problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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One, (20 points) For the four-digit number $\overline{a b c d}$, the sum of its digits $a + b + c + d$ is a perfect square. Reversing the digits forms the four-digit number $\overline{d c b a}$, which is 4995 greater than the original number. Find all such four-digit numbers. | $-, \Pi(1000 d+100 c+10 b+a)-(1000 a+100 b$ $+10 c+d)=4995$ simplifies to
$$
111(d-a)+10(c-b)=555 \Rightarrow 5 \mid(d-a) .
$$
If $d-a$ is $0, -5$, the equation cannot hold, so it must be $d-a=5 \cdot 11 c-b=0$.
$$
\begin{array}{l}
\text { Given } a+b+c+d=n^{2}, \text { then } \\
a+b+b+a+5=n^{2} .
\end{array}
$$
Thus... | 2007, 1116, 1996, 2887, 3778, 4669 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,101 |
Two circles intersect at $P$ and $Q$. Through point $P$, draw three arbitrary lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$, which intersect the two circles at: $A, B, C$ and $A^{\prime}, B^{\prime}, C^{\prime}$. The extensions of $A B$ and $A^{\prime} B^{\prime}$ intersect at $M$, the extensions of $A C$ and $A^{\p... | $$
\begin{array}{l}
\text { II. Connect } P Q, A Q, A^{\prime} Q \text {. } \\
\because \angle P A Q= \\
\frac{1}{2} \overparen{P n Q}, \angle P A^{\prime} Q= \\
\frac{1}{2} \overparen{P m Q} \text {, } \\
\end{array}
$$
$$
\begin{array}{c}
\therefore \angle A Q A^{\prime}=\pi \\
-\frac{1}{2}(\overparen{P n Q}+
\end{ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,102 |
1. Complex numbers
$$
\begin{array}{l}
z_{1}=\frac{1}{1+\cos \theta+i \sin \theta}+\frac{1}{1+\cos \theta-i \sin \theta}, \\
z_{2}=\frac{\cos \theta+i \sin \theta}{1+\cos \theta+i \sin \theta}+\frac{\cos \theta-i \sin \theta}{1+\cos \theta-i \sin \theta}
\end{array}
$$
The relationship between them is ().
(A) $z_{1}>z... | -1. (B).
Let $z=\cos \theta+i \sin \theta$, then $z \bar{z}=1$. We have
$$
\begin{array}{l}
z_{1}=\frac{z \bar{z}}{z \bar{z}+z}+\frac{1}{1+\bar{z}}=\frac{\bar{z}}{1+\bar{z}}+\frac{1}{1+\bar{z}}=1, \\
z_{2}=\frac{z}{z \bar{z}+z}+\frac{\bar{z}}{1+\bar{z}}=\frac{1}{1+\bar{z}}+\frac{\bar{z}}{1+\bar{z}}=1 .
\end{array}
$$
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,104 |
Example 9 There are $n$ players participating in a chess tournament, with the scoring method being: the winner of each game gets 2 points, the loser gets 0 points, and in the case of a draw, both players get 1 point. Midway through the tournament, the player with the highest score has $k$ points. Prove: At this point, ... | Analysis: Let the sum of all players' scores be $S$, then $S \leqslant k n$. Suppose at this point the player with the fewest matches has played $\mathrm{m}$ matches, then the total number of matches played should be no less than $\frac{m n}{2}$. Each match increases the total score by 2 points, thus,
$$
S \geqslant 2 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,105 |
2. Given the function $y=2 x^{2}$ on $[a, b](a<b)$, its value range is $[0,2]$. Then the trajectory of the point $(a, b)$ is ( ).
(A) line segments $A B, B C$
(B) line segments $A B, O C$
(C) line segments $O A, B C$
(D) line segments $O A, O C$ | 2. (A).
From $y=2 x^{2} \geqslant 0$, we know that when $y=0$, $x=0$.
Therefore, $0 \in[a, b]$.
When $y=2$, $x=1$ or -1. There are two possibilities:
If $1 \in[a, b]$, then by $a<b$ and the monotonicity of the function on $R^{+}$, we have $b=1$ and $-1 \leqslant a \leqslant 0$
(see figure); the corresponding trajector... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,106 |
3. Let the triangle: the sides opposite to the interior angles be $a$, $b$, $c$. If $a^{2}=b(b+c)$, $b^{2}=c(c+a)$, then, which of the following is true?
(A) $\angle A=2 \angle B=4 \angle C$
(B) $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$
(C) $\cos C-\cos B-\cos A=\frac{1}{2}$
(D) $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2$ | 3. (D).
There are multiple methods to prove: $a^{2}=b(b+c) \Rightarrow \angle A=2 \angle B$.
Below is a plane geometry proof.
As shown in the figure, extend $C A$ to $D$,
such that $A D=A B$, and connect $B D$. Then
$C D=b+c$. Also, from $a^{2}=b(b+c)$ we get
$$
\frac{B C}{A C}=\frac{D C}{B C}.
$$
Therefore, $\trian... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,107 |
4. ${ }^{1} \cdot \mathrm{j}$ The expression $2\left(\cos 2 \theta+\cos \theta+\frac{1}{2}\right)(\cos 4 \theta$ $-\cos 3 \theta+\cos \theta-\frac{1}{2}$ ) is equivalent to ( )
(A) $\cos 6 \theta+\cos 3 \theta+\frac{1}{2}$
(B) $\cos 6 \theta+\cos 3 \theta-\frac{1}{2}$
(C) $\cos 5 \theta+\cos 4 \theta+\frac{1}{2}$
(D) $... | 4. (A).
Let $x=\cos \theta+i \sin \theta$. Then $\frac{1}{x}=\cos \theta-i \sin \theta$. Therefore,
$$
\begin{array}{l}
\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right), \\
\cos n \theta=\frac{1}{2}\left(x^n+\frac{1}{x^n}\right).
\end{array}
$$
$\therefore$ The original expression $=2\left[\frac{1}{2}\left(x^{2}+\fra... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,108 |
5. Given a quadrilateral pyramid $S$ $A B C D$, where the base quadrilateral $A B C D$ is not a parallelogram, using a plane to intersect the quadrilateral pyramid to get the quadrilateral $A_{1} B_{1} C_{1} D_{1}$ (outer figure), let the set $M$ $=\left\{\right.$ quadrilateral $A_{1} B_{1} C_{1} D_{1}$ is a parallelog... | (5) (B).
Let the plane $SAB$
intersect with the plane $SCD$
at the line $l_{1}^{\prime}$ passing through $S$,
the plane $SAD$ intersect with
the plane $SBC$ at the line
$t_{2}$ passing through $S$. The two intersecting
lines $l_{1}, l_{2}$ determine a
plane $\alpha$ (as shown).
