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4. Let positive integers $a, b$ satisfy $1 \leqslant a < b \leqslant 100$. If there exists a positive integer $k$, such that $a b \mid\left(a^{k}+b^{k}\right)$, then the pair $(a, b)$ is called a "good pair". Find the number of all good pairs.
（Xiong Bin's problem） | 4. Let $(a, b)=d, a=s d, b=t d,(s, t)=1(t>s)$.
Thus, $s t d^{2} \mid d^{k}\left(s^{k}+t^{k}\right)$.
Since $\left(s t, s^{k}+t^{k}\right)=1$, it follows that $s t \mid d^{k}$.
Therefore, all prime factors of $st$ can divide $d$.
If $s$ or $t$ has a prime factor $p$ not less than 11, then $p \mid d$. Thus, $p^{2} \mid ... | 96 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,168 |
Example 7 Given a convex polygon that cannot cover any triangle of area 1. Prove: this convex polygon can be covered by a triangle of area 4.
(1992, Huanggang City, Hubei Province Junior High School Mathematics Competition) | Proof As shown in Figure 5, let the convex polygon be $Q$. Then among the triangles formed by connecting the vertices of $Q$, we can find the one with the largest area (let it be $\triangle A B C$). Clearly, $S_{\triangle A B C}S_{\triangle A B C}$, which contradicts the fact that $S_{\triangle A B C}$ is the largest. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,169 |
5. As shown in Figure $2, \triangle A B C$ is a right triangle, $\angle A C B$ $=90^{\circ}, M_{1} 、 M_{2}$ are any two points inside $\triangle A B C$, $M$ is the midpoint of segment $M_{1}$ $M_{2}$, lines $B M_{1} 、 B M_{2} 、 B M$ intersect side $A C$ at points $N_{1} 、 N_{2} 、 N$ respectively. Prove:
$\frac{M_{1} N_... | 5. Let $H_{1}, H_{2}, H$ be the projections of points $M_{1}, M_{2}, M$ onto line $BC$. Then
$$
\begin{array}{l}
\frac{M_{1} N_{1}}{B M_{1}}=\frac{H_{1} C}{B H_{1}}, \frac{M_{2} N_{2}}{B M_{2}}=\frac{H_{2} C}{B H_{2}}, \\
\frac{M N}{B M}=\frac{H C}{B H}=\frac{H_{1} C+H_{2} C}{B H_{1}+B H_{2}} .
\end{array}
$$
Assume $... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,170 |
6. Let $\mathbf{Z}_{+}$ be the set of positive integers, and define: $a_{1}=2$,
$$
a_{n+1}=\min \left\{\lambda \left\lvert\, \sum_{i=1}^{n} \frac{1}{a_{i}}+\frac{1}{\lambda}<1\right., \lambda \in \mathbf{Z}_{+}\right\}(n=1,2, \cdots) \text {. }
$$
Prove: $a_{n+1}=a_{n}^{2}-a_{n}+1$.
(Li Shenghong) | 6. Given $a_{1}=2$ and equation (1), consider $\frac{1}{a_{1}}+\frac{1}{\lambda}2$.
Thus, $a_{2}=3$, which means the conclusion holds for $n=1$. Assume the conclusion holds for all $n \leqslant k-1(k \geqslant 2)$. For $n=k$, by equation (1), consider
$$
\sum_{i=1}^{k} \frac{1}{a_{i}}+\frac{1}{\lambda}\left(1-\sum_{i=1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,171 |
7. Let $n$ be a positive integer, and real numbers $a_{1}, a_{2}, \cdots, a_{n}$ and $r_{1}, r_{2}, \cdots, r_{n}$ satisfy:
$$
\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \text { and } 0 \leqslant r_{1} \leqslant r_{2} \leqslant \cdots \leqslant r_{n} . \\
\text { Prove: } \sum_{i=1}^{n} \su... | 7. Construct the $n \times n$ numerical table
$$
A_{1}=\left(\begin{array}{cccc}
a_{1} a_{1} r_{1} & a_{1} a_{2} r_{1} & \cdots & a_{1} a_{n} r_{1} \\
a_{2} a_{1} r_{1} & a_{2} a_{2} r_{2} & \cdots & a_{2} a_{n} r_{2} \\
\cdots & \cdots & \cdots & \cdots \\
a_{n} a_{1} r_{1} & a_{n} a_{2} r_{2} & \cdots & a_{n} a_{n} r... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,172 |
8. Given 8 points $A_{1}, A_{2}, \cdots, A_{8}$ on a circle. Find the smallest positive integer $n$, such that among any $n$ triangles with these 8 points as vertices, there must be two triangles that share a common side.
(Tao Pingsheng, problem contributor) | 8. First, consider the maximum number of triangles with no common edges.
Connecting every pair of eight points yields $\mathrm{C}_{8}^{2}=28$ chords.
If each chord belongs to only one triangle, then these chords can form at most $r \leqslant\left[\frac{28}{3}\right]=9$ triangles with no common edges. However, if ther... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,173 |
1. Let $a, b, c$ be positive real numbers, satisfying $a^{2}=b(b+c), b^{2}=c(c+a)$.
Then ( ).
(A) $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$
(B) $\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$
(C) $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$
(D) $\frac{2}{c}=\frac{1}{a}+\frac{1}{b}$ | - 1. C.
Multiplying the two equations given in the problem, we get
$$
a^{2}(b+c)(b-c)=a b c(b+c) \text {. }
$$
Since $a>0, b>0, c>0$, it follows that $b+c>0$.
Thus, $a(b-c)=b c \Rightarrow a b=b c+a c$.
Therefore, $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,174 |
2. For each $x$, the function $y$ is
$$
y_{1}=2 x, y_{2}=x+2, y_{3}=-2 x+12
$$
the minimum of these three functions. Then the maximum value of the function $y$ is ( ).
(A) 4
(B) 6
(C) $\frac{28}{3}$
(D) $\frac{16}{3}$ | 2. D.
In the same coordinate system, draw the graphs of $y_{1}$, $y_{2}$, and $y_{3}$, and thus obtain the graph of the function $y$.
It is easy to see that when $x=\frac{10}{3}$, $y$ has a maximum value of $\frac{16}{3}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,175 |
3. As shown in Figure 1, the three vertices of the equilateral $\triangle D E F$ with side length $n$ are exactly on the sides of the equilateral $\triangle A B C$ with side length $m$. Then the inradius of $\triangle A E F$ is ( ).
(A) $\frac{\sqrt{3}}{6}(m-n)$
(B) $\frac{\sqrt{3}}{4}(m-n)$
(C) $\frac{\sqrt{3}}{3}(m-n... | 3. A.
It is easy to prove that $\triangle A E F, \triangle B D E, \triangle C F D$ are congruent.
Thus, $S_{\triangle A B F}=S_{\triangle B D E}=S_{\triangle C F D}, A F=B E$.
Since $S_{\triangle A B C}=\frac{\sqrt{3}}{4} m^{2}, S_{\triangle D E F}=\frac{\sqrt{3}}{4} n^{2}$, we have
$$
S_{\triangle A E F}=\frac{1}{3}\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,176 |
4. In an equilateral $\triangle A B C$, $D$ and $E$ are points on $B C$ and $A C$ respectively, and $A E=C D$. Connecting $A D$ and $B E$ intersect at point $P$, and connecting $P C$. If $P B=2 P A$, then the degree of $\angle C P D$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 4. A.
It is easy to know that $\triangle A B E \cong \triangle C A D$.
Therefore, $\angle A B E=\angle C A D$.
$$
\begin{array}{l}
\text { Also, } \angle B P D=\angle A B E+\angle B A P \\
=\angle C A D+\angle B A P=\angle B A C=60^{\circ} .
\end{array}
$$
Draw $B Q \perp P D$ at point $Q$. Then
$$
\angle P B Q=30^{\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,177 |
6. As shown in Figure $3, A B$ is the diameter of the semicircle $\odot O$, $C$ is a moving point on the semicircle arc (excluding the midpoint of the semicircle arc and the two endpoints of the diameter), connect $A C, B C$, draw $C D \perp A B$ at point $D$, draw $\odot O_{1}$ tangent to $A D$ at point $E$, tangent t... | 6. B.
Connect $O O_{1}$ and extend it, then this line segment must pass through the tangent point G. Thus,
$$
A O=O B=O G=R \text {. }
$$
Let $A D=m, D B=n, m+n=2 R$.
Connect $O_{1} E$, then $O_{1} E \perp A D$.
In the right triangle $\triangle O_{1} E O$,
$$
O_{1} O^{2}=O_{1} E^{2}+O E^{2} \text {. }
$$
Also, $O_{1... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,179 |
Example 8 Place four circular faces $\odot C_{k}(k=1,2,3,4)$ on top of each other, such that any three of them have a common point. If $\odot C_{1}, \odot C_{2}, \odot C_{3}$ have a common point $P, \odot C_{1}, \odot C_{2}, \odot C_{4}$ have a common point $Q, \odot C_{1}, \odot C_{3}, \odot C_{4}$ have a common point... | Solve: These four circular regions have a common point.
Below, we discuss three positional relationships of points \( P, Q, R, S \).
