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2. Let $x, y, z$ be pairwise distinct real numbers, and
$$
\sqrt{z(x-z)}+\sqrt{z(y-z)}=\sqrt{x-z}-\sqrt{z-y} \text {, }
$$
then $\frac{3 x^{2}+x y-y^{2}}{x^{2}-x y+y^{2}}=$ $\qquad$ | 2. $\frac{1}{3}$.
From the given, we have
$$
\left\{\begin{array}{l}
z(x-z) \geqslant 0, \\
z(y-z) \geqslant 0, \\
x-z \geqslant 0, \\
z-y \geqslant 0 .
\end{array}\right.
$$
From (9) and (1), we get $z \geqslant 0$;
From (3) and (2), we get $z \leqslant 0$.
Thus, $z=0$.
Substituting $z=0$ into the given equation, we... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,470 |
4. As shown in the figure, on a semicircle with center $C$ and diameter $M N$, there are two different points $A$ and $B$. Point $P$ is on $C N$, and $\angle C A P = \angle C B P = 10^{\circ}$. If $\overparen{M A} = 40^{\circ}$, then $\overparen{B N}$ equals . $\qquad$ | 4. $20^{\circ}$.
From the given information, we can obtain $\angle A P C=30^{\circ}$.
In $\triangle A C P$, we have $\frac{\sin 10^{\circ}}{C P}=\frac{\sin 30^{\circ}}{A C}$.
In $\triangle B C P$, we have $\frac{\sin 10^{\circ}}{C P}=\frac{\sin \angle C P B}{B C}$.
Since $A C=B C$,
Therefore, $\sin 30^{\circ}=\sin \an... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,472 |
One, (20 points) If the vertex of the parabola $y=-x^{2}+2 a x+b$ moves on the line $m x-y-2 m+1=0$, and it has a common point with the parabola $y=x^{2}$, find the range of changes for $m$.
| Given the parabola $y=-x^{2}+2 a x+b$ has its vertex at $\left(a, a^{2}+b\right)$, then $\left(a, a^{2}+b\right)$ satisfies the linear equation. We have
$$
\begin{array}{l}
m a-a^{2}-b-2 m+1=0, \\
b=m a-a^{2}-2 m+1 .
\end{array}
$$
If $y=-x^{2}+2 a x+m a-a^{2}-2 m+1=0$,
the discriminant is
$$
\Delta=4 a^{2}-8 m a-8 a^... | m \leqslant 2-\sqrt{2} \text { or } m \geqslant 2+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,473 |
II. (20 points) As shown, $AB$ intersects $CD$ at $E$, $DA // EF // BC$, and $AE: EB=1: 2$, $S_{\triangle ADE}=1$. Find $S_{\triangle AEF}$. | $$
\text { II. } \because A D / / B C, \therefore \triangle A E D \backsim \triangle B E C \text {, }
$$
Thus, $S_{\triangle B E C}: S_{\triangle A E D}=4: 1$.
Let $S_{\triangle A E F}=m$, since $\triangle A E F \backsim \triangle A B C, A E$
: $A B=1: 3$, we get
$$
m: S_{\triangle A B C}=1: 9, m: S_{\text { quadrilat... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,474 |
Three, (20 points) Place a real number in each cell of a $4 \times 4$ grid paper,
such that the sum of the four numbers in each row, each column, and each diagonal equals a constant $k$. Find the sum of the numbers in the four corners of this $4 \times 4$ grid paper. | Three, we use $a_{ij}$ to denote the number in the $i$-th row and $j$-th column of a $4 \times 4$ grid $(1 \leqslant i, j \leqslant 4)$. From the given information, we have
$$
\left\{\begin{array}{l}
\sum_{i=1}^{4} a_{i 2}+\sum_{i=1}^{4} a_{i 3}=2 k, \\
\sum_{j=1}^{4} a_{1 j}+\sum_{j=1}^{4} a_{4 j}=2 k .
\end{array}\ri... | k | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,475 |
1. Among the numbers $3^{8}, 4^{7}, 5^{6}, 6^{5}$, the largest one is ( ).
(A) $3^{8}$
(B) $4^{7}$
(C) $5^{6}$
(D) $6^{5}$ | -1. (B)
$\because 4^{7}>3^{8} \Leftrightarrow\left(\frac{4}{3}\right)^{7}>3$, and indeed $\left(\frac{4}{3}\right)^{7}=\left(1+\frac{1}{3}\right)^{7}>1+\frac{7}{3}>3$.
$4^{7}>5^{6} \Leftrightarrow\left(\frac{5}{4}\right)^{6}6^{5} \Leftrightarrow\left(\frac{5}{6}\right)^{6}>\frac{1}{6}$, and $\left(\frac{5}{6}\right)^{3... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,476 |
2. Let $z_{1}=1+i, z_{2}=2+i, z_{3}=3+i$. Then $\arg \left(-10 \sqrt{3}+z_{1} z_{2} z_{3}\right)$ equals ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{3}$
(C) $\frac{2 \pi}{3}$
(D) $\frac{5 \pi}{6}$ | 2. (D)
$$
\begin{array}{l}
\text { Since } z_{1} z_{2} z_{3}=(1+i)(2+i)(3+i)=10 i, \\
\therefore \arg \left(-10 \sqrt{3}+z_{1} z_{2} z_{3}\right) \\
=\arg (-10 \sqrt{3}+10 i)=\frac{5 \pi}{6} .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,477 |
3. Proposition 甲: Plane $\alpha \perp$ plane $\beta$, plane $\rho \perp$ plane $\gamma$, then plane $\alpha / /$ plane $\gamma$
Proposition 乙: Three non-collinear points on plane $\alpha$ are equidistant from plane $\beta$, then plane $\alpha / /$ plane $\beta$.
Then ( ).
(A) Proposition 甲 is true, Proposition 乙 is tr... | 3. (D)
For proposition A, construct a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, with the plane of $A_{1} B_{1} C_{1} D_{1}$ being $\alpha$, and the plane of $B B_{1} D_{1} D$ being $\beta$. It is easy to see that $\alpha \perp \beta$. The plane of $A A_{1} C_{1} C$ is $\gamma$, and it is easy to see that $\beta \perp \g... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,478 |
Example 9. Determine all triples of real numbers $(x, y, z)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
2 x+x^{2} y=y, \\
2 y+y^{2} z=z, \\
2 z+z^{2} x=x .
\end{array}\right.
$$ | Solving by rearranging and factoring, the original system of equations can be transformed into
$$
\left\{\begin{array}{l}
y=\frac{2 x}{1-x^{2}}, \\
z=\frac{2 y}{1-y^{2}}, \\
x=\frac{2 z}{1-z^{2}} .
\end{array}\right.
$$
Let \( x = \tan \theta \left( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \right) \), then the original... | x = \tan \frac{k\pi}{7}, \quad y = \tan \frac{2k\pi}{7}, \quad z = \tan \frac{4k\pi}{7}. \quad (k = 0, \pm 1, \pm 2, \pm 3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,480 |
5. The solution set of the equation $x^{2}+x-1=x e^{x^{2}-1}+\left(x^{2}-1\right) e^{x}$ is $A$ (where, $e$ is an irrational number, $e=2.71828$ $\cdots$). Then the sum of the squares of all elements in $A$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 5. (C)
Let $y=x^{2}-1$. The original equation becomes
$$
x+y=x e^{y}+y e^{x} \text {, }
$$
which is $x\left(e^{y}-1\right)+y\left(e^{x}-1\right)=0$.
Since $x>0$ implies $e^{x}-1>0$, i.e., $x$ and $e^{x}-1$ always have the same sign, $y$ and $e^{y}-1$ also always have the same sign.
Therefore, $x\left(e^{y}-1\right) ... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,481 |
6. In the Cartesian coordinate system $x-o-y$, the figure defined by the inequalities
$$
y^{100}+\frac{1}{y^{100}} \leqslant x^{100}+\frac{1}{x^{100}}, x^{2}+y^{2} \leqslant 100
$$
has an area equal to ( ).
(A) $50 \pi$
(B) $40 \pi$
(C) $30 \pi$
(D) $20 \pi$ | 6. (A)
The region we are examining is symmetric with respect to the coordinate axes, so we only need to consider the part where $x>0, y>0$.
