \frac{A}{2} \cdot \cos \frac{C}{2}+\sin \frac{A}{2} \cdot \sin \frac{C... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,772 |
Four. (20 points) Given in the rectangular prism $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, the edge lengths are $A B=B C=3, B B_{1}=4$. Connect $B_{1} C$, and draw a perpendicular from point $B$ to $B_{1} C$ intersecting $C C_{1}$ at point $E$ and $B_{1} C$ at point $F$.
(1) Prove: $A_{1} C \perp$ plane $E B D$;
(2) Let $A... | (1) Taking $D$ as the origin, the lines $D A$, $D C$, and $D D_{1}$ as the $x$-axis, $y$-axis, and $z$-axis respectively, we establish a spatial rectangular coordinate system, then
$$
\begin{array}{l}
D(0,0,0), A(3,0,0), C(0,3,0), B(3,3,0), \\
A_{1}(3,0,4), D_{1}(0,0,4), C_{1}(0,3,4), B_{1}(3,3,4).
\end{array}
$$
Let ... | \frac{25 \sqrt{34}}{34} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,773 |
Five. (20 points) Given three positive real numbers $x, y, z$ satisfying $x+y+z+\frac{1}{2} \sqrt{xyz}=16$. Prove:
(1) $\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{1}{8} \sqrt{xyz} \leqslant 7$;
(2) $\sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant 4+\frac{1}{4} \sqrt{xyz}$.
Six. (20 points) Given the quadratic function $f(x)=ax^2+bx+a$ sati... | Five, (1) Since $x+4 \geqslant 4 \sqrt{x}, y+4 \geqslant 4 \sqrt{y}, z+4 \geqslant 4 \sqrt{z}$,
Therefore, $x+y+z+12 \geqslant 4(\sqrt{x}+\sqrt{y}+\sqrt{z})$.
Combining $x+y+z=16-\frac{1}{2} \sqrt{x y z}$, it is easy to get $\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{1}{8} \sqrt{x y z} \leqslant 7$.
(2) From $x+y+\frac{1}{2} \sq... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,774 |
Example 1 As shown in Figure $1, A D, B E, C F$ are the three altitudes of $\triangle A B C$. If $E F$ bisects $A D$, then the Euler line of $\triangle A B C$ is parallel to side $B C$. | As shown in Figure 1, let the orthocenter of $\triangle ABC$ be point $H$, and take the centroid $G$. Then the line $GH$ is the Euler line of $\triangle ABC$.
Let $EF \cap AD = S$, and connect $DE$. Connect and extend $AG$ to intersect side $BC$ at point $K$. It is easy to see that $B, C, E, F$ and $H, D, C, E$ are re... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,775 |
Example 2 As shown in Figure $2, \triangle A B C$ has midpoints of each side as $A^{\prime} 、 B^{\prime} 、 C^{\prime}$. Prove: The Euler line of $\triangle A^{\prime} B^{\prime} C^{\prime}$ coincides with that of $\triangle A B C$.
Translate the above text into English, please keep the original text's line breaks and ... | As shown in Figure 2, take the circumcenter \( O \) of \( \triangle ABC \) and connect \( AA' \) and \( BB' \). Their intersection point is the centroid \( G \), so the line \( OG \) is the Euler line of \( \triangle ABC \). It is easy to prove that
\( OA' \perp BC \),
\( OB' \perp CA \), \( OC' \perp AB' \).
Furthermo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,776 |
Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }... | Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$
By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,777 |
Given $a, b > 0, \lambda \geqslant 2$. Prove:
$$
\sqrt{\frac{a}{a+\lambda b}}+\sqrt{\frac{b}{\lambda a+b}} \leqslant \frac{\lambda}{\sqrt{\lambda^{2}-1}} \text {. }
$$
Equality holds if and only if $x=\frac{b}{a}$ is a solution to the equation $x^{2}-\lambda\left(\lambda^{2}-3\right) x+1=0$. | Proof: Let $x=\frac{b}{a}, y=\frac{a}{b}$, then $x>0, y>0, xy=1$.
Thus, inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{1}{\sqrt{1+\lambda x}}+\frac{1}{\sqrt{1+\lambda y}} \leqslant \frac{\lambda}{\sqrt{\lambda^{2}-1}} \\
\Leftrightarrow \sqrt{\lambda^{2}-1} \cdot(\sqrt{1+\lambda x}+\sqrt{1+\lambda y}) \\
\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,778 |
Given a convex quadrilateral $ABCD$ inscribed in a circle, point $P$ is inside it. If $\angle PAB = \angle PBC = \angle PCD = \angle PDA$, prove: $AB \cdot CD = BC \cdot DA$.
---
The translation is provided as requested, maintaining the original format and line breaks. | Prove: As shown in Figure 3, construct
$\triangle K B A \backsim \triangle P C D$,
connect $P K$, then
$$
\angle A K B=\angle D P C .
$$
Notice that
$$
\begin{array}{l}
\angle A P B=180^{\circ}-(\angle P A B \\
+\angle P B A) \\
= 180^{\circ}-(\angle P B C+ \\
\angle P B A) \\
= 180^{\circ}-\angle A B C=\angle A D C,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,779 |
Example 5 Prove:
$$
S_{m, n}=\sum_{k=m}^{n}(-1)^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{k}^{m}=(-1)^{m} \delta_{m, n}
$$
$(m \leqslant n)$. Here
$$
\delta_{m, n}=\left\{\begin{array}{ll}
1, & m=n ; \\
0, & m \neq n .