(1) If points \( P, Q, R, S \) are collinear, as shown in Figure 6, assume without loss of generality that points \( S, P \) lie on segment \( QR \).
Since points \( P, Q \) are both insi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,180 |
2. Robots A and B start from the starting point at the same time, moving uniformly along a hundred-meter track, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the fin... | 2.1.
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙}}$, respectively. Thus,
$$
\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}=\frac{x}{x-1.01} \text {. }
$$
Solving for $x$ yields $x=101$.
Therefore, this track is $1 \mathrm{~m}$ lon... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,182 |
3. As shown in Figure 6, given the quadratic function $\boldsymbol{y}=$ $(x+m)^{2}+k-m^{2}$, the graph intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and intersects the $y$-axis at point $C$. Then the other intersection point $D$ of the circumcircle of $\triangle A... | 3. $(0,1)$.
It is easy to find that point $C(0, k)$.
From the given condition, $x_{1}$ and $x_{2}$ are the roots of the equation $(x+m)^{2}+k-m^{2}$
$=0$, which simplifies to $x^{2}+2 m x+k=0$. Therefore,
$$
x_{1}+x_{2}=-2 m, x_{1} x_{2}=k \text {. }
$$
By the intersecting chords theorem, we have
$$
O D=\frac{O A \cd... | (0,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,183 |
4. For the non-negative real number $x^{4}$ rounded to the nearest integer, denote it as $\langle x\rangle$, that is, when $n$ is a non-negative integer, if
$$
\begin{array}{l}
n-\frac{1}{2} \leqslant x<n+\frac{1}{2}, \\
\text { then }\langle x\rangle=n(\text { for example, }\langle 0\rangle=\langle 0.48\rangle=0,\lang... | 4. (1), (2).
Since $\langle 2 x-1\rangle=3$, we have
$$
3-\frac{1}{2} \leqslant 2 x-1<3+\frac{1}{2} \Rightarrow \frac{7}{4} \leqslant x<\frac{9}{4} \text {. }
$$
Therefore, conclusion (1) is correct.
Let $\langle x\rangle=n$. Then
$n-\frac{1}{2} \leqslant x<n+\frac{1}{2}$ (where $n$ is a non-negative integer).
Thus, ... | (1), (2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,184 |
One, (20 points) Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+c+\frac{3}{4}\right)\left(c^{2}+a+\frac{3}{4}\right) \\
\geqslant\left(2 a+\frac{1}{2}\right)\left(2 b+\frac{1}{2}\right)\left(2 c+\frac{1}{2}\right) .
\end{array}
$$ | $$
\begin{array}{l}
a^{2}+b+\frac{3}{4}=\left(a-\frac{1}{2}\right)^{2}+a+b+\frac{1}{2} \\
\geqslant a+b+\frac{1}{2} .
\end{array}
$$
Similarly, $b^{2}+c+\frac{3}{4} \geqslant b+c+\frac{1}{2}$,
$$
\begin{array}{l}
c^{2}+a+\frac{3}{4} \geqslant c+a+\frac{1}{2} . \\
\text { Therefore, }\left(a^{2}+b+\frac{3}{4}\right)\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,185 |
II. (25 points) As shown in Figure 7, in the square $ABCD$, $F$ and $E$ are points on sides $BC$ and $CD$ respectively, such that the perimeter of $\triangle CEF$ is equal to half the perimeter of the square $ABCD$. $AE$ and $AF$ intersect $BD$ at points $M$ and $N$ respectively. Prove that $\angle BNF = \angle DEM$ | Let the side length of the square be $1, BF=a, DE=b$. Then $CF=1-a, CE=1-b$. According to the problem, $EF=BF+DE$, which means $\sqrt{(1-a)^{2}+(1-b)^{2}}=a+b$.
Simplifying, we get $ab+a+b=1 \Rightarrow \frac{a}{1+a}=\frac{1-b}{2}$.
Draw $NG \perp BC$.
Since $NB$ bisects $\angle ABF$, we have
$$
\begin{array}{l}
\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,186 |
Three. (25 points) Given that $a$, $b$, and $c$ are positive integers, and $a < b < c$. If
$$
a b c \mid (a b - 1)(b c - 1)(c a - 1),
$$
then, can three segments of lengths $\sqrt{a}$, $\sqrt{b}$, and $\sqrt{c}$ form a triangle? If so, find the area of the triangle; if not, explain the reason. | Three, notice that
$(a b-1)(b c-1)(c a-1)$
$=a b c(a b c-a-b-c)+a b+b c+c a-1$.
From $a b c \mid(a b-1)(b c-1)(c a-1)$, we get
$a b c \mid(a b+b c+c a-1)$,
which means $\frac{a b+b c+c a-1}{a b c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c}$
is an integer.
Since $a b+b c+c a-1>0$, we know that
$$
\frac{1}{a}+\f... | \frac{\sqrt{6}}{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,187 |
1. Among the seven points consisting of the center and the six vertices of a regular hexagon, if any $n$ points are taken, and among them, there must be three points that form the vertices of an equilateral triangle, then the minimum value of $n$ is $\qquad$ | -1.5 .
Consider the regular hexagon $A B C D E F$ as shown in Figure 1, with its center at $O$.
When $n=4$, take $A, C, D, F$, among which no three points can form the three vertices of an equilateral triangle.
When $n=5$, consider the following two cases:
(1) $O$ is among the five points. Consider the three pairs of ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,188 |
3. In $\triangle A B C$, it is known that $\angle A=120^{\circ}$. Let the vectors be
$$
\begin{aligned}
\alpha & =\frac{\overrightarrow{B A}}{|\overrightarrow{B A}| \cos A}+\frac{\overrightarrow{B C}}{|\overrightarrow{B C}| \cos C}, \\
\beta & =\frac{\overrightarrow{C A}}{|\overrightarrow{C A}| \cos A}+\frac{\overright... | 3. $60^{\circ}$.
Notice that $\boldsymbol{\alpha} \cdot \overrightarrow{C A}=\boldsymbol{\beta} \cdot \overrightarrow{B A}=0$, i.e., $\boldsymbol{\alpha} \perp \overrightarrow{C A}, \boldsymbol{\beta} \perp \overrightarrow{B A}$.
Thus, the angle between $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ is equal to or sup... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,190 |
1. Divide a circular piece of paper into several smaller pieces of any size using straight lines. If the number of smaller pieces is no less than 2004, then the minimum number of straight lines needed is ( ).
(A) 61
(B) 62
(C) 63
(D) 64 | Hint: First归纳出induce that $k$ lines can divide a circle into at most $f(k)=\frac{k(k+1)}{2}+1$ pieces of paper, and $f(62)<2004<f(63)$. Answer: $\mathrm{C}$.
Note: The word "induce" in the translation should be "induce that" to better fit the context in English. Here is the corrected version:
Hint: First induce that ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,191 |
4. In the tetrahedron $V-ABC$, it is known that the base $ABC$ is an isosceles right triangle with $\angle B$ as the right angle, the plane $VAC \perp$ plane $ABC$, $AC=4$, $VA=\sqrt{14}$, and the tangent value of the dihedral angle $V-AB-C$ is $\frac{\sqrt{10}}{3}$. Then the angle formed by $VB$ and the base $ABC$ is ... | 4. $45^{\circ}$.
As shown in Figure 2, construct VH
$\perp$ plane $ABC$, $HD \perp$
$AB$. Then $AB \perp VD$.
Therefore, $\angle VDH$ is
the plane angle of the dihedral angle $V-AB-C$, i.e.,
$$
\tan \angle VDH = \frac{\sqrt{10}}{3} \text{.}
$$
Figure 2
Let $VH = \sqrt{10} x$, $HD = 3x$, $AH = 3\sqrt{2} x$.
In the ri... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,192 |
5. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with left and right foci $F_{1}$ and $F_{2}$, respectively, and $P$ as any point on the ellipse not coinciding with the left or right vertices, points $I$ and $G$ are the incenter and centroid of $\triangle P F_{1} F_{2}$, respectively. When $I G$ ... | 5. $\frac{1}{3}$.
Let $\left|P F_{1}\right|=n,\left|P F_{2}\right|=m$. Then $m+n=2 a$.
According to the properties of the orthocenter and incenter, we have
$$
\begin{array}{l}
\overrightarrow{P G}=\frac{1}{3}\left(\overrightarrow{P F_{1}}+\overrightarrow{P F_{2}}\right), \\
\overrightarrow{P I}=\frac{m \overrightarrow... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,193 |
6. Given $a, b, c \in [0,1]$. Then
$$
\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}
$$
the range of values is | 6. $[0,2]$.
Obviously, the minimum value of the required algebraic expression is 0, which is achieved if and only if $a=b=c=0$.
When $a+b+c>0$, $a+b+c \leqslant 1+b+c \leqslant 2(1+b c)$.
Thus, $\frac{a}{b c+1} \leqslant \frac{2 a}{a+b+c}$.
Similarly, $\frac{b}{c a+1} \leqslant \frac{2 b}{a+b+c}, \frac{c}{a b+1} \leq... | [0,2] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,194 |
7. A die (a uniform cube with faces numbered $1,2,3,4$, 5,6) is rolled three times. What is the probability that the three numbers on the top faces can form the side lengths of a triangle whose perimeter is divisible by 3? $\qquad$ | 7. $\frac{11}{72}$.