Let $z=x^{100}, t=y^{100}$, then we have
$$
\begin{array}{l}
z+\frac{1}{z} \geqslant t+\frac{1}{t} \Leftrightarrow\left(z^{2}+1\right) t \geqslant\left(t^{2}+1\right) z \\
\Leftri... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,482 |
1. If $a, b, c$ are all integers, and satisfy
$$
\begin{array}{l}
\sqrt{9-8 \sin 50^{\circ}}=a+b \sin C^{\circ} . \\
\text { then } \frac{a+b}{c}=
\end{array}
$$ | $$
\text { II. 1. } \frac{1}{2} \text {. }
$$
Since $9-8 \sin 50^{\circ}$
$$
\begin{aligned}
= & 9+8 \cos 80^{\circ} \\
& -8 \cos 80^{\circ}-8 \cos 40^{\circ} \\
= & 9+8 \sin 10^{\circ} \\
& -16 \cos 60^{\circ} \cos 20^{\circ} \\
= & 9+8 \sin 10^{\circ} \\
& -8 \cos 20^{\circ} \\
= & 1+8 \sin 10^{\circ} \\
& +8\left(1... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,483 |
2. If the solution set of the inequality
$|x-a|<|x|+|x+1|$ is all real numbers, then the range of values for $a$ is $\qquad$ | 2. $-1<a<0$.
Since the inequality $|x-a|<|x|+|x+1|$ means that the graph of $y=|x-a|$ must be below the graph of $y=|x|+|x+1|$, it is necessary and sufficient that $-1<a<0$. | -1<a<0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,484 |
3. Given a cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with edge length 1, the distance between the skew lines $A C_{1}$ and $B_{1} C$ is $\qquad$ | 3. $\frac{\sqrt{6}}{6}$.
From the conditions of the cube, it is easy to know that $B_{1} C \perp$ plane $\left(A B C_{1}\right)$. Therefore, $B_{1} C$ is perpendicular to any line in $\left(A B C_{1}\right)$. Draw $Q P \perp$ $A C_{1}$ at $P$ through the intersection point $Q$ of $B_{1} C$ and $B C_{1}$. Then $B_{1} C... | \frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,485 |
4. The numbers $2 x, 1, y-1$ form an arithmetic sequence in the given order, and $y+3,|x+1|+|x-1|$, $\cos (\arccos x)$ form a geometric sequence in the given order. Then $x+y+x y=$ | 4. 3.
Since $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
2 x+y-1=2 \Rightarrow y=3-2 x \text {. }
$$
And $\cos (\arccos x)=x,-1 \leqslant x \leqslant 1$, it follows that,
$$
|x+1|+|x-1|=2 \text {. }
$$
From the second condition, we get $\quad(y+3) x=4$.
Substituting (1) into (2) yields
$$
2 x^{2}-6 x+4=0 \t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,486 |
5. From 30 people with distinct ages, select two groups, the first with 12 people and the second with 15 people, such that the oldest person in the first group is younger than the youngest person in the second group. How many ways are there to select these groups? | 5. 4060.
Assume the ages of these 30 people, from smallest to largest, are
$$
a_{1}<a_{2}<a_{3}<\cdots<a_{29}<a_{30} .
$$
Let the first group of 12 people be $a_{i_{1}}<a_{i_{2}}<\cdots<a_{i_{12}}$,
and the second group of 15 people be $a_{j_{1}}<a_{j_{2}}<\cdots<a_{j_{15}}$,
which are the ages of the two selected gr... | 4060 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,487 |
6. If $x$ is a real number, then the solution set of the equation
$$
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right]+\cdots+\left[\frac{x}{1993}\right]+
$$
$\left.〔 \frac{x}{1994}\right\rceil$ is $\qquad$ . (where $[x]$ denotes the greatest integer not exceeding $x$) | 6. $\{0,4,5\}$.
Since $[a] \leqslant a3 .
$$
$\therefore$ The necessary condition for the above equation to hold is $0 \leqslant x<7$.
Note that the equation itself indicates that $x$ is an integer. Testing with $0,1,2,3,4,5$, 6, we find $x=0,4,5$.
$\therefore$ The solution set is $\{0,4,5\}$. | \{0,4,5\} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,488 |
One, (Full marks 30 points) Let $f(x)=x^{2}-(4 a-2) - x-6 a^{2}$ have the minimum value $m$ on the interval $[0,1]$. Try to write the expression for $m$ in terms of $a$, $m=F(a)$. And answer: For what value of $a$ does $m$ achieve its maximum value? What is this maximum value?
---
The translation maintains the origin... | $$
-, f(x)=[x-(2 a-1)]^{2}-10 a^{2}+4 a-1 .
$$
(i) When $2 a-1 \leqslant 0$, i.e., $a \leqslant \frac{1}{2}$, $f(x)$ is \. on $[0,1]$. At this time, $m=f(0)=-6 a^{2}$.
(ii) When $2 a-1 \geqslant 1$, i.e., $a \geqslant 1$, $f(x)$ is L. on $[0,1]$. At this time, $m=f(1)=-6 a^{2}-4 a+3$.
(iii) When $0<2 a-1<1$, i.e., $\fr... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,489 |
II. (Full marks 30 points) In the tetrahedron $A-BCD$, $\angle DAB + \angle BAC + \angle DAC = 90^{\circ}$, $\angle ADB = \angle BDC = \angle ADC = 90^{\circ}$.
It is also known that $DB = a$, $DC = b$.
Try to find the radius of the largest sphere that can fit inside the tetrahedron $A-BCD$. | In the tetrahedron with a volume of 6, the sphere inscribed is its inner sphere.
Given $D B:=a, D C=6$, we set $D A=x$.
Cutting $D A, D B, D C$ and laying them out on a plane, $\angle D_{1} A D_{2}=90^{\circ}$, $D_{1} A=D_{2} A=x$. Extending $D_{1} B, D_{2} C$ to intersect at $K$, it is easy to see that $A D_{1} K D_{... | r=\frac{a b}{2(a+b)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,490 |
8 Utilize the discriminant
Example 10. Try to determine all real solutions of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.
$$ | Solving: Since $x^{2}+y^{2}=(x+y)^{2}-2xy=(3-z)^{2}-2xy=3-z^{2}$, we have, from the first two equations,
$$
x+y=3-z, \quad xy=z^{2}-3z+3.
$$
From this, we know that $x, y$ are the two real roots of the quadratic equation in $t$
$$
t^{2}-(3-z)t+\left(z^{2}-3z+3\right)=0.
$$
By the discriminant theorem, we get
$$
\begi... | x=1, y=1, z=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,491 |
Four, (Full marks 30 points) Arrange $m \times n$ non-negative real numbers into the following $m$ row $n$ column rectangular number table:
$$
\begin{array}{cccccc}
a_{11}, & a_{12}, & a_{13}, & \cdots \cdots, & a_{1 n} & G_{1} \\
a_{21}, & a_{22}, & a_{23}, & \cdots \cdots, & a_{2 n} & G_{2} \\
\cdots, & \cdots, & \cd... | If at least one of $A_{1}, A_{2}, \cdots, A_{n}$ is zero, without loss of generality, let $A_{1}=0$, then $a_{11}=a_{21}=\cdots=a_{m 1}=0$. Thus, $G_{1}=G_{2}=\cdots=G_{n}=0$. Therefore,
\[ G\left(A_{1}, A_{2}, \cdots, A_{n}\right)=A\left(G_{1}, G_{2}, \cdots G_{n}\right)=0 \]
holds.
If $A_{1}, A_{2}, A_{1}, \cdots, A_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,493 |
Given $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, $\odot O_{1}, \odot O_{2}$ are the incircles of $\triangle A D C, \triangle B D C$ respectively, $A C$ is tangent to $\odot O_{1}$ at point $P$, $B C$ is tangent to $\odot O_{2}$ at point $Q$, $A O_{1}, B O_{2}$ intersect at po... | Proof $\because O$ is the intersection of $A O_{1}$ and $B O_{2}$, $A O_{1}$ bisects $\angle B A C$, and $B O_{2}$ bisects $\angle A B C$.
$\therefore O$ is the incenter of $\triangle A B C$, and $\angle A C O=45^{\circ}$.
It is easy to prove that the acute angle formed by $\mathrm{O}_{1} \mathrm{O}_{2}$ and $A C$ is $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,494 |
24. As shown in the figure, $P$ is any point inside $\triangle ABC$, and $AP, BP, CP$ intersect $BC, CA, AB$ at points $D$, $E, F$ respectively. Then, $S_{\triangle DEF}=2 \cdot \frac{PD}{PA} \cdot \frac{PE}{PB} \cdot \frac{PF}{PC} S_{\triangle ABC}$ | Let $B D=x, C E=y, A F=z, u=\frac{x}{a}$, $v=\frac{y}{b}, w=\frac{z}{c}$
$\because S_{\triangle A B C}=\frac{1}{2} \operatorname{casin} B$,
$\therefore \frac{1}{2} \sin B=\frac{S_{\triangle A B C}}{c a}$,
$S_{\triangle B D F}=\frac{1}{2} x(c-z) \sin B=\frac{x(c-z)}{c a} S_{\triangle A B C}$.
