problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 8 Let $f$ and $g$ be real functions defined on $\mathbb{R}$, and for all $x$ and $y$ satisfy the equation:
$$
f(x+y)+f(x-y)=2 f(x) g(y) .
$$
Prove: If $f(x)$ is not identically zero, and $|f(x)| \leqslant 1$ for all $x$, then $|g(y)| \leqslant 1$ for all $y$. | Proof: Since $|f(x)|$ has an upper bound, $|f(x)|$ must have a least upper bound, denoted as $M$. Clearly, $0 < |g(y_0)| \leq 1$, then for all $x$ we have
$$
\begin{array}{l}
2|f(x)|\left|g\left(y_{0}\right)\right| \\
=\left|f\left(x+y_{0}\right)+f\left(x-y_{0}\right)\right| \\
\leqslant\left|f\left(x+y_{0}\right)\righ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,407 |
1. In the real domain, the value of the expression
||$\sqrt{-(x-4)^{2}}-1|-2|$ is ( ).
(A) can only be 3
(B) can only be 1
(C) can only be 2
(D) none of the above answers is correct | $-1 .(\dot{B})$
$$
\begin{array}{l}
\because-(x-4)^{2} \leqslant 0, \text { and }-(x-4)^{2} \geqslant 0, \\
\therefore(x-4)^{2}=0 . \\
\therefore \text { original expression }=|| 0-1|-2|=|1-2|=|-1|=1 .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,408 |
3. If $a, b, c$ are real numbers, $x=a^{2}-2 b+\frac{\pi}{3}, y=$ $b^{2}-2 c+\frac{\pi}{6}, z=c^{2}-2 a+\frac{\pi}{2}$, then among $x, y, z$
(A) at least one is greater than 0
(B) at least one is not less than 0
(C) at least one is not greater than 0
(D) at least one is less than 0 | 3. (A).
$$
\begin{array}{l}
x+y+z \\
=\left(a^{2}-2 b+\frac{\pi}{2}\right)+\left(b^{2}-2 c+\frac{\pi}{3}\right)+\left(c^{2}-2 a\right. \\
\left.+\frac{\pi}{6}\right) \\
=\left(a^{2}-2 a+1\right)+\left(b^{2}-2 b+1\right)+\left(c^{2}-2 c+1\right) \\
+(\pi-3) \\
=(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+\pi-3>0 . \\
\text { If } x ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,410 |
Example 13 Given that $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-(k-2) x + (k^{2}+3 k+5)=0$ (where $k$ is a real number). Then the maximum value of $x_{1}^{2}+x_{2}^{2}$ is $\qquad$ | According to Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=k-2, \quad x_{1} x_{2}=k^{2}+3 k+5 . \\
\therefore \quad x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=(k-2)^{2}-2\left(k^{2}+3 k+5\right) \\
=9-\left(k+5\right)^{2} .
\end{array}
$$
From the discriminant theorem, we get
$$
\be... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,411 |
4. In quadrilateral $A B C D$, diagonal $A C$ is equal to the adjacent sides $A B$ and $A D$. If $\angle D A C$ is $k$ times $\angle C A B$ ($k$ is a real number), then $\angle D B C$ is ( ) of $\angle B D C$.
(A) $k$ times
(B) $2 k$ times
(C) $3 k$ times
(D) None of the above answers is correct | 4. (A).
Given, as shown in the figure, points $B$, $C$, and $D$ are on a circle with center $A$ and radius $AB$.
$$
\begin{array}{l}
\because \angle D A C=k \angle C A B, \\
\text { and } \angle D B C=\frac{1}{2} \angle D A C, \\
\angle B D C=\frac{1}{2} \angle B A C, \\
\therefore \angle D B C=k \angle B D C,
\end{ar... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,412 |
6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$ | 6. (A).
Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$
Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,414 |
2. Given $|x| \leqslant 2$, the sum of the maximum and minimum values of the function $y=x-|1+x|$ is $\qquad$ . | 2. -4 .
$$
\begin{array}{l}
\because|x| \leqslant 2, \text { i.e., }-2 \leqslant x \leqslant 2, \\
\therefore y=x-|1+x| \\
\quad=\left\{\begin{array}{ll}
2 x+1, & \text { when }-2 \leqslant x<-1, \\
-1, & \text { when }-1 \leqslant x \leqslant 2.
\end{array}\right.
\end{array}
$$
As shown in the figure.
Indeed, when $... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,416 |
3. When $a$ equals $\qquad$, the equations $x^{2}+a x-6$ $=0$ and $x^{2}-6 x+a=0$ have at least one common root. | 3. -6 or 5.
Let the common root of the two equations be $x_{0}$, then
$$
\begin{array}{l}
x_{0}^{2} + a x_{0} - 6 = 0, \\
x_{0}^{2} - 6 x_{0} + a = 0.
\end{array}
$$
Subtracting the two equations,
$$
(a+6) x_{0} = a+6 \text{. }
$$
(i) When $a+6 \neq 0, a \neq -6$, $x_{0} = 1$. Substituting into the equations gives $a... | -6 \text{ or } 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,417 |
4. In $\triangle A B C$, $\angle B C A=90^{\circ}$, a perpendicular $C D \perp A B$ is drawn from $C$ intersecting $A B$ at $D$. Suppose the side lengths of $\triangle A B C$ are all integers, and $B D=29^{3}, \cos B=\frac{m}{n}$, where $m$ and $n$ are coprime positive integers. Then $m+n=$ $\qquad$ | 4. 450 .
Let the three sides of $\triangle ABC$ be $a, b, c$. It is easy to prove that $BC^2 = BD \cdot BA$. Therefore, $a^2 = 29^3 c$.
Since 29 is a prime number, $29^2 \mid a$.
Let $a = 29^2 k$ ($k$ is a positive integer), then $c = 29 k^2$.
Since $b = \sqrt{c^2 - a^2} = 29 k \sqrt{k^2 - 29^2}$ is an integer, we kno... | 450 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,418 |
One, (20 points) Let $A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, B=$ $\frac{c^{2}+a^{2}-b^{2}}{2 c a}, C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
(1) When $A+B+C=1$, prove: $A^{1996}+$ $B^{1996}+C^{1996}=3$.
(2) When $A+\pi+C=1$, determine whether the three numbers $a, b, c$ can form the three sides of a triangle, and explain the rea... | $$
\begin{array}{c}
-(1) \because A+B+C=1, \\
\therefore\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}+1\right)+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}-1\right) \\
+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}-1\right)=0, \\
\text { i.e. } \frac{(b+c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 c a}+\frac{(a-b)^{2}-c^{2}}{2 a b}=0 .
\end... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,419 |
II. (25 points) In $\triangle ABC$, from point $A$ draw perpendiculars to the angle bisectors of $\angle B$ and $\angle C$, with the feet of the perpendiculars being $P$ and $Q$ respectively; from point $B$ draw a perpendicular to the angle bisector of $\angle C$, with the foot of the perpendicular being $E$; from poin... | Extend $A P$ and $A Q$ to intersect $B C$ at $M$ and $N$ respectively.
$$
\because B P \text{ bisects } \angle A B C, A P \perp B P,
$$
$$
\therefore A P=P M \text{. }
$$
Similarly, $A Q=Q N$.
$$
\therefore P Q \parallel B C, \angle Q P B=\angle P B C \text{. }
$$
Connect $E F$.
$$
\because \angle B E C=90^{\circ}=\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,420 |
Three. (25 points) Let $a$, $b$, and $c$ be the lengths of the three sides of a right triangle. Prove: If $a$, $b$, and $c$ are integers, then $abc$ is divisible by 30.
---
Please note that the translation retains the original formatting and structure of the text. | For integers $a, b, c$ being the sides of a right-angled triangle, then $c^{2}=a^{2}+b^{2}$ (assuming $c$ is the hypotenuse).
