problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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33. The function $f: \mathbf{N} \rightarrow \mathbf{N} (\mathbf{N}$ being the set of non-negative integers), satisfies:
(1) For any non-negative integer $n$,
$$
f(n+1)>f(n) \text{; }
$$
(2) For any $m, n \in \mathbf{N}$,
$$
f(n+f(m))=f(n)+m+1 \text{. }
$$
Find the value of $f(2001)$.
(2001, British Mathematical Olym... | Solution: Let $f(0)=k$, where $k$ is a non-negative integer. From (2), we have
$$
f(n+k)=f(n)+1 \text {. }
$$
If $k=0$, then $f(n)=f(n)+1$, which is a contradiction. Therefore, $k \neq 0$.
According to (1), we have
$$
f(n+k-1) \cdot 1 \text{, if } k>1, \text{ then } n+k-1 \geqslant n+1. \text{ Thus, by (1), we get }
$... | 2002 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,000 |
34. Prove: The sequence $y_{0}=1, y_{n+1}=\frac{1}{2}\left(3 y_{n}+\right.$ $\left.\sqrt{5 y_{n}^{2}-4}\right) \quad(n \geqslant 0)$ is a sequence of integers.
(2001, British Mathematical Olympiad (Round 2)) | Prove: From $y_{n+1}=\frac{1}{2}\left(3 y_{n}+\sqrt{5 y_{n}^{2}-4}\right)$, we get
$$
2 y_{n+1}-3 y_{n}=\sqrt{5 y_{n}^{2}-4} \text {. }
$$
Rearranging gives
$$
y_{n+1}^{2}-3 y_{n+1} y_{n}+y_{n}^{2}+1=0 \text {. }
$$
Replacing $n$ with $n-1$, we get
$$
y_{n}^{2}-3 y_{n} y_{n-1}+y_{n-1}^{2}+1=0 \text {. }
$$
(1) - (2) ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,001 |
35. Find the function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for $\forall x \in \mathbf{R}, \forall y \in$ $\{f(x) \mid x \in \mathbf{R}\}$. We have $f(x-y)=f(x)+x y+f(y)$, where $\mathbf{R}$ is the set of real numbers.
(15th Korean Mathematical Olympiad) | Solution: First, assume $f(x)=c$ is a constant, then
$$
c=f(x-c)=f(x)+c x+f(c)=c x+2 c, x \in \mathbf{R} \text {. }
$$
Therefore, $c=0$, hence $f(x)=0$.
Second, assume $f(x)$ is not a constant. Then there exist $a, b \in \mathbf{R}$, such that $f(a)=b \neq 0$. Thus for all $x \in \mathbf{R}$, we have
$$
\begin{array}{... | f(x)=0 \text{ or } f(x)=-\frac{x^{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,002 |
Example 2 Let $x \in \mathbf{R}, n \in \mathbf{N}$. Prove that:
$$
\sum_{i=1}^{n} \frac{[i x]}{i} \leqslant[n x] \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding $x$.
(10th United States of America Mathematical Olympiad) | Explanation: From the left side of the equation to be proven, it seems we can use
$$
\begin{aligned}
\sum_{i=1}^{n} \frac{[i x]}{i}= & \frac{1}{n} \sum_{i=1}^{n}[i x]+ \\
& \sum_{k=1}^{n-1}\left(\frac{1}{k}-\frac{1}{k+1}\right) \sum_{i=1}^{k}[i x] .
\end{aligned}
$$
However, it is difficult to proceed further, the key... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,004 |
37. For all integers $k(k \geqslant 4)$, prove: if the polynomial $F(x)$ with integer coefficients satisfies the condition $0 \leqslant F(c) \leqslant k, c=0,1, \cdots, k+1$, then, $F(0)=F(1)=\cdots=F(k+1)$.
(4th China Hong Kong Mathematical Olympiad) | Solution: It suffices to prove that $F(x)-F(0)$ has roots $1,2, \cdots, k+1$.
Since $F(k+1)-F(0)$ is a multiple of $k+1$, and its absolute value is less than $k$, we have $F(k+1)=F(0)$. Therefore, there exists an integer-coefficient polynomial $G(x)$ such that
$$
\begin{array}{l}
F(x)-F(0)=x(x-k-1) G(x) . \\
\text { Th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,005 |
38. Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers greater than or equal to 1, $n \geqslant 1, A=1+a_{1}+a_{2}+\cdots+a_{n}$. Define $x_{0}=1, x_{k}=$ $\frac{1}{1+a_{k} x_{k-1}}, 1 \leqslant k \leqslant n$. Prove:
$$
x_{1}+x_{2}+\cdots+x_{n}>\frac{n^{2} A}{n^{2}+A^{2}} \text {. }
$$
(2002, Ukrainian Mathematical Oly... | Proof: Let $y_{k}=\frac{1}{x_{k}}$, thus,
$$
\frac{1}{y_{k}}=\frac{1}{1+\frac{a_{k}}{y_{k-1}}} \Leftrightarrow y_{k}=1+\frac{a_{k}}{y_{k-1}} \text {. }
$$
Given $y_{k-1} \geqslant 1, a_{k} \geqslant 1$, we have
$$
\left(\frac{1}{y_{k-1}}-1\right)\left(a_{k}-1\right) \leqslant 0 \Leftrightarrow 1+\frac{a_{k}}{y_{k-1}} ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,006 |
39. Find the function $f:(0,+\infty) \rightarrow(0,+\infty)$, such that $f(x)^{\Re(y)}=f\left(x^{y}\right)$, for $\forall x 、 y>0$. (2002, Croatian County Mathematical Competition) | Solution: Clearly, $f(x)=1$ and $f(x)=x$ satisfy the conditions.
Assume there exists $a>0$, such that $f(a) \neq 1$, then
$$
\begin{array}{l}
f(a)^{\int(x y)}=f\left(a^{x y}\right)=f\left(\left(a^{x}\right)^{y}\right) \\
=f\left(a^{x}\right)^{\mu(y)}=f(a)^{\pi(x) f(y)} \\
\Rightarrow f(x y)=f(x) f(y), \forall x, y>0 .
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,007 |
40. Find all positive integers $n$ and non-negative integers $x_{1}, x_{2}, \cdots, x_{n}$, such that $\sum_{i=1}^{n} x_{i}^{2}=1+\frac{4}{4 n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}$.
(2002, Taiwan Mathematical Olympiad) | Let $X=x_{1}+x_{2}+\cdots+x_{n}$. If $x_{i} \leqslant 1, i=1,2,3, \cdots, n$, denote
$$
|I\{i \mid x_{i}=1\}|=a,|\{i \mid x_{1}=0\}|=n-a .
$$
Then the equation can be rewritten as
$$
a=1+\frac{4 a^{2}}{4 n+1} .
$$
It is easy to see that $a \neq 1$, so
$4 n+1=4(a+1)+\frac{4}{a-1}$.
Solving this, we get $a=5, n=6$.
Ass... | (1) n=6, (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})=(0,1,1,1,1,1), (1,0,1,1,1,1), (1,1,0,1,1,1), (1,1,1,0,1,1), (1,1,1,1,0,1), ( | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,008 |
Example 3 Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} \cdot x_{k} x_{j}=1$.
Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.
(2001, National High School Mathematics Competition) | Explanation: It is easy to see that the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1 (this is achieved when one of the $x_{i}$ is 1 and the rest are all zero). To find the maximum value, note that
$$
\begin{array}{l}
1=\sum_{i=1}^{n} i\left(\frac{x_{i}}{\sqrt{i}}\right)^{2}+2 \sum_{1 \leq k<j<n} k\left(\frac{x_{k}}{\s... | \sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,009 |
Example 4 Real numbers $x_{1}, x_{2}, \cdots, x_{2} 0001$ satisfy
$$
\begin{array}{c}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001, \\
\text { let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2, \cdots,
\end{array}
$$
2001 . Find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\r... | Explanation: Since the relationship between $x_{k}$ and $x_{k+1}$ is unknown, the difference $x_{k}-x_{k+1}$ can be treated as a whole, and the condition $\sum_{k=1}^{2000} |x_{k}-x_{k+1}|=2001$ can be viewed as a constraint on $x_{k}-x_{k+1}$. By making the substitution $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,010 |
Example 5 Given that $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots$, $b_{n}$ are real numbers. Prove: The necessary and sufficient condition for the inequality $\sum_{i=1}^{n} a_{i} x_{i} \leqslant \sum_{i=1}^{n} b_{i} x_{i}$ to hold for any real numbers satisfying $x_{1} \leqslant x_{2} \leqslant \cdots \leq... | Let $S_{k}=\sum_{i=1}^{k} a_{i}, T_{k}=\sum_{i=1}^{k} b_{i}, S_{0}=T_{0}=0$, then the condition $\sum_{i=1}^{n} a_{i} x_{i} \leqslant \sum_{i=1}^{n} b_{i} x_{i}$ can be transformed into
$$
\begin{array}{l}
S_{n} x_{n}+\sum_{k=1}^{n-1} S_{k}\left(x_{k}-x_{k+1}\right) \\
\leqslant T_{n} x_{n}+\sum_{k=1}^{n-1} T_{k}\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,011 |
Example 6 Given $c \in\left(\frac{1}{2}, 1\right)$. Find the smallest constant $M$ such that for any integer $n \geqslant 2$ and real numbers $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, if
$$
\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k} \text {, }
$$
then $\sum_{k=1}^{n} a_{k} \leqslant M ... | Explanation: First, according to formula (1), use the special value method to find a lower bound for $M$. The simplest method is to set all $a_{k}$ equal. However, since $c$ is given in advance, setting all $a_{k}$ equal may not satisfy the condition. Therefore, take a step back and set $a_{1}=\cdots=a_{m}$, and $a_{m+... | \frac{1}{1-c} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,012 |
Proposition 1 The necessary and sufficient condition for two vertices of a triangle and the two of its centers (incenter, circumcenter, orthocenter) to be concyclic is that the internal angle at the other vertex is $60^{\circ}$. | Proof: When two of the three centers coincide, or the triangle is a right triangle, the conclusion is obviously true. Below, we discuss the case where the three centers are pairwise distinct and the triangle is not a right triangle, as shown in Figure 1.