\end{array}\right.
$$ | Prove: When $m=n$, $S_{m, n}$ has only one term, and
$$
S_{m, n}=(-1)^{m} \mathrm{C}_{m}^{m} \mathrm{C}_{m}^{m}=(-1)^{m} \text {. }
$$
When $m<n$, from equation (IV) we get
$$
\begin{array}{l}
S_{m, n}=\sum_{k=m}^{n}(-1)^{k} \mathrm{C}_{n}^{m} \mathrm{C}_{n-m}^{k-m} \\
=\mathrm{C}_{n}^{m} \sum_{k=m}^{n}(-1)^{k} \mathr... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,780 |
Example 6 Prove:
$$
\sum_{k=0}^{n-m}(-1)^{k} \mathrm{C}_{n-m}^{k} \cdot \frac{1}{m+k}=\frac{1}{m \mathrm{C}_{n}^{m}} .
$$ | Proof: (1) When $n-m=0$, equation (1) obviously holds.
(2) Suppose equation (1) holds for $n-m$. Notice that
$$
\begin{array}{l}
\sum_{k=0}^{n-m+1}(-1)^{k} \mathrm{C}_{n-m+1}^{k} \cdot \frac{1}{m+k}- \\
\sum_{k=0}^{n-m}(-1)^{k} \mathrm{C}_{n-m}^{k} \cdot \frac{1}{m+k} \\
=\sum_{k=0}^{n-m+1}(-1)^{k}\left(\mathrm{C}_{n-m... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,781 |
Example 7 Prove:
\[
\begin{array}{l}
(1-x)^{-n} \\
=\mathrm{C}_{n-1}^{n-1}+\mathrm{C}_{n}^{n-1} x+\cdots+\mathrm{C}_{n+k-1}^{n-1} x^{k}+\cdots
\end{array}
\] | Proof: (1) When $n=1$, equation (1) is
$$
(1-x)^{-1}=1+x+x^{2}+\cdots \text {. }
$$
By the formula for the sum of an infinite geometric series, equation (1) holds.
(2) Suppose when $n=k$, we have
$$
(1-x)^{-k}=\sum_{l=0}^{\infty} \mathrm{C}_{k+l-1}^{k-1} x^{l} \text {. }
$$
Thus, when $n=k+1$, using the rule for mult... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,782 |
Example 8 Find:
(1) $\mathrm{C}_{n}^{1}-3 \mathrm{C}_{n}^{3}+5 \mathrm{C}_{n}^{5}-7 \mathrm{C}_{n}^{7}+\cdots$;
(2) $2 C_{n}^{2}-4 C_{n}^{4}+6 C_{n}^{6}-8 C_{n}^{8}+\cdots$. | Solution: From the binomial expansion and (III), we have
$$
\begin{array}{rl}
n & n(1+\mathrm{i})^{n-1} \\
= & n\left(\mathrm{C}_{n-1}^{0}+\mathrm{i} \mathrm{C}_{n-1}^{1}+\mathrm{i}^{2} \mathrm{C}_{n-1}^{2} \cdots+\right. \\
& \left.\mathrm{i}^{n-1} \mathrm{C}_{n-1}^{n-1}\right) \\
= & n\left(\mathrm{C}_{n-1}^{0}-\math... | \begin{array}{l}
\text{(1) } \mathrm{C}_{n}^{1}-3 \mathrm{C}_{n}^{3}+5 \mathrm{C}_{n}^{5}-7 \mathrm{C}_{n}^{7}+\cdots = n \times 2^{\frac{n-1}{2}} \cos \frac{(n-1) \pi}{4} \\
\text{(2) } | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,783 |
Example 9 Let $n=1990$. Then
$$
\frac{1}{2^{n}}\left(1-3 \mathrm{C}_{n}^{2}+3^{2} \mathrm{C}_{n}^{4}-\cdots+3^{99} \mathrm{C}_{n}^{108}-3^{90} \mathrm{C}_{n}^{900}\right) \text {. }
$$
(1990, National High School Mathematics Competition) | Solution: Original expression
$$
\begin{aligned}
= & \left(\frac{1}{2}\right)^{1990}-C_{1990}^{2}\left(\frac{1}{2}\right)^{1988}\left(\frac{\sqrt{3}}{2}\right)^{2}+\cdots+ \\
& C_{1990}^{1998}\left(\frac{1}{2}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)^{1988}-C_{1990}^{1990}\left(\frac{\sqrt{3}}{2}\right)^{1990} .