Let the numbers obtained from rolling the three dice be $a$, $b$, and $c$ ($a \leqslant b \leqslant c$). Then the condition is equivalent to
$$
a+b>c \text {, and } 31(a+b+c) \text {. }
$$
If the triangle formed is equilateral, there are 6 cases.
If the triangle formed is isosceles (but not equil... | \frac{11}{72} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,195 |
8. From the numbers $1,2, \cdots, n$, if two different numbers are chosen and multiplied, the arithmetic mean of the products is $\qquad$ | 8. $\frac{(n+1)(3 n+2)}{12}$.
The arithmetic mean of the products is equal to
$$
\begin{array}{l}
\frac{1}{\mathrm{C}_{n}^{2}} \sum_{i<i<j<n} i j=\frac{1}{n(n-1)}\left[\left(\sum_{i=1}^{n} i\right)^{2}-\sum_{i=1}^{n} i^{2}\right] \\
=\frac{(n+1)(3 n+2)}{12} \text {. } \\
\end{array}
$$ | \frac{(n+1)(3 n+2)}{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,196 |
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, \frac{a_{n-1}}{a_{n}}=\frac{2 n a_{n-1}+1}{1-a_{n}}(n \geqslant 2) .
$$
Find the value of $\sum_{n=2}^{\infty} n\left(a_{n}-a_{n+1}\right)$. | $$
\begin{array}{l}
\frac{1}{a_{n}}-\frac{1}{a_{n-1}}=2 n+1 . \\
\begin{array}{l}
\text { Then } \frac{1}{a_{n}}= \frac{1}{a_{1}}+\sum_{i=2}^{n}\left(\frac{1}{a_{i}}-\frac{1}{a_{i-1}}\right) \\
=3+\sum_{i=2}^{n}(2 i+1)=n(n+2) .
\end{array}
\end{array}
$$
Thus, $a_{n}=[n(n+2)]^{-1}$.
Therefore, $\sum_{n=2}^{\infty} n\... | \frac{13}{24} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,197 |
10. (20 points) Try to find the maximum value of the function defined on $(0, \pi)$:
$$
\begin{aligned}
f(x)= & \log _{\sqrt{2}-1}\left[\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)+1\right]+ \\
& \log _{\sqrt{2}+1} \frac{\sin ^{2} x}{2 \sin x-\cos x+1}+ \\
& \cos ^{3} x+\sin ^{2} x-\cos x
\end{aligned}
$$ | 10. Notice that
$$
\begin{aligned}
f(x)= & \log _{\sqrt{2-1}} \frac{\sin x+\cos x+1}{\sin x} \cdot \frac{2 \sin x-\cos x+1}{\sin x}+ \\
& (1-\cos x) \sin ^{2} x \\
= & \log _{\sqrt{2}-1}\left(1+\cot \frac{x}{2}\right)\left(2+\tan \frac{x}{2}\right)+ \\
& 8 \sin ^{2} \frac{x}{2} \cdot \sin ^{2} \frac{x}{2} \cdot \cos ^{... | -\frac{22}{27} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,198 |
11. (20 points) In the Cartesian coordinate system $x O y$, a line $l$ is drawn through the point $M(m, 0) (m>0)$, intersecting the parabola $y^{2}=2 p x (p>0)$ at points $A$ and $B$. A line $l^{\prime}$ perpendicular to the $x$-axis is drawn through the point $M^{\prime}$, which is the symmetric point of $M$ with resp... | 11. (1) Lines $A D$ and $B C$ are both parallel to the $x$-axis. Let the slope of line $l$ be $k$, and assume $k>0$. Then the equation of line $l$ is $y=k(x-m)$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$,
Clearly, $y_{1}$ and $y_{2}$ have opposite signs.
First, we prove: $\angle A M^{\prime} O=\... | \pi-\arccos \frac{11}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,199 |
One. (40 points) Given that quadrilateral $ABCD$ is circumscribed around $\odot O$, the incircle $\odot O_{1}$ of $\triangle ACB$ touches sides $AB$ and $BC$ at points $P$ and $Q$ respectively, and the incircle $\odot O_{2}$ of $\triangle ACD$ touches sides $CD$ and $DA$ at points $R$ and $S$ respectively. Prove that t... | Let the points of tangency of $\odot O_{1}$ and $\odot O_{2}$ with $AC$ be $M$ and $M^{\prime}$, respectively.
From the fact that $\odot O_{1}$ and $\odot O_{2}$ are the incircles of $\triangle ACB$ and $\triangle ACD$, respectively, we have
$$
AM = \frac{AC + AB - CB}{2}, \quad AM^{\prime} = \frac{AC + AD - CD}{2}.
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,200 |
II. (40 points) Given positive real numbers $a, b, c$ satisfying $a+b+c \geqslant abc$. Prove that in the inequalities
$$
\frac{2}{a}+\frac{3}{b}+\frac{6}{c} \geqslant 6, \frac{2}{b}+\frac{3}{c}+\frac{6}{a} \geqslant 6, \frac{2}{c}+\frac{3}{a}+\frac{6}{b} \geqslant 6
$$
at least two of them are true. | Assume that at most one of the following inequalities holds. Without loss of generality, let
$$
\frac{2}{a}+\frac{3}{b}+\frac{6}{c}<6, \frac{2}{b}+\frac{3}{c}+\frac{6}{a}<6 .
$$
Combining the conditions, we get
$$
\begin{array}{l}
\left(\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\right)^{2} a b c<36(a+b+c), \\
\left(\frac{2}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,201 |
2. Divide each face (a total of six faces) of a $3 \times 3 \times 3$ cube into 9 equal smaller squares (a total of 54 smaller squares). Now, color these smaller squares using red, yellow, and blue such that the maximum number of smaller squares sharing a common edge is ( ).
(A) 28
(B) 27
(C) 22
(D) 20
(1989, "Zu Chong... | To consider that small squares on the same face with a common edge cannot be the same color, and also to consider that small squares on adjacent faces with a common edge cannot be the same color. Since each small cube can only have one face painted red, the maximum number of small squares that can be painted red is 26,... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,202 |
Three. (50 points) Color some nodes (vertices of unit squares) on an $n \times n$ chessboard red, such that the boundary of any $k \times k (1 \leqslant k \leqslant n)$ sub-chessboard composed of unit squares has at least one red point. Let the minimum number of red points that satisfy this condition be $l(n)$. Try to ... | For any red point $P$, assign a value to it as follows: if there are $m$ red points on the boundary of a unit square containing $P$, then $P$ gets a score of $\frac{1}{m}$ from that unit square. The value of point $P$ is the sum of the scores it gets from all unit squares containing it.
For red points on the boundary ... | \frac{2}{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,203 |
(1) If a positive integer $n$ can be expressed in the form $a^{b} (a, b \in \mathbf{N}, a, b \geqslant 2)$, then $n$ is called a "good number." Find all good numbers adjacent to positive integer powers of 2.
(2) Find all non-negative integer solutions $(x, y, z)$ of the indeterminate equation $\left|2^{x}-3^{y} \times ... | (1) Let the required good number be $n, n=a^{b}(a, b \in \mathbf{N}_{+}, a \geqslant 2, b \geqslant 2)$.
Thus, there exists a positive integer $t(t>1)$, such that
$2^{t}=a^{b} \pm 1$.
Clearly, $a$ is an odd number.
If $b$ is odd, then
$$
2^{t}=(a \pm 1)\left(a^{b-1} \mp a^{b-2}+\cdots \mp a+1\right).
$$
And $a^{b-1} \... | (1,0,0),(1,1,0),(2,1,0),(3,2,0),(4,1,1),(2,0,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,204 |
Initially 285 Test: How many positive integers $T$ are there such that between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$ (not including $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$) there are exactly 5 different positive integers? | Solution: Obviously, when $T \geqslant 2010$,
$$
\frac{2010}{T} \leqslant 1, \frac{2010+T}{2 T} \leqslant 1,
$$
which does not meet the requirement.
Therefore, $T < 2010$ and
$$
\begin{array}{l}
\frac{2010}{T}>\frac{2010+T}{2 T} \\
\Rightarrow \frac{2010}{T}>\frac{2010+T}{2 T} .
\end{array}
$$
From this, we know
$$
\... | 33 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,205 |
285 On a plane, there are 12 circles. If any three circles have at least two circles that are disjoint, prove: among the 12 circles, there must exist four circles that are pairwise disjoint. | Prove (1) Take any six circles, and represent the six circles with six points.
If two circles are disjoint, connect them with a red line; if two circles are not disjoint, connect them with a blue line. It is easy to know that six points connected with two colors must contain a monochromatic triangle.
Given that there... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,206 |
Example 1 As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=$ $90^{\circ}$, point $D$ is on side $CA$, such that $CD=1, DA=3$, and $\angle BDC=3 \angle BAC$. Find the length of $BC$. ${ }^{[1]}$
(2009, Xin Zhi Cup Shanghai Junior High School Mathematics Competition) | From the problem, we know
$$
\angle A B D=2 \angle B A C \text {. }
$$
Draw the angle bisector of $\angle A B D$ through point $B$, intersecting $D A$ at point $E$. Then $\triangle A E B$ is an isosceles triangle.
