problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. Prove:
$$
a \cos \frac{A}{2}+b \cos \frac{B}{2}+c \cos \frac{C}{2} \leqslant \frac{\sqrt{3}}{2}(a+b+c) .
$$ | Prove that the original inequality is equivalent to
$$
\frac{a \cos \frac{A}{2}+b \cos \frac{B}{2}+c \cos \frac{C}{2}}{a+b+c} \leqslant \frac{\sqrt{3}}{2} \text {. }
$$
By the weighted power mean inequality, we have
$$
\begin{array}{l}
\frac{a \cos \frac{A}{2}+b \cos \frac{B}{2}+c \cos \frac{C}{2}}{a+b+c} \\
\leqslant... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,087 |
As shown in Figure 1, $M$ is a moving point on the circumcircle of quadrilateral $ABCD$, $AB$ intersects $DC$ at point $E$, $DA$ intersects $CB$ at point $F$, $FM$ intersects lines $AB$ and $CD$ at points $R$ and $S$ respectively, $EM$ intersects lines $AD$ and $BC$ at points $P$ and $Q$ respectively, and the intersect... | Proof Let $M X$ and $E F$ intersect at point $Y$. Then
$$
\begin{aligned}
\frac{E Y}{Y F} & =\frac{E M}{M F} \cdot \frac{\sin \angle X M E}{\sin \angle X M F}=\frac{E M}{M F} \cdot \frac{Q X}{X R} \cdot \frac{R M}{M Q} \\
= & \frac{E M}{M F} \cdot \frac{R M}{M Q} \cdot \frac{S M}{R S} \cdot \frac{P Q}{M P}=\frac{E M}{M... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,088 |
For any given positive number $\lambda$, do there exist infinitely many positive integers $n$ such that the largest prime factor of $3 n^{2}+1$ is less than $\lambda \sqrt[3]{n^{2}}$? | Prove that for $n=3m^2$, we have
$$
3 n^{2}+1=\left(3 m^{2}+1\right)\left(3 m^{2}+3 m+1\right)\left(3 m^{2}-3 m+1\right).
$$
First, prove that the prime factors of $3 m^{2}+3 m+1$ can be infinitely large.
If not, the prime factors of $3 m^{2}+3 m+1$ form a finite set $\left\{p_{1}, p_{2}, \cdots, p_{s}\right\}$.
Cons... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,089 |
There are $4 n\left(n \in \mathbf{Z}_{+}\right)$ circles with equal radii on a plane, where any two circles do not intersect with the other three circles. Let the number of intersection points of these circles be $f(n)$. Find the minimum value of $f(n)$. | Let $M$ be the set of these $4 n$ circles $C_{1}, C_{2}, \cdots, C_{4 n}$, and $N$ be the set of the intersection points $P_{1}, P_{2}, \cdots, P_{f(n)}$ of these $4 n$ circles.
Let $C \in M$ and $P \in N$.
If point $P$ is not on circle $C$, define $F(C, P)=0$;
If point $P$ is on circle $C$, define $F(C, P)=\frac{1}{m}... | 4n | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,090 |
Example 8 Given an integer $n \geqslant 4$, let $A, B \subseteq \{1,2, \cdots, n\}$ such that for any $a \in A, b \in B$, $a b+1$ is a perfect square. Prove: $\min \{|A|,|B|\} \leqslant \log _{2} n$. $[6]$ | Proof of the lemma first.
Lemma Let $a, a', b, b'$ be positive integers such that $a < a'$ and $b < b'$.
Proof: Note that,
$$
\begin{array}{l}
(a b + 1)(a' b' + 1) > (a b' + 1)(a' b + 1) \\
\left(\Leftrightarrow (a' - a)(b' - b) > 0\right).
\end{array}
$$
Thus, $\sqrt{(a b + 1)(a' b' + 1)} > \sqrt{(a b' + 1)(a' b + 1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,091 |
1. Given positive integers $a, b$ such that $\frac{a+1}{b}+\frac{b+1}{a}$ is an integer. Prove: $(a, b) \leqslant \sqrt{a+b}$. | Let $(a, b)=d$.
From $\frac{a+1}{b}+\frac{b+1}{a}=\frac{a^{2}+b^{2}+a+b}{a b}$, it can be proven that $d^{2} \mid (a+b)$.
Therefore, $d^{2} \leqslant a+b$. | d^{2} \leqslant a+b | Number Theory | proof | Yes | Yes | cn_contest | false | 727,092 |
2. For a given positive integer $n$, prove: there do not exist positive integers $a$, $b$, $c$, $d$, such that
$$
n^{2} \leqslant a<b<c<d \leqslant(n+1)^{2} \text {, and } a d=b c \text {. }
$$ | Prompt: Proof by contradiction.
Assume there exist $a, b, c, d$ that satisfy the conditions, and
$$
b=a+x, c=a+y(0 \leqslant x2 \sqrt{a}$.
Thus, $d>a+2 \sqrt{a}+1 \geqslant(n+1)^{2}$, contradiction. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,093 |
3. For positive integers $n>1$, let $d(n)$ denote the number of positive divisors of $n$, and $\sigma(n)$ denote the sum of these positive divisors. Prove:
$$
d(n) \sqrt{n}<\sigma(n)<\sqrt{2 d(n)} n .
$$ | Let the positive divisors of $n$ be
$$
1=d_{1}\sum_{i=1}^{k} 2 \sqrt{d_{i} d_{k+1-i}}=k \cdot 2 \sqrt{n},
\end{array}
$$
and
$$
\begin{array}{l}
(\sigma(n))^{2} \leqslant k\left(d_{1}^{2}+d_{2}^{2}+\cdots+d_{k}^{2}\right) \\
=k\left(\frac{n^{2}}{d_{k}^{2}}+\frac{n^{2}}{d_{k-1}^{2}}+\cdots+\frac{n^{2}}{d_{1}^{2}}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,094 |
4. Let positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
\begin{array}{l}
1<a_{1}<a_{2}<\cdots<a_{n} . \\
\text { Let } S=\sum_{k=1}^{n-1}\left[a_{k}, a_{k+1}\right]+\sum_{k=1}^{n} a_{k}, \\
t=\max \left\{a_{1}, a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}\right\} .
\end{array}
$$
Prove: $\frac{1}{t} \sum_... | The right inequality is estimated using
$$
\left[a_{k}, a_{k+1}\right] \leqslant a_{k} a_{k+1} \leqslant\left(a_{k+1}-1\right) a_{k+1}
$$
The equality condition is $a_{k}=k+1(k=1,2, \cdots, n)$.
The left inequality is estimated using
$$
\left[a_{k}, a_{k+1}\right]=\frac{a_{k} a_{k+1}}{\left(a_{k}, a_{k+1}\right)} \geq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,095 |
Example 3 As shown in Figure 4, given that the side length of square $A B C D$ is $1, P$ and $Q$ are two points inside it, and $\angle P A Q = \angle P C Q = 45^{\circ}$. Find the value of $S_{\triangle P A B} + S_{\triangle P C Q} + S_{\triangle Q A D}$. ${ }^{[3]}$ | As shown in Figure 4, rotate $\triangle A Q D$ 90° clockwise around point $A$ to $\triangle A Q^{\prime} B$, and rotate $\triangle C Q D$ 90° counterclockwise around point $C$ to $\triangle C Q^{\prime \prime} B$. Connect $P Q^{\prime}$ and $P Q^{\prime \prime}$. Then,
$$
\begin{array}{l}
\triangle A P Q^{\prime} \cong... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,096 |
Example 1 Let $n, m, k$ be positive integers, and
$$
\frac{1}{2} n(n+1)=m k, m \geqslant n \text {. }
$$
Prove: $(1,2, \cdots, n) \rightarrow\left(m^{k}\right)$. | Prove the statement for $n$ using mathematical induction.
When $n=1$, the conclusion is obviously true.
Assume $n \geqslant 2$, and the conclusion holds for all values less than $n$.
(1) If $m=n$, then $n$ is odd. Group $1,2, \cdots, n$ as $\{n\},\{1, n-1\},\{2, n-2\}, \cdots,\left\{\frac{n-1}{2}, \frac{n+1}{2}\right\}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,097 |
Proposition Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$ be positive integers, $m$ and $k$ be positive integers, $m \geqslant n$, and satisfy $\sum_{i=1}^{n} i a_{i}=m k$. Then there exists a positive integer $M$, such that
$$
\left(1^{a, M}, 2^{a, M}, \cdots, n^{a, M}\right) \rightarrow\left(m^{k M}\... | Prove: Use mathematical induction on $n$.
When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for all values less than $n$, consider the case for $n$.