\end{... | -\frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,784 |
Question 4 As shown in Figure 2, in $\triangle ABC$, $AB=AC$, $AD$ is the angle bisector of $\angle A$, point $M$ is on $BC$, and $DB \perp AB$, $DM \perp EF$, point $E$ is on the extension of $AB$, and point $F$ is on $AC$. Prove: $ME=MF$.
untranslated text remains unchanged. | Proof: As shown in Figure 2, connect $DE$, $DC$, and $DF$. Since $DB \perp AB$, $AB = AC$, and $AD$ is the angle bisector of $\angle A$, we know that $DC \perp AC$.
Notice that $DM \perp$
$EF$, so, $B$, $E$, $D$,
$M$ and $M$, $D$, $C$, $F$ are
respectively concyclic. Therefore,
$$
\begin{aligned}
& \angle MED = \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,785 |
Question 5 As shown in Figure 3, in $\triangle ABC$, $AB=AC$, $AD$ is the angle bisector of $\angle A$, point $M$ is on $BC$, and $DB \perp AB$, $ME=MF$, point $E$ is on the extension of $AB$, and point $F$ is on $AC$. Prove: $DM \perp EF$.
untranslated text remains the same as the original, only the question stateme... | Prove: As shown in Figure 3, draw $EG \parallel AC$ intersecting the extension of $CB$ at point $G$. Thus,
$\angle EGB = \angle FCM$.
Noting that $EM = MF$, therefore,
$\triangle EMG \cong \triangle FMC$.
So, $EG = CF$.
But $\angle ACB = \angle ABC = \angle GBE$, hence
$EB = EG = CF$.
Considering $DC = DB$, then we hav... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,786 |
Example 1 Solve the system of equations
$$
\left\{\begin{array}{l}
x+a y+a^{2} z+a^{3}=0 \\
x+b y+b^{2} z+b^{3}=0 \\
x+c y+c^{2} z+c^{3}=0
\end{array}\right.
$$ | Analysis: Observing the shape characteristics of the three equations in the system, it inspires the construction of a cubic equation
$$
t^{3}+z t^{2}+y t+x=0,
$$
where $a, b, c$ are its three roots, and $a, b, c$ are also the roots of the equation
$$
(t-a)(t-b)(t-c)=0.
$$
Expanding the latter equation yields
$$
t^{3}... | x=-a b c, y=a b+b c+c a, z=-(a+b+c) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,787 |
Example 2 As shown in Figure 1, given that $P$ is a point inside $\triangle ABC$, and $D$, $E$, $F$ are the feet of the perpendiculars from $P$ to $BC$, $CA$, and $AB$ respectively. Find the point $P$ that minimizes $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$.
(22nd IMO) | Let $B C+C A+A B=l$, and the area of $\triangle A B C$ be $S$. Then
$$
B C \cdot P D+C A \cdot P E+A B \cdot P F=2 S .
$$
Construct the function $f(x)=\frac{1}{x}$.
Since $\frac{B C}{l}+\frac{C A}{l}+\frac{A B}{l}=1$, by Jensen's inequality, we have
$$
\begin{array}{l}
\frac{B C}{l} \cdot \frac{1}{P D}+\frac{C A}{l} \... | \text{incenter of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,788 |
Example 3 As shown in Figure 3, $l$ is a fixed line. Construct $\triangle A B C$ such that line $l$ is its Euler line. How many triangles can be constructed with $l$ as the Euler line? Why? | Explain: Divide the following
into several steps:
(1) Take any two points $O, G$ on the line $l$;
(2) Take point $H$ such that $GH = 2OG$;
(3) Draw any line $l_{1}$ through point $H$, and draw line $l_{2} \parallel l_{1}$ through point $O$;
(4) Draw any line $l_{3}$ through point $G$ intersecting lines $l_{1}$ and $l_{... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,789 |
Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-7 x+8=0$, and $\alpha>\beta$. Without solving the equation, find the value of $\frac{2}{\alpha}+3 \beta^{2}$. | Solution: By Vieta's formulas, we know $\alpha+\beta=7, \alpha \beta=8$. Let $M=\frac{2}{\alpha}+3 \beta^{2}, N=\frac{2}{\beta}+3 \alpha^{2}$, it is easy to find that $M+N=100 \frac{3}{4}$,
$M-N=-\frac{85 \sqrt{17}}{4}$.
(1) + (2) gives
$$
M=\frac{2}{\alpha}+3 \beta^{2}=\frac{1}{8}(403-85 \sqrt{17}) \text {. }
$$ | \frac{1}{8}(403-85 \sqrt{17}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,790 |
Example 4 Given the polynomial
$$
f(x)=a x^{3}+b x^{2}-c x-d,
$$
and $f(-1), f(0), f(1), f(2)$ are all integers. Prove that when $x$ is an integer, $f(x)$ is an integer. | Solution: Construct the polynomial using the Lagrange interpolation formula
$$
\begin{aligned}
f(x)= & \frac{(x-2)(x-1) x}{-6} \cdot f(-1)+ \\
& \frac{(x-2)(x-1)(x+1)}{2} \cdot f(0)+ \\
& \frac{(x-2) x(x+1)}{-2} \cdot f(1)+ \\
& \frac{(x-1) x(x+1)}{6} \cdot f(2) .