(i) If at least one of $a, b$ is even, then $2 \mid a b$. Otherwise, $a=2 k_{1}+1, b=2 k_{2}+1$ ($k_{1}, k_{2}$ are natural numbers),
\[
\begin{aligned}
c^{2} & =\left(2 k_{1}+1\right)^{2}+\left... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,421 |
The perimeter of a right triangle is 20, then its maximum area is $\qquad$ .
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let the legs of the right triangle be $a, b$, and the hypotenuse be $c$, then
$$
a+b+c=20, a^{2}+b^{2}=c^{2} .
$$
From this, we get $a+b=20-c, ab=200-$ $20c$. By the inverse of Vieta's theorem, $a, b$ are the two roots of the equation
$$
t^{2}-(20-c) t+(200-20c)=0
$$
By the discriminant theorem, we get
$$
\... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,422 |
1. For $x_{1}>x_{2}>0,1>a>0$, let
$$
y_{1}=\frac{x_{1}}{1+a}+\frac{a x_{2}}{1+a}, y_{2}=\frac{a x_{1}}{1+a}+\frac{x_{2}}{1+a} \text {. }
$$
Then the relationship between $x_{1} x_{2}$ and $y_{1} y_{2}$ is ( ).
(A) $x_{1} x_{2}>y_{1} y_{2}$
(B) $x_{1} x_{2}=y_{1} y_{2}$
(C) $x_{1} x_{2}<y_{1} y_{2}$
(D) Cannot be deter... | -1. (C).
Path 1: (Arithmetic Mean Inequality)
$$
\begin{aligned}
y_{1} y_{2} & =\frac{a}{(1+a)^{2}}\left(x_{1}^{2}+x_{2}^{2}\right)+\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2} \\
& >\frac{a}{(1+a)^{2}}\left(2 x_{1} x_{2}\right)+\frac{1+a^{2}}{(1+a)^{2}} x_{1} x_{2} \\
& =x_{1} x_{2} .
\end{aligned}
$$
Path 2: (Vicsek Inequa... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,423 |
2. Given the function $f(x)=x^{2}-2 a x+2 a+4$ with the domain $R$ and the range $[1,+\infty]$. Then the value of $a$ is ).
(A) interval $[-1,3$ ]
(B) interval $(1-\sqrt{5}, 1+\sqrt{5})$
(C) interval $(-\infty,-1) \cup(3,+\infty)$
(D) set $\{-1,3\}$ | 2. (D). Given $y=1$, we have $x^{2}-2 a x+2 a+4=1$. Its discriminant:
$\Delta=4\left(a^{2}-2 a-3\right)=0$. Solving this yields $a=-1$ or 3.
Note: If $\Delta \leqslant 0$ is used to calculate (A), it only ensures that the points of the parabola are not below the line $y=1$. For example, if $0 \in[-1,3]$, then $f(x)=x^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,424 |
3. As shown in Figure 1, in the geometric body $A B C-A_{1} B_{1} C_{1}$, $A B = A_{1} B_{1}$. The necessary and sufficient condition for $A A_{1}$, $B B_{1}$, and $C C_{1}$ to be concurrent is ( ).
(A) $B C \parallel B_{1} C_{1}$ and $A C \parallel A_{1} C_{1}$
(B) $B C \neq B_{1} C_{1}$ and $A C \neq A_{1} C_{1}$
(C)... | 3. (C).
Three planes intersect each other pairwise, resulting in three intersection lines $A A_{1}$, $B B_{1}$, and $C C_{1}$, which either intersect at one point or are parallel to each other. When $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect at one point, from $A B=A_{1} B_{1}$, we know that plane $A B C$ and plane... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,425 |
4. $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of $d$, then the necessary and sufficient condition for the sum of any (different) two terms in the sequence to still be a term in this sequence is ( ).
(A) There exists an integer $m \geqslant-1$, such that $a_{1}=m d$
(B) $a_{n}=0(n=1,2, \cd... | 4. (A). We get $a_{4}+a_{t}=u_{k}$, so
$$
\begin{array}{l}
2 a_{1}+(s+t-2) d=a_{1}+(k-1) d . \\
\therefore a_{1}=(k-s-t+1) d .
\end{array}
$$
Therefore, there exists an integer $m=k-s-t+1$, such that $a_{1}=m d$.
Next, we prove that $m \geqslant-1$.
When $d \neq 0$, it is obviously true. For $d \neq 0$, if $m$ is less... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,426 |
5. $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}\right.$, $\left.y_{4}\right)$ are four points on the parabola $y=a x^{2}+b x+c(a \neq 0)$, and satisfy $x_{1}+x_{4}=x_{2}+x_{3}$. Then ( ).
(A) $A D / / B C$
(B) $A D \perp B C$
(C) $A D{ }_{-j} B C$ intersect but are n... | 5. (A).
The parabola has 4 points with distinct x-coordinates.
$$
\begin{aligned}
\because k_{A D} & =\frac{y_{4}-y_{1}}{x_{4}-x_{1}}=a\left(x_{4}+x_{1}\right)+b \\
& =a\left(x_{3}+x_{2}\right)+b=\frac{a\left(x_{3}^{2}-x_{2}^{2}\right)+b\left(x_{3}-x_{2}\right)}{x_{3}-x_{2}} \\
& =\frac{\left(a x_{3}^{2}+b x_{3}+c\rig... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,427 |
6. Given that the real part of the expansion of $(1+i x)^{n+2}(x \in I)$ is a polynomial in $x$, then the sum of the coefficients of this polynomial is (
(A) $(-1) * 2^{2 \pi+1}$
(B) 0
(C) $2^{2 n+1}$
(D) $-2^{2 x+1}$ | 6. (B).
Let the real part of the expansion be $f(x)$, and the imaginary part be $g(x)$, then $f(1)$ and $g(1)$ are the sums of the coefficients of the real part polynomial and the imaginary part polynomial, respectively.
$$
\begin{array}{l}
f(1)+i g(1)=(1+i)^{4 n+2} \\
=2^{2 n+1}\left(\cos \frac{\pi}{4}+i \sin \frac{\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,428 |
1. Given $M=\cos 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \sin 35^{\circ}, N=$ $\sin 5^{\circ} \cos 15^{\circ} \cos 25^{\circ} \cos 35^{\circ}$. Then $\frac{M}{N}=$ $\qquad$ . | $=1.1$.
$$
\begin{aligned}
\frac{M}{N} & =\frac{\frac{1}{2}\left(\sin 20^{\circ}+\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}-\cos 60^{\circ}\right)}{\frac{1}{2}\left(\sin 20^{\circ}-\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}+\cos 60^{\circ}\right)} \\
& =\frac{\left(\sin 20^{\circ}+... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,429 |
2. The solution set of the inequality $\sqrt{x^{2}+4 x} \leqslant 4-\sqrt{16-x^{2}}$ is $\qquad$ . | 2. $\{-4,0\}$.
For the expression under the square root to be meaningful in the inequality, we have
$$
\left\{\begin{array}{l}
x^{2}-4 x \geqslant 0, \\
16-x^{2} \geqslant 0 .
\end{array}\right.
$$
This gives $x=-4$ or $x \in[0,4]$.
Clearly, $x=-4$ is a solution to the inequality. For $x \in[0,4]$, we have
$$
\begin{... | \{-4,0\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 709,430 |
3. As shown in Figure 2, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base is an isosceles right triangle, $\angle A_{1} C_{1} B_{1}=90^{\circ}, A_{1} C_{1}=$ $1, A A_{1}=\sqrt{2}$, find $A B_{1}$, $A C_{1}$, and the plane angle between the plane $A A_{1} B_{1}$ and the plane $A B_{1} C_{1}$. (Express u... | 3. $\operatorname{arctg} \sqrt{2}$.
