problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 9 Given $\triangle A B C$, construct $\triangle C B D, \triangle C A E, \triangle A B F$ outside the triangle such that $\angle B A F=\angle C A E, \angle A B F=\angle C B D, \angle A C E=$ $\angle B C D$. Prove that $A D, B E, C F$ are concurrent.
(26th IMO Shortlist) | Proof: As shown in Figure 9, let
$$
\begin{array}{l}
\angle B A F=\angle C A E=\alpha, \\
\angle A B F=\angle C B D=\beta, \\
\angle B C D=\angle A C E=\gamma .
\end{array}
$$
Applying the trigonometric form of Ceva's theorem to $\triangle A B C$ with respect to points $D$, $E$, and $F$, we have
$$
\begin{array}{c}
\f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,583 |
Example 10 As shown in Figure 10, in $\triangle A B C$, $A B>A C$. A tangent line $l$ is drawn through point $A$ to the circumcircle of $\triangle A B C$,
and a circle is drawn with $A$ as the center and $A C$ as the radius, intersecting
line segment $A B$ at point $D$, and
line $l$ at points $E$ and $F$.
Prove: Lines ... | Proof: (1) First, prove that $DE$ passes through the incenter of $\triangle ABC$.
Let $AA'$ and $CC'$ be the angle bisectors of $\angle BAC$ and $\angle ACB$ in $\triangle ABC$, respectively, and connect $CD$. Then,
$$
\begin{array}{l}
\angle DAA' = \angle A'AC = \frac{1}{2} \angle BAC, \\
\angle ACC' = \angle C'CB = \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,584 |
Example 1 Given $0$ 0. Prove: $\frac{a^{2}}{t^{3}}+\frac{b^{2}}{1-t^{3}} \geqslant(a+b)^{2}$.
Analysis: The denominator in the expression to be proven implicitly contains the condition $t^{3}+\left(1-t^{3}\right)$ $=1$. Making full use of 1, the problem can be quickly proven. | Prove: $\frac{a^{2}}{t^{3}}+\frac{b^{2}}{1-t^{3}}=1 \times\left(\frac{a^{2}}{t^{3}}+\frac{b^{2}}{1-t^{3}}\right)$
$$
\begin{array}{l}
=\left[t^{3}+\left(1-t^{3}\right)\right]\left[\frac{a^{2}}{t^{3}}+\frac{b^{2}}{1-t^{3}}\right] \\
=a^{2}+b^{2}+\frac{1-t^{3}}{t^{3}} a^{2}+\frac{t^{3}}{1-t^{3}} b^{2} \\
\geqslant a^{2}+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,585 |
Example 2 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Try to prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}.
$$
Analysis: The condition contains $abc=1$, which should be fully utilized. | Prove: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then $x y z=$
1. Thus, the problem is transformed into proving
$$
\begin{array}{l}
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geqslant \frac{3}{2} . \\
\text { and }[(y+z)+(z+x)+(x+y)] . \\
{\left[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,586 |
Example $3 a_{i}, b_{i}, c_{i}>0(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\sqrt[n]{\left(a_{1}+b_{1}+c_{1}\right)\left(a_{2}+b_{2}+c_{2}\right) \cdots\left(a_{n}+b_{n}+c_{n}\right)} \\
\geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}}+\sqrt[n]{b_{1} b_{2} \cdots b_{n}}+\sqrt[n]{c_{1} c_{2} \cdots c_{n}} .
\end{array}
... | Prove: The original inequality is equivalent to
$$
\begin{array}{l}
\sqrt[n]{\frac{a_{1} a_{2} \cdots a_{n}}{\left(a_{1}+b_{1}+c_{1}\right)\left(a_{2}+b_{2}+c_{2}\right) \cdots\left(a_{n}+b_{n}+c_{n}\right)}}+ \\
\sqrt[n]{\frac{b_{1} b_{2} \cdots b_{n}}{\left(a_{1}+b_{1}+c_{1}\right)\left(a_{2}+b_{2}+c_{2}\right) \cdot... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,587 |
Example 4 Let $a, b, c > 0$. Prove:
$$
\begin{array}{l}
\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^{2} \\
\geqslant 4 \sqrt{3 a b c(a+b+c)} .
\end{array}
$$ | Proof 1: Without loss of generality, let $a+b+c=1$.
Then, the original inequality is transformed into proving
$$
\sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})+1 \geqslant 4 \sqrt{3} \cdot \sqrt{a b c} \text {, }
$$
which is equivalent to $\sqrt{a}+\sqrt{b}+\sqrt{c}+\frac{1}{\sqrt{a b c}} \geqslant 4 \sqrt{3}$.
And $\sqrt{a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,589 |
Example 5 Let $x, y, z$ be positive real numbers. If $xyz \geqslant 1$, prove:
$$
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+x^{2}+z^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0 \text {. }
$$
(46th IMO) | Prove: Since $y^{4}+z^{4}-y z\left(y^{2}+z^{2}\right)$
$$
=(y-z)^{2}\left(y^{2}+y z+z^{2}\right) \geqslant 0 \text {, }
$$
then $y^{4}+z^{4} \geqslant y z\left(y^{2}+z^{2}\right)$.
Since $y^{2}+z^{2} \leqslant x y z\left(y^{2}+z^{2}\right) \leqslant x\left(y^{4}+z^{4}\right)$,
$$
\begin{array}{l}
\text { then } \frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,590 |
Example 1 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}=$ 1. Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+3$. | Proof: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}+3$
$$
\begin{aligned}
\Leftrightarrow & \frac{1-a^{2}}{a^{2}}+\frac{1-b^{2}}{b^{2}}+\frac{1-c^{2}}{c^{2}} \geqslant \frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c} \\
\Leftrightarrow & \frac{\left(a^{2}+b^{2}+c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,591 |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1.
$$
Analysis: The inequality to be proven is a non-homogeneous inequality. Given $abc=1$, we can consider the substitution $a=\frac{x}{y}, b=\frac{y}{z}, c... | Prove: Let \( a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x} \) (where \( x, y, z > 0 \)). Then, the original inequality can be transformed into a homogeneous inequality:
$$
(x-y+z)(y-z+x)(z-x+y) \leqslant x y z .
$$
Let \( u=x-y+z, v=y-z+x, w=z-x+y \). Since the sum of any two of these numbers is positive, \( u, v, w \)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,593 |
Example 4 Given $x, y, z \in \mathbf{R}_{+}, xyz=1$, and
$$
x(1+z)>1, y(1+x)>1, z(1+y)>1 \text{. }
$$
Prove: $2(x+y+z) \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$. | Prove: Let \( x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a} \) (\( a, b, c \in \mathbf{R}_{+} \)), then the known inequality becomes
\[
a+c>b, b+a>c, c+b>a.
\]
Thus, the inequality to be proven becomes
\[
\begin{array}{l}
2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geqslant \frac{b}{a}+\frac{c}{b}+\frac{a}{c}+3 \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,594 |
Let positive numbers $a, b, c, x, y, z$ satisfy $c y + b z = a$, $a z + c x = b$, $b x + a y = c$.
Find the minimum value of the function $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$.
(2005, National High School Mathematics Competition) | Solution 1: From the known three equations, by eliminating $y$ and $z$, we get
$x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}>0$.
Thus, $b+c=\sqrt{b^{2}+c^{2}+2 b c}>\sqrt{b^{2}+c^{2}}>c$
Similarly, by eliminating $x, z$ or $x, y$, we can also obtain
$\left\{\begin{array}{l}y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}>0, \\ a+c>b\end{array}\... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,596 |
Proposition In a bicentric quadrilateral $ABCD$, if its circumradius is $R$, area is $S$, and inradius is $r$, then
$$
\frac{16 r^{2}}{S} \leqslant \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}+\cot \frac{D}{2} \leqslant \frac{8 R^{2}}{S} .
$$ | Prove: Let $A B=a, B C=b, C D=c, D A=$ $d, p=\frac{1}{2}(a+b+c+d)$. Then
$$
\begin{array}{l}
r \cot \frac{A}{2}+r \cot \frac{B}{2}=a, r \cot \frac{B}{2}+r \cot \frac{C}{2}=b, \\
r \cot \frac{C}{2}+r \cot \frac{D}{2}=c, r \cot \frac{D}{2}+r \cot \frac{A}{2}=d .
