problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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5. Let the base of the right quadrilateral prism $A^{\prime} B^{\prime} C^{\prime} D^{\prime}-A B C D$ be a rhombus, with an area of $2 \sqrt{3} \mathrm{~cm}^{2}, \angle A B C=60^{\circ}, E$ and $F$ be points on the edges $C C^{\prime}$ and $B B^{\prime}$, respectively, such that $E C=B C=$ $2 F B$. Then the volume of ... | 5.A.
On the base $ABCD$, take the midpoint $K$ of $BC$, and connect $AK$. From the given information and the properties of a right prism, we get $AK \perp$ plane $BCEF$. Therefore, $AK$ is the height of the pyramid $A-BCFE$.
$$
\begin{array}{l}
BC=AB=AC=2, AK=\sqrt{3}, CE=2, BF=1 . \\
\text { Thus, } V_{A-BCFE}=\frac{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,604 |
6. If the inequality $x^{2}+|x-a|<2$ has at least one negative solution, then the range of the parameter $a$ is ( ).
(A) $\left(-\frac{5}{4}, 2\right)$
(B) $\left(-\frac{7}{4}, 2\right)$
(C) $\left(-\frac{9}{4}, 2\right)$
(D) $\left(-\frac{7}{4}, 3\right)$ | 6.C.
From the original inequality, we know
$$
\begin{array}{l}
|x-a|<2-x^{2} \\
\Rightarrow x^{2}-2<x-a<2-x^{2} .
\end{array}
$$
As shown in Figure 4, the curve segments $C_{1}$
and $C_{2}$ are the parts of the parabola $y=2-x^{2}$ above the x-axis and the parabola $y=x^{2}-2$ below the x-axis, respectively. When equ... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,606 |
1. If $\sin ^{2}\left(x+\frac{\pi}{12}\right)-\sin ^{2}\left(x-\frac{\pi}{12}\right)=\frac{1}{4}$, and $x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, then the value of $\tan x$ is . $\qquad$ | ii. $1.2+\sqrt{3}$.
From the given equation, we get
$\frac{1}{2}\left[1-\cos \left(2 x+\frac{\pi}{6}\right)\right]-\frac{1}{2}\left[1-\cos \left(2 x-\frac{\pi}{6}\right)\right]=\frac{1}{4}$,
which simplifies to $\cos \left(2 x-\frac{\pi}{6}\right)-\cos \left(2 x+\frac{\pi}{6}\right)=\frac{1}{2}$.
Therefore, $2 \sin 2 x... | 2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,607 |
2. If $x \in(-\infty,-1]$, the inequality
$$
\left(m-m^{2}\right) 4^{x}+2^{x}+1>0
$$
always holds, then the range of real number $m$ is $\qquad$ | 2. $-2-\frac{2^{x}+1}{4^{x}}$.
Let $t=\left(\frac{1}{2}\right)^{x}$.
Since $x \in(-\infty,-1)$, then $t \geqslant 2$. Therefore, we have $-\frac{2^{x}+1}{4^{x}}=-t^{2}-t=-\left(t+\frac{1}{2}\right)^{2}+\frac{1}{4} \leqslant-6$.
From $m-m^{2}>-6$, solving gives $-2<m<3$. | -2 < m < 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,608 |
3. In a triangular pyramid, two of the three sides are isosceles right triangles, and the other is an equilateral triangle with a side length of 1. Then, such a triangular pyramid has
保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence "保留源文本的换行和格式，直接输出翻译结果。" is in Chinese and is not part of the translation. It is an inst... | 3.3.
Regarding the tetrahedron $ABCD$, there are the following 3 cases:
(1) $AB=AC=BC=AD=1, CD=BD=\sqrt{2}$;
(2) $BC=BD=CD=1, AB=AC=AD=\frac{\sqrt{2}}{2}$;
(3) $AB=BC=AC=AD=BD=1, CD=\sqrt{2}$. | 3.3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,609 |
4. Fill the $4 \times 4$ grid with the integers from 1 to 16 such that the sum of the numbers in each row and each column is a multiple of $n$ (where $n>1$ and $n \in \mathbf{N}$), and these sums are 8 different multiples of $n$. Then the value of $n$ is $\qquad$. | 4. 2,4 .
First, note that the sum of the integers from $1 \sim 16$ is 136. Let $s_{i}$ and $t_{i}$ be the sums of the numbers in the $i$-th row and the $i$-th column, respectively, where $i=1,2,3,4$. By the problem, $s_{i}=a_{i} n, t_{i}=b_{i} n$, where $a_{i} 、 b_{i}(i=1,2,3,4)$ are 8 different positive integers. The... | 2,4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,610 |
5. For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1(0<b<4)$, the left focus is $F$, and line $l$ intersects the ellipse at points $A$ and $B$, with $|A B|$ $=6$. Then the maximum value of $|F A| \cdot|F B|$ is $\qquad$ | 5. 25 .
Let the right focus of the ellipse be $F^{\prime}$, and connect $A F^{\prime}$ and $B F^{\prime}$. Then,
$$
|A B| \leqslant\left|A F^{\prime}\right|+\left|B F^{\prime}\right| \text {. }
$$
When the chord $A B$ passes through the right focus $F^{\prime}$, the equality holds.
Thus, $|F A|+|F B|+|A B|$
$$
\begin... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,611 |
6. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, the sum of the first $n$ terms is $S_{n}$, for any $n \geqslant 2\left(n \in \mathbf{N}_{+}\right), S_{n-1}-8, a_{n}$, $8 S_{n}-4$ are always in arithmetic progression. Then $\lim _{n \rightarrow \infty}\left(a_{n}{ }^{-}+S_{n}\right)=$ $\qquad$ | 6. $\frac{4}{3}$.
From the problem, we get
$$
2 a_{n}=S_{n-1}-8+8 S_{n}-4=9 S_{n}-a_{n}-12 \text{. }
$$
Thus, $a_{n}=3 S_{n}-4(n \geqslant 2)$.
Given $a_{1}=1$, we have $a_{2}=\frac{1}{2}$.
When $n \geqslant 2$, $3 S_{n}=a_{n}+4,3 S_{n+1}=a_{n+1}+4$.
Subtracting the two equations, we get
$3 a_{n+1}=a_{n+1}-a_{n}$.
Th... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,612 |
Three. (20 points) Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & \text { when } x \in(-\infty, a) \text {, } \\
\left(\frac{x-a}{a-b}\right)^{2}, & \text { when } x \in[a, b] \text {, } \\
1, & \text { when } x \in(b,+\infty) \text {, }
\end{array}\right.
$$
(1) When $x \in\left[\frac{a+b}{2},+\infty\right)$... | (1) When $x \in\left[\frac{a+b}{2},+\infty\right)$.
(i) If $x \in\left[\frac{a+b}{2}, b\right]$, then
$$
f(x)=\frac{1}{(a-b)^{2}}(x-a)^{2}
$$
is an increasing function. Therefore,
$$
f(x) \geqslant \frac{1}{(a-b)^{2}}\left(\frac{a+b}{2}-a\right)^{2}=\frac{1}{4} ;
$$
(ii) If $x \in(b,+\infty)$, then $f(x)=1>\frac{1}{4}... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,613 |
Four. (20 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1 \text{. }
$$
Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 12$. | Given, we have
$$
\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}+\frac{1}{1+\frac{1}{c}}=1 \text {. }
$$
Let $x=\frac{1}{1+\frac{1}{a}}, y=\frac{1}{1+\frac{1}{b}}, z=\frac{1}{1+\frac{1}{c}}$. Then
$$
x+y+z=1 \text {. }
$$
From $\frac{1}{a}=\frac{1}{x}-1, \frac{1}{b}=\frac{1}{y}-1, \frac{1}{c}=\frac{1}{z}-1$, we get
... | 12 | Inequalities | proof | Yes | Yes | cn_contest | false | 715,614 |
Five. (20 points) Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F, A B$ be a focal chord of the parabola, point $M$ be on the directrix of the parabola, and $O$ be the origin. Prove:
(1) The slopes of the lines $M A, M F, M B$ form an arithmetic sequence;
(2) When $M A \perp M B$,
$\angle M F O=|\angle B M F-\an... | (1) Let the slopes of $M A$, $M F$, and $M B$ be $k_{1}$, $k$, and $k_{2}$, respectively, and the coordinates of points $A$, $B$, and $M$ be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $M\left(-\frac{p}{2}, m\right)$. The equation of line $A B$ is $x=t y+\frac{p}{2}$.