First, explain that there exists a positive integer $t$, and non-negative integers $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}$, such t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,098 |
Example 2 Let $x$ be a positive real number, and $n$ be a positive integer. Prove:
$$
[n x] \geqslant \frac{[x]}{1}+\frac{[2 x]}{2}+\cdots+\frac{[n x]}{n} .
$$ | To prove that multiplying both sides of equation (1) by $n!$ is equivalent to proving:
$$
\sum_{i=1}^{n} \frac{n!}{i}[i x] \leqslant n![n x] \text {. }
$$
By the corollary, we need to prove that here $a_{i}=\frac{n!}{i}, k=n!, m=n$. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,099 |
Example 3 Let $n$ be a positive integer, real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, $y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}$, satisfy
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i} .
$$
Prove: For any real number $\alpha$ we have
$$
\sum_{i=1}^{n} x_{i}[i \alpha] \geqslant... | Proof: Let $z_{i}=x_{i}-y_{i} (1 \leqslant i \leqslant n)$.
Then $z_{1} \leqslant z_{2} \leqslant \cdots \leqslant z_{n}$, and $\sum_{i=1}^{n} i z_{i}=0$.
It suffices to prove: $\sum_{i=1}^{n} z_{i}[i \alpha] \geqslant 0$.
Let $z=\left(z_{1}, z_{2}, \cdots, z_{n}\right)$, $u=(1,2, \cdots, n)$,
$v=([\alpha],[2 \alpha], ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,100 |
Example 1 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=5, a_{1}=8, a_{n+1}=a_{n}^{2}-a_{n-1}(n \geqslant 1) \text {. }
$$
Prove: The sequence has infinitely many terms that are multiples of 13. | Prove that for a sequence where each term $a_{n}$ is taken modulo 13, the remainder $\overline{a_{n}}\left(\overline{a_{n}} \in\{0,1, \cdots, 12\}\right)$ is considered. Since $\overline{a_{n}}$ has only 13 different possible values, the pair $\left(\overline{a_{n}}, \overline{a_{n+1}}\right)$ has at most $13^{2}$ diff... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,101 |
Example 2 Figure 1 is a triangular network, from top to bottom, the first row has $n$ nodes, the second row has $n-1$ nodes, $\cdots \cdots$ the $n$-th row has 1 node. Now, each node is filled with a number 0 or 1 according to the following rule: for each inverted unit small equilateral triangle “ $\nabla$ ”, the numbe... | Let's consider a regular triangle with $k$ nodes in the first row, which is called a $k(k=2,3, \cdots)$-order regular triangle, and the maximum sum of the numbers filled in it is denoted as $f(k)$.
Obviously, $f(2)=2, f(3)=4$.
If we also consider the case with only one point as a degenerate regular triangle, then $f(1)... | f(n) \leqslant\left[\frac{n^{2}+n+1}{3}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,102 |
Example 3 A philanthropist recruits members for his club in the following way: each member can introduce two others to join, where these two are not introduced by anyone else; and each new member can also introduce two others to join. For a member $A$, the members introduced by $A$ and the members introduced by those i... | Solve using reverse thinking.
For this, let $200=k$, and denote the minimum number of members when exactly $r$ people receive coupons as $f(r)$. Members who join without being introduced (or without a superior) are called "bosses", and the members they introduce are called "followers". Each boss along with all their fo... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,103 |
Example 4 Find all pairs of coprime positive integers $\langle x, y\rangle$, such that $x\left|\left(y^{2}+210\right), y\right|\left(x^{2}+210\right)$. | $$
\begin{array}{l}
\text { For any pair of } x, y \text { that satisfy the conditions, let } \\
\left\{\begin{array}{l}
x^{2}+210=a y, \\
y^{2}+210=b x,
\end{array}\right. \\
\text { where } a, b \in \mathbf{Z}_{+} \text {. } \\
\text { Then } x^{4}+420 x^{2}+210^{2}=a^{2} y^{2}=a^{2}(b x-210) \\
\Rightarrow x\left(a^... | x_{n}= \frac{1}{107+\sqrt{11235}}(106+\sqrt{11235})^{n}+ \frac{1}{107-\sqrt{11235}}(106-\sqrt{11235})^{n}(n \in \mathbf{N}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,104 |
Example 4 (1). Exploration and Discovery
As shown in Figure 5, with the two sides of $\triangle A B C$, $A B$ and $A C$, squares $A B D E$ and $A C F G$ are constructed outward. Let $H$, $K$, and $M$ be the midpoints of segments $B E$, $C G$, and $B C$, respectively. Conjecture the quantitative and positional relations... | (1) Solution Conjecture: $M H \perp M K$, and $M H=M K$.
In fact, as shown in Figure 5, if $\triangle A E C$ is rotated $90^{\circ}$ counterclockwise around point $A$ to get $\triangle A B G$.
Then $E C \perp B G$, and $E C=B G$.
By the given conditions, we know that $M H$ and $M K$ are the midlines of $\triangle B C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,106 |
Example 2 If $m, n, p \in \mathbf{Z}$, and $m+n+p$ is a multiple of 6, then $m^{3}+n^{3}+p^{3}$ is also a multiple of 6. | Notice,
\[
\begin{array}{l}
m^{3}+n^{3}+p^{3}-3 m n p \\
=(m+n+p)\left(m^{2}+n^{2}+p^{2}-m n-n p-p m\right) .
\end{array}
\]
Since \( m+n+p \) is a multiple of 6, at least one of \( m, n, p \) must be even.
Thus, \( 3 m n p \) is a multiple of 6.
Therefore, \( m^{3}+n^{3}+p^{3} \) is a multiple of 6. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,107 |
Example 3 If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+x y(x+y), \\
f^{\prime}(0)=1,
\end{array}
$$
find the analytical expression of the function $f(x)$. | Notice that, $-x-y+(x+y)=0$.
From formula (1) we get
$$
\begin{array}{l}
(-x)^{3}+(-y)^{3}+(x+y)^{3} \\
=3(-x)(-y)(x+y)=3 x y(x+y) . \\
\text { By } f(x+y)=f(x)+f(y)+x y(x+y) \\
\Rightarrow f(x+y)=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] \\
\Rightarrow f(x+y)-\frac{1}{3}(x+y)^{3} \\
\quad=f(x)-\frac{1}{3... | f(x)=\frac{1}{3} x^{3}+x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,108 |
Example 4 Let $a, b, c$ be three distinct positive integers, and the positive integer $k$ satisfies $a b+b c+c a \geqslant 3 k^{2}-1$. Prove:
$$
\frac{1}{3}\left(a^{3}+b^{3}+c^{3}\right) \geqslant a b c+3 k .{ }^{[1]}
$$ | To prove that the required inequality is equivalent to
$$
a^{3}+b^{3}+c^{3}-3 a b c \geqslant 9 k \text {. }
$$
Given that the integers $|a-b|, |b-c|, |c-a|$ are not all 1, we have
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
\geqslant \frac{1}{2}\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,109 |
Example 5 If the positive integer $m$ makes it true that for any set of positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the pos... | Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
Verification shows that $m=1, m=2$ do not meet the... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,110 |
$$
\begin{array}{l}
\text { Example } 6 \text { Let } u=1+\frac{x^{3}}{3!}+\frac{x^{6}}{6!}+\cdots, \\
v=\frac{x}{1!}+\frac{x^{4}}{4!}+\frac{x^{7}}{7!}+\cdots, w=\frac{x^{2}}{2!}+\frac{x^{5}}{5!}+\frac{x^{8}}{8!}+\cdots
\end{array}
$$
Prove: $u^{3}+v^{3}+w^{3}-3 u v w=1$. | Let $\lambda=\mathrm{e}^{\frac{2 \pi}{3}}$ be a unit cube root.
Then $1+\lambda+\lambda^{2}=0$.
Thus $u^{3}+v^{3}+w^{3}-3 u v w$
$$
=(u+v+w)\left(u+\lambda v+\lambda^{2} w\right)\left(u+\lambda^{2} v+\lambda w\right)
$$
(By Corollary 3)
$$
=\mathrm{e}^{x} \mathrm{e}^{\lambda x} \mathrm{e}^{\lambda^{2} x}=\mathrm{e}^{0}... | 1 | Algebra | proof | Yes | Yes | cn_contest | false | 727,111 |
Example 7 explains that there are only three distinct points $A, B, C$ on the curve $x^{3}+3 x y+y^{3}=1$ that can form an equilateral triangle, and find the area of this equilateral triangle. | From formula (1), we know that $x^{3}+3 x y+y^{3}-1=0$ is actually reducible.
