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$\mathrm{B}-4$. Let $p$ be an odd prime. Prove:
$$
\sum_{j=0}^{p}\binom{p}{j}\binom{p+j}{j} \equiv 2^{p}+1 \quad\left(\bmod p^{2}\right) .
$$
$\left(\binom{m}{m}\right.$ is equivalent to $C_{m}^{n}$. Note by Yize) | $$
\begin{array}{l}
\text { Let } \sum_{j=0}^{p}\binom{p}{j}\binom{p+j}{j} \\
=\sum_{j=0}^{p} \frac{(p+j)!}{(j!)^{2}(p-j)!} \\
=1+\sum_{j=1}^{p-1} \frac{(p+j)!}{(j!)^{2}(p-j)!}+\frac{(2 p)!}{(p!)^{2}} \text {. } \\
\text { Then } \frac{(2 p) i}{(p 1)^{2}}=2 \frac{(p+1)(p+2) \cdots(p+p-1)}{1 \cdot 2 \cdot \cdots \cdot(p... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,133 |
$\mathbf{\Sigma}-5$. Let $p$ be an odd prime, $Z_{p}$ denote the residue class field modulo $p$. How many elements does the following set contain
$$
\left\{x^{2}: x \in Z_{p}\right\} \cap\left\{y^{2}+1: y \in Z_{p}\right\} ?
$$ | This intersection has $[(p+3) / 4]$ elements. First, consider the solution set of the following equation:
$$
x^{2}=y^{2}+1.
$$
Rewriting it as $(x+y)(x-y)=1$, we see that for any non-zero element $r$ in $Z_{p}$, there is exactly one solution to the above equation, i.e., $x+y=r, x-y=r^{-1}$ or $x=\left(\frac{p+1}{2}\ri... | [(p+3) / 4] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,134 |
1. A triangular paper piece with side lengths of $\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}$ is folded along a line perpendicular to the side of length $\frac{3}{2}$. What is the maximum area of the overlapping part? | Solution 1: Let's assume in $\triangle ABC$, $a=\frac{3}{2}, b=\sqrt{2}$, $c=\frac{\sqrt{5}}{2}$. As shown in the figure, let the midpoint of $BC$ be $D$, and $AE \perp BC$. When $MN$ is folded, the overlapping area reaches its maximum. It is easy to see that $M$ is between $D$ and $E$. Let the symmetric point of $C$ w... | \frac{9}{28} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,136 |
2. Given $v_{0}=0, v_{1}=1$,
$$
\begin{array}{l}
v_{n+1}=8 v_{n}-v_{n-1}, \\
n=1,2, \cdots .
\end{array}
$$
Prove: In the sequence $\left\{v_{n}\right\}$, there are no terms of the form $3^{\alpha} \cdot 5^{\beta}(\alpha, \beta$ being positive integers). | Prove that the first few terms of $v$ are directly calculated as:
$0,1,8,63,496,3905,30744, \cdots$. Noting that $v_{3}=63, v_{6}=30744$ are both multiples of 3 and 7, and the remaining terms $v_{1}, v_{2}, v_{4}, v_{5}$ are not multiples of 3 or 7, we conjecture
$$
3\left|v_{n} \Leftrightarrow 7\right| v_{n} .
$$
We ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,137 |
Example 6. There are exactly 35 consecutive natural numbers whose integer parts of the arithmetic square roots are the same. Then, what is this identical integer?
---
The translation maintains the original text's format and line breaks as requested. | Analyzing to determine the integer part of a real number $a$ actually means finding two adjacent integers $n$ and $n+1$, such that $n \leqslant a < n+1$, at this point the integer part of $a$ is $n$. Using the property $2^{\circ}$ above, the answer to this question is 17. | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,139 |
5. Let $N=\{1,2,3,4, \cdots\}$. Is there a function $f: N \rightarrow N$, such that for every $n \in N$, we have $f^{(1880)}(n)=2 n$? Justify your answer. (where $f^{(k)}(n)=f\left(f^{(n-1)}(n)\right), f^{(1)}(n)=f(n))$ | Such a function exists. We replace 1988 with any natural number $m$ and prove that the function $f: N \rightarrow N$ satisfying
$$
f^{(m)}(n)=2 n
$$
exists.
First, let's discuss (1). From (1), we get
$$
\begin{array}{l}
f(2 n)=f\left(f^{(m)}(n)\right)=f^{(m+1)}(n) \\
=f^{(m)}(f(n))=2 f(n) .
\end{array}
$$
That is, $f(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,141 |
6. Given that $A D$ is the altitude of $\triangle A B C$, $B C+A D-A B$ $-A C=0$. Find the range of $x=\angle B A C$.
---
The translation is provided as requested, maintaining the original formatting and line breaks. | Solution: Clearly, $A D \leqslant A B, A D \leqslant A C$. Therefore, from $B C + A D - A B - A C = 0$, we get $B C \geqslant A C, B C \geqslant A B$. Thus, $\angle B, \angle C$ are both acute angles.
If $\angle A$ is a right angle, then $A B \cdot A C = A D \cdot B C$. Therefore, $A B + A C + A B + A C' + C' C > A B ... | \left[2 \arcsin \frac{3}{5}, \frac{\pi}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,142 |
Circular paper pieces. Ask how many different colors are needed at least to ensure that any two tangent circular pieces have different colors?
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Obviously, the three circular pieces $A$, $B$, and $E$ can only be colored with two colors, 1 and 3, and $A$ is one color, while $B$ and $E$ are another color. If there are only three colors, then the circular pieces $C$ and $D$ on the road must be different colors. Therefore, all four different colors are needed. When... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,143 |
3. $a, b, c$ are real numbers, and $a^{2}-b c-8 a+7=0$, $b^{2}+c^{2}+b c-6 a+6=0$. Then the range of values for $a$ is ( ).
(A) $-\infty1$.
(C) $1<a<13$.
(D) $1 \leqslant a \leqslant 9$. | \begin{tabular}{c}
3 \\
\hline $\mathrm{D}$
\end{tabular} | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,147 |
Example 7. The number of natural numbers $n$ that make $n^{2}-19 n+91$ a perfect square is?
Will the above text be translated into English, please keep the original text's line breaks and format, and output the translation result directly. | Notice that $n^{2}-19 n+91=(n-9)^{2}+(10-$ $n$ ). When $n>10$,
$$
(n-10)^{2}<(n-9)^{2}+(10-n)<(n-9)^{2} .
$$
The integers between two consecutive perfect squares cannot be perfect squares, therefore the natural numbers $\pi$ that make $n^{2}-10 n+91$ a perfect square can only be from the ten numbers $1,2,3, \cdots, 9,... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,150 |
1. The graph of a quadratic function passes through $(1,0),(5, 0)$, the axis of symmetry is parallel to the $y$-axis, but does not pass through points above the line $y=2x$. Then the product of the maximum and minimum values of the coordinates of its vertex is $\qquad$ . | 1. 4.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,154 |
Three, (Full marks 20 points) Solve the equation:
$$
|x-| 2 x+1||=3 .
$$ | Three, from the original equation, we get
$$
x^{2}+(2 x+1)^{2}-2 x|2 x+1|=9 \text {. }
$$
When $x \geqslant-\frac{1}{2}$, equation (1) becomes
$$
x^{2}+2 x-8=0 .
$$
Solving, we get $x_{1}=-4, x_{2}=2$.
$$
\because x \geqslant-\frac{1}{2} \text {, }
$$
$\therefore$ Therefore, $x_{2}=2$ is a root of the original equati... | x=-\frac{4}{3}, x=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,158 |
Four. (This question is worth 20 points) In the cyclic hexagon $A B C D E F$, $A B=B C=C D=3 \text{~cm}, D E=E F=$ $F A=5 \text{~cm}$. Find the area of this hexagon.
. | $$
\begin{array}{l}
=\frac{1}{2}(\overparen{B C}+\overparen{A B}+\overparen{A F}+\overparen{E F}) \\
=\frac{1}{2} \cdot \frac{2}{3} \cdot 360^{\circ}=120^{\circ} .
\end{array}
$$
Similarly, $\angle B A F=120^{\circ}$.