When the section is parallel to $\alpha$... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,109 |
1. Given $(-1)^{n} 2^{n} \equiv b_{n}(\bmod 9), 0 \leq b_{n} \leq 8$. Then the range of values for $b_{n}$ is \{ $\qquad$ \}. | $\begin{array}{l}\text { II.1.1. } \\ \text { \| }(-1)^{n} 2^{3 n}=(-8)^{n}=(1-9)^{n} \\ \quad=C_{n}^{n}-C_{n}^{1} 9+C_{n}^{2} 9^{2}+\cdots+(-1)^{n} C_{n}^{n} 9^{n}, \\ \text { know }(-1)^{n} 2^{3 n} \equiv 1(\bmod 9) .\end{array}$ | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,111 |
3. The intersection of the hyperbola $y^{2}-x^{2}=1$ and the line $y=\frac{4 x+\sqrt{2}}{x+\sqrt{2}}$ is . $\qquad$ | $$
\begin{array}{l}
\quad \text { 3. }\left\{\begin{array} { l }
{ x _ { 1 } = 0 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \left\{\begin{array}{l}
x_{2}=2 \sqrt{2} \\
y_{2}=3 ;
\end{array},\left\{\begin{array}{l}
x_{3}=-2 \sqrt{2}+5, \\
y_{3}=2-\sqrt{10} ;
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
x_{4}... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,113 |
4. For $S_{n}=\{1,2, \cdots, n\}$, we pick a non-empty subset $A$. We multiply each element $k(1 \leqslant k \leqslant n)$ in $A$ by $(-1)^{k}$ and then find the sum. The total sum of all these sums is . $\qquad$ | 4. $(-1)^{n}\left[n+\frac{1-(-1)^{n}}{2}\right] 2^{n-2}$
The problem can be transformed into finding the sum of all elements in all subsets of the set $\{-1,2,-3,4, \cdots, (-1)^{n} n\}$. Since in this total sum, each element appears $2^{n-1}$ times, we get
$$
\begin{aligned}
N & =\left[-1+2-3+4-\cdots+(-1)^{n} n\righ... | (-1)^{n}\left[n+\frac{1-(-1)^{n}}{2}\right] \cdot 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,114 |
5. The solution set of the inequality $|\lg x|>|\lg 4 x|>|\lg 2 x|$ is . $\qquad$ | 5. $\left\{x \left\lvert\, \frac{\sqrt{2}}{4}\lg x>\lg 2 x$
$>\lg 4 x$.
$$
\begin{aligned}
\therefore & |\lg x||\lg 4 x|=\lg 4 x \\
\text { pos } 0 & >\lg x+\lg 4 x=\lg 4 x^{2},
\end{aligned}
$$
Thus, $4 x^{2}<1$.
Combining (1) and (2), we get
$$
\frac{1}{4}<x<\frac{1}{2} \text {. }
$$
Therefore, $\frac{1}{2}<2 x<1<4... | \frac{\sqrt{2}}{4}<x<\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 709,115 |
Example 1 As shown in the figure, the incircle $O$ of rhombus $ABCD$ touches the sides at $E, F, G, H$ respectively. On $\overparen{EF}$ and $\overparen{GH}$, draw tangents to $\odot O$ intersecting $AB$ at $M$, $BC$ at $N$, $CD$ at $P$, and $DA$ at $Q$. Prove that $MQ \parallel NP$. (1995, National High School Mathema... | Gong Xin: Establish a rectangular coordinate system as shown in the figure, and let the points of tangency of the tangents $PQ$ and $MN$ be $I$ and $J$, respectively. Since the coordinates of $M$ and $Q$ depend on the positions of $I$ and $J$, we can introduce the central angles of $I$ and $J$ as parameters to express ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,116 |
Three, for the $\mathrm{J}$ curve $C_{1}: 3\left(x^{2}+2 y^{2}\right)^{2}=$ $2\left(x^{2}+4 y^{2}\right)$ at a point $P$ outside the origin, prove that there exists a line ${ }^{1} \mathrm{j}$ passing through $P$ that intersects the ellipse $C_{2}: x^{2}+2 y^{2}=2$.
Translate the text above into English, please ret... | Three, the explanation of the kidney dust, all points on $C_{1}$ are inside $C_{2}$. The equation transforms into
$$
\left(x^{2}+2 y^{2}-2\right)\left(x^{2}+4 y^{2}\right)+2\left(x^{2}+y^{2}\right)\left(x^{2}+2 y^{2}\right)=0 \text {. }
$$
$$
\begin{array}{l}
x^{2}+2 y^{2}-2 \\
=-\frac{2\left(x^{2}+y^{2}\right)\left(x^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,118 |
The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is ... | Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,120 |
In the acute triangle $\triangle A B C$, $B D$ is the altitude on $A C$, $E$ is a point on side $A B$ such that $\angle A E C=45^{\circ}$, and $B D=2 C E$. Prove that the necessary and sufficient condition for $D E \parallel B C$ is $C E = A C + A D$. | Let $A D = x, C E = a (0 < x < a)$, then $B D = 2a$.
In the right triangle $\triangle A B D$,
$$
A B = \sqrt{4 a^{2} + x^{2}}.
$$
Draw $C F \perp A B$ at $F$. Since $\triangle A B C$ is an acute triangle, $\triangle A C E$ is also an acute triangle, and $F$ must be between $A$ and $E$. Therefore,
$$
A E = A F + F E,
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,121 |
Given the complex number $z$ has a modulus of 1. Find
$$
u=\frac{(z+4)^{2}-(\bar{z}+4)^{2}}{4 i} \text {. }
$$
the maximum value. | $$
\begin{array}{l}
\text { Given }|z|=1 \text {, we can set } z=\cos x+i \sin x \text {. Then, } \\
u=(4+\cos x) \sin x \text {. } \\
\text { By } u^{2}=(4+\cos x)^{2} \sin ^{2} x \\
=\frac{(4+\cos x)(4+\cos x)(\sqrt{6}+1)(1+\cos x) \cdot(\sqrt{6}+3)(1-\cos x)}{(\sqrt{6}+1)(\sqrt{6}+3)} \\
\leqslant \frac{1}{9+4 \sqrt... | \frac{\sqrt{9+24 \sqrt{6}}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,122 |
Three, color $a$ points on a line red, $b$ points only yellow, and $c$ points blue, such that no two adjacent points have the same color. The number of such coloring methods is denoted as $F(a, b, c)$. Obviously, we have:
(1) Prove: For natural numbers $a, b, c$ not all equal to 1, we have
$$
\begin{array}{l}
F(a, b, c... | The number of ways to color the 1st point yellow and blue.
$$
F(a, b, c)=\sum_{i=1}^{3} F_{i}(a, b, c) .
$$
And
$$
\begin{aligned}
& F_{1}(a, b, c)=F_{2}(a-1, b, c)+F_{3}(a-1, b, c) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\
& +F_{1}(a-1, b, c-1)+F_{2}(a-1, b, c-1) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\... | 10160640 | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,123 |
The sum and product of the first 51.1995 integers are both equal to F $\cdot$ 1996. What is the sum of the absolute values of these 1995 integers? Please prove your conclusion.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Let these 1995 integers be denoted as $x_{1}, x_{2}, \cdots, x_{1995}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{1995}=1996, \\
x_{1} x_{2} \cdots \cdots x_{1995}=1996=2^{2} \times 499 .
