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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
2. Using geometric figures, the value of $\operatorname{tg} 75^{\circ}$ can be found as
保留源文本的换行和格式，直接输出翻译结果如下：
2. Using geometric figures, the value of $\operatorname{tg} 75^{\circ}$ can be found as | $2.2+\sqrt{3}$.
In the right triangle $\triangle ABC$, $\angle C = 90^{\circ}, \angle B = 75^{\circ}, \angle A = 15^{\circ}, BC = 1$. On the side $AC$: take a point $D$ such that $\angle ABD = 15^{\circ}$, then $\angle BDC = 30^{\circ}, \angle DBC = 60^{\circ}, AD = DB = 2 \cdot BC = 2, CD = BC \cdot \tan 60^{\circ} = ... | 2 + \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,276 |
3. If three integers $a, b, c (a \neq 0)$ make the equation $a x^{2}$ $+b x+c=0$ have two roots $a$ and $b$, then $a+b+c$ equals. $\qquad$ | 3. 18 .
From the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
a+b=-\frac{b}{a}, \\
a b=\frac{c}{a} .
\end{array}\right.
$$
From (1), we have $a^{2}+a b+b=0, (a+1) b=-a^{2}$.
Since $a \neq 0$, then $a+1 \neq 0$, $b \neq 0$. Thus,
$$
b=-\frac{a^{2}}{a+1}=-\frac{(a+1-1)^{2}}{a+1}=-a+1-\... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,277 |
4. Let the real number $x$ be not equal to 0 and 1. Then the range of $y=$ $\frac{x^{4}-x^{2}+1}{x^{3}-x}$ is $\qquad$ . | 4. $|y| \geqslant 2$.
Then $y=\frac{\left(x^{2}-1\right)^{2}+x^{2}}{x\left(x^{2}-1\right)}=\frac{x^{2}-1}{x}+\frac{x}{x^{2}-1}$, let $t_{1}=$ $\frac{x^{2}-1}{x}$, then $x^{2}-t_{1} x-1=0$. Its discriminant $\Delta_{1}=t_{1}^{2}+4>0$, which means $t_{1}$ can take any real number.
Let $t_{1}+\frac{1}{t_{1}}=t_{2}\left(... | |y| \geqslant 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,278 |
One, (Full marks 20 points) Let $a, b, c$ be the lengths of the sides of a triangle. Prove that:
$$
\frac{b-c}{b^{2} c}+\frac{c-a}{c^{2} a}+\frac{a-b}{a^{2} b} \leqslant 0,
$$
with equality holding if and only if $a=b=c$. | Obviously, the original inequality is equivalent to the following inequality:
$$
a^{2} c(b-c)+b^{2} a(c-a)+c^{2} b(a-b) \leqslant 0 .
$$
As shown in the figure, construct the incircle $I$ of $\triangle ABC$, and let $\odot I$ touch the sides at points $D, E, F$, respectively. Denote
$$
\begin{array}{l}
A E=A F=x, B F=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,279 |
II. (Full marks 25 points) Given as shown, $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $A^{\prime}$, the tangents to the two circles at $A$ intersect the tangents at points $E$ and $F$ at point $P$. Prove that $A$, $A^{\prime}$, and $P$ are collinear. | II. Draw $A A^{\prime}, A^{\prime} P$ and other lines.
Since the common chord of two intersecting circles is bisected perpendicularly by their line of centers, we have
$O)_{1} \perp A E, O O_{1} \perp$
$A F$. Also, $A O \perp$
$A E, A O_{1} \perp A F$, so
$O O_{1} / / A O_{2}$,
$O O_{2} / / A O_{1}$, which means
$A O_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,280 |
Three, (Total 25 points) Given a real number $t$ and a real number $k$ greater than $t$, let $a, b$ be positive numbers, and $\frac{1}{a+t}-\frac{1}{b+t}=\frac{1}{k}$. Determine the range of $b$ - $a$. | Three, from the given conditions we get
$b+t=\frac{1}{\frac{1}{a+t}-\frac{1}{k}}=\frac{k(a+t)}{k-a-t}$.
Therefore, from $a0$ we get
$k-t>x>0$.
From (2) we get $b-a=\frac{(k-x)^{2}}{x}$.
Let $\frac{(k-x)^{2}}{x}=y$, clearly $y>0$, then
$x^{2}-2 k x+k^{2}=y x$,
i.e., $x^{2}-(2 k+y) x+k^{2}=0$.
By the quadratic formula, ... | \frac{t^{2}}{k-t} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,281 |
1. The number of real roots of the equation $1-\lg \sin x=\cos x$ is ( ).
(A) 0 .
(B) 1
(C) 2
(D) greater than 2 | -1 (A).
Since for all $x$, $\sin x \leqslant 1$, it follows that $\lg \sin x \leqslant 0$ and $1-\lg \sin x \geqslant 1$. However, for any $x$, $\cos x \leqslant 1$, so for all $x$, $1-\lg \sin x \geqslant \cos x$, with equality holding if and only if $\sin x=\cos x=1$, which is impossible. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,282 |
2. The tangent line of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ intersects the $x$-axis at $A$ and the $y$-axis at $B$, then the minimum value of $|A B|$ is ( ).
(A) $2 \sqrt{a^{2}+b^{2}}$
(B) $a+b$
(C) $\sqrt{2 a b}$
(D) $4 \sqrt{a b}$ | 2. (B).
A circle is a special case of an ellipse. Let's first examine the case when $a=b=1$, i.e., the unit circle. Suppose $AB$ is tangent to $\odot O$ at $C$, and let $\angle AOC = \alpha (\alpha \in (0, \frac{\pi}{2}))$. Then, $|AC| = \tan \alpha$ and $|BC| = \cot \alpha$, so $|AB| = |AC| + |BC| = \tan \alpha + \co... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,283 |
3. In $\triangle A B C$, $\lg \operatorname{tg} A+\lg \operatorname{tg} C=$ $2 \lg \operatorname{tg} B$. Then the range of $\angle B$ is () .
(A) $0<\angle B \leqslant \frac{\pi}{3}$
(B) $\frac{\pi}{3} \leqslant \angle B<\frac{\pi}{2}$
(C) $0<\angle B \leqslant \frac{\pi}{6}$
(D) $\frac{\pi}{6} \leqslant \angle B<\frac... | 3. (B).
From the given, we have $\operatorname{tg}^{2} B=\operatorname{tg} A \operatorname{tg} C \leqslant\left(\frac{\operatorname{tg} A+\operatorname{tg} C}{2}\right)^{2}$. Therefore, $2 \operatorname{tg} B \leqslant \operatorname{tg} A+\operatorname{tg} C=\operatorname{tg}(A+C)(1-\operatorname{tg} A \operatorname{t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,285 |
4. Let $X=\{-1,0,1\}, Y=\{-2,-1,0,1,2\}$, and for all elements $x$ in $X$, $x+f(x)$ is even. Then the number of mappings $f$ from $X$ to $Y$ is ( ).
(A) 7
(B) 10
(C) 12
(D) 15. | 4. (C).
$x$ and $f(x)$ are both even, or both odd.
When $x=0$, there are three possibilities: $-2, 0, 2$;
When $x=-1, 1$, there are two possibilities each for -1 and 1, making 4 possibilities.
Therefore, the total is $3 \times 4=12$. | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 709,286 |
5. The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|=$ $\left|z_{3}\right|=\left|z_{4}\right|=1$, and $z_{1}+z_{2}+z_{3}+z_{4}=0$. Then the quadrilateral formed by the points corresponding to these four complex numbers must be ( ).
(A) Trapezoid
(B) Square
(C) Parallelogram... | 5. (D).
Since $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|z_{4}\right|=1$, these points all lie on the unit circle. From $z_{1}+z_{2}+z_{3}+z_{n}=0$, we get
$$
\frac{z_{1}+z_{2}}{2}=-\frac{z_{3}+z_{4}}{2} \text {. }
$$
Let the four points be $A_{1}, A_{2}, A_{3}, A_{1}$, then the midpoint $M$ of c... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,287 |
6. A monkey is on a ladder with $n$ steps: climbing up and down, it either ascends 16 steps or descends 9 steps each time. If it can climb from the ground to the very top step, and then return to the ground, the minimum value of $n$ is ( ).