Sufficiency. Suppose \(\angle A = 60^\circ\), and \(I\), \(O\), ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,013 |
Example 5 For a quadrilateral, there can be 2 ways to dissect it into triangles, $a_{4}=2$ (Figure 15); for a pentagon, there can be 5 ways to dissect it into triangles, $a_{5}=$ 5 (Figure 16). How many ways are there to dissect a hexagon into triangles? | Solution: Consider the triangles containing the edge $A_{1} A_{2}$, using $\triangle A_{1} A_{2} A_{k}(k=3,4,5,6)$ as a standard, the figure can be divided into 4 categories.
(1) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{3}$, we can take out $\triangle A_{1} A_{2} A_{3}$, leaving the pentagon ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,015 |
Example 2 As shown in Figure 3, let $I$ be the incenter of $\triangle A B C$, and draw a line perpendicular to $A I$ through $I$ that intersects sides $A B$ and $A C$ at points $P$ and $Q$, respectively. Prove that the circle $\odot L$ that is tangent to $A B$ and $A C$ at $P$ and $Q$, respectively, must also be tangen... | Prove: Extend $A I$ to intersect $\odot O$ at point $M$. Let the radius of $\odot O$ be $R$. Then the power of point $L$ with respect to $\odot O$ is $R^{2}-L O^{2}=$ $L A \cdot L M$. Therefore,
$$
\begin{array}{l}
L O^{2}=R^{2}-L A \cdot L M=R^{2}-L A(I M-L I) \\
=R^{2}-L A \cdot I M+L A \cdot L=R^{2}-L A \cdot I M+L ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,016 |
3. In the Cartesian coordinate plane, a point whose both horizontal and vertical coordinates are integers is called an integer point: Let $n$ be a positive integer, then the number of integer points inside (including the boundary) the square $G_{n}$ as shown in Figure 1 is ( ).
(A) $4 n+1$
(B) $8 n-3$
(C) $n^{2}+2 n+2$... | 3.D.
Let the number of integer points inside the square $G_{n}$ be $a_{n}$. Clearly, $a_{1}=5$. When $n$ increases to $n+1$, the integer points added in the first quadrant (including the positive half-axis of the $x$-axis) are $(1, n),(2, n-1),\left(3, n-n_{n}^{\circ}, \cdots\right.$, $(n, 1),(n+1,0)$, a total of $n+1... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,017 |
4. Given a tetrahedron $ABCD$ with the lengths of its 6 edges being $4$, $7$, $20$, $22$, $28$, and $x$ (unit: $\mathrm{cm}$). Then the integer part of the minimum value that $x$ can take is ( ).
(A) 10
(B) 9
(C) 8
(D) 7 | 4.C.
When $x \leqslant 4$, the two side triangles with a side length of $28 \mathrm{~cm}$ must simultaneously satisfy the condition that the sum of any two sides is greater than the third side. Only $22+7>28$, $22+20>28$. At this time, the edge with a length of $22 \mathrm{~cm}$ cannot appear on both sides that have a... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,018 |
5. If the positive number $x$ satisfies $x^{5}-x^{3}+x=2$, then the range of $x$ is ( ).
(A) $\sqrt[6]{3}\sqrt[3]{2}$ | 5.A.
Obviously, $x \neq 0, x^{6}+1=\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)=$ $\frac{x^{2}+1}{x}\left(x^{5}-x^{3}+x\right)=2\left(x+\frac{1}{x}\right)$.
Since $x>0$ and $x \neq 1$, we have
$$
x^{6}+1=2\left(x+\frac{1}{x}\right)>4, x>\sqrt[6]{3} \text {. }
$$
From the given condition, we get $\frac{2}{x^{3}}+1=x... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,019 |
6. There are 9 points on a circle, and 3 triangles are formed using these 9 points as vertices. When these 3 triangles have no common vertices and their sides do not intersect, we call it a configuration. The number of configurations that satisfy this condition is ( ).
(A) 3
(B) 6
(C) 9
(D) 12 | 6.D.
Let these 9 points be denoted as $A_{1}, A_{2}, \cdots, A_{9}$. We consider two cases:
(1) When the 9 points are divided into 3 groups, with each adjacent 3 points forming a triangle (as shown in Figure 5), then these 3 triangles have no common points and their sides do not intersect. Since there are only 3 ways ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,020 |
1. Let the left and right vertices of the hyperbola $x^{2}-y^{2}=6$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be a point on the right branch of the hyperbola such that $\angle P A_{2} x=3 \angle P A_{1} x+10^{\circ}$. Then the degree measure of $\angle P A_{1} x$ is $\qquad$. | $=、 1.20^{\circ}$.
Let $P\left(x_{0}, y_{0}\right)$, then $x_{0}^{2}-y_{0}^{2}=6$. Since $\tan \angle P A_{1} x=\frac{y_{0}}{x_{0}+\sqrt{6}}, \tan \angle P A_{2} x=\frac{y_{0}}{x_{0}-\sqrt{6}}$, therefore, $\tan \angle P A_{1} x \cdot \tan \angle P A_{2} x=\frac{y_{0}^{2}}{x_{0}^{2}-6}=1$. Hence $\angle P A_{2} x=90^{\... | 20^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,021 |
3. Given planar vectors $\boldsymbol{a}=(\sqrt{3},-1), \boldsymbol{b}=$ $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. If there exist non-zero real numbers $k$ and angle $\alpha, \alpha \in$ $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $\boldsymbol{c}=\boldsymbol{a}+\left(\tan ^{2} \alpha-3\right) \boldsymb... | 3. $\frac{1}{4}\left(\tan ^{3} \alpha-3 \tan \alpha\right), \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
From $\boldsymbol{a} \cdot \boldsymbol{b}=(\sqrt{3},-1) \cdot\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)=0$, we get $\boldsymbol{a} \perp \boldsymbol{b}$.
Also, $\boldsymbol{c} \perp \boldsymbol{d}$, the... | \frac{1}{4}\left(\tan ^{3} \alpha-3 \tan \alpha\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,023 |
4. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$. Then the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{200 B}+1}$ is $\qquad$ | 4. 1 .
From the problem, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and $x_{3}>1$. From $\frac{1}{x_{k+1}}=\frac{1}{x_{k}\left(x_{k}+1\right)}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, we get $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$.
Then $S=\sum_{k=1}^{2008} \frac{1}{x_{k}+1}=\sum_{k=1}^{20... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,024 |
5. The equation $x^{3}+p x+q=0(p, q \in$ C) has three complex roots, and the points corresponding to these roots in the complex plane are the three vertices of an equilateral triangle with side length $\sqrt{3}$. If the principal value of the argument of the complex number $q$ is $\frac{2 \pi}{3}$, then $p+q=$ $\qquad$... | 5. $-\frac{1}{2}+\frac{\sqrt{3}}{2}$ i.
The circumradius of an equilateral triangle with side length $\sqrt{3}$ is 1, so the complex numbers corresponding to the three vertices satisfy the equation
$$
(x+a)^{3}=b \quad(a, b \in \mathbf{C},|b|=1),
$$
which is $x^{3}+3 a x^{2}+3 a^{2} x+a^{3}-b=0$.
Comparing with the g... | -\frac{1}{2}+\frac{\sqrt{3}}{2} \text{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,025 |
6. Students $M_{1}, M_{2}, M_{3}, M_{4}$ go to buy 10 different books numbered $1, 2, 3, \cdots, 10$. To save money and facilitate sharing, they agree that each person will only buy 5 of these books, and no two students can buy all 10 books, while any three students must buy all 10 books. When $M_{1}$ buys books number... | 6. $4,6,7,8,10$.
Let the set of book numbers bought by $M_{i}$ be $A_{i}, i=1,2,3,4$. Let the universal set $U=\{1,2,3, \cdots, 10\}$.