Because the left side of the equation can be factored as
$$
(x+y-1)\left(x^{2}-x y+y^{2}+x+y+1\right) \text {. }
$$
From formula (1), we know that the second factor can be written as
$$
\frac{1}{2}\left[(x+1)^{2}+(y+1)^{2}+(x-y)^{2}\right] \... | \frac{3 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,112 |
1. Factorize $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}$ | Let $x=b-c, y=c-a, z=a-b$.
$$
\begin{array}{l}
\text { By } x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \text {. }
\end{array}
$$
When $x+y+z=0$, $x^{3}+y^{3}+z^{3}=3 x y z$.
Therefore, the original expression $=3(a-b)(b-c)(c-a)$. | 3(a-b)(b-c)(c-a) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,113 |
2. Given $x, y, z \in \mathbf{R}_{+}$, satisfying $x^{2}+y^{2}+z^{2}=1$. Then $\min \left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=$ $\qquad$ . | Notice,
$$
\begin{array}{l}
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)\left(x^{2}+y^{2}+z^{2}\right) \\
\geqslant 3 \sqrt[3]{\frac{1}{x^{2}} \cdot \frac{1}{y^{2}} \cdot \frac{1}{z^{2}}} \times 3 \sqrt[3]{x^{2} y^{2} z^{2}}=9,
\end{array}
$$
when and onl... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,114 |
3. Let $a, b, c \in \mathbf{R}_{+}$. Prove the inequality
$$
2\left(\frac{a+b}{2}-\sqrt{a b}\right) \leqslant 3\left(\frac{a+b+c}{3}-\sqrt{3 a b c}\right) \text {, }
$$
and specify when equality holds. | Notice,
$$
\begin{array}{l}
3\left(\frac{a+b+c}{3}-\sqrt[3]{a b c}\right)-2\left(\frac{a+b}{2}-\sqrt{a b}\right) \\
=c+2 \sqrt{a b}-3 \sqrt[3]{a b c}, \\
c+2 \sqrt{a b}=c+\sqrt{a b}+\sqrt{a b} \\
\geqslant 3 \sqrt[3]{\sqrt{a b} \cdot \sqrt{a b} \cdot c}=3 \sqrt[3]{a b c} .
\end{array}
$$
Therefore, $c+2 \sqrt{a b}-3 \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,115 |
4. Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{6}=1, x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} . \\
\text { If } \max \left\{x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+\right. \\
\left.x_{5} x_{6} x_{1}+x_{6... | Let $r=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}$,
$$
\begin{aligned}
s= & x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+ \\
& x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2} .
\end{aligned}
$$
By the AM-GM inequality,
$$
\begin{array}{l}
r+s=\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\ri... | 559 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,116 |
Example 5 Given that $P$ is a point inside an acute triangle $\triangle ABC$ and minimizes $PA+PB+BC$. Determine the position of point $P$ and prove your conclusion.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve As shown in Figure 7, construct equilateral triangles $\triangle A C B^{\prime \prime}$ and $\triangle B C A^{\prime \prime}$ outwardly on sides $A C$ and $B C$, respectively. Connect $B B^{\prime}$ and $A A^{\prime}$, intersecting at point $P$. Then $P$ is the desired point.
The proof is as follows.
It is easy t... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,117 |
For any $\triangle ABC$, let its area be $S$, perimeter be $l$, and $P$, $Q$, $T$ be the points where the incircle of $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ respectively. Prove:
$$
\left(\frac{AB}{PQ}\right)^{3}+\left(\frac{BC}{QT}\right)^{3}+\left(\frac{CA}{TP}\right)^{3} \geqslant \frac{2}{\sqrt{3}} \cdot... | Proof As shown in Figure 1. Let the circumradius and inradius of $\triangle ABC$ be $R$ and $r$ respectively. It is easy to see that,
$$
\begin{array}{l}
BC = 2R \sin A, \\
TQ = AO \sin A \\
= \frac{r}{\sin \frac{A}{2}} \sin A = 2r \cos \frac{A}{2}. \\
\text{Therefore, } \frac{BC}{TQ} = \frac{2R}{r} = \sin \frac{A}{2}.... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,118 |
Let $a, b, c \in \mathbf{R}_{+}, ab + bc + ca \geqslant 3$. Prove:
$$
\begin{array}{l}
a^{5} + b^{5} + c^{5} + a^{3}(b^{2} + c^{2}) + b^{3}(c^{2} + a^{2}) + \\
c^{3}(a^{2} + b^{2}) \geqslant 9 .
\end{array}
$$ | To prove the original proposition is equivalent to proving
$$
\left(a^{3}+b^{3}+c^{3}\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant 9 .
$$
Also, $\left(a^{3}+b^{3}+c^{3}\right)^{2} \geqslant 9\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{3}$, so it suffices to prove $a^{2}+b^{2}+c^{2} \geqslant 3$.
Using the given condit... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,119 |
Question As shown in Figure 1, a unit $\mathrm{L}$ shape is composed of three unit squares. Prove: for any positive integer $k$, a similar $k$ times larger $\mathrm{L}$ shape can be divided into several unit $\mathbf{L}$ shapes. | Proof For convenience, let $m[\mathbf{L}]$ denote an $m$ times larger $\mathrm{L}$-shape similar to the unit $\mathrm{L}$.
The cases for $k=1,2,3$ are shown in Figure 2.
Assume the proposition holds for $k \leqslant n-1(n \geqslant 4)$.
Consider the case $k=n$.
(1) If $n$ is even, by the induction hypothesis, $n[\mathr... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,120 |
$$
\begin{array}{c}
1.2013+2012-2011-2010+2009+2008- \\
2007-2006+\cdots+5+4-3-2+1=(\quad) .
\end{array}
$$
(A) 2013
(B) 2012
(C) 1
(D) 0 | -,1. A.
It is evident that the symbols in the expression follow a periodic pattern.
Starting from the first number 2013, every four numbers form a group, resulting in 503 groups, with each group yielding a result of 4. The remaining number is 1, so the original expression $=4 \times 503+1=2013$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,121 |
2. Simplify $\sqrt{\frac{3+\sqrt{5}}{2}}-\sqrt{\frac{3-\sqrt{5}}{2}}=(\quad)$.
(A) $2 \sqrt{5}$
(B) 2
(C) 1
(D) $\sqrt{5}$ | 2. C.
Let $a=\sqrt{\frac{3+\sqrt{5}}{2}}-\sqrt{\frac{3-\sqrt{5}}{2}}$. It is clear that $a>0$.
Thus, $a^{2}=\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}-2 \sqrt{\frac{3+\sqrt{5}}{2}} \times \sqrt{\frac{3-\sqrt{5}}{2}}$
$=3-2 \sqrt{\left(\frac{3+\sqrt{5}}{2}\right)\left(\frac{3-\sqrt{5}}{2}\right)}$
$=3-2 \sqrt{\frac{9-5}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,122 |
3. Four candidates $A, B, C, D$ are participating in the student council election. It is known that candidate $D$ received more votes than candidate $B$, the sum of votes for candidates $A$ and $B$ exceeds the sum of votes for candidates $C$ and $D$, and the sum of votes for candidates $A$ and $C$ is equal to the sum o... | 3. D.
As shown in Figure 4, represent using a circle.
Since the total votes for candidates $A$ and $C$ are equal to the total votes for candidates $B$ and $D$, draw a diameter to divide the circle into two halves, with the upper half for $B$ and $D$, and the lower half for $A$ and $C$.
Given that candidate $D$ receiv... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,123 |
4. If in a certain month the number of Mondays is greater than the number of Tuesdays, then the year in which this occurs is ( ).
(A) 2010
(B) 2012
(C) 2014
(D) 2016 | 4. D.
From the problem, we know that the number of days for Wednesday, Thursday, Friday, Saturday, and Sunday in the month are not more than the number of days for Tuesday. A month can have 30 days, 31 days, 28 days, or 29 days.
We will discuss the four scenarios below.
(1) A month has 30 days. If the 29th of this mon... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 727,124 |
5. As shown in Figure 1, the diagonal $BD$ of rectangle $ABCD$ passes through the origin $O$, and the sides of the rectangle are parallel to the coordinate axes. The graph of the inverse proportion function $y=\frac{k}{x} (k>0)$ intersects $BC$ and $CD$ at points $M$ and $N$, respectively. If point $A(-2, -2)$, and the... | 5. B.
From the fact that the diagonal of a rectangle bisects the area of the rectangle, we know
Let $O H=a, O G=b$. Then
$$
a b=4, H M=\frac{k}{a}, G N=\frac{k}{b} \text {. }
$$
Thus, the points $M\left(\frac{k}{a}, a\right), N\left(b, \frac{k}{b}\right)$.
