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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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2. The number of all integer pairs $(x, y)$ that satisfy the equation $y^{4}+2 x^{4}+1=4 x^{2} y$ is ( ).
(A) 2
(B) 3
(C) 4
$(\Gamma) 5$ | 2. (A).
From the given, we have
$$
2\left(x^{2}-y\right)^{2}+\left(y^{2}-1\right)^{2}=0 \text {. }
$$
And $2\left(x^{2}-y\right)^{2} \geqslant 0, y^{2}-1 \geqslant 0$,
$$
\therefore x^{2}-y=0, y^{2}-1=0 \text {. }
$$
The integer pairs that satisfy the conditions are $(1,1),(-1,1)$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,680 |
3. If a straight line $l$ passes through three different points $A(a, b)$, $B(b, a)$, and $C(a-b, b-a)$, then the line $l$ passes through ( ) quadrants.
(A) II, IV
(B) I, III
(C) II, III, IV
(D) I, III, IV | 3. (A).
The equation of the line passing through $A(a, b)$ and $B(b, a)$ is $y=-x+a+b$.
Substituting the coordinates of point $C(a-b, b-a)$ into the above equation yields $a+b=0$.
Therefore, the equation of line $l$ is $y=-x$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,681 |
4. Given that the three sides $a$, $b$, and $c$ of $\triangle A B C$ satisfy $\frac{3}{a}=\frac{2}{b}+\frac{1}{c}$. Then $\angle A()$.
(A) is acute
(B) is right
(C) is obtuse
(D) is not a right angle | 4. (A).
From $\frac{3}{a}=\frac{2}{b}+\frac{1}{c}$, it is easy to deduce that
$$
2 c(b-a)=b(a-c) \text {. }
$$
If $a < c$, then $b < a$. In this case, $b^{2}+c^{2}>a^{2}$;
If $a \geqslant c$, then $b \geqslant a$. In this case, $b^{2}+c^{2}>a^{2}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,683 |
5. As shown in Figure 1, two circles are concentric, with radii of $2 \sqrt{6}$ and $4 \sqrt{3}$, respectively. The sides $A B$ and $C D$ of rectangle $A B C D$ are chords of the two circles. When the area of the rectangle is maximized, its perimeter is equal to ( ).
(A) $22+6 \sqrt{2}$
(B) $20+8 \sqrt{2}$
(C) $18+10 \... | 5. (D).
As shown in Figure 4, connect $O A$, $O B$, $O C$, $O D$, draw $O E \perp A D$, and extend $E O$ to intersect $B C$ at $F$. Clearly, $O F \perp B C$. It is easy to see that
$$
\triangle O A D \cong \triangle O B C \text {, and } S_{\text {rhombus } A B C D}=4 S_{\triangle O A D} \text {. }
$$
When $S_{\text {... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,684 |
1. Given $a, b, c$ are all positive integers, and satisfy the conditions
$$
a^{2}-b^{2}-c^{2}=a b c, \quad a^{2}=2(b+c) .
$$
Find $a=$ $\qquad$ ,$b=$ $\qquad$ ,$\dot{c}=$ $\qquad$ | $$
\begin{array}{l}
\text { II. 1. } 2,1,1 . \\
\because a>0, b>0, c>0, \\
\therefore a b c>0, \text { i.e., } a^{2}-b^{2}-c^{2}>0 . . \\
\text { Thus, } a>b, a>c . \\
\text { Also, } \because a^{2}=2(b+c)<4 a, \\
\therefore a<4 .
\end{array}
$$
From $a^{2}=2(b+c)$, we know that $a$ is an even number,
$$
\therefore a=... | 2,1,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,686 |
2. When $m$ takes all real values from 0 to 5, the number of integer $n$ that satisfies $3 n=m(3 m-8)$ is $\qquad$ . | 2. 13 .
Let $n$ be a quadratic function of $m$, i.e.,
$$
n=m^{2}-\frac{8}{3} m \text {, }
$$
then
$$
n=\left(m-\frac{4}{3}\right)^{2}-\frac{16}{9} \text {. }
$$
From the graph, when $0 \leqslant m \leqslant 5$,
$$
\begin{array}{l}
-\frac{16}{9} \leqslant n \leqslant \frac{35}{3} . \\
\therefore n=-1,0,1,2,3,4,5,6,7,... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,687 |
3. The three sides of one triangle are $a, a, b$; the three sides of another triangle are $b, b, a (a \neq b)$, and the smallest interior angle of both triangles is equal to $\alpha$. Then $\alpha=$ $\qquad$ $\frac{a}{b}=$ $\qquad$ | 3. $36 \cdot, \frac{\sqrt{5}-1}{2}$ or $\frac{\sqrt{5}+1}{2}$.
As shown in Figure 6, two triangles are combined to form a quadrilateral $ABCD$, with $AC$ intersecting $BD$ at $K$. It is easy to see that $ABCD$ is an isosceles trapezoid, $\angle A = \angle B = 2\alpha$. At this point, $\angle ADB = 2\alpha$.
From $\tri... | 36^{\circ}, \frac{\sqrt{5} - 1}{2} \text{ or } \frac{\sqrt{5} + 1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,688 |
One, (20 points) Find all positive real numbers $a$ such that the equation $x^{2}-a x+4 a=0$ has only integer roots.
untranslated text is preserved in the response, including the formatting and structure. | Let $\Delta=(-a)^{2}-4 \times 4 a \geqslant 0$, and $a>0$, then $a \geqslant 16$.
Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-a x+4 a=0$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=a, x_{1} x_{2}=4 a \text{. }
$$
Thus, $x_{1} x_{2}-4 x_{1}-4 x_{2}=0$,
or $\left(x_{1}-4\right)\left(x_{2}-4\right)=16$.
Consi... | a=25,18,16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,690 |
II. (25 points) As shown in Figure 3, on the sides $AB$ and $AD$ of square $ABCD$, there are moving points $E$ and $F$ such that $EF = BE + DF$. Lines $EG \perp CD$ and $FH \perp BC$ are drawn. Is the intersection point $P$ of $EH$ and $FG$ a moving point or a fixed point? Prove your conclusion. | II. Point $P$ is always a fixed point.
Auxiliary lines are shown in Figure 8.
Let $DM = x, PM = y$, the side length of the square is 1,
$$
\begin{array}{l}
BE = m, DF = n, EF \\
= m + n.
\end{array}
$$
From the right triangle $\triangle AEF$,
we get
$$
\begin{array}{ll}
& (1-m)^{2}+(1-n)^{2}=(m+n)^{2}, \\
\therefore ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,691 |
2. Solve the equation: $[x]\{x\}+x=2\{x\}+10$.
(1991, Recommended by the Soviet Ministry of Education for Competition Problems) | (Given that the equation can be factored as $([x]-1)(\{x\}+1)$ $=9$. Since $[x]-1$ is an integer, we know that $\{x\}+1$ is a rational number, and we can set $\{x\}$ $=\frac{n}{k}, 0 \leqslant n<k$. Then simplify the equation. $x=10,7 \frac{1}{2}, 9 \frac{1}{8}$, $\left.8 \frac{2}{7}, 6 \frac{4}{5}.\right)$ | x=10,7 \frac{1}{2}, 9 \frac{1}{8}, 8 \frac{2}{7}, 6 \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,693 |
1. Let the sets be
$$
\begin{aligned}
S= & \left\{(x, y) \left\lvert\, x-\frac{1}{2} y^{2}\right. \text { is odd, } x, y \in \mathbf{R}\right\}, \\
T= & \left\{(x, y) \mid \sin (2 \pi x)-\sin \left(\pi y^{2}\right)=\right. \\
& \left.\cos (2 \pi x)-\cos \left(\pi y^{2}\right), x, y \in \mathbf{R}\right\} .
\end{aligned... | $-1 .(\mathrm{A})$.
When $x=\frac{1}{2} y^{2}+$ odd number, it is obvious that
$$
\sin (2 \pi x)-\sin \left(\pi y^{2}\right)=\cos (2 \pi x)-\cos \left(\pi y^{2}\right)
$$
holds.
Therefore, when $(x, y) \in S$, $(x, y) \in T$. Thus $S \subseteq T$.
Moreover, points $(x, y) \in T$ that satisfy $x=\frac{1}{2} y^{2}$ do n... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,694 |
2. Given $\alpha \in\left(0, \frac{\pi}{4}\right), t_{1}=(\tan \alpha)^{\tan \alpha}, t_{2}=$ $(\tan \alpha)^{\cot \alpha}, t_{3}=(\cot \alpha)^{\tan \alpha}, t_{4}=(\cot \alpha)^{\cot \alpha}$. Then the size relationship of $t_{1}, t_{2}, t_{3}, t_{4}$ is ( ).
(A) $t_{1}<t_{2}<t_{3}<t_{4}$
(B) $t_{2}<t_{1}<t_{3}<t_{4}... | 2. (B).
$$
\begin{array}{l}
\because \alpha \in\left(0, \frac{\pi}{4}\right), \\
\therefore 0<\tan \alpha<1<\cot \alpha . \\
\therefore(\tan \alpha)^{\tan \alpha}<(\cot \alpha)^{\tan \alpha}, \\
(\cot \alpha)^{\tan \alpha}<(\cot \alpha)^{\cot \alpha}, \\
(\tan \alpha)^{\cos \alpha}<(\tan \alpha)^{\tan \alpha} .
