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Example 7 Let $a, b, c$ be positive real numbers, and $abc=1$. Prove that:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} .
$$
(36th IMO) | Prove: Using $a b c=1$, the original inequality can be transformed into
$$
\frac{(b c)^{2}}{a b+c a}+\frac{(c a)^{2}}{b c+a b}+\frac{(a b)^{2}}{c a+b c} \geqslant \frac{3}{2} \text {. }
$$
Let $S=a b+b c+c a$. Since the function $f(x)=$ $\frac{x}{S-x}$ is increasing on $(0, S)$, for any $x$ $\in(0, S)$, we always have... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,137 |
Example 8 Given $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} .
$$ | Prove: Let $S=x+y+z$.
Since the function $f(t)=\frac{1}{S+t}$ is decreasing on $(0, S)$, for any $t \in(0, S)$, we always have
$$
\left(t-\frac{S}{3}\right)\left[f(t)-f\left(\frac{S}{3}\right)\right] \leqslant 0 .
$$
From this, it is easy to obtain
$$
\frac{t}{S+t} \leqslant \frac{9 t}{16 S}+\frac{1}{16} \text {. }
$$... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,138 |
Let $\triangle A B C$ have side lengths $a, b, c$, and define
$$
f(\lambda)=\frac{a}{\lambda a+b+c}+\frac{b}{\lambda b+c+a}+\frac{c}{\lambda c+a+b} \text {. }
$$
Prove: When $-11$, we have
$$
\frac{2}{\lambda+1}<f(\lambda) \leqslant \frac{3}{\lambda+2} .
$$ | Proof: Let $S=a+b+c$. When $-12 a$ and $-11$, from $S>2 a$ we get
\[
\frac{a}{\lambda a+b+c}>\frac{2 a}{(\lambda+1) S}.
\]
Similarly, we have
\[
\frac{b}{\lambda b+a+c}>\frac{2 b}{(\lambda+1) S},
\]
\[
\frac{c}{\lambda c+a+b}>\frac{2 c}{(\lambda+1) S}.
\]
Adding these three inequalities, we get $f(\lambda)>\frac{2}{\la... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,139 |
Example 1 If $a, b, c$ are positive numbers, prove:
$$
\begin{array}{l}
\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \\
\geqslant \frac{a+b+c}{3} .
\end{array}
$$ | Proof: Let the left side of the inequality be $M$, and construct the pairing expression
$$
N=\frac{b^{3}}{a^{2}+a b+b^{2}}+\frac{c^{3}}{b^{2}+b c+c^{2}}+\frac{a^{3}}{c^{2}+c a+a^{2}} \text {. }
$$
Then $M-N=0, M=N$.
$$
\begin{array}{l}
\text { Also, } M+N \\
=(a+b) \cdot \frac{a^{2}-a b+b^{2}}{a^{2}+a b+b^{2}}+ \\
(b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,140 |
Example 2 Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$. Prove:
$$
\sum_{k=1}^{n} \frac{a_{k}^{2}}{a_{k}+b_{k}} \geqslant \frac{1}{2} \sum_{k=1}^{n} a_{k} \text {. }
$$
(1991, Asia Pacific Mathematical Olympiad) | Prove: Let the left side of the inequality be $M$, and construct the pairing expression $N=\sum_{k=1}^{n} \frac{b_{k}^{2}}{a_{k}+b_{k}}$. Then
$$
\begin{array}{l}
M-N=\sum_{k=1}^{n}\left(a_{k}-b_{k}\right) \\
=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} b_{k}=0 .
\end{array}
$$
Thus, $M=N$.
$$
\text { Also, } M+N=\sum_{k=1}^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,141 |
Example 3 Given that $p$ is a three-digit prime number, and the hundreds digit of $p$ is $a$, the tens digit is $b$, and the units digit is $c$. Prove: The quadratic equation $a x^{2}+b x+c=0$ has no integer solutions.
Analysis: If we use the quadratic formula $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ to verify that the... | Proof: It is clear that $a \neq 0, c \neq 0$.
Assume the equation $a x^{2} + b x + c = 0$ has an integer solution $x_{0}$.
When $b=0$, we have $x_{0}^{2} = -\frac{c}{a} \leqslant 0$, which implies $x_{0} = 0$, $c = 0$, a contradiction. Therefore, $b \neq 0$, and $x_{0} < 0$.
Substituting $c = p - 100a - 10b$ into the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,143 |
Example 4 If $a_{1}+a_{2}+\cdots+a_{n}=1, a_{i} \in \mathbf{R}_{+}$ $(i=1.2, \cdots, n)$, prove:
$$
\begin{array}{c}
\frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+ \\
\overline{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+ \\
\frac{a_{n}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^... | Proof: Let the left side of the inequality be denoted as $M$, and construct the pairing expression
$$
\begin{array}{c}
N=\frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+ \\
\frac{a_{3}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+ \\
\frac{a_{1}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,144 |
Example 5 Let $x, y, z$ be positive numbers. Prove that:
$$
\frac{z^{2}-x^{2}}{x+y}+\frac{x^{2}-y^{2}}{y+z}+\frac{y^{2}-z^{2}}{z+x} \geqslant 0 .
$$ | Proof: Let the left side of the inequality be denoted as $M$, and construct the pairing expression
$$
N=\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z}+\frac{y^{2}-x^{2}}{z+x} \text {. }
$$
Then $M-N=\frac{y^{2}-x^{2}}{x+y}+\frac{z^{2}-y^{2}}{y+z}+\frac{x^{2}-z^{2}}{z+x}$
$$
=(y-x)+(z-y)+(x-z)=0 \text {. }
$$
Thus, $... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,145 |
Example 6 Given $x \geqslant y \geqslant z>0$. Prove:
$$
\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2} \text {. }
$$
(31st IMO Shortlist) | Proof: Let the left side of the inequality be $M$, and construct the pairing expression
$$
N=\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x} \text {. }
$$
By the Cauchy-Schwarz inequality, we have
$$
M \cdot N \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2} \text {. }
$$
Also, $M-N$.
$$
\begin{aligned}
= & \frac{1}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,146 |
Example 7 Prove: For any real numbers $a>1, b>1$, the inequality
$$
\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8
$$
holds.
(26th Commonwealth of Independent States Mathematical Olympiad) | Proof: Let the left side of the inequality be $M$. Construct the pairing expression
$$
N=\frac{b^{2}}{b-1}+\frac{a^{2}}{a-1} \text {. }
$$
Then $M-N=\frac{a^{2}-b^{2}}{b-1}+\frac{b^{2}-a^{2}}{a-1}$
$$
=\frac{(a+b)(a-b)^{2}}{(a-1)(b-1)} \geqslant 0 \text {. }
$$
Thus, $M \geqslant N$.
$$
\begin{array}{l}
\text { and }... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,147 |
Example 8 If $a, b, c, d$ are all positive numbers, and $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove:
$$
\begin{array}{l}
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+ \\
(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant 6 .
\end{array}
$$
(28th IMO Preliminary Problem) | Proof: Let the left side of the inequality be $M$, and construct the pairing expression
$$
\begin{aligned}
N= & (a-b)^{4}+(a-c)^{4}+(a-d)^{4} \\
& +(b-c)^{4}+(b-d)^{4}+(c-d)^{4} .
\end{aligned}
$$
Then $M+N$
$$
\begin{aligned}
= & 6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+\right. \\
& \left.2 a^{2} d^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,148 |
Example 9 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=$
1. Prove: For every $n \in \mathbf{N}$, we have
$$
(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} .
$$
(1988, National High School Mathematics Competition) | Proof: Let the left side of the inequality be $M$. Then
$$
\begin{array}{l}
M=(a+b)^{n}-a^{n}-b^{n} \\
=\mathrm{C}_{n}^{1} a^{n-1} b+\mathrm{C}_{n}^{2} a^{n-2} b^{2}+\cdots+\mathrm{C}_{n}^{n-1} a b^{n-1} .