\end{array}\right.
$$
(1) From (1), we know that among these 1995 numbers, there are an even number of odd num... | 2496 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,124 |
Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$. | Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$
Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\be... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,125 |
$$
\begin{array}{c}
\text { Example } 1 \text { Calculate } \frac{a}{a^{3}+a^{2} b+a b^{2}+b^{3}}+ \\
\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}}+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
(1995, Tianjin City, Grade 7 "Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\ +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\ =\frac{a^{2}+b^{2}}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\ =\frac{1}... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,127 |
Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition) | $\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,128 |
Example 11 Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Find the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$.
(1995, National Junior High School Mathematics League) | Solution: From the given, we have $\alpha^{2}+\alpha=\frac{1}{4}$.
To form the required expression containing $\alpha^{2}+\alpha$, we have
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}=\frac{(\alpha-1)\left(\alpha^{2}+\alpha+1\right)}{(\alpha-1)\left(\alpha^{2}+\alpha\right)^{2}}=20 .
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,129 |
7. If $m$ and $n$ are positive integers, and $m \leqslant 1996, r=2-\frac{m}{n}$ $>0$, then the minimum value of $r$ is | 7. $\frac{1}{998}$ | \frac{1}{998} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,131 |
8. The system of equations $\left\{\begin{array}{l}x+x y+y=1, \\ x^{2}+x^{2} y^{2}+y^{2}=17\end{array}\right.$ has the real solution $(x, y)=$ | $\begin{array}{l}\text { 8. }\left(\frac{3+\sqrt{17}}{2},\right. \\ \left.\frac{3-\sqrt{17}}{2}\right) \text { or }\left(\frac{3-\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right)\end{array}$ | \left(\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\right) \text{ or } \left(\frac{3-\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,132 |
12. As shown in the figure, $A D$ is the angle bisector of $\triangle A B C$, and the perpendicular bisector of $A D$ intersects the extension of $B C$ at $F$. If $\frac{A C}{A B} = \frac{3}{4}$, then $\frac{C F}{B F}=$ $\qquad$ | 12. $\frac{9}{16}$ | \frac{9}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,136 |
13. $a, b$ are the roots of the equation $x^{2}-4 x+$ $1=0$, $c, d$ are the roots of the equation $x^{2}$ $-5 x+2=0$, and let
$$
\begin{aligned}
t= & \frac{a}{b+c+d}+\frac{b}{a+c+d} \\
& +\frac{c}{a+b+d}+\frac{d}{a+b+c} .
\end{aligned}
$$
Then express $\frac{a^{2}}{b+c+d}+\frac{b^{2}}{a+c+d}+\frac{c^{2}}{a+b+d}+\frac{... | 13. $9 t-9$ | 9 t-9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,137 |
15. Write a group of 4 consecutive natural numbers, such that they are in ascending order, a multiple of 5, a multiple of 7, a multiple of 9, and a multiple of 11: $\qquad$ | 15. General Solution
For $3465 k+1735, 3465 k+1736, 3465 k+1737, 3465 k$ +1738 ( $k$ is a non-negative integer) | 3465 k+1735, 3465 k+1736, 3465 k+1737, 3465 k+1738 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,139 |
Example 12 Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} .
\end{array}\right.
$$
(1996, Beijing Junior High School Mathematics Competition) | Solving: By taking the reciprocal and finding a common denominator, we get
$$
\left\{\begin{array}{l}
\frac{x(y+z)}{x+y+z}=2, \\
\frac{y(z+x)}{x+y+z}=3, \\
\frac{z(x+y)}{x+y+z}=4 .
\end{array}\right.
$$
Adding the three equations yields $\frac{x y+y z+z x}{x+y+z}=\frac{9}{2}$.
(4) Subtracting (1), (2), and (3) respect... | x=\frac{23}{10}, y=\frac{23}{6}, z=\frac{23}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,140 |
2. In $\triangle A B C$, the side lengths are $a=3, b=4, c=6, h_{a}$ is the length of the altitude from $a$, and $h_{b}, h_{c}$ have similar meanings. Then
$$
\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)
$$
is equal to | 2. $\frac{39}{4}$ | \frac{39}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,142 |
9. $1000,1001,1002, \cdots, 1996$ These 997 natural numbers appear in all the digits whose sum is
The sum of all the digits that appear in the 997 natural numbers $1000,1001,1002, \cdots, 1996$ is | 9. 14419 | 14419 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,149 |
Example 13 If the equation $\frac{M x+N}{x^{2}+x-2}$ $=\frac{2}{x+a}-\frac{c}{x+b}$, where $\frac{M x+N}{x^{2}+x-2}$ is in its simplest form, and $a>b, a+b=c$, then $N=$ | $$
\begin{array}{l}
\frac{M x+N}{x^{2}+x-2}=\frac{(2-c) x+(2 b-a c)}{x^{2}+(a+b) x+a b} . \\
\left\{\begin{array} { l }
{ a + b = 1 , } \\
{ a b = - 2 , } \\
{ 2 - c = M , } \\
{ 2 b - a c = N , } \\
{ a > b , a + b = c }
\end{array} \quad \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-1, \\
c=1, \\
M=1, \\
N=-4 .
\en... | N=-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,151 |
In trapezoid $A B C D$, $A D / / B C, \frac{A D}{B C}=\frac{1}{2}$, point $M$ is on side $A B$, such that $\frac{A M}{M B}=\frac{3}{2}$, point $N$ is on side $C D$: such that line segment $M N$ divides the trapezoid into two parts with an area ratio of $3: 1$. Find $\frac{C N}{N D}$. | $$
\begin{array}{c}
=, \because \frac{S_{\triangle M B C}}{S_{\triangle A B C}}=\frac{2}{5}, \\
\frac{S_{\triangle A B C}}{S_{\text {trapezoid }}}=\frac{2}{3}, \\
\therefore \frac{S_{\triangle M B C}}{S_{\text {trapezoid }}}=\frac{4}{15}>\frac{1}{4},
\end{array}
$$
Therefore, it can only be that
$S_{\text {quadrilater... | \frac{29}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,152 |
Three, if the equation with respect to $x$
$$
a x^{2}+2(a-3) x+(a-2)=0
$$
has at least one integer solution, and $a$ is an integer, find $a$.
| When $a=0$, the known equation becomes $-6 x-2=0$, which has no integer solutions.
When $a \neq 0$, the known equation has at least one integer solution, which requires the discriminant $\Delta=4(a-3)^{2}-4 a(a-2)=4(9-4 a)$ to be a perfect square, thus $9-4 a$ must be a perfect square.
Let $9-4 a=s^{2}(s$ be a positiv... | 2, -4, -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,153 |
Four, there are cards labeled with integers from 1 to $2 n+1$ in sequence. What is the maximum number of cards that can be selected so that in the selected cards, the label of any card is not equal to the sum of the labels of any other two selected cards?
Translate the above text into English, please keep the line b... | Four, the number of cards $k$ selected first cannot exceed $n+1$.