And $S_{\text {CHOG }}-S_{\triangle O G N}-S_{\triangle O H ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,125 |
1. Calculate:
$$
\frac{2013^{2}+2011}{2011^{2}-2013} \times \frac{4020^{2}-8040}{2011 \times 2014-4}=
$$
$\qquad$ | $=1.4$.
Let $a=2$ 011. Then
$$
\begin{array}{l}
\text { Original expression }=\frac{(a+2)^{2}+a}{a^{2}-a-2} \cdot \frac{(2 a-2)^{2}-4(a-1)}{a(a+3)-4} \\
=\frac{(a+1)(a+4)}{(a-2)(a+1)} \cdot \frac{4(a-1)(a-2)}{(a+4)(a-1)}=4 .
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,126 |
2. A sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is constructed according to the following rule: $a_{1}=7, a_{k}=a_{k-1}^{2}$'s digit sum $+1(k=2,3, \cdots)$. For example, $a_{2}=14, a_{3}=17$, and so on.
Then $a_{2013}=$ $\qquad$ . | 2.8.
From the problem, we know
$$
\begin{array}{l}
a_{1}=7, a_{2}=14, a_{3}=17, a_{4}=0, \\
a_{5}=5, a_{6}=8, a_{7}=11, a_{8}=5 .
\end{array}
$$
Thus, $a_{8}=a_{5}$, meaning from $a_{5}$ onwards, the sequence repeats with a period of 3.
Therefore, $a_{2013}=a_{6+3 \times 669}=8$. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,127 |
3. In $\triangle A B C$, it is known that $\angle A=2 \angle B, C D$ is the angle bisector of $\angle C$, $A C=16, A D=8$. Then $B C=$ | 3. 24 .
From $\angle A=2 \angle B$, we know $B C>A C$.
As shown in Figure 5, take a point $E$ on side $B C$ such that $E C=A C$,
and connect $D E$.
Then $\triangle C E D \cong \triangle C A D$
$$
\begin{array}{l}
\Rightarrow E D=A D, \\
\angle C E D=\angle C A D . \\
\text { Hence } \angle B D E=\angle C E D-\angle D... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,129 |
4. Given prime numbers $p$ and $q$, such that $p^{3}-q^{5}=(p+q)^{2}$. Then $\frac{8\left(p^{2013}-p^{2010} q^{5}\right)}{p^{2011}-p^{2009} q^{2}}=$ $\qquad$. | 4. 140.
If $p$ and $q$ have the same remainder when divided by 3, and since $p$ and $q$ are both prime numbers:
If the remainder is 0 when divided by 3, then $p=q=3$, at this time,
$$
p^{3}-q^{5}=0 \Rightarrow 3^{3}>q^{5} \Rightarrow q^{5}<27,
$$
such a prime number $q$ does not exist.
Therefore, it can only be $q=3$... | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,130 |
5. As shown in Figure 2, on the two legs $AC$ and $BC$ of the right triangle $\triangle ABC$, squares $ACDE$ and $CBFG$ are constructed, respectively, and line segment $DG$ is connected. The midpoints of line segments $AB$, $BF$, $FG$, $GD$, $DE$, and $EA$ are $P$, $L$, $K$, $I$, $H$, and $Q$, respectively. If $AC=14$,... | 5. 1004.5.
As shown in Figure 6, connect $D F$, $F A$, $A D$, $E G$, $G B$, and $B E$.
$$
\begin{array}{l}
\text { Then } S_{\text {hexagon } A F F G E}=1372, \\
S_{\triangle B I P}=S_{\triangle F K L}=S_{\triangle G K I}=\frac{1}{8} \times 28^{2}=98, \\
S_{\triangle D I H}=S_{\triangle E H Q}=S_{\triangle A Q P}=\fra... | 1004.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,131 |
Three, (10 points) (1) Given that $a$ and $b$ are positive integers. Prove that $(a+b) \mid \left(a^{3}+b^{3}\right)$;
(2) Let $N=1^{3}+2^{3}+\cdots+2012^{3}$. Prove:
$$
(2012 \times 2013) \mid N \text {. }
$$ | $$
\begin{array}{l}
\text { (1) From } a^{3}+b^{3}=a^{3}+a^{2} b-a^{2} b-a b^{2}+a b^{2}+b^{3} \\
=\left(a^{3}+a^{2} b\right)-\left(a^{2} b+a b^{2}\right)+\left(a b^{2}+b^{3}\right) \\
=a^{2}(a+b)-a b(a+b)+b^{2}(a+b) \\
=(a+b)\left(a^{2}-a b+b^{2}\right) \\
\Rightarrow(a+b) \text { divides }\left(a^{3}+b^{3}\right) .
\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,132 |
Four. (15 points) As shown in Figure 3, in isosceles $\triangle ABC$, it is known that the vertex angle $\angle A = 30^{\circ}$. Points $Q$ and $P$ are taken on sides $AB$ and $AC$ respectively, such that $\angle QPC = 45^{\circ}$, and $PQ = BC$. Prove: $BC = CQ$.
---
The translation preserves the original text's lin... | As shown in Figure 7, translate line segment $QP$ to $BO$, and connect $PO$. Then quadrilateral $QP OB$ is a parallelogram.
Thus, $BO = PQ = BC$.
Given $\angle QBO = \angle AQP$
$$
= 45^{\circ} - 30^{\circ} = 15^{\circ},
$$
we know $\angle OBC = \frac{180^{\circ} - 30^{\circ}}{2} - 15^{\circ}$
$$
= 60^{\circ},
$$
whi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,133 |
Five. (15 points) On the City Science Popularization Day, each middle school student can sign up for one subject competition. A reporter held a discussion with 33 contestants who had signed up, asking each of them the same two questions: How many people in the room share your school affiliation? How many people in the ... | Five, divide these 33 contestants into groups based on their school affiliation and the subject they are participating in (some groups may consist of only one contestant, for example, only one person from a certain school). Each person belongs to two groups, one divided by school affiliation and the other by subject. A... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,134 |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$ | 2.9.
From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$ | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,136 |
3. Let the function be
$$
f(x)=\left\{\begin{array}{ll}
2^{x}, & x \leqslant 0 ; \\
\left|\log _{2} x\right|, & x>0 .
\end{array}\right.
$$
Then the solution set of the equation $f(x)=\frac{1}{2}$ is $\qquad$ | 3. $\left\{-1, \frac{\sqrt{2}}{2}, \sqrt{2}\right\}$.
When $x \leqslant 0$, the equation is $2^{x}=\frac{1}{2} \Rightarrow x=-1$;
When $x>0$, the equation is
$\left|\log _{2} x\right|=\frac{1}{2} \Rightarrow \log _{2} x= \pm \frac{1}{2} \Rightarrow x=\sqrt{2}$ or $\frac{\sqrt{2}}{2}$.
Therefore, the solution set of th... | \left\{-1, \frac{\sqrt{2}}{2}, \sqrt{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,137 |
4. The function
$$
y=\frac{1}{|\sin x|}+\frac{1}{|\cos x|}+\frac{1}{|\tan x|}+\frac{1}{|\cot x|}
$$
has a minimum value of | 4. $2(\sqrt{2}+1)$.
It is easy to know that the given function is an even function, and $\pi$ is one of its periods. Therefore, we only need to find the minimum value when $x \in\left(0, \frac{\pi}{2}\right)$.
When $x \in\left(0, \frac{\pi}{2}\right)$,
$y=\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\tan x}+\frac{1}{\co... | 2(\sqrt{2}+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,138 |
5. Let $0<x<y<\frac{\pi}{2}$. Then
$$
P=\cos 2 x-\cos 2 y-4 \cos x+4 \cos y
$$
the range of values for $P$ is $\qquad$ | 5. $(-2,0)$.
From the given, we have
$$
P=2\left[\left(\cos ^{2} x-2 \cos x\right)-\left(\cos ^{2} y-2 \cos y\right)\right] \text {. }
$$
Let $f(t)=t^{2}-2 t$. Then
$$
P=2[f(\cos x)-f(\cos y)] \text {. }
$$
Given $0<x<y<\frac{\pi}{2}$, we know $0<\cos y<\cos x<1$.
Combining with the graph of the quadratic function $... | (-2,0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,140 |
6. Let $F$ be the right focus of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. A tangent line is drawn from a point $P$ outside the ellipse $C$, touching the ellipse at point $M$. If $\angle P F M=90^{\circ}$, then the equation of the locus of point $P$ is $\qquad$ | 6. $x=4$.