\end{aligned}
$$
Since \( f(-1), f(0), f(1), f(2) \) a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,791 |
Example 5 Given that $a, b, x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$. | Given the conditions, construct the recurrence relation:
$$
\begin{array}{l}
s_{n}=a x^{n}+b y^{n}, n=1,2, \cdots, \\
\text { then } s_{n+2}=a x^{n+2}+b y^{n+2} \\
=\left(a x^{n+1}+b y^{n+1}\right)(x+y)- \\
\quad x y\left(a x^{n}+b y^{n}\right) .
\end{array}
$$
Therefore, \( s_{n+2}=(x+y) s_{n+1}-x y s_{n} \). Substit... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,792 |
Example 6 Given $a>0$, and $\sqrt{b^{2}-4 a c}=b-2 a c$. Find the minimum value of $b^{2}-4 a c$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let $y=a x^{2}+b x+c$.
From $a<0$, we know
$$
\Delta=b^{2}-4 a c>0 \text {. }
$$
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points
$$
A\left(x_{1}, 0\right), B\left(x_{2}, 0\right) \text {. }
$$
Since $x_{1} x_{2}=$
$\frac{... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,793 |
Example 7 Let $x, y \in \mathbf{R}$, and $x^{2}+y^{2}=1$. Prove that for any positive numbers $a, b$, we have
$$
\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \geqslant a+b .
$$ | Solution: Let $z_{1}=a x+b y \mathrm{i}, z_{2}=b x+a y \mathrm{i}$.
By $\left|z_{1}\right|+\left|z_{2}\right| \geqslant\left|z_{1}+z_{2}\right|$, then
$$
\begin{array}{l}
\sqrt{a^{2} x^{2}+b^{2} y^{2}}+\sqrt{a^{2} y^{2}+b^{2} x^{2}} \\
\geqslant \sqrt{(a x+b x)^{2}+(b y+a y)^{2}} \\
=\sqrt{(a+b)^{2}\left(x^{2}+y^{2}\ri... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,794 |
Let $n$ be a positive integer, and $\{A, B, C\}$ be a partition of the set $\{1,2,3, \cdots, 3 n\}$ such that $|A|=\mid B$ $|=| C \mid=n$, where $|S|$ denotes the number of elements in set $S$. Prove that there exist $x \in A, y \in B, z \in C$, such that one of $x, y, z$ is the sum of the other two. | Proof 1: Let the smallest elements of sets $A, B, C$ be $a, b, c$, respectively, and $a < b < c$. If $d = c - b > 1$, then $c_1 - 1, c_2 - 1, \cdots, c_n - 1$ these $n$ numbers have at most $n-1$ belonging to $A$, and at least one $c_i - 1 \in B$. Also, $(c_i - 1) + 1 = c_i$, so the proposition holds.
If $d = 1$, let ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,796 |
Let $a$, $b$, $c$ be the lengths of the three sides of a triangle with a perimeter not exceeding $2 \pi$. Prove: $\sin a$, $\sin b$, $\sin c$ can form the lengths of the three sides of a triangle.
(2004, IMO China National Training Team Selection Exam) | Proof: (1) From the given, we know $0c$, i.e., $\frac{a+b}{2}>\frac{c}{2}$.
Since $0<\frac{c}{2}<\frac{\pi}{2}$, we have $\sin \frac{a+b}{2}>\sin \frac{c}{2}>0$.
If $\pi>\frac{a+b}{2}>\frac{c}{2}>\frac{\pi}{2}$, then $\sin \frac{a+b}{2}>\sin \frac{c}{2}>0$ as well.
$$
(3) From the given, we know $|a-b|<c$, i.e., $\left... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,797 |
Given real numbers $a, b, c, x, y, z$ satisfying
$$
\begin{array}{l}
(a+b+c)(x+y+z)=3, \\
\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4 .
\end{array}
$$
Prove: $a x+b y+c z \geqslant 0$.
(2004, China National Team Training Problem) | Analysis: Direct proof of this problem is quite difficult, and the original solution method is the substitution method, which is hard to think of. In fact, a more natural approach to solving the problem is to handle it as a whole.
$$
\text { Let } A=a x+b y+c z, B=a y+b z+c x \text {, }
$$
$C=a z+b x+c y$, we can conje... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,798 |
$$
\begin{array}{l}
\text { 1. } \sqrt{5-\sqrt{5}-\sqrt{3-\sqrt{29-12 \sqrt{5}}}} \text { and } \\
\sqrt{3-\sqrt{29-12 \sqrt{5}} \text { have a ( ). }}
\end{array}
$$
1. $\sqrt{5-\sqrt{5}-\sqrt{3-\sqrt{29-12 \sqrt{5}}}}$ and
(A) sum of 1
(B) difference of 1
(C) product of 1
(D) quotient of 1
| - 1.D.