(A) 22
(B) 23
(C) 24
(D) greater than 24. | 6. (C).
Assume the monkey climbs as follows: 0 $\nearrow 16 \searrow 7 \nearrow 23 \searrow 14 \searrow 5$ $\nearrow 21 \searrow 12 \searrow 3 \nearrow 19 \searrow 10 \searrow 1, \nearrow 17 \searrow 8 \nearrow 24 \searrow 15 \searrow 6 \nearrow$ $22 \searrow 13 \searrow 4$ Л $20 \searrow 11 \searrow 2$ オ゙ $18 \searro... | 24 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,288 |
$\begin{array}{l}\text { 1. } \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+ \\ \frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+ \\ \left.\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=\end{array}$ | $$
\approx .1 .1 \text {. }
$$
Since
$$
\begin{array}{l}
\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+ \\
\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
=\frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)} . \\
=\sqrt[3]{n}-1,
\end{array}
$$
Therefore, the required limit is $\l... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 709,289 |
2. Let $m, n \in N$, and $m>n$, set $A=\{1$, $2,3, \cdots, m\}, B=\{1,2,3, \cdots, n\}$, and $C \subset A$. Then the number of $C$ that satisfies $B \cap C \neq \varnothing$ is $\qquad$ . | 2. $2^{m-n}\left(2^{n}-1\right)$.
Since the subsets of $A$ that consist only of numbers chosen from $n+1, n+2, \cdots, m$ are the only ones that can satisfy $B \cap C = \varnothing$, and there are $2^{m-n}$ such subsets, the number of $C$ that satisfy $B \cap C \neq \varnothing$ is $2^{m}-2^{m-n}=2^{m-n}\left(2^{n}-1\... | 2^{m-n}\left(2^{n}-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,290 |
3. As shown in the figure,
$A B C D$ is a square,
$E$ is the midpoint of $A B$. If
$\triangle D A E$ and
$\triangle C B E$ are folded along the dotted lines $D E$ and $C E$ respectively, so that $A E$ and $B E$ coincide, and the point where $A$ and $B$ coincide is denoted as $P$, then the dihedral angle between plane $... | 3. $30^{\circ}$.
As shown in the figure, draw $P F \perp C D$, and connect $E F$. Since $F$ is the midpoint of $C D$, we know that $E F \perp C D$. Therefore, $\angle P F E$ is the plane angle of the dihedral angle formed by the planes $P C D$ and $E C D$.
Let the side length of the square $A B C D$ be $a$. Then
$$
\b... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,291 |
4. Let $M$ be the set of all integers $x$ that satisfy $|x-a|<a+\frac{1}{2}$, and $N$ be the set of all integers $x$ that satisfy $|x|<2 a(a \in$ $N$). Then the sum of the integers that belong to $M \cap N$ is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | $$
\begin{array}{c}
\text { Given }\left\{\begin{array}{l}
|x-a|<a+\frac{1}{2}, \\
|x|<2 a
\end{array}\right. \\
\text { we have }\left\{\begin{array}{l}
-a-\frac{1}{2}<x-a<a+\frac{1}{2}, \\
-2 a<x<2 a .
\end{array}\right.
\end{array}
$$
Solving, we get $-\frac{1}{2}<x<2 a$.
Then $M \cap N=\{0,1,2, \cdots, 2 a-1\}$.
T... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 709,292 |
5. On each face of an opaque cube, a natural number is written. If several (one, two, or three) faces of the cube can be seen at the same time, then find the sum of the numbers on these faces. Using this method, the maximum number of different sums that can be obtained is _.
翻译结果如下：
5. On each face of an opaque cube,... | 5.26 .
There are 6 cases where a single face can be seen, 12 cases where two faces sharing a common edge can be seen simultaneously, and 8 cases where three faces sharing a common vertex can be seen simultaneously, thus yielding 26 sums. The numbers on the faces can be filled in such a way that all 26 sums are distinc... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,293 |
6. Let the sequence of positive integers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ be a geometric sequence, with the common ratio $r$ not being an integer and $r>1$. The smallest value that $a_{4}$ can take in such a sequence is $\qquad$ . | 6. 27 .
According to the problem, $r$ is a rational number, so there exist coprime positive integers $p$ and $q (q > p \geqslant 2)$, such that $r=\frac{q}{p}$. Therefore, $a_{4}=a_{1} r^{3}=\frac{a_{1} q^{3}}{p^{3}}$.
Since $a_{4}$ is an integer, $a_{1}$ must be a multiple of $p^{3}$.
Thus, let $a_{1}=k p^{3}$ ($k$ i... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,294 |
8. Given the equation $\left(a x+a^{2}-1\right)^{2}+\frac{a^{2}}{(x+a)^{2}}+2 a^{2}-1$ $=0$ has real solutions. Find the range of real values for $a$.
untranslated text remains the same as the source text. | (Solution: $-\frac{1}{2} \leqslant a \leqslant \frac{1}{2}, a \neq 0$ ) | -\frac{1}{2} \leqslant a \leqslant \frac{1}{2}, a \neq 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,295 |
Three, the base of the triangular pyramid $S-ABC$ is a regular $\triangle ABC$, with the side length of this triangle being 4. It is also known that $AS=BS=\sqrt{19}$, and $CS=3$. Find the surface area of the circumscribed sphere of this triangular pyramid. | Three, let $C K \perp A B, L$ be the intersection of its extension with the sphere, then $S K$ is perpendicular to $A B$, and the center of the sphere lies in the plane of $\triangle L S C$. The problem is reduced to finding the radius $R$ of its circumcircle. Since $C K=2 \sqrt{3}, L C=\frac{8}{\sqrt{3}}$ (the diamete... | \frac{268}{11} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,296 |
Four, the function $f_{n}(x)(n=1,2,3, \cdots)$ is defined as follows:
$$
\begin{array}{l}
f_{1}(x)=4\left(x-x^{2}\right)(0 \leqslant x \leqslant 1), \\
f_{n+1}(x)=f_{n}\left(f_{1}(x)\right)(n=1,2, \cdots) .
\end{array}
$$
Let the number of $x$ in $[0,1]$ that makes $f_{n}(x)$ reach its maximum value be $a_{n}$, and th... | When $y=f_{1}(x)$ takes its maximum value 1 at $x=\frac{1}{2}$, and its minimum value 0 at $x=0$, 1, then $a_{1}=1, b_{1}=2$. And when $x=0,1$, for any natural number $n$, they are the minimum value points of $f_{n}(x)$. Thus, $f_{n}(x)$ has $a_{n}$ maximum value points in $0<x \leqslant 1$. Assuming that $f_{n}(x)$ ha... | a_{n}=2^{n-1}, b_{n}=2^{n-1}+1 | Algebra | proof | Yes | Yes | cn_contest | false | 709,297 |
Five, let $S=\left\{\left.\frac{m n}{m^{2}+n^{2}} \right\rvert\, m, \dot{n} \in N\right\}$. Prove that if $x, y \in S$, and $x<y$, then there must exist $z \in$ $S$, such that $x<z<y$. | Let $x, y \in S, x=\frac{m n}{m^{2}+n^{2}}, y=\frac{a b}{a^{2}+b^{2}}, x<y$. Without loss of generality, assume $m \leqslant n, a \leqslant b$.
Consider the function $f(x)=\frac{x}{1+x^{2}}$, it is easy to prove that $f(x)$ is strictly increasing on $[0,1]$. Therefore, for all $c, d \in[0,1]$, we have $f(c)<f(d) \Left... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,298 |
Let $ABCD$ be a cyclic quadrilateral, the angle bisectors of $\angle A$ and $\angle B$ intersect at $E$, and a line through $E$ parallel to $CD$ intersects $AD$ at $L$ and $BC$ at $M$. Prove that $|AL| + |BM| = |LM|$. | Let $\angle L A B=2 \alpha, \angle A B M=2 \beta$. From the cyclic quadrilateral and $M L \parallel C D$, we easily get
$$
\begin{array}{l}
\angle A L M=180^{\circ}-2 \beta, \\
\angle L M B=180^{\circ}-2 \alpha .