Since $A_{1}=\{1,2,3,4,5\}, A_{2}=\{5,6,7,8,9\}$,
$$
\begin{array}{l}
A_{3}=\{1,2,3,9,10\}, \overline{A_{1} \cup A_{2}}=\{10\}, \\
\overline{A_{2} \cup A_{3}}=\{4\}, \overline{A_{3} \c... | 4,6,7,8,10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,026 |
Example 3 As shown in Figure 4, let $\odot O_{1}$
and $\odot O_{2}$ be disjoint,
draw one of their
external common tangents touching
$\odot O_{1}$ at point $A$,
and touching $\odot O_{2}$ at point $C$, then draw one of their internal common tangents touching
$\odot O_{1}$ at point $B$, and touching $\odot O_{2}$ at poi... | Proof: Let the extension of line $AB$ and $CD$ intersect at point $K$, the intersection of the extension of line $AC$ and $DB$ be point $E$, and connect $O_{1}E, O_{2}E$. Then $AB \perp O_{1}E, CD \perp O_{2}E$.
Since $O_{1}E$ bisects $\angle AEB$ and $O_{2}E$ bisects $\angle CED$, we know that $O_{1}E \perp O_{2}E$. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,027 |
Three. (20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, draw two perpendicular chords $A C$ and $B D$ through the center $O$. Let the eccentric angles of points $A$ and $B$ be $\theta_{1}$ and $\theta_{2}$, respectively. Prove:
$$
\left|\sin \left(\theta_{2}-\theta_{1}\right)\right| \ge... | When $A C, B D$ do not coincide with the coordinate axes, let $\angle x O A=\varphi_{1}, \angle x O B=\varphi_{2}$, then
$$
\varphi_{2}=\varphi_{1} \pm \frac{\pi}{2}, \tan \varphi_{1} \cdot \tan \varphi_{2}=-1 .
$$
Since $\theta_{1}$ is the eccentric angle of point $A$, then $A\left(a \cos \theta_{1}, b \sin \theta_{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,028 |
Four, (20 points) Given that $a$ and $b$ are constants not equal to 0, and the variable $\theta$ satisfies the inequality system
$$
\left\{\begin{array}{l}
a \sin \theta + b \cos \theta \geqslant 0, \\
a \cos \theta - b \sin \theta \geqslant 0 .
\end{array}\right.
$$
Try to find the maximum value of $\sin \theta$. | Let $x=\cos \theta, y=\sin \theta$, then from the given conditions we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1, \\
b x+a y \geqslant 0, \\
a x-b y \geqslant 0 .
\end{array}\right.
$$
where $|x| \leqslant 1,|y| \leqslant 1$.
When equations (2) and (3) are equal, the lines
$$
b x+a y=0
$$
and $a x-b y=0$
are perpe... | \frac{a}{\sqrt{a^{2}+b^{2}}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,029 |
Five. (20 points) Given $f(x)=\frac{1}{4^{x}+2}(x \in \mathbf{R})$, $P_{1}\left(x_{1}, y_{1}\right)$ and $P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the function $y=f(x)$, and the x-coordinate of the midpoint $P$ of the line segment $P_{1} P_{2}$ is $\frac{1}{2}$.
(1) Prove that the y-coordinate of ... | (1) From $\frac{x_{1}+x_{2}}{2}=\frac{1}{2}$, we know $x_{1}+x_{2}=1$. Then
$$
\begin{array}{l}
y_{1}+y_{2}=\frac{1}{4^{x_{1}}+2}+\frac{1}{4^{x_{2}}+2} \\
=\frac{\left(4^{x_{2}}+2\right)+\left(4^{x_{1}}+2\right)}{4^{x_{1}+x_{2}}+2 \times 4^{x_{1}}+2 \times 4^{x_{2}}+4} \\
=\frac{4^{x_{1}}+4^{x_{2}}+4}{2\left(4^{x_{1}}+... | a>\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,030 |
One, (50 points) As shown in Figure 3, in the cyclic quadrilateral $ABCD$, $BC=CD$. Points $E$ and $F$ are on sides $AB$ and $AC$ respectively, with $EF \parallel BD$, and $EF$ intersects $AC$ at $G$. Prove:
$$
\begin{array}{c}
\angle GBD = \angle FCD, \\
\angle GDB = \angle ECB.
\end{array}
$$ | Extend $A B$ to $M$, such that $B M = F D$, and connect $F M, C M$.
Since $\angle M B C = \angle F D C, B C = D C$, we have
$$
\triangle M B C \cong \triangle F D C.
$$
Therefore, $M C = F C, \angle B C M = \angle D C F$.
From $B C = D C$, we know $\angle B A C = \angle D A C$. Thus, $\frac{E G}{G F} = \frac{A E}{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,031 |
II. (50 points) Let $n$ be a given positive integer, and $n \geqslant 3$. For $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, denote the minimum value of $\left|x_{i}-x_{j}\right|$ $(1 \leqslant i<j \leqslant n)$ as $m$. If $x_{1}^{2}+x_{2}^{2}+$ $\cdots+x_{n}^{2}=1$, try to find the maximum value of $m$. | Let's assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, then
$$
x_{2}-x_{1} \geqslant m, x_{3}-x_{2} \geqslant m, \cdots, x_{n}-x_{n-1} \geqslant m \text {. }
$$
Thus, $x_{j}-x_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant n)$.
$$
\begin{array}{l}
\text { Hence } \sum_{1 \leqslant i<j \leqslant n}\left... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,032 |
Three. (50 points) There are 800 points on a circle, labeled $1, 2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Now, choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ ... | Consider a circle with $2n$ points.
There are two cases:
(1) On the circle with $2n$ points, if the first red point is an even-numbered point, for example, the $2k$-th point, then according to the coloring rule, each red point that is subsequently colored will also be an even-numbered point. At this time, if the 2nd, 4... | 25 | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,033 |
Given an infinite sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and the sum of any consecutive terms is not equal to 100. Find the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$. | Solution: The minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
First, we prove: If all terms of the sequence $\left\{a_{n}\right\}$ are 1 or 2, then there must exist a continuous subsequence whose sum equals 100.
Consider the first 100 terms of the sequence $\left\{a_{n}\right\}$, and let
$$
S_{0}=0... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,034 |
Find all such real numbers $k$ for which the roots of the equation $\left(\frac{1}{2}+k^{2}\right) x^{2}+(\sqrt{2}+k) x-\frac{5}{4}(\sqrt{2}+1) k=0$ are both rational. | Solution: Such a real number $k$ does not exist. We use proof by contradiction.
Assume the equation has two rational roots $x_{1}, x_{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{\sqrt{2}+k}{\frac{1}{2}+k^{2}}, \\
x_{1} x_{2}=-\frac{\frac{5}{4}(\sqrt{2}+1) k}{\frac{1}{2}+k^{2}} .
\end... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,035 |
131 Given that the function $f(n)$ is a strictly increasing function defined on $\mathbf{N}_{+}$, with its range also in $\mathbf{N}_{+}$, and satisfies $f(f(n))=3n$. Find $f(2003)$. | Solution: (1) First, $f(1) \neq 1$. Otherwise, on one hand, $f(f(1))=f(1)=1$, and on the other hand, according to the given condition, $f(f(1))=3 \times 1=3$. This is a contradiction.
Second, $f(1) \neq 3$. This contradicts the known condition $f(f(1))=3 \times 1$.
In summary, $f(1)=2$ (since $f(n) \in \mathbf{N}_{+}$... | 3822 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,036 |
Given $n(n>3)$ points on a plane, where no three points are collinear. Some points are arbitrarily connected by line segments (these line segments are called edges), resulting in $x$ edges.
(1) If it is ensured that the figure contains a triangle with the given points as vertices, prove:
$$
x \geqslant \frac{n(n-1)(n-2... | Proof: (1) First, construct drawers: each drawer contains three distinct points, for a total of $\mathrm{C}_{n}^{3}$ drawers. Since the same edge will appear in $\mathrm{C}_{n-2}^{1}$ drawers (i.e., the two points of the edge plus any other point can form a drawer), according to the pigeonhole principle, when $x \cdot ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,037 |
Example 4 As shown in Figure 5, given two circles $\odot O_{1}$ and $\odot O_{2}$ with unequal radii intersecting at points $M$ and $N$, and $\odot O_{1}$, $\odot O_{2}$ are internally tangent to $\odot O$ at points $S$ and $T$ respectively. Prove that $O M \perp M N$ if and only if points $S$, $N$, and $T$ are colline... | Prove: Connect $O S$, $O T$, $S T$, draw common tangents $S P$ and $T P$ intersecting at point $P$, then $P S = P T$. Therefore, point $P$ lies on the radical axis of $\odot O_{1}$ and $\odot O_{2}$, and we have
$$
P S^{2} = P N \cdot P M \text{. }
$$
Connect $O P$ intersecting $S T$ at point $Q$, then $O P \perp S T$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,038 |
Example 5 As shown in Figure 6, in $\triangle ABC$, $O$ is the circumcenter, the three altitudes $AD, BE, CF$ intersect at point $H$, line $ED$ and $AB$ intersect at point $M$, $FD$ and $AC$ intersect at point $N$. Prove:
(1) $OB \perp DF, OC \perp DE$;
(2) $OH \perp MN$.