Since $\sqrt{29-12 \sqrt{5}}=\sqrt{(3-2 \sqrt{5})^{2}}=2 \sqrt{5}-3$, then
$$
\begin{array}{l}
\sqrt{3-\sqrt{29-12 \sqrt{5}}}=\sqrt{3-(2 \sqrt{5}-3)} \\
=\sqrt{6-2 \sqrt{5}}=\sqrt{(\sqrt{5}-1)^{2}}=\sqrt{5}-1
\end{array}
$$
And $\sqrt{5-\sqrt{5}-\sqrt{3-\sqrt{29-12 \sqrt{5}}}}$
$$
=\sqrt{5-\sqrt{5}-(\sqrt{5}-1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,799 |
Example 4 As shown in Figure 4, in the acute triangle $\triangle ABC$, side $BC$ is the shortest side. Points $O, G, I, H$ are the circumcenter, centroid, incenter, and orthocenter of the triangle, respectively. When $OG = GI = IH$, then $AB : AC : BC =$ | Explanation: From the definition of the Euler line, the circumcenter $O$, centroid $G$, and orthocenter $H$ are collinear. By the properties of the Euler line, we have
$$
O G: G H=1: 2 \text {. }
$$
Given $O G=G I=I H$, it follows that $I$ is the midpoint of $G H$, meaning point $I$ lies on the Euler line $O H$.
As s... | 2: 2: 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,800 |
2. The sum of all natural numbers $x$ that make the algebraic expression $y=\frac{x^{2}+11}{x+1}$ an integer is ( ).
(A) 5
(B) 6
(C) 12
(D) 22. | 2.D.
$y=\frac{x^{2}+11}{x+1}=x-1+\frac{12}{x+1}$. To make $\frac{12}{x+1}$ an integer, the natural number $x$ can take the values $0,1,2,3,5,11$, a total of 6 values, their sum is 22. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,801 |
3. The face of a clock is circular, with 12 equally spaced points on its circumference to indicate the hours. The number of rectangles that can be formed using these points as vertices is ( ).
(A) 6
(B) 12
(C) 15
(D) 24 | 3.C.
In the lines connecting the equal division points, there are 6 diameters. Each pair of diameters forms the diagonals of 1 rectangle, resulting in $\frac{6 \times 5}{2}=15$ pairs, hence yielding 15 rectangles. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,802 |
4. As shown in Figure 1, in rectangle $A B C D$, $A B=2, B C=$ $\sqrt{3}, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Line segments $D E$, $B F$, and $E F$ intersect diagonal $A C$ at points $M$, $N$, and $P$ respectively. The number of right triangles formed by the line segments in Figure 1 is ( ).
(... | 4.B.
Let the notation $[A]$ represent the number of right-angled triangles with $A$ as the right-angle vertex, then $[A]=2$,
Similarly, $[B]=1,[C]=2,[D]=1,[E]=3$,
$$
[F]=3,[M]=[N]=[P]=0 \text {. }
$$
Therefore, the number of right-angled triangles is
$$
2+1+2+1+3+3=12 \text {. }
$$ | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,803 |
5. $a, b, c, d$ are positive numbers, $a>b>c>d$, and
$$
\begin{array}{l}
x=\sqrt{(a b+c d)(a-b)(c-d)}, \\
y=\sqrt{(a c+b d)(a-c)(b-d)}, \\
z=\sqrt{(a d+b c)(a-d)(b-c)} .
\end{array}
$$
Then the sides $x, y, z$ ( ).
(A) must form an acute triangle
(B) must form an obtuse triangle
(C) must form a right triangle
(D) may ... | 5.C.
Let $a b+c d=\alpha, a c+b d=\beta, a d+b c=\gamma$. Then
$$
\begin{array}{l}
(a-b)(c-d)=\beta-\gamma, \\
(a-c)(b-d)=\alpha-\gamma, \\
(a-d)(b-c)=\alpha-\beta .
\end{array}
$$
The left side of the above three equations are all positive, so the right side is also positive.
Therefore, we have $\alpha>\beta>\gamma$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,804 |
6. Use red, yellow, and blue to color each square of a $3 \times 3$ grid, such that:
(1) each row contains all three colors,
(2) each column contains all three colors,
(3) adjacent squares (squares sharing a common edge) are different colors.
Then the number of different coloring methods is ( ) kinds.
(A) 12
(B) 18
(C)... | 6.A.
Let $A$, $B$, and $C$ represent the colors red, yellow, and blue, respectively. For convenience, we number the cells $1, 2, \cdots, 9$ (as shown in Figure 2). Consider the three coloring methods for the top-left corner $B$, $C$. First, assume that cell 1 is colored $A$: $A$, then the colors of cells 2 and 4 are d... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,805 |
1. The side length of square $A B C D$ is 4. A semicircle is drawn inside the square with $A B$ as its diameter. $C M$ and $D N$ are tangents to the semicircle, with $M$ and $N$ being the points of tangency. If $C M$ and $D N$ intersect at a point $P$ inside the square, then the area of $\triangle P M N$ is | $$
\text { II. } 1 . \frac{27}{25} \text {. }
$$
As shown in Figure 3, let $C M$ intersect $A D$ at point $E$, and $D N$ intersect $B C$ at point $F$. Then quadrilateral $E F C D$ is a rectangle.