Let point $P\left(x_{0}, y_{0}\right), M\left(x_{1}, y_{1}\right)$. Then the equation of the tangent line $P M$ of the ellipse $C$ is $\frac{x_{1} x}{4}+\frac{y_{1} y}{3}=1$.
$$
\text { Hence } \frac{x_{1} x_{0}}{4}+\frac{y_{1} y_{0}}{3}=1 \text {. }
$$
Given point $F(1,0)$, then
$$
\overrightarrow{F P}=\le... | x=4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,141 |
7. From the set $A=\{1,2, \cdots, 30\}$, select five different numbers such that these five numbers form an arithmetic sequence. The number of different arithmetic sequences obtained is $\qquad$ . | 7. 196.
Let the five numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, with a common difference of $d$.
From $a_{5}-a_{1}=4d$, we know $a_{1} \equiv a_{5}(\bmod 4)$.
Divide the set $A=\{1,2, \cdots, 30\}$ into four categories based on the remainder modulo 4:
$$
\begin{array}{l}
B=\{1,5,9, \cdots, 29\}, \\
C=\{2,6,10, \c... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,142 |
8. Given that the volume of tetrahedron $P-ABC$ is $1$, $G$ and $K$ are the centroids of $\triangle ABC$ and $\triangle PBC$ respectively. A line through $G$ intersects $AB$ and $AC$ at points $M$ and $N$. Then the maximum volume of the quadrilateral pyramid $K-MNCB$ is | 8. $\frac{5}{27}$.
In $\triangle ABC$, let
$$
\overrightarrow{AM}=x \overrightarrow{AB}, \overrightarrow{AN}=y \overrightarrow{AC},
$$
$D$ is the midpoint of $BC$.
Then $\overrightarrow{AG}=\frac{2}{3} \overrightarrow{AD}=\frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC})$.
Thus, $\overrightarrow{MG}=\overrightarrow... | \frac{5}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,143 |
9. Given three distinct real numbers $a, b, c$ form a geometric sequence, and $\log _{c} a, \log _{b} c, \log _{a} b$ form an arithmetic sequence with a common difference of $d$. Then $d=$ $\qquad$ . | 9. $\frac{3}{2}$.
Since $a, b, c$ form a geometric sequence, we have $b^{2}=a c$.
Thus, $\log _{b} a+\log _{b} c=2$.
Let $\log _{b} a=x, \log _{b} c=y$. Then $x+y=2$.
Given that $\log _{c} a, \log _{b} c, \log _{a} b$ form an arithmetic sequence,
$\log _{c} a+\log _{a} b=2 \log _{b} c \Rightarrow \frac{x}{y}+\frac{1}{... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,144 |
10. Given $a, b, c, d \in[-1,+\infty)$, and $a+b+c+d=0$.
Then the maximum value of $a b+b c+c d$ is | 10. $\frac{5}{4}$.
Assume $b \geqslant c$ (otherwise, replace $(a, b, c, d)$ with $(d, c, b, a)$). Then
$$
\begin{array}{l}
a b+b c+c d=b(a+c+d)+c d-b d \\
=-b^{2}+(c-b)(d+1)-(c-b) \\
\leqslant-b^{2}+b-c \leqslant-b^{2}+b+1 \\
=\frac{5}{4}-\left(b-\frac{1}{2}\right)^{2} \leqslant \frac{5}{4} .
\end{array}
$$
Equality... | \frac{5}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,145 |
11. Find the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$. | 11. The domain of the function is $\{x \mid x \geqslant 1$ or $x \leqslant-1\}$.
(1) It is easy to see that the function $y=x^{2}+x \sqrt{x^{2}-1}$ is an increasing function on $[1,+\infty)$. Therefore, when $x \geqslant 1$, $y \geqslant 1$.
(2) When $x \leqslant-1$,
$$
y=x\left(x+\sqrt{x^{2}-1}\right)=\frac{x}{x-\sqrt... | \left(\frac{1}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,146 |
12. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}\left(n \in \mathbf{Z}_{+}\right)$ satisfy
$$
\begin{array}{l}
a_{1}=1, b_{1}=3, \\
a_{n+1}=2+\frac{27 a_{n}}{9 a_{n}^{2}+4 b_{n}^{2}}, b_{n+1}=\frac{27 b_{n}}{9 a_{n}^{2}+4 b_{n}^{2}} .
\end{array}
$$
(1) Prove: For all $n \in \mathbf{Z}_{+}$, we ... | 12. (1) Prove by mathematical induction.
When $n=1$, $a_{1}=1, b_{1}=3$, the conclusion is obviously true.
Assume when $n=k$, $\frac{\left(a_{k}-1\right)^{2}}{4}+\frac{b_{k}^{2}}{9}=1$ holds.
Then, $9 a_{k}^{2}+4 b_{k}^{2}=9\left(2 a_{k}+3\right)$.
Thus, $b_{k+1}=\frac{3 b_{k}}{2 a_{k}+3}$,
$a_{k+1}-1=1+\frac{27 a_{k}... | a_{n}=\frac{3 \times 9^{n-1}-1}{9^{n-1}+1} | Algebra | proof | Yes | Yes | cn_contest | false | 727,147 |
13. Let $P\left(x_{0}, y_{0}\right)$ be a fixed point inside the ellipse $\frac{x^{2}}{4}+y^{2}=1$ (not on the coordinate axes), and let two lines passing through $P$ intersect the ellipse at points $A, C$ and $B, D$, respectively, with $A B \parallel C D$.
(1) Prove that the slope of line $A B$ is a constant;
(2) Draw... | $$
\begin{array}{l}
C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right), \overrightarrow{A P}=\lambda \overrightarrow{P C}. \text{ Then } \\
x_{0}-x_{1}=\lambda\left(x_{3}-x_{0}\right), y_{0}-y_{1}=\lambda\left(y_{3}-y_{0}\right). \\
\text { Hence } \left\{\begin{array}{l}
x_{3}=\frac{(1+\lambda) x_{0}-x_{1}}{\lambd... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,148 |
1. Let the function $f(x)=\sqrt{1-x^{2}}\left(|x-2|+\frac{4}{2^{x}-1}\right)$.
Consider the proposition $p: f(x)$ is an odd function; proposition $q: f(x)$ is an even function. Then the correct conclusion is ( ).
(A) $p 、 q$
(B) $p \backsim \neg q$
(C) $\neg p 、 q$
(D) $\neg p \vee \neg q$ | - 1. B.
From the domain of $f(x)$ being $[-1,0) \cup(0,1]$, we know
$$
\begin{array}{l}
f(x)=\sqrt{1-x^{2}}\left(2-x+\frac{4}{2^{x}-1}\right) \\
=\sqrt{1-x^{2}}\left(-x+2 \cdot \frac{2^{x}+1}{2^{x}-1}\right)
\end{array}
$$
is an odd function, not an even function. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,149 |
3. As shown in Figure 11, a shape is formed by square $ABCD$ and $\triangle BEC$, where $\angle BEC$ is a right angle. Let the length of $CE$ be $a$ and the length of $BE$ be $b$. Then the distance from point $A$ to line $CE$ is $\qquad$ | As shown in Figure 11, rotate Rt $\triangle B E C$ 90 degrees clockwise around point $B$ to get Rt $\triangle A B G$.
Extend $A G$, intersecting $C E$ at point $F$. Then $A F \perp C E, A G=C E=a, B G=B E$.
Therefore, quadrilateral $B G F E$ is a square.
$$
\begin{array}{l}
\Rightarrow G F=B E=b \\
\Rightarrow A G+G F=... | a+b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,150 |
2. Let $B$ and $C$ be fixed points, neither of which lies on the plane $\alpha$, and let the moving point $A$ be on the plane $\alpha$, such that $\sin \angle A B C=\frac{1}{2}$. Then the locus of point $A$ is ( ).
(A) ellipse
(B) parabola
(C) hyperbola
(D) any of the above is possible | 2. D.
The locus of point $A$ that satisfies $\sin \angle A B C=\frac{1}{2}$ is a conical surface with $B$ as the vertex and $B C$ as the axis (the angle between the generatrix and the axis is $30^{\circ}$). Now, point $A$ also needs to lie on plane $\alpha$, so its locus is the intersection of the conical surface and ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,151 |
3. In $\triangle A B C$, it is known that $\overrightarrow{B C} \cdot \overrightarrow{B A}=\overrightarrow{C B} \cdot \overrightarrow{C A}$. Then $\triangle A B C$ is ( ).
(A) Isosceles triangle
(B) Right triangle
(C) Isosceles right triangle
(D) None of the above | 3. A.
From the problem, we know that $\overrightarrow{B C} \cdot(\overrightarrow{A B}+\overrightarrow{A C})=0$.