\end{array}
$$
Construct $\left|M B^{\prime}\right|=|M B|$ on $M L$ and connect $A B^{\prime}, B B^{\prime... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,299 |
Given two parabolas $L_{1}$ and $L_{2}$ with parallel axes of symmetry, they intersect at two points $A_{0}$ and $B_{0}$. On $L_{1}$, take $2 n$ points $A_{1}, A_{2}, \cdots, A_{2 n}$, and on $L_{2}$, take $2 n$ points $B_{1}, B_{2}, \cdots, B_{2 n}$ such that $A_{0} A_{1} / / B_{0} B_{1}, A_{1} A_{2} / / B_{1} B_{2}, ... | $$
\begin{array}{l}
L_{1}: y=p_{1} x^{2}+q_{1} x+r_{1}, \\
L_{2}: y=p_{2} x^{2}+q_{2} x+r_{2},
\end{array}
$$
where $p_{1}, p_{2} \neq 0$. Suppose the x-coordinates of $A_{i}$ and $B_{i}$ are $a_{i}$ and $b_{i} (i=0,1,2, \cdots, 2 n)$, respectively, then the slope of the line $A_{i} A_{j}$ is
$$
\begin{array}{l}
\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,300 |
Three, Proof: For any $n \in N, n \geqslant 2$, there exists a set $M$ consisting of $n$ distinct natural numbers such that for any $a \in M$ and $b \in M, |a-b|$ divides $a+b$. | (1) When $n=2$, take $M_{2}=\{1,2\}$, then $(2-1) \mid(2+$ $1)$, the proposition holds.
When $n=3$, take $M_{3}=\{4,5,6\}$, then $(5-4) \mid(5+$ 4), (6-4) $|(6+4),(6-5)|(6+5)$, the proposition also holds.
(2) Suppose when $n=k$, the proposition holds, i.e., there exists $M_{k}=\left\{a_{1}\right.$, $\left.a_{2}, \cdot... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,301 |
Does there exist a positive integer $a$, such that the sum of the digits of $a$ is $1997$, and the sum of the digits of $a^{2}$ is $1997^{2}$? | Proof: We prove the generalized proposition: For any $k \in \mathbb{N}$, there exists a positive integer $a$ such that the sum of the digits of $a$ is $k$, and the sum of the digits of $a^2$ is $k^2$.
For any $k \in \mathbb{N}$, take
$$
a=10^{2^{1}}+10^{2^{2}}+10^{2^{3}}+\cdots+10^{2^{k}} \text{. }
$$
The sum of the d... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,302 |
Initial 54. Given point $P$ inside $\triangle ABC$, satisfying
$$
\begin{array}{l}
\angle B P C - \angle B A C = \angle C P A - \angle C B A \\
= \angle A P B - \angle A C B .
\end{array}
$$
Prove: $P A \cdot B C = P B \cdot C A = P C \cdot A B$. | $$
\begin{array}{r}
\text { Proof: Draw } P A^{\prime} \perp B C, P B^{\prime} \\
\perp C A, P C^{\prime} \perp A B, A^{\prime}, B^{\prime}, C^{\prime} \text { are }
\end{array}
$$
the feet of the perpendiculars, forming $\triangle A^{\prime} B^{\prime} C^{\prime}$.
$$
\text { Let } \begin{aligned}
x & =\angle B P C-\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,303 |
Given 53. Let $n$ be a positive integer that is not an integer power of 2. Prove that there exists a permutation $a_{1}, a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$ such that $\sum_{k=1}^{n} a_{k} \cos \frac{2 k \pi}{n}=0$. | Proof: First, when $n$ is an odd number greater than 1, the conclusion holds. Construct the permutation:
$1,2, \cdots, \frac{n-1}{2}, \frac{n+3}{2}, \cdots, n-1, n, \frac{n+1}{2}$.
Let $S=\sum_{k=1}^{n} a_{k} \cos \frac{2 k \pi}{n}$.
Then $S=\sum_{k=1}^{n-1} a_{n-k} \cos \frac{2(n-k) \pi}{n}+a_{n} \cos \frac{2 n \pi}{n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,304 |
54. As shown in the figure, $AB$ is a non-diameter chord of $\odot O$, $C$ is its midpoint, the line $OC$ intersects $\odot O$ at points $M$ and $N$, $P$ is any point on the minor arc $\overparen{AB}$ except $A, B, M$, the lines $AP$ and $BP$ intersect the line $MN$ at points $E$ and $F$ respectively. Prove:
$$
\sqrt{C... | Prove: Connect $O A, O B$, let $\angle P A B=\alpha, \angle P B A=\beta$. Then $\angle A O C=\angle B O C=\alpha+\beta$. Let $O C=d$, then $C N=2 d+$ $C M$. From $\triangle A E C, \triangle B F C$ and $\triangle B O C$ being right triangles, we have $\operatorname{tg} \alpha=\frac{C E}{A C}=\frac{C E}{B C}, \operatorna... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,305 |
9. Simplify $\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}$ $+\frac{(x+a)(x+b)}{(c-a)(c-b)}$. | (Answer: 1).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,306 |
Example 1 As shown in the figure, the incircle $O$ of rhombus $ABCD$ touches each side at $E$, $F$, $G$, and $H$. On $\overparen{EF}$ and $\overparen{GH}$, draw the tangents to $\odot O$ intersecting $AB$ at $M$, $BC$ at $N$, $CD$ at $P$, and $DA$ at $Q$.
Prove that $MQ \parallel NP$.
(1995, High School Mathematics Lea... | $$
\begin{array}{l}
\text{Proof, let } \angle D O H=\theta \text{ (constant), } \angle D O Q=\theta_{1}, \angle D O M=\pi-\theta_{2}, \text{ and } \angle P O Q=\theta, \text{ then } \angle D O P=\theta-\theta_{1}. \text{ Similarly, } \angle B O N=\theta-\theta_{2}. \text{ Then} \\
z_{Q}=\frac{1}{\cos \left(\theta-\thet... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,308 |
Example 2 Let $O$ be the circumcenter of $\triangle ABC$, $D$ the midpoint of side $AB$, and $E$ the centroid of $\triangle ACD$. Prove that if $AB = AC$, then $OE \perp CD$.
(1983, British Mathematical Olympiad) | Proof, as shown in the figure, let the radius of the circle be $1$, and $\angle B O X=\theta$.
$$
\begin{array}{l}
\because A B=A C, \\
\therefore z_{A}=i, \\
z_{B}=\cos \theta-i \sin \theta, \\
z_{C}=-\cos \theta-i \sin \theta . \\
\therefore \overrightarrow{O E}=z_{K} \\
=\frac{1}{3}\left(z_{A}+z_{C}+z_{n}\right) \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,309 |
Example 4 Let $x=b y+c z, y=c z+a x, z=a x$ $+b y$. Find the value of $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$. | Solution: $\frac{a}{a+1}=\frac{a x}{a x+x}=\frac{a x}{a x+b y+c z}$.