This is because, if $k \geqslant n+2$. Let the numbers on these $k$ cards be $a_{1}, a_{2}, \cdots, a_{k}$, and assume $a_{1}$ is the smallest. Among the following $2 k-2$ natural numbers
$$
a_{2}, a_{3}, \cdots, a_{k}, a_{2}-a_{1}, a_{3}-a_{1}, \cdots, ... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,154 |
2. Given $c>1, x=\sqrt{c}-\sqrt{c-1}, y=\sqrt{c+1}-$ $\sqrt{c}, z=\sqrt{c+2}-\sqrt{c+1}$. Then the size relationship of $x, y, z$ is ( ).
(A) $x>y>z$
(B) $z>x>y$
(C) $y>x>z$
(D) $z>y>x$ | 2. A
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 | 709,156 |
3. As shown in the figure, $\triangle A B C$ is an acute triangle, $D$ is any point on side $B C$, $E$ and $F$ are on sides $A B$ and $A C$ respectively, and satisfy $D E = D B, D F = D C$. If $\angle A = 55^{\circ}$, then $\angle E D F$ equals ( ).
(A) $55^{\circ}$
(B) $60^{\circ}$
(C) $65^{\circ}$
(D) $70^{\circ}$ | 3. D
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 | 709,157 |
4. As shown in the figure, two national defense highways, each 40 m wide, intersect at an angle of $\alpha$. Then, the area of the common part (the shaded area in the figure) of the road surface at the intersection is ( ).
(A) $\frac{1600}{\sin \alpha}\left(\mathrm{m}^{2}\right)$
(B) $\frac{1600}{\cos \alpha}\left(\mat... | 4. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,158 |
5. To green the environment and beautify the city, three lawns in the shape of an equilateral triangle $\left(S_{1}\right)$, a square $\left(S_{2}\right)$, and a circle $\left(S_{3}\right)$ were laid out in a residential area. If the perimeters of these three lawns are the same, then the size relationship of their area... | 5. D
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 | 709,159 |
6. Let the three sides of $\triangle A B C$ be $a, b, c$, and the heights on these sides be $h_{a}, h_{b}, h_{c}$. For all triangles where the sides satisfy $2 b=a+c$, the heights satisfy
(A) $2 h_{b}=h_{a}+h_{c}$
(B) $\frac{2}{h_{b}}=\frac{1}{h_{a}}+\frac{1}{h_{c}}$
(C) $\frac{h_{b}}{h_{a}}=\frac{h_{c}}{h_{b}}$
(D) No... | 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 | 709,160 |
Example 14 Real numbers $a, b$ satisfy $ab=1$, let $M = \frac{1}{1+a} + \frac{1}{1+b}$, $N = \frac{a}{1+a} + \frac{b}{1+b}$. Then the relationship between $M$ and $N$ is $\qquad$
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain
(1996, National Junior High School Mathematics Competition) | Solution: Replacing 1 in $N$ with $a b$ yields $N=\frac{a}{a b+a} +\frac{b}{a b+b}=\frac{1}{1+a}+\frac{1}{1+b}=M$. Therefore, the correct choice is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,162 |
8. There are 1000 numbers arranged in a row, where among any three consecutive numbers, the middle number equals the sum of the numbers before and after it. If the first number and the second number are both 1, then the sum of these 1000 numbers is ( ).
(A) 1000
(B) 1
(C) 0
(D) -1 | 8. B
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 | 709,163 |
Three, (Full marks 20 points) In the right triangle $\triangle A B C$, $\angle C=90^{\circ}$. If $\sin A$ and $\sin B$ are the two roots of the equation $x^{2}-\sqrt{2} x-k=0$, find the degree measures of $\angle A$ and $\angle B$ and the value of $k$.
---
Please note that the translation preserves the original forma... | Three, from the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
\sin A+\sin B=\sqrt{2}, \\
\sin A \sin B=-k .
\end{array}\right.
$$
$$
\begin{array}{l}
\because \angle A+\angle B=90^{\circ}, \\
\therefore \sin B=\cos A .
\end{array}
$$
Thus, we have $\left\{\begin{array}{l}\sin A+\cos A=... | \angle A=\angle B=45^{\circ}, k=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,168 |
Four. (Full marks 20 points) Using sides $AC$ and $BC$ of $\triangle ABC$ as one side each, construct squares $ACDE$ and $CBFG$ outside $\triangle ABC$. Point $P$ is the midpoint of $EF$ (as shown in the figure). Prove that the distance from point $P$ to side $AB$ is half of $AB$. | Four, draw perpendiculars from $E$, $F$, $C$, and $P$ to $A B$, with the feet of the perpendiculars being $R$, $S$, $T$, and $Q$ respectively. Then $P Q$ is the distance from point $P$ to $A B$, and $E R$ $/ / P Q / / C T / / F S$. Therefore, quadrilateral $E R S F$ is a right trapezoid, and $P Q$ is its midline. Thus,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,169 |
Five. (Full marks 20 points) A city encourages water conservation by setting the following收费标准 for tap water: For each household, the first 10 tons of water used in a month are charged at 0.45 yuan per ton; the next 10 tons (from 10 to 20 tons) are charged at 0.80 yuan per ton; and any amount exceeding 20 tons is charg... | Five, let the water usage of household C be $x$ tons ($x$ is an integer, and $0 \leqslant x \leqslant 10$), and the water usage of household B be $(10+y)$ tons ($y$ is an integer, and $0 \leqslant y \leqslant 10$). Since household A pays $7.10 \overrightarrow{\mathrm{J}}$ more than household B, we get
$$
\begin{array}{... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,170 |
1. $\sqrt{2}-\sqrt{8}+\sqrt{18}$ equals ( ).
(A) 0
(B) $\sqrt{2}$
(C) $2 \sqrt{2}$
(D) $3 \sqrt{2}$ | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,171 |
2. Given $a<0$, then $\sqrt{(2 a-|a|)^{2}}$ equals ( ).
(A) $a$
(B) $-a$
(C) $3 a$
(D) $-3 a$ | 2. 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 | 709,172 |
Example 15 Simplify $m=\frac{a^{2}(x-b)(x-c)}{(a-b)(a-c)}$ $+\frac{b^{2}(x-c)(x-a)}{(b-c)(b-a)}+\frac{c^{2}(x-a)(x-b)}{(c-a)(c-b)}$. | Solution: Let $f(x)=m-x^{2}$, then $f(x)$ is a polynomial in $x$ of degree not exceeding 2, and $f(a)=f(b)$ $=f(c)=0$, which indicates that $a$, $b$, and $c$ are all roots of $f(x)=0$. Since $f(x)$ has a degree not exceeding 2, $f(x)=0$ holds true constantly. Therefore, $m=x^{2}$. | m=x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,173 |
6. In quadrilateral $A B C D$, $A D=D C, \angle A D C=\angle A B C=$ $90^{\circ}, D E \perp A B$ at $E$ (as shown in the figure). If the area of quadrilateral $A B C D$ is 8, then the length of $D E$ is ( ).
(A) 2
(B) $2 \sqrt{2}$
(C) 3
(D) $3 \sqrt{2}$ | 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 | 709,177 |
9. Add 1 kilogram of water to the saltwater, the new saltwater concentration is $20 \%$, then add 1 kilogram of salt to the new saltwater, the resulting saltwater concentration is $40 \%$. Then the original concentration of the saltwater is ( ).