\end{ar... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,695 |
3. Given a function $f(x)$ defined on $\mathbf{R}$ with period $T$ that satisfies $f(1+x)=f(1-x)$ and $f(8+x)=f(8-x)$. Then the maximum value of $T$ is ( ).
(A) 16
(B) 14
(C) 8
(D) 2 | 3. (B).
$$
\begin{array}{l}
f(x)=f(1+(x-1))=f(1-(x-1)) \\
=f(2-x)=f(8+(-6-x)) \\
=f(8-(-6-x))=f(14+x) .
\end{array}
$$
According to the definition of the period, we know that $T \leqslant 14$. Figure 1 is the graph of a function that meets the conditions with a period of 14. Therefore, the maximum value of $T$ is 14. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,696 |
4. Given four non-zero complex numbers $a, b, c, d$ satisfying $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$. Then the number of different values that $\frac{a-b+c+d}{a+b-c+d}$ can take is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. (D).
Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then $a=b k=c k^{2}=d k^{3}=a k^{4}$.
That is $\square$
$$
\begin{array}{l}
b=a k^{3}, c=a k^{2}, d=a k, a=u k^{4} . \\
\therefore k^{4}=1 . \text { Hence } k= \pm 1, \pm \mathrm{i} . \\
\therefore \frac{a}{a} \mp \frac{b+c+d}{b-c+d}=\frac{a+(-b)+c+d}{b+... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,697 |
5: For any function $f(x)$, in the same Cartesian coordinate system, the graphs of the functions $y_{1}=f(x-19)$ and $y_{2}=$ $f(99-x)$ are always symmetric with respect to ( ).
(A) $y$-axis
(B) line $x=19$
(C) line $x=59$
(D) line $x=99$ | 5. (C).
$$
\begin{array}{l}
\text { Let } x^{\prime}=x-59, y^{\prime}=y . \text { Then } \\
y_{1}^{\prime}=f\left(x^{\prime}+40\right), y_{2}^{\prime}=f\left(40-x^{\prime}\right) .
\end{array}
$$
Since the function $y=f(a+x)$ is always symmetric to the function $y=f(a-x)$ about the $y$-axis $(x=0)$,
then $y_{1}^{\prim... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,698 |
6. The number of elements in the set $\left\{(x, y, z) \left\lvert\, \log _{\frac{1}{4}}\left(x^{4}+y^{4}+z^{4}+1\right)\right.\right.$ $\left.\geqslant \log _{\frac{1}{4}} x+\log _{\frac{1}{4}} y+\log _{\frac{1}{4}} z-1\right\}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) More than 2 | 6. (B).
From the known inequality, we get $x^{4}+y^{4}+z^{4}+1 \leqslant 4 x y z$ and $x, y, z$ are all positive numbers. Also, from the important inequality, we have $x^{4}+y^{4}+z^{4}+1 \geqslant 4 x y z$ (the equality holds if and only if $x=y=z=1$).
$$
\therefore x^{4}+y^{4}+z^{4}+1=4 x y z \text {. }
$$
Thus, $x... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 711,699 |
1. The range of the function
$$
f(x)=12 \operatorname{arccot}|x|+2 \arccos \frac{1}{2} \sqrt{2 x^{2}-2}
$$
is $\qquad$ . | $$
\begin{array}{l}
=1 \cdot[2 \pi, 4 \pi] . \\
\because 0 \leqslant \frac{1}{2} \sqrt{2 x^{2}-2} \leqslant 1, \\
\therefore 1 \leqslant|x| \leqslant \sqrt{3} .
\end{array}
$$
Thus $\operatorname{arccot} \sqrt{3} \leqslant \operatorname{arccot}|x| \leqslant \operatorname{arccot} 1$, $\arccos 1 \leqslant \arccos \frac{... | 2 \pi \leqslant f(x) \leqslant 4 \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,700 |
2. Given the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \geqslant b>0)$ passes through the point $(\sqrt{3}, 1)$. All these curves that satisfy $y \geqslant 1$ are $\qquad$ . | 2. $\frac{4 \pi}{3}-\sqrt{3}$.
From the problem, we have $\frac{3}{a^{2}}+\frac{1}{b^{2}}=1$,
$$
\begin{array}{l}
\therefore b^{2}=\frac{a^{2}}{a^{2}-3} . \\
\because a \geqslant b, \therefore a^{2} \geqslant \frac{a^{2}}{a^{2}-3}, \text { so } a^{2} \geqslant 4 . \\
\because \frac{x^{2}}{a^{2}}+\frac{\left(a^{2}-3\rig... | \frac{4 \pi}{3}-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,701 |
4. In a planar convex quadrilateral with an area of $18 \mathrm{~cm}^{2}$, the sum of one pair of opposite sides and one diagonal is $12 \mathrm{~cm}$. Then the minimum value of the sum of the other diagonal and the other pair of sides is $\qquad$
Translate the text above into English, please keep the original text's ... | $4.6(\sqrt{2}+\sqrt{5}) \text{ cm}$.
Consider a plane convex quadrilateral $ABCD$, where the diagonal $AC$ and the sum of the lengths of one pair of opposite sides $AB$ and $CD$ is $12 \text{ cm}$. Let $AB = b$, $AC = a$, $CD = c$, $\angle BAC = \alpha$, and $\angle ACD = \beta$ (as shown in Figure 3). Then,
$$
a + b +... | 6(\sqrt{2} + \sqrt{5}) \text{ cm} | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 711,703 |
3. Solve the equation: $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
(1972, Kyiv Mathematical Olympiad) | (Note that $\left[x^{3}\right] 、\left[x^{2}\right] 、[x] 、 1$ are all integers, thus $\{x\}=0$, and hence $x$ is an integer; solving yields $x=-1$.) | x=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,704 |
5. In the expansion of $(a+b)^{n}$, there are $n+1$ different terms. Then, the expansion of $(a+b+c+d)^{21}$ has different terms. | 5.2024 .
$$
\begin{array}{l}
\because(a+b+c)^{n} \\
=(a+b)^{n}+\mathrm{C}_{n}^{1}(a+b)^{n-1} \cdot c+\cdots+\mathrm{C}_{n}^{n} \cdot c^{n}
\end{array}
$$
There are $n+1$ terms, but $(a+b)^{i}$ has $i+1$ terms, $\therefore(a+b+c)^{n}$ has the number of different terms as
$$
\begin{array}{l}
\sum_{i=0}^{n}(i+1)=\frac{1}... | 2024 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,705 |
6. In the tetrahedron $S-ABC$, four edges are of length $\sqrt{2}$, and the other two edges are of lengths $\sqrt{3}$ and 2. Then the angle between the longer two edges is $\qquad$ . | 6. $\frac{\pi}{2}$ or $\arccos \frac{5 \sqrt{3}}{12}$.
(1) When the two longer edges are opposite, let $S A=\sqrt{3}, R C$ $=2$. As shown in Figure 5, $D$ is the midpoint of $B C$, connect $S D$ and $A D$.
$\because A B=A C=S B=S C$,
$\therefore S D \perp B C, A D \perp B C$.
$\therefore B C \perp$ plane $S A D$.
$\the... | \frac{\pi}{2} \text{ or } \arccos \frac{5 \sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,706 |
$\begin{array}{c}\text { Three, (20 points) In } \triangle A B C \text {, the real number } x \text { satisfies } \\ \sec ^{2} x=\csc ^{2} A+\csc ^{2} B+\csc ^{2} C . \\ \text { Prove: } \cos \left[x+(-1)^{n} A\right] \cdot \cos \left[x+(-1)^{n} B\right] \cdot \\ \cos \left[x+(-1)^{n} C\right]+\cos ^{3} x=0 .(n \in \ma... | $$
\begin{array}{l}
\text { Three, in } \triangle A B C \text {, it is always true that } \\
\cot A \cdot \cot B+\cot B \cdot \cot C+\cot C \cdot \cot A=1 \text {. } \\
\because \sec ^{2} x=\csc ^{2} A+\csc ^{2} B+\csc ^{2} C \text {, } \\
\therefore \tan ^{2} x=\cot ^{2} A+\cot ^{2} B+\cot ^{2} C+2 \\
=\cot ^{2} A+\co... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,707 |
Four. (20 points) Given a fixed circle $\odot P$ with radius 1, the distance from the center $P$ to a fixed line $l$ is 2. $Q$ is a moving point on $l$, and $\odot Q$ is externally tangent to $\odot P$. $\odot Q$ intersects $l$ at points $M$ and $N$. For any diameter $MN$, there is a fixed point $A$ in the plane such t... | $$
\begin{array}{rl}
& \text{Let } l \text{ be the } x \text{-axis, and the perpendicular from point } P \text{ to } l \text{ be the } y \text{-axis, as shown in Figure 6.} \\
& \text{Suppose the coordinates of } Q \text{ are } (x, 0), \text{ and point } A \text{ is at } (k, h). \\
& \text{The radius of } \odot Q \tex... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,708 |
Five. (20 points) Let the functions $f(x)$ and $g(x)$ be defined as
$$
\begin{array}{l}
f(x)=12^{x}, g(x)=2000^{x} . \\
a_{1}=3, a_{n+1}=f\left(a_{n}\right)(n \in \mathbf{N}), \\
b_{1}=2000, b_{n+1}=g\left(b_{n}\right)(n \in \mathbf{N}) .