By the cosine rule
$$
\begin{array}{l}
C E^{2}=C D^{2}+D E^{2}-2 C D \cdot D E \cos 120^{\circ} \\
=3... | \frac{47}{2} \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,159 |
Five. (This question is worth 20 points) Prove: From $1,2,3,4$, $\cdots, 3900$, by selecting any 1993 numbers, there must be two numbers whose difference is exactly 93.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since $1993=21 \times 93+40$, among any 159 numbers chosen, there must be 22 numbers that have the same remainder when divided by 93, let the remainder be $r$.
If among these 22 numbers, no two numbers have a difference of 93, then the smallest of these 22 numbers is
$$
\begin{array}{l}
a_{1}=0 \times 93+r, \\
a_{2}=2... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 706,160 |
Example 8. Prove: For any natural number $n$,
$$
n^{4}+2 n^{3}+2 n^{2}+2 n+1
$$
is not a perfect square. | Analyzing whether a polynomial is a perfect square can be done by one method, which is factorization. The original expression is $=(n+1)^{2}\left(n^{2}+1\right)$.
One method is to sandwich the polynomial to be discussed between two consecutive perfect squares, that is,
$$
\begin{aligned}
\left(n^{2}+n\right)^{2} & <n^... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,161 |
1. The range of the function $y=\sqrt{x-4}+\sqrt{15-3 x}$ is ( ).
(A) $[1,2]$.
(B) $(0,2)$.
(C) $(0, \sqrt{3})$.
(D) None of the above. | 1. (A).
Obviously, the domain of the function is $4 \leqslant x \leqslant 5$. Then we can set $x$ $=4+\sin ^{2} \alpha \quad\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$. Thus,
$$
\begin{aligned}
y= & \sqrt{\sin ^{2} \alpha}+\sqrt{3 \cos ^{2} \alpha} \\
& =\sin \alpha+\sqrt{3} \cos \alpha \\
& =2 \sin \left... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,162 |
2. Given a rhombus $A B C D$ with side length $a$ and $\angle A=\frac{\pi}{3}$. The rhombus $A B C D$ is folded along the diagonal to form a dihedral angle $\theta$, where $\theta \in \left[\frac{\pi}{3}, \frac{2 \pi}{3}\right]$. Then the maximum distance between the two diagonals is
(A) $\frac{3}{2} a$.
(B) $\frac{\sq... | 2. (D).
As shown in the figure, fold along $AC$ to form a dihedral angle. Since $AC \perp OB$, $AC \perp OD$ (in the plane figure), then $AC \perp$ plane $BOD$.
Let $\angle A=60^{\circ}$, then $OB=OD=\frac{a}{2}$. Since $\angle BOD$ is the plane angle of the dihedral angle, when $\angle BOD=60^{\circ}$, $BD=\frac{a}{... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,163 |
3. The number of real roots of the equation $x=\sin x+1993$ is ( ).
(A) 0 .
(B) 1 .
(C) 2 .
(D) Greater than 2 . | 3. (B).
Since the roots of the equation are the x-coordinates of the intersection points of the line $y=x$ and the curve $y=\sin x +1993$, the equation has at least one real root.
Assume the equation has two distinct roots $x_{1}, x_{2}, x_{1} \neq x_{2}$, then
$$
\begin{array}{c}
x_{1}=\sin x_{1}+1993, \quad x_{2}=\s... | B | Calculus | MCQ | Yes | Yes | cn_contest | false | 706,164 |
4. Given the complex sequence $\left\{a_{n}\right\}$ with the general term $a_{n}=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)\left(1+\frac{i}{\sqrt{3}}\right) \cdots$
- $\left(1+\frac{i}{\sqrt{n}}\right)$.
Then $\left|a_{n}-a_{n+1}\right|=$ ( ).
(A) $\frac{\sqrt{2}}{2}$.
(B) $\sqrt{2}$.
(C) 1 .
(D) 2 . | 4. (C).
First, prove that $\left|a_{n}\right|=\sqrt{n+1}$. Use mathematical induction. When $n=1$, $\left|a_{1}\right|=|1+i|=\sqrt{1+1}$, the conclusion holds.
Assume that when $n=k$, $\left|a_{k}\right|=\sqrt{k+1}$ holds. Then when $n=$ $k+1$,
$$
\begin{aligned}
\left|a_{k+1}\right| & =\left|a_{k} \cdot\left(1+\frac{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,165 |
5. For a convex $n$-sided polygon, if circles are constructed with each side as the diameter, the convex $n$-sided polygon must be covered by these $n$ circles. Then the maximum value of $n$ is: ( ).
(A) 3 .
(B) 4 .
(C) 5 .
(D) Greater than 5 . | 5. (B).
In any triangle, if a perpendicular is drawn from the vertex of the largest angle to the opposite side, then the circles with the two smaller sides as diameters can cover the entire triangle.
For any convex quadrilateral $ABCD$, if circles are constructed with each side as a diameter, then these four circles ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,166 |
6. Given an integer-coefficient polynomial, a student obtained the results $f(-2)=-56, f(1)=-2, f(3)=53, f(6)=528$. Then the one he calculated incorrectly is
(A). $f(-2)$.
(B) $f(1)$.
(C) $f(3)$.
(D) $f(6)$. | 6. (C).
Let the polynomial with integer coefficients be $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}$ $+\cdots+a_{1} x+a_{0}$. Clearly, $m-t \mid p(m)-p(t)$. Here, $m$ and $t$ are integers. Using this conclusion to verify each case, it is found that $f(3)=53$ is incorrect. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,167 |
1. The solution set of the inequality $\frac{x}{\sqrt{1+x^{2}}}+\frac{1-x^{2}}{1+x^{2}}>0$ is
$\qquad$ | 1. $\left\{x \left\lvert\, x>-\frac{\sqrt{3}}{3}\right.\right\}$.
Let $x=\operatorname{tg} \alpha, -\frac{\pi}{2} < \alpha < \frac{\pi}{2}$. Then, we have:
$$
2 \sin ^{2} \alpha-\sin \alpha-1-\frac{\sqrt{3}}{3} \cos \alpha > 0.
$$
Rearranging, we get $2 \sin ^{2} \alpha-\sin \alpha-1-\frac{\sqrt{3}}{3}$
which means $... | x>-\frac{\sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,168 |
2. Given three planes $\alpha // \beta // \gamma, \beta$ is between $\alpha$ and $\gamma, \alpha$ and $\beta$ are a distance of $4a$ apart, $\beta$ and $\gamma$ are a distance of $2a$ apart, and there is an equilateral triangle $ABC$ with side length $9a$. $A$ is in plane $\alpha$, $B$ is in plane $\beta$, $C$ is in pl... | 2. $\arcsin \frac{4 \sqrt{21}}{27}$.
As shown in the figure, draw $A O \perp \beta$ at $O$, and $A O^{\prime} \perp \gamma$ at $O^{\prime}$, then $A O=4 a, A O^{\prime}=6 a$. Draw $O E \perp B D$ at $E$, and connect $A E$, then $\angle A E O$ is the plane angle of the dihedral angle to be found.
From $\frac{A D}{A C}=... | \arcsin \frac{4 \sqrt{21}}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,169 |
3. Toss a coin, if it lands heads, point $P$ moves +1 on the number line, if it lands tails, it does not move. The coin is tossed no more than 12 times, and if point $P$ reaches coordinate +10, no more tossing occurs. Then the total number of different ways for point $P$ to reach coordinate +10 is . | 3. 66.
Obviously, the last coin toss must be heads.
(i) To get 10 points with 10 heads, there is 1 way.
(ii) For 11 tosses, since the last one is heads, the first 10 must include 1 tails, which has $C_{10}^{1}=10$ ways.
(iii) For 12 tosses, since the last one is heads, the first 11 must include 2 tails, which has $C_{... | 66 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,170 |
4. Given that the line $l$ passes through the focus of the parabola $C: y^{2}=4 x$, and the slope $k>2, l$ intersects the parabola $C$ at points $A, B$; the midpoint $M$ of $A B$ is at a distance of $\frac{1}{5}$ from the line $L_{m}: 3 x+4 y+m=0(m>$ -3 ）. Then the range of values for $m$ is $\qquad$ . | 4. $-32)$. Substituting this equation into $y^{2}=4 x$, eliminating $y$ yields
$$
k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0 .
$$
Let $M\left(x_{0}, y_{0}\right)$, then
$$
x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{k^{2}+2}{\dot{k}^{2}}, y=k\left(x_{0}-1\right)=\frac{2}{k} \text {. }
$$
The distance from $M$ to the line $L_{m... | -3 < m < -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,171 |
Example 9. Prove: For any natural number $n, n^{2}+3 n+3$ is not equal to the product of two consecutive natural numbers.