By the equal tangent segments, we have
$$
C M=C B=4 \text {. }
$$
Let $x=E A=E M$, then
$$
D E=4-x, C E=4+x \text {. }
$$
... | \frac{27}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,806 |
2. Fill in $n$ distinct numbers on a circle so that for every three consecutive numbers, the middle number is equal to the product of the two numbers on its sides. Then $n=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directl... | 2.6.
According to the problem, $n \geqslant 3$. Take any two adjacent numbers $a, b(a \neq b)$, and by the problem's condition, the numbers on the circle can be written in sequence as
$$
a, b, \frac{b}{a}, \frac{1}{a}, \frac{1}{b}, \frac{a}{b}, a, b, \cdots
$$
Starting from the 7th number, the sequence of numbers has... | 6 | Number Theory | proof | Yes | Yes | cn_contest | false | 715,807 |
3. A certain temple has three bronze bells, A, B, and C. Bell A rings once every $4 \mathrm{~s}$, Bell B rings once every $5 \mathrm{~s}$, and Bell C rings once every $6 \mathrm{~s}$. At the arrival of the New Year, the three bells ring simultaneously and stop ringing at the same time. A person hears a total of 365 bel... | 3.484 .
Let the duration of the bell ringing be $60 k \mathrm{~s}$, then during this period, the number of times the first clock rings is $x=\frac{60 k}{4}+1=15 k+1$; the number of times the second clock rings is $y=\frac{60 k}{5}+1=12 k+1$; the number of times the third clock rings is $z=\frac{60 k}{6}+1=10 k+1$; the... | 484 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,808 |
4. From $1,2, \cdots, 10$ choose 3 different numbers $a, b, c$ as the coefficients of the quadratic equation $a x^{2}+b x=c$. Then the number of equations with different solutions is $\qquad$ | 4.654.
When considering $a1$, for each factor $k$ of $d$ greater than 1, if $\frac{c}{k} \geqslant 3$, then the array $\left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\right)$ has already been considered before and should be removed. The number of remaining arrays is denoted as $f(c)$. Thus, we have
$$
\begin{array}{l}
f(3... | 654 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,809 |
1. (20 points) If positive integers $a$ and $b$ are the roots of the equation $x^{2}-\frac{a^{2}-9}{13} x+10 b+56+5 \sqrt{205+52 b}=0$, find the values of $a$ and $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, 1. From the conditions, we have
$$
a+b=\frac{a^{2}-9}{13} \text {, }
$$
$$
a b=10 b+56+5 \sqrt{205+52 b} \text {. }
$$
From equation (1), we get $a^{2}-13 a-(13 b+9)=0$.
Since $a$ and $b$ are positive integers, then
$$
a=\frac{13+\sqrt{13^{2}+4(13 b+9)}}{2}=\frac{13+\sqrt{205+52 b}}{2} \text {. }
$$
Therefore,... | a=23, b=17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,810 |
Example 5 As shown in Figure 5, in acute $\triangle A B C$, $\angle B=$ $30^{\circ}, A D 、 B E$ are two altitudes, point $H$ is the orthocenter, point $G$ is the centroid of $\triangle A B H$, line $C G$ intersects $A B 、 A D$ at $M 、 N$. Prove: $\triangle A M N$ is an equilateral triangle. | As shown in Figure 5, take the circumcenter $O$ of $\triangle A B H$. It is easy to see that point $C$ is the orthocenter of $\triangle A B H$. Therefore, points $O$, $G$, and $C$ are collinear, and this line is the Euler line of $\triangle A B H$.
Draw the altitude $C F$ and connect $O B$, $O H$. It is easy to see th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,811 |
2. (25 points) In quadrilateral $ABCD$, $\angle ABC = \angle ADC = 90^{\circ}$, $P$ is the intersection of diagonals $AC$ and $BD$, $M$ and $N$ are points on $AB$ and $CD$ respectively, such that $DM \perp AC$ and $BN \perp AC$. Prove that $M$, $N$, and $P$ are collinear. | 2. As shown in Figure 4, connect $PM$ and $PN$. We only need to prove that $\triangle PBN \sim \triangle PDM$.
Since $BN \parallel DM$, then $\angle PBN = \angle PDM$.
And $\frac{BP}{DP} = \frac{S_{\triangle ABC}}{S_{\triangle ADC}}$
$$
= \frac{AB \cdot BC}{AD \cdot DC}.
$$
Given that $\angle DAM$ and $\angle BCN$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,812 |
3. (25 points) Given that $a$ and $b$ are coprime positive integers, satisfying $a+b=2005$. Let $[x]$ denote the integer part of the number $x$, and let
$$
\begin{array}{l}
A=\left[\frac{2005 \times 1}{a}\right]+\left[\frac{2005 \times 2}{a}\right]+\cdots+\left[\frac{2005 \times a}{a}\right], \\
B=\left[\frac{2005 \tim... | 3. It is evident that the last terms of the two sums $A$ and $B$ are both 2005. Consider the remaining terms.
Since $(a, b)=1, a+b=2005$, it follows that
$(a, 2005)=1, (b, 2005)=1$.