\end{array}
$$
Reversing the order of $M$ gives
$$
M=\mathrm{C}_{n}^{n-1} a b^{n-1}+\mathrm{C}_{n}^{n-2} a^{2} b^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,149 |
Example 10 Prove:
$$
\frac{1}{15}<\frac{1 \times 3 \times 5 \times \cdots \times 99}{2 \times 4 \times 6 \cdots \times 100}<\frac{1}{10}
$$ | Proof: Let the middle part of the inequality be denoted as $M$. Construct the pairing expression
$$
N=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \cdots \times \frac{98}{99} \times \frac{100}{101} .
$$
Notice that $\frac{n-1}{n}\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \cdots \times \frac{9... | \frac{1}{15}<M<\frac{1}{10} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,150 |
Example 11 Given $n \in \mathbf{N}$. Prove:
$$
\frac{2}{1} \times \frac{5}{4} \times \frac{8}{7} \times \cdots \times \frac{3 n-1}{3 n-2}>\sqrt[3]{3 n+1} .
$$ | Proof: Let the left side of the inequality be $M$. Construct the pairing expressions
$$
\begin{array}{l}
N=\frac{3}{2} \times \frac{6}{5} \times \frac{9}{8} \times \cdots \times \frac{3 n}{3 n-1}, \\
P=\frac{4}{3} \times \frac{7}{6} \times \frac{10}{9} \times \cdots \times \frac{3 n+1}{3 n} .
\end{array}
$$
From $\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,152 |
Example 4 Proof: There do not exist four positive integers $a, b, c, d$ that satisfy the conditions
$$
a d=b c, n^{2} \leqslant a<b<c<d \leqslant(n+1)^{2},
$$
where $n$ is a positive integer.
Analysis: This problem cannot be considered a simple example from any perspective. To prove that there do not exist positive in... | Proof: Assume there exist positive integers $a, b, c, d$ that satisfy the conditions. Let $a=a_{1} a_{2}$, such that $a_{1} \mid b, a_{2} \mid c$. Let $b=a_{1} d_{2}, c=a_{2} d_{1}$, then $d=d_{1} d_{2}, a_{2}n$ when, $d_{1}>a_{1}>n, d_{2}>a_{2} \geqslant a_{1}>n$, so, $d=d_{1} d_{2}>a_{1} a_{2} \geqslant(n+1)^{2}$, co... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,154 |
Example 9 Let $S$ denote the set of all real numbers greater than -1, determine all functions $f: S \rightarrow S$ that satisfy the following two conditions:
(1) For all $x$ and $y$ in $S$, we have
\[
\begin{array}{l}
f(x+f(y)+x f(y)) \\
=y+f(x)+y f(x) ;
\end{array}
\]
(2) In every interval within $-1 < x < 0$ and $0 <... | Analysis: This is an iteration problem of a univariate function, and the properties of fixed points have a very important status, which can open up our thinking.
Solution: Let $x=y$, we get
$$
f(x+f(x)+x f(x))=x+f(x)+x f(x) \text {. }
$$
Fix $x$, let $x+f(x)+x f(x)=c$, then
$$
f(c)=c \text {. }
$$
In (1), let $x=c$,... | f(x)=-\frac{x}{x+1} | Algebra | proof | Yes | Yes | cn_contest | false | 715,155 |
Example 10 Find all functions defined on the set of positive real numbers, which take positive real values, and satisfy the conditions:
(1) For all positive numbers $x, y, f(x f(y))=y f(x)$;
(2) As $x \rightarrow \infty$, $f(x) \rightarrow 0$.
(24th IMO) | Solution: In (1), let $y=\frac{1}{f(x)}$, we get
$$
f(x f(y))=1 \text {. }
$$
This shows that 1 is a value that $f$ can take.
Now, let $y$ be such that $f(y)=1$. From (1), we get
$$
f(x)=y f(x) \text {. }
$$
Thus, $y=1$. Therefore, $f(1)=1$.
For any $x>0$, let $x f(x)=a$. Taking $y=x$ in (1), we get $f(a)=a$.
If $a>1... | f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,156 |
Example 11 Let $\mathbf{N}_{+}$ be the set of all natural numbers, $k \in$ $\mathbf{N}_{+}$, and $k$ is odd. Prove: There exists a function $f$ that is strictly increasing on $\mathbf{N}_{+}$, such that for every $n \in \mathbf{N}_{+}$, $f(f(n))=k n$.
(31st IMO National Training Team Test Question (iv)) | Solution: Construct the function
$$
f(n)=\left\{\begin{array}{ll}
n+\frac{k-1}{2} \cdot k^{i-1}, & k^{i-1} \leqslant n<\frac{k^{i-1}(1+k)}{2}, \\
n k-\frac{k-1}{2} \cdot k^{i}, & \frac{k+1}{2} \cdot k^{i-1} \leqslant n<k^{i} .
\end{array}\right.
$$
(1) Clearly, for any $n \in \mathbf{N}_{+}$,
$$
f(f(n))=k n \text {. }
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,157 |
Example 12 Let $f(x, y)$ be a binary polynomial, and satisfy the conditions:
(1) $f(1,2)=2$;
(2) $y f(x, f(x, y)) \equiv x f(f(x, y), y) \equiv (f(x, y))^{2}$.
Determine all such $f(x, y)$.
(31st IMO National Training Team Test Question (iv)) | Analysis: We can set up a system of equations for the exponents of $x$ and $y$ based on the conditions.
Solution: From condition (2), we know that $x$ and $y$ are factors of $f(x, y)$, i.e., $\square$
$$
f(x, y) = x y g(x, y),
$$
where $g(x, y)$ is also a bivariate polynomial.
According to (1), let the highest degree... | f(x, y) = x y | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,158 |
Example 13 Let $f$ be a function from $\mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, and $f(1)$ $=1, f(2 n+1)=f(2 n)+1, f(2 n)=3 f(n)$. Find the value set of $f$.
(31st IMO National Training Team Test Question (v)) | Analysis: It is more convenient to represent $n$ in binary for this problem.
Solution: Let the binary representation of $n$ be
$$
n=\left(\overline{a_{s} a_{s-1} \cdots a_{0}}\right)_{2} \text {. }
$$
Then $f(n)=\left(\overline{a_{a} a_{s-1} \cdots a_{0}}\right)_{3}$.
We prove by induction as follows:
(1) When $n=1$, ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,159 |
Example 1 As shown in Figure 1, in Rt $\triangle A B C$, $\angle C=90^{\circ}, \angle A=30^{\circ}$, the angle bisector of $\angle C$ intersects the external angle bisector of $\angle B$ at point $E$, and $A E$ is connected. Then $\angle A E B$ is equal to ()$^{[2]}$.
Figure 1
(A) $50^{\circ}$
(B) $45^{\circ}$
(C) $40^... | Solution 1: As shown in Figure 2, draw $E D \perp C B$ at $D$, and draw $E H \perp A B$ at $H$. Since $B E$ is the bisector of $\angle A B D$, we have
$\angle H B E=\angle D B E$.
Also, $B E=B E$, hence
Rt $\triangle H B E \cong \mathrm{Rt} \triangle D B E$.
Thus, $E H=E D$.
Draw $E F \perp C A$ at $F$, similarly we g... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,160 |
Given $C H$ is the altitude of Rt $\triangle A B C$ ( $\angle C$ $=90^{\circ}$ ), and intersects the angle bisectors $A M$ and $B N$ at points $P$ and $Q$, respectively. Prove: The line through the midpoints of $Q N$ and $P M$ is parallel to the hypotenuse $A B^{[1]}$.
(52nd Belarusian Mathematical Olympiad (Final A Ca... | Proof: As shown in Figure 1, let $E$ and $F$ be the midpoints of $QN$ and $PM$ respectively. Connect $CE$ and $EH$. Since $\angle C = 90^{\circ}$ and $CH \perp AB$, we have $\angle B C H = \angle B A C$.
Thus, we have
$$
\begin{array}{l}
\angle C N B = \angle N A B + \angle A B N \\
= \angle B C H + \angle N B C = \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,161 |
Given that $a, b, c$ are positive real numbers. Prove:
$$
a^{2}+b^{2}+c^{2}+a b c=4 \Rightarrow a+b+c \leqslant 3 \text {. }
$$
(The 20th Iranian Mathematical Olympiad (2nd Round))
Reference [1] provides a proof using trigonometric methods. This article presents an algebraic proof.
Translate the above text into Englis... | Proof: If $a, b, c$ are all greater than 1, or all less than 1, it is clearly not satisfied by the conditions of the problem. Therefore, among $a, b, c$, there must be two that are not both greater than 1 or not both less than 1. Without loss of generality, let these be $a, b$. Then,
$$
(1-a)(1-b) \geqslant 0,
$$
whic... | null | Algebra | proof | Yes | Yes | cn_contest | false | 715,162 |
In a cyclic quadrilateral $ABCD$, $AD$ and $BC$ intersect at point $P$, and $AC$ and $BD$ intersect at point $M$, then $PM^2 = PA \cdot PD - AM \cdot MC$. | Proof: As shown in Figure 1, it is easy to see that $\angle P M D > \angle M B C = \angle M A D$.