Similarly, $\frac{b}{b+1}=\frac{b y}{a x+b y+c z}$,
$$
\frac{c}{c+1}=\frac{c z}{a x+b y+c z} \text {. }
$$
Adding them up, the value of the desired expression is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,310 |
Example 3 In $\triangle A B C$, it is known that $\angle A=60^{\circ}$. Through the incenter $I$ of the triangle, a line parallel to $A C$ intersects $A B$ at $F$. Take a point $P$ on side $B C$ such that $3 B P=B C$. Prove: $\angle B F P=\frac{1}{2} \angle B$. | Proof: Let $|BI|=a$. Then the inradius
$$
\begin{array}{l}
r=a \sin \frac{B}{2}, z_{F}=a\left(\cos \frac{B}{2}+\frac{\sqrt{3}}{3} \sin \frac{B}{2}\right), \\
|BC|=a \cos \frac{B}{2}+r \operatorname{ctg} \frac{C}{2} \\
=\frac{a}{\sin \frac{C}{2}} \sin \frac{B+C}{2}=\frac{\sqrt{3} a}{2 \sin \frac{C}{2}}, \\
\therefore z_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,311 |
Example 4 As shown in the figure, on the three sides of $\triangle ABC$, construct $\triangle BPC$, $\triangle CQA$, and $\triangle ARB$ outwardly such that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, and $\angle ABR = \angle RAB = 15^{\circ}$. Prove that $\angle PRQ = 90^{\circ}$ an... | Proof: Let $z_{A}=-1$. Then
$$
\begin{array}{l}
z_{B}=\cos 30^{\circ}+i \sin 30^{\circ} . \\
\because \frac{B P}{B C}=\frac{A Q}{A C}=\frac{\sin 30^{\circ}}{\sin 105^{\circ}}=\frac{\sqrt{2}}{1+\sqrt{3}}, \\
\begin{aligned}
\therefore z_{P} & =z_{B}+\overrightarrow{B P} \\
& =z_{B}+\frac{\sqrt{2}}{1+\sqrt{3}}\left(z_{C}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,312 |
Example 5 In $\triangle A B C$, $\angle C=30^{\circ}, O$ is the circumcenter, $I$ is the incenter, point $D$ on side $A C$ and point $E$ on side $B C$ are such that $A D=B E=A B$. Prove: $O I \perp D E$ and $O I=D E$.
(1988, 5th National Training Team Selection Test for Mathematical Olympiad) | Prove: $\because z_{B}=2 \sin A, \therefore z_{E}=2 \sin A-1$.
$z_{A}=2 \sin B(\cos C+i \sin C), \overrightarrow{B E}=-1$.
$\therefore \overrightarrow{B A}=\overrightarrow{B E} \cdot e^{-i B}=-e^{-i B}$, so $\overrightarrow{A B}=e^{-i B}$.
Then $\overrightarrow{A D}=\overrightarrow{A B} \cdot e^{-i A}=e^{-150^{\circ} i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,313 |
Example 6 Let the two medians of $\triangle A B C$ intersect at point $O$. Prove: $A B^{2}+B C^{2}+C A^{2}=3\left(O A^{2}+O B^{2}+O C^{2}\right)$. | Prove: Place the triangle in the complex plane, and let $\overline{A B}=z_{1}$, $\overrightarrow{B C}=z_{2}$, $\overrightarrow{C A}=z_{3}$. Then $z_{1}+z_{2}+z_{3}=0$.
And $\overrightarrow{A O}=\frac{1}{3}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{1}{3}\left(z_{1}-z_{3}\right)$.
Similarly, $\overrightarrow{B O}=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,314 |
Example 7 Proof: Let point $O$ be on the side $AB$ of $\triangle ABC$, and not coincide with the vertices. Then $OC - AB < OA \cdot BC + OB \cdot AC$.
(1983, Czech Mathematical Competition) | Proof: Let $\overrightarrow{A O}=x \cdot \overrightarrow{A B}$. Then $\overrightarrow{O B}=(1-x) \cdot \overrightarrow{A B}$, where $0<x<1$. We have
$$
\begin{aligned}
O C & =|\overrightarrow{C A}+\overrightarrow{A O}| \\
& =|\overrightarrow{C A}+x \cdot(\overrightarrow{C B}-\overrightarrow{C A})| \\
& =|(1-x) \cdot \o... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,315 |
Example 8 If $P$ is a point on the arc $A_{1} A_{2 n+1}$ of the circumcircle of a regular $(2 n+1)$-sided polygon $A_{1} A_{2} \cdots A_{2 n+1}$, prove that: $P A_{1}+P A_{3}+\cdots+P A_{2 n+1}=P A_{2}+P A_{4}+\cdots+P A_{2 n}$. | Proof: Let $O$ be the center of the circle, diameter $PQ=1$, $\angle POA_1=2\alpha$, and connect $QA_k$. Then $\triangle PQA_k$ is a right triangle. We have
$$
\begin{array}{l}
PA_k=\sin \left[\frac{(k-1) \pi}{2 n+1}+\alpha\right], \\
(1 \leqslant k \leqslant 2 n+1, k \in \mathbb{N}) \\
\left(PA_1+PA_3+\cdots+PA_{2 n+1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,316 |
Theorem: In the complex plane: Points $z_{1}, z_{2}, z_{3}$ are collinear if and only if: $\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1} \in R$ or $\frac{z_{3}-z_{1}}{z_{2}-z_{1}}$ $\in R$. | Proof: Since $z_{1}, z_{2}, z_{3}$ are collinear, hence $\overrightarrow{z_{1} z_{2}}!_{j} \overrightarrow{z_{1} z_{3}}$ are collinear. Then $\frac{z_{3}-z_{1}}{z_{2}-z_{1}} \in R$ or $\operatorname{Im} \frac{z_{3}-z_{1}}{z_{2}-z_{1}}=0$,
Thus $\operatorname{Im}\left(z_{3}-z_{1}\right)\left(\bar{z}_{2}-\bar{z}_{1}\rig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,317 |
Example 9 Consider on the same plane I: two concentric circles with radii $R^{1} \cdot \mathrm{j} r$ $(R>r)$. Let $P$ be a fixed point on the circumference of the smaller circle, and $B$ be a moving point on the circumference of the larger circle. The line $B P$ intersects the smaller circle at another point $C$. The l... | Solution: (1) As shown in the figure, construct the parallelogram $B C C^{\prime} B^{\prime}$, and let $B^{\prime} C^{\prime}$ pass through point $A$. Suppose the intersection point of $B^{\prime} C^{\prime}$ on the smaller side is $A^{\prime}$. Since the distance from $O$ to $BC$ is equal to the distance from $O$ to $... | 6 R^{2}+2 r^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,318 |
Example $10 \quad \mathrm{I}:$ In $\triangle A B C$, two vertices $A$ and $B$ move along $\odot O_{1}$ and $\odot O_{2}$ respectively at the same angular velocity in the same clockwise direction. Prove: Point $C$ moves along a certain circle at a uniform speed. | Proof: Let $\vec{O}_{1} \mathrm{O}_{3}$ be the vector obtained by rotating the vector $\vec{O}_{1} \vec{O}_{2}$ by $60^{\circ}$, and let $A^{\prime} 、 B^{\prime}$ be the images of points $A 、 B$ when $\odot O_{1}$ undergoes a certain rotation.
When $\overrightarrow{O_{1} A}$ and $\overrightarrow{O_{2} B}$ rotate at th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,319 |
Example 5 If $\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=$ $\frac{t}{x+y+z}$, let $f=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}$. Prove: $f$ is an integer.
(1990, Hungarian Mathematical Competition) | Proof: If $x+y+z+t \neq 0$, by the property of proportion, we have
$$
\begin{array}{l}
\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y} \\
=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3} . \\
\text { Then we have }\left\{\begin{array}{l}
y+z+t=3 x, \\
z+t+x=3 y, \\
t+x+y=3 z, \\
x+y+z=3 t .
\end{array}\right.
\e... | -4 | Algebra | proof | Yes | Yes | cn_contest | false | 709,321 |
Let $a, b, c$ be the three sides of $\triangle ABC$, the largest internal angle of the triangle is less than $120^{\circ}$, its area is $S$, $F$ is the Fermat point, $FA=f_{a}, FB=f_{b}, FC=f_{c}, f=f_{a}+f_{b}+f_{c}$. Then,
$$
f=\frac{\sqrt{2}}{2} \sqrt{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S} .
$$ | Proof: As shown in the figure, take any side of $\triangle ABC$, such as $BC$, and construct an equilateral $\triangle BCD$ outside the triangle. Connect $AD$. Construct the circumcircle of $\triangle BCD$ to intersect $AD$ at $F$, then point $F$ is the Fermat point (proof omitted). Connect $FB$ and $FC$. By the proper... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,322 |
Theorem Let $P$ be the Fermat point of $\triangle A B C$, and let the distances from $P$ to sides $B C$, $C A$, and $A B$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively. Let the inradius of $\triangle A B C$ be $r$. Then we have
$$
r_{1}+r_{2}+r_{3} \leqslant 3 r .
$$ | Proof: Let $BC = a, CA = b, AB = c, PA = R_1, PB = R_2, PC = R_3$, then we have
$$
\begin{array}{l}
a^{2} = R_{2}^{2} + R_{3}^{2} + R_{2} R_{3}, \\
b^{2} = R_{3}^{2} + R_{1}^{2} + R_{3} R_{1}.