\end{array}
$$
Find the smallest positive integer $m$ such that $b_{m}>a_{2000}$... | Obviously, $12^{3}8 b_{n} \text {. }$
When $n=1$, $a_{3}=12^{16^{3}}>8 b_{1}$.
Assume that equation (1) holds for $n=k$, i.e., $a_{k+2}>8 b_{k}$.
When $n=k+1$,
$$
\begin{array}{l}
a_{k+3}=12^{a_{k+2}}>12^{8 b_{k}}=\left(12^{8}\right)^{b_{k}} \\
>(8 \times 2000)^{b_{k}}>8 \times 2000^{b_{k}}=8 b_{k+1} .
\end{array}
$$
... | 1999 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,709 |
II. (50 points) In quadrilateral $ABCD$, the circle $\odot O$ with center $O$ on side $AB$ is tangent to the other three sides, and $AB = AD + BC$. Prove that quadrilateral $ABCD$ is a trapezoid or a cyclic quadrilateral.
Translate the above text into English, please retain the original text's line breaks and format, ... | $$
\begin{array}{l}
\text{Let } \odot O \text{ be tangent to } AD \text{ and } BC \text{ at } E \text{ and } F \text{, respectively. Let the radius of } \odot O \text{ be } r, \angle OAD = \alpha, \\
\angle OBC = \beta, \angle ODA = \varphi, \text{ then} \\
\angle ODC = \varphi, \\
\angle OCB = \frac{1}{2} \angle BCD =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,711 |
Three, (50 points) Given two complex coefficient functions $f(x)$ $=\sum_{i=0}^{n} a_{i} x^{n-i}$ and $g(x)=\sum_{i=0}^{n} b_{i} x^{n-i}$ (where $a_{0}=$ $\left.b_{0}=1\right), \sum_{i=1}^{\left[\frac{n}{2}\right]} b_{2 i}$ and $\sum_{i=1}^{\left[\frac{n+1}{2}\right]} b_{2 i-1}$ are both real numbers, and the product o... | $$
\begin{array}{l}
\text{Let the } n \text{ roots of } g(x)=0 \text{ be } x_{i}(i=1, \cdots, n). \text{ Then} \\
g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right), f(x)=\prod_{i=1}^{n}\left(x+x_{i}^{2}\right) . \\
\therefore f(-1)=\prod_{i=1}^{n}\left(-1+x_{i}^{2}\right) \\
=\prod_{i=1}^{n}\left[\left(-1-x_{i}\right)\left(1-x_... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,712 |
In $\triangle A B C$, $\angle B=2 \angle C, P$ is a point inside the triangle, satisfying $A P=A B, P B=P C$.
Prove: $\angle P A C=\frac{1}{3} \angle B A C$. | Proof: If $\triangle \bar{\nu} C P \cong \triangle A B P$, it is easy to conclude that quadrilateral $A B C D$ is an axially symmetric figure, with its axis of symmetry being the perpendicular bisector of side $B C$ through point $P$.
Obviously, quadrilateral $A B C D$ is an isosceles trapezoid, and it must be cyclic,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,713 |
92. On a Chinese chessboard, the "Knight" is at its starting position.
(1) Prove that no matter how the "Knight" moves, it must take an even number of steps to capture the opponent's "General" at its starting position.
(2) If the river boundary is removed, can the "Knight" jump to every position on the board exactly on... | (1) Proof: By alternately dividing the intersections on the chessboard into two categories, the starting position of the "bird" over the "general" belongs to one category, and with each jump, the "horse" must jump to the other category. Therefore, the "horse" must take an even number of steps to capture the opponent's ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,714 |
4. Solve the equation: $x+\frac{92}{x}=[x]+\frac{92}{[x]}$. (1992, All-Russian Mathematical Olympiad) | (By decomposing the known equation, we get $x=[x]$ or $\frac{92}{x[x]}=$ 1. Then, set $x=[x]+r$, solve $\frac{92}{x[x]}=1$, and obtain $x=n$, where $n$ is a non-zero integer, or $x=-9.2$.) | x=n \text{ or } x=-9.2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,715 |
92. In an obtuse triangle $\triangle A B C$, $\angle A$ is the obtuse angle, $h_{a}$ is the altitude to side $a$. Prove: $a+h_{a}>b+c$. | Proof: $\because$ In $\triangle A B C$, $\frac{\angle A+\angle B}{2}=\frac{\pi-\angle C}{2}$,
$$
\therefore \tan \frac{A+B}{2}=\tan \frac{\pi - C}{2}=\cot \frac{C}{2} \text {. }
$$
That is, $\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$, thus
$$
\tan \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,717 |
Example 1 䒴 $\frac{4 x}{x^{2}-4}=\frac{a}{x+2}+\frac{b}{x-2}$, then the value of $a^{2}+$ $b^{2}$ is $\qquad$
(1996, Hope Cup National Mathematics Invitational Competition) | Solution: From the problem, we know that the given equation is not an equation in $x$, but an identity in $x$. From
$$
\frac{a}{x+2}+\frac{b}{x-2}=\frac{(a+b) x-2(a-b)}{x^{2}-4}
$$
we get the identity
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$
Accord... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,718 |
Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
(1995, National Junior High School Mathematics League) | Solution: The method of solving for the root $a$ by setting up an equation and then substituting to find the value is too cumbersome. It is better to use an overall processing method for this problem. From the given, we have
$$
a^{2}+a=\frac{1}{4} \text { or } a^{2}=\frac{1}{4}-a \text {. }
$$
Using equation (1), we s... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,719 |
Example 11 Suppose $1995 x^{3}=1996 y^{3}=1997 z^{3}$, $x y z>0$, and $\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=$ $\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
(1996, National Junior High School Mathematics League) | Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.
Assume $1995 x^{3}=1996 y^{3}=19... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,720 |
Example 12 Given that $x, y, z$ are 3 non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $s=2x+y-z$, then what is the sum of the maximum and minimum values of $s$?
(1996, Tianjin Junior High School Mathematics Competition) | The given equations and the expression for $s$ can be viewed as a system of equations in $x$, $y$, and $z$:
$$
\left\{\begin{array}{l}
3 x+2 y+z=5, \\
x+y-z=2, \\
2 x+y-z=s .
\end{array}\right.
$$
Solving this system, we get $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,721 |
Example 13 (i) $x, y$ are both integers. If $5 \mid (x + 9y)$, prove: $5 \mid (8x + 7y)$.
(ii) $x, y$ are both integers. If $11 \mid (7x + 2y - 5z)$, prove: $11 \mid (3x - 7y + 12z)$.
(1987, Beijing Junior High School Mathematics Competition) | Proof: (i) Since $5 \mid (x + 9y)$, we can write $x + 9y = 5m$ ($m$ is an integer).
Considering equation (1) as an equation in $x$, we solve for $x$:
$$
x = 5m - 9y.
$$
Thus, $8x + 7y = 8(5m - 9y) + 7y$
$$
= 40m - 65y.
$$
Clearly, $5 \mid (8x + 7y)$.
(ii) Since $11 \mid (7x + 2y - 5z)$, we can write $7x + 2y - 5z = 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,722 |
Example 1 As shown in Figure $1, A$ and $B$ are fixed points on the fixed lines $A X$ and $B Y$, respectively. $P$ and $R$ are two points on the ray $A X$, and $Q$ and $S$ are two points on the ray $B Y$. $\frac{A P}{B Q}=\frac{A R}{B S}$ is a fixed ratio. $M$, $N$, and $T$ are points on the segments $A B$, $P Q$, and ... | Solution: Let $\overrightarrow{A B}=\vec{a}, \overrightarrow{A R}=\vec{b}, \overrightarrow{A S}=\vec{c}, \overrightarrow{A M}=$ $m \overrightarrow{A B}, \overrightarrow{A P}=n \overrightarrow{A R}$. Then
$$
\begin{array}{l}
\overrightarrow{M T}=\overrightarrow{M A}+\overrightarrow{A R}+\overrightarrow{R T} \\
=-m \over... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,723 |
Example 2 As shown in Figure 2, $AC$ and $CE$ are two diagonals of the regular hexagon $ABCDEF$. Points $M$ and $N$ internally divide $AC$ and $CE$ such that $AM$ : $AC = CN : CE = r$. If points $B$, $M$, and $N$ are collinear, find $r$.
(23rd IMO) | Solution: Let $\overrightarrow{A C}=2 \vec{a}, \overrightarrow{A F}=2 \vec{b}$, then
$$
\begin{array}{l}
\overrightarrow{A M}=r \overrightarrow{A C}=2 r \vec{a}, \overrightarrow{A B}=\vec{a}-\vec{b} \text {. } \\
\text { Hence } \overrightarrow{B M}=\overrightarrow{B A}+\overrightarrow{A M}=\vec{b}+(2 r-1) \cdot \vec{a... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,724 |
Example 3 As shown in Figure 3, in $\triangle A B C$, $A B=$ $A C, D$ is the midpoint of $B C$, $E$ is the foot of the perpendicular from $D$ to $A C$, and $F$ is the midpoint of $D E$. Prove: $A F$ is perpendicular to $B E$.