\end{array}
$$
That is, $\cot \frac{A}{2}+\cot \frac{B}{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,597 |
Given $0<\alpha<\beta<\gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+\sin ^{3} \gamma=1$.
Prove: $\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2}$.
(2nd China Southeast Mathematical Olympiad)
The original solution pointed out that the equality cannot be achieved. Then, doe... | Proof: Let $x=\sin ^{3} \alpha, y=\sin ^{3} \beta, z=\sin ^{3} \gamma$.
Then equation (1) is equivalent to
$x, y, z > 0$, and $x+y+z=1$. Prove:
$$
\frac{x^{\frac{2}{3}}}{1-x^{\frac{2}{3}}}+\frac{y^{\frac{2}{3}}}{1-y^{\frac{2}{3}}}+\frac{z^{\frac{2}{3}}}{1-z^{\frac{2}{3}}} \geqslant \frac{3}{\sqrt[3]{9}-1}.
$$
In fact,... | \frac{3}{\sqrt[3]{9}-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,598 |
Example 4 As shown in Figure 4(a), the four sides of convex quadrilateral $ABCD$ are all tangent to $\odot O$, with points of tangency being $P, M, Q, N$ respectively. Let $PQ$ and $MN$ intersect at $S$. Prove: $A, S, C$ are collinear.
Figure 4 | Let's first look at $P Q$. As shown in Figure 4(b), let $P Q$ intersect $A C$ at $S'$. It is easy to prove that $\angle A P S'$ and $\angle C Q S'$ are supplementary.
And $\angle A S' P = \angle C S' Q$, then
$\frac{A S'}{A P} = \frac{\sin \angle A P S'}{\sin \angle A S' P} = \frac{\sin \angle C Q S'}{\sin \angle C S' ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,599 |
Given real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}+a_{2}+\cdots+a_{n}=0$.
Prove: $\max _{1 \leqslant k \leqslant n}\left(a_{k}^{2}\right) \leqslant \frac{n}{3} \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$.
(Zhu Huawei) | For any $1 \leqslant k \leqslant n$, prove
$$
a_{k}^{2} \leqslant \frac{n}{3} \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2} \text {. }
$$
Let $d_{k}=a_{k}-a_{k+1}, k=1,2, \cdots, n-1$. Then
$$
\begin{array}{l}
a_{k}=a_{k}, \\
a_{k+1}=a_{k}-d_{k}, a_{k+2}=a_{k}-d_{k}-d_{k+1}, \cdots \cdots, \\
a_{n}=a_{k}-d_{k}-d_{k+1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,600 |
Two positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (which can be the same) are such that $\frac{a_{1}}{a_{2}}, \frac{a_{2}}{a_{3}}, \cdots, \frac{a_{2005}}{a_{2006}}$ are all distinct. How many different numbers are there at least among $a_{1}$, $a_{2}, \cdots, a_{2006}$?
（Chen Yonggao, problem contributor） | Due to the fact that the pairwise ratios of 45 distinct positive integers are at most $45 \times 44 + 1 = 1981$, the number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is greater than 45.
Below is an example to show that 46 can be achieved.
Let $p_{1}, p_{2}, \cdots, p_{46}$ be 46 distinct prime numbers, an... | 46 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,601 |
In right $\triangle ABC$, $\angle ACB=90^{\circ}$, the incircle $\odot O$ of $\triangle ABC$ is tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $AD$, intersecting the incircle $\odot O$ at point $P$. Connect $BP$ and $CP$. If $\angle BPC=90^{\circ}$, prove that $AE + AP = PD$.
(... | Let $A E=A F=x, B D=B F=y, C D=C E=z$, $A P=m, P D=n$.
Since $\angle A C P+\angle P C B=90^{\circ}=\angle P B C+\angle P C B$, therefore, $\angle A C P=\angle P B C$.
As shown in Figure 1, extend $A D$ to $Q$ such that
$$
\begin{array}{l}
\angle A Q C \\
=\angle A C P \\
=\angle P B C,
\end{array}
$$
Connect $B Q, C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,603 |
Five, the sequence of real numbers $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=\frac{1}{2}, a_{k+1}=-a_{k}+\frac{1}{2-a_{k}}, k=1,2, \cdots \text {. }
$$
Prove: $\left[\frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right]^{n}$
$$
\leqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n}\left(\frac{1}{a_{1}}-1\ri... | $$
\begin{array}{l}
0 < a_{1} < a_{2} < \cdots < a_{n} < \frac{1}{2} \\
\Rightarrow\left(\frac{1}{a_{1}}-1\right)\left(\frac{1}{a_{2}}-1\right) \cdots\left(\frac{1}{a_{n}}-1\right) \geqslant\left(\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}-1\right)^{n} .
\end{array}
$$
Therefore, the proposition holds for $n+1$.
The original p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,604 |
Six, let $X$ be a 56-element set. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any 7 of them has at least $n$ elements, then there must exist 3 of these 15 subsets whose intersection is non-empty.
(Cold Gangsong provided the problem) | Six, the minimum value of $n$ is 41.
First, prove that $n=41$ meets the requirement. Use proof by contradiction.
Assume there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Since each element belongs to at most 2 subsets, we ca... | 41 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,605 |
1. Given $m>1, a=\sqrt{m+1}-\sqrt{m}, b=$ $\sqrt{m}-\sqrt{m-1}$. Then, ( ).
(A) $a>b$
(B) $a<b$
(C) $a=b$
(D) The size of $a$ and $b$ depends on the value of $m$ | -.1.B.
Since $a=\frac{1}{\sqrt{m+1}+\sqrt{m}}, b=\frac{1}{\sqrt{m}+\sqrt{m-1}}$, and $\sqrt{m+1}>\sqrt{m}>\sqrt{m-1}>0$, then $a<b$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,606 |
2. In the expansion of $\left(x^{2}+3 x+2\right)^{5}$, the coefficient of the $x$ term is ( ).
(A) 160
(B) 240
(C) 360
(D) 800 | 2.B.
In the expansion of $\left(x^{2}+3 x+2\right)^{5}$, to obtain the term containing $x$, 4 factors take 2, and one factor takes the $3 x$ term, so the coefficient of $x$ is $\mathrm{C}_{5}^{4} \times 3 \times 2^{4}=240$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,607 |
3. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, points $E$ and $F$ are on $A B_{1}$ and $B C_{1}$ respectively (not coinciding with the endpoints of the segments), and $A E=B F$. Then, among the following 4 conclusions:
(1) $A A_{1} \perp E F$; (2) $A_{1} C_{1} / / E F$; (3) $E F / /$ plane $A_{... | 3.D.
As shown in Figure 2, draw $E P \perp A B$ at point $P$, $F Q \perp B C$ at point $Q$, and connect $P Q$. It is easy to see from the given conditions that $P E \xlongequal{\|} Q F$.
Since $P E \perp$ plane $A C$, then $P E \perp P Q$, which means quadrilateral $P E F Q$ is a rectangle.
Therefore, $A A_{1} \perp... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,608 |
Example 5 As shown in Figure 5, three equal circles
$\odot O_{1}$,
$\odot O_{2}$,
$\odot O_{3}$ inside $\triangle ABC$ intersect each other and all pass through point $P$, and each pair of circles
are tangent to one side of $\triangle ABC$. Given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$... | Connect $\mathrm{O}_{1} \mathrm{O}_{2}, \mathrm{O}_{1} \mathrm{O}_{3}, \mathrm{O}_{2} \mathrm{O}_{3}$. From the given, we have $O_{1} O_{2} / / A B, O_{2} O_{3} / / B C, O_{3} O_{1} / / C A$. It can be concluded that $\triangle O_{1} O_{2} O_{3}$ and $\triangle A B C$ are a pair of homothetic triangles, and it is easy ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,610 |
5. If a positive integer with at least two digits is such that all digits except the leftmost one are smaller than the digit to their left, then such a positive integer is called a "good number". Then, the total number of all such good numbers is ( ).
(A) 1013
(B) 1011
(C) 1010
(D) 1001 | 5.A.