(2001, National High School Mathematics Competi... | Proof: (1) Draw the tangent line $B T$ of the circumcircle of $\triangle A B C$ through $B$. Since $A, F, D, C$ are concyclic, we know $\angle T B A = \angle A C B = \angle B F D$, thus $D F \parallel B T$. Since $O B \perp B T$, it follows that $O B \perp D F$. Similarly, $O C \perp D E$.
(2) Let $V$ be the midpoint ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,039 |
Example 6 As shown in Figure 7, $\odot$ passes through the vertices $A$ and $C$ of $\triangle ABC$, and intersects $AB$ and $BC$ at points $K$ and $N$ (distinct from $K$ and $N$). The circumcircle of $\triangle ABC$ and the circumcircle of $\triangle BKN$ intersect at $B$ and $M$. Prove that $\angle BMO=90^{\circ}$.
(2... | Proof: Let the circumcenters of $\triangle ABC$ and $\triangle BKN$ be $O_1$ and $O_2$, respectively. From the given conditions, we know that $O$, $O_1$, and $O_2$ are not collinear (otherwise, $B$ and $M$ would coincide). The lines $AC$, $KN$, and $BM$ are the radical axes of the circles taken two at a time, so they m... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,040 |
Example 7 As shown in Figure 8, quadrilateral $ABCD$ is inscribed in a circle, the extensions of sides $AB$ and $DC$ intersect at point $P$, and the extensions of sides $AD$ and $BC$ intersect at point $Q$. Two tangents $QE$ and $QF$ are drawn from $Q$ to the circle, with points of tangency at $E$ and $F$ respectively.... | Prove: Connect $P Q$, take a point $M$ on $P Q$ such that points $B, C, M, P$ are concyclic. Therefore, $Q$ is a point on the radical axis of the circumcircles of quadrilaterals $A B C D$ and $B C M P$. Thus,
$$
Q E^{2}=Q C \cdot Q B=Q M \cdot P Q \text {. }
$$
At this point, $\angle P M C=\angle A B C=\angle P D Q$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,041 |
Example 8 As shown in Figure 9, a circle is tangent to sides $AB$ and $BC$ of a convex quadrilateral $ABCD$ at points $G$ and $H$, respectively, and intersects diagonal $AC$ at points $E$ and $F$. What necessary and sufficient condition must $ABCD$ satisfy so that there exists another circle passing through $E$ and $F$... | The required necessary and sufficient condition is
$$
A B+A D=C B+C D \text {. }
$$
Proof: Necessity.
Suppose another circle passing through points $E$ and $F$ is tangent to the extensions of $D A$ and $D C$ at points $J$ and $K$ respectively: Note that point $A$ has equal power with respect to these two circles, i.e.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,042 |
The first question: Through a point $P$ outside a circle, draw two tangents and one secant to the circle, the points of tangency being $A$ and $B$, and the secant intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Take a point $Q$ on the chord $CD$ such that $\angle D A Q = \angle P B C$. Prov... | Proof 1: As shown in Figure 1, connect $A B$. We have
$\angle Q A B = \angle D A B -$
$\angle D A Q = \angle D C B -$
$\angle P B C = \angle Q P B$,
or by $\angle P Q A = \angle D A Q$
$+ \angle A D Q = \angle P B C +$
$\angle A B C = \angle A B P$,
it is known that $A, Q, B, P$ are concyclic. Thus, $\angle P A B = \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,043 |
The second question: Let the three sides of a triangle be integers $l$, $m$, and $n$, and $l>m>n$. It is known that $\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=$ $\left\{\frac{3^{n}}{10^{4}}\right\}$, where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the... | Proof: Since $l, m, n$ are the lengths of the sides of a triangle, hence $m+n>l$.
From the given information,
$$
3^{l-n} \equiv 3^{m-n} \equiv 3^{l-m} \equiv 1\left(\bmod 10^{4}\right).
$$
Without loss of generality, let $\left\{\begin{array}{l}l-m=s, \\ m-n=t\end{array}\left(s, t \in \mathbf{N}_{+}\right)\right.$, th... | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,044 |
In 2002, the first problem of the Chinese National Training Team Selection Examination for the IMO:
Let the convex quadrilateral $ABCD$ have its two pairs of opposite sides intersecting at points $E$ and $F$, respectively, and the intersection of the diagonals be point $P$. Draw $PO \perp EF$ at $O$. Prove that $\angl... | Proof: As shown in Figure 1, it is only necessary to prove that $\angle P O B = \angle P O D$ and $\angle P O C = \angle P O A$. And $\angle P O B = \angle P O D$ is equivalent to $\angle B O E = \angle D O F$.
Construct $B M \perp E F$, $D N \perp E F$, and $A H \perp E F$, with the feet of the perpendiculars being $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,045 |
Example 6 As shown in Figure 22, use 6 different colors to color the 4 different regions $A, B, C, D$, requiring that adjacent regions be colored with different colors. How many different coloring methods are there? | Solution 1: Complete in four steps according to the order of $A-B-C-D$.
(1) Color region $A$ with no restrictions, 6 colors can be used, giving 6 ways to color.
(2) Color region $B$, which must be a different color from $A$, only 5 remaining colors can be used, giving 5 ways to color.
(3) Color region $C$, which must b... | 600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,046 |
9.1 The set $M$ consists of 2003 different numbers. For any two different elements $a, b$ in $M$, the number $a^{2} + b \sqrt{2}$ is rational. Prove: For any number $a$ in $M$, the number $a \sqrt{2}$ is rational. | 9.1 For any $a, b, c \in M, a \neq b \neq c$, then $a^{2}+b \sqrt{2}$, $b^{2}+a \sqrt{2}$, $c^{2}+a \sqrt{2}$, $c^{2}+b \sqrt{2}$ are all rational numbers. Therefore,
$$
\begin{array}{l}
a^{2}+b \sqrt{2}-\left(b^{2}+a \sqrt{2}\right)=(a-b)(a+b-\sqrt{2}) \\
=\frac{1}{2}(a \sqrt{2}-b \sqrt{2})(a \sqrt{2}+b \sqrt{2}-2) \i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,047 |
$9.2 \odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. The tangents to the two circles at point $A$ intersect $B O_{1}$ and $B O_{2}$ at points $K$ and $L$ respectively. Prove: $K L \parallel \mathrm{O}_{1} \mathrm{O}_{2}$. | 9.2 We have
$$
\begin{array}{l}
\angle L A K=\angle B A K+\angle B A L \\
=\frac{1}{2}\left(\angle B O_{2} O_{1}+\angle B O_{1} O_{2}\right)=180^{\circ}-\angle L B K .
\end{array}
$$
Therefore, $A L B K$ is a cyclic quadrilateral. Thus $\angle B O_{2} O_{1}=\angle B A K$ $=\angle B L K$. Hence, $K L \parallel O_{1} O_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,048 |
9.3 On a line, there are $2 k-1$ white segments and $2 k-1$ black segments. It is known that any white segment intersects with at least $k$ black segments, and any black segment intersects with at least $k$ white segments. Prove: it is possible to find a black segment that intersects with all white segments, and a whit... | 9.3 It is only necessary to prove: if any white line segment intersects with at least $k$ black line segments, then it is always possible to find a black line segment that intersects with all white line segments.
Assume the above statement is false, then for each black line segment, there exists a white line segment t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,049 |
9.5 There are $N$ cities in a country. Between any two cities, there is either a road or a railway. A traveler hopes to visit each city exactly once and return to the city from which he started. Prove: The traveler can choose a city as the starting point, not only to fulfill his wish, but also to change the mode of tra... | 9.5 The original problem can be restated in graph theory terms as follows:
Given a complete graph with $N$ vertices, its edges are colored with two different colors. Prove: one can find a cycle passing through all vertices, which can be divided into at most two monochromatic parts.
Prove this proposition using mathema... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,051 |
9.6 Let $a, b, c$ be positive numbers, their sum equals 1. Prove:
$$
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geqslant \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} .
$$ | 9.6 Solution 1: From the inequality $\frac{1}{x}+\frac{1}{y} \geqslant \frac{4}{x+y}$, where $x>0, y>0$, we can obtain
$$
\begin{array}{l}
\frac{1}{a+b}+\frac{1}{b+c} \geqslant \frac{4}{a+2 b+c}, \\
\frac{1}{b+c}+\frac{1}{c+a} \geqslant \frac{4}{b+2 c+a}, \\
\frac{1}{c+a}+\frac{1}{a+b} \geqslant \frac{4}{c+2 a+b} .
\en... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,052 |
9.7 Can we fill each cell of an infinitely large grid paper with a positive integer so that for any positive integers $m, n>$ 100, the sum of the numbers in any $m \times n$ grid on the paper is divisible by $m+n$? | 9.7 Impossible.