Similarly, $S_{\triangl... | S_{\triangle DEF}=2 \cdot \frac{PD}{PA} \cdot \frac{PE}{PB} \cdot \frac{PF}{PC} S_{\triangle ABC} | Geometry | proof | Yes | Yes | cn_contest | false | 707,495 |
$$
\begin{array}{l}
\sqrt[3]{\cos \frac{2 \pi}{9} \cos \frac{4 \pi}{9}}+\sqrt[3]{\cos \frac{4 \pi}{9} \cos \frac{8 \pi}{9}} \\
+\sqrt[3]{\cos \frac{8 \pi}{9} \cos \frac{2 \pi}{9}}=-\sqrt[3]{\frac{3}{4}(\sqrt[3]{9}-1)} .
\end{array}
$$
Prove the trigonometric identity:
$$
\begin{array}{l}
\sqrt[3]{\cos \frac{2 \pi}{9} ... | Let $x=\sqrt[3]{\cos \frac{2 \pi}{9}}, y=\sqrt[3]{\cos \frac{4 \pi}{9}}$,
$z=\sqrt[3]{\cos \frac{8 \pi}{9}}$. It is easy to see that
$$
\left\{\begin{array}{l}
x^{3}+y^{3}+z^{3}=0, \\
(x y)^{3}+(y z)^{3}+(z x)^{3}=-\frac{3}{4}, \\
x y z=-\frac{1}{2} .
\end{array}\right.
$$
Thus,
$$
\left\{\begin{array}{l}
2(x+y+z)^{3}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,496 |
24. Use coins to stack a triangle, with $2^{10}$ coins on the bottom layer, and $2^{9}+1$ coins facing up. The principle for stacking upwards is: (1) each layer has one less coin than the adjacent lower layer; (2) if the two adjacent coins are both facing up or both facing down, the coin stacked between them faces up, ... | Solve for the $j$-th coin of the $i$-th layer with integer $a_{i, j}(1 \leqslant i \leqslant 2^{10}, i \geqslant j)$, where a coin with the head side up is +1, and a coin with the head side down is -1. According to the problem, list the array:
$$
\begin{array}{l}
a_{1.1} \\
\begin{array}{ll}
a_{2,1} & a_{2,2}
\end{arra... | a_{1,1}=-1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,497 |
Example 1. Let the vertices of a convex quadrilateral $ABCD$ lie on the same circle, and the center of another circle lies on side $AB$, and is tangent to the other three sides of the quadrilateral. Prove that $AD + BC = AB$. (26th IMO) | ```
Proof Let the center of another circle be $O$, extend $A D, B C$ to intersect at $E$, and let $\odot O$ be the excircle of $\triangle E C D$. Connect $O E, O C, O D$, and set $E C=a, E D=b, C D=c$, and the radius of $\odot O$ as $R$. Then
\[
\begin{array}{l}
S_{\triangle E C D}=S_{\triangle E D O}+S_{\triangle E O... | AD + BC = AB | Geometry | proof | Yes | Yes | cn_contest | false | 707,498 |
Example 2. In quadrilateral $ABCD$, $M$ is the midpoint of $CD$, and it is known that the area of $\triangle ABM$ is half the area of quadrilateral $ABCD$. Prove that $AD \parallel BC$. (1991, 6th Northeast Mathematics Invitation Competition) | Extend $B M$ to $E$, such that $M E = B M$, and connect $A E, D E$. Then $\triangle A B E$ and $\triangle A B M$ have the same height, but the base $B E = 2 B M$, so we have
$S_{\triangle A B E} = 2 S_{\triangle A B M} = S_{\text {parallelogram }ABCD}$ (given),
By $\triangle D M E \cong \triangle C M B$, we have $S_{\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,499 |
Example 3. There is a convex pentagon, the area of the triangles formed by any three consecutive vertices is 1. Find the area of this pentagon (1972, American Mathematics Competition) | Analyze the figure. Since the area of the triangle formed by any three adjacent vertices is 1, according to the theorem of equal areas of similar triangles, we have $A E / / B D, A D / / B C, B E / / C D, \therefore$ quadrilateral $B C D O$ is a parallelogram. Using the area relationships of parallelograms and trapezoi... | \frac{5+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,500 |
9.5 Prove the identity
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}\left(a_{1}+a_{2}\right)}+\frac{a_{2}}{a_{3}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{n}}{a_{1}\left(a_{n}+a_{1}\right)} \\
=\frac{a_{2}}{a_{1}\left(a_{1}+a_{2}\right)}+\frac{a_{3}}{a_{2}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{1}}{a_{n}\left(a_{n}+a_{1}\rig... | 9.5 We have
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}\left(a_{1}+a_{2}\right)}+\frac{a_{2}}{a_{3}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{n}}{a_{1}\left(a_{n}+a_{1}\right)} \\
=\left(\frac{1}{a_{2}}-\frac{1}{a_{1}+a_{2}}\right)+\left(\frac{1}{a_{3}}-\frac{1}{a_{2}+a_{3}}\right)+\cdots \\
+\left(\frac{1}{a_{1}}-\frac{1}{a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,501 |
9.6 On 1000 cards, write down the natural numbers 1 to 1000 (one number per card), and then use these cards to cover 1000 squares in a $1 \times 1994$ rectangle (the rectangle consists of 1994 $1 \times 1$ squares. The size of the cards is also $1 \times 1$). If the square to the right of the square containing the card... | 9.6 Notice that the card with 1 written on it will not be moved, so the card with 2 written on it can be moved at most once, the card with 3 written on it can be moved at most twice, and so on. Thus, for any $n \leqslant 1000$, the card with $n$ written on it can be moved at most $n-1$ times (because we can only move i... | 500000 | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,502 |
9.7 Let $A B C D$ be a trapezoid $(A B \parallel C D)$, and on its two non-parallel sides $A D$ and $B C$, there exist points $P$ and $Q$ respectively, such that $\angle A P B = \angle C P D$, $\angle A Q B = \angle C Q D$. Prove: The distances from points $P$ and $Q$ to the intersection point of the diagonals of the t... | 9. 7 Let $\triangle A P B$
have the circumcircle $S_{1}$, and $\triangle C P D$ have the circumcircle $S_{2}$. The other intersection point of the two circles, besides $P$, is $Q_{1}$ (as shown in the figure: connect $Q_{1}$ to $P, B, C$. In addition, there are $\angle C Q_{1} P + \angle C D A = 180^{\circ}$, $\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,503 |
9.8 The plane is divided into unit squares by two sets of parallel lines. Consider an $n \times n$ square formed by the divided unit squares. The union of the unit squares that have at least one side on the boundary is called the frame of the square. Now consider the frame question of a $100 \times 100$ square formed b... | 9.8 In the given $100 \times 100$ square $A B C D$, 200 unit squares are distributed on the two diagonals $A C$ and $B D$. Each "edge frame" can cover at most 4 of these unit squares (since each diagonal intersects with an "edge frame" at most at two unit squares). Therefore, each of the 50 "edge frames" we are conside... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,504 |
10. 1 Given? three quadratic trinomials; $P_{1}(x)=x^{2}+p_{1} x$ $+q_{1}, P_{2}(x)=x^{2}-p_{2} x+q_{2}$ and $P_{3}(x)=x^{2}+p_{3} x+q_{3}$. Prove: The equation $\left|P_{1}(x)\right|+\left|P_{2}(x)\right|=\left|P_{3}(x)\right|$ has at most 8 real roots. | 10.1 Each root of the original equation should be a root of a quadratic trinomial of the form $\pm p_{1}: p_{2} \pm p_{3}$, and with different choices of signs, there are 8 such quadratic trinomials. Since the coefficient of $x^{2}$ has the form $\pm 1 \pm 1 \pm 1$, the coefficient of $x^{2}$ will never be 0 regardless... | 8 | Algebra | proof | Yes | Yes | cn_contest | false | 707,505 |
10.3 Let the lengths of the three sides of a triangle be $a, b, c$, the lengths of the medians to these sides be $m_{a}, m_{b}$, and $m_{c}$, and the diameter of the circumcircle be $D$. Prove: $\frac{a^{2}+b^{2}}{m_{c}}+\frac{b^{2}+c^{2}}{m_{a}}+\frac{c^{2}+a^{2}}{m_{b}} \leqslant 6 D$. | 10.3 Extend each median so that they intersect the circumcircle at points \(A_{1}\), \(B_{1}\), and \(C_{1}\) (as shown in the figure). Clearly, we have
\[
A A_{1} \leqslant D, B B_{1} \leqslant D, C C_{1} \leqslant D,
\]
i.e., \(m_{a} + A_{1} A_{2} \leqslant D, m_{b} + B_{1} B_{2} \leqslant D, m_{c} + C_{1} C_{2} \leq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,506 |
10. 4 In a regular $(6n+1)$-gon, $k$ vertices are colored red, and the remaining vertices are colored blue. Prove that the number of isosceles triangles with vertices of the same color does not depend on the coloring method. | 10. 4 For convenience, we simply refer to isosceles triangles with vertices at the vertices of a $(6n+1)$-gon as "triangles".