First, consider the number of good numbers with $k(2 \leqslant k \leqslant 10)$ digits is $\mathrm{C}_{10}^{k}$. Therefore, there are $\sum_{k=2}^{10} \mathrm{C}_{10}^{k}=\sum_{k=0}^{10} \mathrm{C}_{10}^{k}-\mathrm{C}_{10}^{1}-\mathrm{C}_{10}^{0}=1013$ in total. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,611 |
6. Let set $A=\{-2,0,1\}, B=\{1,2,3,4, 5\}$, and the mapping $f: A \rightarrow B$ such that for any $x \in A$, $x + f(x) + x f(x)$ is an odd number. Then the number of such mappings $f$ is ( ).
(A) 45
(B) 27
(C) 15
(D) 11 | 6.A.
When $x=-2$, $x+f(x)+x f(x)=-2-f(-2)$ is odd, then $f(-2)$ can take $1,3,5$, which gives 3 possible values;
When $x=0$, $x+f(x)+x f(x)=f(0)$ is odd, then $f(0)$ can take $1,3,5$, which gives 3 possible values;
When $x=1$, $x+f(x)+x f(x)=1+2 f(1)$ is odd, then $f(1)$ can take $1,2,3,4,5$, which gives 5 possible ... | 45 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,612 |
7. The inverse function of $y=f(x)$ is $y=f^{-1}(x), y=f(x-1)$ passes through the point $(3,3)$. Then the graph of the function $y=f^{-1}(x+2)$ must pass through the point | $$
=.7 .(1,2) \text {. }
$$
From the problem, we have $f^{-1}(3)=2$, which means $f^{-1}(1+2)=2$. Therefore, it passes through the point $(1,2)$. | (1,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,613 |
8. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the right vertex is $A$, the upper vertex is $B$, the left focus is $F$, and $\angle A B F=$ $90^{\circ}$. Then the eccentricity of the ellipse is . $\qquad$ | 8. $\frac{\sqrt{5}-1}{2}$.
From $B F^{2}+A B^{2}=A F^{2}$, we have
$$
\begin{array}{l}
c^{2}+b^{2}+b^{2}+a^{2}=(a+c)^{2} . \\
\text { Then }\left(\frac{c}{a}\right)^{2}+\frac{c}{a}-1=0 .
\end{array}
$$
Solving, we get $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$. | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,614 |
9. In a regular tetrahedron $S-ABC$, $M$ and $N$ are the midpoints of $SB$ and $SC$ respectively. If the section $AMN \perp$ side face $SBC$, then the dihedral angle between the side face and the base of the tetrahedron is $\qquad$ | 9. $\arccos \frac{\sqrt{6}}{6}$.
Since plane $A M N \perp$ plane $S B C$, take the midpoint $G$ of $M N$, and connect $A G$, then $A G \perp M N$. Take the midpoint $D$ of $B C$, and connect $S D, A D$, then $\angle S D A$ is the plane angle of the dihedral angle to be found. Let the side length of the base be $a$, t... | \arccos \frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,615 |
10. Given the set $M=\left\{x \left\lvert\, x=\lim _{n \rightarrow \infty} \frac{2^{n+1}-2}{\lambda^{n}+2^{n}}\right.\right.$, $\lambda$ is a constant, and $\lambda+2 \neq 0\}$. Then the sum of all elements of $M$ is $\qquad$ . | 10.3.
When $|\lambda|>2$, $x=\lim _{n \rightarrow \infty} \frac{2\left(\frac{2}{\lambda}\right)^{n}-2\left(\frac{1}{\lambda}\right)^{n}}{1+\left(\frac{2}{\lambda}\right)^{n}}=0$;
When $\lambda=2$, $x=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^{n}}\right)=1$;
When $|\lambda|<2$, $x=\lim _{n \rightarrow \infty} \fra... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,616 |
11. Let the function
$$
f(x)=4 \sin x \cdot \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+\cos 2 x .
$$
If $|f(x)-m|<2$ holds for $\frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3}$, then the range of the real number $m$ is $\qquad$ | 11. $(1,4)$.
$$
\begin{array}{l}
f(x)=4 \sin x \cdot \frac{1-\cos \left(\frac{\pi}{2}+x\right)}{2}+\cos 2 x \\
=2 \sin x(1+\sin x)+1-2 \sin ^{2} x=1+2 \sin x .
\end{array}
$$
When $\frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3}$, $|f(x)-m|<2$ always holds, which means $f(x)-2<m<f(x)+2$ always holds. Therefore, we... | (1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,617 |
Three, 13. (20 points) Given the function
$$
f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}),
$$
when $x \in[-1,1]$, $|f(x)| \leqslant 1$.
(1) Prove: $|b| \leqslant 1$;
(2) If $f(0)=-1, f(1)=1$, find the value of $a$. | $$
\text { Three, 13. (1) Since } f(1)=a+b+c, f(-1)=a-b+
$$
$c$, then $b=\frac{1}{2}(f(1)-f(-1))$.
By the problem, $|f(1)| \leqslant 1,|f(-1)| \leqslant 1$.
Thus, $|b|=\frac{1}{2}|f(1)-f(-1)|$
$$
\leqslant \frac{1}{2}(|f(1)|+|f(-1)|) \leqslant 1 \text {. }
$$
(2) From $f(0)=-1, f(1)=1$, we get $c=-1, b=2$ $\boldsymbol{... | a=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,619 |
14. (20 points) Given a constant $a>0$, vectors $\boldsymbol{p}=(1, 0)$, $\boldsymbol{q}=(0, a)$, a line passing through the fixed point $M(0, -a)$ with direction vector $\lambda \boldsymbol{p} + \boldsymbol{q}$ intersects with a line passing through the fixed point $N(0, a)$ with direction vector $\boldsymbol{p} + 2\l... | 14. (1) Let $R(x, y)$, then
$\boldsymbol{M R}=(x, y+a), \boldsymbol{N R}=(x, y-a)$.
Also, $\lambda \boldsymbol{p}+\boldsymbol{q}=(\lambda, a), \boldsymbol{p}+2 \lambda \boldsymbol{q}=(1,2 \lambda a)$, and
$\boldsymbol{M R} / /(\lambda p+q), N R / /(p+2 \lambda q)$,
thus $\left\{\begin{array}{l}\lambda(y+a)=a x, \\ y-a=... | \boldsymbol{F A} \cdot \boldsymbol{F B} \in\left(-\infty,-\frac{1}{2}\right] \cup\left(\frac{1}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,620 |
Example 6 As shown in Figure 6, in trapezoid $A B C D$, $D C / / A B$. For three points $P_{1}, P_{2}$, and $P_{3}$ inside the trapezoid, if the sum of the distances to the four sides is equal, then $P_{1}, P_{2}$, and $P_{3}$ are collinear. Prove it.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 6 As shown in Figure 6, in trapezo... | Consider points $P_{1}$ and $P_{2}$. Let the line $P_{1} P_{2}$ intersect $A D$ and $B C$ at points $M$ and $N$, respectively. Let $P_{1} E_{1} \perp B C$ at $E_{1}$, $P_{2} E_{2} \perp B C$ at $E_{2}$, $P_{1} F_{1} \perp A D$ at $F_{1}$, and $P_{2} F_{2} \perp A D$ at $F_{2}$.
Since $D C \parallel A D$, the sum of th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,621 |
1. Condition $p:|x+1|>2$, Condition $q: \frac{1}{3-x}>1$. Then $\neg p$ is $\neg q$'s ( ).
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient nor necessary condition | - 1.A.
From condition $p$, we get $\neg:-3 \leqslant x \leqslant 1$;
From condition $q$, we get ᄀ $q: x \leqslant 2$ or $x \geqslant 3$. | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,623 |
2. A table of positive integers is shown in Table 1 (the number of elements in the next row is twice the number of elements in the previous row). Then the 5th number in the 8th row is ( ).
(A) 68
(B) 132
(C) 133
(D) 260
Table 1
\begin{tabular}{|c|cccc|}
\hline First Row & 1 & & & \\
\hline Second Row & 2 & 3 & & \\
\hl... | 2.B.