Let $A E=x$. Then $B E=x, D E=3-x$. Applying properties to $\triangle E B C$ and $\triangle A B C$ respec... | \frac{4 \sqrt{11}}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,207 |
Example 2 Given in $\triangle A B C$, $\angle A C B=90^{\circ}$.
(1) As shown in Figure 2, when point $D$ is on side $A B$ (excluding endpoints), prove:
$$
\begin{array}{l}
\frac{C D^{2}-B D^{2}}{B C^{2}} \\
=\frac{A D-B D}{A B} .
\end{array}
$$
(2) When point $D$ coincides with $A$, does the equation in (1) hold? Plea... | (1) Draw $C E \perp B D$ at point $E$, then by the projection theorem we have
$$
B C^{2}=B A \cdot B E \text {. }
$$
For the right triangle $\triangle C B E$ and point $D$, applying the property we get
$$
\begin{array}{l}
C D^{2}=B C^{2}+B D^{2}-2 B D \cdot B E . \\
\text { Therefore, } \frac{C D^{2}-B D^{2}}{B C^{2}}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,208 |
Example 5 Find positive integers $a, b, c$, such that
$$
[a, b, c]=a+b+c .{ }^{[2]}
$$
(2005, Austrian Mathematical Olympiad) | Let's assume $a \leqslant b \leqslant c$, where the equality cannot hold simultaneously.
Therefore, $a+b<2 c \Rightarrow c<a+b+c<3 c$.
Combining $l c=[a, b, c]=a+b+c$, we get $l=2$, i.e., $a+b=c$.
On the other hand, $k b=[a, b, c]=a+b+c=2 a+2 b$. Hence $b \leqslant 2 a$.
Combining $2 a \leqslant 2 b$, we get $b=a$ or $... | b=2a, c=3a | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,209 |
$$
\begin{array}{l}
\text { Three, (25 points) Let three distinct prime numbers } a, b, c \text { satisfy } \\
a \text { divides }(3 b-c), b \text { divides }(a-c), c \text { divides }(2 a-7 b), \\
20<c<80 .
\end{array}
$$
Find all values of $a^{b} c$. | When $a>b, a>c$, we have:
$$
-7 c<-7 b<2 a-7 b<2 a<2 c \text {. }
$$
Let $2 a-7 b=k c$,
where, $k=-6,-5,-4,-3,-2,-1,0,1$.
Then $2 a=7 b+k c \Rightarrow 7 b=2 a-k c$,
$$
\begin{array}{l}
a|(3 b-c) \Leftrightarrow a|(3 \times 7 b-7 c) \\
\Leftrightarrow a \mid[3(2 a-k c)-7 c] \\
\Leftrightarrow a \mid[6 a-(3 k+7) c] \\
... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,210 |
$$
\begin{array}{l}
\text { 1. Let } f(x)=k x+b \text {, and let } \\
A=\{x \mid f(x)=x, x \in \mathbf{R}\}, \\
B=\{x \mid f(f(x))=x, x \in \mathbf{R}\} .
\end{array}
$$
If $A \neq B$, then in the $x O y$ plane, the figure formed by the points $(k, b)$ is $\qquad$ | 1. The line $x=-1$. From $f(x)=x$, we get $(1-k) x=b$.
When $k=1, b=0$, $A=\mathbf{R}$; when $k=1, b \neq 0$, $A=\varnothing$; when $k \neq 1$, $A=\left\{\frac{b}{1-k}\right\}$. From $f(f(x))=x$, we get $\left(1-k^{2}\right) x=k b+b$. When $k=1, b=0$, $B=\mathbf{R}$; when $k=1, b \neq 0$, $B=\varnothing$; when $k=-1$, ... | x=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,211 |
2. Let $k$ be a constant. If for all $x, y \in(0,1)$, we have
$$
x^{k}+y^{k}-x^{k} y^{k} \leqslant \frac{1}{x^{k}}+\frac{1}{y^{k}}-\frac{1}{x^{k} y^{k}},
$$
then the range of the real number $k$ is $\qquad$ . | 2. $(-\infty, 0]$.
Notice that
$$
\begin{array}{l}
x^{k}+y^{k}-x^{k} y^{k} \leqslant \frac{1}{x^{k}}+\frac{1}{y^{k}}-\frac{1}{x^{k} y^{k}} \\
\Leftrightarrow\left(1-x^{k}\right)\left(1-y^{k}\right) \geqslant\left(1-\frac{1}{x^{k}}\right)\left(1-\frac{1}{y^{k}}\right) \\
\Leftrightarrow x^{k} y^{k} \geqslant 1 \Leftrig... | (-\infty, 0] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,212 |
3. Let $x, y, z \geqslant 0, x+y+z=1$. Then
$$
\sqrt{2011 x+1}+\sqrt{2011 y+1}+\sqrt{2011 z+1}
$$
The sum of the maximum and minimum values is $\qquad$ L. | $$
\begin{array}{l}
\text { 3. } \sqrt{6042}+2 \sqrt{503}+2 \text {. } \\
\text { Let } a=2011 x+1, b=2011 y+1 \text {, } \\
c=2011 z+1 \text {. }
\end{array}
$$
Then \( a, b, c \geqslant 1, a+b+c=2014 \).
First, by the Cauchy-Schwarz inequality, we have
$$
\sqrt{a}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3(a+b+c)}=\sqrt{60... | \sqrt{6042}+2 \sqrt{503}+2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,213 |
4. Let the acute angle $\alpha$ satisfy
$$
3 \sqrt{41-40 \cos \alpha}+4 \sqrt{34-30 \sin \alpha}=25 \text {. }
$$
Then $\sin \alpha=$ . $\qquad$ | 4. $\frac{3}{5}$.
As shown in Figure 4, construct two triangles $\triangle ACD$ and $\triangle BCD$ with a common side $CD$, such that
$$
\begin{array}{l}
\angle ACD = \alpha, \\
\angle BCD \\
= 90^{\circ} - \alpha, \\
AC = 15, CD = 12, BC = 20.
\end{array}
$$
By the Law of Cosines, we have
$$
\begin{array}{l}
AD = ... | \frac{3}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,214 |
5. As shown in Figure 1, in the cube $A C_{1}$ with edge length 1, points $P$ and $Q$ are moving points on edges $A D$ and $A_{1} B_{1}$, respectively. If the skew lines $B D_{1}$ and $P Q$ are perpendicular to each other, then $A P + A_{1} Q =$ $\qquad$ | 5. 1 .
As shown in Figure 5, establish a spatial rectangular coordinate system, and set
$$
\begin{array}{l}
B(1,0,0), \\
D_{1}(0,1,1), \\
P(0, a, 0), \\
Q(b, 0,1) . \\
\text { Then } \overrightarrow{B D_{1}}=(-1,1,1), \\
\overrightarrow{P Q}=(b,-a, 1) .
\end{array}
$$
$$
\text { Therefore, } 0=\overrightarrow{B D_{1}... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,215 |
6. Let $A B$ be a chord of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passing through its left focus $F$, with point $A$ above the $x$-axis. If $F$ divides $A B$ in the ratio $\lambda(\lambda \neq 1)$, then the slope of $A B$ $k=$ $\qquad$ | 6. $\frac{\sqrt{4 \lambda a^{2}-(\lambda+1)^{2} b^{2}}}{(\lambda-1) a}$.
Let the left directrix of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ be $l$, and the projections of points $A$ and $B$ on $l$ be $A_{1}$ and $B_{1}$, respectively.
$$
\begin{array}{l}
\text { Then } \frac{\left|A A_{1}\right|}{\left|... | \frac{\sqrt{4 \lambda a^{2}-(\lambda+1)^{2} b^{2}}}{(\lambda-1) a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,216 |
7. Given positive integers $n, p$ greater than 1, for any positive integer, if its $p$-ary representation contains no digit zero, then it is called “$p$-full”. Then, the number of $p$-full natural numbers among $1,2, \cdots, p^{n}$ is $\qquad$ | 7. $\left\{\begin{array}{ll}n, & p=2 ; \\ \frac{(p-1)^{n+1}-p+1}{p-2}, & p>2 .\end{array}\right.$
Obviously, the number of $p$-full natural numbers with $k(1 \leqslant k \leqslant n)$ digits in $p$-ary representation is $(p-1)^{k}$. Therefore, the number of $p$-full natural numbers we are looking for is
$$
\sum_{k=1}^... | \left\{\begin{array}{ll}n, & p=2 ; \\ \frac{(p-1)^{n+1}-p+1}{p-2}, & p>2 .\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,217 |
8. For a positive integer $x$, let $S(x)$ denote the sum of the digits of $x$. Then the maximum value of $S(x)-9[\lg x]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 8. 9 .
Let $x=\overline{a_{n} a_{n-1} \cdots a_{0}}\left(a_{0} \neq 0\right)$. Then
$$
\begin{array}{l}
10^{n} \leqslant x<10^{n+1} \\
\Rightarrow n \leqslant \lg x<n+1 \Rightarrow[\lg x]=n . \\
\text { Hence } S(x)=a_{0}+a_{1}+\cdots+a_{n} \leqslant 9(n+1) \\
=9([\lg x]+1) .