Therefore, for $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1, \frac{2005 m}{a}$ and $\frac{2005 n}{b}$ are not integers, and t... | 2011016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,813 |
1. (16 points) Given the equation in $x$
$$
x^{2}-\left(m^{2}+2 m-3\right) x+2(m+1)=0
$$
has two real roots that are opposites of each other.
(1) Find the value of the real number $m$;
(2) If the equation in $x$ $x^{2}-(k+m) x-3 m-k-5=0$ has roots that are all integers, find all real numbers $k$ that satisfy the condi... | Three, 1. (1) From the problem, we know that $m^{2}+2 m-3=0$. Solving this, we get $m=-3, m=1$.
When $m=-3$, $\Delta=16>0$; when $m=1$, $\Delta=-16<0$. Therefore, the value of the real number $m$ is -3.
(2) From (1), $m=-3$, then the equation becomes $x^{2}-(k-3) x+4-k=0$. Let the two roots of the equation be $x_{1}$ a... | k=4 \text{ or } k=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,814 |
2. (16 points) A warehouse has 50 containers of the same specification, which are to be delivered to the dock by a transportation company. The transportation company has three types of trucks that can carry 1, 2, and 3 containers per trip, respectively. The charges for each trip are 120 yuan, 160 yuan, and 180 yuan, re... | 2. Let the number of trucks needed to transport 1, 2, and 3 containers be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=20, \\
x+2 y+3 z=50 .
\end{array}\right.
$$
(1) $\times 3$ - (2) gives $2 x+y=10$.
Then $\left\{\begin{array}{l}y=10-2 x, \\ z=10+x .\end{array}\r... | 3300 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,815 |
3. (18 points) As shown in Figure 6, in trapezoid $ABCD$, $AD \parallel BC$, $AB \perp BC$, $AD = 8 \, \text{cm}$, $BC = 16 \, \text{cm}$, $AB = 6 \, \text{cm}$. Moving points $M$ and $N$ start from points $B$ and $C$ simultaneously, moving along $BC$ and $CD$ respectively,
B
The speeds of points $M$ and $N$ are $2 \, ... | 3. (1) Suppose points $M$ and $N$ have moved for $x \, \text{s}$, such that $M N \parallel B C$. Then
$$
B M=2 x(\text{cm}), C M=16-2 x(\text{cm}), C N=x(\text{cm}).
$$
From the given information, $B D=10(\text{cm})$.
Draw $D H \perp B C$ at point $H$, then $B H=A D=8$.
Thus, $C H=8(\text{cm}), C D=B D=10(\text{cm})$.... | \frac{40}{9} \text{s}, \frac{93}{5} \text{cm}^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,816 |
9.1. Given a parallelogram $A B C D(A B<B C)$, take any two points $P 、 Q$ on its sides $B C$ and $C D$, respectively, such that $C P=C Q$. Prove: For all different choices of $P 、 Q$, the circumcircle of $\triangle A P Q$ passes through a common point other than point $A$. | 9.1. As shown in Figure 1, construct the angle bisector of $\angle C$, and let the reflection of point $A$ over this bisector be $A^{\prime}$. Observe point $A^{\prime}$. Since $C P = C Q$, the angle bisector of $\angle C$ is the perpendicular bisector of segment $P Q$. Therefore, quadrilateral $P Q A^{\prime} A$ is an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,817 |
9.2. Liza writes the positive integers from 1 to $22^{2}$ in the squares of a 22 $\times$ 22 grid (each square contains one integer). Is it possible for Alex to choose 2 squares that share a common side or vertex, such that the sum of the numbers written in them is a multiple of 4? | 9.2. OK.
Assume Alek cannot select the required squares. Now, replace each number with its remainder when divided by 4, so the grid contains 121 each of 0, 1, 2, and 3. Divide the grid into 121 \(2 \times 2\) squares. Since Alek cannot select the required squares, in each \(2 \times 2\) square, there is at most 1 zero... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,818 |
9.3. Let $a_{1}>1, a_{2}>1, a_{3}>1, a_{1}+a_{2}+a_{3}=S$. Given that for $i=1,2,3$, we have $\frac{a_{i}^{2}}{a_{i}-1}>S$. Prove: $\frac{1}{a_{1}+a_{2}}+\frac{1}{a_{2}+a_{3}}+\frac{1}{a_{3}+a_{1}}>1$ : | 9.3. It is known that
$$
\begin{array}{l}
\frac{a_{1}^{2}}{a_{1}-1}>S \\
\Leftrightarrow a_{1}^{2}>\left(a_{1}+a_{2}+a_{3}\right)\left(a_{1}-1\right) \\
\Leftrightarrow a_{1}+a_{2}+a_{3}>a_{1}\left(a_{2}+a_{3}\right) . \\
\Leftrightarrow \frac{1}{a_{2}+a_{3}}>\frac{a_{1}}{a_{1}+a_{2}+a_{3}} .
\end{array}
$$
Adding the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,819 |
9.4. On the table, there are 365 cards, each with a different number written on its back. Vasya can pay 1 ruble to choose any 3 cards and ask Borya to arrange them from left to right in increasing order of the numbers on their backs. Can Vasya achieve the following goal by paying 2000 rubles: to arrange all 365 cards i... | 9.4. It can be done.