From the problem, we know that the number of digits in the $n$-th row is $2^{n-1}$, and starting from the second row, the first number of each row is $2^{n-1}$. Therefore, the first number in the eighth row is $2^{7}=128$. Hence, the fifth number in this row is 132. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,624 |
3. If the solution set of the inequality $a x^{2}+b x+4>0$ is $\{x \mid-2<x<1\}$, then the maximum and minimum values of the quadratic function $y=b x^{2}+4 x+$ $a$ on the interval $[0,3]$ are ( ).
(A) $0,-8$
(B) $0,-4$
(C) 4,0
(D) 8,0 | 3. A.
From the problem, we know that $a<0$ and the quadratic equation $a x^{2}+b x+4=0$ has two roots, -2 and 1. Therefore, $-\frac{b}{a}=-1, \frac{4}{a}=-2$. Hence, $a=-2, b=-2$. So, the quadratic function $y=b x^{2}+4 x+a$ has a maximum value of 0 and a minimum value of -8 on the interval $[0,3]$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,625 |
4. As shown in Figure $1, f(x)$ is a function defined on the interval $[-4$, 4] and is an odd function. Let $g(x)=a f(x)+b$. Then, among the following statements about $g(x)$, the correct one is ( ).
(A) If $a<0$, then the graph of the function $g(x)$ is symmetric with respect to the origin
(B) If $a=-1,-2<b<0$, then t... | 4.B.
If $a=-1$, the graph of $f(x)$ is reflected over the $x$-axis to get $-f(x)$; if $-2<b<0$, the graph of $-f(x)$ is translated downward along the $y$-axis by $|b|$ units, then there are real roots less than 2. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,626 |
In $\triangle A B C$, let
$$
\begin{array}{l}
x=\cos A+\cos B+\cos C, \\
y=\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} .
\end{array}
$$
Then the relationship between $x$ and $y$ is ( ).
(A) $x=y$
(B) $x \geqslant y$
(C) $x \leqslant y$
(D) cannot be determined | 5.C.
From the conditions, we have
$$
\begin{array}{l}
\cos A+\cos B=2 \cos \frac{\pi-C}{2} \cdot \cos \frac{A-B}{2} \\
=2 \sin \frac{C}{2} \cdot \cos \frac{A-B}{2} \leqslant 2 \sin \frac{C}{2} .
\end{array}
$$
Similarly, $\cos B+\cos C \leqslant 2 \sin \frac{A}{2}$,
$$
\cos C+\cos A \leqslant 2 \sin \frac{B}{2} \text... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,627 |
6. Three people, A, B, and C, practice passing a ball, making a total of $n$ passes. The ball starts from A and the $n$-th pass is still to A, with a total of $a_{n}$ different ways to pass the ball. Let the sequence of passing methods be $\left\{a_{n}\right\}$. Then the general term formula for the sequence $\left\{a_... | 6.C.
The number of ways the ball is passed to A on the $n$th pass is $a_{n}$. Let the number of ways the ball is not passed to A on the $n$th pass be $b_{n}$. For A, since there are 2 ways to pass the ball each time, there are $2^{n}$ ways for $n$ passes in total. Therefore, $a_{n}+b_{n}=2^{n}$.
Additionally, if the ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,628 |
1. If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is $\qquad$ | $=\sqrt{1} \cdot \sqrt{3}$.
From the problem, we have
$$
\left\{\begin{array}{l}
x + 2 y > 0, \\
x - 2 y > 0, \\
(x + 2 y)(x - 2 y) = 4.
\end{array} \text { Then } \left\{\begin{array}{l}
x > 2|y| \geqslant 0, \\
x^{2} - 4 y^{2} = 4.
\end{array}\right.\right.
$$
By its symmetry, we only need to consider $y \geqslant 0... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,629 |
2. The function $f(x)=x^{2}-t x+2$ has an inverse function on $[-1,2]$. Then the range of all possible values of the real number $t$ is $\qquad$ . | $$
\text { 2. }(-\infty,-2] \cup[4,+\infty) \text {. }
$$
According to the given conditions, the axis of symmetry is $\frac{t}{2} \notin(-1,2)$. | (-\infty,-2] \cup[4,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,630 |
3. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}(n \geqslant 2) \text {. }
$$
Then the value of $a_{2006}$ is $\qquad$ . | 3.2.
Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=$ 2, we conjecture that $a_{n}$ is a periodic sequence with a period of 4. And
$$
a_{n+4}=\frac{1+a_{n+3}}{1-a_{n+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=a_{n} .
$$
Therefore, $a_{2008}=a_{4 ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,631 |
Example 7 As shown in Figure 7, the convex quadrilateral $ABCD$ is inscribed in a circle. Extend $AD$ and $BC$ to intersect at point $P$, draw $PE$ and $PF$ tangent to the circle at $E$ and $F$, and let $AC$ and $BD$ intersect at $K$. Prove: $E$, $K$, and $F$ are collinear. | Connect $A E$, $E D$, $C F$, $F B$, to get the convex hexagon $A B F C D E$.
To prove that points $E$, $K$, $F$ are collinear, it is sufficient to show that $A C$, $B D$, and $E F$ are concurrent, which only requires proving
$$
A B \cdot F C \cdot D E = B F \cdot C D \cdot E A \text{. }
$$
Notice that $\triangle P A B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,632 |
$$
\begin{array}{l}
\text { 4. Let } \theta \in\left[0, \frac{\pi}{2}\right] \text {, and } \\
\cos ^{2} \theta+2 m \sin \theta-2 m-1<0
\end{array}
$$
Always holds. Then the range of $m$ is $\qquad$ | 4. $m>0$.
Taking $m$ as the main variable. From the condition, $2 m(\sin \theta-1)<\frac{\sin ^{2} \theta}{\sin \theta-1}$ always holds, it is only necessary that $2 m>t$, where $t$ is the maximum value of $\frac{\sin ^{2} \theta}{\sin \theta-1}$.
And $\frac{\sin ^{2} \theta}{\sin \theta-1}=\sin \theta-1+\frac{1}{\sin... | m>0 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,633 |
5. In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5, 5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$, such that for all integers $n$, $a_{n}=$ $b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. The... | 5.2.
Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$
Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, sol... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,634 |
6. Let set $M$ be the collection of all functions $f(x)$ that satisfy the following property: there exists a non-zero constant $T$, such that for any $x \in \mathbf{R}$, $f(x+T)=T f(x)$ holds. If the function $f(x)=\sin k x \in M$, then the range of the real number $k$ is $\qquad$ | 6. $k=m \pi, m \in \mathbf{Z}$.
When $k=0$, $f(x)=0 \in M$.
When $k \neq 0$, since $f(x)=\sin k x \in M$, there exists a non-zero constant $T$, for any $x \in \mathbf{R}$, such that $\sin (k x+k T)=T \sin k x$. When $x=0$, $\sin k T=0$, then for any $x \in \mathbf{R}$, $\cos k T \cdot \sin k x=T \sin k x$.
Thus $T=\co... | k=m \pi, m \in \mathbf{Z} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,635 |
Three. (20 points) Let the domain of the function $f(x)$ be $(0,+\infty)$, and for any positive real numbers $x, y$, the following always holds:
$$
f(x y)=f(x)+f(y)
$$
It is known that $f(2)=1$, and when $x>1$, $f(x)>0$.
(1) Find the value of $f\left(\frac{1}{2}\right)$;
(2) Determine the monotonicity of $y=f(x)$ on $... | (1) Let $x=y=1$, then $f(1)=f(1)+f(1)$, so $f(1)=0$.
Let $x=2, y=\frac{1}{2}$, then $f(1)=f(2)+f\left(\frac{1}{2}\right)$, so $f\left(\frac{1}{2}\right)=-1$.
(2) Suppose $0 < x_1 < x_2$.
Then $f\left(\frac{x_{1}}{x_{2}}\right)>0$, which means $f\left(x_{1}\right)>f\left(x_{2}\right)$.
Therefore, $y=f(x)$ is monotonical... | S_{n}=\frac{n(n+1)}{2}, a_{n}=n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,636 |
Four. (20 points) Given acute angles $\alpha, \beta$ satisfy
$\sin \beta=m \cos (\alpha+\beta) \cdot \sin \alpha\left(m>0, \alpha+\beta \neq \frac{\pi}{2}\right)$.