Extend $P M$ to $H$, and connect $A H$, such that $\angle P A H = \angle D M P$. Then, $\triangle P D M \sim \triangle P H A$. Therefore,
$$
\frac{P D}{P H} = \frac{P M}{P A},
$$
which implies $P A \cdot P D = P M \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,163 |
Proposition Let $h_{a}$ be the altitude of $\triangle A B C$ on side $B C$, $D$ be any point on side $B C$, and $r 、 r_{1} 、 r_{2}$ be the inradii of $\triangle A B C 、 \triangle A B D 、 \triangle A C D$ respectively; let $r^{\prime} 、 r_{1}^{\prime} 、 r_{2}^{\prime}$ be the exradii of the triangles opposite to $\angle... | Proof: As shown in Figure 1, it is easy to see that
$$
\begin{array}{l}
r=\frac{S}{p}, \\
r^{\prime}=\frac{S}{p-a}.
\end{array}
$$
Here, \( S \) and \( p \) represent the area and the semi-perimeter of \( \triangle ABC \), respectively.
Therefore, we have
$$
\begin{array}{l}
h_{a}\left(r^{\prime}-r\right)=h_{a}\left(\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,164 |
Example 5 On a plane, there are $n(n \geq 3)$ points. If any line passing through any two of these points must also pass through a third point among these $n$ points, then these $n$ points all lie on the same line.
Analysis: It is unclear how to construct a solution directly using the given condition. However, if we c... | Proof: Assume that these $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ are not on the same line. Let the line passing through any two points $P_{i}, P_{j} (i \neq j)$ be $l_{ij}$. Then, there must be a known point outside $l_{ij}$. Let the shortest distance from all known points outside $l_{ij}$ to $l_{ij}$ be $d_{ij}$, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,165 |
Let $\theta_{i} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), i=1,2,3,4$. Prove: There exists $\theta \in \mathbf{R}$, such that the following two inequalities
$$
\begin{array}{l}
\cos ^{2} \theta_{1} \cdot \cos ^{2} \theta_{2}-\left(\sin \theta_{1} \cdot \sin \theta_{2}-x\right)^{2} \geqslant 0, \\
\cos ^{2} \theta_{... | Certainly, here is the translation of the provided text into English, preserving the original formatting:
---
Clearly, equations (1) and (2) are equivalent to
$$
\begin{array}{l}
\sin \theta_{1} \cdot \sin \theta_{2}-\cos \theta_{1} \cdot \cos \theta_{2} \\
\leqslant x \leqslant \sin \theta_{1} \cdot \sin \theta_{2}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,166 |
II. A circle intersects the three sides $BC$, $CA$, $AB$ of $\triangle ABC$ at points $D_{1}$, $D_{2}$, $E_{1}$, $E_{2}$, $F_{1}$, $F_{2}$, respectively. The line segments $D_{1} E_{1}$ and $D_{2} F_{2}$ intersect at point $L$, the line segments $E_{1} F_{1}$ and $E_{2} D_{2}$ intersect at point $M$, and the line segme... | II. As shown in Figure 2, draw perpendiculars from point $L$ to $AB$ and $AC$, with the feet of the perpendiculars being $L'$ and $L''$ respectively.
Let $\angle L A B=\alpha_{1}, \angle L A C=\alpha_{2}, \angle L F_{2} A=\alpha_{3}$, $\angle L E_{1} A=\alpha_{4}$. Then we have
$$
\frac{\sin \alpha_{1}}{\sin \alpha_{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,167 |
Three, as shown in Figure 1, a circular pond is divided into $2 n(n \geqslant 5)$ "cells". We call cells that share a common partition (common edge or arc) adjacent, thus, each cell has three neighboring cells.
A total of $4 n+1$ frogs have jumped into the pond, and the frogs find it difficult to coexist peacefully. A... | Three, we refer to an event where three frogs simultaneously jump to three adjacent cells from a single cell as an "outbreak" in that cell; a cell is said to be in a "balanced state" if it either contains a frog or all three of its adjacent cells contain frogs.
It is easy to see that once a frog jumps into a cell, tha... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,168 |
Given the sequence $\left\{a_{n}\right\}$ satisfies the conditions:
$$
a_{1}=\frac{21}{16}, 2 a_{n}-3 a_{n-1}=\frac{3}{2^{n+1}}, n \geqslant 2 \text {. }
$$
Let $m$ be a positive integer, $m \geqslant 2$. Prove: when $n \leqslant m$, we have
$$
\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left[m-\left(\frac{2}{3... | From equation (1), we have
\[2^{n} a_{n}=3 \cdot 2^{n-1} a_{n-1}+\frac{3}{4}.\]
Let \(b_{n}=2^{n} a_{n}, n=1,2, \cdots\), then
\[b_{n}=3 b_{n-1}+\frac{3}{4}, \quad b_{n}+\frac{3}{8}=3\left(b_{n-1}+\frac{3}{8}\right).\]
Since \(b_{1}=2 a_{1}=\frac{21}{8}\), we have
\[
b_{n}+\frac{3}{8}=3^{n-1}\left(b_{1}+\frac{3}{8}\rig... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,169 |
Six, find all non-negative integer solutions $(x, y, z, w)$ of the equation $2^{x} \cdot 3^{y}-5^{x} \cdot 7^{w}=1$.
(Supplied by Chen Yonggao) | Given that $5^{x} \cdot 7^{x}+1$ is even, we know $x \geqslant 1$.
Case 1 If $y=0$, then
$$
2^{x}-5^{x} \cdot 7^{x}=1 \text {. }
$$
If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, which implies $41 x$.
Therefore, $31\left(2^{x}-1\right)$, which contradicts $2^{x}-5^{x} \cdot 7^{x}=1$.
If $z=0$, then $2^{x}-7^{x}=1$.
Fo... | (1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,171 |
2. As shown in Figure 1, $C D / / B E$. Then $\angle 2+\angle 3-\angle 1=$ ( ).
(A) $90^{\circ}$
(B) $120^{\circ}$
(C) $150^{\circ}$
(D) $180^{\circ}$ | 2.D. Extend $D C$ to intersect $A B$ at $F$. Since $C D / / B E$, we have $\angle B F C=\angle 3$. Then
$$
\begin{array}{l}
\angle 2-\angle 1=\angle A F C \\
=180^{\circ}-\angle B F C=180^{\circ}-\angle 3 .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,173 |
3. As shown in Figure 2, fold and press the paper piece $\triangle A B C$ along $D E$. Then $\angle A=(\quad)$.
(A) $\angle 1+\angle 2$
(B) $\frac{1}{2}(\angle 1+\angle 2)$
(C) $\frac{1}{3}(\angle 1+\angle 2)$
(D) $\frac{1}{4}(\angle 1+\angle 2)$ | 3. B.
From the sum of the interior angles of a quadrilateral being $360^{\circ}$, we get
$$
\begin{array}{l}
\angle B+(\angle 1+\angle A D E)+(\angle A E D+\angle 2)+\angle C \\
=360^{\circ} .
\end{array}
$$
And $\angle B+\angle C=\angle A D E+\angle A E D=180^{\circ}-\angle A$, thus
$$
\angle A=\frac{1}{2}(\angle 1+\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,174 |
4. If $a+2 b+3 c=12$, and $a^{2}+b^{2}+c^{2}$ $=a b+b c+c a$, then $a+b^{2}+c^{3}=(\quad)$.
(A) 12
(B) 14
(C) 16
(D) 18 | 4.B.
$$
\begin{array}{l}
\text { From } a^{2}+b^{2}+c^{2}=a b+b c+c a \text {, we get } \\
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0 \\
\Rightarrow a=b=c \text {. }
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,175 |
Example 6: There are 6 circles on a plane, and the center of each circle is outside the other 5 circles. Prove: There does not exist a point on the plane that is inside all 6 circles.
Analysis: This problem is to prove that these 6 circles have no common point. Directly using the condition that the center of each circ... | Proof: Suppose there exists a point $M$ inside 6 circles simultaneously. Let the centers of the 6 circles be $O_{1}, O_{2}, \cdots, O_{6}$, and their radii be $r_{1}, r_{2}, \cdots, r_{6}$, then $O_{i} M < r_{i}$.
Let $M O_{1}, M O_{2}, \cdots, M O_{6}$ be arranged counterclockwise (some may coincide), and assume $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,176 |
5. A toy has 2 buttons (1 yellow, 1 red) and 100 small wooden figures that can stand or sit. Pressing the red button once will make 1 standing figure sit down; pressing the yellow button once will double the number of standing figures. Now, only 3 figures are standing, and you need to make 91 figures stand. What is the... | 5.C.