From $\left\{\begin{array}{l}x... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,615 |
$-、$ (50 points) As shown in Figure 2, two equal circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P$ and $Q$, with $O_{1}$ and $O_{2}$ not in the common part of the two circles. A secant line $AB$ and $CD$ are drawn through point $P$, with points $A$ and $C$ on $\odot O_{1}$, and points $B$ and $D$ on $\odo... | Take the midpoint $E$ of $AB$ and the midpoint $F$ of $CD$, and draw auxiliary lines as shown in Figure 6.
Since $\odot O_{1}$ and $\odot O_{2}$ are congruent circles and the inscribed angles subtended by the minor arc $\overparen{Q Q}$ and the arc $\widehat{B Q}$ are both $\angle B P Q$, we have $A Q = B Q$.
Similarl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,617 |
II. (50 points) Determine all positive integer triples $(x, y, z)$ such that $x^{3}-y^{3}=z^{2}$, where $y$ is a prime, $y \chi_{z}, 3 \nmid z$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | From the problem, we have
$$
(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2} \text {. }
$$
Since $y$ is a prime number, and $y \gamma_{z}, 3 Y_{z}$, combining with equation (1), we know
$$
(x, y)=1,(x-y, 3)=1 \text {. }
$$
Then $\left(x^{2}+x y+y^{2}, x-y\right)=(3 x y ; x-y)=1$.
From equations (1) and (2), we get
$$
x-y=m^{... | (8,7,13) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,618 |
$$
\begin{array}{l}
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\} \text { satisfies } \\
a_{0}=1, a_{1}=0, a_{2}=2005, \\
a_{n+2}=-3 a_{n}-4 a_{n-1}+2008, \\
n=1,2, \cdots. \\
\text { Let } b_{n}=5\left(a_{n+2}-a_{n}\right)\left(502-a_{n-1}-a_{n-2}\right) \\
+4^{n} \times 2004 \times 501. \\
\end{... | Let's assume
$$
a_{n+2}+\alpha=-3\left(a_{n}+\alpha\right)-4\left(a_{n-1}+\alpha\right) \text {. }
$$
Then we have $a_{n+2}=-3 a_{n}-4 a_{n-1}-8 \alpha$.
From the given information, $-8 \alpha=2008$, which means $\alpha=-251$.
Let $a_{n}-251=d_{n}$, then we have
$$
d_{n+2}=-3 d_{n}-4 d_{n-1} \text {. }
$$
At this poi... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,619 |
As shown in Figure 1, in $\triangle A B C$, $\angle C=90^{\circ}$, points $D_{1}$ and $D_{2}$ are on side $B C$, and $\angle C A D_{1}=\angle B A D_{2}$. Prove:
$$
\frac{A C^{2}}{A D_{1} \cdot A D_{2}} \leqslant \frac{B C}{2 \sqrt{B D_{1} \cdot B D_{2}}} .
$$ | Prove: As shown in Figure 2, draw $C E_{1} \perp A D_{1}$ at $E_{1}$, intersecting $A B$ at $F_{1}$, draw $E_{1} G_{1} / / B C$, intersecting $A B$ at $G_{1}$, draw $C E_{2} \perp A D_{2}$ at $E_{2}$, intersecting $A B$ at $F_{2}$, draw $E_{2} G_{2} / / B C$, intersecting $A B$ at $G_{2}$. It is easy to see that
Rt $\t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,620 |
Given a convex 2005-gon whose all vertices are inside a square with side length 2005. Prove: there must be two sides whose lengths sum to no more than 8. | Proof: As shown in Figure 3, let the square be $O A B C$, and establish a Cartesian coordinate system with the lines $O A$ and $O C$ as the $x$ and $y$ axes, respectively.
For a convex 2005-sided polygon, project each point on its sides onto the $x$ and $y$ axes.
Since it is a convex polygon, the projections of the p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,621 |
153 If $x, y, z$ are all greater than 1, prove:
$$
\frac{x^{4}}{(y-1)^{2}}+\frac{y^{4}}{(z-1)^{2}}+\frac{z^{4}}{(x-1)^{2}} \geqslant 48 .
$$ | Prove: $\frac{x^{4}}{(y-1)^{2}}+16(y-1)+16(y-1)+16$
$$
\geqslant 4 \sqrt[4]{16^{3} x^{4}}=32 x \text {. }
$$
That is, $\frac{x^{4}}{(y-1)^{2}} \geqslant 32(x-y)+16$.
Similarly, $\frac{y^{4}}{(z-1)^{2}} \geqslant 32(y-z)+16$;
$$
\frac{z^{4}}{(x-1)^{2}} \geqslant 32(z-x)+16 \text {. }
$$
Adding the three inequalities, ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,622 |
Place $n$ balls numbered $1,2, \cdots, n(n \geqslant 6)$ in sequence on a circle. Starting from the 1st ball, remove 1 ball every 5 balls, i.e., remove the balls numbered $6,12, \cdots, 6\left[\frac{n}{6}\right], 6\left(\left[\frac{n}{6}\right]+1\right)-n$, $\cdots$ (where $[x]$ represents the integer part of $x$), unt... | Solution: Place $n$ balls numbered $1, 2, \cdots, n$ in a circular arrangement, and remove the balls according to the specified method. Let the original numbers of the 5 remaining balls be $F_{i}(n), i=1,2,3,4,5$.
First, we find the relationship between $F_{1}(n+1)$ and $F_{i}(n)$.
Place $n+1$ balls numbered $1, 2, \cd... | 5, 16, 26, 35, 43; 11, 22, 32, 41, 49; 17, 28, 38, 47, 55; 1, 23, 34, 44, 53 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,623 |
Example 1 Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$. | Solution: Let the prime factorization of $n$ be
$$
n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}},
$$
where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers.
Thus, the number of distinct positive divisors of $n$ is
$$
\left(r_{1}+1\rig... | 32400 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,624 |
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=99$. Find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{10}^{2}$. | Solution: Since there are only a finite number of ways to write 99 as the sum of 10 positive integers, there must exist a way that maximizes the sum of the squares of these 10 positive integers: and there must also exist a way that minimizes the sum of the squares of these 10 positive integers.
Assume $x_{1}, x_{2}, \... | 8109 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,625 |
5. From $1,2, \cdots, 50$ choose $k$ numbers, among which there must be two numbers that are coprime. Find the minimum value of $k$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (It is known that among any 26 numbers chosen from $1,2, \cdots, 50$, there must be two that are coprime. Also, any two numbers from $2,4, \cdots, 50$ these 25 numbers are not coprime. Therefore, the minimum value of $k$ is 26.) | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,626 |
1. The graph of the function $y=f(x)$ is translated by the vector $a=$ $\left(\frac{\pi}{4}, 2\right)$, resulting in the graph with the equation $y=$ $\sin \left(x+\frac{\pi}{4}\right)+2$. Therefore, the equation of $y=f(x)$ is
(A) $y=\sin x$
(B) $y=\cos x$
(C) $y=\sin x+2$
(D) $y=\cos x+4$ | 1. B.
$y=\sin \left[\left(x+\frac{\pi}{4}\right)+\frac{\pi}{4}\right]$, i.e., $y=\cos x$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,627 |
2. If the positive root of the quadratic equation $x^{2}-p x-q=0$ $\left(p, q \in \mathbf{N}_{+}\right)$ is less than 3, then the number of such quadratic equations is ( ).
(A) 5
(B) 6
(C) 7
(D) 8 | 2.C.
From $\Delta=p^{2}+4 q>0,-q<0$,
it follows that $3 p+q<9$.
Given $p, q \in \mathbf{N}_{+}$, then $p=1, q \leqslant 5$ or $p=2, q \leqslant 2$.
Therefore, there are 7 pairs $(p, q)$ that satisfy the conditions. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,628 |
3. Let $a>b>0$. Then, the minimum value of $a^{2}+\frac{1}{b(a-b)}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 3.C.
Given $a>b>0$, we know that $0<b(a-b) \leqslant \frac{1}{4} a^{2}$. Therefore, $a^{2}+\frac{1}{b(a-b)} \geqslant a^{2}+\frac{4}{a^{2}} \geqslant 4$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,629 |
4. Let the base of the quadrilateral pyramid $P-ABCD$ not be a parallelogram. If a plane $\alpha$ cuts this quadrilateral pyramid such that the cross-section is a parallelogram, then the number of such planes $\alpha$ ( ).
(A) does not exist
(B) there is only 1
(C) there are exactly 4
(D) there are infinitely many | 4. D.