Take the midpoint $D$ of $A_{1} B_{1}$, and connect $C_{1} D$. In the isosceles right $\triangle A_{1} B_{1} C_{1}$, since $C_{1} D$ is the median to the hypotenuse, we know that $C_{1} D \perp A_{1} B_{1}$. Also, by the properties of a right prism, $C_{1} D \perp A A_{1}$. Therefor... | \operatorname{arctg} \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,431 |
4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$ | 4. 10.
Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$
Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,432 |
6. There are 5 boxes, each box has a key, and the keys cannot be used. If a key is randomly placed inside one of the boxes, and then all the boxes are locked. Now, if you are allowed to open one box, so that you can subsequently use the keys to open the remaining 4 boxes, then the number of ways to place the keys is $\... | 6. 4!.
First, in the opened box, there must be the key of another positive box $i_{1}$, in box $i_{1}$ there is the key of box $i_{2}$, in box $i_{2}$ there is the key of box $i_{3}$, in box $i_{3}$ there is the key of box $i_{4}$, and the opened box is $i_{5}$. This is equivalent to $1,2,3,4,5$ forming a circular per... | 4! | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,435 |
Three, given the line $l: y=k \cdot x+h(k, h \neq 0)$ intersects the $x$-axis at point $A$, the $y$-axis at point $B$, and the ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ has no common points. Prove: $|A B|>a+b$.
---
Note: The text "r_{-j} y" and "H. ${ }^{j}$" in the original text seem to be incorrectly... | $$
\left\{\begin{array}{l}
y=k x+h, \\
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\end{array}\right.
$$
has no real solutions. Eliminating $y$, we get
$$
\begin{array}{l}
\left(a^{2} k^{2}+b^{2}\right) x^{2}+2 a^{2} k h x+a^{2} h^{2}-a^{2} b^{2}=0 . \\
\Delta=4 a^{2} b^{2}\left(a^{2} k^{2}+b^{2}-h^{2}\right)\left(a^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,436 |
Four, as shown in Figure 3, in a cube $A B C D-A_{1} B_{1} C_{1} D$ with edge length $a$, cut off its two corners along $A C B_{1}, A_{1} B I$, and then cut off two more corners along the faces $B D C_{1}, A C D_{1}$. Find the volume of the remaining geometric body. | The geometric body intercepted is shown in Figure 6. The base is a square, and there are 4 isosceles right triangles congruent to $\triangle M A_{1} B_{1}$, as well as 4 trapezoids congruent to $N M A_{1} G$.
See the original Figure 3, the cube has 4 tetrahedra of the form $B_{1}-A B C$ cut off, each with a volume of $... | \frac{1}{2} a^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,437 |
Five, given $a_{0}$, $a_{1}, a_{2}, \cdots, a_{n}$ are constants, $f(x)=\frac{b^{x}}{b^{x}+\sqrt{b}}$ $(b>0)$. Prove: $\sum_{k=0}^{n} a_{k} C_{n}^{*}[f(x)]^{k}[f(1-x)]^{n-k}$ $=a_{0} f(1-x)+a_{n} f(x)$ | Given $f(1-x)=\frac{b^{1-x}}{b^{1-x}+\sqrt{b}}=\frac{b}{b+\sqrt{b} b^{x}}$
$$
=\frac{\sqrt{b}}{\sqrt{b}+b^{x}},
$$
we know that $f(x)+f(1-x)=1$.
Also, from the arithmetic sequence, we have $a_{k}=a_{0}+k\left(a_{1}-a_{0}\right)$. Therefore,
$$
\begin{array}{l}
\sum_{k=0}^{n} a_{k} C_{n}^{*}[f(x)]^{k}[f(1-x)]^{n-k} \\... | a_{0} f(1-x)+a_{n} f(x) | Algebra | proof | Yes | Yes | cn_contest | false | 709,438 |
$$
\left\{\begin{aligned}
a_{1}= & \frac{1}{2}, \\
a_{n}= & \frac{1}{3 n-1}\left(a_{1} a_{n-1}+a_{2} u_{n \cdots 2}\right. \\
& \left.+\cdots+a_{n-2} a_{2}+a_{n-1} a_{1}\right) .
\end{aligned}\right.
$$
Prove: $a_{n+1}<a_{n}$. | Second, the second mathematical induction for electricity.
$$
a_{2}=\frac{1}{5} a_{1}^{2}=\frac{1}{10} a_{1}<a_{1}, a_{3}=\frac{2}{8} a_{1} a_{2}=\frac{1}{8} a_{2}<a_{2} .
$$
Assume $a_{k}<a_{k-1}<\cdots<a_{3}<a_{2}<a_{1}$, then for $n=k+1$,
$$
\begin{aligned}
a_{k+1}= & \frac{1}{3 k+2}\left(a_{1} a_{k}+a_{2} a_{k-1}+... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,440 |
Three, in the interior angles of a 66-sided polygon, there must exist two angles $\alpha, \beta$, such that $|\cos \alpha - \cos \beta| < \frac{1}{1997}$. | Three, let the interior angles of a polygon be denoted in descending order as $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{66}$, then $0<\alpha_{66} \leqslant \alpha_{65} \leqslant \cdots \leqslant \alpha_{2} \leqslant \alpha_{1}<\pi$. It must be that $a_{23} \geqslant \frac{21 \pi}{22}$. If not, $\alpha_{23}<\frac{21 \pi}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,441 |
On September 4, 1996, scientists used a supercomputer to find the 33rd Mersenne prime, which is the largest prime number known to humans so far. It is: $2^{125787}-1$ (378632 digits). Try to find the last two digits of this prime number. | Solution: First, find the unit digit.
$$
\begin{array}{l}
\because 2^{1257787}-1=2 \cdot 4^{628893}-1=8 \cdot 16^{31446}-1 \\
\quad \equiv 8 \times 6-1(\bmod 10) \equiv 7(\bmod 10),
\end{array}
$$
$\therefore 2^{1257787}-1$ has a unit digit of 7.
Next, find the tens digit.
$$
\begin{aligned}
\because & \frac{1}{10}\lef... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,442 |
Given in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, $D$ is on $A C$ and $E$ is on $A B$. If $\angle A B D=10^{\circ}, \angle B D E$ $=20^{\circ}$, find the degree measure of $\angle A C E$. | $$
\begin{array}{l}
\text { Solution: } \because \angle A=20^{\circ}, A B=A C, \\
\therefore \angle A B C=\angle A C B=80^{\circ} .
\end{array}
$$
As shown in the figure, take point $O$ as the circumcenter of $\triangle B D E$, and connect $O D, O E, O B, O C$. We have
$$
\begin{array}{c}
\angle B O E=2 \angle B D E=4... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,443 |
In the plane: There are 212 points all located within or on the boundary of a unit circle. Connect any two points to form a line segment. Prove: The number of line segments with length not greater than 1 is at least 1996. | Proof: (1) First, prove that in a circle or on any $\epsilon$ points, there must be two points whose distance is not greater than:
Let these 6 points be $A_{1}, A_{2}, \cdots, A_{6}$, and the center be $O$. If one point is at the center, the proposition holds. Otherwise, connect $O A_{i},(i=1,2, \cdots, 6)$, to get 6 ... | 1996 | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,445 |
56. Let $a, b, c, d \in \mathbb{R}$, and $a+b+c+d=0$. Prove that,
(i) $3\left(a^{3}+b^{3}+c^{3}+d^{3}\right)^{2} \leqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3}$;
(ii)
$$
\begin{array}{l}
108\left(a^{5}+b^{5}+c^{5}+d^{5}\right)^{2} \\
\leqslant 25\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{5}
\end{array}
$$ | Prove: Let $S_{k}=a^{k}+b^{k}+c^{k}+d^{k}(k \in N)$.