\end{array}
$$
Assume $a \geqslant b \geqslant c$. Then we can prove:
$$
R_{1} \leqslant R_{2} \leqslant R_{3},
$$
$$
\text{H.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,323 |
Proposition: Let $\max (A, B, C) < 120^{\circ}$, $P$ be the Fermat point of $\triangle ABC$, and $R$, $r$ be the circumradius and inradius of $\triangle ABC$, respectively. Then we have
$$
\begin{array}{l}
\frac{\sqrt{3}}{3}(3 R + 2 r) < P A + P B + P C \\
\leqslant \frac{2 \sqrt{3}}{3} \sqrt{4 R^{2} + 5 R r + r^{2}}.
... | $$
\begin{array}{l}
\text { Prove: (1) } \max (A, B, C)0 \\
\Leftrightarrow \cos 3 A+\cos 3 B+\cos 3 C-13 R+r$.
(2) By the Wlombier-Doncet inequality
$$
3 s^{2} \leqslant(4 R+r)^{2}
$$
we get $\sqrt{3} s \leqslant 4 R+r$.
$$
\begin{array}{l}
\text { (3) } f(s)=P A+P B+P C \\
=\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}+4 \sqr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,324 |
Proposition: Let $\triangle A B C=$ side lengths, median lengths be $a, b, c, m_{a}, m_{b}, m_{c}$, and the sum of distances from the Fermat point to the vertices be $l$. Then when $\max (A, B, C)<\frac{2}{3} \pi$,
$$
l \leqslant \frac{2 \sqrt{3}}{3} \sqrt{m_{a} m_{b}+m_{b} m_{c}+m_{c} m_{a}} .
$$ | Prove: When $\max (A, B, C)<\frac{2}{3} \pi$, we have
$$
l=\sqrt{\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+4 \sqrt{3} \triangle\right) .}
$$
By Feuerbach's inequality, we have
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} \triangle+(a-b)^{2}+(b- \\
c)^{2}+(c-a)^{2} .
\end{array}
$$
Thus, $l \leqslant \sqrt{a b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,325 |
Proposition Let $F$ be the Fermat point of $\triangle A B C$, and let $F A=u, F B=v, F C=w$, and the inradii of $\triangle F B C$, $\triangle F C A$, and $\triangle F A B$ be $r_{a}$, $r_{b}$, and $r_{c}$, respectively. Then
$$
r_{a}+r_{b}+r_{c} \leqslant \frac{2 \sqrt{3}-3}{2}(u+v+w) .
$$ | $$
\begin{array}{l}
S_{\triangle F B C}=\frac{1}{2}(v+w \\
+a) \cdot r_{a}, \\
S_{\triangle F B C}=\frac{1}{2} v \omega \sin 120^{\circ} \\
=\frac{\sqrt{3}}{4} v w, \\
\therefore \frac{1}{2}(v+w+a) \cdot r_{a}=\frac{\sqrt{3}}{4} v w. \\
\therefore r_{a}=\frac{\sqrt{3}}{2} \cdot \frac{v w}{v+w+a}. \\
\text { Also, } \be... | r_{a}+r_{b}+r_{c} \leqslant \frac{2 \sqrt{3}-3}{2}(u+v+w) | Inequalities | proof | Yes | Yes | cn_contest | false | 709,327 |
Let $P$ be the Fermat point inside $\triangle ABC$, and let $PA = u, PB = v, PC = w$, with the three sides of $\triangle ABC$ being $a, b, c$. Then,
$$
u + v + w \leqslant \sqrt{ab + bc + ac}.
$$
Inequality (1) improves the result of Zhou Hong from Sichuan (see *High School Mathematics* 1993, Issue 1). | Proof: As shown in the figure, $\angle A P B$, $\angle B P C$, and $\angle C P A$ are all $120^{\circ}$. Let the area of $\triangle A B C$ be $\Delta$. Then,
$$
\begin{aligned}
\Delta= & \frac{\sqrt{3}}{4}(u v+v w \\
& +w u) .
\end{aligned}
$$
That is, $3(u v+v w+u w)=4 \sqrt{3} \Delta$.
By the Law of Cosines, we have... | u + v + w \leqslant \sqrt{ab + bc + ac} | Inequalities | proof | Yes | Yes | cn_contest | false | 709,328 |
Proposition Let $P$ be the Fermat point of $\triangle A B C$, and $P$ is inside $\triangle A B C$. $O_{1}, O_{2}, O_{3}$ are the circumcenters of $\triangle A P B$, $\triangle A P C$, $\triangle B P C$, respectively. Then
(1) $\triangle O_{1} O_{2} O_{3}$ is an equilateral triangle;
(2) $P O_{1}+P O_{2}+P O_{3} \geqsla... | Proof: (1) As shown in the figure, from the given information, we have $\angle A P B=\angle A P C=\angle B P C=120^{\circ}$. Also, $A P$, $B P$, and $C P$ are the common chords of $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$, respectively. Therefore, $\mathrm{O}_{1} \mathrm{O}_{2}$, $\mathrm{O}_{2} \mathrm{O}_{3}$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,329 |
1. Among the following four propositions:
(1) If the reciprocal of a number is equal to itself, then this number is 1;
(2) A quadrilateral with diagonals that are perpendicular and equal is a square;
(3) The square root of $a^{2}$ is $\pm|a|$;
(4) An angle greater than a right angle is definitely an obtuse angle.
Amon... | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,330 |
2. Given $\frac{4}{\sqrt{3}+\sqrt{2}}<x<\frac{4}{\sqrt{5}-\sqrt{3}}$. Then, the number of integers $x$ that satisfy the above inequality is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 709,331 |
Example 6 Find the real solutions of the equation $\left(1+x+x^{2}\right)(1+x+\cdots$ $\left.+x^{10}\right)=\left(1+x+\cdots+x^{6}\right)^{2}$.
(54th Moscow Mathematical Olympiad) | Solution: Clearly, $x=0$ is a solution to the original equation.
When $x \neq 0$, we have
$$
\begin{array}{c}
\frac{1+x+\cdots+x^{10}}{1+x+\cdots+x^{6}}=\frac{1+x+\cdots+x^{6}}{1+x+x^{2}}, \\
\text { i.e., } \frac{1+x+\cdots+x^{6}}{1+x+x^{2}}=\frac{x^{7}+x^{8}+x^{9}+x^{10}}{x^{3}+x^{4}+x^{5}+x^{6}}=x^{4},
\end{array}
$... | x=0 \text{ or } x=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,332 |
4. Given $n$ points on a plane, it is known that $1, 2, 4, 8, 16, 32$ are all distances between some pairs of these points. Then, the minimum possible value of the number of points $n$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,334 |
5. In trapezoid $A B C D$, $A D / / B C, \angle B=30^{\circ}, \angle C$ $=60^{\circ}, E, M, F, N$ are the midpoints of $A B, B C, C D, D A$ respectively. Given that $B C=7, M N=3$. Then $E F$ is ().
(A) 4
(B) $4 \frac{1}{2}$
(C) 0
(1) 6 | 5. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,335 |
6. As shown in the figure, it is known that $\angle A=\angle B, A A_{1} 、 P P_{1} 、 B B_{1}$ are all perpendicular to $F_{1} B_{1}, A A_{1}=17, P P_{1}=16, B B_{1}=20, A_{1} B_{1}=12$. Then the value of $A P+P B$ is ( ).