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
The translation:
Example 9. Prove: For any natural number $n, n^{2}+3 n+3$ is not equal to the product of two consecutive natural numbers. | To prove this problem, we introduce a conclusion and make full use of the tool of adjacent square numbers.
Conclusion: If a natural number $M$ can be written as the product of two consecutive natural numbers, then $4M+1$ must be a perfect square. Conversely, this is also true.
If $M=n(n+1)$, then $4M+1=(2n+1)^2$;
If $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,172 |
5. Let $0 \leqslant x \leqslant 8$, then the range of the function $f(x)$ $=\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1}$ is | 5. $\{y \mid 0 \leqslant y \leqslant 4\}$.
Clearly $y \geqslant 0$, and when $x=0$, $y=0$.
$$
\text{Also } \begin{aligned}
y & =\frac{\sqrt{\left(x^{2}+8\right)\left(8 x-x^{2}\right)}}{x+1} \\
& \leqslant \frac{1}{x+1} \cdot \frac{x^{2}+8+8 x-x^{2}}{2} \\
& =4 .
\end{aligned}
$$
Equality holds if and only if $x^{2}+8... | 0 \leqslant y \leqslant 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,173 |
6. Given $f(x)=(\sin x+4 \sin \theta+4)^{2}+(\cos x$ $-5 \cos \theta)^{2}$, the minimum value of $f(x)$ is $g(\theta)$. Then the maximum value of $g(\theta)$ is | 6. 49.
For $f(x)$, expanding and rearranging yields
$$
\begin{aligned}
f(x) & =8(1+\sin \theta) \sin x-10 \cos \theta \cdot \cos x \\
& -9 \sin ^{2} \theta+32 \sin \theta+42 \\
& =\sqrt{64(1+\sin \theta)^{2}+100 \cos ^{2} \theta} \\
& \cdot \sin (x+\varphi)-9 \sin ^{2} \theta+32 \sin \theta \\
& +42 . \\
g(\theta) & =... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,174 |
Three. (This question is worth 20 points) Given $0<a<1, \alpha \geqslant 0$, $\beta \geqslant 0$, and $x=a^{a}+a^{\beta}, y=a^{i \alpha}+a^{2 \beta}$. Try to plot the region of the point $(x, y)$ in the Cartesian coordinate system. | Three, from the known we get $x^{2}-2 a^{*} a^{\beta}=y$. Therefore, from
$$
a^{a}+a^{\beta}=x, a^{a} a^{\beta}=\frac{1}{2}\left(x^{2}-y\right)
$$
we know that $a^{*}, a^{\beta}$ are the two roots of the equation
$$
t^{2}-x t+\frac{1}{2}\left(x^{2}-y\right)=0
$$
and the two roots satisfy $00, \\
f(1)=1-x+\frac{1}{2}\... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,175 |
Four. (This question is worth 20 points) Given the circle $C: x^{2}+(y-d)^{2}=r^{2}(d>r)$. Take two points $M, N$ on the $x$-axis such that the circle with $MN$ as its diameter is externally tangent to circle $C$. Find the coordinates of point $A$ such that for all $M, N$ satisfying the condition, $\angle MAN$ is a con... | Let the center of the circle with diameter $MN$ be $T(h, 0)$, and the radius be $s$. Then $M(h-s, 0), N(h+s, 0)$. Let the center of circle $C$ be $K(0, d)$.
Then $K T^{2}=(r+s)^{2}=h^{2}+d^{2}$.
By symmetry, the point $A$ we are looking for lies on the $y$-axis, so we can set $A(0, t)$, assuming $t>0$.
Let the slope o... | (0, \pm \sqrt{d^{2}-r^{2}}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,176 |
Five. (This question is worth 20 points) Given an odd function $f(x)$ defined on $(-\infty, 0) \cup(0,+\infty)$, and it is an increasing function on $(0, +\infty)$, with $f(1)=0$. Also, the function
$g(\theta)=\sin ^{2} \theta+m \cos \theta-2 m, \theta \in\left[0, \frac{\pi}{2}\right]$, and the sets
$M, N$ satisfy $M=\... | Five, since the odd function $f(x)$ is increasing on $(0,+\infty)$, then $f(x)$ is also increasing on $(-\infty, 0)$.
Furthermore, because $f(1)=0$, we have
$$
f(-1)=-f(1)=0 .
$$
Therefore, simultaneously satisfying $g(\theta)0 .
\end{array}
$$
Let $\cos \theta=t$, then $t \in[0,1]$. Suppose
$$
\varphi(t)=t^{2}-m t+2... | m>4-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,177 |
One. (This question is worth 35 points) Given that the circumcircle of $\triangle ABC$ is circle $O$, and $P, Q, R$ are the midpoints of $\widehat{BC}$, $\overparen{AC}$, and $\overparen{AB}$ respectively. Chord $PR$ intersects $AB$ at $D$, and chord $PQ$ intersects $AC$ at $E$.
Prove: $DE \parallel BC$. | Connect $A P, B Q, C R$.
Since $P, Q, R$ are the midpoints of $\widehat{B C}, \widehat{C A}, \overparen{A B}$, then $A P, B Q, C R$ concur at the incenter $I$ of $\triangle A B C$.
It can be proven that $\angle A I R = \angle A D R$. Therefore, $A, I, D, R$ are concyclic.
Thus, $\angle A I D + \angle A R D = 180^{\cir... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,178 |
II. (This question is worth 35 points) Prove that if and only if $a = 4m^4$ (where $m > 1$), then for all natural numbers $n$, $n^4 + a$ is a composite number. | When $a=4 m^{4}$,
$$
\begin{aligned}
n^{4}+a & =n^{4}+4 m n^{4} \\
& =\left(n^{2}+2 m^{2}\right)-(2 m n)^{2} \\
& =\left[(n+m)^{2}+m^{2}\right] \\
& \cdot\left[(n-m)^{2}+m^{2}\right] .
\end{aligned}
$$
When $m>1$, $(n+m)^{2}+m^{2},(n-m)^{2}+m^{2}$ are both integers greater than 1 for all natural numbers $n$, so $n^{4}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,179 |
Three. (This question is worth 35 points) There are 10 boys, each of whom is no more than 20 years old, and 20 girls, each of whom is no more than 10 years old. Prove that there must be a certain number of girls whose ages sum up to the sum of the ages of a certain number of boys. The ages are all positive integers.
-... | Three, let the ages of 10 boys be $a_{1}, a_{2}, \cdots, a_{10}$. Where $a_{i}$ satisfies $1 \leqslant a_{i} \leqslant 20, a_{i} \in N$.
And let the ages of 20 girls be $b_{1}, b_{2}, \cdots, b_{20}$, where $b_{j}$ satisfies $1 \leqslant b_{j} \leqslant 10, b_{j} \in N$.
Let $S_{i}=a_{1}+a_{2}+\cdots+a_{i} \quad(i=1,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,180 |
Example 10. If $x$ and $y$ are both natural numbers, prove that $x^{2}+y+1$ and $y^{2}+4x+3$ cannot both be perfect squares at the same time. | Prove that when $x \geqslant y$,
$$
x^{2}<x^{2}+y+1 \leqslant x^{2}+x+1<(x+1)^{2} .
$$
Thus, $x^{2}+y+1$ is not a perfect square.
When $x<y$,
$$
y^{2}<y^{2}+4 x+3<y^{2}+4 y+3<(y+2)^{2} .
$$
This shows that $y^{2}+4 x+3$ can only be a perfect square when $y^{2}+4 x+3=(y+1)^{2}$, i.e., $y=2 x+1$. However, in this case,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,181 |
Example 1. (1983, National High School Competition) As shown in Figure 1, in quadrilateral $A B C D$, the area ratios of $\triangle A B D, \triangle B C D, \triangle A B C$ are $3: 4: 1$. Points $M, N$ are on $A C, C D$ respectively, satisfying $A M: A C=C N: C D$, and $B, M, N$ are collinear. Prove that $M$ and $N$ ar... | Proof as shown in Figure 1,
Let \(\frac{A M}{A C} = \frac{C N}{C D} = r (0 < r < 1)\).