(1962, All-Russian Mathematical Olympiad) | $$
\begin{array}{l}
\text { Prove: } \overrightarrow{A F} \cdot \overrightarrow{B E}=\overrightarrow{A F} \cdot(\overrightarrow{B C}+\overrightarrow{C E}) \\
=\left(\overrightarrow{A D}+\frac{1}{2} \overrightarrow{D E}\right) \cdot \overrightarrow{B C}+\left(\overrightarrow{A E}+\frac{1}{2} \overrightarrow{E D}\right) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,725 |
Example 4 As shown in Figure 4, on the arc $\overparen{A B}$ of the circumcircle of rectangle $A B C D$, take a point $M$ different from vertices $A$ and $B$. Points $P, Q, R, S$ are the projections of $M$ onto lines $A D, A B, B C, C D$ respectively. Prove that lines $P Q$ and $R S$ are perpendicular to each other.
(1... | Proof: Let $\overrightarrow{O H}=\vec{a}, \overrightarrow{H A}=\vec{b}, \overrightarrow{O G}=\vec{c}$, $\overrightarrow{G M}=\vec{d}$, and $\vec{a}, \vec{c}$ are perpendicular to $\vec{b}, \vec{d}$. Therefore, their dot products are zero.
$$
\begin{array}{l}
\text { Also, since } \overrightarrow{O A}=\vec{a}+\vec{b}, \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,726 |
Example 5 As shown in Figure 6, on the extensions of the sides $AB, BC, CD, DA$ of the convex quadrilateral $ABCD$, take points $A_{1}, B_{1}, C_{1}, D_{1}$ such that $\overrightarrow{A_{1} A}=\overrightarrow{A B}$, $\overrightarrow{B_{1} B}=\overrightarrow{B C}$, $\overrightarrow{C_{1} C}=\overrightarrow{C D}$, $\over... | Proof: Let $\overrightarrow{A B}=\vec{a}, \overrightarrow{A C}=\vec{b}, \overrightarrow{A D}=\vec{c}$. Then
$2 S_{\text {quadrilateral } A_{1} B_{1} C_{1} D_{1}}$
$$
\begin{array}{l}
=\left|\overrightarrow{A_{1} B_{1}} \times \overrightarrow{A_{1} C_{1}}+\overrightarrow{A_{1} C_{1}} \times \overrightarrow{A_{1} D_{1}}\... | S_{\text {quadrilateral } A_{1} B_{1} C_{1} D_{1}}=5 S_{\text {quadrilateral } A B C D} | Geometry | proof | Yes | Yes | cn_contest | false | 711,727 |
Example 6 Divide the sides $AB$ and $CD$ of the convex quadrilateral $ABCD$ into $m$ equal parts, and connect the corresponding division points. Then divide the sides $AD$ and $BC$ into $n$ equal parts, and connect the corresponding division points, where $m$ and $n$ are both odd numbers greater than 1. Find the ratio ... | Solution: As shown in Figure 7, from Example 1, it is easy to know that each intersection point equally divides the horizontal line $m$ and the vertical line $n$.
Let $\overrightarrow{A B}=\vec{a}$, $\overrightarrow{A C}=\vec{b}$, $\overrightarrow{A D}=\vec{c}$. Then the horizontal line where $\overrightarrow{A_{0} B_... | \frac{1}{m n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,728 |
Example 7 As shown in Figure $8, A B C D$ is any quadrilateral. Construct squares $A D S M$ and $B C F E$ outwardly on sides $A D$ and $B C$ respectively. Then construct squares $A C G P$ and $B D R Q$ on the diagonals $A C$ and $B D$ of the quadrilateral respectively. Prove: Quadrilateral $M E Q P$ is a parallelogram. | $$
\begin{array}{l}
\overrightarrow{A D}=\vec{a}+\vec{b}+\vec{c}, \overrightarrow{B E}=-\overrightarrow{\mathrm{k}} \times \vec{b}, \\
\text { then } \overrightarrow{A M}=\overrightarrow{\mathrm{k}} \times(\vec{a}+\vec{b}+\vec{c}), \\
\overrightarrow{E M}=\overrightarrow{\mathrm{k}} \times(\vec{a}+\vec{b}+\vec{c})-\vec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,729 |
Example 3 If $a^{2}+11 a=-16, b^{2}+11 b=$ $-16(a \neq b)$, then the value of $\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}}$ is $\qquad$
(1990, Shaoxing City Mathematics Competition) | Given the equation, we know that $a$ and $b$ are the two distinct real roots of the quadratic equation $x^{2} + 11x + 16 = 0$. According to the relationship between the roots and coefficients of a quadratic equation, we have
$$
\begin{array}{l}
a + b = -11, \quad ab = 16. \\
\text { Therefore, } \sqrt{\frac{b}{a}} + \... | \frac{11}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,730 |
Example 8 Point $O$ is inside a convex polygon $A_{1} A_{2} \cdots A_{n}$, consider all $\angle A_{i} O A_{j}$, where $i, j$ are different natural numbers from $1 \sim n$. Prove: among them, at least $n-1$ are not acute.
Translate the above text into English, please keep the original text's line breaks and format, and... | Proof: When $n=3$, the proposition is obviously true. And it is known that there exists a set of positive numbers $k_{1}, k_{2}, \cdots, k_{n}$ such that the vectors $\overrightarrow{O A_{i}}$ satisfy
$$
\sum_{i=1}^{n} k_{i} \overrightarrow{O A}_{i}=0 \text {. }
$$
Let $k_{i} \overrightarrow{O A}_{i}=\overrightarrow{a... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,731 |
Example 1 Write an $n$-digit number using the digits 1 or 2, where any two adjacent positions are not both 1. Let the number of $n$-digit numbers be $f(n)$. Find $f(10)$.
(Jiangsu Province Second Mathematical Correspondence Competition) | Solution: The $n$-digit numbers that meet the conditions can be divided into two categories:
I. The first digit is 2, then the number of $(n-1)$-digit numbers that meet the conditions is $f(n-1)$;
II. The first digit is 1, then the second digit should be 2, and the number of $(n-2)$-digit numbers is $f(n-2)$. Therefore... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,732 |
Example 2 Use $1$, $2$, and $3$ to write $n$-digit numbers, with the requirement that no two $1$s are adjacent. How many $n$-digit numbers can be formed? | Solution: Let the number of $n$-digit numbers that meet the condition be $a_{n}$. By dividing according to the first digit, we can categorize them as follows:
I. If the first digit is 2 or 3, then the following $n-1$ digits each have $a_{n-1}$ possibilities, totaling $2 a_{n-1}$;
II. If the first digit is 1, the second... | a_{n}=\frac{\sqrt{3}+2}{2 \sqrt{3}}(1+\sqrt{3})^{n}+\frac{\sqrt{3}-2}{2 \sqrt{3}}(1-\sqrt{3})^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,733 |
Example 3 Let $a_{n}$ be the number of natural numbers $N$ such that the sum of the digits of $N$ is $n$, and each digit can only be $1$, $3$, or $4$. Prove that $a_{2 n}$ is a perfect square, $n=1,2, \cdots$.
(1991, National High School League) | Proof: Let $N=\overline{x_{1} x_{2} \cdots x_{k}}$, where $x_{1}, x_{2}, \cdots, x_{k} \in\{1,3,4\}$, and $x_{1}+x_{2}+\cdots+x_{k}=n$. Suppose $n>4$, and remove $x_{1}$, leaving $x_{2}, \cdots, x_{k}$. Then, when $x_{1}$ takes the values $1,3,4$ respectively, $x_{2}+\cdots+x_{k}$ equals $n-1, n-3, n-4$. When $n>4$,
$$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,734 |
Example 4 There are $n$ squares arranged in a row, to be painted with red, yellow, and blue. Each square is painted one color, with the requirement that no two adjacent squares are the same color, and the first and last squares are also different colors. How many ways are there to paint them?
(1991, Jiangsu Mathematics... | Solution: Let there be $a_{n}$ ways of coloring, it is easy to see that $a_{1}=3, a_{2}=$ $6, a_{3}=6$, and when $n \geqslant 4$, after numbering the $n$ cells in sequence, cell 1 and cell $(n-1)$ are not adjacent.