For any $200 \times 200$ grid $A$, assume it is located at a corner of some $200 t \times 200 t$ grid $B$, where $t$ is a positive integer that cannot divide the sum of all numbers in grid $A$. Divide the shape $B / A$ into a series of $200 \times 200(t-1)$ rectangles. According to the problem, the sum... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,053 |
9.8 On the sides $AP$ and $PD$ of the acute triangle $\triangle APD$, take points $B$ and $C$ respectively. The diagonals of quadrilateral $ABCD$ intersect at point $Q$. The orthocenters of $\triangle APD$ and $\triangle BPC$ are $H_{1}$ and $H_{2}$ respectively. Prove that if the line $H_{1} H_{2}$ passes through the ... | 9.8 Construct circles $\omega_{1}$ and $\omega_{2}$ with diameters $AC$ and $BD$ respectively. Let $BB_{1}$, $CC_{1}$, $AA_{1}$, and $DD_{1}$ be the altitudes of $\triangle BPC$ and $\triangle APD$, where points $A_{1}$ and $C_{1}$ lie on circle $\omega_{1}$, and points $B_{1}$ and $D_{1}$ lie on circle $\omega_{2}$. T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,054 |
10.1 The set $M$ consists of 2003 different positive numbers. For any three distinct elements $a, b, c$ in $M$, the number $a^{2}+bc$ is rational. Prove: there exists a positive integer $n$ such that for any number $a$ in $M$, the number $a \sqrt{n}$ is rational. | 10.1 From the set $M$, take 4 different numbers $a, b, c, d$. Since $d^{2}+a b \in \mathbf{Q}, d^{2}+b c \in \mathbf{Q}$, it follows that $b c-a b \in \mathbf{Q}$. Therefore, $a^{2}+a b=a^{2}+b c+(a b-b c) \in \mathbf{Q}$.
Similarly, we know that $b^{2}+a b \in \mathbf{Q}$. Thus, for any two different numbers $a$ and ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,055 |
10.2 In a cyclic quadrilateral $ABCD$ inscribed in a circle, the diagonals intersect at point $O$. Let the circumcircles of $\triangle ABO$ and $\triangle CDO$ be circles $S_{1}$ and $S_{2}$, respectively, and their intersection points be $O$ and $K$. Draw lines through point $O$ parallel to $AB$ and $CD$, which inters... | 10.2 Without loss of generality, let $\angle A B O \geqslant \angle B A O$. In this case, $A B O L$ and $D C O M$ are both isosceles trapezoids (as shown in Figure 1). Therefore, we have
$$
\begin{array}{l}
\angle L O A=\angle O A B=\angle O D C \\
=\angle D O M=\angle D C M=\angle C M O, \\
\angle K M C=\pi-\angle K O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,056 |
10.3 Given a "tree" with $n(n \geqslant 2)$ vertices (i.e., a graph with $n$ vertices and $n-1$ edges, from any vertex one can reach any other vertex along the edges, and there are no loops formed by the edges), numbers $x_{1}, x_{2}, \cdots, x_{n}$ are placed on each vertex, and the product of the numbers at its two e... | 10.3 Consider the edge $l$ connecting vertices $x_{i}$ and $x_{j}$. Let $k_{i}(l)$ denote the number of vertices that cannot be reached from vertex $x_{i}$ after removing edge $l$. Similarly, define $k_{j}(l)$. Clearly,
$$
1 \leqslant k_{i}(l), k_{j}(l) \leqslant n-1, \quad k_{i}(l)+k_{j}(l)=n,
$$
and for each vertex ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,058 |
10.4 In the plane, a finite point set $X$ and a regular triangle $T$ are given. It is known that any subset $X^{\prime}$ of $X$ consisting of no more than 9 points can be covered by two translated copies of $T$. Prove: the point set $X$ can be covered by two translated copies of $T$. | 10.4 First, we prove two lemmas.
Lemma 1 If two triangles $T_{1}$ and $T_{2}$ are positively homothetic, and $T_{2}$ intersects the lines containing the sides of $T_{1}$, then $T_{1}$ is located inside $T_{2}$.
This proposition is obviously true.
Lemma 2 For any finite point set $M$ and any triangle $T$, there exists ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,059 |
10.6 The sequence of positive integers $\left\{a_{n}\right\}$ is constructed as follows: $a_{0}$ is some positive integer. If $a_{n}$ is divisible by 5, then $a_{n+1}=\frac{a_{n}}{5}$; if $a_{n}$ is not divisible by 5, then $a_{n+1}=\left[\sqrt{5} a_{n}\right]$ (where $[x]$ denotes the greatest integer not exceeding $x... | 10.6 The conclusion of the problem is equivalent to “from some $n$ onwards, $a_{n}$ is never a multiple of 5”. We will prove this.
First, we prove that there exist two consecutive terms in the sequence that are not multiples of 5. Otherwise, for any $n$, either $a_{n+1}=\frac{a_{n}}{5}$ or $a_{n+2}=\frac{a_{n+1}}{5}$.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,060 |
10.7 Let the circumcenter and incenter of $\triangle A B C$ be $O$ and $I$, respectively. The excircle $\omega_{c}$ is tangent to the extensions of sides $A B$ and $A C$ at points $K$ and $M$, respectively, and to side $B C$ at point $N$. Given that the midpoint $P$ of segment $K M$ lies on the circumcircle of $\triang... | 10.7 Solution 1: If $B C / / K M$ (i.e., $A B=A C$), then the conclusion is obviously true. Therefore, assume that $B C$ is not parallel to $K M$.
As shown in Figure 2, let the incircle of $\triangle A B C$ touch side $B C$ at point $T$, the midpoint of side $B C$ be $E$, and the projection of point $N$ on $K M$ be $Q... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,061 |
10.8 Try to find the largest positive integer $N$, such that no matter how the positive integers from 1 to 400 are filled into the cells of a $20 \times 20$ grid, one can always find two numbers in the same row or the same column whose difference is not less than $N$. | 10.8209.
First, we illustrate that $N \leqslant 209$. Divide the grid into two $20 \times 10$ grids using a vertical line through the center. Fill the numbers from 1 to 200 into the left grid in increasing order row by row, and fill the numbers from 201 to 400 into the right grid in the same manner. This way, the maxi... | 209 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,062 |
11.1 Let $\alpha, \beta, \gamma, \tau$ be positive numbers, and for all real numbers $x$, we have $\sin \alpha x + \sin \beta x = \sin \gamma x + \sin \tau x$.
Prove: $\alpha = \gamma$ or $\alpha = \tau$. | 11.1 Without loss of generality, let $\alpha-\beta \geqslant 0, \gamma-\tau \geqslant 0$. Let $a=\frac{\alpha+\beta}{2}, b=\frac{\alpha-\beta}{2}, c=\frac{\gamma+\tau}{2}, d=\frac{\gamma-\tau}{2}$.
Thus, the condition in the problem is transformed into
$\sin a x \cdot \cos b x=\sin c x \cdot \cos d x$,
where $a>b \geqs... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,063 |
On 11.4, person A and person B each have a long strip of paper. One strip has the letter $A$ written on it, and the other strip has the letter $B$ written on it. Every minute, one of the two (not necessarily taking turns in order) copies the word on the other person's strip of paper to the left or right of the word on ... | 11.4 If a word reads the same from left to right as it does from right to left, it is called "symmetric." Below, we use induction to prove: after $n$ minutes, each long strip of paper can be divided into two "symmetric" words. Thus, by simply reversing these two "symmetric" words, the resulting words will be the same a... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 714,065 |
11.5 The lengths of the three sides of a triangle are the roots of a cubic equation with rational coefficients. Prove: The altitude of the triangle is a root of a sextic equation with rational coefficients. | 11.5 Let the three sides of a triangle be $a$, $b$, and $c$. By the rationality of the coefficients of the polynomial, it can be considered irreducible, i.e.,
$$
f(x)=(x-a)(x-b)(x-c)
$$
$=x^{3}-A x^{2}+B x-C$ (by the problem and Vieta's formulas), which means $2 p=a+b+c=A \in \mathbf{Q}$,
$$
ab+bc+ca=B \in \mathbf{Q}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,066 |
11.7 In a certain country, there are 100 cities, some of which are connected by roads. Among any four cities, there are at least two roads connecting them. It is known that there is no road passing through each city exactly once. Prove: There exist two cities in the country such that any other city is connected by a ro... | 11.7 Draw a graph with the cities of the country as vertices and the roads as edges. In this graph, take the longest non-self-intersecting path $S$, and assume its two endpoints are $A$ and $B$. By the problem statement, the number of vertices on path $S$ is no more than 99, and $A$ and $B$ cannot be connected to any v... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,067 |
11.8 The inscribed sphere of tetrahedron \(ABCD\) touches its faces \(ABC\), \(ABD\), \(ACD\), and \(BCD\) at points \(D_1\), \(C_1\), \(B_1\), and \(A_1\) respectively. Consider the plane equidistant from point \(A\) and the plane \(B_1C_1D_1\), and similarly the other three planes. Prove that the circumcenter of the ... | 11.8 Let the four vertices of the tetrahedron formed by the four planes under consideration be denoted as \(A_{2}, B_{2}, C_{2}, D_{2}\). Thus, the tetrahedron \(A_{1} B_{1} C_{1} D_{1}\) is homothetic to the tetrahedron \(A_{2} B_{2} C_{2} D_{2}\). The corresponding homothety transformation maps the incenter of the te... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,069 |
1. In the Cartesian coordinate system, consider the hyperbola
$$
\Gamma=\left\{M(x, y) \in \mathbf{R}^{2}\left|\frac{x^{2}}{4}-y^{2}=1\right|\right.