3(yellow)6(yellow)12(yellow)24(red)23(yellow)46(yellow)92(red) 91. A total of 7 button presses are required. | C | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,177 |
6. As shown in Figure 3, in quadrilateral $A B C D$, the angle bisectors of $\angle A$, $\angle B$, $\angle C$, and $\angle D$ intersect at a single point $P$. Let the areas of $\triangle A P D$, $\triangle A P B$, $\triangle B P C$, and $\triangle D P C$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively. Then ( ... | 6. A.
From the point on the angle bisector to the two sides of the angle, the distances are equal, so the distances from point $P$ to each side of the quadrilateral are equal, denoted as $h$ (as shown in Figure 11). It is easy to see that
$\triangle A P E \cong \triangle A P F$,
$\triangle B P E \cong \triangle B P H$... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,178 |
1. Calculate $\frac{20042003^{2}+1}{20042002^{2}+20042004^{2}}=$ | $$
\text { II. } 1 . \frac{1}{2} \text {. }
$$
Let $a=20042003$, then
$$
\text { Original expression }=\frac{a^{2}+1}{(a-1)^{2}+(a+1)^{2}}=\frac{a^{2}+1}{2\left(a^{2}+1\right)}=\frac{1}{2} \text {. }
$$ | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,179 |
2. Given that $x$ and $y$ are positive integers, and satisfy $2 x^{2}+3 y^{2}$ $=4 x^{2} y^{2}+1$. Then $x^{2}+y^{2}=$ $\qquad$ . | 2. From the given, we have $4 x^{2} y^{2}-2 x^{2}-3 y^{2}=-1$. Multiplying both sides by 4 and rearranging, we get
$$
\left(4 x^{2}-3\right)\left(4 y^{2}-2\right)=2=1 \times 2 \text {. }
$$
Solving, we get $\left\{\begin{array}{l}x^{2}=1, \\ y^{2}=1\end{array}\right.$ or $\left\{\begin{array}{l}x^{2}=1.25, \\ y^{2}=0.7... | x^2 + y^2 = 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,180 |
3. As shown in Figure 4, all quadrilaterals are squares, and all triangles are right triangles, where the side length of the largest square is $13 \mathrm{~cm}$. Then the sum of the areas of the four shaded squares is $\qquad$ | 3.169.
As shown in Figure 12, by repeatedly applying the Pythagorean theorem, we get
$$
\begin{array}{l}
S_{A}+S_{B}+S_{C}+S_{D} \\
=S_{E}+S_{F}=S_{G} \\
=13^{2}=169 .
\end{array}
$$ | 169 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,181 |
4. Given $\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}$ $=p$. Then $p^{3}+p^{2}+p=$ $\qquad$ | 4.1 .
From the given, we have
$$
\begin{array}{l}
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y}=\frac{x+y-z}{x+y+z}, \\
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z} .
\end{array}
$$
Then \( p^{3}+p^{2}+p=1 \). | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,182 |
6. If the sum of the areas of three square pieces of paper with integer side lengths is 2004, and the area of the largest square piece of paper is $S_{1}$, and the area of the smallest square piece of paper is $S_{2}$, then the maximum value of $\frac{S_{1}}{S_{2}}$ is $\qquad$ | 6.484.
Let the side lengths of the squares be $a$, $b$, and $c$, with $a<b<c$. Then, $a^{2}+b^{2}+c^{2}=2004$.
Obviously, $S_{2}=a^{2}, S_{1}=c^{2} \Rightarrow \frac{S_{1}}{S_{2}}=\frac{c^{2}}{a^{2}}$.
It is easy to see that $40^{2}+20^{2}+2^{2}=2004$, $44^{2}+8^{2}+2^{2}=2004$.
To maximize $\frac{S_{1}}{S_{2}}$, $S_... | 484 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,184 |
8. As shown in Figure 5, the area of rectangle $A B C D$ is $35 \mathrm{~cm}^{2}$, the area of shaded $\triangle A B E$ is $5 \mathrm{~cm}^{2}$, and the area of shaded $\triangle A D F$ is $7 \mathrm{~cm}^{2}$. What is the area of $\triangle A E F$ in square centimeters? | 8.15.5.
As shown in Figure 13, draw rectangle $A D F P$ and rectangle $A B E T$. Let the intersection of $P F$ and $T E$ be $O$. Given $S_{\triangle B E}=5 \mathrm{~cm}^{2}$, we know
Thus, $S_{\text {rectangleECor }}=35-10=25\left(\mathrm{~cm}^{2}\right)$.
Given $S_{\triangle A D F}=7 \mathrm{~cm}^{2}$, we know
$S_{\... | 15.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,186 |
Example 7 There is an $n \times n$ grid paper, from which you are allowed to arbitrarily select $n-1$ cells to be colored black, and then gradually color black those cells that are adjacent to at least two already blackened cells. Prove: No matter how the initial $n-1$ cells are chosen, it is impossible to color all ce... | Proof: Assuming it is possible to color all squares black, then for the black area, the boundary of the grid paper is the boundary of the region. If the side length of each small square is denoted as 1, then the total boundary length of the black area is $4 n$.
However, at the beginning, the black area only includes $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,187 |
3. As shown in Figure 8, point $C$ is on line segment $A B$, $D A \perp A B, E B \perp$ $A B, F C \perp A B$, and $D A=$ $B C, E B=A C, F C=A B$, $\angle A F B=51^{\circ}$. Then $\angle D F E=$ $\qquad$ | 3. $39^{\circ}$.
As shown in Figure 15, connect $A E$ and $B D$. Then
$\triangle A B D \cong \triangle C F B$,
$\triangle A B E \cong \triangle F C A$.
Therefore, $D B=F B$,
$$
A E=A F \text {. }
$$
Thus, $\triangle D B F$ and $\triangle E A F$ are both
isosceles right triangles. Hence,
$$
\angle B F D=45^{\circ}, \a... | 39^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,190 |
4. There are 8 consecutive positive
integers, the sum of which can be expressed as the sum of 7
consecutive positive integers, but cannot be expressed as the sum of 3 consecutive positive integers. Then, the minimum value of the largest number among these 8 consecutive positive integers is $\qquad$ | 4.21.
Let these 8 consecutive positive integers be
$$
a, a+1, a+2, a+3, a+4, a+5, a+6, a+7 \text {. }
$$
Their sum is $S=8a+28$, which means $S$ leaves a remainder of 4 when divided by 8.
Since $S$ is divisible by 7, we have $8a+28=7n$.
Therefore, $41n$.
Let $n=4k$, then $2a+7=7k$.
Thus, $a=\frac{7(k-1)}{2}$.
For $a$... | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,191 |
5. As shown in Figure 9, in $\triangle A B C$, $\angle C=$ $90^{\circ}, I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$ | 5.24
As shown in Figure 16, construct the symmetric point $F$ of $E$ with respect to $AD$, and the symmetric point $G$ of $D$ with respect to $BE$. Then $F$ and $G$ lie on $AB$, and $AF = AE$, $BG = BD$. Therefore, $\angle AIB = \angle EID = 135^{\circ}$, $\angle DIB = \angle EIA = \angle AIF = \angle BIG = 45^{\circ}... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,192 |
II. (15 points) Given that $a$ is a positive integer, and $a^{2}+$ $2004 a$ is a perfect square of a positive integer. Find the maximum value of $a$. | Let $a^{2}+2004 a=m^{2}$, where $m$ is a positive integer.
By completing the square and constant transformation, we get
$$
(a+1002)^{2}-m^{2}=1002^{2}=2^{2} \times 3^{2} \times 167^{2} \text{. }
$$
It is easy to see that $a+1002+m$ and $a+1002-m$ are both even numbers, and
$$
a+1002+m>a+1002-m>0 \text{. }
$$
To find ... | 250000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,193 |
Three, (15 points) In $\triangle A B C$, $B C=a, A C=$ $b, A B=c$, and it satisfies
$$
a^{4}+b^{4}+\frac{1}{2} c^{4}=a^{2} c^{2}+b^{2} c^{2} \text {. }
$$
Determine the shape of $\triangle A B C$. | Three, from $a^{4}+b^{4}+\frac{1}{2} c^{4}=a^{2} c^{2}+b^{2} c^{2}$, we get $\left(a^{2}-\frac{1}{2} c^{2}\right)^{2}+\left(b^{2}-\frac{1}{2} c^{2}\right)^{2}=0$.
It is easy to see that $a^{2}=\frac{1}{2} c^{2}$ and $b^{2}=\frac{1}{2} c^{2}$.