Case 1: The color of cell $(n-1)$ is different from that of cell 1. In this case, cell $n$ has only one ... | a_{n}=2^{n}+2(-1)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,735 |
Example 5 Toss a coin $n$ times, what is the probability of getting two consecutive heads during the tossing process? | Solution: Let $Q_{n}$ denote the probability of getting two consecutive heads in $n$ coin tosses, and $P_{n}$ denote the probability of not getting two consecutive heads. Clearly, $P_{1}=1, P_{2}=\frac{3}{4}$. If $n>2$, there are two scenarios:
I. If the first toss is tails, the probability of not getting two consecut... | Q_{n}=1-\frac{F_{n+2}}{2^{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,736 |
Example 6 There are two rivers, A and B, each with a flow rate of $300 \mathrm{~m}^{3} / \mathrm{s}$, which converge at a certain point and continuously mix. Their sediment contents are $2 \mathrm{~kg} / \mathrm{m}^{3}$ and $0.2 \mathrm{~kg} / \mathrm{m}^{3}$, respectively. Assume that from the convergence point, there... | Solution: Let the sand content of two water flows be $a \, \text{kg} / \text{m}^{3}$ and $b \, \text{kg} / \text{m}^{3}$, and the water volumes flowing through in a unit time be $p \, \text{m}^{3}$ and $q \, \text{m}^{3}$, respectively. Then the sand content after mixing is
$$
\frac{a p + b q}{p + q} \, \text{kg} / \te... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,737 |
Example 7 As shown in Figure 1, let $A$ and $E$ be opposite vertices of a regular octagon. There is a kangaroo at vertex $A$. Except for vertex $E$, the kangaroo can jump to any one of the two adjacent vertices from any vertex of the octagon, and the kangaroo stops when it lands on vertex $E$. Let $a_{n}$ be the number... | Solution: Clearly, $a_{4}=2, a_{1}=a_{2}=a_{3}=0$, and any odd number of jumps cannot reach $E, a_{2 m-1}=0 .(m \in \mathbf{N})$
From $A$, jumping 2 times can reach $C$ and $G$. Let the number of ways to jump $n$ steps from point $C$ to $E$ be $b_{n}$, then $b_{2}=1$. Thus,
$$
b_{n}=a_{n-2}+2 b_{n-2} .
$$
(indicating ... | a_{2 m}=\frac{1}{\sqrt{2}}(2+\sqrt{2})^{m-1}-\frac{1}{\sqrt{2}}(2-\sqrt{2})^{m-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,738 |
Example 8 Using $0,1,2,3,4$ how many $n$-digit numbers can be formed where each pair of adjacent digits differ by exactly 1? | Let the total number of $n$-digit numbers that meet the conditions be $x_{n}$, and the numbers starting with $0,1,2,3,4$ be $y_{n}, z_{n}, u_{n}, v_{n}, w_{n}$ respectively.
Then $x_{n}=y_{n}+z_{n}+u_{n}+v_{n}+w_{n}$.
By symmetry,
$y_{n}=w_{n}, z_{n}=v_{n}$.
Thus, $x_{n}=2 y_{n}+2 z_{n}+u_{n}$.
$1^{\circ}$ When startin... | x_{2 n}=8 \cdot 3^{n-1}, \quad x_{2 n+1}=14 \cdot 3^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,739 |
Example 9 In a convex $n$-sided polygon, drawing $(n-3)$ non-intersecting diagonals, how many ways are there to do this? | Solution: It is easy to know that a convex $n$-sided polygon has at most $n-3$ non-intersecting diagonals. Let the number of different connection methods be $a_{n}$. Obviously, $a_{4}=2$, and it is agreed that $a_{3}=a_{2}=1$. For the convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, suppose the diagonal $A_{1} A_{k... | \frac{(2 n-4)!}{(n-1)!(n-2)!} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,740 |
$$
\begin{array}{l}
\text { Example } 4 \text { Given } \frac{x-a-b}{c}+\frac{x-b-c}{a}+ \\
\frac{x-c-a}{b}=3 \text {, and } \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0 \text {. Then } x-a- \\
b-c=
\end{array}
$$
Example 4 Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+$ $\frac{x-c-a}{b}=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{... | Solution: The given equation contains four letters $x, a, b, c$, and can certainly be treated as an equation in $x$, solving for $x$ and then finding the value, but this is a bit cumbersome. Instead, it is better to treat the expression to be evaluated, $x-a-b-c$, as the unknown. Thus, we get the following solution:
By... | x-a-b-c=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,741 |
Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{arr... | Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.
When $1$ 0, if we reduce the number of balls in each drawer by 1, we ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,742 |
Proposition: Let $\max (A, B, C) < 120^{\circ}$, and $F$ be the Fermat point of $\triangle ABC$. Extend $AF$, $BF$, and $CF$ to intersect the opposite sides at $A'$, $B'$, and $C'$, respectively. Let $AA' = x$, $BB' = y$, $CC' = z$, and $R$, $r$ be the circumradius and inradius of $\triangle ABC$, respectively. Then
$$... | $$
\begin{array}{l}
\text { Proof: }(1) \max (A, B, C)0 \\
\Leftrightarrow \cos 3 A+\cos 3 B+\cos 3 C-1\frac{3 R+r}{\sqrt{3}}$.
(2) By the Wlombier-Doncet inequality $3 p^{2} \leqslant$ $(4 R+r)^{2}$, we have
$$
p \leqslant \frac{4 R+r}{\sqrt{3}} .
$$
Equality holds if and only if $\triangle A B C$ is an equilateral t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,744 |
Lemma Let $D$ be any point on side $A B$ of $\triangle A B C$, $D E / / B C$, and $E$ is on side $A C$, then there exists a unique point $Z$ on segment $D E$ such that $\triangle A B Z$ and $\triangle C A Z$ have equal incircles. | Proof: As shown in Figure 1, let
$$
D Z=t, D E=T \text {, }
$$
the inradii of $\triangle A B Z$ and $\triangle C A Z$ are $r_{C}(t)$ and $r_{B}(t)$, respectively, and set $r(t)=r_{C}(t)-r_{B}(t)$. By the formula for the inradius of a triangle
$$
r=\frac{\triangle}{p}=\sqrt{\frac{(p-a)(p-b)(p-c)}{p}}
$$
it is known tha... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,745 |
1. Let the universal set be the set of real numbers. If $A=\{x \mid \sqrt{x-2} \leqslant 0\}$, $B=\left\{x \mid 10^{x^{2}-2}=10^{x}\right\}$, then $A \cap \bar{B}$ is ( ).
(A) $\{2\}$
(B) $\{-1\}$
(C) $\{x \mid x \leqslant 2\}$
(D) $\varnothing$ | $-、 1 .(\mathrm{D})$
From $\sqrt{x-2} \leqslant 0$ we get $x=2$, so $A=\{2\}$. From $10^{x^{2}-2}=$ $10^{x}$ we get $x^{2}-x-2=0$. Therefore, $B=\{-1,2\}$. Then $A \cap \bar{B}=\varnothing$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,748 |
3. Given point $A$ is the left vertex of the hyperbola $x^{2}-y^{2}=1$, points $B$ and $C$ are on the right branch of the hyperbola, and $\triangle A B C$ is an equilateral triangle. Then the area of $\triangle A B C$ is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{3 \sqrt{3}}{2}$
(C) $3 \sqrt{3}$
(D) $6 \sqrt{3}$ | 3. (C).
Let point $B$ be above the $x$-axis. Since $\triangle ABC$ is an equilateral triangle, the slope $k$ of line $AB$ is $k=\frac{\sqrt{3}}{3}$.
Since the line passes through point $A(-1,0)$, the equation is
$$
y=\frac{\sqrt{3}}{3} x+\frac{\sqrt{3}}{3} \text {. }
$$
Substituting into the hyperbola equation $x^{2}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,750 |
4. Given positive numbers $p, q, a, b, c$, where $p \neq q$. If $p, a, q$ form a geometric sequence, and $p, b, c, q$ form an arithmetic sequence, then the quadratic equation $b x^{2}-2 a x+c=0(\quad)$.
(A) has no real roots
(B) has two equal real roots
(C) has two distinct real roots with the same sign
(D) has two rea... | 4. (A).
From the problem, we know that $p q=a^{2}, 2 b=p+c, 2 c=q+b$. From the last two equations, we get $b=\frac{2 p+q}{3}, c=\frac{p+2 q}{3}$. Therefore, we have
$$
\begin{aligned}
b c & =\frac{p+p+q}{3} \cdot \frac{p+q+q}{3} \\
& \geqslant \sqrt[3]{p^{2} q} \cdot \sqrt[3]{p q^{2}}=p q=a^{2} .
\end{aligned}
$$
Sin... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,751 |
Example 5 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-$ $a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics Competition) | Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.
Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$
Its ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,752 |
5. The minimum distance from lattice points (points with integer coordinates) to the line $y=\frac{5}{3} x+\frac{4}{5}$ is ( ).
(A) $\frac{\sqrt{34}}{170}$
(B) $\frac{\sqrt{34}}{85}$
(C) $\frac{1}{20}$
(D) $\frac{1}{30}$ | 5. (B).
Let the integer point be $\left(x_{0}, y_{0}\right)$, then its distance to the line $25 x-15 y+12$ $=0$ is
$$
d=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{\sqrt{25^{2}+(-15)^{2}}}=\frac{\left|25 x_{0}-15 y_{0}+12\right|}{5 \sqrt{34}} .
$$
Since $x_{0} y_{0} \in \mathbf{Z}$, hence $25 x_{0}-15 y_{0}$ is a multip... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,753 |
6. Let $\omega=\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$. Then the equation with roots $\omega, \omega^{3}, \omega^{7}, \omega^{9}$ is ( ).