$$
and the conic section $\Gamma^{\prime}$, where $\Gamma^{\prime}$ does not intersect $\Gamma$. Let $n\left(\Gamma, \Gamma^{\prime}\right)$ be the maximum number of poin... | Solution: (1) $p=0$. Let $\Gamma^{\prime}: \frac{x^{2}}{4}-y^{2}=2$. Consider point $B\left(n, \sqrt{\frac{n^{2}}{4}-1}\right) \in \Gamma, B^{\prime}\left(n, \sqrt{\frac{n^{2}}{4}-2}\right) \in \Gamma^{\prime}$, then
$$
\begin{array}{c}
B B^{\prime}=\sqrt{\frac{n^{2}}{4}-1}-\sqrt{\frac{n^{2}}{4}-2} \\
=\frac{1}{\sqrt{\... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,070 |
2. Let $M$ and $N$ be points on the sides $AC$ and $BC$ of $\triangle ABC$ respectively, and $\angle ACB=90^{\circ}$. Let $AN$ and $BM$ intersect at point $L$. Prove that the orthocenters of $\triangle AML$ and $\triangle BNL$ and point $C$ are collinear.
(2002, Bulgarian Winter Mathematical Competition) | Prove: As shown in Figure 1, let the orthocenters of $\triangle AML$ and $\triangle BNL$ be $H_{1}$ and $H_{2}$, respectively. The line $AH_{1}$ intersects the extension of $BM$ at point $P$, the line $AC$ intersects $LH_{1}$ at point $Q$, the line $BC$ intersects $LH_{2}$ at point $R$, and the line $BH_{2}$ intersects... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,071 |
3. Let $E$ and $F$ be points on the sides $AD$ and $CD$ of square $ABCD$, such that $\angle AEB = \angle AFB = 90^\circ$, and $EG \parallel AB$, where $EG$ intersects $BF$ at point $G$. If $AF$ intersects $BE$ at point $H$, and $DH$ intersects $BC$ at point $I$, prove that $FI \perp GH$.
(2002, Bulgarian National Mathe... | Proof: As shown in Figure 2, let $AF$ intersect $BC$ at point $J$, and let $K$ be the projection of $I$ on $AF$.
Then $\frac{KF}{JF}=\frac{IB}{JB}=\frac{DE}{AE}=\frac{FG}{BG}$, $\frac{KF}{KJ}=\frac{FG}{FB}$.
Also, $\angle JKI=\angle BFH=90^{\circ}$,
$\angle IJK=90^{\circ}-\angle FBJ=\angle HBF$,
Therefore, $\triangle I... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,072 |
4. Given a fixed circle with two fixed points $A$ and $B$ and a moving point $C$ forming $\triangle ABC$ which is an acute triangle, the midpoint $M$ of $AB$ has projections on sides $AC$ and $BC$ at points $E$ and $F$ respectively. Prove that the perpendicular bisector of $EF$ passes through a fixed point.
(2002, Bulg... | Proof: As shown in Figure 3, let $K$ and $L$ be the midpoints of $AM$ and $BM$ respectively, and let $D$ be a point on the perpendicular bisector of $AB$, with $C$ and $D$ on opposite sides of line $AB$,
$$
\begin{array}{l}
\angle D K L=\angle A C B . \text { If } \\
\angle B A C=\angle A B C,
\end{array}
$$
the concl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,073 |
5. As shown in Figure 4, given $\triangle A B C$ with the external angle bisectors of $\angle A C B$, $\angle B A C$, and $\angle A B C$ being $A_{1} C$, $B_{1} A$, and $C_{1} B$ respectively, and the projections of points $A$, $B$, and $C$ on $A_{1} C$, $B_{1} A$, and $C_{1} B$ being $A_{1}$, $B_{1}$, and $C_{1}$ resp... | Proof: Let the midpoints of the three sides of $\triangle ABC$ be $A_{2}, B_{2}, C_{2}$, then
$$
\begin{array}{l}
\angle B C_{1} A_{2}=\angle A_{2} B C_{1}=90^{\circ}-\frac{\angle B}{2}, \\
\angle B A_{2} C_{1}=\angle B.
\end{array}
$$
Also, $\angle B_{2} A_{2} C=\angle B$, so, $B_{2}, A_{2}, C_{1}$ are collinear. Let... | r^{2}+p^{2}=d^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 714,074 |
6. Find the smallest number $k$ such that $\frac{t_{a}+t_{b}}{a+b}<k$, where $a$ and $b$ are two sides of a triangle, and $t_{a}$ and $t_{b}$ are the lengths of the angle bisectors corresponding to these sides.
(2002, Bulgaria National Mathematical Olympiad (Final)) | When $\triangle A B C$ satisfies $a=b$, and the base angle $\angle A \rightarrow 0$,
$$
\begin{array}{l}
c \rightarrow 2 b, \cos \frac{A}{2} \rightarrow 1, \\
t_{a}=\frac{b c \sin A}{(b+c) \sin \frac{A}{2}}=\frac{2 b c}{b+c} \cos \frac{A}{2} .
\end{array}
$$
Thus, $\frac{t_{a}+t_{b}}{a+b}=\frac{t_{a}}{a} \rightarrow \... | \frac{4}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,075 |
7. Let $AB$ be the diameter of $\odot O$, and the tangents at points $A$ and $B$ be $l_{a}$ and $l_{b}$, respectively. $C$ is any point on the circumference, and $BC$ intersects $l_{a}$ at $K$. The angle bisector of $\angle CAK$ intersects $CK$ at $H$. Let $M$ be the midpoint of arc $\overparen{CAB}$, and $HM$ intersec... | Proof: As shown in Figure 5, let $A H$ intersect $\odot O$ at point $R$, then $\angle A R B=90^{\circ}$, and $B R$ is the angle bisector of $\angle C B A$, so $\triangle H A B$ is an isosceles triangle, and $H B=A B$.
Since $M$ is the midpoint of arc $\overparen{C A B}$, $M T$ $/ / B C$. Because $A H$ bisects $\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,076 |
8. Let the unit disk be $D=\left\{(x, y) \in \mathbf{R}^{2} \mid x^{2}+y^{2} \leqslant\right.$ 1\}, and define the width of a strip formed by two parallel lines as the distance between these lines. If the unit disk can be covered by some strips, prove: the sum of the widths of these strips is at least 2.
(19th Iranian ... | Solution: Consider the unit sphere that contains this unit circle. Let the unit circle belong to plane II. For each strip, replace the two parallel lines that form this strip with two planes that pass through these lines and are perpendicular to plane $I$, then for each strip, we can obtain two parallel planes, which w... | 2 | Geometry | proof | Yes | Yes | cn_contest | false | 714,077 |
9. Let $M$ and $N$ be points on the sides $AC$ and $BC$ of $\triangle ABC$, respectively. Let $K$ be the midpoint of segment $MN$, and let $D$ be the second intersection point of the circumcircles of $\triangle CAN$ and $\triangle BCM$. Prove that $CD$ passes through the circumcenter of $\triangle ABC$ if and only if t... | Proof: Let $O$ be the circumcenter of $\triangle ABC$, and $CO$ intersects $AB$ at point $C_{1}$. $CD$ intersects $AB$ at point $P$. In Figure 6,
$$
\begin{array}{l}
\frac{A C_{1}}{B C_{1}}=\frac{\sin \angle 1}{\sin \angle 2} \\
=\frac{\sin \angle A O C}{\sin \angle B O C}=\frac{\sin 2 B}{\sin 2 A},
\end{array}
$$
the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,078 |
10. As shown in Figure 8, given two circles with unequal radii intersecting at points $A$ and $B$, the common tangents $ST$ and $MN$ touch the two circles at points $S$, $T$ and $M$, $N$ respectively. Prove: The orthocenters of $\triangle A M N$, $\triangle A S T$, $\triangle B M N$, and $\triangle B S T$ are the four ... | Proof: Let the orthocenters of $\triangle A M N, \triangle A S T, \triangle B M N, \triangle B S T$ be $H_{1}, H_{2}, H_{3}, H_{4}$ respectively. By symmetry, quadrilateral $H_{1} H_{2} H_{3} H_{4}$ is an isosceles trapezoid, where $H_{1} H_{4} \parallel A B \parallel H_{2} H_{3}$. Therefore, we only need to prove that... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,080 |
11. Find all tetrahedra that have exactly four edges of length \( a \), and when unfolded into a plane, form a kite that is not a rhombus.
(51st Czech and Slovak Mathematical Olympiad (Open Book)) | Solution: Let each edge length be as shown in Figure 9, and we discuss in two cases.