Therefore, $a=b$ and $a^{2}+b^{2}=c^{2}$.
Thus, $\triangle A B C$ is an isos... | \triangle ABC \text{ is an isosceles right triangle} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,194 |
Four, (15 points) Can any 8 consecutive positive integers be divided into two groups such that the sum of the squares of the 4 numbers in each group is equal? If so, please provide a grouping method and verify it; if not, please explain the reason.
| That is,
$$
\begin{array}{l}
(a+1)^{2}+(a+2)^{2}+(a+4)^{2}+(a+7)^{2} \\
=a^{2}+(a+3)^{2}+(a+5)^{2}+(a+6)^{2} .
\end{array}
$$
Thus, any 8 consecutive positive integers can be divided into the following two groups:
$$
\{a+1, a+2, a+4, a+7\},\{a, a+3, a+5, a+6\} \text {, }
$$
which satisfy
$$
\begin{array}{l}
(a+1)^{2}... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,195 |
Five. (15 points) As shown in Figure 10, let $A_{1}, B_{1}, C_{1}, D_{1}, E_{1}, F_{1}$ be the midpoints of the sides $AB, BC, CD, DE, EF,$ and $FA$ of the convex hexagon $ABCDEF$, respectively. It is known that
the areas of $\triangle ABC_{1}, \triangle BCD_{1},$
$\triangle CDE_{1}, \triangle DEF_{1},$
$\triangle EFA_... | $$
\begin{array}{l}
S_{\triangle \triangle C_{1}}=\frac{S_{\triangle \triangle B C}+S_{\triangle A B D}}{2}, \\
2 S_{\triangle B B C_{1}}=S_{\triangle \triangle B C}+S_{\triangle B B D} .
\end{array}
$$
Similarly, by connecting $B E$ and $C F$, we have
$$
\begin{array}{l}
2 S_{\triangle B C D_{1}}=S_{\triangle B C D}+... | S=\frac{2}{3} m | Geometry | proof | Yes | Yes | cn_contest | false | 715,196 |
Example 8 If $n$ is even, prove: in the $n \times n$ table as shown in Table 1, there does not exist a set of $n$ numbers $\{1,2$, $\cdots, n\}$, such that any two numbers in this set are in different rows and different columns.
Analysis: To prove this problem, if the condition "any two numbers among the $n$ numbers a... | Proof: Suppose there exists $\{1,2, \cdots, n\}$, any two numbers are in different rows and columns. Let the coordinates of each number be $(i, j)$, where $i$ and $j$ represent the $i$-th row and $j$-th column, respectively, then the total sum of the row and column coordinates of all numbers in $\{1,2, \cdots, n\}$ is
... | null | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,198 |
II. (20 points) As shown in Figure 4, in circle $\odot O$, chord $AB$ divides the circle into two segments with areas in the ratio $1:3$. Find the size of the central angle $\angle AOB$ (require precision to $1'$ or 0.001 radians).
| Let $\odot O$ have a radius of $r$, and let $M$ be a point on the minor arc $\overparen{A B}$. The radian measure of $\overparen{A M B}$ is $\alpha$. Then
$$
\begin{array}{l}
=\frac{1}{2} a r^{2}-\frac{1}{2} r^{2} \sin \alpha \text {. } \\
\end{array}
$$
Then $\frac{1}{2} a r^{2}-\frac{1}{2} r^{2} \sin \alpha=\frac{1}... | 2.31 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,206 |
Three. (20 points) Place the 2004 positive integers 1, 2, $\cdots$, 2004 randomly on a circle. By counting the parity of all adjacent 3 numbers, it is found that there are 600 groups where all 3 numbers are odd, and 500 groups where exactly 2 numbers are odd. How many groups have exactly 1 odd number? How many groups h... | Three, let there be $x$ groups with exactly 1 odd number. Then, all are not odd in $2004-600-500-x=904-x$.
Starting from a certain number on the circle, sequentially denote them as $x_{1}, x_{2}$, $\begin{array}{l} \cdots, x_{2000} \text {, let } \\ y_{i}=\left\{\begin{array}{cc}-1, & \text { when } x_{i} \text { is o... | 206, 698 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,207 |
$$
\begin{array}{l}
A=\left\{(x, y) \mid x^{2}+y^{2}-10 x-16 y+81<0\right\}, \\
B=\{(x, y)|y \geqslant| x-t \mid+8\} .
\end{array}
$$
(1) If $A \cap B \neq \varnothing$, find the range of real number $t$;
(2) Suppose point $P(t, 8) \in A$, and the boundaries of the two planar regions represented by sets $A$ and $B$ int... | (1) Set
$$
A=\left\{(x, y) \mid(x-5)^{2}+(y-8)^{2}<(2 \sqrt{2})^{2}\right\} \text {. }
$$
When the ray $y=x-t+8(x \geqslant t)$ is tangent to the circle $(x-5)^{2}+(y-8)^{2}=(2 \sqrt{2})^{2}$, we have
$$
\frac{|5-8-t+8|}{\sqrt{2}}=2 \sqrt{2} \text {. }
$$
Solving this, we get $t=1$ or $t=9$ (discard).
When the ray $y... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,208 |
One, (20 points) Given
$$
f(x)=(x-1)(x-2) \cdots \cdots(x-2004) \text {. }
$$
Ask: How many real numbers $x$ satisfy $|f(x)|=1$? | 一、As shown in Figure 7, draw a sketch of the function $f(x)$.
From the graph, we know that between 1 and 2004, for every 2 consecutive integers, there are 2 real numbers such that $|f(x)|=1$.
In the intervals $(0,1)$ and $(2004,+\infty)$, there is 1 real number $x$ in each interval such that $|f(x)|=1$.
Therefore, th... | 4008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,210 |
Sure, here is the translated text:
---
II. (20 points) Let the set $M=\left\{1^{2}, 2^{2}, \cdots, 1000^{2}\right\}$. Question: Can the set $M$ be partitioned into 2 non-empty subsets $A$ and $B$, such that:
(1) $A \cup B=M, A \cap B=\varnothing$;
(2) The sum of the elements in set $A$ is equal to the sum of the elem... | II. The conclusion is affirmative.
Because $1^{2}+2^{2}+\cdots+1000^{2}$
$$
=\frac{1}{6} \times 1000 \times 1001 \times 2001=333833500 \text {, }
$$
Therefore, $\frac{333833500}{2}=166916750$.
$$
\begin{array}{l}
\text { and } 1^{2}+2^{2}+\cdots+793^{2}<166916750 \\
<1^{2}+2^{2}+\cdots+794^{2}=167170745,
\end{array}
$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,211 |
Three. (20 points) As shown in Figure 5, there are two rectangular metal parts $ABCD$ and $ECFG$ placed on the Cartesian coordinate plane. A laser beam emitted from the origin $O$ will cut the parts. If the laser $(l)$ cuts two metal parts (both above $l$) from the two rectangles with exactly equal areas in one go, fin... | Three, it is known that $k_{O C}=\frac{2}{5}, k_{O E}=\frac{4}{7}, k_{O F}=\frac{3}{5}, k_{O G}=\frac{6}{7}$.
Let the equation of line $l$ be $y=k x$.
(1) When $k_{O F} \leqslant k \leqslant k_{O G}$, i.e., $\frac{3}{5} \leqslant k \leqslant \frac{6}{7}$, let $l$ intersect the boundaries of the two metal parts at point... | y=\frac{2}{3} x \text{ or } y=\frac{31-\sqrt{33}}{58} x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,212 |
1. The three interior angles of $\triangle A B C$ satisfy $\sin A \cdot \cos B$ $-\sin B=\sin C-\sin A \cdot \cos C$. Then ( ).
(A) $\angle A=90^{\circ}$
(B) $\angle B=90^{\circ}$
(C) $\angle C=90^{\circ}$
(D) $\triangle A B C$ is not necessarily a right triangle | 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,213 |
2. Let $\left\{z_{n}\right\}$ be a sequence of complex numbers, defined as $z_{n}=(1+\mathrm{i})\left(1+\frac{\mathrm{i}}{\sqrt{2}}\right) \cdots\left(1+\frac{\mathrm{i}}{\sqrt{n}}\right)$. Then $\sum_{n=1}^{2004}\left|z_{n}-z_{n+1}\right|=(\quad$.
(A) 2004
(B) 1
(C) 0
(D) $\sqrt{2004}$ | 2.A
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 | 715,214 |
3. Given that the equations of the lines on which two sides of a rectangle lie are
$$
(m+1) x+y-2=0
$$
and $4 m^{2} x+(m+1) y-4=0$.