(i) $S_{3}=a^{3}+b^{3}+c^{3}-(a+b+c)^{3}$.
Since when $a=\cdots b$ or $b=-c$ or $c=-a$, $0=0$.
Assume $S_{3} \equiv \mu(a+b)(b+c)(c+a)$. When $a=b=c=1$, we get $\mu=-3$.
$$
\begin{aligned}
\therefore \quad S_{3} & =-3(a+b)(b+c)(c+a) . \\
\text { Also, } S_{2} & =a^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,446 |
3. In $\triangle A B C$, one side is 5, and the other two sides are exactly the two roots of the equation $2 x^{2}-12 x+m=0$. Then the range of values for $m$ is $\qquad$ . | (Solution: $\frac{11}{2}<m \leqslant 18$ ) | \frac{11}{2}<m \leqslant 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,447 |
4. Given the quadratic equation $x^{2}-x+1-m=0$ with two real roots $\alpha, \beta$ satisfying $|\alpha|+|\beta| \leqslant 5$. Then the range of the real number $m$ is $\qquad$ | (Answer: $\frac{3}{4} \leqslant m \leqslant 7$. ) | \frac{3}{4} \leqslant m \leqslant 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,448 |
Example 3 If the two roots of the equation $(x-1)\left(x^{2}-2 x+m\right)=0$ can serve as the lengths of the three sides of a triangle, then the range of the real number $m$ is ( ).
(A) $0 \leqslant m \leqslant 1$
(B) $m \geqslant \frac{3}{4}$
(C) $\frac{3}{4}<m \leqslant 1$
(D) $\frac{3}{4} \leqslant m \leqslant 1$ | Solution: It is obvious that the original equation has a root of 1. Let $\alpha, \beta$ be the two roots of the equation $x^{2}-2 x+m=0$. By Vieta's formulas, we have
$$
\alpha+\beta=2, \alpha \beta=m .
$$
Since the three sides of the triangle are $1, \alpha, \beta$, then
$$
\begin{aligned}
& |\alpha-\beta|<1 \\
& 1+\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,451 |
7. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5$ $=0, b^{2}+2 b-5=0$. Then $a^{2} b+a b^{2}=$ $\qquad$ . | （Answer:10） | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,452 |
12. For the equation $x^{2}+(2 m-1) x+(m-6)=0$, one root is no greater than -1, and the other root is no less than 1.
(1) Find the range of values for $m$;
(2) Find the maximum and minimum values of the sum of the squares of the roots. | (Solution: $-4 \leqslant m \leqslant 2$; the maximum value is 101, the minimum value is $\frac{43}{4}$.)
| -4 \leqslant m \leqslant 2; \text{ maximum value is } 101, \text{ minimum value is } \frac{43}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,455 |
For $x \in R, f(x)=\min \{4 x+2$, $\left.-12 x^{2}+12,-2 x+4\right\}$, find the expression of $f(x)$ and $f(x)_{\max } \cdot$ | Solution:
As shown in the figure, in
the same Cartesian
coordinate system,
draw the graphs of $y_{1}=$
$$
\begin{array}{l}
4 x+2, y_{2} \\
=-12 x^{2} \\
+12, y_{3}= \\
-2 x+4
\end{array}
$$
According to the definition of $f(x)$, we can obtain the piecewise expression of $f(x)$
$$
\left\{\begin{array}{ll}
& f(x)= \\
-... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,457 |
Example 3 In the quadratic function $x^{2}+$ $p x+q$ with the leading coefficient of 1, find the function expression that makes $M=\max \left|x^{2}+p x+q\right|(-1$ $\leqslant x \leqslant 1$ ) take the minimum value. | Analyzing this problem, we can first obtain the inequalities of $p$ and $q$ through the value of the function, and then use the method of magnification and reduction to get the lower bound.
$$
\text{Let } g(x)=\left|x^{2}+p x+q\right|(-1 \leqslant x \leqslant
$$
1), then $g(x)_{\max }$ can only be $|f(1)|$ or $|f(-1)|$... | f(x)=x^{2}-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,458 |
Let $s=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be a permutation of the natural numbers $1,2, \cdots, n$ over the past $n$ months. $f(s)$ is defined as the minimum value of the absolute difference between any two adjacent elements in $s$. Find the maximum value of $f(s)$. | Solution: We discuss in two cases.
(1) If $n=2 k$ is even, then $f(s) \leqslant k$ and the absolute difference between it and its adjacent numbers is no more than $k$. On the other hand, in
$$
\begin{array}{l}
s=(k+1,1, k+2,2, \cdots, 2 k, k) \text {, we have } \\
f(s)=k=\left[\frac{n}{2}\right] .
\end{array}
$$
(2) If... | \left[\frac{n}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,460 |
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most bought the book that was purchased by the fewest people? | Solution: Let the number of people who bought the most popular book be $x$. Among the 10 people, person A bought three books. Since the other 9 people each have at least one book in common with A, and $9 \div 3=3$, it follows that among A's three books, the most popular one must have been bought by at least 4 people, s... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,461 |
Example 4 Given that $a, b, c$ satisfy $a+b+c=0, abc=8$. Then the range of values for $c$ is $\qquad$ | Given the conditions, we have
$$
a+b=-c, a b=\frac{8}{c} \text {. }
$$
By the inverse of Vieta's formulas, $a, b$ are the two real roots of the equation
$$
t^{2}+c t+\frac{8}{c}=0
$$
Applying the discriminant theorem, we get
$$
\Delta=c^{2}-4 \times \frac{8}{c} \geqslant 0 .
$$
Solving for $c$, the range of $c$ is $... | c<0 \text{ or } c \geqslant 2 \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,462 |
Example 7 Given a line $l$ and two points $A$ and $B$ in the plane, how should a point $P$ be chosen on line $l$ so that $\max \{A P, B P\}$ is minimized? | Solution: We can assume that points $A$ and $B$ are on the same side of line $l$. If not, we can use the symmetric point $B'$ of point $B$ with respect to line $l$ for discussion.