(1) If in the unfolded diagram there is one triangle adjacent to the other three triangles, as shown in Figure 10. Since $\angle D_{1} 、 \angle D_{2} 、 \angle D_{3}$ cannot be straight angles, assume without loss of generality that $\a... | A D= B D= C D= B C= a, A B= \frac{a}{2}, A C= \frac{\sqrt{6}}{2} a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,081 |
12. $B_{1}, B_{2}, \cdots, B_{N}$ are $N$ unit spheres, and each sphere is externally tangent to exactly two other spheres, with the points of tangency denoted as $C_{1}, C_{2}, \cdots, C_{N}$. $P$ is a point outside all the spheres. Tangents are drawn from point $P$ to the sphere $B_{i}(1 \leqslant i \leqslant N)$, an... | Proof: Consider the sphere $B_{l}$ and sphere $B_{m}$ tangent at point $C_{r}$. It is only necessary to consider the part of the plane containing point $P$, the centers $O_{1}$, and $O_{m}$. Since point $C$ lies on the line connecting the centers $O_{l}$ and $O_{m}$, $C$ is also in this plane. The tangent line through ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,082 |
13. Given $\triangle K L M, A$ is on the extension of $I K$. Construct a rectangle $A B C D$ such that $B, C, D$ are on the lines containing sides $K M, K L, L M$ respectively.
(51st Czech and Slovak Mathematical Olympiad (Final)) | Solution: As shown in Figure 17, let quadrilateral $ABCD$ be the desired rectangle. Translate it along $BA$ to rectangle $A'B'C'D'$, where $B'=A$ and $C'=D$. Then $A'$ lies on the line $K'M'$, which is the reflection of line $KM$ about point $A$, and $K'M'$ intersects $LK$ at point $K'$ and $LM$ at point $M'$. Since $A... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,083 |
14. Given that $C H$ is the altitude of the right triangle $\triangle A B C$ $\left(\angle C=90^{\circ}\right)$, and intersects the angle bisectors $A M$ and $B N$ at points $P$ and $Q$, respectively. Prove that the line passing through the midpoints of $Q N$ and $P M$ is parallel to the hypotenuse $A B$.
(52nd Russian... | Proof: As shown in Figure 19, let $E$ and $F$ be the midpoints of $QN$ and $PM$ respectively, $a = BC, b = AC, c = AB$. From the given, we have $\frac{AN}{NC} = \frac{c}{a}$. Therefore,
$$
CN = \frac{ab}{a+c}.
$$
Also, $\frac{QH}{CQ} = \frac{HB}{a} = \frac{a}{c}$, so,
$$
CQ = \frac{ab}{a+c}.
$$
Thus, $CN = CQ, CE \pe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,084 |
15. Given three fixed points $A, B, C$ on a plane, let the moving point $D$ satisfy that $A, B, C, D$ are concyclic. $I_{A}, I_{B}, I_{C}, I_{D}$ are the Simson lines of points $A, B, C, D$ with respect to $\triangle BCD, \triangle ACD, \triangle ABD, \triangle ABC$ respectively. When point $D$ moves, find the locus of... | Solution: As shown in Figure 20, let points $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively. If $A'$ and $C'$ are the midpoints of $BC$ and $AB$, respectively, then $AH \parallel OA'$, $CH \parallel OC'$, and $AC \parallel A'C'$. Therefore, $\triangle AHC \sim \triangle A'OC'$, and we h... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,085 |
$1 . x$ and $y$ are positive numbers and are inversely proportional. If $x$ increases by $p \%$, let the percentage decrease in $y$ be $q$ \%. Then the value of $q$ is ( ).
(A) $\frac{100 p}{1+p \%}$
(B) $\frac{100}{p \%}$
(C) $\frac{p}{100+p}$
(D) $\frac{100 p}{100+p}$ | -、1.D.
Since $x, y$ are inversely proportional, we can set $y=\frac{k}{x}$, then we have
$$
(1-q \%) y=\frac{k}{(1+p \%) x}, q=\frac{100 p}{100+p} .
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,086 |
3. Given the three sides of $\triangle A B C$ are $a, b, c$, and they satisfy the equation $a^{2} x^{2}-\left(c^{2}-a^{2}-b^{2}\right) x+b^{2}=0$. Then the nature of the roots of the equation is ( ).
(A) has two equal real roots
(B) has two distinct real roots
(C) has no real roots
(D) cannot be determined | 3.C.
$$
\begin{array}{l}
\Delta=\left(c^{2}-a^{2}-b^{2}\right)^{2}-4 a^{2} b^{2} \\
=(c+a-b)(c-a+b)(c+a+b)(c-a-b) .
\end{array}
$$
Since $c+a-b>0, c-a+b>0$,
$$
c+a+b>0, c-a-b<0 \text {, }
$$
Therefore, $\Delta<0$, the original equation has no real roots. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,088 |
4. If $\frac{t_{1}}{\left|t_{1}\right|}+\frac{t_{2}}{\left|t_{2}\right|}+\frac{t_{3}}{\left|t_{3}\right|}=1$, then the value of $\frac{\left|t_{1} t_{2} t_{3}\right|}{t_{1} t_{2} t_{3}}$ is ( ).
(A) -1
(B) 1
(C) $\pm 1$
(D) Uncertain | 4. A.
Since $t_{1}, t_{2}, t_{3}$ are all non-zero, then $\frac{t_{1}}{\left|t_{1}\right|}, \frac{t_{2}}{\left|t_{2}\right|}, \frac{t_{3}}{\left|t_{3}\right|}$ must have two values of 1 and one value of -1. That is, $t_{1}, t_{2}, t_{3}$ have two positive numbers and one negative number.
Therefore, $t_{1} t_{2} t_{3}<... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,089 |
4. (1) In $\triangle ABC$, there are $n$ points inside. Using these $n+3$ points as vertices, connect $y$ lines to divide $\triangle ABC$ into $x$ non-overlapping small triangles. Prove: $x=2n+1, y=3n$.
If $\triangle ABC$ is changed to polygon $A_{1} A_{2} \cdots A_{k}$, what would be the values of $x$ and $y$?
(2) In... | (Solution: $x=2 n-2+k, y=3 n-3+k$ )
(Hint: If the two endpoints of an edge are the same color, then merge the two endpoints, which does not affect the number of small triangles with three vertices of different colors.) | x=2n-2+k, y=3n-3+k | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,090 |
5. In the Cartesian coordinate system, given points $A(2,2), B(2$, $-3)$, point $P$ is on the $y$-axis, and $\triangle A P B$ is a right triangle. Then the number of points $P$ is ( ).
(A) 2
(B) 3
(C) 1
(D) 4 | 5.D.
As shown in Figure 6. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,091 |
6. The same product at the same price underwent two price adjustments in three malls. Mall A: The first price increase percentage is $a$, and the second price increase percentage is $b$; Mall B: The price increase percentage for both times is $\frac{a+b}{2}(a>0, b>0)$; Mall C: The first price increase percentage is $b$... | 6.B.
After two price increases, the price at Store A is
$$
\begin{array}{l}
(1+a)(1+b) \\
=1+a+b+a b ;
\end{array}
$$
After two price increases, the price at Store B is
$$
\left(1+\frac{a+b}{2}\right)\left(1+\frac{a+b}{2}\right)=1+(a+b)+\left(\frac{a+b}{2}\right)^{2} \text {; }
$$
After two price increases, the pric... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,092 |
7. Given that $a$ is a positive root of the equation $x^{2}-x-2000=0$. Then the value of the algebraic expression $3+\frac{2000}{1+\frac{2000}{1+2000}}$ is $\qquad$ . | $$
=.7 . \frac{5+3 \sqrt{889}}{2}
$$
Given that $a$ is a positive root of the equation $x^{2}-x-2000=0$, we have $a^{2}-a-2000=0$, and $a=\frac{1+3 \sqrt{889}}{2}$. From $a^{2}-a-2000=0$, we get $a=1+\frac{2000}{a}$. Therefore, the original expression $=3+\frac{2000}{1+\frac{2000}{a}}=2+1+\frac{2000}{a}$ $=2+a=\frac{5... | \frac{5+3 \sqrt{889}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,093 |
8. As shown in Figure $3, A C=$ $B C, A C \perp B C$ at point $C, A B=A D=B D$, $C D=C E=D E$. If $A B=\sqrt{2}$, then $B E=$ | 8. 1 .
In $\triangle A D C$ and $\triangle B D C$, given $A D=D B, D C=D C$, $A C=B C$, we can conclude that
$\triangle A D C \cong \triangle B D C, \angle A D C=\angle B D C$.
Since $\angle A D B=60^{\circ}$, it follows that,
$\angle A D C=\angle B D C=\angle E D B=30^{\circ}$.
Therefore, $D B \perp C E, B C=B E$.
Mo... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,094 |
10. Given that the real number $a$ satisfies $a^{2}-a-1=0$. Then the value of $a^{8}+7 a^{-4}$ is - $\qquad$ | 10.48 .
From $a^{2}-a-1=0$, we get $a-a^{-1}=1$, then, $a^{2}+a^{-2}$
$$
\begin{aligned}
=3, & a^{4}+a^{-4}=7 . \\
& a^{8}+7 a^{-4}=a^{4}\left(a^{4}+a^{-4}\right)+7 a^{-4}-1 \\
& =7\left(a^{4}+a^{-4}\right)-1=7 \times 7-1=48 .