Then the value of $m$ is ( ).
(A) -1
(B) $-\frac{1}{3}$ or -1
(C) $\frac{1}{3}$ or 1
(D) $-\frac{1}{3}$ | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,215 |
4. Two vectors $a, b$ satisfy
$$
|a-2 b|=1,|2 a+3 b|=\frac{1}{3} \text {. }
$$
Then the value of $(5 a-3 b) \cdot(a-9 b)$ is $(\quad)$.
(A) 0
(B) 10
(C) $\frac{80}{9}$
(D) $\frac{80}{3}$ | 4.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. The text "4.C" does not require translation as it is already in a form that is the same in both Chinese and ... | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,216 |
5. Let the distance from $P(x, y)$ to the fixed point $M\left(-\frac{27}{2}, 0\right)$ be $d_{1}$, and the distance from point $P$ to the $y$-axis be $d_{2}$. If $5 d_{1}+2 d_{2} = 48$, then ( ).
(A) The locus of the moving point $P$ is one branch of a hyperbola
(B) The eccentricity of the locus curve of the moving poi... | 5.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,217 |
Example 2 Let $a, b, c, d$ be non-negative real numbers, and
$$
a b+b c+c d+d a=1 .
$$
Prove: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+$
$$
\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3} \text {. }
$$
(31st IMO Shortlist) | Prove: Let $S=a+b+c+d, S>0$.
Construct the function $f(x)=\frac{x^{2}}{S-x}$, since this function is increasing on $[0, S)$, therefore, for any $x \in[0, S)$, we have
$$
\left(x-\frac{S}{4}\right)\left[f(x)-f\left(\frac{S}{4}\right)\right] \geqslant 0 .
$$
Thus, $\frac{x^{3}}{S-x}-\frac{S x^{2}}{4(S-x)}-\frac{S x}{12}... | \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,220 |
8. Given that the molecular formula of a series of compounds is $C_{m} \mathrm{H}_{n}$ (where $m, n$ are positive integers), the calculation formula for the ratio of carbon atoms is $\frac{m}{m+n}$. A series of molecular formulas for these compounds are given: $\mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6}, \cdots, \m... | 8. $\frac{1}{5} \leqslant x<\frac{1}{3}$ | \frac{1}{5} \leqslant x < \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,221 |
9. Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} \geqslant 1 \text {. }
$$
Analysis: To prove the inequality in the conclusion, we can first use the known condition and the mean inequality to prove the following 3 inequalities:
$$
\begin{array}{l}
\frac{1}{1+2a} ... | 9. $-\frac{2}{3}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,222 |
10. Let $f(x)=x^{2}-2 a x-a^{2}-\frac{3}{4}$. If for any $x \in[0,1]$, we have $|f(x)| \leqslant 1$, then the range of real number $a$ is $\qquad$ . | 10. $-\frac{1}{2} \leqslant a \leqslant \frac{\sqrt{2}}{4}$ | -\frac{1}{2} \leqslant a \leqslant \frac{\sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,223 |
Three. (20 points) Let $\Gamma$ be the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0), A(x, y)$ be a point on $\Gamma$, and $B, C, D$ be the points symmetric to $A$ with respect to the $y$-axis, the origin, and the $x$-axis, respectively. Let $E$ be a point on $\Gamma$ such that $A E \perp A C$, and let $... | Three, let $E$ be $\left(x_{1}, y_{1}\right)$, from $k_{A C} \times k_{A E}=-1$, we have
$$
\frac{y}{x} \cdot \frac{y_{1}-y}{x_{1}-x}=-1 \text {. }
$$
Rearranging gives $y_{1}=\frac{x^{2}+y^{2}}{y}-\frac{x}{y} \cdot x_{1}$.
Also, $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$, substituting $y_{1}$ into equation (... | \frac{x^{\prime 2}}{\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot a\right)^{2}}+\frac{y^{\prime 2}}{\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot b\right)^{2}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,226 |
Five. (20 points) Find all positive integers $n$ such that $n=$ $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}$, where $p_{1}, p_{2}, p_{3}, p_{4}$ are the four smallest distinct positive divisors of $n$. | If $n$ is odd, then all factors of $n$ are odd, i.e., $n \not\equiv 0(\bmod 4)$.
But $n=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2} \equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $21 \nmid n$.
If $4 \mid n$, then $p_{1}=1, p_{2}=2$.
Thus, $n=1+0+p_{3}^{2}+p_{4}^{2} \not\equiv 0(\bmod 4)$, which is a contradiction... | 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,227 |
Six, (20 points) Find the summation formula for $\sum_{k=0}^{n} k^{4}$, and provide a proof.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Six, from $(n+3)(n+2)(n+1) n$ $=n^{4}+6 n^{3}+11 n^{2}+6 n$,
we have $n^{4}=(n+3)(n+2)(n+1) n-6 n^{3}-11 n^{2}-6 n$.
Notice that
$$
\begin{array}{l}
\sum_{k=1}^{n}(k+3)(k+2)(k+1) k \\
=\sum_{k=1}^{n} \frac{1}{5}\{(k+3)(k+2)(k+1) k[(k+4)- \\
(k-1)]\} \\
=\frac{1}{5} \sum_{k=1}^{n}[(k+4)(k+3)(k+2)(k+1) k- \\
(k+3)(k+2)(k... | \frac{1}{5} n^{5}+\frac{1}{2} n^{4}+\frac{1}{3} n^{3}-\frac{1}{30} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,228 |
Four. (20 points) Let $a_{1} \in \mathbf{R}_{+}, i=1,2, \cdots, 5$. Find
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}+3 a_{3}+5 a_{4}+7 a_{5}}+\frac{a_{2}}{a_{3}+3 a_{4}+5 a_{5}+7 a_{1}}+ \\
\cdots+\frac{a_{5}}{a_{1}+3 a_{2}+5 a_{3}+7 a_{4}}
\end{array}
$$
the minimum value. | Let the original expression be $A$. By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
A \cdot\left[a_{1}\left(a_{2}+3 a_{3}+5 a_{4}+7 a_{5}\right)+a_{2}\left(a_{3}+3 a_{4}+\right.\right. \\
\left.\left.\quad 5 a_{5}+7 a_{1}\right)+\cdots+a_{5}\left(a_{1}+3 a_{2}+5 a_{3}+7 a_{4}\right)\right] \\
\quad \geqslant\... | \frac{5}{16} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,229 |
1. Let $M=\left(\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2+\sqrt{3}}}-2\right)\left(2005^{2}-2005 \times 2004 + 2004^{2}\right)$. Then among the following conclusions, the correct one is ( ).
(A) $M1$
(C) $M=0$
(D) $0<M<2005 \times 2004$ | - 1.C. Since $\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2+\sqrt{3}}}=\frac{\sqrt{2}(1+\sqrt{3}) \times \sqrt{2}}{\sqrt{4+2 \sqrt{3}}}=2$, therefore, $M=0$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,230 |
Example 1 Given that $p$ is a prime number, such that the quadratic equation in $x$, $x^{2}-2 p x+p^{2}-5 p-1=0$, has two integer roots. Find all possible values of $p$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Since this is a quadratic equation with integer coefficients and it has integer roots, we have
$$
\Delta=4 p^{2}-4\left(p^{2}-5 p-1\right)=4(5 p+1)
$$
is a perfect square. Therefore, $5 p+1$ is a perfect square.
Let $5 p+1=n^{2}$. Since $p \geqslant 2$, we have $n \geqslant 4$.
Notice that $5 p=(n-1)(n+1)$, ... | p=3 \text{ or } p=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,231 |
Example 2 Given that $a$, $b$, and $c$ are all integers, and for all real numbers $x$, we have
$$
(x-a)(x-2005)-2=(x-b)(x-c)
$$
holds. Find all such ordered triples $(a, b, c)$.