(A) $x^{4}+x^{3}+x^{2}+x+1=0$
(B) $x^{4}-x^{3}+x^{2}-x+1=0$
(C) $x^{4}-x^{3}-x^{2}+x+1=0$
(D) $x^{4}+x^{3}+x^{2}-x-1=0$ | 6. (B).
From $\omega=\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}$, we know that $\omega, \omega^{2}, \omega^{3}, \cdots, \omega^{10}$. $(=1)$ are the 10 10th roots of 1:
$$
\begin{array}{l}
(x-\omega)\left(x-\omega^{2}\right)\left(x-\omega^{3}\right) \cdots\left(x-\omega^{10}\right) \\
=x^{10}-1 .
\end{array}
$$
Al... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,754 |
8. Let $a_{n}$ be the coefficient of the $x$ term in the expansion of $(3-\sqrt{x})^{n}$ $(n=2,3,4, \cdots)$. Then
$$
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right)=
$$
$\qquad$ | 8.18
By the binomial theorem, $a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}$. Therefore,
$$
\begin{array}{l}
\frac{3^{n}}{a_{n}}=\frac{3^{2} \times 2}{n(n-1)}=18\left(\frac{1}{n-1}-\frac{1}{n}\right) . \\
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right) \\
=\lim _{n... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,755 |
10. In the ellipse $\frac{x^{2}}{a^{2}} \div \frac{y^{2}}{b^{2}}=1(a>b>0)$, let the left focus be $F$, the right vertex be $A$, and the upper endpoint of the minor axis be $B$. If the eccentricity of the ellipse is $\frac{\sqrt{5}-1}{2}$, then $\angle A B F=$ $\qquad$ . | $10.90^{\circ}$.
Let $c$ be the semi-focal distance of the ellipse, then from $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$ we get
$$
\begin{array}{l}
c^{2}+a c-a^{2}=0 . \\
\text { Also }|A B|^{2}=a^{2}+b^{2},|B F|^{2}=a^{2}, \\
\text { so }|A B|^{2}+|B F|^{2}=2 a^{2}+b^{2}=3 a^{2}-c^{2} . \\
\text { And }|A F|^{2}=(a+c)^{2}=a^{... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,757 |
12. If:
(1) $a, b, c, d$ all belong to $\{1,2,3,4\}$;
(2) $a \neq b, b \neq c, c \neq d, d \neq a$;
(3) $a$ is the smallest value among $a, b, c, d$.
Then, the number of different four-digit numbers $\overline{a b c d}$ that can be formed is
$\qquad$ | 12.28.
When $\overline{a b c d}$ has exactly 2 different digits, it can form $C_{4}^{2}=6$ different numbers.
When $\overline{a b c d}$ has exactly 3 different digits, it can form $\mathrm{C}_{3}^{\mathrm{l}} \mathrm{C}_{2}^{\mathrm{d}} \mathrm{C}_{2}^{1}+$ $\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}=12+4=16$ different numb... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,758 |
For three numbers $a, b, c$ not equal to 0, satisfying $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$. Prove: $a, b, c$ include at least two numbers that are opposites of each other.
(1999, Beijing Junior High School Mathematics Competition) | Solution: Consider the given equation as a quadratic equation in $a$, and rearrange it into a polynomial equation in $a$:
$$
(b+c) a^{2}+(b+c)^{2} a+b c(b+c)=0 \text {. }
$$
If $b+c \neq 0$, equation (1) can be simplified to
$$
a^{2}+(b+c) a+b c=0 .
$$
Solving this, we get $a=-b$ or $a=-c$. This indicates that $a$ an... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,760 |
15. Given $C_{0}: x^{2}+y^{2}=1$ and $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Find the conditions that $a$ and $b$ must satisfy such that for any point $P$ on $C_{1}$, there exists a parallelogram with $P$ as a vertex, which is externally tangent to $C_{0}$ and internally tangent to $C_{1}$. Prove ... | 15. The required condition is $\frac{1}{a^{2}}+\frac{1}{b^{2}}=1$. The center of the circle is also the center of the rhombus.
Assuming the conclusion is correct for the point $(a, 0)$, there is a rhombus with $(a, 0)$ as a vertex, inscribed in $C_{1}$ and circumscribed around $C_{0}$. The opposite vertex of $(a, 0)$ ... | \frac{1}{a^{2}}+\frac{1}{b^{2}}=1 | Geometry | proof | Yes | Yes | cn_contest | false | 711,761 |
一、(满分 50 分) As shown in Figure 1, in the acute triangle $\triangle ABC$, there are two points $E$ and $F$ on side $BC$ such that $\angle BAE = \angle CAF$. Draw $FM \perp AB$ and $FN \perp AC$ (where $M$ and $N$ are the feet of the perpendiculars). Extend $AE$ to intersect the circumcircle of $\triangle ABC$ at point $... | -、As shown in Figure 4, connect
$M N, B D$.
$\because F M \perp A C$,
$F N \perp A C$,
$\therefore A, M, F, N$ are
concyclic.
Thus, $\angle A M N=$
$\angle A F N$,
$\angle A M N+\angle B A E$
$=\angle A F N+\angle C A F=90^{\circ}$, which means $M N \perp A D$.
Therefore, $S_{\text {quadrilateralAMDN }}=\frac{1}{2} A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,762 |
Sure, here is the translated text:
```
II. (Full marks 50 points)
Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$a_{0}=1, b_{0}=0$, and
\[
\left\{\begin{array}{l}
a_{n+1}=7 a_{n}+6 b_{n}-3, \\
b_{n+1}=8 a_{n}+7 b_{n}-4,
\end{array} \quad n=0,1,2, \cdots .\right.
\]
Prove: $a_{n} (n=0,1,2... | Proof 1: From the assumption, we have $a_{1}=4, b_{1}=4$, and for $n \geqslant 1$,
$$
\begin{array}{l}
\left(2 a_{n+1}-1\right)+\sqrt{3} b_{n+1} \\
=\left(14 a_{n}+12 b_{n}-7\right)+\sqrt{3}\left(8 a_{n}+7 b_{n}-4\right) \\
=\left[\left(2 a_{n}-1\right)+\sqrt{3} b_{n}\right](7+4 \sqrt{3}) .
\end{array}
$$
By induction... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,763 |
Three. (Full marks 50 points) There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ of them is equal and is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
Please note that the translation preserv... | Obviously, $n \geqslant 5$. Let the $n$ people be $A_{1}, A_{2}, \cdots, A_{n}$.
Suppose the number of calls made by $A_{i}$ is $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i, j}, 1 \leqslant i, j \leqslant n$. Then
$$
m_{i}+m_{j}-\lambda_{i, j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c .
$$... | n=5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,764 |
In $\triangle A B C$, $A B=A C$. There is a point $D$ on line segment $A B$, and a point $E$ on the extension of line segment $A C$, such that $D E=A C$. Line segment $D E$ intersects the circumcircle of $\triangle A P C$ at point $T$, and $P$ is a point on the extension of line $A T$. Prove: Point $P$ satisfies $P D+P... | First, we prove the sufficiency.
Method 1: As shown in Figure 2, take a point $F$ on line segment $A T$ such that $\angle A B F = \angle E D P$.
Since $P$ is on the circumcircle of $\triangle A D E$, we have $\angle B A F = \angle D A P = \angle D E P$. Also, $A B = A C = D E$, so $\triangle A B F \cong \triangle E D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,765 |
Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the derivation process.
(Xu Yichao, provided) | II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,766 |
For a positive integer $a \geqslant 2$, let $N_{a}$ be the number of positive integers $k$ that satisfy the following property: the sum of the squares of the digits of $k$ in base $a$ equals $k$: Prove:
(1) $N_{a}$ is odd;
(2) For any given positive integer $M$, there exists a positive integer $a \geqslant 2$ such that... | Three, (1) Let the base-$a$ representation of $k$ be $k=x_{n} x_{n-1} \cdots x_{1} x_{0}$, where $x_{n} \neq 0, 0 \leqslant x_{i} \leqslant a-1, i=0,1, \cdots, n$.
According to the problem, we have
$$
x_{n} a^{n}+\cdots+x_{1} a+x_{0}=x_{n}^{2}+\cdots+x_{1}^{2}+x_{0}^{2} .
$$
That is,
$$
x_{n}\left(a^{n}-x_{n}\right)+\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,767 |
Let $f(x)$ be a polynomial with integer coefficients, and $f(x)=1$ has an integer root. We agree to denote the set of all such $f$ as $F$.
For any given integer $k>1$, find the smallest integer $m(k)>1$, which guarantees the existence of $f \in F$, such that
$$
f(x)=m(k)
$$
has exactly $k$ distinct integer roots.
(Zh... | Suppose $f \in F$ such that $f(x)=m(k)$ has exactly $k$ distinct integer roots, denoted as $\beta_{1}, \beta_{2}, \cdots, \beta_{k}$. Then there exists an integer polynomial $g(x)$ such that
$$
\begin{array}{l}
f(x)-m(k) \\
=\left(x-\beta_{1}\right)\left(x-\beta_{2}\right) \cdots\left(x-\beta_{k}\right) g(x) .