\(\because S_{\triangle A R D} : S_{\triangle A C D} : S_{\triangle A B C} = 3 : 4 : 1\),
\[
\begin{array}{l}
\therefore \frac{B E}{B D} = \frac{1}{7}, \frac{A E}{A C} = \frac{3}{7}. \\
\frac{E M}{M C} = \frac{A M - A E}{M C} = \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,182 |
List 2. (Pascal's Theorem) Prove: The three pairs of opposite sides of a hexagon inscribed in a circle (if the opposite sides are not parallel) intersect at points that are collinear.
| Proof As shown in Figure 2, in the cyclic hexagon $A B C D E F$, let $A B$ and $D E$ intersect at $P$, $B C$ and $E F$ intersect at $Q$, and $C D$ and $A F$ intersect at $R$.
Also, let $A B$ and $D C$ intersect at $V$, $F E$ and $C D$ intersect at $U$, and $E F$ and $P V$ intersect at $W$. For $\triangle U V W$ and the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,183 |
Example 2. (Burgess, 1985, see Figure 5) Let the three sides of $\angle A_{1} A_{2} A_{3}$ be $A_{2} A_{3} + A_{3} A_{1}$ and $A_{1} A_{2}$, denoted as $a_{1}, a_{2}, a_{3}$, and $a_{1} > a_{2} > a_{3}$. Let $C_{1}, C_{2}, C_{3}$ be the midpoints of sides $A_{2} A_{3}$, $A_{3} A_{1}$, and $A_{1} A_{2}$, respectively. $... | Proof As shown in the figure. Extend $A_{1} A_{2}, C_{2} B_{3}$ to intersect at $B_{3}^{\prime}$. In the theorem, replace $B_{3}$ with $B_{3}^{\prime}$, then extend $A_{3} A_{2}$ to $A_{3}^{\prime}$, such that $A_{3}^{\prime} B_{3}=B_{3} A_{3}$. Thus, $A_{3}^{\prime} A_{1}$ is parallel to $B_{3} C A_{2}$. Since $A_{1} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,184 |
Three. (This question is worth 35 points) Divide the set $M=\{1, 2, \cdots, 89\}$ into two subsets $A$ and $B$. If $A \cap B=\varnothing, A \cup B=M$, then it is called a partition of set $M$. Prove that for any partition of set $M$ into two subsets $A$ and $B$, the equation $x_{1}+x_{2}+\cdots+x_{9}=x_{10}$ must have ... | Three, prove using the proof by contradiction.
Suppose the set is divided into two subsets $A$ and $B$, in both $A$ and $B$, the equation $x_{1} + x_{2} + \cdots + x_{9} = x_{10}$ has no solution.
If $1 \in A$, to make the equation unsolvable, since
$$
x_{1} + x_{2} + \cdots + x_{9} \geqslant 9 \text {, }
$$
then $9 \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,185 |
In the letter摘登 of the sixth issue of "Middle School Mathematics" in 1992, Comrade Yang Xuezhi from Fuzhou No. 24 High School mentioned a conjecture by a senior high school student: "Given $\triangle P_{1} P_{2} P_{3}$ and a point $P$ inside it. The lines $P_{1} P, P_{2} P, P_{3} P$ intersect the opposite sides at $Q_{... | First, it should be noted that the above "conjecture" has long been proposed as a correct proposition and has been proven. Now, we introduce a proof method as follows (figure omitted).
Proof: Let the areas of $\triangle P_{2} P P_{3}, \triangle P_{3} P P_{1}, \triangle P_{1} P P_{2}$ be $S_{1}, S_{2}, S_{3}$, respecti... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false | 706,187 |
Example: In the complex plane, it is known that the vertices $A$, $B$ of the equilateral triangle $\triangle ABC$ represent the complex numbers 2 and $\frac{1}{2}+\frac{\sqrt{3}}{2} i$, respectively. Find the complex number that $C$ represents. | Let $C$ correspond to $z$, by the theorem
$$
\begin{array}{l}
\left\{\begin{array}{l}
2+\frac{1}{2}+\frac{\sqrt{3}}{2} i+z=3 z_{0}, \\
2\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)+2 z+\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) z=3 z_{0}^{2},
\end{array}\right. \\
\text { Solving, we get }\left\{\begin{array}{l}
z=... | C\left(\frac{1}{2}-\frac{\sqrt{3}}{2} i\right) \text{ or } C(2+\sqrt{3} i) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,188 |
Example. Find the equation of the quadrilateral with vertices $A(-3,-1), B(0,-4)$, $C(4,0), D(-1,1)$. | Solving, by the two-point form, the equations of $A C$ and $B D$ are
$$
x-7 y-4=0, \quad 5 x+y+4=0 \text { . }
$$
Substituting (*) we get
$$
\begin{array}{l}
|x-7 y-4|+r|5 x+y+4|+a x+b y+ \\
c=0 .
\end{array}
$$
Substituting the coordinates of the vertices, we solve to get $r=1, a=-2, b=2$, $c=-16$. Thus, the equatio... | |x-7 y-4|+|5 x+y+4|-2 x-2 y-16=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,189 |
Example 1. Let $n \in N$. Prove: $: 14 \mid 3^{\operatorname{tn}+2}+5^{2 n+1}$. | Consider the function
$$
f(x)=(x-5)^{2 n+1}+5^{2 n-1}.
$$
Since $f(0)=0$, $f(x)$ can be divided by $(x-0)$, particularly $14 \mid f(14)$, that is
$$
14 \mid f(14)=9^{2 n+1}+5^{2 n+1}=3^{2 n+1}+5^{2 n+1}.
$$ | 14 \mid 3^{2 n+1}+5^{2 n+1} | Number Theory | proof | Yes | Yes | cn_contest | false | 706,190 |
Let $n$ be an odd number. Prove: there exist $2 n$ integers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$, such that for any integer $k, 0<k<n$, the following $3 n$ numbers
$a_{i}+a_{i+1}, a_{i}+b_{i}, b_{i}+b_{i+k}$
$\left(i=1,2, \cdots, n\right.$; where $a_{n+1}=a_{1}, b_{n+j}=b_{j}, 0$ $<j<n)$ have dist... | Prove that taking $a_{i}=3 i-2, b_{i}=3 i-3(i=1,2$, $\cdots, n)$, we have
$$
\begin{aligned}
a_{i}+a_{i+1} & =3 i-2+3(i+1)-2 \\
& =6 i-1=6(i-1)+5, \\
a_{i}+b_{i} & =3 i-2+3 i-3 \\
& =6 i-5=6(i-1)+1, \\
b_{i}+b_{i+k} & =3 i-3+3(i+k)-3 \\
& =6(i-1)+3 k .
\end{aligned}
$$
(1), (2), (3) three sets of numbers are different ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,192 |
Given $k \in N$ and a real number $a>0$, under the following conditions,
$$
k_{1}+k_{2}+\cdots+k_{r}=k, k_{i} \in N, 1 \leqslant r \leqslant k .
$$
Find the maximum value of $a^{k_{1}}+a^{k_{2}}+\cdots+a^{k}$.
(Pan Fu supplied the problem) | For $a>0$ and $s, t \in \mathbb{N}$, it is obvious that
$$
a\left(a^{t-1}-1\right)\left(a^{t-1}-1\right) \geqslant 0,
$$
thus we have
$$
a^{t}+a^{t} \leqslant a+a^{t+t-1}.
$$
Using this inequality, we can obtain
$$
a^{k_{1}}+a^{k_{2}}+\cdots+a^{k} \leqslant (r-1) a+a^{k-(r-1)}.
$$
Since the function $a x+a^{k-x}$ is... | \max \left\{k a, a^{k}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,193 |
Three, let circles $K$ and $K_{1}$ be concentric, with radii $R$ and $R_{1}$ respectively, where $R_{1} > R$. Quadrilateral $ABCD$ is inscribed in circle $K$, and quadrilateral $A_{1}B_{1}C_{1}D_{1}$ is inscribed in circle $K_{1}$. Points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ lie on the rays $CD, DA, AB$, and $BC$ respect... | Let's denote $S_{A B C D}$ as $S$, and $S_{A_{1} B_{1} C_{1} D_{1}}$ as $S_{1}$. Let $a, b, c, d$ represent the lengths of segments $A B, B C, C D, D A$, and $x, y, z, w$ represent the lengths of segments $A B_{1}, B C_{1}, C D_{1}, D A_{1}$, respectively.