Analysis: $(x-a)(x-2005)-2=(x-b)$. $(x-c)$ always holds, which means $x^{2}-(a+2005) x+$ $2005 a-2=(x-b)(x-c)$ always holds, indicating that
... | Given by the problem, we have
$$
\begin{array}{l}
x^{2}-(a+2005) x+2005 a-2 \\
=(x-b)(x-c)
\end{array}
$$
always holds, hence
$$
x^{2}-(a+2005) x+2005 a-2=0
$$
has two integer roots $b$ and $c$. Therefore,
$$
\begin{array}{l}
\Delta=(a+2005)^{2}-4(2005 a-2) \\
=(a-2005)^{2}+8
\end{array}
$$
is a perfect square. Let ... | (2004,2003,2006),(2004,2006,2003),(2006,2004,2007),(2006,2007,2004) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,232 |
Example 1 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd "Hope Cup" National Mathematics Invitational Competition) | Explanation: The original formula is transformed to
$$
u(x, y)=\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}+(x-y)^{2}-2 \text {. }
$$
Consider the points $P_{1}\left(x, \frac{9}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$. When $x \in \mathbf{R}(x \neq 0)$, point $P_{1}$ lies on a hyperbola with the coordinate axes... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,233 |
7.1. Can the positive integers $1 \sim 16$ be placed in the 16 squares of a $4 \times 4$ grid so that the sums of the numbers in each row and each column are exactly 8 consecutive integers? | 7.1. Not possible.
Assuming it can be done, the sum of the sums of each row and each column should be twice the sum of the numbers in the table, which is $16 \times 17$. However, the equation $k+(k+1)+\cdots+(k+7)=16 \times 17$ has no integer solution. | Not possible | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,234 |
7.2. A city held three rounds of Go tournaments, with the same participants in each round. It is known that every 2 participants in the three rounds of the tournament won 1 time each, drew 1 time each. A person came in last place in the first two rounds. Question: What place did he get in the third round? | 7.2. Calculate the sum of scores for each participant in the three rounds of the competition, and find that they are all equal. Since someone scored less than any other participant in the first two rounds, he scored more than any other participant in the third round. Therefore, he came in first in the third round. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,235 |
7.3. A four-digit number can be divided by the sum of the two two-digit numbers formed by its first two digits and last two digits. Question: Can the sum of these two two-digit numbers be equal to 94?
---
Translating the text as requested, retaining the original formatting and line breaks. | 7.3. Cannot.
Let the four-digit number be $\overline{a b c d}$. Suppose $\overline{a b}+\overline{c d}=94$, then $\overline{a b c d}$ can be divisible by 94, and $\overline{a b c d}=100 \overline{a b}+\overline{c d}=99 \overline{a b}+94$. Therefore, $99 \overline{a b}$ can be divisible by 94, which is only possible wh... | Cannot | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,236 |
7.4. Given point $O$ inside a convex pentagon $A B C D E$, such that $\triangle O A B, \triangle O B C, \triangle O C D, \triangle O D E,$ and $\triangle O E A$ are all congruent. Prove: These triangles are either all isosceles triangles or all right triangles. | 7.4. Consider the 5 interior angles at vertex $O$ of the triangles. Since the 5 triangles are congruent to each other, there are at most three different values among these angles.
If there are 3 angles that are all different, then they must correspond to the 3 interior angles of 1 triangle, so their sum is $180^{\circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,237 |
7.5. If $a, b, c, d, e, f, g, h, k$ are all 1 or -1, try to find the maximum possible value of
$$
a e k - a f h + b f g - b d k + c d h - c e g
$$ | 7.5. Since each term in the expression $a e k - a f h + b f g - b d k + c d h - c e g$ is either 1 or -1, the value of the expression is even. However, the expression cannot equal 6. If it did, then $a e k$, $b f g$, and $c d h$ would all have to be 1, making their product 1, while $a f h$, $b d k$, and $c e g$ would a... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,238 |
7.6. If a person has fewer than 10 acquaintances, then this person is called "寡合的" (solitary); if all of a person's acquaintances are "寡合的" (solitary), then this person is called "古怪的" (eccentric). Prove: "eccentric" people will not outnumber "solitary" people. | 7.6. Referring to the "solitary" odd people as normal people. Suppose there are $m$ other "sociable" people, and $k$ other "odd" people. Calculate the number of acquaintanceship pairs among them.
On one hand, the number of such acquaintanceship pairs is no less than $10 k$, because each "odd person" only knows "sociab... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 715,239 |
7.7. In some cells of a rectangular grid, stars are drawn. It is known that in each cell with a star, the number of cells with stars in its row is equal to the number of cells with stars in its column. Prove that the number of rows with stars is equal to the number of columns with stars. | 7.7. In each cell marked with an asterisk, write a fraction whose value is the reciprocal of the number of asterisks in the column of that cell. Thus, in each column with asterisks, the sum of the numbers written is 1. By the problem statement, in each row with asterisks, the sum of the numbers written is also 1. The n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,240 |
8.3. Let $\triangle A B C$ be an equilateral triangle, point $M$ on side $A C$, and point $N$ on the extension of side $B C$ beyond point $C$. Given that $B M=M N$. Prove: $A M=C N$. | 8.3. Take a point $K$ on side $A B$ such that $K M / / B C$. It is easy to see that $\triangle M C N \cong \triangle B K M$. Therefore, $C N=K M=A M$. | A M=C N | Geometry | proof | Yes | Yes | cn_contest | false | 715,242 |
8.4. Let \( x, y \) be positive integers, and \(\frac{x^{2}-1}{y+1}+\frac{y^{2}-1}{x+1}\) is also a positive integer. Prove: the two terms in the sum are also integers. | 8.4. Let $\frac{x^{2}-1}{y+1}=u, \frac{y^{2}-1}{x+1}=v$. Since $u+v$ and $w w$ are both positive integers, $u, v$ are roots of a certain quadratic equation with integer coefficients $z^{2}+m z+n=0$. Because $u, v$ are rational numbers, the discriminant $m^{2}-4 n$ of the equation has a rational square root. Since $m^{2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,243 |
Example 2 Let non-negative real numbers $a, b, c$ satisfy $ab + bc + ca = 1$. Find the minimum value of $u = \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$. | Given:
$$
(b+c)(c+a)=1+c^{2} \text {. }
$$
Taking the reciprocal and transforming, we get
$$
\frac{1}{b+c}+\frac{1}{c+a}=\frac{2 c+(a+b)}{1+c^{2}} \text {. }
$$
Thus, \( u=\frac{1}{a+b}+\frac{a+b}{1+c^{2}}+\frac{2 c}{1+c^{2}} \).
From \( a b+b c+c a=1 \),
$$
\begin{array}{l}
\Rightarrow \frac{1}{4}(a+b)^{2}+c(a+b) \g... | \frac{5}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,244 |
8.6. Let quadrilateral $ABCD$ be a convex quadrilateral, and point $M$ be the midpoint of side $BC$. It is known that $\angle AMD=120^{\circ}$. Prove:
$$
AB+\frac{BC}{2}+CD \geqslant DA \text{. }
$$ | 8.6. Let the symmetric point of $B$ with respect to $A M$ be $B^{\prime}$, and the symmetric point of $C$ with respect to $M D$ be $C^{\prime}$. It is easy to see that $\angle B^{\prime} M C^{\prime}=60^{\circ}$, and $\triangle B^{\prime} M C^{\prime}$ is an equilateral triangle. Therefore, the length of the broken lin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,245 |
8.7. Squares in a grid that share a common edge are called adjacent. In a $7 \times 7$ grid, one Go stone is placed in one of the squares. It is allowed to place new stones in empty squares, but only in those that have at most one adjacent square occupied by a stone. How many stones can be placed in the grid at most? | 8.7. Place chess pieces according to the rules while calculating the number of sides of the squares that have pieces (each side is counted only once, even if both squares on either side of the side have pieces. After placing the 1st piece, 4 sides are obtained, and each subsequent piece can add at most 3 new sides. The... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,246 |
Question 11 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2},$ $\cdots, b_{n}$ satisfying the conditions $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+$ $\cdots+b_{n}=1$. Find
$$
\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}
$$
the minimum value.
(2003, Belar... | Solution: By Cauchy-Schwarz inequality, we have
$$
\sum_{t=1}^{n} \frac{a_{t}^{2}}{a_{t}+b_{t}} \geqslant \frac{\left(\sum_{t=1}^{n} a_{t}\right)^{2}}{\sum_{t=1}^{n}\left(a_{t}+b_{t}\right)}=\frac{1}{2},
$$
with equality if and only if $a_{1}=a_{2}=\cdots=a_{n}=b_{1}=b_{2}=\cdots=$ $b_{n}$, in which case the minimum v... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,247 |
1. Given real numbers $x, y, z$ satisfy $x+y+z=0$ and $x y z>0$. Let $T=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, then ().