(A) $25 \%$
(B) $27.5 \%$
(C) $30 \%$
(D) $32.5 \%$ | 9. 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 | 709,180 |
1. The result of factoring $(x+y)(x-y)+4(y-1)$ is $\qquad$ . | 1. $(x+y+2)(x-y+2)$ | (x+y+2)(x-y+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,182 |
Example 16 Consider the equation in $x$
$$
x^{2}+\left(\frac{a}{x}\right)^{2}-7 x-\frac{7 a}{x}+2 a+12=0
$$
has two equal real roots. Find the value of $a$.
(1995, Huanggang Region, Hubei Province Junior High School Mathematics Competition) | (1) The equation can be rewritten as $\left(x+\frac{a}{x}\right)^{2}-7\left(x+\frac{a}{x}\right)+12$ $=0$, which gives $\left(x+\frac{a}{x}-4\right)\left(x+\frac{a}{x}-3\right)=0$, meaning $x+\frac{a}{x}$ $-4=0$ or $x+\frac{a}{x}-3=0$. Therefore, we have $x^{2}-4 x+a=0$ or $x^{2}-3 x+a=0$. Clearly, when $a=4$ or $a=\fr... | a=4 \text{ or } a=\frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,184 |
16. In $\triangle A B C$, one side is 5, and the other two sides are exactly the roots of the equation $2 x^{2}-12 x+m=0$. Then, the range of values for $m$ is $\qquad$. | 10. $\frac{11}{2}<m \leqslant 18$ | \frac{11}{2}<m \leqslant 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,192 |
Example 17 Given $x+y+z \neq 0$, and $\frac{x}{y+z}=a$, $\frac{y}{z+x}=b, \frac{z}{x+y}=c$. Prove: $\frac{3 a}{1+a}, \frac{3 b}{1+b}, \frac{3 c}{1+c}$ at least one is not less than 1, and at least one is not greater than 1. | Proof: $\frac{3 a}{1+a}+\frac{3 b}{1+b}+\frac{3 c}{1+c}=3\left(\frac{x}{x+y+z}\right.$
$$
\left.+\frac{y}{x+y+z}+\frac{z}{x+y+z}\right)=3 .
$$
Therefore, the proposition holds. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,195 |
ニ、(Full score 20 points) In $\triangle A B C$, $A B=A C$, line $l$ passes through $A$ and $l / / B C$, the angle bisector of $\angle B$ intersects $A C$ and $l$ at $D$ and $E$ respectively, the angle bisector of $\angle C$ intersects $A B$ and $l$ at $F$ and $G$ respectively. Prove: $D E=F G$. | From the given, it is obvious that $\triangle B C D \cong \triangle C B F$, thus $C D = B F, \angle B D C = \angle C F B$, hence $A D = A C - C D = A B - B F = A F, \angle A D E = \angle B D C = \angle C F B = \angle A F G$.
Since $l / / B C$, then $\angle E A D = \angle C = \angle B = \angle G A F$.
Therefore, $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,199 |
Three. (Full marks 20 points) Given the equation $x^{2}+m x-m+1=0$ (where $m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
---
The translation maintains the original format and line breaks as requested. | Three, let the two distinct positive integer roots be $\alpha, \beta(\alpha<\beta)$. By Vieta's formulas, we have $\left\{\begin{array}{l}\alpha+\beta=-m, \\ \alpha \beta=-m+1 .\end{array}\right.$ Eliminating $m$, we get $\alpha \beta-\alpha-\beta=1$.
That is, $(\alpha-1)(\beta-1)=2$.
Then $\left\{\begin{array}{l}\alph... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,200 |
Four. (Full marks 25 points) Let the four vertices of the convex quadrilateral $A B C D$ satisfy the condition: the sum of the distances from each point to the other three points is equal. What kind of quadrilateral is this? Please prove your conclusion.
---
Translate the above text into English, please retain the or... | Four, it is a rectangle.
According to the conditions, we have $A B+A C+A D=m, B A+B C+B D=m, C A+C B+C D=m, D A+D B+D C=m$. Here $m$ is a constant. By subtracting the first equation from the subsequent three equations, we get
$$
\left\{\begin{array}{l}
A C+A D=B C+B D, \\
A B+A D=B C+C D, \\
A B+A C=B D+C D .
\end{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,201 |
Five. (Full marks 25 points) Solve the equation
$$
x=\left(x^{2}+3 x-2\right)^{2}+3\left(x^{2}+3 x-2\right)-2 \text {. }
$$ | Let $y=x^{2}+3 x-2$, then $x=y^{2}+3 y-2$.
Subtracting the two equations, we get
$$
\begin{aligned}
y-x & =\left(x^{2}+3 x-2\right)-\left(y^{2}+3 y-2\right) \\
& =(x-y)(x+y)+3(x-y) .
\end{aligned}
$$
That is, $(x-y)(x+y+4)=0$,
so $x-y=0$ or $x+y+4=0$.
If $x-y=0$, then $x=y$, substituting into $y=x^{2}+3 x-2$, we get $... | x = -1 \pm \sqrt{3} \text{ or } x = -2 \pm \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,202 |
5. Among the following propositions,
(1) $\sqrt{0.31}$ is in its simplest radical form;
(2) The final result of factoring $x^{3}+3 x^{2}-4$ is $(x+2)$
- $\left(x^{2}+x-2\right)$;
(3) If $a+b=1$, then $a^{3}+b^{3}+3 a b=1$;
(4) Rationalizing the denominator of $\frac{a b}{a \sqrt{b}+b \sqrt{a}}$ results in $\frac{a \sqr... | $5 . A$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,208 |
7. Given that $x^{2}-a x+3-b=0$ has two distinct real roots, $x^{2}+(6-a) x+6-b=0$ has two equal real roots, and $x^{2}+(4-a) x+5-b=0$ has no real roots. Then the range of values for $a$ and $b$ is ( ).
(A) $2<a<4,2<b<5$
(B) $1<a<4,2<b<5$
(C) $1<a<4,1<b<5$
(D) $2<a<4,1<b<5$ | 7.
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 | 709,210 |
8. Let $a$ and $b$ be any two different three-digit positive integers. Then, among the numbers from 12345678000 to 12345678999, ( ).