(A) 12
(B) 13
(C) 14
(D) 15 | 6. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,336 |
3. If $a, b$ satisfy $3 \sqrt{a}+5|b|=7$, then the range of $s=2 \sqrt{a}-$ $3|b|$ is | 3. $-\frac{21}{5} \leqslant s \leqslant \frac{14}{3}$ | -\frac{21}{5} \leqslant s \leqslant \frac{14}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,339 |
One. (20 points) Let $P$ be any point on the hypotenuse $AB$ of an isosceles right triangle $ACB$, $PE$ perpendicular to $AC$ at point $E$, $PF$ perpendicular to $BC$ at point $F$, $PG$ perpendicular to $EF$ at point $G$, extend $GP$ and take a point $D$ on its extension such that $PD = PC$. Prove: $BC \perp BD$, and $... | $\begin{array}{c}-\because \angle E P G= \\ \angle E F P=\angle C P F, \\ \therefore \angle D P B=\angle A P G \\ =45^{\circ}+\angle E P G \\ =45^{\circ}+\angle C P F \\ =\angle B P F+\angle C P F \\ =\angle B P C . \\ \chi \because P C=P D, P B \text { is common, }\end{array}$
$\begin{array}{l}\therefore \triangle P D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,341 |
II. (25 points) Given that $a$ and $b$ are integers, and $a > b$, the equation $3 x^{2} + 3(a+b) x + 4 a b = 0$ has two roots $\alpha, \beta$ that satisfy the relation
$$
\alpha(\alpha+1) + \beta(\beta+1) = (\alpha+1)(\beta+1).
$$
Find all integer pairs $(a, b)$. | From the equation, we get
$$
\alpha+\beta=-(a+b), \alpha \beta=\frac{4}{3} a b \text {. }
$$
From the condition $\alpha(\alpha+1)+\beta(\beta+1)=(\alpha+1)(\beta+1)$, we have $(\alpha+\beta)^{2}-3 \alpha \beta=1$.
Substituting (1) into (2), we get
$$
(a+b)^{2}-4 a b=1 \text {. }
$$
That is, $(a-b)^{2}=1$.
Since $a>b$... | (1,0) \text{ or } (0,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,342 |
Example 7 Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0(x y z \neq 0)$. Find the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$.
(1992, Sichuan Province Junior High School Mathematics League Preliminary) | Solution: Let $y=k_{1} x, z=k_{2} x$. Then
$$
\left\{\begin{array} { l }
{ 4 - 3 k _ { 1 } - 6 k _ { 2 } = 0 , } \\
{ 1 + 2 k _ { 1 } - 7 k _ { 2 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k_{1}=\frac{2}{3}, \\
k_{2}=\frac{1}{3} .
\end{array}\right.\right.
$$
Then $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,343 |
Three. (25 points) Given the theorem: "If three prime numbers greater than 3, $a, b$, and $c$, satisfy the equation $2a + 5b = c$, then $a + b + c$ is a deficient number of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion. | Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false | 709,344 |
3. In an equilateral $\triangle ABC$, $P$ is a point on side $AB$, $Q$ is a point on side $AC$, and $AP = CQ$. It is measured that the distance between point $A$ and the midpoint $M$ of line segment $PQ$ is $19 \mathrm{~cm}$. Then the distance from point $P$ to point $C$ is $\qquad$ $\mathrm{cm}$. | 3. 38 . | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,347 |
II. (Full marks 15 points) A natural number $a$ is exactly equal to the square of another natural number $b$, then the natural number $a$ is called a perfect square (for example, $64=8^{2}$, so $64$ is a perfect square).
If $a=1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2}$, prove that $a$ is a perfect square, and write dow... | Let $x=1995$, then $x+1=1996$.
$$
\begin{aligned}
a= & 1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2} \\
= & x^{2}+x^{2}(x+1)^{2}+(x+1)^{2} \\
= & (x+1)^{2}-2 x(x+1)+x^{2}+2 x^{2}(x+1) \\
& +x^{2}(x+1)^{2} \\
= & (x+1-x)^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & 1^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & {[1+x(x+1)]^{2} } \\
= & (1+1995 \... | 3982021 | Number Theory | proof | Yes | Yes | cn_contest | false | 709,349 |
Three, (Full marks 15 points) In the convex quadrilateral $A B C D$, $\angle A B C = 30^{\circ}, \angle A D C = 60^{\circ}, A D = D C$ (as shown in the right figure). Prove: $B D^{2} = A B^{2} + B C^{2}$.
---
The translation maintains the original text's formatting and line breaks. | Connect $A C . \because A D=D C$, $\angle A D C=60^{\circ}$,
$\therefore \triangle A D C$ is an equilateral triangle, so $D C=C A=A D$. Construct an equilateral $\triangle B C E$ outward with $B C$ as a side, i.e., $B C=B E=C E$, then $\angle B C E=$ $\angle E B C=60^{\circ}, \angle A B E=\angle A B C+\angle E B C=90^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,350 |
Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning. | Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,351 |
Example 8 Given
$$
\begin{array}{l}
\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1, \\
\frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0,
\end{array}
$$
Calculate the value of $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}$.
(1996, Tianjin City Junior High School Mathematics Competition) | Solving: Squaring (1) yields
$$
\left(\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}\right)+2\left(\frac{x y}{m n}+\frac{y z}{n p}+\frac{z x}{m p}\right)=1 .
$$
From (2), removing the denominators gives
$$
x y p+y z m+z x n=0 \text {. }
$$
That is, $\frac{x y}{m n}+\frac{y z}{n p}+\frac{z x}{m p}=0$.
The... | \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,354 |
3. If $a+b=4, a^{3}+b^{3}=28$, then the value of $a^{2}+b^{2}$ is ( ).
(A) 8
(B) 10
(C) 12
(D) 14 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,356 |
5. As shown in the figure, in $\triangle A B C$, $\angle B=\angle C, D$ is on $B C$. $\angle B A D=50^{\circ}$, take a point $E$ on $A C$ such that $\angle A D E$ $=\angle A E D$. Then the degree of $\angle E D C$ is ( ).
(A) $15^{\circ}$
(B) $25^{\circ}$
(C) $30^{\circ}$
(D) $50^{\circ}$ | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,358 |
6. If the solution to the equation $9 x-17=k x$ with respect to $x$ is a positive integer, then the value of $k$ is ( ).
(A) 8
(B) 2,10
(C) $6,-10$
(D) $\pm 8$ | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,359 |
1. Factorize $\left(x^{4}+x^{2}-1\right)^{2}+\left(x^{4}+x^{2}-1\right)$ $-2=$ $\qquad$ | 1. $(x-1)(x+1)\left(x^{2}+2\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$ | (x-1)(x+1)\left(x^{2}+2\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,360 |
2. For $a>0, b<0$, the solution set of the equation $|x-a|+|x-b|=a-b$ is $\qquad$ . | 2. $b \leqslant x \leqslant a$ | b \leqslant x \leqslant a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,361 |
Example 9 If $x, y, z$ are real numbers, and
$$
\begin{aligned}
(y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\
= & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\
& +(x+y-2 z)^{2},
\end{aligned}
$$
find the value of $M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}$. | Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$
Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,365 |
Three, (Full marks 16 points) It is known that the ages of A, B, and C are all positive integers. A's age is twice B's age, and B is 7 years younger than C. The sum of the three people's ages is a prime number less than 70, and the sum of the digits of this prime number is 13. Try to find the ages of A, B, and C.
Tran... | Three, let the ages of A, B, and C be $x$ years, $y$ years, and $z$ years, respectively.
According to the problem, we have
$\left\{\begin{array}{l}x=2 y, \\ y=z-7,\end{array}\right.$ and $x+y+z$ is a prime number.
$y+x+z<70$.
Also, $\because 13=9+4=8+5=7+6$,
$\therefore$ the only prime number less than 70 with a digit ... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,367 |
Four, (Full marks 20 points) Given that $B D$ and $C E$ are altitudes of $\triangle A B C$, point $P$ is on the extension of $B D$, $B P=A C$, and point $Q$ is on $C E$, $C Q=A B$. Prove:
(1) $A P=A Q$;
(2) $A P \perp A Q$.
保留源文本的换行和格式，直接输出翻译结果。 | (1) $\because B D \perp C A, C E \perp A B$,
$$
\therefore \angle B E F=\angle C D F=90^{\circ} \text {. }
$$
While $\angle B F E=\angle C F D$, thus $\angle A B P=\angle Q C A$.