If we denote the midpoint of side $B C$ as $D$, then $\overrightarrow{A B}+\overrightarrow{A C}=2 \overrightarrow{A D}$. Therefore, $\overrightarrow{B C} \cdot \overrightarrow{A D}=0$. This indicates that $B... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,152 |
4. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $S_{n}$, and $\frac{S_{25}}{a_{23}}=5, \frac{S_{45}}{a_{33}}=25$. Then $\frac{S_{65}}{a_{43}}=(\quad$.
(A) 125
(B) 85
(C) 45
(D) 35 | 4. C.
Since $S_{25}=25 a_{13}$, then $\frac{a_{13}}{a_{23}}=\frac{1}{5}$.
Similarly, $\frac{a_{23}}{a_{33}}=\frac{5}{9}$.
Thus, $\frac{a_{33}}{a_{43}}=\frac{9}{13}$.
$$
\text { Therefore, } \frac{S_{65}}{a_{43}}=\frac{65 a_{33}}{a_{43}}=45 \text {. }
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,153 |
5. If the two perpendicular tangents of the curve $y=2 \sin \frac{x}{2}$ intersect at point $P$, then the coordinates of point $P$ cannot be ( ).
(A) $(\pi, \pi)$
(B) $(3 \pi,-\pi)$
(C) $(5 \pi,-\pi)$
(D) $(7 \pi,-\pi)$ | 5. C.
Notice that the slope of the tangent line to the function $y=2 \sin \frac{x}{2}$ at $x=x_{i}$ is $\cos \frac{x_{i}}{2}$.
Therefore, if the tangent lines at $x_{1}$ and $x_{2}$ are perpendicular to each other, then $\cos \frac{x_{1}}{2} \cdot \cos \frac{x_{2}}{2}=-1$.
Assume without loss of generality that $\cos ... | C | Calculus | MCQ | Yes | Yes | cn_contest | false | 727,154 |
6. If the solution set of the inequality $x^{2}<|x-1|+a$ is a subset of the interval $(-3,3)$, then the range of the real number $a$ is ( ).
(A) $(-\infty, 7)$
(B) $(-\infty, 7]$
(C) $(-\infty, 5)$
(D) $(-\infty, 5]$ | 6. D.
When $x \geqslant 1$, the original inequality becomes
$$
x^{2}-x+1-a<0 \text {, }
$$
its solution set does not contain any number greater than or equal to 3.
Therefore, when $x \geqslant 3$, we always have
$$
x^{2}-x+1-a \geqslant 0 \text {, }
$$
which leads to $a \leqslant 7$.
When $x<1$, the original inequal... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,155 |
1. If $\log _{2} \log _{8} x=\log _{8} \log _{2} x$, then $\log _{4} x=$ $\qquad$ | 2. 1. $\frac{3 \sqrt{3}}{2}$.
From the given, we know $\left(\log _{8} x\right)^{3}=\log _{2} x>0$.
Let $\log _{2} x=y$. Then $\left(\frac{y}{3}\right)^{3}=y \Rightarrow y=3 \sqrt{3}$.
Therefore, $\log _{4} x=\frac{3 \sqrt{3}}{2}$. | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,156 |
2. Let $M$ be a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. Given points $F(1,0)$ and $P(3,1)$. Then the maximum value of $2|M F|-|M P|$ is $\qquad$. | 2.1.
Notice that $F$ is the right focus of the ellipse, and the right directrix of the ellipse is $l: x=4$. Then $2|M F|$ is the distance from point $M$ to $l$.
Draw a perpendicular line $M A$ from point $M$ to $l$, and draw a perpendicular line $P B$ from point $P$ to $M A$, where $A$ and $B$ are the feet of the per... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,157 |
3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ . | 3. 1 .
Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,158 |
4. Let $x_{n}=\sum_{k=1}^{2013}\left(\cos \frac{k!\pi}{2013}\right)^{n}$. Then $\lim _{n \rightarrow \infty} x_{n}=$ | 4. 1953.
Notice that, $2013=3 \times 11 \times 61$.
Therefore, when $1 \leqslant k \leqslant 60$,
$$
\frac{k!}{2013} \notin \mathbf{Z}, \cos \frac{k!\pi}{2013} \in(-1,1).
$$
Hence $\lim _{n \rightarrow \infty} \cos ^{n} \frac{k!\pi}{2013}=0$.
And when $k \geqslant 61$, $\frac{k!}{2013}$ is an integer, and always an e... | 1953 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,159 |
5. If the sum of all elements in a non-empty subset of $\{1,2, \cdots, 9\}$ is a multiple of 3, the subset is called a "Jin state subset". Then the number of such Jin state subsets is $\qquad$ . | 5. 175.
If a proper subset is a peculiar subset, then its complement is also a peculiar subset. Therefore, we only need to consider peculiar subsets with fewer than or equal to 4 elements.
Thus, there are 3 peculiar subsets with only 1 element; there are $\mathrm{C}_{3}^{2}+3 \times 3=12$ peculiar subsets with exactl... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,160 |
Example 1 Given that $d$ and $d^{\prime}\left(d^{\prime}>d\right)$ are two positive divisors of the positive integer $n$. Prove: $d^{\prime}>d+\frac{d^{2}}{n}$. | Prove that since $f=\frac{n}{d}, f^{\prime}=\frac{n}{d^{\prime}}$ are integers, and $f>f^{\prime}$, we have $1 \leqslant f-f^{\prime}=\frac{n}{d}-\frac{n}{d^{\prime}}=\frac{\left(d^{\prime}-d\right) n}{d d^{\prime}}d+\frac{d^{2}}{n}$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,161 |
6. In $\triangle A B C$, it is known that $b c=b^{2}-a^{2}$, and $\angle B-$ $\angle A=80^{\circ}$. Then $\angle C=$ $\qquad$ (answer in degrees). | 6. $60^{\circ}$.
From $b c=b^{2}-a^{2}$
$$
\begin{aligned}
\Rightarrow & \sin B \cdot \sin C=\sin ^{2} B-\sin ^{2} A \\
& =\frac{1}{2}(\cos 2 A-\cos 2 B) \\
& =\sin (B+A) \cdot \sin (B-A) \\
\Rightarrow & \sin B=\sin (B-A)=\sin 80^{\circ} \\
\Rightarrow & \angle B=100^{\circ}, \angle A=20^{\circ} \\
\Rightarrow & \ang... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,162 |
1. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $D$ is the midpoint of side $A C$.
(1) Prove: $A B_{1} / /$ plane $B D C_{1}$.
(2) When $\frac{A A_{1}}{A B}$ takes what value, $A B_{1} \perp B C_{1}$? | Three, 1. (1) Take the midpoint $D_{1}$ of edge $A_{1} C_{1}$. Then $A D_{1} / / D C_{1}$. Therefore, $A D_{1} / /$ plane $B D C_{1}$.
Also, $B_{1} D_{1} / / B D$, so $B_{1} D_{1} / /$ plane $B D C_{1}$.
Hence, plane $A B_{1} D_{1} / /$ plane $B D C_{1}$.
Thus, $A B_{1} / /$ plane $B D C_{1}$.
(2) Suppose the base edge... | \frac{\sqrt{2}}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 727,163 |
2. In the Cartesian coordinate system, let $A$, $B$, $C$ be three distinct points on the curve $xy=1$, and let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Prove that the circumcircle of $\triangle DEF$ passes through the origin $O$. | 2. Without loss of generality, assume point $O$ does not coincide with any of $D$, $E$, or $F$.
Let point $A\left(a, \frac{1}{a}\right), B\left(b, \frac{1}{b}\right), C\left(c, \frac{1}{c}\right)$. Then the midpoint of side $B C$ is $D\left(\frac{1}{2}(b+c), \frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)\right)$, and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,164 |
3. Let $0 \leqslant \beta \leqslant \alpha \leqslant \frac{\pi}{4}(\alpha, \beta \in \mathbf{R}), n>1\left(n \in \mathbf{Z}_{+}\right)$.
Prove:
(1) $\frac{\tan \alpha - \tan \beta}{1 + \tan^2 \alpha} \leqslant \alpha - \beta$, and specify the conditions under which equality holds;
(2) $\sum_{k=1}^{n} \frac{1}{n^2 + k^... | 3. (1) Proof 1 Let
$$
\begin{array}{l}
f(\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan ^{2} \alpha}-(\alpha-\beta) \\
\Rightarrow f^{\prime}(\beta)=-\frac{1+\tan ^{2} \beta}{1+\tan ^{2} \alpha}+1 \geqslant 0 \\
\Rightarrow f(\beta) \leqslant f(\alpha)=0 .
\end{array}
$$
Since $f$ is strictly increasing on $[0, \alpha)$,... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,165 |
1. The range of the function $f(x)=|x+1|+|x-1|+\sqrt{4-x^{2}}$ is $\qquad$ . | - 1. $[2+\sqrt{3}, 2 \sqrt{5}]$.
Notice that, $f(x)$ is an even function with the domain $[-2,2]$.