(A) no number is a multiple of $a$, nor is it a multiple of $b$
(B) there is a number that is a multiple of $a$, but no number is a multiple of $b$
(C) there is a number that is a multipl... | 8. I))
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,211 |
3. Let the hypotenuse of a right triangle be $c$, and the radius of its inscribed circle be $r$. Then the ratio of the area of the inscribed circle to the area of the triangle is $\qquad$ . | 3. $\frac{\pi r}{c+r}$ | \frac{\pi r}{c+r} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,214 |
Three, (Full marks 20 points) Let $\triangle A B C$ have three inscribed squares on its sides (with two vertices on one side of the triangle, and the other two vertices on the other two sides of the triangle) with equal areas. Prove: $\triangle A B C$ is an equilateral triangle. | Let the area of $\triangle ABC$ be $S$, the altitude from $A$ to $BC$ be $h_{a}$, and the side length of the inscribed square on $BC$ be $x$ (as shown in the figure). Then, from $\triangle A B^{\prime} C^{\prime} \propto \triangle A B C$, we have $\frac{x}{a}=\frac{h_{a}-x}{h_{a}}$, which leads to $x=\frac{2 S}{a+h_{a}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,216 |
Four, (Full marks 20 points, (1) 10 points, (2) 10 points) A total of 240 students participating in a large group performance stand in a row facing the coach, and report numbers in sequence from left to right as $1,2,3,4,5, \cdots$. The coach asks all students to remember their reported numbers and perform the followin... | (1) Because all students who report numbers that are multiples of $15$, $21$, and $35$ but not multiples of $105$ turn twice, still facing the coach. Students who report numbers that are multiples of $3$ but not multiples of $15$ or $21$, multiples of $5$ but not multiples of $15$ or $35$, and multiples of $7$ but not ... | 118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,218 |
Five. (Full marks 20 points, (1) 12 points, (2) 8 points)
If the sines of the two acute angles $A, B$ of a right triangle are the two roots of the equation $x^{2}+p x+q=0$:
(1) What conditions should the real numbers $p, q$ satisfy?
(2) If $p, q$ satisfy these conditions, are the two roots of the equation $x^{2}+p x+q=... | $$
\begin{array}{l}
(1) \text{ Let } A, B \text{ be the two acute angles of a right-angled triangle, and } \sin A, \sin B \text{ are the roots of the equation } x^{2}+p x+q=0, \text{ then } \\
\Delta=p^{2}-4 q \geqslant 0 . \\
\sin A+\sin B=-p, \sin A \sin B=q . \\
\because \sin A>0, \sin B>0, \\
\therefore p<0 . \\
\b... | p<0, \quad p^{2}-2 q=1, \quad q \leqslant \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,219 |
2. As shown in the figure, point $P$ inside the equilateral triangle $A B C$ draws perpendiculars to the three sides, $P Q=6, P R=8, P S$ $=10$. Then the area of $\triangle A B C$ is
(A) $190 \sqrt{3}$
(B) $192 \sqrt{3}$
(C) $194 \sqrt{3}$
(D) $196 \sqrt{3}$ | 2. 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 | 709,221 |
3. If real numbers $a, b, c$ satisfy $a+b+c=0, abc=2, c>0$, then ( ).
(A) $ab<0$
(B) $|a|+|b| \geqslant 2$
(C) $|a|+|b| \geqslant 4$
(D) $0<|a|+|b| \leqslant 1$ | 3. B
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 | 709,222 |
5. $p, q$ are prime numbers and are roots of the equation $x^{2}-13 x+m=0$. Then, the value of $\frac{q}{p}+\frac{p}{q}$ is ( ).
(A) $\frac{121}{22}$
(B) $\frac{123}{22}$
(C) $\frac{125}{22}$,
(D) $\frac{127}{22}$ | 5. 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 | 709,224 |
2. Given $a, b, c$ are distinct. Find the value of $\frac{2 a-b-c}{(a-b)(a-c)}$ $+\frac{2 b-c-a}{(b-c)(b-a)}+\frac{2 c-a-b}{(c-a)(c-b)}$. | (Answer: 0) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,228 |
Three. (Full marks 12 points) If the two legs of a right triangle are integers and are roots of the equation $m x^{2}-2 x-m+1=0$ ($m$ is an integer), does such a triangle exist? If it exists, find the lengths of all sides of the triangle that meet the conditions; if not, explain the reason.
---
Translate the above te... | Three, if $m=0$, then $x=\frac{1}{2}$. At this point, the original equation should have no integer roots, so $m \neq 0$. Therefore, $x_{1,2}=\frac{1 \pm \sqrt{m^{2}-m+1}}{m}$.
When $m=1$, $x=2$ or 0, and such a right-angled triangle does not exist.
Assume there exists an integer $m$ other than 0 or 1, such that the e... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,232 |
Four, (Full marks 14 points) As shown in the figure, line $AB$ intersects $\odot O$ at points $E$ and $F$, where $EF$ is the diameter of $\odot O$, and $AE = EF = FB$. Line $AP$ is perpendicular to the radius $OD$ of $\odot O$ at $D$. Prove: $\angle ADE = \angle PDB$ | Extend $D O$ to intersect
$\odot O$ at $M$, connect $A M$, extend $D E$ to intersect $A M$ at $N$, then $\triangle O A M \cong \triangle O B D$, so $\angle O A M=\angle O B D$, which means $A M \parallel B D$, hence $\angle P D B$ $=\angle D A N$.
Since $A E=E F, O$ is the midpoint of $E F$ and $D M$, then $E$ is the c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,233 |
Five. (Full marks 14 points) Find the non-negative integer solutions $x, y, z$ that satisfy the equation $2^{x}+3^{y}=z^{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | If $y=0, 2^{x}=(z-1)(z+1)$, when $z-1=2$, $z=3, z+1=4, x=3$. Thus, we get a solution $(x, y, z)=(3,0,3)$.
If $y>0$, let $x=2a+1$ ($a \geqslant 0$ be an integer), $3^{y}=z^{2}-2^{2a+1}$. Then $z^{2} \equiv 0,1(\bmod 3)$.
$$
2^{2a+1}=\left(2^{2}\right)^{a} \cdot 2^{1} \equiv 1^{a} \cdot 2=2(\bmod 3).
$$
Then $z^{2}-2^{... | (x, y, z) = (3, 0, 3), (0, 1, 2), (4, 2, 5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,234 |
1. If positive numbers $a, b, c$ satisfy the inequality system
$$
\left\{\begin{array}{l}
\frac{11}{6} c<a+b<2 c, \\
\frac{3}{2} a<b+c<\frac{5}{3} a, \\
\frac{5}{2} b<a+c<\frac{11}{4} b .
\end{array}\right.
$$
then the size relationship of $a, b, c$ is ().
(A) $a<b<c$
(B) $b<c<a$
(C) $c<a<b$
(D) cannot be determined | 1. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 709,235 |
2. Given that $a$, $b$, and $c$ are positive numbers. If the quadratic equation $a x^{2}+b x+c=0$ has two real roots, then the equation $a^{2} x^{2}+b^{2} x+c^{2}=0$ ( ).
(A) has two distinct positive roots
(B) has one positive root and one negative root
(C) has two distinct negative roots
(D) does not necessarily have... | 2. 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 | 709,236 |
4. In the acute triangle $\triangle A B C$, it is known that two sides are $a=1, b=3$. Then, the range of the third side is ( ).
(A) $2<c<4$
(B) $2<c \leqslant 3$
(C) $2<c<\sqrt{10}$
(D) $2 \sqrt{2}<c<\sqrt{10}$ | 4. D
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 | 709,238 |
Example 3 Simplify:
$$
\frac{a^{3}(c-b)+b^{2}(a-c)+c^{3}(b-a)}{a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)} \text {. }
$$ | Let $c-b=x, a-c=y, b-a=z$. Then $x+y+z=0, z=-(x+y)$.