First, we prove the following conclusions:
$$
a b + c d \leqsl... | \frac{S_{A_{1}B_{1}C_{1}D_{1}}}{S_{ABCD}} \geqslant \frac{R_{1}^{2}}{R^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 706,194 |
$$
S=\left\{z_{1}, z_{2}, \cdots, z_{1993}\right\},
$$
where $z_{1}, z_{2}, \cdots, z_{1993}$ are non-zero complex numbers (which can be viewed as non-zero vectors on the plane). Prove that the elements of $S$ can be divided into several groups such that
(1) each element of $S$ belongs to and only to one of these grou... | For the non-empty subsets of $S$, we call the modulus of the sum of all complex numbers in the subset the sum modulus of the subset. Among all non-empty subsets of $S$, choose the one with the largest sum modulus and denote it as $A_{1}$, and let $a_{1}$ be the sum of all complex numbers in $A_{1}$.
We can prove that ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 706,195 |
Five, 10 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 most purchased book? Explain your reasoning.
(Na Chengzhang provided the question) | Solution 1: Let's assume that each person buys at most one copy of the same book. Thus, by condition (1), 10 people buy a total of 30 books.
Let the minimum number of people who buy the most popular book be $n$. Assume $A$ buys books 甲, 乙, and 丙.
Clearly, $n > 3$. Because if $n \leqslant 3$, then at most two more peo... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,196 |
Example 3. (29th IMO Preliminary Question) In tetrahedron $ABCD$, the midpoints of edges $AB$ and $CD$ are $K$ and $L$ respectively. Prove that any plane passing through $KL$ divides the tetrahedron into two parts of equal volume. | Proof As shown in Figure 3, let the plane $KEL$ passing through $K, L$ intersect $BC$ at $E$, $AD$ at $F$, and the extension of $BD$ at $G$.
Since $L$ is the midpoint of $CD$, the distances from $C$ and $D$ to the plane $KEL$ are equal.
Thus, $V_{C-KELF} = V_{D-KELF}$.
Let the volume of tetrahedron $ABCD$ be $V$:
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,197 |
Six, let the function $f: (0,+\infty) \rightarrow (0,+\infty)$ satisfy $f(x y) \leqslant f(x) f(y)$ (for any $x>0, y>0$).
Prove: $f\left(x^{n}\right) \leqslant f(x) f\left(x^{2}\right)^{\frac{1}{2}} \cdots f\left(x^{n}\right)^{\frac{1}{n}}$, for any $x>0, n \in N$.
(Supplied by Zhang Zhuxian) | Prove that for $F_{n}(x)=\prod_{k=1}^{n} f\left(x^{k}\right)^{\frac{1}{k}}, n=1,2$, $\cdots$.
Thus, $F_{1}(x)=f(x)$,
$$
F_{i}(x)=f\left(x^{i}\right)^{\frac{1}{i}} \cdot F_{i-1}(x), i=2,3, \cdots .
$$
Therefore, for any $n \in \mathbb{N}$, we have
$$
\begin{array}{l}
{\left[F_{n}(x)\right]^{n}=f\left(x^{n}\right)\left[... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,198 |
2. If the equation $\sin ^{2} x+\cos x+m=0$ has real solutions, then the range of values for $m$ is $\qquad$ | 2. $-\frac{5}{4} \leqslant m \leqslant 1$.
The above text has been translated into English, retaining the original text's line breaks and format. | -\frac{5}{4} \leqslant m \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,200 |
4. Arrange the numbers $\{2 n+1\}$ in a cycle of: one, two, three, four terms: (3), $(5,7),(9,11,13)$, $(15,17,19,21),(23),(25,27),(29,31$, $33),(35,37,39,41),(43), \cdots$, then the sum of the numbers in the 100th parenthesis is $\qquad$. | 4. The 100th parenthesis
is the fine one in the
25th group,
the subsequence of parentheses in.
Accordingly, it forms
an arithmetic sequence
with the first term being 72 and the
common difference being 80. Therefore, its sum is $72+$ (25-
$$
\text { 1) } \times 80=1992 \text {. }
$$ | 1992 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,202 |
5. $n$ parabolas $y=a x^{2}+b x+c(a \neq 0)$, divide the plane $O x y$ into at most $\qquad$ parts. | 5. $n^{2}+1$ (Hint: Apply mathematical induction).
| n^{2}+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,203 |
II. Given that $AB$ and $CD$ are the diameter and chord of a unit circle, respectively, and $AB \perp CD, \overparen{BD}=2 \overparen{AD}, E$ is the midpoint of $BD$, connect $AE$, intersecting $CD$ at point $P$, extend $AE$ to intersect the circle at point $F$; then connect $CF$, intersecting $AB$ at point $Q$.
(1) Pr... | (1) Proof: Connect $A D, C E, B F$, and $B C$.
$\because A B$ is the diameter, and $A B \perp C D$, we have $\overparen{A D}=$ $\overparen{A C}$,
$$
\therefore \angle A F C=\angle D B A \text {. }
$$
Therefore, points $B, F, E, Q$ are concyclic, then
$$
\angle E Q R=180^{\circ}-\angle E F B=90^{\circ} \text {. }
$$
T... | \frac{3}{8} \sqrt{3} | Geometry | proof | Yes | Yes | cn_contest | false | 706,204 |
Three, let $a>1$ be a positive real number, and $n \geqslant 2$ be a natural number, and the equation $[a x]=x$ has exactly $n$ different solutions. Try to find the range of values for $a$. ( $[x]$ denotes the greatest integer less than or equal to $x$) | Three, Solution From the given, we know that $x$ must be an integer, and $x \geqslant 0$. Let $\{a\}=a-[a]$, then
$$
a x=[a] x+\{a\} x,
$$
thus the original equation becomes
$$
x=[a x]=[a] x+[\{a\} x] .
$$
$\because[a] \geqslant 1, \therefore(1)$ holds if and only if
$$
[a]=1 \text { and }\{a\} x<1 \text {. }
$$
$\bec... | 1+\frac{1}{n} \leqslant a<1+\frac{1}{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,205 |
Let the integer $n \geqslant 4, a_{1}, a_{2}, \cdots, a_{n}$ be distinct integers in the interval $(0,2 n)$. Prove: The set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ has a subset such that the sum of all its elements is divisible by $2 n$.
---
The translation maintains the original text's formatting and structure,... | (1) If $n \in \{a_1, a_2, \cdots, a_n\}$, then the $2n$ integers $a_1, a_2, \cdots, a_n, 2n - a_1, 2n - a_2, \cdots, 2n - a_n$ all lie in $(0, 2n)$. Therefore, at least two of these numbers are equal, let $a_i = 2n - a_j$.
Since $n \in \{a_1, a_2, \cdots, a_n\}$, it must be that $i \neq j$.
Thus, $a_i + a_j$ is divisib... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,206 |
6. Determine the smallest positive integer A with the following property: for any permutation of all positive integers from 1001 to 2000, it is possible to find 10 consecutive terms whose sum is greater than or equal to $A$.
| (Continued from page 28)
$S_{i}=b_{i}+b_{i+1}+\cdots+b_{i+9}(i=1,2, \cdots, 1991)$.
Then
$$
\begin{array}{l}
S_{1}+S_{11}+S_{21}+\cdots S_{1991} \\
=b_{1}+b_{2}+\cdots+b_{2000} \\
=\frac{(1001+2000) \times 1000}{2} \\
=1500500 .