Let the projections of points $A$ and $B$ on line $l$ be $A_1$ and $B_1$, respectively. Without loss of generality, assume $AA_1 \geqslant ... | C | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,463 |
Example 8 Determine a point $P$ on the plane of $\triangle A B C$ such that $\max \{A P, B P, C P\}$ is minimized. | Analysis: It is obvious that to minimize $\max \{A P, B P, C P\}$, point $P$ must be inside or on the boundary of $\triangle A B C$. By moving point $P$ for observation, as shown in Figure 4, when $\triangle A B C$ is an acute or right triangle, the circumcenter is the desired point $P$. As shown in Figure 5, when $\tr... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,464 |
Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,465 |
Example 2 Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, then the total number of different coloring methods is $\qquad$ | Solution: Vertex $S$ can be colored with any of the $m$ colors, and the color on $S$ cannot appear on the vertices of the polygon $A_{1} A_{2} \cdots A_{n}$. The problem is then reduced to coloring the vertices of the polygon with $m-1$ colors, ensuring that adjacent vertices have different colors. Let there be $a_{n}$... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,466 |
Example 4 The permutation of integers $1,2, \cdots, n$ satisfies: each number is greater than all the numbers before it or less than all the numbers before it. How many such permutations are there? | Explanation: The original solution is calculated through a recursive relationship. Let the number sought be $a_{n}$, then
$$
a_{1}=1 \text {. }
$$
For $n>1$, if $n$ is placed in the $i$-th position, then the $n-i$ numbers after it are completely determined, and can only be $n-i, n-i-1$, $\cdots, 2,1$. The $i-1$ number... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,467 |
1. Let the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $M$. Draw a line through $M$ parallel to $AD$, intersecting $AB$ and $CD$ at points $E$ and $F$ respectively, and intersecting the extension of $BC$ at point $O$. Let $P$ be a point on the circle with center $O$ and radius $OM$ (as... | Prove: The line $O C B$ intersects the extensions of the three sides of $\triangle D M F$ and $\triangle A E M$, by Menelaus' theorem we have
$$
\begin{array}{l}
\frac{D B}{M B} \cdot \frac{M O}{F O} \cdot \frac{F C}{D C}=1, \\
\frac{A B}{E B} \cdot \frac{E O}{M O} \cdot \frac{M C}{A C}=1 . \\
\therefore \frac{O F}{O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,469 |
3. Construct a semicircle with the base $BC$ of $\triangle ABC$ as its diameter, intersecting sides $AB$ and $AC$ at points $D$ and $E$ respectively. Draw perpendiculars from points $D$ and $E$ to $BC$, with feet of the perpendiculars being $F$ and $G$ respectively. The line segments $DG$ and $EF$ intersect at point $M... | Proof 1: Let the line $A M$ intersect $B C$ at point $M$, and connect $B E, C D$. We have $\angle B E C=\angle B D C=90^{\circ}$. The line $F M E$ intersects $\triangle A H C$, and the line $G M D$ intersects $\triangle A B H$. By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{A M}{M H} \cdot \frac{H F}{F C} \cdo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,471 |
4. Quadrilateral $ABCD$ is inscribed in a circle, the extensions of its sides $AB$ and $DC$ intersect at point $P$, and the extensions of $AD$ and $BC$ intersect at point $Q$. Two tangents are drawn from $Q$ to the circle, touching it at points $E$ and $F$. Prove that points $P$, $E$, and $F$ are collinear. | Proof 1: First, we prove a lemma.
Lemma: Through a point $Q$ outside the circle $\odot O$, draw two tangents $QE$ and $QF$ and a secant $QDA$. The line segment $EF$ and $AD$ intersect at point $M$ (as shown in the figure), then $\frac{AM}{DM}=\frac{AQ}{DQ}$.
Connect auxiliary lines as shown in the figure. By the secan... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,472 |
Example 5 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)(n$ is an integer). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) $1991^{-1}$
(B) $-1991^{-1}$
(C) $(-1)^{n} 1991$
(D) $(-1)^{n} 1991^{-1}$ | Solution: Let $a=1991^{\frac{1}{n}}, b=-1991^{-\frac{1}{n}}$, then $a+b=2 x, a b=-1$.
By the inverse of Vieta's formulas, $a$ and $b$ are the two real roots of the equation
$$
t^{2}-2 x t-1=0
$$
Solving this equation, we get
$$
t=x \pm \sqrt{1+x^{2}} .
$$
Since $b<a$, we have $b=x-\sqrt{1+x^{2}}$.
$$
\begin{array}{l}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,473 |
Example 1 Let the function $f:\{x \mid x \neq 0,1, x \in R\} \rightarrow$ $R$, and satisfies $f(x)+f\left(\frac{x-1}{x}\right)=1+x$. Find $f(x)$. | Solution: Let $\frac{x-1}{x}=y$, then $x=\frac{1}{1-y}$. Substituting into (1), we get $f\left(\frac{1}{1-y}\right)+f(y)=\frac{y-2}{y-1}$,
i.e., $f(x)+f\left(\frac{1}{1-x}\right)=\frac{x-2}{x-1}$. Let $x=\frac{u-1}{u}$. Substituting into (1), we get $f\left(\frac{u-1}{u}\right)+f\left(\frac{1}{1-u}\right)=\frac{2 u-1}{... | f(x)=\frac{x^{3}-x^{2}-1}{x(x-1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,474 |
Example 2 Find all real-valued functions $f$ defined on the set of rational numbers, satisfying $f(x+y)=f(x)+f(y)+2xy$.
untranslated text remains unchanged. | Let $y=1, x=n \in N$. Then we have
$$
f(n+1)=f(n)+f(1)+2 n,
$$
which means $f(n+1)-f(n)=2 n+f(1)$.
$$
\begin{array}{c}
\therefore \sum_{i=1}^{n-1}[f(i+1)-f(i)] \\
=\sum_{i=1}^{n-1}[2 n+f(1)],
\end{array}
$$
which implies $f(n)=n^{2}+[f(1)-1] n=n^{2}+c n$
(let $f(1)-1=c$).
Also, $f\left(\frac{k+1}{n}\right)=f\left(\fr... | f(x)=x^{2}+c x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,475 |
Example 3 Let the domain of $f(x)$ be $R$, and for all $x, y \in R$ it satisfies $f(x+y)+f(x-y)=2 f(x) - \cos y, f(0)=A, f\left(\frac{\pi}{2}\right)=B$.
| Let $x=0$, we get
$$
f(y)+f(-y)=2 f(0) \cos y,
$$
which means $f(x)+f(-x)=2 A \cos x$.
Let $x=\frac{\pi}{2}, y=\frac{\pi}{2}+u$, we get
$$
\begin{array}{l}
f(\pi+u)+f(-u) \\
=2 f\left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{2}+u\right),
\end{array}
$$
which means $f(\pi+x)+f(-x)=-2 B \sin x$.
Let $x=\frac{\pi}{2}+... | f(x)=B \sin x+A \cos x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,476 |
Example 4 Prove that there exists a unique function $f: N \rightarrow$ $N$ satisfying for all $m, n \in N$, $f[m+f(n)]=n$ $+f(m+95)$. Find the value of $\sum_{k=1}^{19} f(k)$. | Solution: Observe and verify that $f(n)=n+95$ meets the requirement, at this time, $\sum_{k=1}^{19} f(k)=1995$.
Now prove the uniqueness.
When $n_{1} \neq n_{2}$,
$$
\begin{array}{l}
f\left[m+f\left(n_{1}\right)\right]=n_{1}+f(m+95) \\
\neq n_{2}+f(m+95)=f\left[m+f\left(n_{2}\right)\right],
\end{array}
$$
i.e., $f\lef... | 1995 | Algebra | proof | Yes | Yes | cn_contest | false | 709,477 |
Example 5 The domain of the function $f(x)$ is symmetric about the origin, and it satisfies the following three conditions:
(i) $x_{1} 、 x_{2}$ are numbers in the domain of $f(x)$, and
$$
f\left(x_{1}-x_{2}\right)=\frac{f\left(x_{1}\right) f\left(x_{2}\right)+1}{f\left(x_{2}\right)-f\left(x_{1}\right)} \text {; }
$$
(i... | Proof: (1) $\because f\left(x_{2}-x_{1}\right)$.
$$
\begin{array}{l}
=\frac{f\left(x_{2}\right) f\left(x_{1}\right)+1}{f\left(x_{1}\right)-f\left(x_{2}\right)}=-\frac{f\left(x_{1}\right) f\left(x_{2}\right)+1}{f\left(x_{2}\right)-f\left(x_{1}\right)} \\
=-f\left(x_{1}-x_{2}\right),
\end{array}
$$
$\therefore f(x)$ is a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,478 |
Example 6 Given that $f(x)$ is a function with domain $R$, and $f(x+2)[1-f(x)]=1+f(x), f(1)=2+\sqrt{3}$. Find $f(1989)$.