Let the intersection lines of two sets of non-adjacent lateral faces of a quadrilateral pyramid be $m$ and $n$. The lines $m$ and $n$ determine the plane $\beta$. Construct a plane $\alpha$ parallel to plane $\beta$ that intersects the lateral faces of the quadrilateral pyramid. The resulting quadrilateral is a ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,630 |
5. Let the sequence $\left\{a_{n}\right\}: a_{0}=2, a_{1}=16, a_{n+2}=$ $16 a_{n+1}-63 a_{n}(n \in \mathbf{N})$. Then the remainder when $a_{2008}$ is divided by 64 is $(\quad)$.
(A) 0
(B) 2
(C) 16
(D) 48 | 5.C.
The sequence $\left\{a_{n}\right\}$ has a remainder of $2,16,-2,-16,2,16$, $-2,-16, \cdots$ when divided by 64, with a period of 4. Since 2005 leaves a remainder of 1 when divided by 4, the solution is obtained. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,631 |
6. A corridor is 2 m wide and 8 m long. It is to be paved with 6 different colors of 1 m x 1 m whole tiles (each tile is a single color, and there are enough tiles of each color), with the requirement that adjacent tiles must be of different colors. The total number of different color schemes is ( ).
(A) $30^{8}$
(B) $... | 6. D.
First, lay the first column (two tiles) with 30 ways. Then lay the second column, assuming the two cells in the first column are covered with colors $A$ and $B$ (as shown in Figure 1). The upper cell of the second column cannot be covered with color $A$. If it is covered with color $B$, there are 6-1 ways to lay... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,632 |
1. Let vector $O A$ rotate counterclockwise around point $O$ by $\frac{\pi}{2}$ to get vector $O B$, and $2 O A+O B=(7,9)$. Then vector $O B$ $=$ . $\qquad$ | 1. $\left(-\frac{i 1}{5}, \frac{23}{5}\right)$.
Let $\boldsymbol{O A}=(m, n)$, then $\boldsymbol{O B}=(-n, m)$. Therefore,
$2 O A+O B=(2 m-n, 2 n+m)=(7,9)$,
i.e., $\left\{\begin{array}{l}2 m-n=7, \\ m+2 n=9 .\end{array}\right.$
Solving, we get $m=\frac{23}{5}, n=\frac{11}{5}$.
Thus, $O A=\left(\frac{23}{5}, \frac{11}{... | \left(-\frac{11}{5}, \frac{23}{5}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,633 |
2. Let the infinite sequence $\left\{a_{n}\right\}$ have all positive terms, and let $S_{n}$ be the sum of its first $n$ terms. For any positive integer $n$, the arithmetic mean of $a_{n}$ and 2 equals the geometric mean of $S_{n}$ and 2. Then the general term formula of the sequence is $\qquad$ . | 2. $a_{n}=4 n-2\left(n \in \mathbf{N}_{+}\right)$.
From the problem, we know that $\frac{a_{n}+2}{2}=\sqrt{2 S_{n}}$, which means
$$
S_{n}=\frac{\left(a_{n}+2\right)^{2}}{8} \text {. }
$$
From equation (1), we have $\frac{a_{1}+2}{2}=\sqrt{2 a_{1}}$, which gives $a_{1}=2$.
Also from equation (1), we get
$S_{n-1}=\fra... | a_{n}=4 n-2\left(n \in \mathbf{N}_{+}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,634 |
3. The minimum value of the function $y=|\cos x|+|\cos 2 x|(x \in \mathbf{R})$ is $\qquad$ . | 3. $\frac{\sqrt{2}}{2}$.
Let $t=|\cos x| \in[0,1]$, then $y=t+\left|2 t^{2}-1\right|$. When $\frac{\sqrt{2}}{2} \leqslant t \leqslant 1$,
$$
y=2 t^{2}+t-1=2\left(t+\frac{1}{4}\right)^{2}-\frac{9}{8} \text {. }
$$
We get $\frac{\sqrt{2}}{2} \leqslant y \leqslant 2$.
When $0 \leqslant t<\frac{\sqrt{2}}{2}$,
$$
y=-2 t^{... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,635 |
4. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B$ $=2, A A_{1}=A D=1$, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. Then, the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$ . | 4. $\frac{3}{8}$.
Take a point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{4}$. It is easy to prove that $H E / / B_{1} G, H E / /$ plane $B_{1} F G$. Therefore,
$S_{\triangle B_{1} F H}=\frac{9}{8}$, and the distance from $G$ to plane $B_{1} F H$ is 1. Hence $V_{B_{1}-E R C}=\frac{3}{8}$. | \frac{3}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,636 |
6. Among the numbers $1,2, \cdots, 2005$, what is the minimum number of numbers that need to be removed so that in the remaining numbers, no number is equal to the product of any two other numbers? | (Cue: Strike out the 43 numbers $2,3, \cdots, 44$. Among the remaining numbers (excluding 1), the product of any two numbers is greater than $45^{2}=2025>2005$. Also, for the following 43 triples:
$$
(2,87,2 \times 87),(3,86,3 \times 86), \cdots,(44,45,44 \times 45),
$$
these 129 numbers are all distinct, greater than... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,637 |
5. Among the five-digit numbers formed by the 3 digits $1,2,3$, $1,2,3$ each appear at least once. There are $\qquad$ such five-digit numbers. | 5.150.
Among the five-digit numbers,
if 1 appears only once, there are $\mathrm{C}_{5}^{1}\left(\mathrm{C}_{4}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{3}\right)=70$;
if 1 appears twice, there are $C_{5}^{2}\left(C_{3}^{1}+C_{3}^{2}\right)=60$;
if 1 appears three times, there are $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=2... | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,638 |
6. Given two sets of points on a plane
$$
\begin{aligned}
M= & \left\{(x, y)|| x+y+1 \mid \geqslant \sqrt{2\left(x^{2}+y^{2}\right)},\right. \\
& x, y \in \mathbf{R}\}, \\
N= & \{(x, y)|| x-a|+| y-1 \mid \leqslant 1, x 、 y \in \mathbf{R}\} .
\end{aligned}
$$
If $M \cap N \neq \varnothing$, then the range of values for... | 6. $[1-\sqrt{6}, 3+\sqrt{10}]$.
From the problem, $M$ is the set of points on a parabola with the origin as the focus and the line $x+y+1=0$ as the directrix, and its concave side. $N$ is the set of points in a square and its interior with center $(a, 1)$ (as shown in Figure 2).
Consider the case when $M \cap N=\varn... | [1-\sqrt{6}, 3+\sqrt{10}] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,639 |
1. (15 points) Given that point $M$ is a point on the median $A D$ of $\triangle A B C$, the line $B M$ intersects side $A C$ at point $N$, and $A B$ is the tangent to the circumcircle of $\triangle N B C$. Let $\frac{B C}{B N}=\lambda$. Try to find $\frac{B M}{M N}$ (expressed in terms of $\lambda$). | 1. As shown in Figure 3, in $\triangle B C N$, by Menelaus' theorem, we have
$$
\frac{B M}{M N} \cdot \frac{N A}{A C} \cdot \frac{C D}{D B}=1 .
$$
Since $B D=D C$, it follows that $\frac{B M}{M N}=\frac{A C}{A N}$.
Given $\angle A B N=\angle A C B$, we know $\triangle A B N \backsim \triangle A C B$. Therefore,
$$
\f... | \lambda^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,640 |
2. (15 points) Find all positive integers $n(n \geqslant 2)$ such that:
For any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, when $\sum_{i=1}^{n} x_{i}=0$, it always holds that $\sum_{i=1}^{n} x_{i} x_{i+1} \leqslant 0$ (where $x_{n+1}=x_{1}$ ). | 2. When $n=2$, from $x_{1}+x_{2}=0$, we get
$$
x_{1} x_{2}+x_{2} x_{1}=-2 x_{1}^{2} \leqslant 0 \text {. }
$$
Thus, the proposition holds for $n=2$.
When $n=3$, from $x_{1}+x_{2}+x_{3}=0$, we get
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1} \\
=\frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}-\left(x_{1}^{2}+x_{2}^... | 2,3,4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,641 |
3. (24 points) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let segment $P Q$ be a focal chord passing through the left focus $F$ and not perpendicular to the $x$-axis. If there exists a point $R$ on the left directrix such that $\triangle P Q R$ is an equilateral triang... | 3. As shown in Figure 4, let the midpoint of line segment $P Q$ be $M$, and draw perpendiculars from points $P$, $M$, and $Q$ to the directrix, with the feet of the perpendiculars being points $P$, $M^{\prime}$, and $Q^{\prime}$, respectively. Then
$$
\begin{array}{l}
\left|M M^{\prime}\right| \\
=\frac{1}{2}\left(\lef... | \left(\frac{\sqrt{3}}{3}, 1\right), \pm \frac{1}{\sqrt{3 e^{2}-1}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,642 |
4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2... | 4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.
Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\e... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,643 |
1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2 | 1. A.
From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$. | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,644 |
2. Given $\sqrt{x}=\frac{1-a}{2}, \sqrt{x+a}-$ $\sqrt{x-a+2}=-2$. Then the range of values for $a$ is ( ).
(A) $a \leqslant 1$
(B) $-1 \leqslant a \leqslant 1$
(C) $a \leqslant-1$
(D) $-1 \leqslant a \leqslant 0$ | 2.C.
Since $\sqrt{x}=\frac{1-a}{2} \geqslant 0$, it follows that $a \leqslant 1$, and $x=\frac{(1-a)^{2}}{4}$.
Thus, $\sqrt{x+a}=\frac{|1+a|}{2}, \sqrt{x-a+2}=\frac{3-a}{2}$.
Therefore, $\frac{|1+a|}{2}-\frac{3-a}{2}=-2$, which means
$$
|1+a|=-(1+a) .
$$
Hence, $1+a \leqslant 0, a \leqslant-1$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,645 |
3. In $\triangle A B C$, $\angle C=90^{\circ}$, point $D$ is on side $A B$, and $C D^{2}=A D \cdot D B$. Then the position of point $D$ exists in ( ) cases.
(A) 1
(B) 2
(C) 3
(D) infinitely many | 3. B.
As shown in Figure 2, draw $C D^{\prime} \perp$ $A B$ at $D^{\prime}$.
Let $A D=a, D D^{\prime}=$ $b, D^{\prime} B=c$. Therefore, we have
$$
\begin{aligned}
& C D^{2}-C D^{\prime 2} \\
= & a(b+c)-(a+b) c \\
= & b(a-c)=D D^{\prime 2}=b^{2} .
\end{aligned}
$$
Thus, $b(a-c)-b^{2}=0$, which gives $b(a-b-c)=0$.
(1)... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,646 |
4. The real root situation of the equation $\frac{1}{x}-2=x^{2}-2 x$ is ( ).
(A) It has three real roots
(B) It has only two real roots
(C) It has only one real root
(D) It has no real roots | 4.C.
The original equation is transformed into $\frac{1}{x}=(x-1)^{2}+1$.
Let $y_{1}=\frac{1}{x}, y_{2}=(x-1)^{2}+1$.
In the same coordinate system, draw the graphs of the inverse proportion function $y_{1}=\frac{1}{x}$ and the parabola $y_{2}=(x-1)^{2}+1$. At this point, the solution to the equation is converted to t... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,647 |
Example 1 If the equations of the two asymptotes of a hyperbola are $y= \pm \frac{2}{3} x$, and it passes through the point $M\left(\frac{9}{2},-1\right)$, try to find its equation. | Solution: The equation of the hyperbola system with asymptotes $y= \pm \frac{2}{3} x$, i.e., $\frac{x^{2}}{3^{2}}-\frac{y^{2}}{2^{2}}=0$, is $\frac{x^{2}}{3^{2}}-\frac{y^{2}}{2^{2}}=\lambda(\lambda \neq 0)$.
Given that point $M\left(\frac{9}{2},-1\right)$ lies on it, substituting gives $\lambda=2$.
Therefore, the equat... | \frac{x^{2}}{18}-\frac{y^{2}}{8}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,648 |
5. Let $a>b>0, \sqrt[6]{a b}=4, \sqrt[6]{a}+\sqrt[6]{b}=2 \sqrt{7}$. If $[x]$ denotes the greatest integer less than or equal to $x$, then $[a]$ is ( ).
(A) 7037
(B) 7038
(C) 7039
(D) 7040 | 5.C.
According to the problem, $\sqrt[5]{a} 、 \sqrt[6]{b}$ are the two real roots of the equation $t^{2}-2 \sqrt{7} t+4=0$.
Solving this equation, we get
$$
a=(\sqrt{7}+\sqrt{3})^{6}, b=(\sqrt{7}-\sqrt{3})^{6} \text {. }
$$
Let $m=\sqrt{7}+\sqrt{3}, n=\sqrt{7}-\sqrt{3}$, then we have
$$
m+n=2 \sqrt{7}, m n=4 \text {.... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,649 |
6. In $\triangle A B C$, $A B=A C$, semicircle $O$ is tangent to $A B$ and $A C$, and the center $O$ is on side $B C$. A line is tangent to the semicircle and intersects $A B$ and $A C$ at points $E$ and $F$, respectively. Then $A O$ passes through the ( ) of $\triangle O E F$.
(A) circumcenter
(B) incenter
(C) centroi... | $$
\begin{array}{l}
\text { 6.A. } \\
\text { Let } \angle B=\angle C=\alpha, \\
\angle B E F=2 \beta, \angle C F E=
\end{array}
$$
$2 \gamma$. Since the sum of the interior angles of quadrilateral $B C F E$ is $360^{\circ}$, that is,
$$
\alpha+\alpha+2 \beta+2 \gamma=360^{\circ},
$$
hence $\alpha+\beta+\gamma=180^{\c... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,650 |
1. All integer pairs $(x, y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y-\left|x^{2}-2 x\right|+\frac{1}{2}>0, \\
y+|x-1|<2
\end{array}\right.
$$
are $\qquad$ | 1. $(0,0),(1,1),(2,0)$.
From $y-\left|x^{2}-2 x\right|+\frac{1}{2}>0$ we get
$$
y>\left|x^{2}-2 x\right|-\frac{1}{2} \geqslant-\frac{1}{2} \text {; }
$$
From $y+|x-1|<2$ we get
$$
y<2-|x-1| \leqslant 2 \text {. }
$$
Therefore, $-\frac{1}{2}<y<2$.
So $y=0$ or 1.
When $y=0$, we have $\left\{\begin{array}{l}\left|x^{2}... | (0,0),(1,1),(2,0) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,651 |
2. In trapezoid $A B C D$, $D C / / A B$ and $A B=2 D C$, point $P$ is on $A B$. If the perimeters of $\triangle P B C$, $\triangle P C D$, and $\triangle P D A$ are all equal, then $A P: P B=$ $\qquad$ | 2.1:1.
As shown in Figure 4, extend $C D$ to $E$ and extend $A B$ to $F$, such that $D E=D P$, $B F=B C$. Connect $E P$ and $C F$. Since $\triangle P B C$ and $\triangle P C D$ have equal perimeters, we have
$$
P B+B C=P D+D C .
$$
Thus, $P F=E C$.
Also, $D C \parallel A B$, which means $E C \parallel P F$, so quadri... | 1:1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,652 |
3. Point $D$ is inside $\triangle A B C$, given $A B=a b$, $B C=b c$, $C A=c a$, $A D=a d$, $B D=b d$, $C D=c d$ $(a, b, c, d$ are all positive numbers). Then $\angle A B D+\angle A C D=$ $\qquad$ | $3.60^{\circ}$
As shown in Figure 5, construct $\triangle A C K \sim \triangle A B D$, then $\triangle A D K \sim \triangle A B C$.
At this point, $\frac{C K}{b d}=\frac{c a}{a b}$, which means
$$
C K=c d=C D \text {; }
$$
Similarly, $\frac{D K}{b c}=\frac{a d}{a b}$, which means
$$
D K=c d=C D \text {. }
$$
Therefor... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,653 |
4. Place the natural numbers $1,2, \cdots, 2 n$ randomly on a circle. It is found that among all sets of three consecutive numbers, there are $a$ sets where all three numbers are odd, $b$ sets where exactly two numbers are odd, $c$ sets where only one number is odd, and $d$ sets where all three numbers are even. If $a ... | 4. -3 .
If each number is counted 3 times, a total of $6 n$ numbers are counted ($3 n$ odd numbers, $3 n$ even numbers), then the system of equations can be set up as
$$
\left\{\begin{array}{l}
3 a+2 b+c=3 n, \\
2 c+b+3 d=3 n .