\end{array}
$$
Thus, among $S_{1}, S_{11}, S_{21}, \cdots, S_{1991}$, there is at least on... | 15005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,207 |
Example 4. (1990, Suzhou City High School Competition) As shown in Figure 4, in $\triangle A B C$, $\angle B A C=90^{\circ}$, and $G$ is a given point on $A B$ (where $G$ is not the midpoint of segment $A B$). Let $D$ be any point on line $G C$ different from $C$ and $G$, and let the lines $A D$ and $B C$ intersect at ... | Proof As shown in Figure 4, let $G$ divide $AB$ in the ratio $\lambda_{1}$, and $H$ divide $AB$ in the ratio $\lambda_{2}$. Prove that $\lambda_{2}$ is determined by $\lambda_{1}$, and when points $A, B$ are given, the position of point $Y$ is uniquely determined by point $G$. In $\triangle ABC$, by applying Ceva's The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,208 |
1. If the combination number $C_{10}^{k}=45$, then the non-negative integer $k=$ | 1. 2 or 8 | 2 \text{ or } 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,209 |
3. Let $A, B, C$ be the three interior angles of $\triangle ABC$, then the imaginary part of the complex number
$$
\frac{(1+\cos 2B+i \sin 2 B)(1+\cos 2 C+i \sin 2 C)}{1+\cos 2 A-i \sin 2 A}
$$
is . $\qquad$ | 3. 0 .
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "3. 0 ." is already in a numerical and punctuation format that is universal and does not require translation, the output remains the same:
3. ... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,211 |
4. If there exists a real number $\theta$ that satisfies $\cos \theta-\sin ^{2} \theta=\alpha$, then the range of real number $\alpha$ is | 4. $-\frac{5}{4} \leqslant \alpha \leqslant 1$.
The above text has been translated into English, retaining the original text's line breaks and format. | -\frac{5}{4} \leqslant \alpha \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,212 |
5. Let $x, y$ be coprime natural numbers, and $xy=$ 1992. Then the number of different ordered pairs $(x, y)$ is $\qquad$ | 5. 8.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,213 |
6. If the maximum value of the two numbers $19 x+1,92 x+74$ is non-negative, then the range of real number $x$ is $\qquad$ | 6. $\left[-\frac{37}{46},+\infty\right)$ | \left[-\frac{37}{46},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,214 |
7. In a convex quadrilateral $ABCD$, the four interior angles satisfy $\angle A<\angle B<\angle C<\angle D$, and $\angle A, \angle B, \angle C, \angle D$ form an arithmetic sequence. Then the range of the common difference $d$ is $\qquad$ | ․ $\left(0, \frac{\pi}{3}\right)$ | \left(0, \frac{\pi}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,215 |
9. If the equation $x^{2}-5 x \log _{2} a+$ $6\left(\log _{2} a\right)^{2}=0$ has two roots and only the smaller root lies in the interval $(1,2)$, then the range of the real number $a$ is $\qquad$ | 9.
$$
(\sqrt[3]{4}, 2)
$$ | (\sqrt[3]{4}, 2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,217 |
10. If $x, y \in R$, and $x^{2}+2 \sqrt{3} x y-y^{2}=$ 3 , then the minimum value of $x^{2}+y^{2}$ is $\qquad$ | 10. $\frac{3}{2}$. | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,218 |
Example 5. On the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>b>0)$. A point $C$ connects the two foci $E_{1}, F_{2}$, which intersect the ellipse at another point $A, B$ respectively. The tangents through points $A, B$ intersect at $F$, and intersect with the tangent through point $C$ at points $D, E$. Prove:
(i) $... | Prove (i) Make the symmetric point $F_{3}$ of $F_{1}$ with respect to $E F$. From the optical property of the ellipse, we get that $F_{2}, B, F_{3}$ are collinear. Connect $F F_{2}$, $F F_{1}$, $F F_{3}$, and denote the area of the triangle with side lengths $a, b, c$ as $S(a, b, c)$. Thus, we have
$$
\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,219 |
11. If the length of the major axis of an ellipse is 4, the left vertex is on the parabola $y^{2}=x-1$, and the left directrix is the $y$-axis, then the maximum possible value of the eccentricity of such an ellipse is $\qquad$ . | 11. $\frac{2}{3}$. | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,220 |
12. If the positive number $a$ makes the inequality $\sqrt{x}+\sqrt{y} \leqslant$ $a \sqrt{x+y}$ hold for all positive numbers $x, y$, then the minimum possible value of $a$ is $\qquad$. | 12. $\sqrt{2}$. | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,221 |
13. In the parallelepiped $A E C D-A_{1} B_{1} C_{1} D_{1}$, it is known that the diagonals $A_{1} C=4, B_{1} D=2$. If a point $P$ in space satisfies $P A_{1}=3, P C=5$, then $P B_{1}^{2}+P D^{2}=$ | 13. 28 . | 28 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,222 |
15. For the right square prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a base edge length of 1. If the dihedral angle $A-B D_{1}-C$ is $\frac{2 \pi}{3}$, then $A A_{1}=$ $\qquad$ | 15. 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,224 |
16. For a plane region $D$, let $N(D)$ denote the number of all integer points (i.e., points on the $X O Y$ plane where both coordinates $x, y$ are integers) within $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10, y=1$ (including the boundaries); $B$ represents ... | 16. 1010 . | 1010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,225 |
17. Let the universal set $I=\{(x, y) \mid x, y \in R\}$, and the set $A=\{(x, y) \mid x \cos \theta+y \sin \theta-2=0, x, y, \theta \in R\}$. Then on the $X O Y$ plane, the points corresponding to the elements of the set $\bar{A}$ form a figure with an area of $\qquad$. | 17. $4 \pi$ | 4 \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,226 |
20. For $n \in N$, let $S(n)$ denote the number of distinct ordered pairs $(x, y)$ of integers such that $x \geqslant 0, y \geqslant 0, 2x + y \leqslant n$. Let $n, m$ be natural numbers with $n > m$. Then express $S(n+m) - S(n-m) =$ $\qquad$ | 20. $m(n+2)$ | m(n+2) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,229 |
Example 6. (1989, National High School Competition) Given in $\triangle ABC$, $AB > AC$, the bisector of an exterior angle at $\angle A$ intersects the circumcircle of $\triangle ABC$ at point $E$. A perpendicular line $EF$ is drawn from $E$ to $AB$, with the foot of the perpendicular being $F$. Prove: $2 AF = AB - AC$... | Prove as shown in Figure 6, take a point $D$ on $FB$ such that $FD = FA$. Connect $ED$ and extend it to intersect the circle at point $G$. Connect $AC$ and $EC$. Then
$$
\begin{array}{l}
\angle ACE = \angle AGD. \\
\angle ADG \\
= \pi - \angle ADE \\
= \pi - \angle EAH \\
= \angle EAC.
\end{array}
$$
Thus, $\triangle ... | 2AF = AB - AC | Geometry | proof | Yes | Yes | cn_contest | false | 706,230 |
2. $x, y$ are real numbers, $z_{1}=x+\sqrt{11}+y i, z_{2}=$ $x-\sqrt{11}+y i$ ( $i$ is the imaginary unit), $\left|z_{1}\right|+\left|z_{2}\right|=$ 12 , let $u=|5 x-6 y-30|$. Then the maximum value of $u$ is $\qquad$ , the minimum value of $u$ is $\qquad$ . | 2. $30(\sqrt{2}+1), 0$ | 30(\sqrt{2}+1), 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,232 |
4. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $P(1)=10, P(2)$ $=20, P(3)=30$, then $P(10)+P(-6)=$ $\qquad$ . | 4.
$$
8104 .
$$ | 8104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,234 |
II. On a plane, there are seven wooden balls $O, O_{1}, O_{2}, \cdots, O_{6}$ with a radius of $R$. Balls $O_{1}, O_{2}, \cdots, O_{6}$ are all tangent to ball $O$. Can three more wooden balls $O_{7}, O_{8}, O_{9}$ with a radius of $R$ be placed on these balls so that $O_{7}, O_{8}, O_{9}$ are all tangent to ball $O$? ... | $O, O_{1}, \cdots, O_{6}$ are on the same plane $\pi$. Using this plane to intersect these spheres results in 7 circles, $\odot O_{1}, \cdots, \odot O_{6}$ all being tangent to $\odot O$. $O_{1} O_{2} \cdots O_{6}$ form a regular hexagon with side length $2R$, and $O$ is their center. Without loss of generality, assume... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,236 |
Three, let $n$ be a given natural number, find all positive pairs $a, b$, such that $x^{2}+a x+b$ is a factor of $a x^{2 x}+(a x+b)^{2 n}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, to solve $x_{2}+a x+b$ is a factor of $a x^{2 n}+(a x+b)^{2 n}$, the necessary and sufficient condition is that the roots of $x^{2}+a x+b=0$ are also roots of $a x^{2 n}+(a x+b)^{2 n}=0$.