保留源文本的换行和格式，直接输出翻译结果。 | $$
\begin{array}{l}
\text { Solution: } \because f(x+2)=\frac{1+f(x)}{1-f(x)}, \\
\therefore f(x+4)=\frac{1+f(x+2)}{1-f(x+2)} \\
=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)} . \\
\text { Also, } \because f(x)=\frac{f(x+2)-1}{f(x+2)+1}, \\
\therefore f(x-4)=\frac{f(x-2)-1}{f(x-2)+1} \\
=\frac... | \sqrt{3}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,479 |
Example 7 The function $f(x)$ is defined on the set of real numbers, and for all real numbers $x$ it satisfies the equations: $f(2+x)=f(2-x)$ and $f(x+7)=f(7-x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ denote the number of roots of $f(x)=0$ in the interval $[-1000,1000]$. Find the minimum value of $N$. | $$
\begin{array}{l}
\text { Solution: } \because f(x+2)=f(2-x), \\
\begin{array}{l}
\therefore f(x+4)=f[(x+2)+2] \\
\quad=f[2-(x+2)]=f(-x) . \\
\because f(x+7)=f(7-x), \\
\therefore f(x+14)=f[(x+7)+7]
\end{array} \\
\quad=f[7-(x+7)]=f(-x) .
\end{array}
$$
Thus, $f(x+14)=f(4+x)$,
which means $f(x+10)=f(x)$.
Therefore, ... | 401 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,480 |
1. On the coordinate plane, points with integer coordinates form the vertices of unit squares, which are colored in two alternating colors (like a chessboard).
For any pair of positive integers $m$ and $n$, consider a right-angled triangle whose vertices have integer coordinates, with the lengths of the two legs being... | Solution: (a) Let $\triangle A B C$ be a right-angled triangle, with its vertices having integer coordinates, and the two legs lying on the sides of these square grids. Let $\angle A=90^{\circ}, A B=m, A C=n$. Consider the rectangle $A B C D$ as shown in Figure 1.
For a polygon
$P$, let $S_{1}(P)$ be the total
area of ... | f(2 k+1,2 k)=\frac{2 k-1}{6} | Geometry | proof | Yes | Yes | cn_contest | false | 709,481 |
2. Let $\angle A$ be the smallest interior angle of $\triangle ABC$. Points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be a point on the arc not containing $A$ such that the arc $B U C$ is equal to $B C$.
The perpendicular bisectors of segments $AB$ and $AC$ intersect segment $AU$ at $V... | Proof: As shown in Figure 4, because point $V$ lies on the perpendicular bisector of segment $AB$, we have
$$
\begin{array}{l}
\angle VAB \\
=\angle VBA.
\end{array}
$$
Since $\angle A$ is the smallest interior angle of $\triangle ABC$ and $\angle VAB = \angle UAB < \angle CAB$, it follows that
$$
\angle VBA = \angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,482 |
3. Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers satisfying the following conditions:
$$
\left|x_{1}+x_{2}+\cdots+x_{n}\right|=1
$$
and $\left|x_{i}\right| \leqslant \frac{n+1}{2}, i=1,2, \cdots, n$.
Prove: There exists a permutation $y_{1}$, $y_{2}, \cdots, y_{n}$ of $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
\l... | Prove: For any permutation $\pi = y_{1}, y_{2}, \cdots, y_{n}$ of $x_{1}, x_{2}, \cdots, x_{n}$, let $S(\pi)$ be the value of the sum $y_{1} + 2 y_{2} + \cdots + n y_{n}$. Let $r = \frac{n+1}{2}$. We need to show that there exists a permutation $\pi$ such that $|S(\pi)| \leqslant r$.
Let $\pi_{0} = x_{1}, x_{2}, \cdot... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,483 |
Example 6 If $x=\frac{\sqrt{5}-1}{2}$, then $x^{4}+x^{2}+2 x-$
$$
1=
$$ | Let $x=\frac{\sqrt{5}-1}{2}$ be a root of the rational coefficient equation $x^{2}+p x+q=0$, then the other root of this equation is
$$
\begin{array}{l}
\frac{-\sqrt{5}-1}{2} . \\
\because p=-\left(\frac{\sqrt{5}-1}{2}+\frac{-\sqrt{5}-1}{2}\right)=1, \\
q=\frac{\sqrt{5}-1}{2} \cdot \frac{-\sqrt{5}-1}{2}=-1, \\
\therefo... | 3-\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,484 |
4. An $n \times n$ matrix (square matrix) is called an $n$-order "silver matrix" if its elements are taken from the set
$$
S=\{1,2, \cdots, 2 n-1\},
$$
and for each $i=1,2, \cdots, n$, the elements in the $i$-th row and the $i$-th column together are exactly all the elements of $S$. Prove:
(a) There does not exist a s... | Proof: (a) Let $n>1$ and there exists an $n$-order silver matrix $A$. Since all $2n-1$ numbers in $S$ must appear in matrix $A$, and $A$'s main diagonal has only $n$ elements, there must be at least one $x \in S$ that is not on $A$'s main diagonal. Choose such an $x$, and for each $i=1,2, \cdots, n$, let the set of all... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,485 |
5. Find all integer pairs $(a, b)$, where $a \geqslant 1, b \geqslant 1$, and satisfy the equation $a^{b^{2}}=b^{a}$.
$$
b=3, a=b^{k}=3^{3}=27 \text {. }
$$
In summary, all positive integer pairs that satisfy the equation are
$$
(a, b)=(1,1),(16,2),(27,3) .
$$ | Obviously, when $a, b$ have one equal to 1, $(a, b)=(1, 1)$. Now assume $a, b \geqslant 2$.
Let $t=\frac{b^{2}}{a}$, then from the equation in the problem we get $b=a^{t}, a t=a^{2 t}$, thus $t=a^{2 t-1}$. Therefore, $t>0$. If $2 t-1 \geqslant 1$, then $t=a^{2 t-1} \geqslant(1+1)^{2 t-1} \geqslant 1+(2 t-1)=2 t>t$, wh... | (a, b)=(1,1),(16,2),(27,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,486 |
6. For each positive integer $n$, express $n$ as a sum of non-negative integer powers of 2. Let $f(n)$ be the number of different representations of the positive integer $n$.
If the only difference between two representations is the order in which the numbers are added, these two representations are considered the sam... | Prove: For any odd number $n=2k+1, n$'s any representation must contain a "1". Removing this 1 yields a representation of $2k$. Conversely, adding a "1" to any representation of $2k$ yields a representation of $2k+1$. This is clearly a one-to-one correspondence between the representations of $2k+1$ and $2k$. Thus, we h... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,487 |
1. Given two lines $x-y=2$ and $c x+y=3$ intersect in the first quadrant. Then the range of the real number $c$ is | 1. $\left(-1, \frac{3}{2}\right)$ | \left(-1, \frac{3}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,488 |
4. Let $\alpha$ be an interior angle of a triangle, and $(\lg 2 + \lg 3)^{\sin 4 \alpha} > 1$. Then the range of the real number $\alpha$ is $\qquad$ | $\begin{array}{l}\text { 4. }\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ \cup\left(\frac{3 \pi}{4}, \pi\right)\end{array}$ | \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \cup \left(\frac{3 \pi}{4}, \pi\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,490 |
Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ . | Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,494 |
11. In $\triangle A B C$, it is known that $B C=4, A C=3$, $\cos (A-B)=\frac{3}{4}$. Then the area of $\triangle A B C$ is | 11. $\frac{3 \sqrt{7}}{2}$ | \frac{3 \sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,497 |
12. In the Cartesian coordinate system, among the lines passing through the point $(1,2)$ with a slope less than 0, the slope of the line with the smallest sum of intercepts on the two coordinate axes is $\qquad$ . | 12. $-\sqrt{2}$ | -\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,498 |
13. Given 10 points of the World Team, where 5 of these points lie on a straight line, and no three points lie on another straight line besides these, the number of distinct rays that can be drawn through any 2 of these 10 points is $\qquad$. | 13. 78 | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,499 |
15. Let the complex numbers $z_{1}, z_{2}$ satisfy $z_{1} z_{2}=1, z_{1}^{3}+z_{2}^{3}=$ $0, H z_{1}+z_{2} \neq 0, z_{1}, z_{2}$ correspond to points $Z_{1}, Z_{2}$ in the complex plane, and $O x_{1}$ is the origin. Then the area of $\triangle Z_{1} O Z_{2}$ is | 15.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,501 |
18. Let $x>1, y>1$, then the solution to the equation $x+y+\frac{3}{x-1}$ $+\frac{3}{y-1}=2(\sqrt{x+2}+\sqrt{y+2})$ is $(x, y)=$ | 18. $\left(\frac{3+\sqrt{13}}{2}, \frac{3+\sqrt{13}}{2}\right)$ | \left(\frac{3+\sqrt{13}}{2}, \frac{3+\sqrt{13}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,504 |
Example 8 The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Then $1993+5 a^{2}+$ $9 a^{4}=$ $\qquad$ | Solution: By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=a, x_{1} x_{2}=-a . \\
\therefore x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3} \\
=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)+\left(x_{1} x_{2}\right)^{3} \\
=a^{3}-3(-a) a+(-a)^{3}=3 a^{2} .