\end{array}\right.
$$
By eliminating $n$ from equations (1) and (2), we get
$$
\frac{b-c}{... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,654 |
II. (25 points) As shown in Figure 1, in the cyclic quadrilateral $ABCD$, from the midpoint $P$ of $AB$, draw $PE \perp BC, PF \perp CD, PG \perp DA$ (where $E$, $F$, and $G$ are the feet of the perpendiculars). Prove: $S_{\triangle PEF} = S_{\triangle ACF}$. | As shown in Figure 6, connect $E G$ intersecting $P F$ at $K$. Draw a perpendicular from $G$ to $C D$ (with the foot of the perpendicular at $M$) intersecting the ray $E P$ at $N$, and connect $A N$. It is easy to see that
$$
\angle B=\angle A D M \text {, }
$$
thus
$$
\angle B P E=\angle D G M \text {. }
$$
At this ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,656 |
Three. (25 points) Let $a_{n}$ denote the integer closest to $\sqrt{n}$ (where $n$ is a positive integer). Find the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{200 s}}$.
untranslated part:
s should be 00, making it 20000. The problem should read $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2... | Three, let $a_{n}=k$, then $|\sqrt{n}-k|<\frac{1}{2}$, which means
$$
\begin{array}{l}
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2} \\
\Leftrightarrow k^{2}-k+\frac{1}{4}<n<k^{2}+k+\frac{1}{4} \\
\Leftrightarrow k^{2}-k+1 \leqslant n \leqslant k^{2}+k .
\end{array}
$$
This indicates that when $n$ takes the values $k^{2}-k+1, ... | 88 \frac{5}{9} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,657 |
1. Given $x \in \mathbf{R}$, and $x \neq \frac{k \pi}{2}(k \in \mathbf{Z})$. Then $M=$ $4^{\sin ^{2} x}+4^{\cos ^{2} x}$ the integer part is ( ).
(A) 4
(B) 5
(C) 6
(D) 8 | 1. A.
On the one hand,
$$
M \geqslant 2 \sqrt{4^{\sin ^{2} x} \times 4^{\cos ^{2} x}}=2 \sqrt{4^{\sin ^{2} x+\cos ^{2} x}}=4.
$$
On the other hand, since $0<\sin ^{2} x<1$, then $1<4^{\sin ^{2} x}<4$, hence
$$
\left(4^{\sin ^{2} x}-1\right)\left(4^{\sin ^{2} x}-4\right)<0,
$$
which means $\left(4^{\sin ^{2} x}\right)... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,658 |
Example 2 Draw two tangents from point $P(-2,-2)$ to circle $\odot M$:
$$
(x-1)^{2}+(y-2)^{2}=9
$$
Let $A$ and $B$ be the points of tangency. Find the equation of the line containing the chord $AB$. | Solution 1: Since the radius $r=3$, the center of the circle is $M(1,2)$, so $|PM|=5$.
Also, $PA$ is a tangent, then $PA \perp AM$. Therefore,
$$
|PA|=\sqrt{|PM|^{2}-|AM|^{2}}=4 \text{. }
$$
Similarly, $|PB|=4$.
Thus, $A$ and $B$ are the intersection points of the circle $(x+2)^{2}+(y+2)^{2}=$ 16 and the circle $(x-1)... | 3 x+4 y-2=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,659 |
2. If the line $y=k x+1$ intersects the circle $x^{2}+y^{2}+k x+$ $m y-4=0$ at points $P$ and $Q$, and points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, then the area of the plane region represented by the system of inequalities
$$
\left\{\begin{array}{l}
k x-y+1 \geqslant 0, \\
k x-m y \leqslant 0, \\... | 2.D.
From the fact that points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, we know that the line $y=kx+1$ is perpendicular to $x+y=0$, and the center $\left(-\frac{k}{2},-\frac{m}{2}\right)$ of the circle lies on the line $x+y=0$, which gives $k=1$, and $\left(-\frac{k}{2}\right)+\left(-\frac{m}{2}\rig... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,660 |
3. Given that $P$ is a point inside $\triangle A B C$ and satisfies $2 P A+3 P B+4 P C=0$.
Then, $S_{\triangle P B C}: S_{\triangle P C A}: S_{\triangle P A B}$ equals ( ).
(A) $1: 2: 3$
(B) $2: 3: 4$
(C) $3: 4: 2$
(D) $4: 3: 2$ | 3. B.
As shown in Figure 4, extend $PA$ to $D$ such that $PD=2PA$; extend $PB$ to $E$ such that $PE=3PB$; extend $PC$ to $F$ such that $PF=4PC$. Then,
$$
PD + PE + PF = 0.
$$
Thus, $P$ is the centroid of $\triangle DEF$. Therefore, we have
$$
\begin{array}{l}
S_{\triangle PBC} = \frac{1}{3 \times 4} S_{\triangle PEF}... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,661 |
4. In modern society, the difficulty of deciphering passwords is increasing. There is a type of password that breaks down English plaintext (real text) into letters, where the 26 English letters $\mathrm{a}, \mathrm{b}, \cdots, \mathrm{z}$ (regardless of case) correspond to the 26 natural numbers $1,2, \cdots, 26$, as ... | 4.C.
The inverse transformation formula is
$$
x=\left\{\begin{array}{ll}
2 x^{\prime}-1, & 1 \leqslant x^{\prime} \leqslant 13, x^{\prime} \in \mathbf{N} ; \\
2 x^{\prime}-26, & 14 \leqslant x^{\prime} \leqslant 26, x^{\prime} \in \mathbf{N} .
\end{array}\right.
$$
According to the above transformation, we have
$$
\b... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 715,662 |
5. On the 4 surfaces of a regular tetrahedron, the numbers $1,2,3,4$ are written respectively. Four such uniform regular tetrahedrons are simultaneously thrown onto a table, and the product of the 4 numbers on the 4 surfaces that touch the table can be divided by 4. The probability of this is ( ).
(A) $\frac{1}{8}$
(B)... | 5.C.
The probability of the event "all 4 numbers are odd" is
$$
P_{1}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \text {; }
$$
The probability of the event "3 numbers are odd, 1 number is 2" is
$$
P_{2}=\mathrm{C}_{4}^{3}\left(\frac{1}{2}\right)^{3} \times \frac{1}{4}=\frac{1}{8} \text {. }
$$
Therefore, the probabil... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,663 |
6. Given that $a, b, c, d$ are all even numbers, and $0<a<b<c<d, d-a=90$. If $a, b, c$ form an arithmetic sequence, and $b, c, d$ form a geometric sequence, then the value of $a+b+c+d$ is ( ).
(A) 194
(B) 284
(C) 324
(D) 384 | 6. A.
According to the problem, we can set $a, b, c, d$ as $b-m, b, b+m, \frac{(b+m)^{2}}{b}$ (where $m$ is a positive even number, and $m<b$).
From $d-a=90$, we get $\frac{(b+m)^{2}}{b}-(b-m)=90$, which simplifies to
$$
m^{2}+3 b m-90 b=0 \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know th... | 194 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,664 |
1. Given a triangle $\triangle A B C$ with side lengths $4,5,6$ respectively, the circumcircle of $\triangle A B C$ is a great circle of sphere $O$, and $P$ is a point on the sphere. If the distances from point $P$ to the three vertices of $\triangle A B C$ are all equal, then the volume of the tetrahedron $P-A B C$ is... | 1.10 .
Since $P A=P B=P C$, the projection of point $P$ on the plane $A B C$ is the circumcenter $O$ of $\triangle A B C$, i.e., $P O \perp$ plane $A B C$, and $P O$ equals the radius $R$ of sphere $O$. Therefore,
$$
V_{P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{1}{3} \cdot \frac{a b c}{4 R} \cdot R=10 \... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,665 |
2. Let $x, y, z$ be real numbers, not all zero. Then the maximum value of the function $f(x, y, z)=\frac{x y+y z}{x^{2}+y^{2}+z^{2}}$ is $\qquad$ | 2. $\frac{\sqrt{2}}{2}$.
Introduce positive parameters $\lambda, \mu$.
Since $\lambda^{2} x^{2}+y^{2} \geqslant 2 \lambda x y, \mu^{2} y^{2}+z^{2} \geqslant 2 \mu y z$. Therefore, $x y \leqslant \frac{\lambda}{2} \cdot x^{2}+\frac{1}{2 \lambda} \cdot y^{2}, y z \leqslant \frac{\mu}{2} \cdot y^{2}+\frac{1}{2 \mu} \cdot... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,666 |
3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.
Then the value of $a-2 b+c$ is $\qquad$. | 3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$
When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,667 |
4. As shown in Figure 1, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, respectively, and $P$ is a point on the right branch of the hyperbola (excluding the right vertex). $\odot A$ is tangent to the side $P F_{2}$ of $\triangle P F_{1} F_{2}$, and also tangent to... | 4. $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5)$.