Let $x=x_{0}$ be a root of $x^{2}+a x+b=0$, then we have $a x_{0}+b=-x_{0}^{2}$, and
$$
\begin{aligned}
& a x_{0}^{2 n}+\lef... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,237 |
2. Every positive integer can be expressed as the sum of one or more consecutive positive integers. Try to find, for each positive integer $n$, how many different ways $n$ can be represented as such a sum. | Let $n$ be expressible as the sum of $m$ consecutive positive integers. Let
$$
n=k+(k+1)+\cdots+[k+(m-1)].
$$
Then
$$
\begin{aligned}
n & =m k+\frac{m(m-1)}{2} \\
& =m\left(\frac{2 k+(m-1)}{2}\right).
\end{aligned}
$$
1) If $m$ is odd, then $m-1$ is even, and from (2) we know $m \mid n$, and
$$
\frac{m(m-1)}{2}<n.
$$
... | f(n)=f_{1}(n)+f_{2}(n) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,240 |
Example 7. (1991, Issue 11 of "Mathematics Bulletin" problem for solution) Given the inscribed acute $\triangle A B C$ in $\odot O$, the distances from point $O$ to the three sides $a, b, c$ of $\triangle A B C$ are $H_{a}, H_{b}$, $H_{c}$. Try to prove: the radius $R$ of $\odot O$ is a root of the equation $x^{3}-\lef... | As shown in Figure 7, the extensions of $A O, B O, C O$ intersect $O$ at $M, N, P$ respectively, and connect $A P, P B, B M, M C$, $N C, N A$.
Since $O$ is inside $\triangle A B C$, then
$$
\begin{array}{l}
B M=2 H_{c}, M C=2 H_{b}, \\
N C=2 H_{a}, N A=2 H_{c}, \\
P A=2 H_{b}, P B=2 H_{a} .
\end{array}
$$
Applying Pto... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,241 |
4. Let $r$ be a positive integer. Define the sequence $\left\{a_{n}\right\}$ as follows: $a_{1}=1$, and for each positive integer $n$,
$$
a_{n+1}=\frac{n a_{n}+2(n+1)^{2 r}}{n+2} \text {. }
$$
Prove that each $a_{n}$ is a positive integer, and determine for which $n, a_{n}$ is even. | Given the conditions, we have
$$
(n+2) a_{n+1}=n a_{n}+2(n+1)^{2 r} \text {. }
$$
Multiplying both sides by \( n+1 \), we get
$$
\begin{array}{l}
(n+2)(n+1) a_{n+1} \\
\quad=(n+1) n a_{n}+2(n+1)^{2 r+1} .
\end{array}
$$
Let \( b_{n}=(n+1) n a_{n} \) (for \( n=1,2, \cdots \)), then we have
$$
b_{n+1}=b_{n}+2(n+1)^{2 r... | a_{n} \text{ is even if and only if } n \equiv 0 \text{ or } 3(\bmod 4) | Algebra | proof | Yes | Yes | cn_contest | false | 706,243 |
5. Let $\bar{i}$ be the incenter of $\triangle A B C$, and draw a perpendicular from $I$ to $A J$, intersecting sides $A B$ and $A C$ at $P$ and $Q$ respectively. Prove that the circle $L$ which is tangent to $A B$ and $A C$ at $P$ and $Q$ respectively must also be tangent to the circumcircle $O$ of $\triangle A B C$. | Prove that extending $A I$ to intersect $\odot O$ at $M$, with $\odot O$ having a radius of $R$ (as shown in the figure), the power of $L$ with respect to $\odot O$ is
$$
\begin{aligned}
& R^{2}-L O^{2} \\
= & L A \cdot L M .
\end{aligned}
$$
Therefore, $L O^{2}=R^{2}$
$$
\begin{array}{l}
-L A \cdot L M \\
\quad=R^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,244 |
1. Try to use a function of $n$ to represent the product
$$
9 \times 99 \times 9999 \times \cdots \times\left(10^{2^{n}}-1\right)
$$
the sum of the digits in decimal representation. | First, we prove the following lemma: Let the number of digits of the natural number $m$ be no more than $d$, and $M=\left(10^{*}-1\right) m(b \geqslant \geqslant i)$. Then $S(M)-S$ S.
Here $S(M)$ represents the sum of the digits of $M$.
Let $M=\left(10^{k}-1\right) m=p+q+r$, where
$$
\begin{array}{l}
p=10^{k}(m-1), q=... | 9 \cdot 2^{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,245 |
$$
\begin{array}{l}
\frac{1}{\cos 0^{\circ} \cos 1^{\circ}}+\frac{1}{\cos 1^{\circ} \cos 2^{\circ}}+\cdots+ \\
\frac{1}{\cos 88^{\circ} \cos 89^{\circ}}=\frac{\cos 1^{\circ}}{\sin ^{2} 1^{\circ}}
\end{array}
$$
Prove:
$$
\begin{array}{l}
\frac{1}{\cos 0^{\circ} \cos 1^{\circ}}+\frac{1}{\cos 1^{\circ} \cos 2^{\circ}}+\... | To prove for the sake of writing convenience, we agree that all angles in the following triangles are in degrees, i.e., $\sin 1$ refers to $\sin 1^{\circ}$, and $\sin (x+1)$ refers to $\sin (x+1)^{\circ}$. Thus, the right side of the original equation is:
$$
\frac{\cos 1}{\sin ^{2} 1}=\frac{\operatorname{ctg} 1}{\sin 1... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,246 |
3. Let $\sigma(S)$ denote the sum of all elements in a non-empty set of integers $S$. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a set of positive integers, and $a_{1}<a_{2}<\cdots<a_{n}$. For each positive integer $n \leqslant$ 1500, there exists a subset $S$ of $A$ such that $\sigma(S)=n$. Find the minimum... | Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k} \quad(1 \leqslant k \leqslant 11)$. By the given conditions, there exists an index $m$ such that $S_{m-1}S_{k-1}+1$, and $a_{1}+\cdots+a_{k-1}=S_{k-1}$. Thus, there does not exist $S \simeq A$ such that $\sigma(S)=S_{k-1} +1$. Therefore,
$$
S_{k}=S_{k-1}+a_{k} \leqslant 2 S_{k-1}+1 .... | 248 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,247 |
4. Let $A A^{\prime}, B B^{\prime}, C C^{\prime}$ be three chords of a sphere that do not lie in the same plane, intersecting at a point $P$ inside the sphere. If the sphere passing through $A, B, C, P$ is tangent to the sphere passing through $A^{\prime}, B^{\prime}, C^{\prime}, P$, prove that $A A^{\prime}=B B^{\prim... | Proof: From the given, the plane passing through points $A, A^{\prime}, B$, and $B^{\prime}$ intersects the three spheres mentioned in the problem into three circles. Among these, two circles are the circumcircles of $\triangle A B P$ and $\triangle A^{\{\prime\}} B^{\{\prime\}} P$, respectively, and these two circles ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,248 |
5. Let $P(z)$ be a complex polynomial of degree 1992 with distinct roots. Prove: there exist numbers $a_{1}, a_{2}, \cdots, a_{1991}, a_{1992}$, such that $P(z)$ divides the polynomial
$$
\left\{\cdots\left[\left(z-a_{1}\right)^{2}-a_{2}\right]^{2}-\cdots-a_{1991}\right\}^{2}-a_{1992} .
$$ | Let $n=1992, r_{1}, r_{2}, \cdots, r_{n}$ be $n$ distinct roots of $P$ ( $\approx$ ). Define non-empty sets $S_{i}$ and complex numbers $a_{i}$ inductively as follows: Let
$$
S_{0}=\left\{r_{1}, r_{2}, \cdots, r_{n}\right\} .
$$
For $i=1$, define $a_{i}$ as the average of any two numbers in $S_{i-1}$ (if $S_{i-1}$ con... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 706,249 |
2. A river has a constant flow rate of 3 km per hour. A boat travels 4 km downstream and then returns to the starting point at a constant speed, taking a total of 1 hour (ignoring the time to turn the boat around). What is the ratio of the boat's downstream speed to its upstream speed?