If $x=\tan \alpha, y=\tan \beta$,
(1) Find the expression for $y=f(x)$;
(2) Under (1), when $\alpha \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$, find the ... | (1) From $\sin \beta=m \cos (\alpha+\beta) \cdot \sin \alpha(m>0$, $\alpha+\beta \neq \frac{\pi}{2}$ ), we have
$$
\sin [(\alpha+\beta)-\alpha]=m \cos (\alpha+\beta) \cdot \sin \alpha,
$$
which means
$$
\sin (\alpha+\beta) \cdot \cos \alpha=(m+1) \cos (\alpha+\beta) \cdot \sin \alpha.
$$
Since $\alpha, \beta$ are acu... | \frac{m}{m+2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,637 |
Five. (20 points) $\triangle A B C$ is an inscribed triangle in $\odot O$, with $A B > A C > B C$. Point $D$ is on the minor arc $\overparen{B C}$, and perpendiculars from point $O$ to $A B$ and $A C$ intersect $A D$ at points $E$ and $F$, respectively. The rays $B E$ and $C F$ intersect at point $P$. If $P B = P C + P... | Five, connect $O B, O C$, we know that $\triangle E A B, \triangle F A C$ are both isosceles triangles, and
$$
\begin{array}{l}
\angle B P C=\angle A E P+\angle C F D=2 \angle B A D+2 \angle C A D \\
=2 \angle B A C=\angle B O C .
\end{array}
$$
Therefore, $B, C, P, O$ are concyclic.
By Ptolemy's theorem, we have
$$
P... | 30^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 716,638 |
Six. (20 points) Let $M=\{f(x) \mid f(x)=a \cos x + b \sin x, a, b$ are constants $\}, F$: maps any point $(a, b)$ in the plane to the function $a \cos x + b \sin x$.
(1) Prove: There do not exist two different points corresponding to the same function;
(2) Prove: When $f_{0}(x) \in M$, $f_{1}(x) = f_{0}(x+t) \in M, t$... | (1) Suppose there are two different points $(a, b),(c, d)$ corresponding to the same function, i.e.,
$$
\begin{array}{l}
F(a, b)=a \cos x+b \sin x, \\
F(c, d)=c \cos x+d \sin x
\end{array}
$$
are the same, which means
$$
a \cos x+b \sin x=c \cos x+d \sin x
$$
holds for all $x \in \mathbf{R}$.
Let $x=0$, we have $a=c$... | m^{2}+n^{2}=a_{0}^{2}+b_{0}^{2} | Algebra | proof | Yes | Yes | cn_contest | false | 716,639 |
6.2. In a $2 \times 2$ grid, 4 positive integers are filled in, where the difference between any two numbers in the same column is 6, and the difference between any two numbers in the same row is 1 times (i.e., one number is twice the other). Try to determine: which numbers are filled in the grid? Can there be differen... | 6.2, there is only one way to fill:
In the grid, there are only 6 and 12, as shown in Figure 2(a).
\begin{tabular}{|c|c|}
\hline 6 & 12 \\
\hline 12 & 6 \\
\hline
\end{tabular}
(a)
\begin{tabular}{|c|c|}
\hline$x$ & $2 x$ \\
\hline$x \pm 6$ & $2 x \pm 6$ \\
\hline
\end{tabular}
(b)
Figure 2
The following proves that ... | 6 \text{ and } 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,641 |
6.3. Along the riverbank, there are eight types of plants growing, and the number of berries on adjacent plants differs by 1. Can the eight plants have a total of 225 berries? Explain your reasoning. | 6.3. Since the number of berries on any two adjacent plants differs by 1, the sum of the berries on any two adjacent plants is always odd. This means that the total number of berries on the eight plants is the sum of 4 odd numbers, which must be even. Therefore, it is impossible for them to have a total of 225 berries. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,642 |
Example 1 Let square $P Q R S$ be inscribed in $\triangle A B C$, with its vertices $P$ and $Q$ on side $B C$, and vertices $R$ and $S$ on sides $C A$ and $A B$, respectively, and let its center be $A_{1}$. Similarly, define the centers of the inscribed squares with two vertices on sides $C A$ and $A B$ as $B_{1}$ and ... | Proof: As shown in Figure 1, connect $S A_{1}$ and $R A_{1}$. Thus, $S A_{1} = R A_{1}$. By the Law of Sines, we have
$$
\begin{array}{l}
\frac{\sin \angle S A A_{1}}{\sin \angle A S A_{1}} \\
=\frac{S A_{1}}{A A_{1}}=\frac{R A_{1}}{A A_{1}} \\
=\frac{\sin \angle A_{1} A R}{\sin \angle A R A_{1}} . \\
\text { Then } \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,643 |
6.4. A five-digit number is a multiple of 54, and none of its digits are 0. After deleting one of its digits, the resulting four-digit number is still a multiple of 54; after deleting one of the digits of this four-digit number, the resulting three-digit number is still a multiple of 54; after deleting one of the digit... | 6.4. The original five-digit number is 59994.
Since the number faced after each deletion of the front or back is a multiple of 9, the number deleted each time is a multiple of 9. This means that only a 9 can be deleted each time, and the last remaining two-digit number can only be 54. Among the three-digit numbers 549... | 59994 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,644 |
7.1. Proof: Any positive integer can be written as the quotient of a perfect square and a perfect cube. | 7.1. For example: $n=\frac{\left(n^{2}\right)^{2}}{n^{3}}$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,645 |
7.2. The sum of three positive integers (not necessarily distinct) is 100. By subtracting them pairwise (the larger minus the smaller), three difference numbers can be obtained. What is the maximum possible value of the sum of these three difference numbers? | 7.2. The maximum possible value of the sum of these three differences is 194.
Let the three positive integers be $a$, $b$, and $c$, and assume without loss of generality that $a \geqslant b \geqslant c$. The sum of their pairwise differences is
$$
(a-b)+(a-c)+(b-c)=2(a-c).
$$
Since $b \geqslant 1$ and $c \geqslant 1$... | 194 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,646 |
7.3. In quadrilateral $A B C D$, $A B=B C, C D=$ $D A$, points $K$ and $L$ are located on segments $A B$ and $B C$ respectively, such that $B K=2 A K, B L=2 C L$, points $M$ and $N$ are the midpoints of segments $C D$ and $D A$ respectively. Prove: $K M=L N$. | 7.3. From $\triangle A B D \cong \triangle C B D, \triangle A D M \cong \triangle C D N$, we have
$$
\begin{array}{l}
\angle K A M=\angle B A D-\angle M A D \\
=\angle B C D-\angle N C D=\angle L C N .
\end{array}
$$
Thus, $\triangle K A M \cong \triangle L C N$.
Therefore, $K M=L N$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,647 |
7.4. Toma thought of a positive integer and found the remainders when it is divided by 3, 6, and 9. It is known that the sum of these three remainders is 15. Try to find the remainder when the number is divided by 18. | 7.4. The remainder when this number is divided by 18 is 17.