Connecting $P A$ and extending it to intersect the $x$-axis at point $B$, by the properties of the angle bisector, the ratio theorem, and the definition of a hyperbola, we get
$$
\begin{array}{l}
\frac{|P A|}{|A B|}=\frac{\left|P F_{1}\right|}{\left|F_{1} B\right|}=\fra... | \frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,668 |
5. Now define an operation * :
When $m$ and $n$ are both positive odd numbers or both positive even numbers,
$$
m * n=m+n \text {; }
$$
When one of $m$ and $n$ is a positive odd number and the other is a positive even number,
$$
m * n=m \cdot n \text {. }
$$
Then, the number of elements in the set $M=\{(a, b) \mid a... | 5.41.
Since $36=1+35=3+33=5+31=7+29=\cdots=$ $33+3=35+1$ (a total of 18) $; 36=2+34=4+32=\cdots=32$ $+4=34+2$ (a total of 17) ; $36=1 \times 36=3 \times 12=4 \times 9=9$ $\times 4=12 \times 3=36 \times 1$ (a total of 6), therefore, the number of elements in set $M$ is $18+17+6=41$. | 41 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,669 |
Example 3 In quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$. Take a point $E$ on $CD$, and let $BE$ intersect $AC$ at $F$. Extend $DF$ to meet $BC$ at $G$. Prove that:
$$
\angle GAC = \angle EAC \text{.}
$$
(1999, National High School Mathematics Competition) | Proof: As shown in Figure 1, with $A$ as the origin and the line $A C$ as the $x$-axis, establish a Cartesian coordinate system. Let $C(m, 0)$, $F(n, 0)$, and $D\left(x_{1}, k x_{1}\right)$.
Since $A C$ bisects $\angle B A D$, let $B\left(x_{2},-k x_{2}\right)$.
Therefore, the equation of $C D$ is
$$
\left(x-x_{1}\rig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,670 |
6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$
Then $u_{\text {max }}+u_{\text {min }}=$ | 6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,671 |
Four. (20 points) Let $A$, $B$, and $C$ be three distinct points on the parabola $x^{2}=y$. If the lines $AB$ and $BC$ are both tangent to the circle $x^{2}+(y-2)^{2}=1$, prove: The line $AC$ is also tangent to this circle. | Let $A\left(t_{1}, t_{1}^{2}\right) 、 B\left(t_{2}, t_{2}^{2}\right) 、 C\left(t_{3}, t_{3}^{2}\right)$, then the equation of line $A B$ is
$$
\left(t_{1}+t_{2}\right) x-y-t_{1} t_{2}=0 .
$$
Since line $A B$ is tangent to the circle $x^{2}+(y-2)^{2}=1$, we have
$$
\frac{\left|-2-t_{1} t_{2}\right|}{\sqrt{\left(t_{1}+t_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,673 |
Five. (20 points) Given the function
$$
\begin{array}{l}
f(x)=\frac{a(x-1)^{2}+1}{b x+(c-b)}(a, b, c \in \mathbf{N}), \\
f(2)=2, f(3)=0, \text{ and } a > 0, \text{ prove:}
\end{array}
$$
$$
\begin{array}{l}
{[f(x+1)]^{n}-f\left(x^{n}+1\right)} \\
\geqslant 2^{n}-2\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$ | The function expression of the obtained image is $f(x+1)=\frac{a x^{2}+1}{b x+c}$.
Since the image of $f(x+1)$ is centrally symmetric about the origin, we have
$$
f(-x+1)=-f(x+1)
$$
always holds, that is,
$$
\frac{a(-x)^{2}+1}{b(-x)+c}=-\frac{a x^{2}+1}{b x+c} \text {. }
$$
Therefore, $-b x+c=-b x-c$ always holds, y... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,674 |
一、(50 points) As shown in Figure 2, let $H$ be the orthocenter of acute $\triangle A B C$. Draw a tangent $A E$ from vertex $A$ to the circle $\odot O$ with diameter $B C$, and let the point of tangency be $E$. Connect $E H$ to intersect $A O$ at point $G$, and draw an arbitrary chord $P Q$ of $\odot O$ through $G$. Pr... | As shown in Figure 5, let $A D$ and $C F$ be two altitudes of $\triangle A B C$. Therefore, $F$ lies on $\odot O$. Since
$$
\begin{array}{l}
\angle B F C=\angle A D B \\
=90^{\circ},
\end{array}
$$
thus, points $B, F, H, D$ are concyclic.
By the secant theorem, we have
$$
A F \cdot A B=A H \cdot A D \text {. }
$$
Con... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,675 |
Given the sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and $1<a_{1}<a_{2}<\cdots<a_{n}<\cdots$. Let
$$
\begin{array}{l}
S=\sum_{k=1}^{n-1}\left[a_{k}, a_{k+1}\right]+\sum_{k=1}^{n} a_{k}, \\
t=\max _{1 \leqslant k \leqslant n}\left\{a_{k}-a_{k-1}\right\},
\end{array}
$$
where $[a, b]$ denotes... | Sure, here is the translation:
---
II. First, prove the right inequality.
Since $\left[a_{k}, a_{k+1}\right] \leqslant a_{k} a_{k+1}, a_{k}+1 \leqslant a_{k+1}$, we have,
$$
\begin{aligned}
S & \leqslant \sum_{k=1}^{n-1} a_{k} a_{k+1}+\sum_{k=1}^{n} a_{k} \\
& =\sum_{k=1}^{n-1} a_{k} a_{k+1}+\sum_{k=1}^{n-1} a_{k+1}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,676 |
Three. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\} \text {, }
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedra. How many triangles can graph $G$ have at most? | Three, first prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Proof: Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the most, say there are $k$ edges: $... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,677 |
Do there exist two irrational numbers $a$ and $b$ ($a > b$) such that their sum and product are equal to the same integer, and both $a$ and $b$ are greater than $\frac{1}{2}$ and less than 4? If they exist, find these two numbers; if not, explain the reason. | Solution: Let $a+b=ab=t, t$ be an integer, then $a, b$ are the two roots of the equation $x^{2}-t x+t=0$.
Let $f(x)=x^{2}-t x+t$.
Since $a, b$ are both greater than $\frac{1}{2}$ and less than 4, then
$$
\left\{\begin{array}{l}
\Delta = t^{2} - 4 t > 0, \\
\frac{1}{2} < \frac{t}{2} < 4, \\
f\left(\frac{1}{2}\right) = \... | a=\frac{5+\sqrt{5}}{2}, b=\frac{5-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,678 |
Given a three-digit positive integer $A$, swap the last two digits and add the resulting number to the given number, thus obtaining a four-digit number starting with 173. Find the number $A$.
Swap the last two digits of the given three-digit positive integer $A$ and add the resulting number to the given number, thus o... | Let $\overline{\alpha \beta \gamma}$ be the given three-digit number, then the new three-digit number is $\overline{\alpha \gamma \beta}$. Therefore, we have
$$
\begin{array}{l}
\overline{\alpha \beta \gamma}+\overline{\alpha \gamma \beta}=100 \alpha+10 \beta+\gamma+100 \alpha+10 \gamma+\beta \\
=200 \alpha+11(\beta+\g... | 839,848,857,866,875,884,893 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,679 |
Given $a, b, c$ are positive numbers satisfying $a+b+c=1$. Prove:
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geqslant \frac{27}{4} .
$$ | Prove: From the binary mean inequality, we have
$$
\begin{array}{l}
\frac{1}{a(1+b)}+\frac{81 a(1+b)}{16} \geqslant \frac{9}{2}, \\
\frac{1}{b(1+c)}+\frac{81 b(1+c)}{16} \geqslant \frac{9}{2}, \\
\frac{1}{c(1+a)}+\frac{81 c(1+a)}{16} \geqslant \frac{9}{2} .
\end{array}
$$
Adding the above three inequalities and rearra... | \frac{27}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,680 |
Example 4 Given the family of curves
$$
2(2 \sin \theta-\cos \theta+3) x^{2}-(8 \sin \theta+\cos \theta+1) y=0
$$
( $\theta$ is a parameter). Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
(1995, National High School Mathematics Competition) | Solution: Given that the family of curves always passes through the origin, and the line $y=2x$ also passes through the origin, we know that the length of the chord intercepted by the family of curves on $y=2x$ depends only on the coordinates of the other intersection point of the family of curves with $y=2x$. Substitu... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,681 |
Given that quadrilateral $ABCD$ is a square, $P$ is a point on side $CD$ (point $P$ does not coincide with the vertices), and the extension of $AP$ intersects the extension of $BC$ at point $Q$. Let the inradii of $\triangle ABQ$, $\triangle PAD$, and $\triangle PCQ$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively.