(A) $5: 2$
(B) 2:1
(C) $5: 3$
(D)... | 2. Let the speed be $x$, then
$$
\frac{4}{x+3}+\frac{4}{x-3}=1 \text{.}
$$
Solving it, we get $x_{1}=9, x_{2}=-1$ (not suitable, discard) At this point $\frac{x+3}{x-3}=\frac{12}{6}=2$, that is, the ratio of the boat's speed downstream to upstream is 2:1.
Choose $\mathrm{B}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,251 |
Example 9. Given: As shown in Figure 8, point $S$ on $\odot O$, draw three chords $SA, SB, SC$ through $S$, and the circles with these chords as diameters intersect at $P, Q, R$.
Prove: Points $P, Q, R$ are collinear. | Proof $\because S A, S B$ are both diameters,
$$
\therefore \angle S P A, \angle S P B
$$
are right angles.
That is, $S P \perp A B$ at $P$.
Similarly, $S Q \perp A C$ at $Q$,
$$
S R \perp B C \text { at } R \text {. }
$$
Thus, $P, Q, R$ are the projections of a point $S$ on the circumcircle $\odot O$ of $\triangle A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,252 |
3. If $n$ is an integer greater than 1, then the value of $p=n+$ $\left(n^{2}-1\right)^{\frac{1-(-1)^{2}}{2}}$ is ( ).
(A) always even
(B) always odd
(C) even, but not 2
(D) can be either even or odd | 3. Solve for $n$ being even, $\frac{i-(-1)^{n}}{2}=0$. $\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}=1$, then $p=n+1$ is odd;
$n$ being odd, $\frac{1-(-1)^{n}}{2}=1$, $\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}=n^{2}-1$, then $p=n+n^{2}-1$ is odd. Choose B. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,253 |
4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is ().
(A) 4
(B) $\pi$
(C) 2
(D) 1 | 4. Solution From the original equation, we get
$$
0 \leqslant|x-1| \leqslant 1,0 \leqslant|y-1| \leqslant 1 \text {, }
$$
which means $0 \leqslant x \leqslant 2,0 \leqslant y \leqslant 2$.
When $\left\{\begin{array}{l}0 \leqslant x \leqslant 1 \\ 0 \leqslant y \leqslant 1\end{array}\right.$, we have $x+y=1$.
When $\le... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,254 |
5. Let quadrilateral $ABCD$ be a cyclic quadrilateral, and now four relations are given.
(1) $\sin A=\sin C$;
(2) $\sin A+\sin C=0$;
(3) $\cos B+\cos D=0$;
(4) $\cos B=\cos D$.
The number of relations that always hold is ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5. Solution Since $A B C D$ is inscribed in a circle, then $\angle A+\angle C=180^{\circ}, \angle B+\angle D=180^{\circ}$.
Thus, $\sin A=\sin C, \cos B=-\cos D$.
That is, equations (1) and (3) hold.
Wash B. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,255 |
6. In $\triangle A B C$, it is known that $a^{4}+b^{4}+c^{4}=2 c^{2} \left(a^{2}+b^{2}\right)$. Then the degree of $\angle C$ is ( ).
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $120^{\circ}$
(D) $45^{\circ}$ or $135^{\circ}$ | 6. Given $a^{4}-b^{4}+c^{6}-2 a^{2} c^{2}-2 b^{2} c^{2}=0$, we have
$$
\begin{array}{r}
\left(u^{2}+\sqrt{2} a b+b^{2}-c^{2}\right)\left(a^{2}-\sqrt{2} a b+b^{2}-c^{2}\right)=0 \\
\quad c^{2}=a^{2}+b^{2}+\sqrt{2} a b, \quad c^{2}=a^{2}+b^{2}-\sqrt{2} a b
\end{array}
$$
Thus, $\cos C= \pm \frac{\sqrt{2}}{2}, \angle C=4... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,256 |
7. Given that $\triangle A B C$ is a right-angled triangle with integer sides, $B C$ is the hypotenuse, and $2 A C=B C+A B$. Which of the following numbers could be a side length of this triangle ( ).
(A) 31
(B) 41
(C) 51
(D) 61 | 7. Solution From the given, we have
$$
\left\{\begin{array}{l}
B C^{2}=A C^{2}+A B^{2}, \\
2 A C=B C+A B .
\end{array}\right.
$$
By solving the equations, we get
$$
B C: A C: A B=5: 4: 3 \text{. }
$$
Therefore, the sides of $\triangle A B C$ are multiples of 5, 4, and 3. Among the four options, only 51 is a multiple ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,257 |
8. If $\odot O$ is inscribed in $\triangle A B C$ and touches the three sides at points $X, Y, Z$. Then, $\triangle X Y Z$ satisfies ( ).
(A) Each angle is equal to $60^{\circ}$
(B) One is an obtuse angle, the other two angles are acute
(C) It is similar to $\triangle A B C$
(D) Each angle is equal to half the sum of t... | 8. It can be proven that points $O, Y, X, B$ are concyclic, thus
$$
\angle O X Y=\angle O B Y=\frac{1}{2} \angle A B C \text {. }
$$
Similarly, $\angle O X Z=\angle O C Z=\frac{1}{2} \angle A C B$.
Therefore, $\angle Y X Z=\frac{1}{2}(\angle A B C+\angle A C B)$.
Similarly, $\angle X Z Y=\frac{1}{2}(\angle C A B+\angl... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,258 |
2. Given that the vertex of the parabola $y=-x^{2}+2 a x+b$ moves on the line $m x-y-2 m+1=0$, and it has a common point with the parabola $y=x^{2}$. Then, the range of values for $m$ is $\qquad$ | 2. The vertex of the parabola $y=-x^{2}+2 a x+b$ is $(a$, $\left.a^{2}+b\right)$, then $\left(a, a^{2}+b\right)$ satisfies the linear equation, which gives
$$
m a-a^{2}-b-2 m+1=0,
$$
i.e., $b=m a-a^{2}-2 m+1$.
If $y=-x^{2}+2 a x+m a-a^{2}-2 m+1$ and $y=x^{2}$ have intersection points, then the discriminant of the equa... | m \leqslant 2-\sqrt{2} \text{ or } m \geqslant 2+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,260 |
4. The chord $AB=18$ of the sector $OAB$, a circle $C$ with radius 6 is exactly tangent to $OA$, $OB$, and the arc $\widehat{AB}$. Another circle $D$ is tangent to circle $C$, $OA$, and $OB$ (as shown in the figure). Then the radius of circle $D$ is $\qquad$. | 4. Solution As shown in the figure, let the radii of $\odot O, \odot D$ be $x, y$ respectively. Then $\alpha C=x-6$.
Since $\triangle O L B \backsim \triangle O M C$.
Then
$$
\frac{O B}{O C}=\frac{L B}{C M},
$$
which means $\frac{x}{x-6}=\frac{9}{6}$.
Solving for $x$ gives $x=18, Q C=12$, $O D=$ $6-y$.
Also, since $\... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,262 |
One, (20 points) Let $x, y$ be real numbers, and $x^{2}+x y+y^{2}=1$. Find the range of values for $x^{2}-x y+y^{2}$. | Given $x^{2}+x y+y^{2}=1$.
Let $x^{2}-x y+y^{2}=\hbar$.
(1) - (2), rearranging gives $x y=\frac{1-k}{2}$.
From (1), we have $(x+y)^{2}=x y+1$.
Substituting (3) into (1), we get
$$
(x+y)^{2}=\frac{3-k}{2} .
$$
Since $x, y$ are real numbers, $(x+y)^{2} \geqslant 0$.
From (5), we get $k \leqslant 3$.
Under the condition... | -\frac{1}{3} \leqslant x^{2}-x y+y^{2} \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,264 |
II. (20 points) From a point $C$ on a circle, draw a perpendicular to the diameter $A B$, and let the foot of the perpendicular be $D$. Construct $\odot O^{\prime}$ tangent to $\widehat{B C}$, $C D$, and $D B$ at $E$, $F$, and $G$ respectively. Then $A C = A G$. | II. Proof As shown in the figure, let the center of the semicircle be $O$, then $O, O^{\prime}$, $E$ are on the same line.
Given $O^{\prime} F \perp C D$.
we have $O^{\prime} F / / A B$.
Connecting $E F, A E$, we get
$$
\begin{array}{l}
\angle F E O^{\prime} \\
=\frac{1}{2} \angle F O^{\prime} O \\
=\frac{1}{2} \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,265 |
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