First, we prove a lemma.
Lemma: Suppose Vasya can arrange a certain \(N-1\) cards in the required order on the table by paying \(x\) rubles, where \(N \leqslant 3^{k}\). Then, he can place the remaining 1 card in the correct position by paying no more than \(k\) rubles, i.e., the total cost will n... | 1825 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,820 |
9.5. Given 10 distinct non-zero numbers, the sum or product of any two of them is a rational number. Prove: the square of each number is a rational number. | 9.5. Proof 1: If all the numbers are rational, the proposition naturally holds.
Now suppose that among the 10 numbers, there is an irrational number $a$, then the other numbers are of the form $p-a$ or $\frac{p}{a}$, where $p$ is a rational number.
We will prove that there are no more than 2 numbers of the form $p-a$.... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,821 |
Example 6 As shown in Figure $6, \odot O$ is the circumcircle of $\triangle A B C$, and point $H$ is the orthocenter of $\triangle A B C$. Let $B C=a, C A=b$, $A B=c$, and the radius of $\odot O$ be $R$. Try to express the length of $O H$ using $a, b, c, R$.
Express the length of $O H$ using $a, b, c, R$. | As shown in Figure 6, take the centroid $G$ of $\triangle ABC$, then point $G$ must lie on $OH$ (by the definition of the Euler line), and
$$
O H=3 O G \text{. }
$$
The ray $AG$ intersects $\odot O$ at point $A'$ and intersects side $BC$ at point $D$, where $D$ is the midpoint of $BC$. The line $OH$ intersects $\odot ... | O H=\sqrt{9 R^{2}-\left(a^{2}+b^{2}+c^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,822 |
9.6. How many ways are there to divide the set
$$
\left\{2^{0}, 2^{1}, 2^{2}, \cdots, 2^{2000}\right\}
$$
into two non-empty disjoint subsets $A$ and $B$, such that the equation $x^{2}-S(A) x+S(B)=0$ has integer roots? Here, $S(M)$ denotes the sum of all elements in the set $M$. | 9.6.1 003 types.
Let $x_{1} \leqslant x_{2}$ be the roots of the equation, then
$$
x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B),
$$
and $x_{1}, x_{2} \in \mathbf{Z}_{+}$. Therefore,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+S(A)+1 \\
=1+2+4+\cdots+2^{200}+1=2^{2006},
\end{array}
$$
which means $x_{1}+1... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,823 |
9.7. In an acute $\triangle ABC$, construct altitudes $AA'$ and $BB'$. Let $D$ be a point on the circumcircle of $\triangle ABC$ on the arc $\overparen{ACB}$. Suppose the line $AA'$ intersects $BD$ at point $P$, and the line $BB'$ intersects $AD$ at point $Q$. Prove: the line $A'B'$ passes through the midpoint of segme... | 9.7. As shown in Figure 2, for definiteness, let point $D$ lie on the arc $\overparen{B C}$ that does not contain point $A$. Let the orthocenter of $\triangle A B C$ be $H$, and let the other intersection points of lines $A H, B H$ with the circumcircle be $H_{A}, H_{B}$, respectively.
Clearly, $\angle C A H_{A}=\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,824 |
9.8. There are 100 representatives from 50 countries sitting around a round table, with 2 representatives from each country. Prove: it is possible to divide them into two groups such that each group consists of 50 representatives from 50 countries, and each person is in the same group as at most one of their neighbors. | 9.8. Divide 100 representatives into 50 "adjacent pairs," and call the two people in the same pair acquaintances. Clearly, to prove the assertion in the problem, it is only necessary to prove that each pair of acquaintances can be separated into two different groups. Below is a specific procedure.
First, place the rep... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,825 |
10.1. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers. | 10.1.11.
From the problem, we have
$$
\begin{array}{l}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}, \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}, \\
2=2 \times 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \times 5=\frac{2^{6}-2^{2}}{2^{3}-2} .
\en... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,826 |
10.2. In a $2 \times n$ grid, each cell contains a positive number such that the sum of the two numbers in each column is equal to 1. Prove: It is possible to delete one number from each column so that the sum of the remaining numbers in each row does not exceed $\frac{n+1}{4}$. | 10.2. Suppose the numbers placed in the first row are $a_{1}, a_{2}, \cdots, a_{n}$. If necessary, by swapping the positions of the columns, we can ensure that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. At this point, the numbers written in the second row are
$$
b_{1}=1-a_{1}, b_{2}=1-a_{2}, \cdots, b_{n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,827 |
10.3. On the back of 2005 cards, there are 2005 different real numbers written. Each time you ask a question, you can point to any three cards and ask about the set of 3 numbers written on them. How many times do you need to ask at minimum to definitely find out what numbers are written on the back of each card? | 10.3.1003 times.
Assume that $N$ questions have been proposed.
Obviously, each card should participate in at least one question; otherwise, it would be impossible to determine the number on that card.
Suppose there are $k$ cards that each participate in exactly one question. Then, in the same question, it is impossib... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,828 |
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