Since the remainders when divided by $3$, $6$, and $9$ do not exceed $2$, $5$, and $8$ respectively, the sum of these three remainders is always no more than $2+5+8=15$. Since their sum is equal to 15, these three remainders must be $2$, $5$, and $8$. Further... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,648 |
8.2. The least common multiple of two positive integers is 16 times their greatest common divisor. Prove: one of these two positive integers is divisible by the other. | 8.2. Solution 1: Let the greatest common divisor of these two positive integers be $d$, then their least common multiple is $16 d$. Since these two positive integers are multiples of $d$ and divisors of $16 d=2^{4} d$, they can be expressed in the form $2^{k} d$. Here, $k$ is a non-negative integer. Among such two posi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,649 |
8.3. In quadrilateral $A B C D$, points $E$, $F$, $G$ are the midpoints of sides $A B$, $B C$, $A D$ respectively. It is known that $G E \perp A B$, $G F \perp B C$, $\angle A B C=96^{\circ}$. Try to find the degree measure of $\angle A C D$. | $$
\text { 8.3. } \angle A C D=90^{\circ} \text {. }
$$
In $\triangle A B G$ and $\triangle B C G$, there is a height coinciding with a median, so they are both isosceles triangles. Thus, $D G=A G=$ $B G=C G$. Therefore, in $\triangle A C D$, the median $C G$ on side $A D$ is half of $A D$, so $\triangle A C D$ is a r... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,650 |
8.4. Try to answer: Among the following two types of ten-digit numbers, which type has a larger sum: the first type of numbers contains the adjacent digits 1 and 2; the second type of numbers contains the adjacent digits 2 and 1? Explain your reasoning. | 8.4. Let the set of the first type of numbers be denoted as $A$, and the set of the second type of numbers as $B$. Since some ten-digit numbers contain both 12 and 21, they belong to both $A$ and $B$. After removing such ten-digit numbers, the remaining sets of the two types of numbers are denoted as $A^{\prime}$ and $... | S_{1} < S_{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,651 |
6.1.100 sailors are loading and unloading large cargo boxes, and it takes 7 people to carry each box. The captain believes that since each sailor has handled 65 boxes, the workload of each sailor is the same. Prove: The captain's calculation is incorrect. | 6.1. The captain gives 1 gold coin to each person involved in moving each box, so he distributes a total of 6500 gold coins. For each box moved, the captain issues 7 gold coins, which should be a multiple of 7. However, 6500 is not a multiple of 7. Therefore, the captain's calculation is incorrect. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,652 |
6.2. The colonel, major, captain, and private Qiongjing stand in a row. Their heights are all different, but the sum of Qiongjing's and the major's heights is exactly equal to the sum of the colonel's and the captain's heights. If the colonel and the second tallest person swap positions, and then the captain and the sh... | 6.2. The Major is in front of the Lieutenant Colonel.
First, it is easy to see that after the first position swap (Colonel with the second tallest person), the Major is at the front of the queue, because after the second position swap (Major with the shortest person), the shortest person is at the front. Below, we dis... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,653 |
Example 2 Given three points $D, E, F$ inside $\triangle A B C$ such that $\angle B A E=\angle C A F, \angle A B D=\angle C B F$. Prove that the necessary and sufficient condition for $A D, B E, C F$ to be concurrent is $\angle A C D=\angle B C E .{ }^{[1]}$ | Proof: As shown in Figure 2, let $\angle B A E = \angle C A F = \alpha$, $\angle A B D = \angle C B F = \beta$, $\angle A C D = x$, $\angle B C E = y$.
Applying the trigonometric form of Ceva's theorem to $\triangle A B C$ with points $D$, $E$, and $F$ respectively, we have
$$
\begin{aligned}
1= & \frac{\sin \angle B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,654 |
6.3. In a $10 \times 10$ grid, each cell contains a positive integer. In each row, the largest number is circled (if there are multiple largest numbers, one of them is circled); in each column, the smallest number is circled (or one of the smallest numbers is circled). It is known that each circled number is circled tw... | 6.3. First, prove that all the circled numbers are equal to each other. Use proof by contradiction.
Assume there are two circled numbers $a > b$. Now consider the number $c$ that is in the same column as $a$ and the same row as $b$. By the problem's condition, $a \leqslant c \leqslant b$, which implies $a \leqslant b$... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,655 |
6.4. Find all such four-digit numbers: they are all 83 times the sum of their digits. | 6.4. There is only one such four-digit number: $1494=18 \times 83$.
Solution 1: Let the number we are looking for be $x$, and the sum of its digits be $a$. From the problem, we know that $x=83a$. Since the remainder when $x$ is divided by 9 is the same as the remainder when $a$ is divided by 9, $x-a=82a$ must be divis... | 1494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,656 |
6.5. Two successful businessmen bought $100 \mathrm{~L}$ of alcohol, which was distributed in bottles with capacities of $0.5 \mathrm{~L}$, $0.7 \mathrm{~L}$, and $1 \mathrm{~L}$. Afterward, they had an argument and decided to split the alcohol evenly. Prove: without breaking any bottles, they can achieve an even split... | 6.5. First, consider the following scenario: at least $50 \mathrm{~L}$ of alcohol is bottled in $0.5 \mathrm{~L}$ and $1 \mathrm{~L}$ bottles. In this case, $50 \mathrm{~L}$ of alcohol can be allocated to one of the merchants: if there are 50 bottles of $1 \mathrm{~L}$, this is obvious; if there are fewer than 50 bottl... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 716,657 |
6.6. On a $9 \times 9$ chessboard, there are 9 rooks, none of which can attack each other. Each rook has moved according to the knight's move. Prove: There must now be at least two rooks that can attack each other. | 6.6. Write down the row number and column number of each rook, then add all these row numbers and column numbers. If the 9 rooks cannot attack each other, the sum must be \(2 \times (1+2+\cdots+9)=90\). This is because, in this case, these rooks are all in different rows and different columns. Therefore, each row numbe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,658 |
7.1. Can several diagonals be drawn in a convex hexagon so that each diagonal intersects exactly three other diagonals inside the hexagon? | 7.1. Not possible.
In a convex hexagon, there are two types of diagonals: one type is the "long diagonals," which connect two opposite vertices; the other type is the "short diagonals," which connect vertices that are one apart. If only long diagonals are drawn, the required property in the problem cannot be satisfied... | Not possible | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,659 |
7.2. A country has 17 cities, and the West Street (a street located in the western part of the city) of each city is named after another city. An invading army has entered the country, occupying 1 city per day. Their principle of attack is: on the first day, they occupy a city whose West Street is named after another c... | 7.2. First, consider the moment when the invading army discovers for the first time that the city they have just occupied had already been occupied before, assuming this happens on the $k+1$-th day after their invasion. After this, they can only repeatedly occupy these $k$ cities, and the number of days between any two... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 716,660 |
7.3. Cut a rectangular grid from a rectangular grid, the cut grid includes the bottom-right corner of the original grid, but does not include the other three corners of the original grid. Prove: In the remaining part, the number of ways to place a corner-shaped piece (obtained by removing one square from a $2 \times 2$... | 7.3. Pair the positions of the corner-shaped placements: In the $2 \times 2$ square where the corner-shaped placements are placed, pair the positions of two corner-shaped placements that are symmetric with respect to the center of the square (as shown in Figure 4).
Only the position of the corner-shaped placement that... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,661 |
7.4. Let $[a, b]$ denote the least common multiple of the positive integers $a, b$. Try to find the positive integer solutions of the equation $\left[x^{2}, y\right]+\left[x, y^{2}\right]=1996$. | 7.4. The equation has no positive integer solutions.
Because $\left[x^{2}, y\right]$ and $\left[x, y^{2}\right]$ are both divisible by $x$ and $y$, their sum 1996 should also be divisible by $x$ and $y$. Since $1996=2^{2} \times 499$, and 499 is a prime number, $x$ and $y$ can only be $1, 2, 4, 499, 998$, and 1996. Ho... | no positive integer solutions | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,662 |
7.5. On a $1 \times 40$ grid paper, a checker is placed in the rightmost square. Two players take turns moving the checker, which can be moved any number of squares to the left or right, but the squares that the checker has already visited cannot be used again. The player who cannot make a move loses. Who has a winning... | 7.5. The first player has a winning strategy.
He can adopt the following strategy: number the 40 squares from left to right. First, move the piece to square 1 (the leftmost square), then pair the remaining 38 squares: $(2,39),(3,38),\cdots,(20,21)$. Whenever the opponent moves the piece to square $k$, he then moves th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,663 |
7.6. Each node (intersection of grid lines) on an infinite large square grid is colored with one of three colors, and there are points of each color. Prove: A right triangle (whose legs do not necessarily lie on the grid lines) can be found, whose three vertices are colored with three different colors. | 7.6. Proof by contradiction.
Assume that there does not exist a right-angled triangle with three vertices colored in three different colors. It is not difficult to see that we can find a horizontal or vertical line \( l \) that contains at least two different colored nodes. For definiteness, let it be horizontal. If \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,664 |
Example 3 A circle intersects the three sides $BC$, $CA$, $AB$ of $\triangle ABC$ at points $D_{1}$, $D_{2}$, $E_{1}$, $E_{2}$, $F_{1}$, $F_{2}$, respectively. The line segments $D_{1} E_{1}$ and $D_{2} F_{2}$ intersect at point $L$, $E_{1} F_{1}$ and $D_{2} E_{2}$ intersect at point $M$, $F_{1} D_{1}$ and $F_{2} E_{2}... | Proof 1: As shown in Figure 3, connect $D_{1} E_{2}$, $E_{1} F_{2}$, and $F_{1} D_{2}$.