We notice that each diagonal and each side of the given polygon $M$ is a side of exactly 3 such "triangles" (this conclusion holds only for polygons with the number of sides leaving a remainder... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,507 |
10. 5 Proof: For natural numbers $k, m$ and $n$, the following inequality holds:
$$
[k, m] \cdot [m, n] \cdot [n, k] \geqslant [k, m, n]^{2} \text{. }
$$
$[a, b, \cdots, z]$ denotes the least common multiple of the natural numbers $a, b, \cdots, z$. | 10.5 Let's compare the exponents (powers) of a prime number $p$ on both sides of the inequality to be proven. Suppose the exponent of $p$ in the prime factorization of the number $k$ is $\alpha$, in the number $m$ is $\beta$, and in the number $n$ is $\gamma$. Without loss of generality,
we can assume $a \leqslant \be... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,508 |
10.6 Functions $f(x)$ and $g(x)$ are defined on the set of integers whose absolute values do not exceed 100. The number of integer pairs $(x, y)$ such that $f(x)=g(y)$ is denoted by $m$, the number of integer pairs $(x, y)$ such that $f(x)=f(y)$ is denoted by $n$, and the number of integer pairs $(x, y)$ such that $g(x... | 10.6 Let $a$ be a value of the function $f(x)$, and denote the number of $x$ such that $f(x)=a$ and $g(x)=a$ as $n_{4}$ and $k_{4}$ ($k_{4}$ may be 0). Then the number of pairs $(x, y)$ that satisfy the equations $f(x)=a, g(y)=a$ will be $n_{a} \cdot k_{a}$, the number of pairs that satisfy the equations $f(x)=a, f(y)=... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,509 |
10.7. Circles $S_{1}, S_{2}, S_{3}$ are all externally tangent to circle $S$ (at points $A_{1}, B_{1}, C_{1}$, respectively), and they are also tangent to two sides of $\triangle ABC$ (as shown in the figure). Prove: $A A_{1}$, $B B_{1}, C C_{1}$ intersect at a single point. | 10.7 First, we prove an auxiliary proposition:
Suppose the transformation $H$ of the plane is the composition of two homotheties $H_{0_{1}}^{t_{1}}$ and $H_{0_{2}}^{t_{2}}$ (with centers at $O_{1}$ and $O_{2}$, and coefficients $k_{1}$ and $k_{2}$, respectively).
Then, when $k_{1} k_{2}=1$, $H$ is a translation, and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,510 |
Example 4. In the right triangle $\triangle ABC$, there are two inscribed squares $MNPQ$ and $DECF$. Prove: the area $A_{1}$ of $DECF$ is greater than the area $A_{2}$ of $MNPQ$.
Keep the original text's line breaks and format, and output the translation result directly. | Analyzing $\triangle A B C \equiv$ side
lengths $a, b, c$.
$\because \triangle D B E \sim \sim$
$\triangle A B C$, from the
similarity ratio, we
establish the
equation, solving for $C E=\frac{a b}{a+b}$. Similarly, from $\triangle A B C \backsim \triangle Q P C \sim \Omega$ $\triangle P B N$, we solve for $N P=\frac{a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,511 |
10. 8 Class
30 students in total, each
student in the class
has the same number of friends.
How many students at most can there be who are
better than the majority of their friends' performance? (Assuming for any two students in the class, their performance can be compared to determine who is better or worse.)
保留源文本... | 10.8 䇾案, 25 个.
We refer to a student who performs better than the majority of their friends as a "good student". Let there be $x$ "good students", and each student has $k$ friends.
The best student is better in all $k$ "friend pairs", while each of the other "good students" is better in at least $\left[\frac{k}{2}\r... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,512 |
11. 1 Let \(a\) and \(b\) be two given natural numbers such that \(\frac{a+1}{b}+\frac{b+1}{a}\) is an integer. Prove: the greatest common divisor of \(a\) and \(b\) does not exceed \(\sqrt{a+b}\). | 11. 1 We have
$$
\frac{a+1}{b}+\frac{b+1}{a}=\frac{a^{2}+b^{2}+a+b}{a b} \text {. }
$$
Assume the greatest common divisor of $a$ and $b$ is $d$. Since $ab$ can be divided by $d^{2}$, $a^{2}+b^{2}+a+b$ can be divided by $d^{2}$. Since $a^{2}+b^{2}$ can be divided by $d^{2}$, $a+b$ can be divided by $d^{2}$, therefore $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,513 |
11.2 Inside a convex 100-gon, $k$ points are marked, $2 \leqslant$ $k \leqslant 50$. Prove: It is possible to select $2 k$ vertices from the vertices of the 100-gon, such that all the marked $k$ points lie inside the $2 k$-gon formed by the selected vertices. | 11.2 We call the smallest convex polygon that contains all points of a finite point set the "convex hull" of that point set. It can be proven that any finite point set has a unique "convex hull".
Let \( M = A_{1} A_{2} \cdots A_{n} \) be the convex hull of the marked \( k \) points \((n \leqslant k)\), and let \( O \)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,514 |
11.3 Circles $S_{1}$ and $S_{2}$ are externally tangent at point $F$, and their external common tangents touch $S_{1}$ and $S_{2}$ at points $A$ and $B$, respectively. A line parallel to $A B$ is tangent to $S_{2}$ at point $C$ and intersects $S_{1}$ at points $D$ and $E$. Prove: The common chord of the circumcircles o... | 11.3 In problem 9.2, it has been proven that points $A, F, C$ are collinear. Now, we will prove that the circumcenters of $\triangle A B C$ and $\triangle B D E$, denoted as $\omega_{1}$ and $\omega_{2}$ respectively, both lie on line $A C$. Note that $\angle B F C=90^{\circ}$, as it is the inscribed angle subtended by... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,515 |
There is a real number filled in each small square. Consider two figures $G_{1}$ and $G_{2}$, each composed of a finite number of small squares. The figures can move along the grid lines by an integer number of squares. It is known that for any position of $G_{1}$, the sum of the numbers covered by it is positive. Prov... | 11.4 Take any small square, and let its center be denoted as $O$. Establish a rectangular coordinate system with $O$ as the origin, such that the coordinate axes are parallel to the grid lines, and use the side length of the square as the unit length. Fix the shapes $G_{1}$ and $G_{2}$ in two positions, and denote the ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,516 |
11.5 Given a sequence of natural numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}$ is not divisible by 5, and for each $n$, the equation
$$
a_{n+1}=a_{n}+b_{n} \text {, }
$$
holds, where $b_{n}$ is the last digit of $a_{n}$. Prove: the sequence contains infinitely many powers of 2. | 11.5 From the problem, we know that $b_{1}$ cannot be 0 or 5, so $b_{2}$ must be one of $2, 4, 6, 8$. This means the sequence $b_{2}, b_{3}, \cdots$ must have a period of 4,
$$
\cdots, 2,4,8,6,2,4,8,6, \cdots .
$$
Thus, for any $n>1$, we have $a_{n+4}=a_{n}+(2+4+8+6)$, and for any $s>1$, we have $a_{n+\mu}=a_{n}+20 s$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,517 |
1. The sum of all irreducible proper fractions with a denominator of 1994 is ( ).
(A) 497.5
(B) 498
(C) 498.5
(D) 997 | $-1 . B$.
From $1994=2 \times 997$, and knowing that 997 is a prime number, the sum of the reduced proper fractions in the problem is
$$
\begin{aligned}
N= & \frac{1}{1994}+\frac{3}{1994}+\cdots+\frac{995}{1994}+\frac{999}{1994}+\cdots \\
& +\frac{1993}{1994} \\
= & \frac{1+1993}{1994}+\frac{3+1991}{1994}+\cdots+\frac{... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,519 |
3. If $\frac{x-y}{a}=\frac{y-z}{b}=\frac{z-x}{c}=a b c<0$, then the number of negative numbers among $a, b, c$ is ( ).
(A) 1
(B) 2
(C) 3
(D) Cannot be determined | 3. A.