\end{arr... | m(k)=\left(\left(\frac{k}{2}\right)!\right)^{2}+1 \text{ for even } k, \text{ and } m(k)=\left(\frac{k-1}{2}\right)!\left(\frac{k+1}{2}\right)!+1 \text{ for odd } k | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,768 |
Five, (1) Let $a, b$ be positive real numbers, and the sequences $\left\{x_{k}\right\}$ and $\left\{y_{k}\right\}$ satisfy $x_{0}=1, y_{0}=0$, and
$$
\left\{\begin{array}{l}
x_{k+1}=a x_{k}-b y_{k}, \\
y_{k+1}=x_{k}+a y_{k},
\end{array} \quad k=0,1,2, \cdots .\right.
$$
Prove:
$$
x_{k}=\sum_{i=0}^{[k / 2]}(-1)^{i} a^{... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Algebra | proof | Yes | Yes | cn_contest | false | 711,769 |
'. Six, let $n$ be a positive integer, and let the set
$M=\{(x, y) \mid x, y$ are integers, $1 \leqslant x, y \leqslant n\}$.
Define a function $f$ on $M$ with the following properties:
(a) $f(x, y)$ takes values in the set of non-negative integers;
(b) When $1 \leqslant x \leqslant n$, we have $\sum_{y=1}^{n} f(x, y)... | Six, we first prove the following general lemma.
Lemma In each cell of an $m$-row $n$-column grid, fill in a non-negative integer, the number filled in the cell at the $i$-th row and $j$-th column is denoted by $a_{ij}$, $r_i (1 \leqslant i \leqslant m)$ and $s_j (1 \leqslant j \leqslant n)$ are non-negative integers, ... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,770 |
Example 7 Real numbers $a$ and $b$ satisfy the equation $\frac{a^{2} b^{2}}{a^{4}-2 b^{4}}=$
1. Then $\frac{a^{2}-b^{2}}{19 a^{2}+96 b^{2}}=$ $\qquad$
(1996, Beijing Junior High School Mathematics Competition) | Solution: The given equation can not only be seen as an equation of $a$, $b$, $a^{2}$, and $b^{2}$, but also as an equation of $\frac{a^{2}}{b^{2}}$ (or $\frac{b^{2}}{a^{2}}$).
In fact, from the given equation, we have
$$
a^{2} b^{2}=a^{4}-2 b^{4} \text {. }
$$
From the given equation, we know $b \neq 0$. Dividing bo... | \frac{1}{134} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,771 |
2. Arrange the $n^{2}$ natural numbers from 1 to $n^{2}$ randomly in an $n \times n$ square grid, where $n \geqslant 2$. For any pair of numbers in the same row or column, compute the ratio of the larger number to the smaller number. The minimum value among these $n^{2}(n-1)$ ratios is called the "characteristic value"... | First, we prove that for any arrangement $A$, its characteristic value
$$
c(A) \leqslant \frac{n+1}{n}.
$$
If the largest $n$ natural numbers $n^{2}-n+1, n^{2}-n+2, \cdots, n^{2}-1, n^{2}$ have two in the same row or column, then
$$
c(A) \leqslant \frac{a}{b} \leqslant \frac{n^{2}}{n^{2}-n+1}b.
$$
If these $n$ large n... | \frac{n+1}{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,772 |
3. $n(n \geqslant 2)$ girls play a game, each holding a ball. For each pair of girls that can be formed from $\mathrm{C}_{n}^{2}$ pairs, they exchange their balls in any order. If in the end no girl holds her original ball, it is called a "good game"; if in the end every girl holds her original ball, it is called a "ba... | Solution: Consider the game played by $n$ girls as a set of $N=\mathrm{C}_{n}^{2}$ distinct element pairs $(i, j)$ in the set $\{1,2, \cdots, n\}$, and each pair swap is denoted as $t_{1}, t_{2}, \cdots, t_{N}$. If the permutation $P=t_{N} \circ t_{N-1} \cdots \cdots \circ t_{2} \circ t_{1}$ has no fixed points, it is ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,773 |
4. Prove that the set of positive integers cannot be divided into three non-empty subsets without common elements, such that taking any two different positive integers $x, y$ from two different subsets, $x^{2}-x y+y^{2}$ belongs to the third subset. | Proof: Let $f(x, y) = x^2 - xy + y^2$. Assume the set of positive integers can be divided into three non-empty subsets:
$$
\mathrm{N}^{+} = A \cup B \cup C.
$$
Without loss of generality, assume $1 \in A, b \in B, c \in C$, where $b < c$. Since $c$ is the smallest number in $C$, $c - r \notin C$. Also, since $r \leq b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,774 |
1. Given that $x, y, z$ are all real numbers, $x>0, y>0$, and $a=\frac{(y-z)^{2}}{x}-\frac{(x-z)^{2}}{y}, b=x-y$. Then, among the following conclusions, the one that must be true is ( ).
(A) If $x<y$, then $a \geqslant b$
(B) $a \leqslant b$
(C) $a \geqslant b$
(D) If $x<y$, then $a<b$ | $$
\begin{array}{l}
=y(x-y)^{2}+2 \frac{y(y-x)(x-z)}{x y} \\
+\frac{y(x-z)^{2}-x(x-z)^{2}-x y(x-y)}{x y} \\
=\frac{(x-y)\left(x y-y^{2}-2 x y+2 y z-x y\right)-(x-y)(x-z)^{2}}{x y} \\
=\frac{-(x-y)\left[y^{2}+2 x y-2 y z+(x-z)^{2}\right]}{x y} \\
=\frac{-(x-y)[y+(x-z)]^{2}}{x y} \text {. } \\
\end{array}
$$
If $x > 0$,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,776 |
2. The range of $a$ for which the equation $\sqrt{a-x^{2}}=\sqrt{2}-|x|(a>0)$ has unequal real roots is ( ).
(A) $0<a<1$
(B) $0<a \leqslant 1$
(C) $a>0$
(D) $a \geqslant 1$ | 2. (D).
Squaring both sides of the given equation, we get
$$
\begin{array}{l}
a-x^{2}=2+x^{2}-2 \sqrt{2}|x|, \\
2 x^{2}+2-a=2 \sqrt{2}|x| .
\end{array}
$$
Squaring again and rearranging, we obtain
$$
4 x^{4}-4 a x^{2}+a^{2}-4 a+4=0 \text {. }
$$
It is easy to see that when $x \neq 0$, if $x$ is a root of equation (1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,777 |
3. In isosceles $\triangle A B C$, vertex angle $\angle B A C=$ $100^{\circ}$, extend $A B$ to $D$, such that $A D=B C$. Then $\angle B C D$ $=(\quad)$.
(A) $10^{\circ}$
(B) $15^{\circ}$
(C) $20^{\circ}$
(D) $30^{\circ}$ | 3. (A).
As shown in Figure 2, draw $DE // BC$, intersecting at $DE = AB$, and connect $AE$. It is easy to prove that $\triangle ABC \cong \triangle EAD$.
Thus, $\angle 2 = 40^{\circ}$,
$$
\angle 3 = 60^{\circ}, AE = EC.
$$
Therefore, $\triangle AEC$ is an equilateral triangle. From $\angle 6 = 40^{\circ}$, we get $\a... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,778 |
4. Given equation A: $x^{2}+p_{1} x+q_{1}=0$, equation B: $x^{2}+p_{2} x+q_{2}=0$, where $p_{1} 、 p_{2} 、 q_{1} 、 q_{2}$ are all real numbers, and satisfy $p_{1} p_{2}=2\left(q_{1}+q_{2}\right)$. Then $(\quad)$.
(A) Both A and B must have real roots
(B) Neither A nor B has real roots
(C) At least one of A and B has rea... | 4. (C).
$$
\Delta_{1}=p_{1}^{2}-4 q_{1}, \Delta_{2}=p_{2}^{2}-4 q_{2} \text {. }
$$
If equations 甲 and 乙 have no real roots, then
$$
\Delta_{1}<0, \Delta_{2}<0, \Delta_{1}+\Delta_{2}<0 \text {, }
$$
which means
$$
\begin{array}{l}
p_{1}^{2}+p_{2}^{2}-4\left(q_{1}+q_{2}\right)<0, \\
p_{1}^{2}+p_{2}^{2}-2 p_{1} p_{2}<0... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,779 |
5. On the plane, there are $n$ points, among which any three points are the vertices of some equilateral triangle. Then the maximum value of $n$ is ( ).
(A) 3
(B) 4
(C) 6
(D) greater than 6 | Among the $n$ points that meet the conditions, there must be two points with the greatest distance, let these two points be $A$ and $B$. Then there is a third point $C$, such that $\triangle ABC$ is an equilateral triangle, and the $n$ points are all within the curvilinear triangle in Figure 3 (the curvilinear triangle... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,780 |
6. In the right triangle $\triangle A B C$, the right-angle sides $A C=3$ and $B C=4$. Line $l$ intersects the two right-angle sides $A C$ and $B C$ at points $S$ and $T$ respectively, and $S$ and $T$ are on the interiors of $A C$ and $B C$, and $l$ bisects the perimeter and area of $\triangle A B C$. Then ().
(A) Ther... | 6. (D).