Thus, we have
$$
\begin{array}{l}
\angle D_{1} E_{1} F_{2} = \angle D_{1} E_{2} F_{2}, \\
\angle D_{2} F_{2} F_{1} = \angle D_{2} D_{1} F_{1}, \\
\angle E_{2} E_{1} D_{1} = \angle E_{2} D_{2} D_{1}, \\
\angle E_{1} F_{2} D_{2} = \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,665 |
7.7. The government of a certain country has decided to change the operating status of state-owned air routes and plans to sell the operating rights of air routes between all 239 cities to private companies. To prevent the loss of state assets, the People's Khural (equivalent to a parliament) passed the following resol... | 7.7. When the number of cities is $n$, the number of companies can be any integer from 1 to $n-1$.
Use mathematical induction to prove: When the number of cities is $n$, the number of companies cannot exceed $n-1$.
For $n=2$, the conclusion is obviously true.
Assume the conclusion holds for $n=k$, prove that when $n=k... | 1 \text{ to } n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,666 |
8.3. In quadrilateral $A B C D$, the midpoints of sides $A B, B C, C D$, and $D A$ are $P, Q, R, S$ respectively. It is known that point $M$ is located inside quadrilateral $A B C D$, such that quadrilateral $A P M S$ is a parallelogram. Prove: Quadrilateral $C R M Q$ is also a parallelogram. | 8.3. Line $S M$ passes through the midpoint of side $A D$ of $\triangle A B D$ and is parallel to side $A B$, so it contains a midline of $\triangle A B D$. Similarly, it can be proven that line $P M$ contains another midline of $\triangle A B D$. Therefore, as the intersection of lines $S M$ and $P M$, $M$ coincides w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,667 |
8.4. In order to commit a crime in a 13-story residential building, a criminal deliberately broke the buttons in the elevator. Now the elevator does not stop at the floor number pressed (i.e., the floor it stops at does not necessarily correspond to the number pressed). An elderly person pressed the number of the floor... | 8.4. The path the old man has walked is a loop. In fact, he is repeatedly circling on a graph with a length of no more than 13. The length of this loop should be a divisor of 1313, which is 13. Therefore, it is possible to reach any level from any level. | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 716,668 |
8.5. Let the number of positive divisors of a positive integer $n$ be denoted by $d(n)$. Prove: For any positive integers $a, b$, we have
$$
d(a b) \geqslant d(a)+d(b)-1 .
$$ | 8.5. Let the positive divisors of the positive integer $a$ be denoted as $a_{1}=1$, $a_{2}, a_{3}, \cdots, a_{d(a)-1}, a_{d(a)}$. It is easy to see that all positive divisors of $b$ and $b a_{2}, b a_{3}, \cdots, b a_{d(a)-1}, b a_{d(a)}$ are distinct positive divisors of $a b$, and there are a total of $d(a)+d(b)-1$ o... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,669 |
8.6. Two players take turns placing stones on a $101 \times 101$ grid, each placing 1 stone per turn. The first player can place a stone in any empty cell where the total number of stones already placed in the row and column of that cell is even; the second player can place a stone in any empty cell where the total num... | 8.6. The first player has a winning strategy.
Let the first player be called Player A, and his opponent be called Player B. Player A's winning strategy is as follows.
First, he occupies the center square. Then he adopts a symmetric strategy: if Player B places a piece in a square $B$ in the center column, then Player... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,670 |
8.7. In some cities of a country, there are direct (two-way) air routes. The aviation administration has decided to change the route layout once a month according to the following rule: a direct (two-way) route will be opened between cities $A$ and $B$ next month if and only if, this month, there exists a route that al... | 8.7. Suppose the layout of air routes begins to change from winter, and by July (half a year later), the air route network covers the entire country, meaning that one can fly from any city to any other city (including with transfers).
Now let's look at the air route layout in June. First, mark the capital on the map; ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,671 |
1. Among the numbers $1,2, \cdots, 100$, add “+” and “ _ ” signs so that the algebraic sum of the resulting expression is 4150. The maximum number of “+” signs that can be added is ( ) .
(A) 92
(B) 93
(C) 94
(D) 95 | - .1.C.
Let the sum of all positive numbers be $x$, and the sum of all negative numbers be $y$. Then
$$
\left\{\begin{array} { l }
{ x + y = 4150 , } \\
{ x + | y | = \frac { 100 \times ( 1 + 100 ) } { 2 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x+y=4150, \\
x-y=5050 .
\end{array}\right.\right.
$$
So... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,672 |
2. The inequality about $x$ $\left\{\begin{array}{l}x-1>0, \\ x^{3}-x^{2}+x \leqslant k\end{array}\right.$ has only the positive integer solutions 2 and 3. Then the value of $| k-21|+| k-52 \mid+$ 1975 is ( ).
(A) 2004
(B) 2005
(C) 2006
(D) 2007 | 2.C.
From the problem, we know that $x>1$.
Since $x=2, 3$ are solutions to $x^{3}-x^{2}+x \leqslant k$, we have,
$$
\left\{\begin{array}{l}
2^{3}-2^{2}+2 \leqslant k, \\
3^{3}-3^{2}+3 \leqslant k
\end{array}\right.
$$
Solving this, we get $k \geqslant 21$.
From the given information, $x=4,5,6, \cdots$ are solutions t... | 2006 | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,673 |
4. Given that $a$ is a real number, the equation about $x$
$$
27 x^{2}+2 a^{2} x+a=0
$$
has real roots. Then the maximum value of $x$ is ( ).
(A) 0
(B) $-\frac{3}{2}$
(C) $-\frac{1}{3}$
(D) $\frac{1}{6}$ | 4.D.
Consider the original equation as an equation in terms of $a$, and transform it into $2 x a^{2}+a+27 x^{2}=0$.
When $x=0$, $a=0$;
When $x \neq 0$, $\Delta=1-216 x^{3} \geqslant 0$, solving this yields $x \leqslant \frac{1}{6}$.
When $a=-\frac{3}{2}$, $x=\frac{1}{6}$.
Therefore, the maximum value of $x$ is $\frac{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,675 |
Example 1 Calculate $(x+y+z)(xy+yz+zx)$.
Analysis: Since both factors in the original expression are cyclic symmetric expressions in $x, y, z$, by property 5, their product is also a cyclic symmetric expression in $x, y, z$. Therefore, we only need to multiply the first letter of the first factor by the second factor, ... | Solution: Since $x(x y+y z+z x)=x^{2} y+x y z+z x^{2}$, therefore,
$$
\begin{array}{c}
\text { the original expression }=x^{2} y+x y z+z x^{2}+y^{2} z+y z x+x y^{2}+ \\
z^{2} x+z x y+y z^{2} \\
=x^{2} y+z x^{2}+y^{2} z+x y^{2}+z^{2} x+y z^{2}+3 x y z .
\end{array}
$$ | x^{2} y+z x^{2}+y^{2} z+x y^{2}+z^{2} x+y z^{2}+3 x y z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,676 |
Example 2 Factorize:
$$
(x y-1)^{2}+(x+y-2)(x+y-2 x y)
$$ | Solution: Let $x+y=u, x y=v$, then
$$
\begin{array}{l}
\text { original expression }=(v-1)^{2}+(u-2)(u-2 v) \\
=v^{2}-2 v+1+u^{2}-2 u-2 u v+4 v \\
=(u-v)^{2}-2(u-v)+1 \\
=(u-v-1)^{2} \\
=(x+y-x y-1)^{2} \\
=(x-1)^{2}(y-1)^{2} .
\end{array}
$$ | (x-1)^{2}(y-1)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,677 |
$$
\begin{array}{l}
\frac{a^{2}\left(\frac{1}{b}-\frac{1}{c}\right)+b^{2}\left(\frac{1}{c}-\frac{1}{a}\right)+c^{2}\left(\frac{1}{a}-\frac{1}{b}\right)}{a\left(\frac{1}{b}-\frac{1}{c}\right)+b\left(\frac{1}{c}-\frac{1}{a}\right)+c\left(\frac{1}{a}-\frac{1}{b}\right)} \\
=
\end{array}
$$ | (Hint: First find a common denominator, set the quotient as $k(a+b+c)$. Answer:
$$
a+b+c .)
$$ | a+b+c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,678 |
4. If the equation in terms of $x$
$$
\frac{1}{x^{2}-x}+\frac{k-5}{x^{2}+x}=\frac{k-1}{x^{2}-1}
$$
has an extraneous root $x=1$, then the value of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 6 | 4.C.