From the given
$$
x-y=a^{2} b c, y-z=a b^{2} c, z-x=a b c^{2} .
$$
Adding them up yields $a b c(a+b+c)=0$. But $a b c \neq 0$, so
$$
a+b+c=0 .
$$
This implies that among $a, b, c$, there is 1 or 2 negative numbers, but since $a b c<0$, there can only be 1 negative number. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,521 |
Example 9: Given $\triangle A B C$ intersects with a line $P Q$ parallel to $A C$ and the area of $\triangle A P Q$ equals a constant $k^{2}$, what is the relationship between $b^{2}$ and the area of $\triangle A B C$? Is there a unique solution? | Given $P Q // A C$, we have $\triangle B P Q \sim \triangle B A C$. Therefore, the area relationship between $\triangle A P Q$ and $\triangle A B C$ can be constructed from the similarity ratio, and the various cases of the equation's solutions can be studied from the discriminant.
Solution: Let $\frac{B P}{A B}=x, S=... | S \geqslant 4 k^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,522 |
4. If $a$ is a real root of the quadratic equation $3 x^{2}-2 x-663=0$, then $\left(3 a^{3}-64 \frac{1}{3} a-443\right)^{3}=(\quad)$.
(A) 1
(B) -1
(C) 8
(D) -8 | 4. B.
$$
\text { Given } \begin{aligned}
& 3 a^{3}-664 \frac{1}{3} a-443 \\
& =\left(a+\frac{2}{3}\right)\left(3 a^{2}-2 a-663\right)-1=-1 .
\end{aligned}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,523 |
5. Given $A\left(\frac{1}{3}, \frac{1}{a}\right), B\left(\frac{1}{4}, \frac{1}{b}\right), C\left(\frac{1}{5}, \frac{1}{c}\right)$ satisfy $\frac{a}{b+c}=\frac{1}{3}, \frac{b}{a+c}=\frac{1}{2}$. Then the positions of points $A, B, C$ are suitable for
(A) on the same straight line
(B) forming an acute triangle
(C) formin... | 5. A.
From $\frac{a}{b+c}=\frac{1}{3}, \frac{b}{a+c}=\frac{1}{2}$, we get the following by combining ratios:
$$
\begin{array}{c}
\frac{a}{a+b+c}=\frac{1}{4}=\frac{3}{12}, \\
\frac{b}{a+b+c}=\frac{1}{3}=\frac{4}{12} . \\
\text { Therefore, }\left\{\begin{array}{l}
a=3 t, \\
b=4 t, \\
a+b+c=12 t .(t \neq 0)
\end{array}\... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,524 |
6. In Rt $\triangle A B C$, $A B=3, A C=4, B C=5$. Now, let the distances from $A, B, C$ to a certain line $l$ be $d_{A}, d_{B}, d_{C}$ respectively. If $d_{A}: d_{B}: d_{C}=1: 2: 3$, then the number of lines $l$ that satisfy the condition is ( ) lines.
(A) 1
(B) 2
(C) 3
(D) 4 | 6. D.
Since the distances from $A, B, C$ to $l$ are all unequal, $l$ is not parallel to $AB$, $BC$, or $CA$. Consider the intersection points of $l$ with the sides (or their extensions) of the triangle. Given that $d_{A}: d_{B}=1: 2$, there are internal division points $X_{1}$ and external division points $X_{2}$ on $... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,525 |
3. Let $a$ be the decimal part of $\sqrt{3}$, $b$ be the decimal part of $\sqrt{2}$: $\frac{a}{(a-b) b}$ has an integer part of $\qquad$ | $$
\begin{array}{l}
\text { 3. } \frac{a}{(a-b) b}=\frac{\sqrt{3}-1}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{(\sqrt{3}-\sqrt{2})+(\sqrt{2}-1)}{(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)} \\
=\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}-\sqrt{2}} \\
=\sqrt{3}+2 \sqrt{2}+1 \\
\approx 5.56 .
\end{array}
$$
The integer part is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,528 |
4. Given the three sides of a triangle $a, b, c$ are integers, and $a+b+c=11$. Then when the product $abc$ takes the minimum value, the area of the triangle is $\qquad$ . | 4. Since $a, b, c$ are the lengths of the three sides of a triangle, when $a=1$, $b=c=5$, $abc=25$; when $a=2$, $b=4$, $c=5$, $abc=40$; when $a=3$, $b=3$, $c=5$, $abc=45$, or $b=c=4$, $abc=48$. It is evident that the product $abc=25$ is the minimum value. In the isosceles $\triangle ABC$, the height from vertex $A$ to ... | \frac{3 \sqrt{11}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,529 |
一、Let $AB$ be a chord of $\odot O$ that is shorter than the diameter. Rotate $\triangle OAB$ clockwise around the center $O$ by an angle $\alpha (0^{\circ} < \alpha < 360^{\circ})$ to get $\triangle OA'B'$. During the rotation process, can the moving chord $A'B'$ pass through every point on the chord $AB$? Prove your c... | Let the midpoint of $AB$ be $C$, then except for $C$, $A^{\prime} B^{\prime}$ can pass through all other points on $AB$.
With $O$ as the center and $OC$ as the radius, draw $\odot K$. Take any point $P$ on the chord $AB$ other than $C$, then $P$ is outside $\odot K$. Two tangents can be drawn from $P$ to $\odot K$, th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,530 |
Given $0<\frac{b}{3}<a<b<\frac{3 c}{4}$. Prove: $a$ is not a root of the quadratic equation $x^{2}-2(2 c-b) x+b^{2}=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $f(x)=x^{2}-(2 b+c) x+b^{2}$, its graph opens upwards, and the x-coordinate of the vertex is $x_{0}=b+\frac{c}{2}$. For
$$
\frac{b}{3}<a<b+\frac{c}{2},
$$
we have $\quad f(a)<f\left(\frac{b}{3}\right)$
$$
\begin{array}{l}
=\frac{b^{2}}{9}-(2 b+c) \cdot \frac{b}{3}+b^{2} \\
=\frac{4}{9} b\left(b-\frac{3 c}{4}\right... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,531 |
Three, there are 6 points on a circle, and 11 line segments are drawn arbitrarily with these points as endpoints. Prove that there must be 3 line segments, each pair of which has no common endpoint.
untranslated text:
三、圆周上有 6 个点,以它们为端点任意作 11 条线段. 求证：其中必有 3 条线段,两两之间无公共端点.
translated text:
Three, there are 6 points ... | Three, let the 6 points on the circumference be $A_{1}, A_{2}, \cdots, A_{6}$, and let the line connecting $A_{i}$ and $A_{j}$ be $l_{i j}$, for a total of 15 lines: $l_{16}, l_{23}, l_{15}, l_{12}, l_{34}, l_{56}$, $l_{13}, l_{25}, l_{65}, l_{15}, l_{24}, l_{36}, l_{14}, l_{24}, l_{35}$.
Divide these 15 line segments... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,532 |
Example. Simplify $\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$. | $$
\begin{array}{c}
\text { Solve the original expression }=\frac{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{5})^{2}}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \\
=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\sqrt{2}+\sqrt{3}+\sqrt{5}} . \\
=\sqrt{2}+\sqrt{3}-\sqrt{5} .
\end{array}
$$
This method is very effective, but as a... | \sqrt{2}+\sqrt{3}-\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,533 |
1. Let $P_{0}\left(x_{0}, y_{0}\right)$ be any point on the circle $x^{2}+(y-1)^{2}=1$, and we want the inequality $x_{0}+y_{0}+c \geqslant 0$ to always hold. Then the range of values for $c$ is ( ).
(A) $[0,+\infty$ )
(B) $[\sqrt{2}-1, \cdots-\infty$ )
(C) $(-\infty, \sqrt{2}+1]$
(D) $(1 \cdots \sqrt{2},+\infty)$ | -1 . B.
Circle $x^{2}+(y-1)^{2}=1$ should be above the line $x+y+c=0$, meaning the line $x+y+c=0$ should be below or tangent to the circle, or separate from it. As shown in the figure, $A(0,1-\sqrt{2})$, hence $c \geqslant \sqrt{2}-1$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,534 |
2. Let's define three functions, the first function is a monotonic function defined on the set of real numbers denoted as $y=f(x)$, its inverse function is the second function, and the third function is symmetrical to the second function about the line $x+y=0$. Then, the third function is ( ).