Also, $\because A B=Q C, B P=C A$,
$\therefore \triangle A B P \cong \triangle Q C A$, hence $A P=Q A$.
$$
\begin{array}{l}
\text { (2) } \b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,368 |
Five. (Full marks 20 points) There is a 14-digit number, the digit in the units place is 3 less than the digit in the tens place, and the new four-digit number formed by reversing its digits differs from the original four-digit number by 8987. Find this four-digit number and write out the reasoning process.
---
Trans... | Five, let the thousands digit of this four-digit number be $a$, the hundreds digit be $b$, the tens digit be $c$, and the units digit be $c-3$.
This number is $1000a + 100b + 10c + (c-3)$.
The new four-digit number is $1000(c-3) + 100c + 10b + a$.
According to the problem, we have
$$
\begin{array}{l}
1001(a+c-3) + 110... | 1996 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,369 |
2. If $a, b, c, d$ are four positive numbers with a product of 1, then the minimum value of the algebraic expression $a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d$ is ( ).
(A) 0
(B) 4
(C) 8
(D) 10 | 2. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,371 |
4. In Rt $\triangle A B C$, $C D$ is the altitude to the hypotenuse $A B$, then the sum of the inradii of the three right triangles ( $\triangle A B C, \triangle A C D$, $\triangle B C D$ ) is equal to ( ).
(A) $C D$
(B) $A C$
(C) $B C$
(D) $A B$ | 4. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,373 |
1. If the equation $x^{2}+(m+2) x+m+$ $5=0$ has two positive roots, then the range of values for $m$ is | 1. $-5 < m \leqslant -4$ | -5 < m \leqslant -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,374 |
Example 10 Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y, \\
\frac{4 y^{2}}{1+4 y^{2}}=z, \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.
$$
and prove that your solution is correct. (1996, Canadian Mathematical Olympiad) | Solution: Clearly, $x=y=z=0$ is a solution to the system of equations. If $x, y, z$ are all non-zero, taking the reciprocal yields
$$
\left\{\begin{array}{l}
\frac{1}{4 x^{2}}-\frac{1}{y}+1=0, \\
\frac{1}{4 y^{2}}-\frac{1}{z}+1=0, \\
\frac{1}{4 z^{2}}-\frac{1}{x}+1=0 .
\end{array}\right.
$$
Adding them up gives $\left... | x=y=z=0 \text{ or } x=y=z=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,376 |
Three. (Full marks 20 points) Given the function $y=x^{2}-|x|-12$ intersects the $x$-axis at two distinct points $A$ and $B$. Another parabola $y=a x^{2}+b x+c$ passes through points $A$ and $B$, with its vertex at $P$, and $\triangle A P B$ is an isosceles right triangle. Find $a$, $b$, and $c$.
---
Translate the ab... | Let $y=x^{2}-|x|-12=0$.
When $x>0$, $x^{2}-x-12=0, x_{1}=4, x_{2}=-3$ (discard);
When $x<0$, $x^{2}+x-12=0, x_{3}=-4, x_{4}=3$ (discard).
Thus, the coordinates of points $A$ and $B$ are $A(4,0), B(-4,0)$.
Since $y=a x^{2}+b x+c$ passes through points $A$ and $B$,
$\therefore y=a(x+4)(x-4)$.
Since $\triangle A P B$ is a... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,379 |
Four, As shown in the figure, in the inscribed $\triangle A B C$ in a circle, $A B>A C, D$ is the midpoint of $\overparen{B A C}$, and $D E \perp A B$ at $E$. Prove: $B D^{2}-A D^{2}=A B \cdot A C$. | Connect $C D$.
$$
\begin{aligned}
\because & B D^{2}=B E^{2}+D E^{2}, A D^{2}=A E^{2}+D E^{2}, \\
\therefore & B D^{2}-A D^{2}=B E^{2}-A E^{2} \\
& =(B E+A E)(B E-A E) \\
& =A B(B E-A E),
\end{aligned}
$$
On $B A$, intercept $B F=A C$, connect $D F$.
$$
\begin{array}{c}
\because B D=D C, \angle D B A= \\
\angle D C A,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,380 |
Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the avera... | Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each ma... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,381 |
Example 1 As shown, in Rt $\triangle ABC$, the hypotenuse $AB=5, CD \perp AB$. It is known that $BC, AC$ are the two roots of the quadratic equation $x^{2}-(2 m-1) x+4(m-1)=0$. Then the value of $m$ is $\qquad$. | Solution: Let $A C=b$, $B C=a$. By Vieta's formulas, we get $a+b=2 m-$
$$
\begin{array}{l}
1, a b=4(m-1) . \\
\begin{aligned}
\therefore A B^{2} & =a^{2}+b^{2}=(a+b)^{2}-2 a b \\
& =(2 m-1)^{2}-2 \times 4(m-1)=5^{2},
\end{aligned}
\end{array}
$$
i.e., $m^{2}-3 m-4=0$.
$$
\therefore m=4 \text { or } m=-1 \text {. }
$$
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,387 |
Example 2 If the equation $\left(x^{2}-1\right)\left(x^{2}-4\right)=k$ has four non-zero real roots, and the four points corresponding to them on the number line are equally spaced, then $k=$ $\qquad$. | Solution: Let $x^{2}=y$, the original equation becomes
$$
y^{2}-5 y+(4-k)=0 \text {. }
$$
Assume this equation has real roots $\alpha, \beta(0<\alpha<\beta)$, then the four real roots of the original equation are $\pm \sqrt{\alpha} 、 \pm \sqrt{\beta}$. Since the four points corresponding to them on the number line are... | \frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,388 |
Example 11 Let $a, b$ be the two roots of the equation $x^{2}-3 x+1=0$, and $c, d$ be the two roots of the equation $x^{2}-4 x+2=0$. Then
$$
\begin{array}{l}
\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c} \\
=B . \\
\text { Prove: (1) } \frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b} \\
+\f... | Proof: (1) By Vieta's formulas, we have
$$
\begin{array}{l}
a+b=3, a b=1 ; c+d=4, c d=2 . \\
\therefore a+b+c+d=3+4=7 . \\
\because a^{2}+b^{2}=(a+b)^{2}-2 a b=3^{2}-2 \times 1=7, \\
a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=18, \\
c^{2}+d^{2}=(c+d)^{2}-2 c d=12, \\
c^{3}+d^{3}=(c+d)^{3}-3 c d(c+d)=40, \\
\therefore a^{2}+b^{2}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,389 |
3. Given $a b c=1, a+b+c=2, a^{2}+b^{2}+c^{2}=3$. Then the value of $\frac{1}{a b+c-1}+\frac{1}{b c+a-1}+\frac{1}{c a+b-1}$ is ( ).
(A) 1
(B) $-\frac{1}{2}$
(C) 2
(D) $-\frac{2}{3}$ | 3. D
3. $a b+c-1=(a-1)(b-1), b c+a-i=(b-1)$
- $(c-1), c a+b-1=(c-1)(a-1)$ Simplify and solve. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,390 |
$4.1997^{2000}$ when divided by 7 leaves a remainder of ( ).
(A) 1
(B) 2
(C) 4
(D) 6 | 4. C
4. From $1997=7 n \div 2$, we have $1997^{3}=(7 n+2)^{3}=7 m + 1$, thus $1997^{1998}=7 k+1$. Therefore, $1997^{2000}=(7 k+1)(7 h + 4)$. ($m, n, k, h$ are integers). | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,391 |
1. Factorize the expression $4 x^{4}-4 x^{3}-14 x^{2}+12 x$ $+6=$ $\qquad$ | $\begin{array}{l}=1 \cdot(x+\sqrt{3})(x-\sqrt{3})(2 x-1+\sqrt{3}) \\ (2 x-1-\sqrt{3})\end{array}$ | (x+\sqrt{3})(x-\sqrt{3})(2 x-1+\sqrt{3})(2 x-1-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,394 |
$=$ 2. Simplify the radical $\sqrt{2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})}$. | $$
\text { 2. } \begin{aligned}
\text { Original expression } & =\sqrt{12-4 \sqrt{3}-4 \sqrt{5}+2 \sqrt{15}} \\
& =\sqrt{12-2 \sqrt{3 \times 4}-2 \sqrt{4 \times 5}+2 \sqrt{3 \times 5}} \\
& =\sqrt{(1 \overline{3}-\sqrt{4}+\sqrt{5})^{2}} \\
& =\sqrt{3}+\sqrt{5}-2 .