\end{array}
$$
Therefore, $S(x)-9[\lg x] ... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,218 |
Example 6 Find $k, n \in \mathbf{N}_{+}$, such that
$$
(n+1)^{k}=n!+1 \text {. }^{[3]}
$$
(2002-2003, Finnish High School Mathematics Competition) | 【Analysis】From the perspective of prime numbers, let $p$ be a prime factor of $n+1$. Then $p!(n!+1)$.
Obviously, when $p \leqslant n$, this equation cannot hold.
Therefore, $n+1$ must be a prime number $p$, i.e., $p^{k}=(p-1)!+1$.
By enumeration, we first study a few simple cases.
When $p=2$ or 3, $k=1$; when $p=5$, $k... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,220 |
10. (20 points) As shown in Figure 2, in the tetrahedron $ABCD$,
Figure 2
it is known that the edges $AB, AC, AD$ are mutually perpendicular, $BC = CD = 10$, and $BD = 12$. A plane $\alpha_{A}$ parallel to the base $BCD$ is constructed such that the distance between $\alpha_{A}$ and the base $BCD$ is 1. Similarly, co... | 10. As shown in Figure 6, let the sections $\alpha_{B}, \alpha_{C}, \alpha_{D}$ intersect at a point $A'$. Similarly, define points $B', C', D'$.
Thus, the tetrahedron $A' B' C' D'$ is similar to the tetrahedron $A B C D$.
Let the similarity ratio be $k$, and the volumes of tetrahedra $A B C D$ and $A' B' C' D'$ be $... | \frac{\sqrt{7}}{9}(6 - \sqrt{7} - \sqrt{2})^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,221 |
11. (20 points) Let $A$ be a moving point on the hyperbola $y=\frac{2011}{x}$. From $A$, draw two tangents $A P$ and $A Q$ to the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, with points of tangency $P$ and $Q$. If the left focus of the ellipse is $F$, find the minimum value of $\frac{|A F|^{2}}{|P F||Q F|}$. | 11. Let $A\left(x^{\prime}, y^{\prime}\right), P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then $x^{\prime} y^{\prime}=2011$.
It is easy to know that the chord of tangents from point $A$ to the ellipse, i.e., the line $P Q$, is given by
$$
\frac{x^{\prime} x}{25}+\frac{y^{\prime} y}{9}=1 .
$$
From $\left\... | \frac{4022}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,222 |
One. (40 points) As shown in Figure 3, in $\triangle ABC$, $BC < AB < AC$. Points $R$ and $Q$ are taken on sides $AB$ and $AC$ respectively, such that $BR = BC$ and $AQ = AB$. Let $BQ$ and $CR$ intersect at point $A_1$. A line through $A$ intersects $BC$, $BQ$, and $CR$ at points $P$, $C_1$, and $B_1$ respectively, and... | Given $B R=B C, A Q=A B$, therefore,
$$
\angle B C R=\angle B R C, \angle B Q A=\angle A B Q \text {. }
$$
From the given $\triangle A_{1} R B \backsim \triangle A_{1} C_{1} B_{1}$, we know that $B, R, B_{1}, P$ are concyclic. Therefore,
$$
\angle B R C=\angle B C_{1} B_{1}, \angle A B Q=\angle C_{1} B_{1} A_{1} .
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,223 |
ii. (40 points) Find all positive integers $n$ such that
$$
\sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n} \geqslant \frac{n-1}{n} .
$$ | When $n=1$, $\sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}=0=\frac{n-1}{n}$.
When $n=2$, $\sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}=0\frac{2}{3}=\frac{n-1}{n} \text {. }$
When $n=4$,
$$
\sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}=\sin \frac{\pi}{2}=1>\frac{3}{4}=\frac{n-1}{n} \text {. }
$$
When $n=5$, since ... | n \text{ is all positive integers except 2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,224 |
Three. (50 points) Prove: For any positive integer $n$, there exists a unique pair of positive odd integers $(x, y)$, such that $2^{n+2}=x^{2}+7 y^{2}$.
保留源文本的换行和格式，直接输出翻译结果。 | Three, Existence.
When $n=1$, take $x_{1}=y_{1}=1$, then
$$
2^{n+2}=x_{1}^{2}+7 y_{1}^{2} \text {. }
$$
Assume for $n\left(n \in \mathbf{N}_{+}\right)$, there exists a pair of positive odd numbers $\left(x_{n}, y_{n}\right)$, such that $2^{n+2}=x_{n}^{2}+7 y_{n}^{2}$.
Let $x_{n+1}=\frac{x_{n}-7 y_{n}}{2}, y_{n+1}=\fra... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,225 |
Four. (50 points) Two people, A and B, play a game. A first writes down any sequence of length $n$ consisting of $\mathrm{L}$ and $\mathrm{R}$ on a piece of paper. Then B places $n$ distinct prime weights on a balance scale one by one (weights already placed are not removed), and writes down a letter in sequence: if th... | Four, Player B has a winning strategy.
Let the masses of the $n$ weights be
$a_{1}<a_{2}<\cdots<a_{n}$.
Suppose the sequence written by Player A is $A$.
We will prove: Player B has a way to write down the sequence $A$.
Without loss of generality, assume the last term of $A$ is $L$, and $n$ is even.
Player B's strategy ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 723,226 |
Given 2011 positive integers, the product of which is equal to their sum. How many 1s are there at least among these 2011 numbers? | Let the 2011 positive integers be
$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2011}$.
From $a_{1} a_{2} \cdots a_{2011}=a_{1}+a_{2}+\cdots+a_{2011}$
$\leqslant 2011 a_{2011}$,
we know $a_{1} a_{2} \cdots a_{2010} \leqslant 2011a b \Rightarrow(a-1)(b-1)a b \Rightarrow(a-1)(b-1)<1,
\end{array}
$$
Contradiction.... | 2004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,227 |
$\left(x^{2}-x+1\right)$ "has the number of coefficients.
Translating the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, for the actual translation of the given text, it would be:
$\left(x^{2}-x+1\right)$ "has the number of coe... | Let polynomials $P(x), Q(x)$ be called similar if they have the same number of terms and the same coefficients, denoted as $P^{\prime}(x) \sim Q(x)$. At this point, the polynomials $P(x)$ and $Q(x)$ have the same number of terms and coefficients, and the number of non-zero coefficients of the polynomial $P(x)$ is denot... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,230 |
Example 7 Let $x, y, z$ be integers greater than 2, and
$$
\begin{array}{l}
x y \equiv 2(\bmod z), y z \equiv 2(\bmod x), \\
z x \equiv 2(\bmod y) .
\end{array}
$$
Find the value of $x+y+z$. | Three congruences are equivalent to
$$
x|(y z-2), y|(z x-2), z \mid(x y-2) \text {. }
$$
Therefore, $x y z \mid[(y z-2)(z x-2)(x y-2)]$.
Notice that
$$
\begin{array}{l}
(y z-2)(z x-2)(x y-2) \\
=A x y z+4(x y+y z+z x)-8 .
\end{array}
$$
Thus, $4(x y+y z+z x)-8=k x y z$, which means
$$
4\left(\frac{1}{x}+\frac{1}{y}+\... | 54 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,231 |
Example 8 Let $a, b, c$ be positive rational numbers, and $a+\frac{1}{b c}$, $b+\frac{1}{c a}$, $c+\frac{1}{a b}$ are all integers. Find the value of $a+b+c$. | $$
M=\left(a+\frac{1}{b c}\right)\left(b+\frac{1}{a c}\right)\left(c+\frac{1}{a b}\right)=\frac{(a b c+1)^{3}}{a^{2} b^{2} c^{2}}
$$
is a positive integer.
Let $a b c$ be the irreducible fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$. Then $M=\frac{(p+q)^{3}}{p^{2} q} \Rightarrow(p+q)^{3}=M p^{2} q$.
From ... | 3, 3.5, 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,232 |
Example 9 Find the rational solutions of the equation $x^{x+y}=(x+y)^{y}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 【Analysis】If $x=0$, then $x^{x+y}=0,(x+y)^{y}=0$. Therefore, $y=0$, but $0^{0}$ is undefined, so $x \neq 0$.
If $y=0$, then $x^{x}=1$, which implies $x=1$.
Clearly, $x=1, y=0$ is a rational solution.
Hereafter, we assume $x \neq 0, y \neq 0$.
Let $y=x z$. Then the original equation is equivalent to
$$
\begin{array}{l}
... | x=(1+z)^{2}, y=z(1+z)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,233 |
Question 1 As shown in Figure 3, in the square $ABCD$, $E$ and $F$ are points on the extensions of $CB$ and $CD$ respectively, and $EF=BE+DF$. The extensions of $EA$ and $FA$ intersect the line $BD$ at points $M$ and $N$ respectively. Prove:
$$
M N^{2}=B M^{2}+D N^{2} .
$$ | Given that $\angle E A F=135^{\circ}$, then
$$
\begin{array}{l}
\angle M A N \\
=135^{\circ} .