(A) $y=-f(x)$
(B) $y=-f(-... | 2. B.
Let $P(x, y)$ be any point on the third function, $Q_{1}\left(x_{1}, y_{1}\right)$ be the symmetric point of $P$ with respect to $x+y=0$, then $Q_{1}\left(x_{1}, y_{1}\right)$ is on the second function and $x_{1}=-y, y_{1}=-x$. Let $Q_{2}\left(x_{2}, y_{2}\right)$ be the symmetric point of $Q_{1}$ with respect t... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,535 |
4. The function $f(x)$ is a real function defined on $R$, it is symmetric about the line $x=1$ and also symmetric about the line $x=3$, then the function $f(x)($ ).
(A) is not a periodic function
(B) is a periodic function with period 1
(C) is a periodic function with period 2
(D) is a periodic function with period 4 | 4. D.
Given $f(x)=f(2-x), f(t)=f(5-t)$, let $t=2-x$, then $f(x)=f(2-x)=f[6-(2-x)]=f(x+4)$, hence
$f(x+4)=f(x)$ for all $x$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,537 |
5. Given a tetrahedron with three lateral edges that are pairwise perpendicular, and the angles formed by each lateral edge with the base are $\alpha, \beta, \gamma$. Then, $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$ | 5. C. The projection of point W onto the plane of $\triangle ABC$ is the orthocenter of $\triangle ABC$, and $AD \perp \mathcal{Z}$. Since $PD \perp BC$, therefore, in the right triangle $\triangle PAD$, $\angle PAD = \alpha$, and since $D = \frac{bc}{\sqrt{b^2 + c^2}}$, $\operatorname{tg} c = \frac{bc}{\sqrt{a^2 b^2 +... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,538 |
6. The numbers $a, \beta$ satisfy the following equations: $\alpha^{3}-3 \alpha^{2}+5 \alpha=1$, $\beta^{3}-3 \beta^{2}+5 \beta=5$. Then, the value of $\alpha+\beta$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 5 | 6. B.
$$
\text { Let } \begin{aligned}
f(x) & =(x-1)^{3}+2(x-1) \\
& =x^{3}-3 x^{2}+5 x-3,
\end{aligned}
$$
Now $g(y)=y^{3}+2 y$.
For $\alpha$, we have $g(\alpha-1)=f(a)=\alpha^{3}-3 \alpha^{2}+5 \alpha-3$
$$
\begin{aligned}
& =1-3=-2, \\
g(\beta-1) & =f(\beta)=\beta^{3}-3 \beta^{2}+5 \beta-3=5-3=2 .
\end{aligned}
$$
... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,539 |
2. Given $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, and $\sin \frac{\alpha}{2}$ $=a \cos \beta$. When $0<\alpha+\beta<\frac{\pi}{2}$, the range of values for $a$ is $\qquad$. | 2. $0<a \leqslant \frac{1}{2}$. | 0<a \leqslant \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,541 |
3. Let $M_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$ be four points on the parabola $y=a x^{2}$ $+b x+c(a \neq 0)$. When $M_{1}, M_{2}, M_{3}, M_{4}$ are concyclic, $x_{1}+x_{2}+x_{3}+x_{4}=$ $\qquad$ . | 3. $-\frac{2 b}{a}$.
$M_{2}, M_{2}, M_{3}, M_{6}$ are concyclic. Let's assume the equation of this circle is $x^{2} + y^{2} + D x + E y + F = 0$. Combining this with $y = a x^{2} + b x + c$ and eliminating $y$, we get $a^{2} x^{4} + 2 a b x^{3} + (2 a c + E a + 1) x^{2} + (2 b c + D + E b) x + c^{2} + E c + F = 0$. It ... | -\frac{2 b}{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,542 |
5. Let the three roots of the cubic equation $x^{3}+p x+1=0$ correspond to points in the complex plane that form an equilateral triangle. Then $p=$ , and the area of this equilateral triangle is $\qquad$ . | 5.
$p=0, S=3 \sqrt{4}$.
From the definition of coordinate translation and the square root of a complex number, it is known that the complex numbers corresponding to these three points must satisfy the equation $(x+a)^{3}=b$ (where $a, b$ are complex numbers). From $x^{3}+p x+1=(x+a)^{3}-b=x^{3}+3 x^{2} a + 3 x a^{2} + ... | p=0, S=\frac{3 \sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,545 |
6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ . | 6. $p+q=8$.
From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,546 |
Given the parabola equation as $y=a x^{2}+b x+c(a \neq 0)$. Inside it, construct a circle $\odot C_{1}$ with radius $\frac{1}{|a|}$ that is internally tangent to the parabola. Construct $\odot C_{2}$ to be externally tangent to $\odot C_{1}$ and internally tangent to the parabola. Construct $\odot C_{3}$ to be external... | $$
\begin{array}{l}
\text{I. The center coordinates of } C_{n} \text{ are } \left(-\frac{b}{2 a}, d_{n}\right), \text{ with radius } r_{n} (n=1,2, \cdots). \text{ Then the equation of the circle } C_{n} \text{ is: } \left(x+\frac{b}{2 a}\right)^{2}+\left(y-d_{n}\right)^{2}=r_{n}^{2}, \text{ combined with } y=a x^{2}+b ... | \frac{\pi}{6 a^{2}} n(n+1)(2 n+1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,547 |
II. Let the convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ have sides $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ in sequence, and its area be $\triangle_{n}$. Prove: $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$ $\geqslant 4 \triangle_{n} \tan \frac{\pi}{n}$. Equality holds if and only if the $n$-sided polygon $A_{1} A_{2} \cd... | By the Cauchy-Schwarz inequality: \(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+\right.\) \(\left.a_{2}+\cdots+a_{n}\right)^{2}\), equality holds if and only if \(a_{1}=a_{2}=\cdots=a_{n}\). Let \(a_{1}+a_{2}+\cdots+a_{n}=l\). By plane geometry, the area of a regular \(n\)-sided polygon with p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,548 |
Three, let the three sides of $\triangle A B C$ be $a, b, c$, and all are numbers. If it satisfies $\angle A=3 \angle B$, try to find the minimum value of the perimeter and provide a proof. | From the Law of Sines, we have \(\frac{a}{b} = \frac{\sin 3B}{\sin B} = (2 \cos B)^2 - 1\), \(\frac{c}{b} = (2 \cos B)^3 - 4 \cos B\). Since \(2 \cos B = \frac{a^2 + c^2 - b^2}{ac} \in \mathbb{Q}\), there must exist positive integers \(p, q\) with \((p, q) = 1\) such that \(2 \cos B = \frac{p}{q}\). Therefore, \(\frac{... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,549 |
There are 1994 points on a circle, which are painted in several different colors, and the number of points of each color is different. Now, take one point from each color set to form a polygon with vertices of different colors inside the circle. To maximize the number of such polygons, how many different colors should ... | Let 1994 points be colored with $n$ colors, and the number of points of each color, in ascending order, is $m_{1}1$. In fact, if $m_{1}=1$, because $m_{1} m_{2} \cdots m_{n}m_{k+1} m_{k}$.
Therefore, when $m^{\prime}{ }_{n+1}, m^{\prime} \star$ replace $m_{\star+1}, m_{\star}$, the value of $M$ increases, leading to a ... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,550 |
25. On the sides $AB, BC$ (or their extensions) of square $ABCD$, take points $M, N$ respectively, such that $\angle MDN=45^{\circ}$. Draw $MP \perp DN$ at $P$. Prove that $\angle BPN=2 \angle ADM$. | Prove: Connect $D B, A P$. It is easy to know that $A, P, D, M$ are concyclic.
(1) When $M$ is on $A B$ and $N$ is on $B C$, using $\angle M A P$ $=\angle M D P=45^{\circ}$, we get
$$
\triangle B A P \cong \triangle D A P \Rightarrow P B=P D \text {. }
$$
$\therefore P B=P M=P D$, point $P$ is the circumcenter of $\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,551 |
Initial 26. Find all natural numbers of the form $\overline{30 x 070 y 03}$ that are divisible by 37. | Solve $\overline{30 x 070 y 03}=300070003+10^{6} x+10^{2} y$
$$
\begin{aligned}
= & 37(8110000+27027 x+2 y) \\
& +(3+x-11 y) .
\end{aligned}
$$
Therefore, from the given conditions, $3+x-11 y$ is a multiple of 37.