(1) ... | Solution: (1) As shown in Figure 3, since
$$
\begin{aligned}
& \mathrm{Rt} \triangle A B Q \backsim \\
& \mathrm{Rt} \triangle P D A \backsim \\
& \mathrm{Rt} \triangle P C Q, \\
& \text { therefore, } r_{1}: r_{2}: r_{3} \\
= & A B: P D: P C .
\end{aligned}
$$
And $A B=C D=P D+P C$, hence
$$
r_{1}=r_{2}+r_{3} \geqsla... | \left(3-2 \sqrt{2}, \frac{1}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,682 |
Example 5 Given the equations of the three sides of a triangle are $x-6=0$, $x+2 y=0$, and $x-2 y+8=0$. Find the equation of the circumcircle of this triangle.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | The equation of the quadratic curve system passing through the three vertices of a triangle is
$$
\begin{array}{l}
(x-6)(x+2 y)+m(x+2 y)(x-2 y+8)+ \\
n(x-6)(x-2 y+8)=0 .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
(1+m+n) x^{2}+(2-2 n) x y-4 m y^{2}+ \\
(8 m+2 n-6) x+(16 m+12 n-12) y- \\
48 n=0 .
\end{array... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,683 |
Example 6 Draw two lines from any point $P$ not on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ intersecting the ellipse at points $A, B$ and $C, D$ respectively. If the inclination angles of lines $AB, CD$ are $\alpha, \beta$ respectively, and $\alpha+\beta=\pi$, prove that points $A, B, C, D$ are concyclic... | Proof: Let the equations of lines $AB$ and $CD$ be $y = kx + m$ and $y = -kx + n$, respectively. Then the equation of the quadratic curve system passing through points $A, B, C, D$ is
$$
\begin{array}{l}
(y - kx - m)(y + kx - n) + \\
\lambda\left(b^{2} x^{2} + a^{2} y^{2} - a^{2} b^{2}\right) = 0,
\end{array}
$$
which... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,684 |
Example 7 Given that $M N$ is a chord of $\odot O$, $R$ is the midpoint of $M N$, a chord $A B$ and a chord $C D$ are drawn through $R$, and a conic section passing through points $A, B, C, D$ intersects $M N$ at points $P, Q$. Prove: $R$ is the midpoint of $P Q$. | Proof: Establish a Cartesian coordinate system with the center $O$ as the origin and the line passing through point $R$ as the $y$-axis. Let the radius of $\odot O$ be $r$, and the coordinates of $R$ be $(0, b)$. The equations of the lines $A B$ and $C D$ are $y=k_{1} x+b$ and $y=k_{2} x+b$, respectively. Then the equa... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,685 |
Example 3 (1) In a $4 \times 4$ grid paper, some small squares are colored red, and then two rows and two columns are crossed out. If no matter how they are crossed out, there is at least one red small square that is not crossed out, how many small squares must be colored at least?
(2) If the “$4 \times 4$” grid paper ... | Solution: (1) If the number of colored small squares is less than or equal to 4, then we can appropriately strike out two rows and two columns to remove all the colored small squares.
If the number of colored small squares is 5, then by the pigeonhole principle, there must be at least one row with 2 or more colored sm... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,687 |
2. Find the equation of the circle passing through the intersection points of the two circles
$$
x^{2}+y^{2}+6 x-4=0 \text { and } x^{2}+y^{2}+6 y-28=0
$$
and whose center lies on the line $x-y-4=0$. | (Tip: Let the equation of the required circle be $x^{2}+y^{2}+6 x-4+$ $\lambda\left(x^{2}+y^{2}+6 x-28\right)=0$. Convert it to the standard form of a circle and substitute the coordinates of the center into the given line equation to find $\lambda=-7$. The equation of the required circle is $\left(x-\frac{1}{2}\right)... | \left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{7}{2}\right)^{2}=\frac{178}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,688 |
3. From the vertex $A$ of $\triangle A B C$, draw a perpendicular to $B C$, with the foot of the perpendicular being $D$. Take any point $H$ on $A D$, and let the line $B H$ intersect $A C$ at $E$, and the line $C H$ intersect $A B$ at $F$. Prove: $A D$ bisects the angle formed by $D E$ and $D F$.
| (Tip: Establish a Cartesian coordinate system with the line $BC$ as the $x$-axis, $AD$ as the $y$-axis, and $D$ as the origin. Let $A(0, a)$, $B(0, b)$, $C(c, 0)$, and $H(0, h)$. Then the equation of the line system passing through point $E$ is
$$
\begin{array}{l}
\lambda\left(\frac{x}{b}+\frac{y}{h}-1\right)+\mu\left(... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,689 |
4. Let $a>b>0$, draw lines $l$ and $m$ through the fixed points $A(a, 0)$ and $B(b, 0)$ respectively, such that they intersect the parabola $y^{2}=x$ at four distinct points. When these four points are concyclic, prove that the intersection point of lines $l$ and $m$ always lies on a straight line. | (The equation of a circle passing through four points is
$$
\left(y-k_{1} x+k_{1} a\right)\left(y-k_{2} x+k_{2} b\right)+\lambda\left(y^{2}-x\right)=0 \text {. }
$$
From the equality of the coefficients of $x^{2}$ and $y^{2}$ and the coefficient of $x y$ being 0, it can be deduced that the intersection point of lines ... | x=\frac{a+b}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 715,690 |
Example 1: Do there exist positive integers $m, n$ that satisfy
$$
5 m^{2}-6 m n+7 n^{2}=2004 \text{?}
$$ | Solution: Completing the square for the original equation, we get
$$
(5 m-3 n)^{2}+26 n^{2}=10020 \text {. }
$$
From the coefficient of $n^{2}$, 26, and the constant term, we know that both sides of the equation are congruent modulo 13, hence
$$
(5 m-3 n)^{2} \equiv 10(\bmod 13) \text {. }
$$
Since for any integer $k... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,691 |
Example 2 Find all positive integers $x, y$ such that $x^{2}+615=2^{y}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: For non-negative integer $k$, we have
$$
2^{2 k+1}=4^{k} \times 2 \equiv(-1)^{k} \times 2 \equiv 2 \text { or } 3(\bmod 5) \text {. }
$$
Since $x^{2}=0$ or 1 or $4(\bmod 5)$, $y$ must be even.
Let $y=2 z$, substituting into the given equation we get
$$
\left(2^{3}-x\right)\left(2^{5}+x\right)=615=3 \times 5 ... | x=59, y=12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,692 |
Example 3 Find all triangles with integer side lengths and an incircle radius of 1.
Find all triangles with integer side lengths and an incircle radius of 1. | Let the three sides of the triangle be $a$, $b$, and $c$, with $a \geqslant b \geqslant c$, and let $p=\frac{1}{2}(a+b+c)$. By Heron's formula and the area formula, we have
$$
p \times 1=\sqrt{p(p-a)(p-b)(p-c)},
$$
which simplifies to $p=(p-a)(p-b)(p-c)$.
Thus, $4(a+b+c)$
$$
=(b+c-a)(c+a-b)(a+b-c) \text {. }
$$
There... | x=1, y=2, z=3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,693 |
Example 4 In a Cartesian coordinate system, points with both coordinates as integers are called integer points. Please design a scheme to color all integer points, with each integer point being colored one of red, yellow, or blue, such that
(1) points of each color appear on infinitely many lines parallel to the x-axis... | Solution: Let the integer point be $(x, y)$, then the integer points in the Cartesian coordinate system can be divided into four categories: (odd, odd), (odd, even), (even, odd), (even, even). Select any two of these four categories and color them red and blue, respectively, and color the remaining two categories yello... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,694 |
Example 5 Let $A$ be the sum of the digits in the decimal representation of $4444^{4444}$, and $B$ be the sum of the digits of $A$. Try to find the sum of the digits in the decimal representation of $B$.
Translate the above text into English, please retain the original text's line breaks and format, and output the tra... | Solution: According to (1), we can obtain the estimate
$$
4444^{444}<\left(10^{4}\right)^{4444}=10^{1776} \text {, }
$$
Thus, $A \leqslant 17776 \times 9=159984$.
Furthermore, $B<1+5 \times 9=46$.
Let the sum of the digits of $B$ be $C$, then $C \leqslant 12$.