From the original equation, we get
$$
\frac{1}{x(x-1)}+\frac{k-5}{x(x+1)}=\frac{k-1}{(x+1)(x-1)}
$$
Eliminating the denominators, we get
$$
x+1+(k-5)(x-1)=(k-1) x \text {. }
$$
Let $x=1$, we get $k=3$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,679 |
5. The graph of the function $y=(x-2004)(x+2005)$ intersects the $x$-axis and the $y$-axis at three points. If there is a circle that passes through these three points, then the other intersection point of this circle with the coordinate axes is ( ).
(A) $\left(0, \sqrt{\frac{2004}{2005}}\right)$
(B) $(0,1)$
(C) $\left... | 5.B.
According to the intersecting chords theorem, it is easy to find that the coordinates of the other intersection point of this circle with the coordinate axis are $(0,1)$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,680 |
6. As shown in Figure $3, P, D, A$ and $P, B, C$ are collinear respectively, $AB$ intersects $CD$ at $Q$, and $PQ$ bisects $\angle A P C$. Let $P A=$ $a, P B=b, P C=c, P D=$ $d$. Then the relationship between $a, b, c, d$ is ( ).
(A) $a b=c d$
(B) $a+b=c+d$
(C) $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$
(D) $\fr... | 6.C.
As shown in Figure 8, draw $Q E / / P C$ intersecting $P A$ at $E$.
Let $Q E=x$, it is easy to see that
$$
P E=x, A E=a-x .
$$
By $\triangle A Q E \backsim \triangle A B P$, we get
$$
\frac{x}{b}=\frac{a-x}{a}=1-\frac{x}{a} .
$$
Rearranging gives $\frac{x}{a}+\frac{x}{b}=1$, which means
$$
\frac{1}{a}+\frac{1}{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,681 |
1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$ | \begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array} | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,682 |
2. Given $0 \leqslant x \leqslant 1$, the inequality $x^{2}+a x+3-a$
$>0$ always holds. Then the range of the real number $a$ is $\qquad$ | 2. $\left\{\begin{array}{l}0 \leqslant -\frac{a}{2} \leqslant 1, \\ f\left(-\frac{a}{2}\right)=3-\frac{a^{2}}{4}-a>0\end{array}\right.$ or $\left\{\begin{array}{l}-\frac{a}{2}>1, \\ f(1)=4>0 .\end{array}\right.$
From this, we get $0<a<3$ or $-2 \leqslant a \leqslant 0$ or $a<-2$. In summary, we have $a<3$. | a<3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,683 |
3. The graph of the equation $6 x^{2}-5 x y+y^{2}-8 x+2 y-8=0$ and the $x$-axis enclose a figure whose area is $\qquad$ | 3. $\frac{64}{3}$.
Given $(2 x-y-4)(3 x-y+2)=0$.
Therefore, the image of the equation with the $x$-axis forms a triangle.
$$
\text { Let }\left\{\begin{array}{l}
2 x-y-4=0, \\
3 x-y+2=0,
\end{array}\right.
$$
The intersection point coordinates are $(-6,-16)$.
The $x$-intercepts of the lines $2 x-y-4=0$ and $3 x-y+2=0... | \frac{64}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,684 |
4. As shown in Figure 4, the diameter $AB = d$ of semicircle $O$. $M$ and $N$ are the midpoints of $OA$ and $OB$, respectively, and point $P$ is on the semicircle. Draw $MC \parallel PN$ and $ND \parallel PM$, with points $C$ and $D$ both on the semicircle. Then $PC^2 + PD^2 + CM^2 + DN^2 =$ $\qquad$ | 4. $\frac{5}{8} d^{2}$.
Obviously, $O M=O N$.
As shown in Figure 9, take the midpoint $K$ of $P C$, and connect $O K$, then we have
$O K \perp P C$.
Since $O K$ is the midline of trapezoid CMNP, we have
$$
C M / / O K / / P N \text {. }
$$
Therefore, $M C \perp P C$.
Similarly, $N D \perp P D$.
Connect $P O$, since $... | \frac{5}{8} d^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,685 |
One, (20 points) In a certain competition, each player plays exactly one game against every other player. The winner of each game gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. After the competition, it is found that each player's score is exactly half from games played against t... | Given $n$ players, the $n$ players scored a total of $\frac{n(n-1)}{2}$ points. The 10 "math players" scored a total of $\frac{10 \times 9}{2}=45$ points through their matches with each other, which is half of their total score, so they scored a total of 90 points. The remaining $n-10$ players scored a total of $\frac{... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,686 |
II. (25 points) As shown in Figure 5, in $\triangle ABC$, points $P$ and $Q$ are both on the perpendicular bisector of side $BC$, and satisfy $\angle BAP = \angle CAQ$. Prove:
$$
\begin{array}{l}
A P \cdot A Q \\
=A B \cdot A C + B P \cdot C Q .
\end{array}
$$ | As shown in Figure 10, take a point $E$ on the perpendicular bisector of side $BC$ such that points $A, P, B, E$ are concyclic. Connect $EA, EB, EC$. Extend $AC$ to $G$ such that $GQ = AQ$, and draw $QH \perp AG$ at $H$. It is easy to see that $AH = CH$.
From $\angle CAQ = \angle BAP = \angle BEP = \angle CEP = \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,687 |
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of ... | (1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,688 |
1. Given $\sin ^{2005} x+\cos ^{2005} x=1$. Then for any $k>0$, we must have ( ).
(A) $\sin ^{k} x+\cos ^{k} x=1$
(B) $\sin ^{k} x+\cos ^{k} x>1$
(C) $\sin ^{k} x+\cos ^{k} x<1$
(D) The value of $\sin ^{k} x+\cos ^{k} x$ is uncertain | -.1.A.
Since $\sin ^{200} x+\cos ^{2008} x \leqslant \sin ^{2} x+\cos ^{2} x=1$, equality holds if and only if $\sin x$ and $\cos x$ take 0 and 1, respectively.
At this time, for any $k>0$, it must be that $\sin ^{k} x+\cos ^{k} x=1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,690 |
2. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have one focus at $F$, and point $P$ be on the $y$-axis. The line $P F$ intersects the ellipse at points $M$ and $N$, with $P M=\lambda_{1} M F$ and $P N=\lambda_{2} N F$. Then the real number $\lambda_{1}+\lambda_{2}=(\quad)$.
(A) $-\frac{2 b^{2}}{a... | 2.C.
Let's assume $F(c, 0)$.
Let $P(0, p)$ and $M(x, y)$. Suppose $P M=\lambda M F$, i.e., $(x, y-p)=\lambda(c-x,-y)$.
Solving this, we get $x=\frac{\lambda c}{1+\lambda}, y=\frac{p}{1+\lambda}$.
Substituting into the ellipse equation and simplifying, we get
$$
b^{4} \lambda^{2}+2 a^{2} b^{2} \lambda+a^{2} b^{2}-a^{2}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,691 |
3. The graphs of the exponential function $y=a^{x}$ and the logarithmic function $y=\log _{a} x$ (where $a>0, a \neq 1$) are denoted as $C_{1}$ and $C_{2}$, respectively. Point $M$ is on curve $C_{1}$, and line segment $O M$ (where $O$ is the origin) intersects curve $C_{1}$ at another point $N$. If there exists a poin... | 3. B.
Let $M\left(x_{1}, a^{x_{1}}\right), N\left(x_{2}, a^{x_{2}}\right)$, then we have $x_{p}=a^{x_{1}}, y_{p}=2 x_{2}=\log _{a} a^{x_{1}}=x_{1}$. Given that points $O, N, M$ are collinear, hence $\frac{x_{1}}{x_{2}}=\frac{a^{x_{1}}}{a^{x_{2}}}=2$,
which means $a^{2 x_{2}}=2 a^{x_{2}}$.
Solving this, we get $x_{2}=\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,692 |
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