\end{array}
$$
In the square $A B C D$, $\angle B A D=90^{\circ}$, so $\angle N A B+\angle D A M=45^{\circ}$.
Since $\angle N+\angle N A B=\angle A B D=45^{\circ}$, we have $\angle N=\angle D A M$.
Similarly, $\angle N A B=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,234 |
Question 2 As shown in Figure 4, outside the square $ABCD$, construct $\angle EAF = 135^{\circ}$, which intersects the extensions of $CB$ and $CD$ at points $E$ and $F$, respectively. The extensions of $BA$ and $DA$ intersect $EF$ at points $M$ and $N$, respectively. Prove:
$$
M N^{2}=M E^{2}+N E^{2}
$$ | Proof: From the proof of the proposition, it is easy to see that $\angle M F A=\angle A F D$.
Since $B M / / C F$, then $\angle M A F=\angle A F D=\angle M F A$. Therefore, $M A=M F$. Similarly, $N A=N E$. In the right triangle $\triangle A M N$, we have $M N^{2}=M A^{2}+N A^{2}$.
Thus, $M N^{2}=M F^{2}+N E^{2}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,235 |
Question 3 As shown in Figure $5, E$ and $F$ are points on the extensions of sides $CB$ and $CD$ of square $ABCD$, respectively, and $EF = BE + DF$. The extensions of $EA$ and $FA$ intersect line $BD$ at points $M$ and $N$. Prove: $S_{\triangle MEF} = 2 S_{\triangle MMN}$. | Given that $\angle E A F=135^{\circ}$. Then $\angle E A N=45^{\circ}$. Also, $\angle E B N=\angle C B D=45^{\circ}$, so $\angle E A N=\angle E B N$. Therefore, points $A$, $B$, $N$, and $E$ are concyclic. Connecting $E N$, we get $\angle E N A=\angle A B E=90^{\circ}$. Thus, $\triangle A E N$ is an isosceles right tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,236 |
Question 4 As shown in Figure $6, E$ and $F$ are points on the extensions of sides $CB$ and $CD$ of square $ABCD$, respectively, and $\angle EAF = 135^{\circ}$. Prove:
$$
\frac{AE^2}{AF^2} = \frac{EF - FC}{EF - EC}.
$$ | Prove that as shown in Figure 6, rotating $\triangle A B E$ counterclockwise by $90^{\circ}$ around point $A$ yields $\triangle A D G$. Then, it is easy to see from the proof of the proposition that
$$
A E=A G, E F=F G \text {. }
$$
Thus, $E F-F C=C G$.
Similarly, rotating $\triangle A D F$ clockwise by $90^{\circ}$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,237 |
Example 3 As shown in Figure 3, given that quadrilateral $ABCD$ is a square, $\odot O$ passes through vertex $A$ of the square and the intersection point $P$ of the diagonals, and intersects $AB, AD$ at points $F, E$ respectively.
(1) Prove:
$$
DE = AF \text{; }
$$
(2) If the radius of $\odot O$ is $\frac{\sqrt{3}}{2}$... | Solution: Clearly, $EF$ is the diameter of $\odot O$, meaning point $O$ lies on side $EF$.
Connecting $EP$ and $FP$, it is easy to see that $\triangle EPF$ is an isosceles right triangle. Therefore, $EP = \sqrt{2} OE = \frac{\sqrt{6}}{2}$.
(1) Since $DP = AP$, $\angle EDP = 45^{\circ} = \angle FAP$, and $\angle DEP = ... | \sqrt{2} \text{ or } \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,239 |
Let $n$ be a positive integer. Prove that $\mathrm{C}_{2^{n}-1}^{0}$, $C_{2^{n-1}}^{1}, \cdots, C_{2^{n}-1}^{2 n-1}$ modulo $2^{n}$ is congruent to some permutation of $1, 3, \cdots, 2^{n}-1$. ${ }^{[1]}$
(49th IMO Shortlist) | Prove that obviously, when $n=1,2,3$, the proposition holds, and for any $n (n \geqslant 2), 0 \leqslant k \leqslant 2^{n-1}-1$, we have
$$
\begin{array}{l}
\mathrm{C}_{2^{n-1}}^{2 k}+\mathrm{C}_{2^{n-1}}^{2 k+1}=\mathrm{C}_{2^{n}}^{2 k+1} \\
=\frac{2^{n}}{2 k+1} \mathrm{C}_{2^{n-1}}^{2 k} \equiv 0\left(\bmod 2^{n}\rig... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,240 |
For an integer $m$, there exists a unique number $t(m)$ in $\{1,2,3\}$ such that $m+t(m)$ is a multiple of 3. The function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $f(-1)=0$, $f(0)=1$, $f(1)=-1$, and for all non-negative integers $m, n$ such that $2^{n}>m$, we have
$$
f\left(2^{n}+m\right)=f\left(2^{n}-t(m)\rig... | Proof: Let $n=\sum_{i=0}^{1} 2^{a_{i}}$, where $a_{0}>a_{1}>\cdots >a_{s} \geqslant 0$. Then
$$
\begin{array}{l}
f(n)=\sum_{j=0}^{l}\left[(-1)^{j} f\left(2^{a_{j}}-t\left(n_{j+1}\right)\right)\right]+ \\
\quad(-1)^{l+1} f\left(n_{l+1}\right) \\
=\sum_{j=0}^{t+1}(-1)^{j} f\left(n_{j}^{\prime}\right),
\end{array}
$$
whe... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,241 |
Example 1 Real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
x_{1}+x_{2}+\cdots+x_{n}=0, x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1 \text {. }
$$
Prove: $x_{1}, x_{2}, \cdots, x_{n}$ contain at least two numbers whose product is less than or equal to $-\frac{1}{n}$. | Proof Without loss of generality, assume that among $x_{1}, x_{2}, \cdots, x_{n}$, $x_{1}$ is the largest and $x_{n}$ is the smallest. Then
$$
\left(x_{1}-x_{i}\right)\left(x_{i}-x_{n}\right) \geqslant 0(i=1,2, \cdots, n) \text {. }
$$
Expanding the left side of (3) and summing over $i$ yields
$$
\text { i.e. } \begin... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,242 |
Example: 2100 real numbers that are not equal to each other are written on a
circle. Prove: there must exist four consecutive numbers where the sum of the two end numbers is greater than the sum of the two middle numbers.
Many students unhesitatingly use proof by contradiction. | Proof by contradiction. Assume the conclusion does not hold. Let 100 real numbers on a circle be $a_{1}, a_{2}, \cdots, a_{100}$. Then,
$$
a_{i}+a_{i+3} \leqslant a_{i+1}+a_{i+2},
$$
where $i=1,2, \cdots, 100$, with the convention $a_{i+100}=a_{i}$, i.e.,
$$
a_{i+3}-a_{i+2} \leqslant a_{i+1}-a_{i} \text {. }
$$
Summi... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,243 |
Example 3 In the sequence of integers $\left\{a_{n}\right\}$,
$$
a_{1}=1, a_{n+1}=a_{n}+S\left(a_{n}\right)(n=1,2, \cdots) \text {, }
$$
$S(a)$ is the sum of the digits of $a$. Is 12345 a term in this sequence? | Prove that 12345 is not a term of the sequence $a_{n+1}$.
Since $a_{n+1}=a_{n}+S\left(a_{n}\right) \equiv 2 a_{n}(\bmod 3)$, and $12345 \equiv 0(\bmod 3)$,
thus, $2 a_{n} \equiv 0(\bmod 3), a_{n} \equiv 0(\bmod 3)$.
Similarly, $a_{n-1} \equiv 0(\bmod 3), \cdots \cdots a_{1} \equiv 0(\bmod 3)$, which contradicts $a_{1}=... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,244 |
Example 4 Given positive integers $m, n$ and a prime $p$ satisfy: for any positive integer $k$, we have
$$
(p k-1, m)=(p k-1, n) .
$$
Prove: There exists an integer $t$, such that $m=p^{t} n$. | Proof Let $q$ be a prime factor of $m n$, $q \neq p$, and $q^{\alpha}\left\|m, q^{\beta}\right\| n, \alpha \neq \beta$. Without loss of generality, assume $\alpha>\beta$.
Since $q \neq p$, we have $(q, p)=1$, and the congruence equation
$$
p k-1 \equiv 0\left(\bmod q^{\alpha}\right)
$$
has a solution.