Therefore, from
$$
4444^{444} \equiv 7(\bmod 9)
$$
we can... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,695 |
Example 6 Let $P(x)$ be the product of the digits in the decimal representation of $x$. Try to find all positive integers $x$ such that
$$
P(x)=x^{2}-10 x-22
$$
holds. | Solution: Let $x=\overline{a_{n} a_{n-1} \cdots a_{1}}$.
When $n \geqslant 2$, according to (1) and (2), we have the estimate
$$
x \geqslant 10^{n-1} \cdot a_{n}>9^{n-1} \cdot a_{n} \geqslant P(x) \text {, }
$$
thus $x^{2}-10 x-22<x$.
Solving this, we get $0<x<13$.
Upon verification, $x=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,696 |
Example 4 Some middle school students in a city participated in a mathematics invitational competition, and this competition had a total of 6 problems. It is known that each problem was solved correctly by exactly 500 students, but for any two students, there is at least one problem that neither of them solved correctl... | Solution: First, each student can answer at most 4 questions correctly.
In fact, according to the problem, it is impossible for any student to answer 6 questions correctly. If a student answers 5 questions correctly, by the problem's condition, all other students must have answered the same question incorrectly, which ... | 1000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,698 |
Example 8 Try to find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$
has positive integer solutions. | Solution: Without loss of generality, let $x \leqslant y \leqslant z$. Clearly, $z^{2} \mid\left(x^{3}+y^{3}\right)$, so, $z^{2} \leqslant x^{3}+y^{3}$.
But $x^{3} \leqslant x z^{2}, y^{3} \leqslant y z^{2}$, hence we have
$$
z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y) \text {. }
$$
Thus, w... | n=3,1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,699 |
Question 6 As shown in Figure 7, given an equilateral $\triangle ABC$, $D$ is any point on side $BC$, the circumcenter and orthocenter of $\triangle ABD$ are $O_{1}$ and $H_{1}$, respectively, and the circumcenter and orthocenter of $\triangle ACD$ are $O_{2}$ and $H_{2}$, respectively. Prove:
(1) $\angle H_{1} O_{1} H... | Proof: (1) Without loss of generality, assume $B D \leqslant C D$. Since $O_{2}$ is the circumcenter of $\triangle A C D$ and $H_{2}$ is the orthocenter of $\triangle A C D$, we have
$$
\begin{array}{l}
\angle A O_{2} D=2 \angle C=120^{\circ}, \\
\angle A H_{2} D=180^{\circ}-\angle C=120^{\circ} .
\end{array}
$$
There... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,700 |
Question 7 As shown in Figure 8, let $D$ be a point on the arc $\overparen{B C}$ of the circumcircle $\odot O$ of the equilateral $\triangle A B C$. The incircles of $\triangle A B D$ and $\triangle A C D$ are $\odot I_{1}$ and $\odot I_{2}$, respectively. One of the external common tangents of these two circles inters... | Proof: (1) As shown in Figure 8, let the line $A I_{1}$ intersect $B D$ at $P$ and $B C$ at $G$, and the line $A I_{2}$ intersect $C D$ at $Q$.
Since $A P$ bisects $\angle B A D$, we have $\frac{A B}{A D}=\frac{B P}{D P}$.
Since $A I_{2}$ bisects $\angle C A D$, we have $\frac{A C}{A D}=\frac{C Q}{D Q}$.
Given $A B=A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,701 |
Question 8 As shown in Figure 9, in $\triangle ABC$, $\angle BAC = 120^{\circ}$, $AD$ is the angle bisector of $\angle BAC$, the circumcenter and incenter of $\triangle ABD$ are $O_{1}$ and $I_{1}$, respectively, and the circumcenter and incenter of $\triangle ACD$ are $O_{2}$ and $I_{2}$, respectively. The line $O_{1}... | Proof: (1) Auxiliary lines as shown in Figure 9.
Since $\angle B O_{1} D=2 \angle B A D=120^{\circ}$, $\angle C O_{2} D=2 \angle C A D=120^{\circ}$, then $\triangle B O_{1} D \sim \triangle C O_{2} D$.
Therefore, $\frac{B D}{C D}=\frac{D O_{1}}{D O_{2}}$.
By $A D$ bisecting $\angle B A C$, we get $\frac{A B}{A C}=\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,702 |
Example 1 If three lines $a$, $b$, and $c$ in space are pairwise skew lines, then the number of lines that intersect with lines $a$, $b$, and $c$ is ( ).
(A)0
(B) 1
(C) more than 1 but a finite number
(D) infinitely many
(1997, National High School Mathematics League Competition) | Solution: First, regardless of the positional relationship between lines $a$, $b$, and $c$, we can always construct a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D D_{1}$ lies on line $c$. Next, take any point $M$ on the extension of $D D_{1}$,... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,703 |
Proposition 2 There exist four pairwise skew lines such that no line can intersect all of them simultaneously.
A direct proof idea is, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, a line that intersects $A B, A_{1} C, D_{1} D$ simultaneously must be a skew line to $B_{1} C_{1}$. Here, we introduce a more concise ind... | Proof: Proof by contradiction.
Take a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, then, the lines $A B, A_{1} C, D_{1} D, B_{1} C_{1}$ are pairwise skew lines (as shown in Figure 6).
Assume there exists
a line $l$ that intersects
lines $A B, A_{1} C$,
$D_{1} D, B C_{1}$ at
points $E$,
$F, G, H$ respectively.
Now rotate th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,705 |
Proposition 4 For pairwise skew lines $a_{1}, a_{2}, \cdots$, $a_{n}(n \geqslant 3)$, there exists a line $l$ that is not perpendicular to the lines $a_{1}, a_{2}, \cdots, a_{n}$. | Proof: Take any point $A$ outside these $n$ lines, and draw a line $a_{i}^{\prime} \parallel a_{i}(i=1,2, \cdots, n)$ through point $A$, then draw a plane $\alpha_{i} \perp a_{i}^{\prime}(i=1,2, \cdots, n)$ through point $A$. Then the lines through $A$ and perpendicular to $a_{i}^{\prime}$ (and thus also to $a_{i}$) ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,707 |
A paper has a circle $\odot O$ with radius $R$ and a fixed point $A$ inside the circle, where $O A=a$. Fold the paper so that a point $A_{1}$ on the circumference coincides exactly with point $A$. Each such fold leaves a crease. Find the set of points on all the crease lines when $A_{1}$ covers all points on the circum... | Solution: Establish a rectangular coordinate system as shown in Figure 1, let $P(x, y)$ be any point on the line $l$ (the crease), and $B$ be the midpoint of $A A_{1}$. Thus, we have
$$
\left\{\begin{array}{l}
O B=\frac{1}{2}\left(O A+O A_{1}\right), \\
P B=O B-O P, \\
A A_{1}=O A_{1}-O A, \\
P B \cdot A A_{1}=0 .
\end... | \frac{\left(x-\frac{a}{2}\right)^{2}}{\left(\frac{R}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{R}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,708 |
Proposition If the incircle of $\triangle A B C$ touches the sides at points $D, E, F$, and the circumradius and inradius of $\triangle A B C$ are $R, r$ respectively, then $S_{\triangle D E F}=\frac{r}{2 R} S_{\triangle A B C}$. | Proof: As shown in Figure 1, connect $O A, O D, O E, O F$, then $O A$ is the perpendicular bisector of $E F$. Let the side lengths of $\triangle A B C$ and $\triangle D E F$ be $a, b, c, d, e, f$ respectively. Therefore,
$$
\begin{array}{l}
E F=2 r \sin \angle A O E=2 r \sin \left(90^{\circ}-\frac{A}{2}\right) \\
=2 r ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,710 |
4. As shown in Figure 1, in $\triangle A B C$, $C D$ and $C E$ are the altitude and median on side $A B$, respectively, with $C E=B E=1$. The perpendicular bisector of $C E$ passes through point $B$ and intersects $A C$ at point $F$. Then $C D+B F=$ $\qquad$ | 4. $\frac{7 \sqrt{3}}{6}$ | \frac{7 \sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,714 |
9. Given a quadratic equation with real coefficients $a x^{2}+2 b x$ $+c=0$ has two real roots $x_{1} 、 x_{2}$. If $a>b>c$, and $a$ $+b+c=0$, then $d=\left|x_{1}-x_{2}\right|$ the range of values is $\qquad$ | 9. $\sqrt{3}<d<2 \sqrt{3}$ | \sqrt{3}<d<2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,719 |
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