Original expression $=\frac{a^{3} x+b^{3} y-c^{3}(x+y)}{a^{2} x+b^{2} y-c^{2}(x+y)}$
$=\frac{x\left(a^{3}-c^{3}\right)+y\left(b^{3}-c^{3}\right)}{x\left(a^{2}-c^{2}\right)+y\left(b^{2}-c^{2}\right)}=$
$\frac{x(a-c)\left(a^{2}+a c+c^{2}\right)+y(b-c)\left(b^{2}+b c+c^... | a+b+c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,239 |
$\begin{array}{c}\text { 3. Let } a>0, b>0, \text { and } \\ \sqrt{a}(\sqrt{a}+2 \sqrt[3]{b}) \\ =\sqrt[3]{b}(\sqrt[3]{a}+6 \sqrt[3]{b}) . \\ \text { Find the value of } \frac{2 a^{4}+a^{3} b-128 a b^{2}-64 b^{3}+b^{4}}{a^{4}+2 a^{3} b-64 a b^{2}-128 b^{3}+2 b^{4}} .\end{array}$ | (Answer: $\frac{1}{2}$ ) | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,240 |
1. Real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$. If $x=$
$$
\begin{array}{l}
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}\right. \\
\left.+\frac{1}{b}\right) \text {. Then, } x+2 y+3 x y=
\end{array}
$$
. | 1. 2
The above text has been translated into English, maintaining the original text's line breaks and format. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,243 |
3. Given that $E$ and $F$ are points on the sides $BC$ and $CD$ of square $ABCD$, respectively, and $AE$ and $AF$ intersect the diagonal $BD$ at $M$ and $N$. If $\angle EAF=$ $50^{\circ}$, then $\angle CME+\angle CNF=$ $\qquad$ | 3. $100^{\circ}$ | 100^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,245 |
Three. (Full marks 12 points) Solve the equation:
$$
\frac{13 x-x^{2}}{x+1}\left(x+\frac{13-x}{x+1}\right)=42 \text {. }
$$ | Three, let $y=\frac{13-x}{x+1}$. The original equation becomes
$x y(x+y)=42$.
$\quad x y+(x+y)=\frac{13 x-x^{2}}{x+1}+\frac{x^{2}+13}{x+1}=13$.
From (1), (2), and Vieta's formulas, we know that $x y$ and $x+y$ are the roots of the equation $z^{2}-13 z+42=0$. Solving this, we get $z_{1}=6, z_{2}=7$.
Thus, we have $\left... | x_{1}=1, x_{2}=6, x_{3}=3+\sqrt{2}, x_{4}=3-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,250 |
Four, (Full marks 14 points) Does there exist a real number $k$ such that the quadratic equation $x^{2}+(2 k-1) x-(3 k+2)=0$ has two real roots, and both roots are between 2 and 4? If so, determine the range of $k$; if not, explain the reason.
Translate the above text into English, please retain the original text's li... | Let the function $y=f(x)=x^{2}+(2 k-1) x-(3 k+2)$. Then its graph is a parabola opening upwards, with its vertex below the $x$-axis, and between $x=2$ and $x=4$. The parabola intersects the $x$-axis at points also between $x=2$ and $x=4$. Therefore, the value of $k$ that meets the requirements should satisfy the follow... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,252 |
Five. (Full marks 14 points) As shown in the figure, $P$ is a point on $BC$ of $\triangle ABC$, and $PC=2PB$. It is known that $\angle ABC=45^{\circ}$, $\angle APC=60^{\circ}$. Find $\angle ACB$.
保留源文本的换行和格式，直接输出翻译结果如下：
Five. (Full marks 14 points) As shown in the figure, $P$ is a point on $BC$ of $\triangle ABC$, an... | Five, make the symmetric point $C_{1}$ of point $C$ with respect to line $A P$, and connect $B C_{1}$, $P C_{1}$, and $A C_{1}$. Then $C_{1} P = C P = \angle B P$, $\angle C_{1} P B = 180^{\circ} - \angle A P C - \angle A P C_{1} = 60^{\circ}$. Therefore, $A C^{\prime} P B$ is similar to a right triangle with a hypoten... | 75^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,253 |
Six, (Full marks 12 points) On a circle, there are
12 points, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Among the convex polygons formed by these points, those whose vertices include both the red and blue points are called bicolored polygons; those that include on... | For any bicolored $n(n \geqslant 5)$-gon, naturally, removing the red and blue vertices results in a colorless $(n-2)$-gon. Different bicolored $n$-gons, after removing the red and blue vertices, yield different colorless $(n-2)$-gons; conversely, for any colorless polygon, adding red and blue vertices can always resul... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,254 |
1. Kenneth starts from 633, calculates the sum of the cubes of each digit, and then repeats this operation on the resulting number. Then,
(A) he will eventually get a three-digit number that equals the sum of the cubes of its digits
(B) he will get a different three-digit number each time
(C) he will eventually get a n... | 1. A.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,255 |
2. Both parents agreed to donate a certain amount, an integer number of dollars, to the children's "allowance reserve fund." Ace pointed out: "The product of their donations is zero." Bie pointed out: “Neither of them donated a single cent.” Sek pointed out: “If what Ace said is true, then what Bie said is also true.” ... | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 709,256 |
3. The left side of the figure below is a - - $5 \times 5$ chessboard, with a 1-shaped hole between 1 | 1. The right side represents a chess piece. Now, it needs to be placed on the chessboard so that it exactly covers 3 out of 20 small squares. The piece can be rotated when placed.
(A) The chess piece has 12 different... | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 709,257 |
4. Each angle of $\triangle A B C$ has a degree measure that is a positive integer. $\angle A$ is not less than the other two angles, $\angle C$ is not greater than the other two angles, and it is known that $\angle C$ equals $30^{\circ}$. Then, ( ).
(A) The maximum degree measure of $\angle A$ is $75^{\circ}$
(B) The ... | 4. C
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 | 709,258 |
3. There is a regular six-sided die. Bernie rolled it once, and Jeremiah rolled it twice. What is the probability that Jeremiah gets at least one higher number than Bernie? | 3. $\frac{125}{216}$ | \frac{125}{216} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,261 |
5. Given $\frac{x+y-z}{z}=\frac{x-y+z}{y}=\frac{-x+y+z}{x}$, and $x y z \neq 0$. Find the value of the fraction $\frac{(x+y)(y+z)(z+x)}{x y z}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | (Answer: -1 or 8 ) | -1 \text{ or } 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,262 |
1. Different letters represent different digits, and the same letter always represents the same digit.
(a) The four-digit number $A D A M$ is the fourth power of some number, and the three-digit number $E V E$ is the cube of some number. Find $A D A M$ and $E V E$.