When $0 \leqslant x \leqslant 1$, $f(x)=2+\sqrt{4-x^{2}}$ is monotonically decreasing, and its range is $[2+\sqrt{3}, 4]$;
When $1 \leqslant x \leqslant 2$, let $\theta=\arccos \frac{x}{2}$. Then
$$
\begin{array}{l}
f(x)... | [2+\sqrt{3}, 2 \sqrt{5}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,166 |
3. Simplify $\sin 12^{\circ} \cdot \sin 48^{\circ} \cdot \sin 54^{\circ}=$ $\qquad$ (answer in numbers). | 3. $\frac{1}{8}$.
From $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$, we get
$$
\begin{array}{l}
\sin 12^{\circ} \cdot \sin 48^{\circ} \cdot \sin 54^{\circ} \\
=\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right) \cos 36^{\circ} \\
=\frac{1}{4}\left(1-\cos 36^{\circ}+\cos 72^{\circ}\right) \\
=\frac{1}{4}\left(2 \sin 18^... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,168 |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=a_{2}=1, a_{n}=\sqrt{3} a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$
Then $a_{2013}=$ $\qquad$ | 4. $1-\sqrt{3}$.
From the given, we know
$$
\begin{array}{l}
a_{3}=\sqrt{3}-1, a_{4}=2-\sqrt{3}, a_{5}=\sqrt{3}-2, \\
a_{6}=1-\sqrt{3}, a_{7}=-1, a_{8}=-1 .
\end{array}
$$
Thus, $a_{n+6}=-a_{n}$.
Therefore, $\left\{a_{n}\right\}$ is a periodic sequence with a period of 12.
Hence, $a_{2013}=a_{9}=1-\sqrt{3}$. | 1-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,169 |
5. Let the circumcenter $P$ of $\triangle A B C$ satisfy
$$
\overrightarrow{A P}=\frac{2}{5}(\overrightarrow{A B}+\overrightarrow{A C}) \text {. }
$$
Then $\cos \angle B A C=$ $\qquad$ | 5. $\frac{1}{4}$.
Let $D$ be the midpoint of $B C$. According to the problem, point $P$ lies on segment $A D$ and does not coincide with $D$, thus $A B=A C$. Therefore, $\cos \angle B A C=\cos \angle B P D=\frac{D P}{B P}=\frac{D P}{A P}=\frac{1}{4}$. | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,170 |
6. Given the complex number $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, satisfying that the ratio of the real part to the imaginary part of $\frac{z+1}{z+2}$ is $\sqrt{3}$. Then the maximum value of $\frac{y}{x}$ is | 6. $\frac{4 \sqrt{2}-3 \sqrt{3}}{5}$.
Notice,
$$
\begin{array}{l}
\frac{z+1}{z+2}=\frac{x+1+y \mathrm{i}}{x+2+y \mathrm{i}} \\
=\frac{(x+1+y \mathrm{i})(x+2-y \mathrm{i})}{(x+2+y \mathrm{i})(x+2-y \mathrm{i})} \\
=\frac{(x+1)(x+2)+y^{2}+y \mathrm{i}}{(x+2)^{2}+y^{2}} .
\end{array}
$$
From the given condition,
$$
\beg... | \frac{4 \sqrt{2}-3 \sqrt{3}}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,171 |
Example 2 Let positive integers $a, b (a \neq b)$ satisfy $\left(a^{2}+a b+b^{2}\right) \mid a b(a+b)$.
Prove: $|a-b|>\sqrt[3]{3 a b}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Notice,
$$
\begin{array}{l}
a^{3}=a\left(a^{2}+a b+b^{2}\right)-a b(a+b), \\
b^{3}=b\left(a^{2}+a b+b^{2}\right)-a b(a+b) .
\end{array}
$$
According to the condition, we know
$$
\left(a^{2}+a b+b^{2}\right)\left|a^{3},\left(a^{2}+a b+b^{2}\right)\right| b^{3} .
$$
Therefore, $\left(a^{2}+a b+b^{2}\right) \mid\left(a^... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 727,172 |
8. Randomly select three different vertices from a regular 11-sided polygon, the probability that they form an acute triangle is $\qquad$ . | 8. $\frac{1}{3}$.
Randomly select three vertices $A, B, C$ from a regular 11-sided polygon. If $\angle B A C$ is an obtuse angle, then the arc $\overparen{B A C}$ is less than a semicircle. In this case, the number of ways to form an obtuse triangle is $11 \sum_{i=1}^{4}(5-i)=110$. The probability of forming an acute ... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,174 |
9. (21 points) Let the side length of the base of a regular tetrahedron be 1, and the length of the lateral edges be 2. Find its volume and the radius of the inscribed sphere.
保留了源文本的换行和格式。 | From the given, the height of the pyramid is
$$
h=\sqrt{2^{2}-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{11}{3}},
$$
its volume $V=\frac{1}{3} \times \frac{\sqrt{3}}{4} h=\frac{\sqrt{11}}{12}$,
its surface area $S=\frac{\sqrt{3}}{4}+\frac{3 \sqrt{15}}{4}$.
Thus, the radius of the inscribed sphere $r=\frac{3 V}{S... | \frac{\sqrt{11}}{\sqrt{3}+3 \sqrt{15}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,175 |
10. (21 points) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any $x, y$ we have
$$
f(x+y)=f(x)+f(y)+2xy,
$$
and $x^{2}-|x|^{\frac{1}{2}} \leqslant f(x) \leqslant x^{2}+|x|^{\frac{1}{2}}$. | 10. Let $g(x)=f(x)-x^{2}$. Then for any $x, y$ we have $g(x+y)=g(x)+g(y)$, and $|g(x)| \leqslant|x|^{\frac{1}{2}}$. Therefore, for any $x$ and positive integer $n$ we have $|g(x)|=\frac{|g(n x)|}{n} \leqslant \sqrt{\frac{|x|}{n}}$.
Let $n \rightarrow+\infty$, we get $g(x)=0$. Thus, $f(x)=x^{2}$. | f(x)=x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,176 |
11. (22 points) Let $a, b, c$ be real numbers not all zero. Find the range of $F=\frac{a b-b c+c^{2}}{a^{2}+2 b^{2}+3 c^{2}}$. Under what conditions on $a, b, c$ does $F$ achieve its maximum and minimum values? | 11. This problem is to find a real number $F$ such that the equation
$$
F a^{2}+2 F b^{2}+(3 F-1) c^{2}-a b+b c=0
$$
has a real solution $(a, b, c) \neq(0,0,0)$.
When $F=0$, equation (1) clearly has a solution
$$
(a, b, c) \neq(0,0,0).
$$
When $F \neq 0$, equation (1) has a solution $(a, b, c) \neq(0,0,0)$ if and onl... | \frac{-1-\sqrt{13}}{12} \leqslant F \leqslant \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,177 |
12. (22 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{2}=2, a_{n}=\frac{\left(1+a_{n-1}\right)^{2}}{a_{n-2}}(n \geqslant 3) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: For any positive integer $k, \sqrt{a_{2 k-1}}$ and $\sqrt{\frac{a_{2 k}... | 12. (1) From
$$
\left\{\begin{array}{l}
a_{n-2} a_{n}=a_{n-1}^{2}+2 a_{n-1}+1, \\
a_{n-1} a_{n+1}=a_{n}^{2}+2 a_{n}+1
\end{array}\right.
$$
subtracting the two equations gives
$$
a_{n-1}\left(a_{n+1}+a_{n-1}+2\right)=\left(a_{n-2}+a_{n}+2\right) a_{n} \text {. }
$$
Given $a_{3}=9$, we have
$$
\begin{array}{l}
\frac{a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,178 |
1. Given sets
$$
\begin{array}{l}
P=\{x|x \in \mathbf{R},| x-1 \mid<1\}, \\
Q=\{x|x \in \mathbf{R},| x-a \mid \leqslant 1\},
\end{array}
$$
and $P \cap Q=\varnothing$. Then the range of real number $a$ is ( ).
(A) $a \geqslant 3$
(B) $a \leqslant-1$
(C) $a \leqslant-1$ or $a \geqslant 3$
(D) $-1 \leqslant a \leqslant ... | 1. C.
From the problem, we know
$$
P=\{x \mid 0<x<2\}, Q=\{x \mid a-1<x<a+1\} .
$$
If $P \cap Q=\varnothing$, then $a-1 \geqslant 2$ or $a+1 \leqslant 0$.