\end{aligned}
$$ | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,096 |
11. Observe the array: $(1),(3,5),(7,9,11),(13,15,17$,
19), $\cdots \cdots$. Then 2003 is in the group. | 11.45 .
Using the trial method.
Assume 2003 is in the 50th group, then the first 50 groups have
$$
1+2+3+\cdots+50=51 \times 25=1275
$$
numbers, the last number in this group is 2549, and the first number is 2451, so 2003 is not in the 50th group;
Assume 2003 is in the 40th group, then the first 40 groups have
$$
1+2+... | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,097 |
12. As shown in Figure 4, given that the two diagonals of rhombus $A B C D$ are $a$ and $b$, respectively, and semicircles are drawn inside the rhombus with each side as the diameter. Then the area of the petal-shaped region (the shaded area) formed by the 4 semicircular arcs is $\qquad$ | 12. $\frac{1}{2} \pi\left(a^{2}+b^{2}\right)-\frac{a b}{2}$.
From the problem, we can find the side length of the rhombus to be
$$
\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}}=\frac{\sqrt{a^{2}+b^{2}}}{2}
$$
Therefore, $S_{\text {field }}=4 S_{\text {half circle }}-S_{\text {rhombus }}$
$$
\begin{... | \frac{1}{2} \pi\left(a^{2}+b^{2}\right)-\frac{a b}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,098 |
14. In the Cartesian coordinate system, there is a square with vertices at $A(-1,-1)$, $B(1$, $-1)$, $C(1,1)$, and $D(-1,1)$. Let the area of the part of the square above the broken line $y=|x-a|+a$ be $S$. Try to find the functional relationship of $S$ with respect to $a$, and draw the graph of $S$. | 14. (1) When $a \geqslant 1$, the graph of $y=|x-a|+a$ has no common parts with the square $A B C D$, so $S=0$.
(2) When $0 \leqslant a \leqslant 1$,
$$
S=\frac{1}{2} \times(1-a) \times 2(1-a)=(1-a)^{2} .
$$
(3) When $-1 \leqslant a<0$,
$$
S=2-\frac{2(1-|a|)(1-|a|)}{2}=2-(1+a)^{2} \text {. }
$$
(4) When $a<-1$, $S=2$.
... | S=\left\{\begin{array}{ll}
0, & a \geqslant 1, \\
(1-a)^{2}, & 0 \leqslant a<1, \\
2-(1+a)^{2}, & -1 \leqslant a<0, \\
2, & a<-1 .
\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,100 |
15. A coin collector wants to exchange a 3.50 yuan banknote for 1 fen, 2 fen, and 5 fen coins. He requires the total number of coins to be 150, and each type of coin must be at least 20. Additionally, the number of 5 fen coins must be more than the number of 2 fen coins. Please design an exchange plan accordingly. | 15. Let the number of 1 cent, 2 cent, and 5 cent coins be \( x \) coins, \( y \) coins, and \( z \) coins, respectively. According to the problem, we have:
\[
\left\{\begin{array}{l}
x+y+z=150, \\
x+2 y+5 z=350, \\
z>y, \\
x \geqslant 20, y \geqslant 20, z \geqslant 20 .
\end{array}\right.
\]
From (1) and (2), we get:... | x=73, y=36, z=41; x=76, y=32, z=42; x=79, y=28, z=43; x=82, y=24, z=44; x=85, y=20, z=45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,102 |
2. Given that $a, b$ are real numbers, $x=a^{2}+b^{2}+20, y=$ $4(2 b-a)$. Then the relationship between $x, y$ is ( ).
(A) $x \leqslant y$
(B) $x \geqslant y$
(C) $xy$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.B.
$$
\begin{array}{c}
\text { Since } x-y=a^{2}+b^{2}+20-4(2 b-a) \\
\quad=(a+2)^{2}+(b-4)^{2} \geqslant 0,
\end{array}
$$
thus $x \geqslant y$. | B | Algebra | proof | Yes | Yes | cn_contest | false | 714,104 |
3. There are four water pools, Jia, Yi, Bing, and Ding, each containing the same amount of water. The following changes are made:
In Jia pool, first add 10% of the pool's water volume, then release 5% of the water volume after the addition;
In Yi pool, first add 9% of the pool's water volume, then release 4% of the w... | 3.D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,105 |
4. Given $-1<a<0,|x-a|=a^{2}$. Then the range of values for $x$ is ( ).
(A) $-1<x<0$
(B) $0<x<1$
(C) $-2<x<0$
(D) $0<x<2$ | 4.C.
From the problem, we know that $x \neq a$.
When $x > a$, we get $x - a = a^2$, which means $x = a^2 + a$.
When $-1 < a < 0$, we have $-\frac{1}{4} < x < 0$.
When $x < a$, we get $a - x = a^2$, which means $x = -a^2 + a$.
When $-1 < a < 0$, we have $-2 < x < 0$.
In summary, we get $-2 < x < 0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,106 |
5. As shown in Figure 1, given $\triangle A B C$, a tangent line to the circumcircle through point $A$ intersects the extension of $B C$ at point $P, \frac{P C}{P A}=\frac{\sqrt{2}}{2}$, point $D$ is on $A C$ such that $\frac{A D}{C D}=\frac{1}{2}$, and the extension of $P D$ intersects $A B$ at point $E$. Then the val... | 5.A.
As shown in Figure 4, since $\angle P A C = \angle B$, then $\triangle P A C \backsim \triangle P B A$.
Therefore, $\frac{S_{\triangle P A C}}{S_{\triangle P B A}} = \frac{P C^{2}}{P A^{2}} = \frac{1}{2}$.
Also, $\frac{A D}{C D} = \frac{1}{2}$, then
$$
\begin{aligned}
\frac{S_{\triangle P A C}}{S_{\triangle P A}}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,107 |
1. When $-1 \leqslant x \leqslant 2$, the function $y=(k-2)x + 2|k|-1$ is always positive for $x$. Then the range of values for $k$ is | $$
\text { II. } 1 . k>\frac{5}{4} \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ 2 - k + 2 | k | - 1 > 0 , } \\
{ 2 k - 4 + 2 | k | - 1 > 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
2|k|-k+1>0, \\
2|k|+2 k-5>0 .
\end{array}\right.\right.
$$
When $k \geqslant 0$, we get $k>\fra... | k>\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,108 |
2. The solution to the equation $1-\frac{2}{x}-\frac{1}{x^{2}}=2 x+x^{2}$ is | 2. $\frac{-3 \pm \sqrt{5}}{2}$.
The original equation is $\left(x+\frac{1}{x}\right)^{2}+2\left(x+\frac{1}{x}\right)-3=0$. Let $x+\frac{1}{x}=t$, then we get $t^{2}+2 t-3=0$, which is $(t+3)(t-1)=0, t_{1}=1, t_{2}=-3$. When $t=1$, we get $x+\frac{1}{x}=1$, which has no solution; when $t=-3$, we get $x+\frac{1}{x}=-3, ... | \frac{-3 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,109 |
3. Given the equation about $x$: $x^{3}+(1-a) x^{2}-2 a x+a^{2}$ $=0$ has only one real root. Then the range of the real number $a$ is $\qquad$. | 3. $a<-\frac{1}{4}$.
The original equation is transformed into $(x-a)\left(x^{2}+x-a\right)=0$.
Thus, $x=a$ or $x^{2}+x-a=0$.
Since $x=a$ is the only real root of the equation, the equation $x^{2}+x-a=0$ has no real roots. Therefore, $\Delta=1+4a<0, a<-\frac{1}{4}$. | a<-\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,110 |
4. Given a right-angled triangle with side lengths as positive integers, its perimeter is 80. Then the lengths of the three sides are $\qquad$ . | 4. $30,16,34$.
Let the three sides of a right triangle be $a, b, c$. From the problem, we have $a^{2}+b^{2}=c^{2}$ and $a+b+c=80$.
Substituting $c=80-a-b$ into $a^{2}+b^{2}=c^{2}$, we get
$$
6400-80 a-80 b+a b=3200,
$$
which simplifies to $(80-a)(80-b)=3200$.
Since $0<a, b<80$, we have $(80-a)(80-b) = 50 \times 64$.
... | 30,16,34 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,111 |
5. As shown in Figure 2, given that the area of $\triangle A B C$ is $S$, $D$ is the trisection point of side $B C$, $E$ is the quadrisection point of side $A C$, and $F$ and $G$ are both the quintisection points of side $A B$. Then the area of quadrilateral $D E F G$ is $\qquad$ | 5. $\frac{29}{60} S$.
As shown in Figure 5, connect $C F$, $C G$, and $A D$. Then
$$
\begin{array}{l}
S_{\triangle M E F}=S \times \frac{1}{5} \times \frac{3}{4}=\frac{3}{20} S, \\
S_{\triangle B D C}=S \times \frac{3}{5} \times \frac{1}{3}=\frac{1}{5} S, \\
S_{\triangle C D E}=S \times \frac{2}{3} \times \frac{1}{4}=... | \frac{29}{60} S | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,113 |
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