\end{aligned}
$$
Note: $12=(\sqrt{3})^{2}+(\sqrt{14})^... | \sqrt{3}+\sqrt{5}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,395 |
3. Given that $AB$ is a chord of the circle $\odot O$ with radius 1, and the length of $AB$ is the positive root of the equation $x^{2}+x-1=0$. Then the degree of $\angle AOB$ is $\qquad$ . | 3. $36^{\circ}$.
From $x^{2}+x-1=0$, we know $x=1-x^{2}<0$, so $AB < OB$. Therefore, on $OB$, we take $OC = AB = x$. Also, from $x^{2}+x-1=0$, we can get $\frac{x}{1-x}=\frac{1}{x}$. As shown in the figure, $\frac{AB}{BC} = \frac{OA}{AB}$, thus $\triangle OAB \sim \triangle ABC$. Since $\triangle OAB$ is an isosceles ... | 36^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,396 |
4. If $0<\theta<30^{\circ}$, and $\sin \theta=k m+\frac{1}{3}(k$ is a constant and $k<0)$, then the range of values for $m$ is | 4. $\frac{1}{6 k}<m<-\frac{1}{3 k}$.
Given $0<\theta<30^{\circ}$, then $0<\sin \theta<\frac{1}{2}$, i.e., $0<k m+\frac{1}{3}<$ $\frac{1}{2}$. And $k<0$, solve accordingly. | \frac{1}{6 k}<m<-\frac{1}{3 k} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,397 |
5. The train is $400 \mathrm{~m}$ long, and it takes 10 minutes to pass through the tunnel (from the front of the train entering the tunnel to the rear of the train leaving the tunnel). If the speed increases by 0.1 kilometers per minute, then it will take 9 minutes, and the length of the tunnel is $\qquad$ $ـ$. | 5. 8600 meters.
Let the full length be $x$ meters, and the original speed of the train be $v$ meters per minute. According to
$$
\left\{\begin{array}{l}
\frac{x+400}{v}=10, \\
\frac{x+400}{v+100}=9 .
\end{array} \text { Solving, we get } v=900\right. \text { (meters/minute.) }
$$ | 8600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,398 |
6. As shown in the figure, given that the area of the equilateral $\triangle A B C$ is $S, D$ is the midpoint of $A B, D E \perp B C, E F \perp A C, F G \perp$ $A B$. Then the area of quadrilateral $D E F G$ is ${ }^{\circ}$ . $\qquad$ | 6. $\frac{51}{128} S$
As shown in the figure, draw the heights $A M, B N$, and $C D$, and let $S_{1}, S_{2}, S_{3}$ be the areas of $\triangle D B E, \triangle E C F, \triangle F A G$ respectively. From $\triangle D B M \backsim \triangle A B M$, we have $\frac{S_{1}}{S}=\frac{1}{8}$. From $\triangle E C F \backsim \t... | \frac{51}{128} S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,399 |
If $a b c \neq 0, \mathrm{H} c$ and $d$ are the real roots of the equation $x^{2}+a x+b=0$, and $a$ and $b$ are the real roots of the equation $x^{2}+c x+d=0$. Prove:
$$
(a+b+c+d)^{2}=a b c d .
$$ | Proof: Applying Vieta's formulas, we get
$$
\begin{array}{l}
c+d=-a, c d=b . \\
\therefore a+b+c+d=c d . \\
\because \because a+b=-c, a b=d, \\
\therefore a+b+c+d=a b .
\end{array}
$$
(1) $\times$ (2), we get $(a+b+c+d)^{2}=a b c d$. | (a+b+c+d)^{2}=a b c d | Algebra | proof | Yes | Yes | cn_contest | false | 709,400 |
One, (Full marks 14 points) Solve the equation
$$
1+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}=\left|\frac{3 x+2}{2 x+1}\right|
$$ | First, find the allowed values of $x$.
$$
\left\{\begin{array} { l }
{ x \neq 0 , } \\
{ 1 + \frac { 1 } { x } \neq 0 , } \\
{ 1 + \frac { 1 } { 1 + \frac { 1 } { x } } \neq 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \neq 0, \\
x \neq-1, \\
x \neq-\frac{1}{2} .
\end{array}\right.\right.
$$
From the left $=... | x \leqslant-\frac{2}{3} \text{ and } x \neq-1 \text{ or } x>-\frac{1}{2} \text{ and } x \neq 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,401 |
II. (Full marks 14 points) Use 100 matchsticks of equal length to form a triangle, such that the length of the longest side is three times the length of the shortest side. Find the number of matchsticks used for each side of the triangle that meets this condition.
Use 100 matchsticks of equal length to form a triangle... | Let the number of matchsticks required for each side of the triangle be $x, y, 3x$. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+3 x=100, \\
x \leqslant y \leqslant 3 x, \\
x+y<3 x .
\end{array}\right.
$$
From (1) and (2), we get $\frac{100}{7} \leqslant x \leqslant 20$, and from (3) we get $x<\fra... | 15,40,45 \text{ or } 16,36,48 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,402 |
Three. (Full marks 14 points) For the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function $y=-2\left|a x^{2}+b x+c\right|-1$;
(2) If for any real number $k$, the line $y=k(x-1)-$ $\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the va... | (1) $\because a>0, c>0$, hence the minimum value of $\left|a x^{2}+b x+c\right|$ is 0.
Then the maximum value of $y=-2\left|a x^{2}+b x+c\right|-1$ is -1.
(2) To make the line $y=k(x-1)-\frac{k^{2}}{4}$ intersect the parabola $y=a x^{2}+b x+c$ at only one point, the system of equations
$$
\left\{\begin{array}{l}
y=k(x-... | -1, a=1, b=-2, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,403 |
Four, (Full marks 14 points) Given that the roots of the equation $x^{2}+p x+q=0$ are 1997 and 1998, when $x$ takes the integer values $0,1,2, \cdots, 1999$, the corresponding values of the quadratic trinomial $y=x^{2}+p x+q$ are $y_{0}$. Find the number of these values that are divisible by 6. | Let $y=(x-1997)(x-1998)$ be divisible by 6. Since when $x$ takes integer values, all $y$ values can be divisible by 2, we only need to examine the cases where the factors are divisible by 3.
$$
\begin{array}{l}
\text { (1) When } x-1997=3 k \text {, then } x-1998=3 k-1 \text {. } \\
\because 0 \leqslant x \leqslant 199... | 1333 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,404 |
Five. (Full marks 13 points) There are 1997 coins, of which 1000 are heads up and 997 are tails up. Now, it is required to flip any 6 coins at a time, making their heads face the opposite direction. Can all the coins be made to have their heads up after a finite number of flips? Provide your conclusion and proof.
---
... | Five, assign "+1" to the national emblem facing up and "-1" to the national emblem facing down. Then the situation of the national emblems facing up or down for 1997 coins can be represented by the product of 1997 numbers. If the product of these numbers is -1 (or +1), it indicates that there is an odd (or even) number... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,405 |
Six. (Full marks 13 points) Given a cyclic quadrilateral $A B C D$, extend $A B$ and $C D$ to intersect at $E$, and extend $A D$ and $B C$ to intersect at $F$. $E M$ and $F N$ are tangents to the circle, with $E$ and $F$ as centers and $E M, F N$ as radii, respectively, draw arcs that intersect at $K$. Prove that $E K ... | Six, connect $E F$. Draw a circle through points $B, C, E$ intersecting $E F$ at $H$, and connect $C H$.
$$
\because B, C, H, E
$$
are concyclic.
$$
\begin{array}{l}
\therefore \angle 1=\angle 2 . \\
\because A, B, C, D
\end{array}
$$
are concyclic,
$$
\therefore \angle 1=\angle 3 \text {, }
$$
thus $\angle 2=\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,406 |
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