As shown in Figure 4, it is easy to know that the perimeter of $\triangle ABC$ is 12, and the area is 6. If line $l$ bisects the perimeter and area of $\triangle ABC$, then
$$
CS + CT = 6, S_{\triangle CST} = 3.
$$
Let $CS = x$, then $CT = 6 - x$. Thus, $\frac{1}{2} x(6 - x) = 3$.
Solving for $x$ gives $x = 3... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,781 |
Example 8 Given $x+y=2, x^{2}+y^{2}=\frac{5}{2}$. Then $x^{4}+y^{4}=$ $\qquad$
(1995, Chongqing and Four Other Cities Junior High School Mathematics League) | $$
\begin{array}{l}
\because x y=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2} \\
\quad=\frac{4-\frac{5}{2}}{2}=\frac{3}{4}, \\
\therefore x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2(x y)^{2} \\
=\left(\frac{5}{2}\right)^{2}-2 \times\left(\frac{3}{4}\right)^{2}=\frac{41}{8} .
\end{array}
$$
The given two equations form ... | \frac{41}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,782 |
1. If positive numbers $m, n$ satisfy $m+4 \sqrt{m n}-$ $2 \sqrt{m}-4 \sqrt{n}+4 n=3$, then $\frac{\sqrt{m}+2 \sqrt{n}}{\sqrt{m}+2 \sqrt{n}+3}=$ | $$
=1 . \frac{1}{2} \text {. }
$$
Let $\sqrt{m}+2 \sqrt{n}=t$, then the given equation can be transformed into
$$
\begin{array}{l}
t^{2}-2 t-3=0 . \\
\because t>0, \therefore t=3 .
\end{array}
$$
Thus, $\frac{\sqrt{m}+2 \sqrt{n}}{\sqrt{m}+2 \sqrt{n}+3}=\frac{t}{t+3}=\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,783 |
2. $A B$ is a chord of $\odot O$, $P$ is a point outside $\odot O$, $P B$ is tangent to $\odot O$ at $B$, $P A$ intersects $\odot O$ at $C$, and $A C = B C$, $P D \perp A B$ at $D$, $E$ is the midpoint of $A B$, $D E = 1000$. Then $P B=$ | 2.2000 .
As shown in Figure 5, take the midpoint $F$ of $PA$, and connect $DF$, $EF$. Then $EF \parallel PB$. Also, $DF = \frac{1}{2} PA$, so $DF = AF$. Since $AC = BC$, $\triangle AFD$ is an isosceles triangle. Therefore, $\angle 2 = \angle 4 = \angle 1$. Since $EF \parallel PB$ and $DF \parallel BC$, $\angle 3 = \an... | 2000 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,784 |
4. A certain product has a purchase price of 80 yuan per unit and a retail price of 100 yuan per unit. To promote sales, it is proposed to give a small gift for each unit of the product purchased. Experiments show that when the gift is priced at 1 yuan, sales volume increases by 10%, and within a certain range, for eve... | 4.9 or 10 yuan.
Let the sales volume without gifts be $m$ units, and the profit when the gift is $n$ yuan be $y_{n}$ yuan. Then
$$
\begin{aligned}
y_{n} & =(100-80-n) \cdot m \cdot(1+10 \%)^{n} \\
& =m \cdot(20-n) \cdot 1 \cdot 1^{n} . \\
(0 & <n<20, n \text { is a positive integer })
\end{aligned}
$$
$(0<n<20, n$ is ... | 9 \text{ or } 10 | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 711,786 |
One, (20 points) A rectangle is divided into several right-angled triangles, with the two legs of these right-angled triangles being 1 and 2. Prove: The total number of these right-angled triangles is even. | In a right-angled triangle with side lengths $1, 2, \sqrt{5}$. Suppose a rectangle has $b$ and $d$ hypotenuses on each pair of opposite sides, and the rest are the legs. Since the opposite sides of a rectangle are equal, there exist positive integers $a$ and $c$ such that
$$
a+b \sqrt{5}=c+d \sqrt{5} \text {. }
$$
Sin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,787 |
II. (25 points) As shown in Figure 1, the parabola opening downwards is given by
$$
y = a x^{2} - 8 a x + 12 a
$$
It intersects the $x$-axis at points $A$ and $B$ (point $A$ is to the left of point $B$). There is another point $C$ on the parabola, located in the first quadrant, such that $\triangle O C A \sim \triangl... | (1) Given that $a<0$, and the equation $a x^{2}-8 a x+12 a=0$ has two roots $x_{1}=2, x_{2}=6$. Thus, $O A=2, O B=6$.
$\because \triangle O C A \backsim \triangle O B C$,
$$
\therefore O C^{2}=O A \cdot O B=12 \text {, }
$$
which means $O C=2 \sqrt{3}$, and
$$
\frac{B C^{2}}{A C^{2}}=\frac{S_{\triangle O O B}}{S_{\tri... | y=-\frac{\sqrt{3}}{3} x^{2}+\frac{8}{3} \sqrt{3} x-4 \sqrt{3}, \quad y=-\frac{\sqrt{3}}{3} x+2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,788 |
Three. (25 points) A well-known proposition is: the midpoints of the three sides $BC$, $CA$, and $AB$ of $\triangle ABC$ are $A_{1}$, $B_{1}$, and $C_{1}$, respectively, then the segments $A_{1}B_{1}$, $B_{1}C_{1}$, and $C_{1}A_{1}$ divide $\triangle ABC$ into four triangles of equal area. Now, please prove its convers... | Three, as shown in Figure 6, let $\triangle ABC$ have sides $BC=a, CA=b, AB=c$, and $AA_1=xc, BB_1=ya, CC_1=zb$.
\[
\frac{S_{\triangle A_1C_1}}{S_{\triangle ABC}}=\frac{xc \cdot b(1-z)}{c \cdot b}=\frac{1}{4}
\]
Thus, $x(1-z)=\frac{1}{4}$.
Similarly, $y(1-x)=\frac{1}{4}$,
$z(1-y)=\frac{1}{4}$.
Multiplying (1), (2), and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,789 |
1. If the coefficient of $x$ in the expansion of $(1+x)(1+3 x) \cdots(1+2001 x)$ is $n$, then $\mathrm{i}^{n}$ equals ( ).
(A)i
(B) $-\mathrm{i}$
(C) 1
(D) -1 | - 1. (A).
$(1+x)(1+3x) \cdots(1+2001x)$ The coefficient of the linear term in the expanded form is $1+3+\cdots+2001=\frac{1}{2} \times 2002 \times 1001=$ $1001^{2}$.
Therefore, $i^{n}=i^{1001^{2}}=i$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,790 |
3. The sequence $\lg 1000, \lg \left(1000 \cdot \cos \frac{\pi}{3}\right)$, $\lg \left(1000 \cdot \cos ^{2} \frac{\pi}{3}\right), \cdots, \lg \left(1000 \cdot \cos ^{n-1} \frac{\pi}{3}\right)$, $\cdots$, when the sum of the first $n$ terms is maximized, the value of $n$ is ( ).
(A) 9
(B) 10
(C) 11
(D) Does not exist | 3. (B).
Given $a_{n}=\lg \left(1000 \cdot \cos ^{n-1} \frac{\pi}{3}\right)=\lg \frac{1000}{2^{n-1}}$, we know that $\left\{a_{n}\right\}$ is a decreasing sequence.
Also, since $2^{9}=512,2^{10}=1024$, it follows that $a_{10}>0, a_{11}<0$, thus $S_{10}$ is the maximum. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,792 |
Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,793 |
4. If the moving point $P(x, y)$ rotates counterclockwise on the unit circle with a constant angular velocity $\omega$, then the motion equation of point $Q\left(-2 x y, y^{2}-x^{2}\right)$ is ( ).
(A) rotates clockwise on the unit circle with angular velocity $\omega$
(B) rotates counterclockwise on the unit circle wi... | 4. (C).
Express the coordinates of the moving point $P(x, y)$ in parametric form:
$$
\left\{\begin{array}{l}
x=\cos \omega t, \\
y=\sin \omega t .
\end{array}(\omega>0)\right.
$$
Then $-2 x y=-\sin 2 \omega t=\cos \left(-2 \omega t+\frac{3}{2} \pi\right)$,
$$
y^{2}-x^{2}=-\cos 2 \omega t=\sin \left(-2 \omega t+\frac{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,794 |
6. For squares $A E C D$ and $A B E F$ with side length $a$, the angle between the planes they lie in is $120^{\circ}$. Points $M$ and $N$ are on the diagonals $A C$ and $B F$ respectively, and $A M = F N$. The range of $M N$ is ( ).
(A) $\left(\frac{\sqrt{2}}{2} a, a\right]$
(B) $\left(\frac{\sqrt{3}}{2} a, a\right]$
... | 6. (D).
As shown in Figure 1, draw $MP \perp AB$ at $P$, and connect $PN$, then $PN \perp AB$. Therefore, $\angle MPN$ is the plane angle of the dihedral angle $ABCD - A \sqrt{-} ABEF$, and $\angle MPN = 120^{\circ}$.
Let $AM = FN = x$, then
$$
\begin{aligned}
MP & = \frac{\sqrt{2}}{2} x, \quad PN = \frac{\sqrt{2} a ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,796 |
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