\end{array}
$$
According to the probl... | 7743 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,505 |
19. Given $\mathrm{Rt} \triangle A B C$ with the two legs $A C=2$, $C B=3$, and $C P$ as the angle bisector of $\angle A C B$ ($P$ is on the hypotenuse $A B$), fold the triangle along $C P$ to form a dihedral angle $A$ $-C P-B$. When $A U=24 / \sqrt{c}$, the size of the dihedral angle $A-C P$ $-B$ is | 19. $\arccos \left(-\frac{1}{6}\right)$ | \arccos \left(-\frac{1}{6}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,506 |
20. For a certain commodity, if you buy 100 (including 100) pieces or less, it is settled at the retail price; if you buy 101 (including 101) pieces or more, it is settled at the wholesale price. It is known that the wholesale price is 2 yuan lower per piece than the retail price. A person originally intended to buy a ... | $\begin{array}{l}20 \\ 840\end{array}$ | 840 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,507 |
2. Given points $A(0,4), B(4,0)$. If the parabola $y=x^{2}-m x+m+1$ intersects the line segment $A B$ (excluding endpoints $A$ and $B$) at two distinct points, then the range of values for $m$ is $\qquad$. | 2. $\left(3, \frac{11}{3}\right)$ | \left(3, \frac{11}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,509 |
Example 9 Solve the equation
$$
(6 x+7)^{2}(3 x+4)(x+1)=6 .
$$ | Solution: Multiply both sides of the original equation by 12, we get
$$
(6 x+7)^{2}(6 x+8)(6 x+6)=72 \text {. }
$$
Let $m=(6 x+7)^{2}, n=(6 x+8)\left(6 x^{2}+6\right)$ $=(6 x+7)^{2}-1$, then
$$
m+(-n)=1, m(-n)=-72 \text {. }
$$
By the inverse of Vieta's theorem, $m$ and $-n$ are the two roots of the equation $y^{2}-y... | x_{1}=-\frac{2}{3}, x_{2}=-\frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,516 |
II. On the hyperbola $x y=1$, the point with the abscissa $\frac{n}{n+1}$ is $A_{n}$, and the point with the abscissa $\frac{n+1}{n}$ is $B_{n}(n \in N)$. The point with coordinates $(1,1)$ is denoted as $M, P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$. Find the coordinates $(a, b)$ ... | II. It is easy to get $A_{n}\left(\frac{n}{n+1}, \frac{n+1}{n}\right), B_{n}\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$.
$\therefore\left|A_{n} M\right|=\left|B_{n} M\right|$, and $k_{A_{n} B_{n}}=-1$.
Thus, $\triangle M A_{n} B_{n}$ is an isosceles triangle with $A_{n} B_{n}$ as the base, and the slope of the altitude ... | (2,2) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,519 |
Three, let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
| Three, the minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$. This is because, if a number in $S$ appears only once in this sequence, since there are 3 two-element subsets containing this number, but in the sequence, the adjacent pairs... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,520 |
Four, find the minimum perimeter of a convex pentagon on a Cartesian plane whose vertices are lattice points (i.e., the x and y coordinates of each vertex are integers).
untranslated text:
求平面直角坐标系中格点凸五边形 (即每个顶点的纵、横坐标都是整数的凸五边形)的周长的最小值.
translated text:
Find the minimum perimeter of a convex pentagon on a Cartesian ... | Let the 5 vertices of this convex pentagon be $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, with coordinates $A_{j}(x_{j}, y_{j})$, and represent vertex $A_{j}$ as $x_{j} + i y_{j}, j=1,2,3,4,5, i$ being the imaginary unit.
Let $z_{j} = A_{j+1} - A_{j}, j=1,2,3,4,5, A_{6} = A_{1}$, then
(1) The real and imaginary parts of $z_{j... | 2 + 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,521 |
1. A math test paper contains only 25 multiple-choice questions. Correct answers earn 4 points, and wrong answers deduct 1 point. A student answered all the questions and scored a total of 70 points. Then he got ( ) questions correct.
(A) 17
(B) 18
(C) 19
(D) 20 | $$
-、 1 . \mathrm{C}
$$
1. Let the number of correct answers be $x$, then the number of wrong answers is $(25-x)$. According to the problem, we have
$$
4 x-(25-x)=70 .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,522 |
2. The value of $\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}$ is ( ).
(A) $\sqrt{6}$
(B) $2 \sqrt{6}$
(C) 6
(D) $\sqrt[4]{6}$ | 2. D
2. $(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}=6$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
2. D
2. $(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}=6$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,523 |
3. The sides of a right-angled triangle are $a-b, a, a+b$, and both $a$ and $b$ are positive integers. Then one of the sides could be ( . .
(A) 61
(B) 71
(C) 81
(D) 91 | $3 . \mathrm{C}$
3. Since only 81 is a perfect square among the 4 options, it is the only one that could possibly be a side of a right triangle. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,524 |
Example 10 Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10 \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$ | The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
\left(x+\frac{9}{x}\right)+\left(y+\frac{4}{y}\right)=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$
By Vieta's Theorem, \( x+\frac{9}{x} \) and \( y+\frac{4}{y} \) are the roots of the equation ... | \left\{\begin{array}{l}x=3, \\ y=2\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,527 |
6. In an isosceles $\triangle ABC$, $AD$ is the altitude to the base $BC$, $BE$ is the altitude to $AC$, $AD$ and $BE$ intersect at $H$, $EF \perp BC$ intersects $BC$ at $F$, $M$ is a point on the extension of $AD$ such that $DM = EF$, and $N$ is the midpoint of $AH$. Let $MN^2 = b$, $BN^2 = m$, $BM^2 = n$. The relatio... | 6. B
6. Without losing generality, we can take an isosceles right $\triangle ABC$ (as shown in the figure). Then points $A, E, H, N$ coincide, and points $D, F$ coincide: It is easy to see that $\triangle BMA$ is the same as $\triangle NBM$, which are isosceles right triangles, $\angle NBM=90^{\circ}$, hence $b=m+n$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,528 |
1. The average score of the participants in a junior high school mathematics competition at a certain school is 75 points. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore,... | II. 1.84.
Let the number of female participants be $x$, then the number of male participants is $1.8 x$; Let the average score of male participants be $y$ points, then the average score of female participants is $1.2 y$ points. According to the problem, we have $\frac{1.8 x y + 1.2 x y}{x + 1.8 x} = 75$, which simplifi... | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,529 |
2. Given $p+q=96$, and the quadratic equation $x^{2}+p x$ $+q=0$ has integer roots. Then its largest root is | 2. 98 .
Let the two roots be $x_{1}$ and $x_{2}$. Then $96=(x_{1}-1)(x_{2}-1)-1$, so $(x_{1}-1)(x_{2}-1)=97$. Since 97 is a prime number, then $x_{1}-1= \pm 1$, $x_{2}-1= \pm 97$, hence the largest root is 98. | 98 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,530 |
3. As shown in the figure, the side lengths of $\triangle A B C$ are $A B=14, B C$ $=16, A C=26, P$ is a point on the angle bisector $A D$ of $\angle A$, and $B P \perp A D, M$ is the midpoint of $B C$. Find the value of $P M$ $\qquad$ | 3. 6 .