And $0 \leqslant x, y \leqslant 9, -96 \leqslant 3+x-11 y \leqslant 12$. Thus, the value of $3+x-11 y$ ca... | 308070103, 304070403, 300070703 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,552 |
甽 2. As shown in the figure, in $\triangle ABC$, it is known that $M, N$ are on sides $AC, BC$ respectively, and $BM$ intersects $AN$ at $O$. If $S_{\triangle OMA}=3 \text{~cm}^{2}, S_{\triangle OAB}$ $=2 \text{~cm}^{2}, S_{\triangle OBN}=1 \text{~cm}^{2}$.
Find $S_{\triangle CMN}$. (1990, Shanghai Junior High School ... | Analysis From the figure, we know that $\triangle A B C$ is composed of $\triangle C M A, \triangle O A B$, $\triangle O B N, \triangle O M N, \triangle C M N$, and the areas of the latter triangles.
First, $\because \triangle A R O$ and $\triangle A O M$ have the same height, $\therefore S_{\triangle O M A}$ 1 $S_{\t... | 22.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,554 |
Example 3. As shown in the figure, in $\triangle ABC$, $DE \parallel BC$, and $S_{\triangle ADE} : S_{\triangle CDE} = 1: 3$. Find the value of $S_{\triangle ADE} : S_{\triangle CDS}$. (1990, Sichuan Province Junior High School Mathematics Joint Competition) | Analyzing, it is known that $S_{\triangle C D E}$ is three times $S_{\triangle A D E}$. According to the problem, it is actually only necessary to find out how many times $S_{\triangle C D B}$ is of $S_{\triangle A D K}$. From the given information, we can get $\frac{A E}{E C}=\frac{1}{3}$, and $\because D E / /$ thus ... | \frac{1}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,555 |
Example 4. As shown in the figure, in $\triangle A B C$, $A D$ and $B E$ are both medians, $M N$ bisects $B E$ and is parallel to $A B$. It is also known that $A D$, $B E$, and $M N$ divide $\triangle A B C$ into six parts, with areas $a, b, c, d, e, 18$ respectively. Try to find the values of $a, b, c, d, e$. (Anqing ... | Analyzing the six blocks where only one is known, using the center of gravity $G$ to divide the three medians into segments of $2:1$ and $H$ bisecting $B E$, and $M N / / A D$: we can obtain: $\frac{N H}{H M}=\frac{A G}{G}=2: 1 \Rightarrow a=9$,
$$
\begin{array}{l}
2 d=18+e, \\
d=a+b, \\
\dot{u}+c=a+b+e+18, \\
\frac{18... | a=9, b=7, c=32, d=16, e=14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,557 |
Example 3. Two equilateral triangles are inscribed in a circle of radius $r$, and let the area of their common part be $S$. Prove that: $2 S \geqslant \sqrt{3} r^{2}$.
(26th IMO Preliminary Problem) | $$
\begin{array}{l}
\text{Prove that since } S \text{ equals the area of the equilateral triangle minus the areas of the three shaded regions, and by symmetry, } B^{\prime} M = B^{\prime} M, A N = A^{\prime} N. \text{ Therefore, } A M + M N + N A = \sqrt{3} r, \text{ showing that the perimeter of } \triangle A M N \tex... | 2 S \geqslant \sqrt{3} r^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 707,559 |
Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition) | Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,560 |
Example 7. As shown in the figure, let $D$ and $E$ be on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at $F$, $AE = EB$, $\frac{AD}{DC} = \frac{2}{3}$, and $S_{\triangle ABC} = 40$. Find $S_{\text{quadrilateral AEFD}}$. (6th National Junior High School Mathematics Correspondence Com... | Analyzing, first connect $A F$ and then set:
$$
S_{\triangle A E P}=x, S_{\triangle A P D}=y, S_{\triangle C D F}=z, S_{\triangle B E P}=t\left(S_{\triangle B C F}\right.
$$
$=u$ ), to find $S_{\text {quadrilateral } A E F D}$, we only need to find the values of $x$ and $y$.
We easily get:
$$
\begin{array}{l}
x+y+z=20,... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,561 |
1. Find all positive integers $k$ such that the set $X=\{1994,1997$, $2000, \cdots, 1994+3k\}$ can be partitioned into two disjoint subsets $A$ and $B$ whose union is $X$, and the sum of the elements in $A$ is 9 times the sum of the elements in $B$. (1st Test, 2nd Question, provided by Huang Yiguo) | Let $X$ contain $k+1$ positive integers, the sum of these $k+1$ positive integers is denoted as $S$. According to the problem,
$$
\begin{aligned}
S & =1994+1997+2000+\cdots+(1994+3 k) \\
& =1990(k+1)+\frac{1}{2}(k-1)(3 k+13) .
\end{aligned}
$$
Let the new positive integer in $B$ be $x$, then the sum of all positive in... | k=20 t-1(t \in \mathbb{N}) \text{ and } k=20 t+4 \text{ (positive integer } t \geqslant 8) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,565 |
2. A car can only carry $L$ liters of gasoline, with which it can travel $a$ kilometers. Now, it needs to travel $d(d>a)$ kilometers to a certain place, with no refueling stations along the way. However, some gasoline can be transported to any point along the road and stored for later use. Suppose there are $k(k \in N)... | Let $d_{n}=a\left(1+\frac{1}{k+2}+\frac{1}{k+4}+\cdots+\frac{1}{k+2 n}\right)$,
$d_{0}=a$. First, we prove by induction: when $d=d_{n}$, at least $(k+n) L$ liters of fuel are required.
As shown in the figure, let $O$ be the starting point, and $X$ be the destination. In the proof, we will use the following fact: if $P$... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,566 |
3. Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ that satisfy: (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k_{i}$ (2) $\mid A_{i} \cap$ $A_{j} \mid \leqslant 2, i, j=1,2, \cdots, k, i \neq j$. Find the maximum value of $k$. (6th test, 1st question, provided by Wu Chang) | First, prove that each element in $S$ belongs to at most 3 of the sets $A_{1}, A_{2}, \cdots, A_{k}$.
By contradiction, suppose there is an element in $S$ that belongs to at least 4 subsets. Without loss of generality, assume $1 \in A_{1}, A_{2}, A_{3}, A_{4}$.
(1) If each of the other elements in $S$ belongs to at mo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,567 |
4. It is known that a total of 12 theater groups are participating in a 7-day drama festival. Each day, some of the groups perform on stage, while the groups not performing watch from the audience. It is required that each group must see the performances of all other groups. What is the minimum number of performances n... | Certainly, here is the translation of the provided text into English, preserving the original formatting:
```
It is impossible for 3 troupes to perform only for 2 days and meet the conditions of the problem. By the pigeonhole principle, on one day, at least two troupes must perform, and on that day, these two troupes ... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,568 |
5. Given a positive integer $n$, how many distinct points must be appropriately selected on a plane so that the following property holds: for each $k \in\{1$, $2, \cdots, n\}$, there always exists a line on the plane that passes through exactly $k$ of the selected points. (2nd Test, 3rd Problem, 1992 Commonwealth of In... | Let $X$ denote the set of all selected points, and let $l_{k}$ represent a line that contains exactly $k$ points from $X$, where $k=1,2, \cdots, n$.
On one hand, since $l_{n}$ contains $n$ points from $X$, $l_{n-1}$ contains $n \cdots 1$ points from $X$, and at least $n-2$ points are not on $l_{n}$. $l_{n-2}$ contains... | \left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,569 |
7. Find all positive integer solutions $(x, r, p, n)$ of the equation $x^{\prime}-1=p^{n}$ that satisfy the following two conditions:
(1) $p$ is a prime; (2) $r \geqslant 2, n \geqslant 2$. (4th test, 2nd question, an example related to Catalan's conjecture) | If $x=2, 2^{r}-1$ is an odd number $(r \geqslant 2)$, then $p$ is an odd number. If $n$ is even, then $p^{n}+1 \equiv 2 \bmod 4$, thus $2^{r} \equiv 2 \bmod 4, r=1$, no solution. If $n$ is odd, then $2^{r}=p^{n}+1=(p+1)\left(p^{n-1}-p^{n-2}+\cdots-p+1\right)$, hence we can set $p+1=2^{t}(t \in N)$. Then, $p^{n}+1=\left... | x=3, r=2, p=2, n=3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,572 |
8. Let $n$ be a positive integer. If a sequence of positive integers $\left(a_{1}, a_{2}, \cdots, a_{k}\right)$ satisfies $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{4}$, and $a_{1}+a_{2}+\cdots+a_{k}=n$, then $\left(a_{1}, a_{2}, \cdots, a_{k}\right)$ is called a partition of $n$. The set of all partitions of... | Prove that the following constructs a mapping $f$ from $D_{n}$ to $O_{n}$.