For this $k, q^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,245 |
Example 1 What is the maximum number of intersection points of $n$ lines on the same plane? | When $n=1$, there are no intersection points; when $n=2$, there is at most one intersection point; as $n$ increases, to ensure the maximum number of intersection points, each new line can be made to intersect with all the lines already on the plane. In this case, $n$ lines can have at most $0+1+\cdots+(n-1)=\frac{n(n-1... | \frac{n(n-1)}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,246 |
Example 2 Divide a circle into $n(n \geqslant 2)$ sectors, each sector is colored with one of $r$ different colors, and it is required that adjacent sectors are colored differently. Question: How many coloring methods are there? | Solution $\mathbf{1}^{[1]}$ Let the sectors of the circle be $S_{1}, S_{2}, \cdots, S_{n}$, and there are $a_{n}$ ways to color them. Then
$$
a_{1}=0, a_{2}=r(r-1) \text {. }
$$
When $n \geqslant 2$, $S_{1}$ has $r$ ways to be colored, $S_{2}$ has $r-1$ ways to be colored, $\cdots \cdots S_{n}$ has $r-1$ ways to be co... | a_{n}=(r-1)(-1)^{n}+(r-1)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,247 |
Example 3 A password lock's password setting involves assigning one of the two numbers, 0 or 1, to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex either red or blue, such that for any two adjacent vertices, at least one of the number or color is the same. How many differ... | Let $S_{n}$ be the desired result, and let $a_{n}, b_{n}, c_{n}$ represent the number of ways to set the password when $A_{n}=A_{1}$, $A_{n} \neq A_{1}$, and $A_{n} \# A_{1}$, respectively.
Then $S_{n}=a_{n}+b_{n}$.
It is easy to see that $\left(a_{1}, b_{1}, c_{1}\right)=(4,0,0)$,
$\left(a_{2}, b_{2}, c_{2}\right)=(4,... | S_{n}=3^{n}+(-1)^{n}+2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,248 |
Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the valu... | Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\begin{array}{l}
g(n)=\frac{1}{2} f(n) . \\
\text { Hence } a_{n-1}=1, a_{n}=2,3,4 ; \\
a_{n-1}=2, a_{n}=3,4 ; \\
a_{n-1}=3, a_{n}=4 .
\end{array}
$$
Let $m(i)$ represent the number of $(n-1)$-dig... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,249 |
Example 4 In $\triangle A B C$, $\angle A=75^{\circ}, \angle B=35^{\circ}$, and $D$ is a point on side $B C$ such that $B D=2 C D$. Prove:
$$
A D^{2}=(A C+B D)(A C-C D) \text {. }{ }^{[2]}
$$
$(2008$, I Love Mathematics Junior High School Camp Mathematics Competition) | Prove that, as shown in Figure 4, extend $B C$ to point $E$, such that
$$
C E=A C \text {. }
$$
From the given, $\angle C E$ $=70^{\circ}$, then
$$
\begin{array}{l}
\angle E=35^{\circ} \\
=\angle B,
\end{array}
$$
which means $\triangle A B E$ is an isosceles triangle.
Draw $A M \perp B E$ at point $M$. Then $M$ is t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,250 |
2. At a certain charity fundraising dinner, each person ate half a plate of rice, one-third of a plate of vegetables, and one-quarter of a plate of meat. The dinner provided a total of 65 plates of food. How many people attended this fundraising dinner? | 2. 60 .
Each person ate $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ of food, and $65 \div \frac{13}{12}=60$.
Therefore, 60 people attended this charity dinner. | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,252 |
3. The number of triples of positive integers $(x, y, z)$ that satisfy $x y z=3^{2010}$ and $x \leqslant y \leqslant z<x+y$ is $\qquad$.
| 3. 336 .
Let $x=3^{a}, y=3^{b}, z=3^{c}$. Then
$$
0 \leqslant a \leqslant b \leqslant c, a+b+c=2010 \text {. }
$$
If $c \geqslant b+1$, then
$$
x+y=3^{a}+3^{b}<3^{b+1} \leqslant 3^{c}=z \text {, }
$$
which contradicts $z<x+y$. Hence, $c=b$.
Thus, $a+2 b=2010$.
Therefore, $670 \leqslant b \leqslant 1005$.
Hence, the ... | 336 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,253 |
4. As shown in Figure 1, given that $E$ is a point on side $BC$ of rectangle $ABCD$, and folding along $AE$ makes vertex $B$ coincide with point $F$ on side $CD$. If $AD=16, BE=10$, then the length of $AE$ is $\qquad$ | 4. $10 \sqrt{5}$.
From the problem, we have
$$
E F=B E=10, E C=A D-B E=6 \text {. }
$$
By the Pythagorean theorem, $C F=\sqrt{E F^{2}-E C^{2}}=8$.
Clearly, Rt $\triangle A F D \backsim \mathrm{Rt} \triangle F E C$.
Thus, $\frac{D F}{D A}=\frac{E C}{C F}=\frac{3}{4}$. Therefore, $D F=12$.
At this point, $A B=C D=C F+D... | 10 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,254 |
5. The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ . | 5.1458.
Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore,... | 1458 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,255 |
6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72 \text {, }
$$
then the last digit of $f(2010)$ is $\qquad$ | 6. 2 .
Since $f(1)=f(2)=f(3)=0$, the quartic polynomial can be set as
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(a x+b) . \\
\text { By } f(4)=6, f(5)=72, \text { we get } \\
\left\{\begin{array} { l }
{ 6 ( 4 a + b ) = 6 , } \\
{ 2 4 ( 5 a + b ) = 7 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-7 .
\e... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,256 |
7. As shown in Figure 2, a circle and two semicircles are pairwise tangent, and their radii are all 1. Then the area of square $A B C D$ is $\qquad$ | 7. $3+\sqrt{2}+\sqrt{3}+\sqrt{6}$.
As shown in Figure 6, let $P$, $Q$, and $R$ be the centers of the circle and the two semicircles, respectively. This figure is symmetric about the diagonal $AC$ of the square, and $AP=\sqrt{2}$.
$$
\begin{array}{l}
\text { It is easy to see that } P S=\sqrt{3}, \\
S C=S R=1, \\
A C=1... | 3+\sqrt{2}+\sqrt{3}+\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,257 |
8. Let $p, q$ be prime numbers, and satisfy $p^{3}+q^{3}+1=p^{2} q^{2}$. Then the maximum value of $p+q$ is | 8.5.
Assume $q \leqslant p$. Notice that
$$
\begin{array}{l}
p^{3}+q^{3}+1=p^{2} q^{2} \\
\Rightarrow q^{3}+1=p^{2} q^{2}-p^{3} \\
\Rightarrow(q+1)\left(q^{2}-q+1\right)=p^{2}\left(q^{2}-p\right) .
\end{array}
$$
Therefore, $p^{2} \mid(q+1)\left(q^{2}-q+1\right)$.
Since $q \leqslant p$, we have $0<q^{2}-q+1<p^{2}$.
T... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,258 |
9. Given $n$ positive integers (not necessarily distinct), their sum is 100, and the sum of any seven of them is less than 15. Then the minimum value of $n$ is $\qquad$ . | 9. 50.
Let these $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{49} \leqslant 14 \times 7=98 \text {. }
$$
Thus $n \geqslant 50$, and when these 50 numbers are all 2, the condition is satisfied. | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,259 |
10. Given point $P$ inside $\triangle A B C$, satisfying $\angle A B P$ $=20^{\circ}, \angle P B C=10^{\circ}, \angle A C P=20^{\circ}$ and $\angle P C B$ $=30^{\circ}$. Then $\angle C A P=$ | 10.20.
As shown in Figure 7, construct a regular $\triangle QBC$ on side $BC$ on the same side as point $A$.
Notice that $BA$ is the angle bisector of the regular $\triangle QBC$, so, $\angle AQC = \angle ACQ = 10^{\circ}$.
Also, $CP$ is the angle bisector of the regular $\triangle QBC$, hence
$$
\angle PQC = \angle... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,260 |
Example 5 As shown in Figure 5, given that $\odot O$ is inside rectangle $ABCD$, and tangents to $\odot O$ are drawn through vertices $A, B, C,$ and $D$, touching points are $A_{1}, B_{1}, C_{1}, D_{1}$ respectively. If $AA_{1} = 3, BB_{1} = 4, CC_{1} = 5$, find the length of $DD_{1}$. | Connect $A O$,
$B O$, $C O$, $D O$, $A_{1} O$, $B_{1} O$, $C_{1} O$, $D_{1} O$. Then
$O A_{1} \perp A A_{1}$, $O B_{1} \perp B B_{1}$,
$O C_{1} \perp C C_{1}$, $O D_{1} \perp D D_{1}$.
Let the radius of $\odot O$ be $r$. Then by the Pythagorean theorem, we have
$A O^{2}=A A_{1}^{2}+r^{2}$, $B O^{2}=B B_{1}^{2}+r^{2}$,
... | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,261 |
11. A farmer has 100 pigs and 100 chickens. He has four adjacent square yards, forming a $2 \times 2$ grid. The farmer wants to distribute the livestock among the yards according to the following requirements: the first row has 120 heads, the second row has 300 feet; the first column has 100 heads, the second column ha... | 11.341.
As shown in Figure 8, label the four courtyards with letters $A_{i} (i=1,2,3,4)$. Let $A_{i}$ courtyard have $x_{i}$ pigs and $y_{i}$ chickens.
From the problem, we get
every 8
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=100, \\
y_{1}+y_{2}+y_{3}+y_{4}=100, \\
x_{1}+y_{1}+x_{2}+y_{2}=120, \\
4\left(x_{3... | 341 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,262 |
12. There are five cages arranged in a row, labeled from left to right as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline \begin{tabular}{c}
Red \\
Wolf
\end{tabular} & \begin{tabular}{c}
Silver \\
Lion
\end{tabular} & \begin{tabular}{c}
Brown \\
Fox
\end{tabular} & \begin{tabular}{c}
White \\
Cow
\en... | 12. White.