From the figure, take $B^{\prime}$ on $A C$ such that $A B^{\prime}=A B=14$, then $B^{\prime} C=12$. Since $\triangle A B B^{\prime}$ is an isosceles triangle, we know that the intersection point of $B B^{\prime}$ and $A D$ is $P$ (concurrency of five lines), so $P$ is the midpoint of $B B^{\prime}$. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,531 |
4. There is a pile of balls in red and white, with several of each. It is known that the number of white balls is less than the number of red balls, but twice the number of white balls is more than the number of red balls. If each white ball is marked as “2”, and each red ball is marked as “3”, then the total is 60. Th... | 4. 9,14 . (3 points each)
Let the number of white balls be $x$, and the number of red balls be $y$. Then $\left\{\begin{array}{l}x<y<2 x, \\ 2 x+3 y=60\end{array} \Rightarrow\right.$ 7. $5<x<12$. Also, $2 x=60-3 y=3(20-y)$, so $x$ is a multiple of 3. | 9,14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,532 |
Three. (Full marks 20 points) If the equation $(x + a)(x + b) + (x + b)(x + c) + (x + c)(x + a) = 0$ (where $a, b, c$ are all positive numbers) has two equal real roots, prove: the segments of lengths $a, b, c$ can form a triangle, and indicate the characteristics of the triangle.
---
Translate the above text into En... | Three, the original equation is transformed into
$$
3 x^{2}+2(a+b+c) x+(a b+b c+c a)=0 \text {. }
$$
By the problem, we have
$$
\Delta=[2(a+b+c)]^{2}-12(a b+b c+c a)=0,
$$
Further transformation yields
$$
2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0,
$$
Thus, $a=b=c$, and $a>0, b>0, c>0$.
Therefore, segments of len... | null | Algebra | proof | Yes | Yes | cn_contest | false | 709,533 |
Four, (Full marks 20 points) As shown in the figure, the side length of the equilateral $\triangle ABC$ is $1$, and points $R, P, Q$ on $AB, BC, CA$ satisfy $AR + BP + CQ = 1$ while moving. Let $BP = x, CQ = y$, $AR = z$, and the area of $\triangle PQR$ be $S$. Try to express $S$ in terms of $x, y, z$.
---
The transl... | \begin{array}{l}\text { RT }, S_{\triangle A B C}=\frac{\sqrt{3}}{4}, S_{\triangle K Q Q}=-\frac{1}{2}(1-y) z \sin 60^{\circ} \\ =\frac{\sqrt{3}}{4}-(1-y) z \text {. } \\ \text { Similarly, } S_{\triangle B P R}=\frac{\sqrt{3}}{4}(1-z) x \text {, } \\ S_{\triangle G Q P}=\frac{\sqrt{3}}{4}(1-x) y \text {. } \\ \text { ... | \frac{\sqrt{3}}{4}(x y + y z + z x) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,534 |
Five. (Full marks 20 points) Given 1996 natural numbers $a_{1}$, $a_{2}, \cdots, a_{1996}$ that satisfy the condition: the sum of any two numbers is divisible by their difference. Let $n=a_{1} \cdot a_{2} \cdot \cdots \cdot a_{1996}$. Prove: $n, n+a_{1}, n+a_{2}, \cdots, n+a_{1996}$ these 1997 numbers still satisfy the... | Five, $\because a_{i}\left|n(i=1,2, \cdots, 1996), \therefore a_{i}\right| 2 n+a_{i}$.
Then $n+a_{i}+n$ can be divided by $a_{i}=n+a_{i}-n$, i.e., the sum of $n+a_{i}$ and $n$ can be divided by their difference.
We need to prove: $\left[n+a_{i}-\left(n+a_{j}\right)\right] \mid\left[\left(n+a_{i}\right)+(n+\right.$ $\l... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,535 |
2. The number of real roots of the equation $(x+1)|x+1|-x|x|+1=6$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. A
2. Discuss in three cases: $x>0, -1 \leqslant x \leqslant 0, x<-1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,537 |
Example 1 Find the smallest natural number with the following properties:
(1) Its decimal representation ends with the digit 6;
(2) If the last digit 6 is deleted and this digit 6 is written in front of the remaining digits, then the resulting number is 4 times the original number.
(4th IM) | Solution: From (1), we know that the last digit of the required smallest natural number is 6. Combining (2) with the multiplication table, it is not difficult to know that its tens digit is 4, and it is also the last digit of the obtained number.
$46 \times 4=184$.
Thus, the tens digit of the obtained number should be ... | 153846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,538 |
For example, $1 \overline{a b c d}$ is a four-digit natural number. It is known that $\overline{a b c d}-\overline{a b c}-\overline{a b}-a=1995$. Try to determine this four-digit number $\overline{a b c d}$
(1995, Beijing Middle School Advanced Mathematics Competition for Junior High School Students) | Solution: For easier analysis, convert the known horizontal expression into a vertical format, we get
$$
\begin{array}{r}
1000 a+100 b+10 c+d \\
-100 a-10 b-c \\
-10 a-b \\
\text { +) } \begin{array}{r}
-a
\end{array} \\
\hline 1000 \times 1+100 \times 9+10 \times 9+5
\end{array}
$$
By the subtraction rule and borrowi... | 2243 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,539 |
High $\mathbf{5 8}$ Two large circles $\odot A$ and $\odot B$ are equal and intersect. Two smaller circles $\odot C$ and $\odot D$ are unequal and also intersect, with intersection points at $P$ and $Q$. If $\odot C$ and $\odot D$ are simultaneously internally tangent to $\odot A$ and externally tangent to $\odot B$, p... | Proof: Let line $PQ$ intersect $AB$ at $M^{(4)}$, connect $AC, MC, BC, AD, MD, BD, PC, PD, CD$. Clearly, $PQ \perp CD$, and let the foot of the perpendicular be point $N$; $\odot A, \odot B$ have radii both $a$; $\odot C, \odot D$ have radii $R, r$ respectively, $(R \neq r)$.
It is easy to see that
\[
\begin{aligned}
A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,541 |
Example 1 Prove: For positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ whose sum is 1, the inequality $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}^{2}}{a_{n-1}+a_{n}}$ $+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$ holds.
(24th All-Soviet Union Mathematical Olympiad for High Sc... | Proof: Let the left side of the inequality be denoted as $A$, and construct the conjugate of $A$, $B=\frac{a_{2}^{2}}{a_{2}+a_{1}}+\frac{a_{3}^{2}}{a_{3}+a_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+\cdots-a_{n-1}}$
$$
+\frac{a_{1}^{2}}{a_{1}+a_{n}}
$$
Then $A-B=\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots$
$$
+\l... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 709,549 |
Example 2 Prove:
$$
\cdot \cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2} \text {. }
$$
(5th IMO) | Proof: Let $A=\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$, and construct its complementary conjugate $B=\sin \frac{\pi}{7}-\sin \frac{2 \pi}{7}+\sin \frac{3 \pi}{7}$. Then $\square$
$$
\begin{array}{c}
A^{2}+B^{2}=3-4 \cos \frac{\pi}{7}+2 \cos \frac{2 \pi}{7} \\
A^{2}-B^{2}=-\cos \frac{\pi}{7}+3 \cos ... | \frac{1}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 709,550 |
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