For any element $\left(a_{1}, a_{2}, \cdots, a_{k}\right)$ in $D_{n}$, write each $a_{i}$ as $a_{i}=2^{s_{i}} t_{i}$, where $t_{i}$ is odd and $s_{i}$ is a non-negative integer. Add all $a_{i}$ with the same $t_{i}$, i.e., $a_{i_{1}}+a_{i_{2}}+\... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,573 |
9. Let there be $n(n \geqslant 3)$ cities, and a certain airline operates direct flights between some of these cities. It is known that:
(1) From any of these $n$ cities, the way to reach any other city by the airline's flights (i.e., with possible layovers, but passing through the same city at most once) is unique;
(2... | Prove that for any city $A$, let the set of cities among these $n$ cities where the airfare to $A$ by the company's flights is an even hundred yuan be $S$, and the set of cities where the airfare is an odd hundred yuan be $T$. Specifically, city $A$ belongs to set $S$. Let the number of elements in $S$ be $x$, and the ... | n = (x - y)^{2} \text{ or } n = (x - y)^{2} + 2 | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,574 |
26. Given that $O$ is a point inside $\angle ABC$, the distances from $O$ to the vertices $A, B, C$ are $d_{A}, d_{B}, d_{C}$, respectively. The perpendicular distances from $O$ to the sides $a, b, c$ are $d_{a}, d_{b}, d_{c}$, respectively. Prove: $\frac{d_{A}}{d_{a}}+\frac{d_{B}}{d_{b}}+\frac{d_{C}}{d_{c}} \geqslant ... | (Upward Poem Base)
Similarly, $\frac{d_{B}}{d_{b}} \geqslant \frac{S_{3}+S_{1}}{S_{2}}, \frac{d_{c}}{d_{c}} \geqslant \frac{S_{1}+S_{2}}{S_{3}}$.
$$
\begin{array}{l}
\therefore \frac{d_{1}}{d_{a}}+\frac{d_{3}}{d_{b}}+\frac{d_{c}}{d_{c}} \\
\geqslant \frac{S_{2}+S_{3}}{S_{1}}+\frac{S_{3}+S_{1}}{S_{2}}+\frac{S_{1}+S_{2}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,575 |
3. Given point $P(4,5)$, two lines $i_{1}: x=3, l_{2}: x=13$. Then the equation of the circle passing through $P$ and tangent to $i_{1}, l_{2}$ is $\qquad$ | $\begin{array}{l}\text { 3. }(x-8)^{2}+(y-5)^{2}=25 \text { or }(x-8)^{2}+(y+1)^{2} \\ =25\end{array}$ | (x-8)^{2}+(y-5)^{2}=25 \text{ or } (x-8)^{2}+(y+1)^{2}=25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,578 |
5. In a regular tetrahedron with edge length $5 \sqrt{6} \mathrm{~cm}$, there is a point $P$, which is at distances of $1 \mathrm{~cm}, 2 \mathrm{~cm}$, and $3 \mathrm{~cm}$ from three of its faces. Then the distance from $P$ to the fourth face is $\qquad$ $\mathrm{cm}$. | $5.4$ | 5.4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,580 |
Example 5. Consider a square with side length 1. Try to find an inscribed equilateral triangle with the maximum area and one with the minimum area within this square, and calculate each area.
untranslated part:
"美殓" seems to be a mistranslation or typo in the original text. It does not have a clear meaning in this c... | Points are on two pairs of opposite sides, so at least one pair of opposite sides contains three points.
Without loss of generality, let $\bar{F}, G$ be on $A B, C D$ respectively, and $E$ be on $x$. Let the midpoint of $F$ be $K$, and connect $A K, D K, K E$.
$$
\because K, G, D, E \text{ are concyclic } (\angle E X ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,581 |
7. If the circle $x^{2}+y^{2}=r^{2}$ covers at least one maximum point and one minimum point of the function $f(x)=\sqrt{30} - \sin \frac{\pi x}{2 \sqrt{r}}$ (in this problem, if $x=x_{0}$, $f(x)$ reaches the maximum value $M$, then $\left(x_{0}, M\right)$ is called a maximum point; similarly for a minimum point), then... | 7. $[6,+\infty)$ | [6,+\infty) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 707,583 |
10. The range of the function $y=\sin x \cos x-\sin x-\cos x+1$ is $\qquad$ . | 10. $0 \leqslant y \leqslant \frac{3+2 \sqrt{2}}{2}$ | 0 \leqslant y \leqslant \frac{3+2 \sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,586 |
11. Let complex numbers $z_{1}, z_{2}$ satisfy $z_{2}=z_{1} i-2$, and $z_{1}$ corresponds to point $Z_{1}$ on the curve $|z-2|+|z+2|=10$ in the complex plane. Then the trajectory equation of $z_{2}$ in the complex plane, corresponding to point $Z_{2}$, is $\qquad$
(in the original Cartesian coordinate system of the com... | 11. $\frac{(x+2)^{2}}{21}+\frac{y^{2}}{25}=1$ | \frac{(x+2)^{2}}{21}+\frac{y^{2}}{25}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,587 |
12. In the Cartesian coordinate system, satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-4 x-6 y+4 \leqslant 0, \\
|x-2|+|y-3|>0
\end{array}\right.
$$
$\qquad$ . | $12.9 \pi-18$ | 9\pi-18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,588 |
A small ball with a radius of $1 \mathrm{~cm}$, no matter how the box is moved, the volume of the space in the box that the ball cannot reach is $\qquad$ $\mathrm{cm}^{3}$ (the thickness of the box is negligible). | $13.56-\frac{40}{3} \pi$ | 13.56-\frac{40}{3} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,589 |
16. In tetrahedron $ABCD$, the lengths of edges $AB$ and $CD$ are $a$ and $b$ respectively, and the distance between the midpoints of these two edges is $d$. Then the maximum volume of tetrahedron $ABCD$ is $\qquad$ | 16. $\frac{1}{6} a b d$ | \frac{1}{6} a b d | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,593 |
2. Let $\triangle A B C$ be an acute triangle, and construct isosceles right triangles $\triangle B C D, \triangle A B E, \triangle C A F$ outside $\triangle A B C$. In these triangles, $\angle B D C, \angle B A E, \angle C F A$ are right angles. Construct another isosceles right triangle $\triangle E F G$ outside quad... | $$
\begin{array}{l}
z_{G}=\overrightarrow{A G}=\overrightarrow{A F}+\overrightarrow{F G} \\
= \overrightarrow{A F}+\overrightarrow{E F} i \\
= \overrightarrow{A F}+(\overrightarrow{E A}+\overrightarrow{A F}) i \\
= \overrightarrow{A F}(1+i)+\overrightarrow{B A}, \\
(-1+i) z_{D} \\
=(-1+i) \overrightarrow{A D} \\
= \ov... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,599 |
4. Let the natural number $n \geqslant 5, n$ different natural numbers $a_{1}, a_{2}, \cdots$, $a_{n}$ have the following property: for any two different non-empty subsets $A, B$ of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, the sum of the numbers in $A$ and the sum of the numbers in $B$ will not be equal.... | 4. Let's assume $a_{1}c_{2}>\cdots >c_{n}$, and $D_{k}=\sum_{i=1}^{t} a_{i}-\left(1+2+\cdots+2^{k-1}\right)=\sum_{i=1}^{k} a_{i}-\left(2^{k}-1\right) \geqslant 0$. Then,
$$
\begin{array}{l}
1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}-\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)=\sum_{i=1}^{n} c_{i} d_{i} \\
= c_{1} ... | 2-\frac{1}{2^{n-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,601 |
1. A line $l$ passes through a tetrahedron, then the maximum number of faces of the tetrahedron that can intersect with $l$ is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | 1. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,602 |
Given $A C, C E$ are two diagonals of a regular hexagon $A B C D E F$, points $M, N$ internally divide $A C, C E$ respectively, such that
$A M \cdot A C=C N: C E=$
$r$. If $B, M, N$ are collinear, find the value of $r$. (23rd IMO) | Solve: Since $B, M, N=$ are collinear points, then we have
$$
S_{\triangle B C C^{2}}=S_{\triangle B C M}+S_{\triangle C M .} .
$$
Let the side length of the equilateral triangle be $a$, then $A C=C E=\sqrt{3} a$.
$$
\begin{array}{l}
\text { Also, } A M: A C=C N: C E=r, \\
\therefore A M=C N=\sqrt{3} a r . \\
\text { ... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,603 |
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