Solving this, we get $a \leqslant-1$ or $a \geqslant 3$. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,179 |
2. If $\alpha, \beta \in \mathbf{R}$, then $\alpha+\beta=90^{\circ}$ is a ( ) condition for $\sin \alpha+\sin \beta>1$.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | 2. D.
When $\alpha=0, \beta=90^{\circ}$, $\sin \alpha+\sin \beta=1$.
When $\alpha=\beta=60^{\circ}$,
$\sin \alpha+\sin \beta=\sqrt{3}>1$, but $\alpha+\beta \neq 90^{\circ}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,180 |
3. Given a geometric sequence $\left\{a_{n}\right\}: a_{1}=3$, and the geometric mean of the first to eighth terms is 9. Then the third term is ( ).
(A) $3 \sqrt[9]{81}$
(B) $3 \sqrt[7]{81}$
(C) $\sqrt[3]{9}$
(D) $3 \sqrt{3}$ | 3. B.
Calculating, we get $q=3^{\frac{2}{7}}$, thus, $a_{3}=3 \sqrt[7]{81}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,181 |
4. Given the complex number $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, and $z^{2}=8 \mathrm{i}$. Then $z=(\quad$.
(A) $2+2$ i
(B) $-2+2 \mathrm{i}$ or $2-2 \mathrm{i}$
(C) $-2-2$ i
(D) $2+2 \mathrm{i}$ or $-2-2 \mathrm{i}$ | 4. D.
From the given, $x^{2}+2 x y \mathrm{i}-y^{2}=8 \mathrm{i} \Rightarrow x=y= \pm 2$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,182 |
Example 2 (Nine-Point Circle Theorem) The feet of the three altitudes, the midpoints of the three sides, and the midpoints of the segments joining the orthocenter to the vertices, these nine points are concyclic.
保持源文本的换行和格式如下：
Example 2 (Nine-Point Circle Theorem) The feet of the three altitudes, the midpoints of th... | Proof As shown in Figure 1, let $D, E, F$ be the feet of the altitudes, $L, M, N$ be the midpoints of the three sides, $H$ be the orthocenter, and $P, Q, R$ be the midpoints of segments $HA, HB, HC$ respectively.
From Rt $\triangle CBF \sim \text{Rt} \triangle ABD \Rightarrow \frac{BC}{BF}=\frac{BA}{BD}$.
Notice that, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,183 |
Example 3 Given that $H$ is the orthocenter of acute $\triangle A B C$, the circle with the midpoint of side $B C$ as its center and passing through point $H$ intersects line $B C$ at points $A_{1}$ and $A_{2}$; the circle with the midpoint of side $C A$ as its center and passing through point $H$ intersects line $C A$... | Prove as shown in Figure 2, let $B_{0}$ and $C_{0}$ be the midpoints of sides $CA$ and $AB$, respectively. Let the other intersection point of the two circles centered at $B_{0}$ and $C_{0}$, both passing through point $H$, be $A'$. Then $A'H \perp C_{0}B_{0}$.
Since $C_{0}B_{0} \parallel BC$, we know $A'H \perp BC$.
T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,184 |
Example 3 As shown in Figure 1, a plane is composed of an infinite triangular network of equilateral triangles with side length 1. The vertices of the triangles become lattice points, and lattice points with a distance of 1 are adjacent lattice points.
"A and B" two frogs are playing a jumping game.
"A single jump" ref... | It is impossible.
Without loss of generality, assume frog $B$ is on the adjacent grid point to the right of $A$.
Now, assign a value to each grid point: first, take the initial grid point of frog $A$, and assign this grid point a value of 1. The subsequent assignment rule is that the value assigned to any grid point is... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,185 |
Given the set
$$
A=\left\{\left(a_{1}, a_{2}, \cdots, a_{7}\right) \mid a_{i} \in\{0,1\}, i=1,2, \cdots, 7\right\} \text {, }
$$
the subset $V \subseteq A$ satisfies $|V|=16,(0,0, \cdots, 0) \in V$, and for any $v_{i} 、 v_{j} \in V(i \neq j), v_{i} 、 v_{j}$ differ in at least three components. Prove: $(1,1, \cdots, 1)... | Prove that an element $a$ in set $A$ covers element $b$ if and only if elements $a$ and $b$ differ in at most one position.
Let $V=\left\{v_{1}, v_{2}, \cdots, v_{16}\right\}$, and suppose the set of all elements covered by element $v_{i}$ in set $V$ is $V_{i}, V_{i} \subseteq A$. Then for any $v_{i}, v_{j} \in V (i \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,186 |
Prove: The equation $x\left(z^{2}-y^{2}\right)=z\left(x^{2}+y^{2}\right)$ has no positive integer solutions $(x, y, z)$. | Assume the original equation has positive integer solutions. Transform the original equation into
$$
x z^{2}-\left(x^{2}+y^{2}\right) z-x y^{2}=0 .
$$
Then the discriminant
$$
\Delta=\left(x^{2}+y^{2}\right)^{2}+(2 x y)^{2}
$$
is a perfect square.
Let $a=x^{2}+y^{2}, b=2 x y$. Then
$a>b$, and $a^{2}-b^{2}=\left(x^{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,187 |
Example 4: There are $n$ bookmarks, each with one side white and the other side black. Arrange them in a row with all the white sides facing up. Each operation (if possible) involves removing a white side-up bookmark (not at the edge) and flipping the adjacent two bookmarks to the other side. Prove: The necessary and s... | Prove that if a bookmark with the white side up is called a "white bookmark," and vice versa is called a "black bookmark," then the parity of the number of black bookmarks remains unchanged after all possible operations. Therefore, if only two bookmarks remain, these two bookmarks will have the same color.
If a bookma... | 3 \nmid (n-1) | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,188 |
Example 1 Calculate
$$
S\left(9 \times 99 \times 9999 \times \cdots \times \underset{2^{\circ} \uparrow}{99 \cdots 9}\right) \text {. }
$$
(1992, USA Mathematical Olympiad)
[Analysis] If you are familiar with Lemma 1 and have a bit of a sense of magnitude, you can get the answer right away. | Let $N=9 \times 99 \times 9999 \times \cdots \times \underset{2^{\cdots} \cdots 9}{99}$. Then
$$
N<10^{1+2+\cdots+2^{\cdots-1}}<10^{2^{n}}-1 \text {. }
$$
Using Lemma 1, we get
$$
\begin{array}{l}
S(9 \times 99 \times 9999 \times \cdots \times \underbrace{99 \ldots 9}_{2^{n} \uparrow}) \\
=S(N(\underbrace{99.9}_{2^{n}... | 9 \times 2^{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,190 |
1. The equation
$$
x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}-2 z^{2} x^{2}=24
$$
The set of all integer solutions $(x, y, z)$ is $\qquad$ | $-1 . \varnothing$.
Notice,
$$
\begin{aligned}
x^{4}+ & y^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}-2 z^{2} x^{2} \\
& =\left(x^{2}-y^{2}-z^{2}\right)^{2}-4 y^{2} z^{2} \\
& =\left(x^{2}-y^{2}-z^{2}+2 y z\right)\left(x^{2}-y^{2}-z^{2}-2 y z\right) \\
& =(x+y-z)(x-y+z)(x+y+z)(x-y-z) \\
& =24 .
\end{aligned}
$$
If the equat... | \varnothing | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,192 |
3. The function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
\begin{array}{c}
f(f(n))+f(n)=2 n+3 \text {, and } f(0)=1 . \\
\text { Then } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=
\end{array}
$$ | 3. 2772 .
Substituting $n=0$ into equation (1) yields
$$
f(1)+1=3 \Rightarrow f(1)=2 \text {. }
$$
Furthermore, let $n=1$.
From equation (1), we get $f(2)=3$.
Similarly, $f(n)=n+1(n=3,4, \cdots, 10)$.
$$
\text { Hence } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=2772 \text {. }
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,194 |
4. As shown in Figure 2, in $\triangle A B C$, it is known that $\angle B=60^{\circ}$, $\angle A C B=75^{\circ}$, and $D$ is a moving point on side $B C$. A circle $\odot O$ is constructed with $A D$ as its diameter, intersecting $A B$ and $A C$ at points $E$ and $F$ respectively. If the minimum length of chord $E F$ i... | 4. $\frac{4 \sqrt{6}}{3}$.
As shown in Figure 3, connect $O E$ and $O F$.
Given $\angle B=60^{\circ}, \angle A C B=75^{\circ}$, we know $\angle B A C=45^{\circ}$. Therefore, $\angle E O F=2 \angle B A C=90^{\circ}$.
Thus, $E F=\sqrt{2} O E$.
Since the minimum value of $E F$ is 2, the minimum value of $O E$ is $\sqrt{2... | \frac{4 \sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,195 |
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