(A) $T>0$
(B) $T=0$
(C) $T<0$
(D) The sign of $T$ cannot be determined | $$
-1 . C .
$$
Since $x y z>0$, then $x, y, z$ are all non-zero.
$$
\begin{array}{l}
\text { Therefore, } x^{2}+y^{2}+z^{2}>0 . \\
\text { Also, }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0,
\end{array}
$$
Thus, $x y+y z+z x<0$.
$$
\text { Therefore, } T=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{x y+y z+z x}{x... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,248 |
2. If the positive real numbers $a_{1}: a_{2}: a_{3}: a_{4}=1: q: q^{2}: q^{3}$ $(q \neq 1)$, and the average of $a_{1}$ and $a_{2}$ is $\frac{1}{2} a_{3}$, then the value of $\frac{a_{1}+a_{2}+a_{3}}{a_{2}+a_{3}+a_{4}}$ is $(\quad$.
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{-2+\sqrt{5}}{2}$
(C) $\frac{-1+\sqrt{5}}{2}$
(D)... | 2.C.
It is evident that $q>0$. From the given condition, we have
$$
\frac{a_{1}+a_{2}+a_{3}}{a_{2}+a_{3}+a_{4}}=\frac{1}{q} \text {. }
$$
Also, $a_{1}+a_{1} q=a_{1} q^{2}$, which simplifies to $1+q-q^{2}=0$.
Therefore, $\frac{1}{q^{2}}+\frac{1}{q}-1=0$.
Solving this, we get $\frac{1}{q}=\frac{-1 \pm \sqrt{5}}{2}$.
Si... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,249 |
3. As shown in Figure 1, quadrilateral DEFG is inscribed in $\triangle A B C$, and it is known that the areas of $\triangle A D E$, $\triangle E F C$, and $\triangle D B G$ are $1$, $3$, and $1$, respectively. Therefore, the area of $\square D E F G$ is ( ).
(A) $2 \sqrt{3}$
(B) 2
(C) 3
(D) 4 | 3. D.
As shown in Figure 4, draw $AO \parallel DG$ intersecting $BC$ at point $O$, then
$$
\begin{array}{l}
\frac{S_{\triangle DBC}}{S_{\triangle OOB}}=\frac{S_{\triangle EFC}}{S_{\triangle AOC}} \\
=\frac{S_{\triangle EFC}+S_{\triangle DOCC}}{S_{\triangle BBC}} .
\end{array}
$$
Therefore, $S_{\triangle \triangle OB}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,250 |
4. Given the parabola $y=a x^{2}(a>0)$ and the line $y=b x+c(b \neq 0)$ have two common points, their x-coordinates are $x_{1} 、 x_{2}$, and the line $y=b x+c$ intersects the x-axis at the point $\left(x_{3}, 0\right)$. Then the relationship satisfied by $x_{1} 、 x_{2} 、 x_{3}$ is ( ).
(A) $x_{1}+x_{2}=x_{3}$
(B) $\fra... | 4. B.
Solving the system of equations $\left\{\begin{array}{l}y=a x^{2}, \\ y=b x+c\end{array}\right.$ yields $a x^{2}-b x-c=0$.
Therefore, $x_{1}+x_{2}=\frac{b}{a}, x_{1} x_{2}=-\frac{c}{a}$.
Substituting $y=0$ into $y=b x+c$, we get $x_{3}=-\frac{c}{b}$.
Thus, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,251 |
5. As shown in Figure 2, point $O$ is inside $\triangle A B C$, and points $P, Q, R$ are on sides $A B, B C, C A$ respectively, with $O P / / B C$, $A B, O P=O Q=O R$ $=x$. Let $B C=a, C A$ $=b, A B=c$. Then $x$ equals ( ).
(A) $\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
(B) $\frac{1}{9}(a+b+c)$
(C) $\frac{1}{3} \s... | 5. A.
As shown in Figure 5, construct $P K \parallel O Q, Q L \parallel O R, R M \parallel O P$, intersecting $B C, C A, A B$ at points $K, L, M$ respectively. It is easy to see that
$$
\begin{array}{l}
P K=K Q=Q L \\
=O R=x .
\end{array}
$$
Let $B K=y, Q C=z$, then $x+y+z=B C=a$.
From $\frac{P K}{A C}=\frac{B K}{B C... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,252 |
6. Given $x, y \neq 0$, and $x, y$ satisfy $x y=\frac{x}{y}=$ $x-y$. Then the value of $x+y$ is ( .).
(A) $-\frac{3}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) $\frac{3}{2}$ | 6. A.
Since $x, y \neq 0$, from $xy = \frac{x}{y}$ we know $y^2 = 1$.
Thus, $y = \pm 1$.
If $y = 1$, from $xy = x - y$, we get $x = x - 1$, which is impossible. Therefore, $y = -1$.
From $xy = x - y$, we get $x = -\frac{1}{2}$.
Therefore, $x + y = -\frac{1}{2} + (-1) = -\frac{3}{2}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,253 |
1. Given real numbers $x, y$ satisfy $x^{3}+y^{3}=2$. Then the maximum value of $x+y$ is $\qquad$ . | Ni.1.2.
Let $x+y=k$, it is easy to know that $k>0$.
From $x^{3}+y^{3}=2$, we get
$$
(x+y)\left(x^{2}-x y+y^{2}\right)=2 \text {. }
$$
Thus, $x y=\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)$.
From this, we know that $x, y$ are the two real roots of the equation about $t$
$$
t^{2}-k t+\frac{1}{3}\left(k^{2}-\frac{2}{k}\ri... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,254 |
Example 3 Given that $x, y, z$ are non-negative real numbers, not all zero. Find
$$
u=\frac{\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z x}}{x+y+z}
$$
the minimum value. | Let $z_{1}=x+\frac{y}{2}+\frac{\sqrt{3}}{2} y \mathrm{i}$,
$$
z_{2}=y+\frac{z}{2}+\frac{\sqrt{3}}{2} z \mathrm{i}, z_{3}=z+\frac{x}{2}+\frac{\sqrt{3}}{2} x \mathrm{i},
$$
where $x \geqslant 0, y \geqslant 0, z \geqslant 0$. Then
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z x} \... | \sqrt{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,255 |
2. In Rt $\triangle A B C$, $\angle C=90^{\circ}, B C=5$, and the other two sides $A B$ and $A C$ are of integer lengths. Extend $A C$ to $D$, and make $\angle A B D=90^{\circ}$ to get $\triangle A B D$. Then the circumradius $R$ of $\triangle A B D$ is $\qquad$ | $2.7 \frac{1}{24}$.
By the Pythagorean theorem, we get $A B^{2}-A C^{2}=25$, which means $(A B+A C)(A B-A C)=25$.
Since the positive integer factors of 25 are $1,5,25$, we have
$$
\left\{\begin{array}{l}
A B+A C=25, \\
A B-A C=1 .
\end{array}\right.
$$
Solving, we get $A B=13, A C=12$.
By the projection theorem, we ge... | 7 \frac{1}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,256 |
3. In $\triangle A B C$, the lengths of the three sides are $a=3$, $b=4, c=6, h_{a}$ represents the altitude on side $B C$, and $h_{b} 、 h_{c}$ are defined similarly. Then
$$
\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)=
$$
$\qquad$ | $3.9 \frac{3}{4}$.
Let the area of $\triangle ABC$ be $S$, then we have
$$
\begin{array}{l}
S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c} \text {. } \\
\text { Therefore, }\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right) \\
=\left(\frac{2 S}{a}+\frac{2 S}{b}+\fr... | 9 \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,257 |
4. Given a three-digit number $\overline{a b c}$, the last three digits of its square are also $\overline{a b c}$. Then, the sum of all such three-digit numbers is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.1001 .
Let $x=\overline{a b c}$, then the last three digits of $x^{2}-x$ are all 0. Therefore, $x(x-1)$ is a multiple of 1000.
Since $1000=2^{3} \times 5^{3}$, and $x$ and $x-1$ are coprime, thus, $x-1$ is a multiple of $5^{3}$ or $x$ is a multiple of $5^{3}$.
When $x-1=125,125 \times 3,125 \times 5,125 \times 7$,... | 1001 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,258 |
(20 points) (1) Prove: If the quadratic trinomial $a x^{2}+b x+c$ always takes integer values when $x$ takes any integer value, then $2a$, $a+b$, and $c$ are all integers;
(2) First write the converse of (1), then determine its truth value, and prove your conclusion. | (1) Let $M=a x^{2}+b x+c$.
When $x=0$, $M$ is an integer, i.e., $c$ is an integer.
When $x=1$, $M$ is an integer, i.e., $a+b+c$ is an integer. Thus, $a+b=(a+b+c)-c$ is the difference of two integers, and is also an integer.
When $x=2$, $M$ is an integer, which implies $4 a+2 b+c$ is an integer. And $2 a=(4 a+2 b+c)-2(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 715,259 |
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