Consider the labels on the middle three cages. From (3), we know that the lion, fox, and ox are not in the middle cages. Since the horse is also not in the middle cages, it can only be the wolf inside.
Since the first label is “Red · Wolf,” the wolf is not red. From (3), we also know it is not silver, brow... | White | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 723,263 |
1. Line segment $A B$ divides a square into two polygons (points $A$ and $B$ are on the sides of the square), each polygon has an inscribed circle, one of which has a radius of 6, while the other has a radius greater than 6. What is the difference between the side length of the square and twice the length of line segme... | If line segment $A B$ divides the square into two triangles, then $A B$ can only be the diagonal of the square. But in this case, the radii of the two incircles are equal, which contradicts the condition.
If one of the polygons is a quadrilateral, then the sum of the lengths of $A B$ and its opposite side is greater t... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,264 |
2. In a candy store, candies are sold in three types of packages: small packs contain 6 candies, medium packs contain 9 candies, and large packs contain 20 candies. If you can only buy whole packs of candies, what is the maximum number of candies you cannot purchase? | 2. Write non-negative integers in the following six-row number table:
\begin{tabular}{ccccccccc}
$\mathbf{0}$ & 6 & 12 & 18 & 24 & 30 & 36 & 42 & 48 \\
1 & 7 & 13 & 19 & 25 & 31 & 37 & 43 & 49 \\
2 & 8 & 14 & 20 & 26 & 32 & 38 & 44 & 50 \\
3 & 9 & 15 & 21 & 27 & 33 & 39 & 45 & 51 \\
4 & 10 & 16 & 22 & 28 & 34 & 40 & 46... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,265 |
3. There is a sequence $a_{1}, a_{2}, \cdots, a_{2010}$. For $1 \leqslant n$ $\leqslant 2010$ ( $n$ being a positive integer), let
$$
S_{n}=a_{1}+a_{2}+\cdots+a_{n} \text {. }
$$
If $a_{1}=2010$ and for all $n$, $S_{n}=$ $n^{2} a_{n}$, then what is $a_{2010}$? | 3. From the given information, we have
$$
a_{n}=S_{n}-S_{_{n-1}}=n^{2} a_{n}-(n-1)^{2} a_{n-1} \text {, }
$$
which means $\left(n^{2}-1\right) a_{n}=(n-1)^{2} a_{n-1}$.
At this point, $a_{n}=\frac{n-1}{n+1} a_{n-1}=\cdots=\frac{2}{(n+1) n} a_{1}$.
Therefore, $a_{2010}=\frac{2 \times 2010}{2011 \times 2010}=\frac{2}{20... | \frac{2}{2011} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,266 |
1. (40 points) Find the real solutions for the system of equations in $w, x, y, z$:
$$
\left\{\begin{array}{l}
w+8 x+3 y+5 z=20 \\
4 w+7 x+2 y+3 z=-20 \\
6 w+3 x+8 y+7 z=20 \\
7 w+2 x+7 y+3 z=-20 .
\end{array}\right.
$$ | (1) + (4) gives $8(w+z)+10(x+y)=0$.
(2) + (3) gives $10(w+z)+10(x+y)=0$.
Thus, $w+z=0$ and $x+y=0$.
Substituting $y=-x$ and $z=-w$ into equations (1) and (2) yields $5 x-4 w=20, w+5 x=-20$.
Eliminating $x$ gives $-5 w=40$, i.e., $w=-8$.
From this, we get $x=-\frac{12}{5}, y=\frac{12}{5}, z=8$. | w=-8, x=-\frac{12}{5}, y=\frac{12}{5}, z=8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,267 |
2. (40 points) In the convex quadrilateral $ABCD$, $AB$ is the shortest side, and $CD$ is the longest side. Prove:
$$
\angle A > \angle C \text{, and } \angle B > \angle D \text{. }
$$ | 2. As shown in Figure 11, connect $A C$ and $B D$. In $\triangle A B C$ and $\triangle A C D$, $B C > B A$, $C D > A D$, thus,
$$
\begin{array}{l}
\angle B A C > \angle B C A, \\
\angle D A C > \angle D C A . \\
\text { Therefore, } \angle A = \angle B A C + \angle D A C \\
> \angle B C A + \angle D C A \doteq \angle C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,268 |
3. (40 points) Let integers $m, n$ satisfy $m \geqslant n$ and $m^{3}+n^{3}+1=4 m n$. Find the maximum value of $m-n$.
保留了原文的换行和格式，如上所示。 | 3. Let $s=m+n, p=mn$. Then $m^{3}+n^{3}=(m+n)^{3}-3m^{2}n-3mn^{2}$ $=s^{3}-3ps$.
From $s^{3}-3ps+1=4p$, we get $p=\frac{s^{3}+1}{3s+4}$.
Since $m, n$ are integers, 27p must also be an integer, and
$27p=9s^{2}-12s+16-\frac{37}{3s+4}$.
Therefore, $3s+4$ must divide 37.
Thus, $3s+4 = \pm 1$ or $\pm 37$.
If $3s+4=-1$ or -3... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,269 |
4. (40 points) 64 points on a plane form an $8 \times 8$ grid. The distance between any two adjacent points in the same row or column is 1. How many rectangles with an area of 12 can be formed using four of these 64 points as vertices? | 4. Connecting points in the same row and column can form a grid.
First, consider rectangles with sides parallel to the grid lines. For a rectangle of shape $2 \times 6$, there are two orientation choices (horizontal and vertical). One pair of opposite sides has $8-6=2$ choices, and the other pair has $8-2=6$ choices. ... | 84 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,270 |
5. (40 points) Find the largest positive integer $n$, such that there exists a unique positive integer $k$ satisfying $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$. | 5. The inequality can be written as $1+\frac{7}{8}>1+\frac{k}{n}>1+\frac{6}{7}$. Or $\frac{98}{112}>\frac{k}{n}>\frac{96}{112}$.
If $n=112$, then the unique value of $k$ is 97.
Assuming $n>112$, then $\frac{98 n}{112 n}>\frac{112 k}{112 n}>\frac{96 n}{112 n}$.
Between $96 n$ and $98 n$, there are at least two numbers t... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,271 |
Example 6 Let $H$ be the orthocenter of acute $\triangle A B C$, and $P$ be any point in the plane of $\triangle A B C$. Construct $H M \perp P B$ at point $M$ and intersect line $A C$ at point $J$, construct $H N \perp P C$ at point $N$ and intersect line $A B$ at point $I$. Prove: $P H \perp I J$. | Proof: Let $BE$ and $CF$ be two altitudes of $\triangle ABC$. Then $H$ is their intersection.
By the properties of the orthocenter, we know that $BH \cdot HE = CH \cdot HF$.
Since $MJ \perp BM, BE \perp EJ$, we know that $B, J, E, M$ are concyclic, i.e., $MH \cdot HJ = BH \cdot HE$.
Since $NI \perp CN, CF \perp FI$, w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,272 |
6. (40 points) In each small square of a $9 \times 9$ grid, fill in a number such that each row and each column contains at most four different numbers. What is the maximum number of different numbers that can be in this grid?
| 6. Suppose this square table contains 29 different numbers. According to the pigeonhole principle, there must be a row with four different numbers, let's assume it is the first row. The remaining 25 numbers are in rows $2 \sim 9$. Similarly, according to the pigeonhole principle, there must be a row with four, let's as... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,273 |
7. (40 points) In the convex quadrilateral $ABCD$, $\angle ABD$ $=16^{\circ}, \angle DBC=48^{\circ}, \angle BCA=58^{\circ}, \angle ACD=$ $30^{\circ}$. Find the measure of $\angle ADB$. | 7. It is known that $\angle B D C=44^{\circ}, \angle B A C=58^{\circ}$.
Therefore, $B A=B C$.
As shown in Figure 15, let $E$ be a point on side
$C D$ such that
$\angle E B C=4^{\circ}$.
Then $B E=B C=B A$.
Also, $\angle A B E=60^{\circ}$, so
$\triangle A B E$ is an equilateral
triangle.
Since $\angle D B E=44^{\circ}$... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,274 |
8. (40 points) Find all ordered triples $(x, y, z)$, such that $x, y, z \in Q_{+}$, and $x+\frac{1}{y}$, $y+\frac{1}{z}$, $z+\frac{1}{x}$ are all integers. | 8. Notice that
$$
\begin{array}{l}
\left(x+\frac{1}{y}\right)\left(y+\frac{1}{z}\right)\left(z+\frac{1}{x}\right) \\
=x y z+\frac{1}{x y z}+x+\frac{1}{y}+y+\frac{1}{z}+z+\frac{1}{x} .
\end{array}
$$
Thus, $h=x y z+\frac{1}{x y z}$ is an integer.
Then $x y z=\frac{h \pm \sqrt{h^{2}-4}}{2}$.
Since the difference between... | (1,1,1), \left(\frac{1}{2}, 2,1\right), \left(\frac{3}{2}, 2, \frac{1}{3}\right), \left(3, \frac{1}{2}, \frac{2}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,275 |
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