(b) Adam divides the four-digit number $E D E N$ by $\... | Three, 1. (a) Since $10^{4}=10000, \sqrt[1]{A D A M}$ - must be a $\therefore$ digit number. $5^{4}=625$ is too small, while $6^{4}=1296,7^{4}=2401,8^{4}=$ 4096 do not fit the given form, therefore, $A D A M=9^{4}=$ 6561. Also, since $10^{3}=1000, \sqrt[3]{E V E}$ must also be a one-digit number. $4^{3}=64$ is too smal... | V | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,264 |
2. Anita, Kieran, and Mitchell form a team to participate in a mixed cycling and running race, which takes place on a 3-kilometer circular track. It is known that each person's running speed is 5 kilometers per hour, and their cycling speed is 20 kilometers per hour. The starting point of the track is marked by a flag,... | 2. (a) It is easy to see that each person takes 12 minutes to run 1 km and 3 minutes to cycle 1 km. At the start, let Anita and Kemai cycle, while Mitchell runs. When they reach the 1 km mark, Kemai gives the bike to Mitchell and starts running. When they reach the 2 km mark, Anita gives the bike to Kemai and starts ru... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 709,265 |
3. Anton has two $4 \times 6$ boxes, one containing 8 $V$-shaped pieces, and the other containing 6 $L$-shaped pieces, as shown in the figure below.
(a) Anton lost some pieces from the first box... and can only replace it with a $1 \times 3$ rectangle. How can Anton fit this new piece and the 7 remaining old pieces int... | 3. (a) The answer is as shown in the figure.
(b) It is impossible. Color the 4 rows of the box alternately with black and white. When placing the new piece in the box, it will cover 4 small squares, of which the number of black squares is even. In addition, when placing the 5 old pieces in the box, each one will also c... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 709,266 |
4. The triangle inequality states that -
the sum of the lengths of any two sides of a triangle is greater
than the length of the third side. In the figure on the right, $E$
is on the line segment $C A$, and $D$ is on the line segment $B E$.
(a) Explain why $A B+A E>D B+D E$;
(b) Explain why $A B+A C>D B+D C$;
(c) Whic... | 4. (a) By the triangle inequality, $A B + A E > B E = D B + D E$.
(b) By the triangle inequality, $D E + E C > D C$, therefore,
$$
\begin{array}{l}
A B + A C = A B + A E + E C \\
> D B + D E + E C > D B + D C .
\end{array}
$$
(c) By the triangle inequality, $D A + D B > A B$, $D B + D C > B C$ and $D C + D A > C A$. Ad... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,267 |
1. Let $n$, $b$, and $k$ be integers, and $\left(k-\frac{b}{2}\right)^{2} \leqslant n$. The equation $x^{2}+b x-k^{2}+k b+n=0$ has real roots. Consider the following four conclusions:
(1) The equation must have two distinct real roots;
(2) The equation has exactly one integer root;
(3) $b$ is divisible by 2;
(4) $n$ is... | -1 . (B).
Since the equation has real roots, its discriminant $\Delta \geqslant 0$, i.e., $\left(k-\frac{b}{2}\right)^{2}=n$. From the given, we have $\left(k-\frac{b}{2}\right)^{2}=n$, which implies that $b$ is an integer multiple of 2. By $\Delta=0$, we know that $n$ is a perfect square or zero, and the equation has ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,268 |
2. The integer part of the number $(\sqrt{2}+\sqrt{3}-\sqrt{6})(-\sqrt{2}$ $+\sqrt{3}+\sqrt{6})(\sqrt{2}-\sqrt{3}+\sqrt{6})$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. (B).
Obviously, the original number equals $\frac{23}{\sqrt{2}+\sqrt{3}+\sqrt{6}}$, and $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is greater than 5.6 but less than 5.7, so the original number is greater than 4 and less than 4.2, and its integer part equals 4. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,269 |
3. The number of positive integers $n$ such that $2 n+1$ divides $n^{3}+2$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. (C).
Since $2 n+1$ is coprime with 2, the problem is equivalent to $\mathrm{2} \frac{2^{3}\left(n^{3}+2\right)}{2 n+1}$ $=\frac{(2 n)^{3}+16}{2 n+1}=\frac{(2 n+1-1)^{3}+16}{2 n+1}$ being an integer, which is also equivalent to $\frac{16-1}{2 n+1}=\frac{3 \times 5}{2 n+1}$ being an integer. Therefore, $2 n+1=3,5,15$... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,270 |
4. As shown in the figure, it is known that $A B C D E F$ is a regular hexagon, $M$ and $N$ are the midpoints of sides $C D$ and $D E$ respectively, and $A M$ intersects $B N$ at point $P$. Then $\angle A P B$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 4. (C).
As shown in the figure, connect $BD$ and $AC$, then $\angle BDN = \angle ACM = 90^{\circ}$, $BD = AC$, $DN = CM$, so $\triangle BDN \cong \triangle ACM$. Therefore, $\angle PND = \angle PMC$, which means points $P$, $N$, $D$, and $M$ are concyclic. Hence, $\angle APB = \angle MPN = 180^{\circ} - \angle MDN = 1... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,271 |
5. The number of right-angled triangles with integer lengths of the two legs and whose perimeter is numerically equal to an integer multiple of its area is ( ).
(A) 1
(B) 2
(C) 3
(D) infinitely many | 5. (C).
Let the two legs of the right triangle be positive integers $a, b(a \leqslant b)$. Then, by the problem, we have $a+b+\sqrt{a^{2}+b^{2}}=k \cdot \frac{1}{2} \cdot a b$. ( $k$ is a positive integer) Rearranging and squaring both sides, we get
$$
\begin{array}{l}
(k a-4)(k b-4)=8 . \\
\text { Therefore }\left\{\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,272 |
6. As shown in the figure, in the convex quadrilateral $A B C D$, $A B=2$, $P$ is the midpoint of side $A B$. If $\angle D A B=\angle A B C$ $=\angle P D C=90^{\circ}$, then the minimum possible value of the area of quadrilateral $A B C D$ is ( ).
(A) 4
(B) $3 \sqrt{2}$
(C) $\frac{9}{2}$
(D) $2+2 \sqrt{21}$ | 6. (A).
If the garden is set as $A D=x, B C$ $=y$, then by the given information we get $P D=\sqrt{1+x^{2}}, \quad P C=$ $\sqrt{1+y^{2}}, C D=\sqrt{y^{2}-x^{2}}$ (using the Pythagorean theorem).
Construct $D Q \perp B C, Q$ as the foot of the perpendicular, then $C Q=y-x, D Q=2$. Thus,
$$
y^{2}-x^{2}=2^{2}+(y-x)^{2},... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,274 |
1. $x^{5}+x+1$ when divided by $x^{2}+1$ yields a remainder of | $=.1 .2 x+1$.
Since $x^{5}+x+1=x\left(x^{4}+1\right)+1=x\left(\left(x^{2}+1\right)^{2}-\right.$
$$
\begin{array}{l}
\left.2 x^{2}\right)+1=x\left(x^{2}+1\right)^{2}-2 x^{3}+1=x\left(x^{2}+1\right)^{2}-2 x\left(x^{2}\right. \\
+1)+2 x+1=\left(x^{3}-x\right)\left(x^{2}+1\right)+2 x+1,
\end{array}
$$
then $x^{\sqrt{1}}+x... | 2x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,275 |
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