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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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5. Let the sequence $\left\{b_{n}\right\}_{n \geqslant 1}$ satisfy
$$
\begin{array}{l}
b_{200 n}=2, b_{2008}=4, \\
b_{n+3}+b_{n+2} \\
=\frac{\left(b_{n+3}+b_{n+2}\right)\left(b_{n+1}+b_{n}\right)}{2}+1(n=1,2, \cdots)
\end{array}
$$
Then $b_{1}+b_{2}+\cdots+b_{1900}=$ | 5.1912 .
Let $a_{n}=b_{n+1}+b_{n}$. Then
$2 a_{n+2}-2=a_{n+2} a_{n}$
$$
\begin{array}{l}
\Rightarrow a_{n+2}=\frac{-2}{a_{n}-2}(n=1,2, \cdots) \\
\Rightarrow a_{n+4}=\frac{-2}{a_{n+2}-2}=\frac{-2}{\frac{-2}{a_{n}-2}-2}=\frac{a_{n}-2}{a_{n}-1} \\
\Rightarrow a_{n+x}=\frac{a_{n+4}-2}{a_{n+4}-1}=\frac{\frac{a_{n}-2}{a_{n... | 1912 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,418 |
6. Let $a, b \in \mathbf{N}_{+}$, and $a+b \sqrt{2}=(1+\sqrt{2})^{100}$. Then the last two digits of $a b$ are $\qquad$ . | 6.64.
Since $a+b \sqrt{2}=(1+\sqrt{2})^{100}$
$$
\begin{array}{l}
=(1+\sqrt{2})\left[\left((1+\sqrt{2})^{3}\right]^{33}\right. \\
=(1+\sqrt{2})(7+5 \sqrt{2})^{33},
\end{array}
$$
Therefore, by the binomial theorem we get
$$
\begin{array}{l}
a= \sum_{i=1}^{16} \mathrm{C}_{33}^{2 i} 7^{33-2 i}(5 \sqrt{2})^{2 i}+ \\
\s... | 64 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,419 |
Three. (20 points) Given a positive number $t$, let two non-zero vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in the same plane satisfy
$$
|\boldsymbol{a}+\boldsymbol{b}|=t|\boldsymbol{a}-\boldsymbol{b}| .
$$
Try to find the range of the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$. | Three, let $|\boldsymbol{a}|=k|\boldsymbol{b}|, k>0, \boldsymbol{a}$ and $\boldsymbol{b}$ have an angle $\theta$ between them. Then from $|\boldsymbol{a}+\boldsymbol{b}|=t|\boldsymbol{a}-\boldsymbol{b}|$ we get
$$
\begin{array}{l}
(\boldsymbol{a}+\boldsymbol{b}) \cdot(\boldsymbol{a}+\boldsymbol{b})=t^{2}(\boldsymbol{a}... | \theta \in\left[0, \arccos \frac{t^{2}-1}{t^{2}+1}\right] \text{ if } t>1; \theta \in\left[\arccos \frac{t^{2}-1}{t^{2}+1}, \pi\right] \text{ if } 0<t<1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,420 |
$$
\begin{array}{l}
\text { Four, (20 points) Let } a_{1}=1, \\
a_{n+1}=2 a_{n}+n^{2}\left(1+3^{n}\right)(n=1,2, \cdots) \text {. }
\end{array}
$$
Find the general formula for the term $a_{n}$. | Consider the following "auxiliary problem":
Let $x_{1}=1, x_{n+1}=2 x_{n}+n^{2}(n=1,2, \cdots)$.
Find the general term $x_{n}$.
$$
\text { Let } \begin{aligned}
& x_{n+1}+s(n+1)^{2}+t(n+1)+r \\
= & 2 x_{n}+n^{2}+s(n+1)^{2}+t(n+1)+r \\
= & 2\left[x_{n}+\frac{s+1}{2} n^{2}+\left(s+\frac{t}{2}\right) n+\frac{s+t+r}{2}\rig... | -23 \times 2^{n-1} + 3^{n}(n^2 - 6n + 15) - n^2 - 2n - 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,421 |
One, (50 points) Find all non-zero integers $a, b$, such that $\frac{(a, b)\left(a b^{2}+1\right)}{a^{2} b+2}$ is an integer. Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$. | (1) If $a=2 b$, then $(a, b)=|b|$,
$$
\frac{d\left(a b^{2}+1\right)}{a^{2} b+2} \in \mathbf{Z} \Leftrightarrow \frac{|b|\left(2 b^{3}+1\right)}{4 b^{3}+2} \in \mathbf{Z}
$$
$$
\Leftrightarrow \frac{|b|}{2} \in \mathbf{Z}.
$$
Let $b=2 t(t \in \mathbf{Z}, t \neq 0)$. Then $a=4 t$. Thus,
$$
\left\{\begin{array}{l}a=4 t, \... | (a, b) = (4t, 2t) \text{ for } t = \pm 1, \pm 2, \cdots; (1, -7), (1, -3), (1, -1), (1, 3); (-1, -5), (-1, -3), (-1, -1), (-1, 1); (2, -1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,423 |
II. (50 points) As shown in Figure 3, given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $O$ and $I$ do not coincide, $I$ is inside or on the boundary of $\triangle AOC$, and either $O$ is inside $\triangle ABC$ or $AB=AC$.
Try to find a necessary and sufficient condition (express... | The required necessary and sufficient condition is
$\angle B A C=60^{\circ}$ or $100^{\circ}$.
Sufficiency.
As shown in Figure 7, if $\angle B A C=60^{\circ}$, then
$\angle B O C=2 \angle B A C=120^{\circ}$,
$\angle B I C=180^{\circ}-\frac{\angle A B C+\angle A C B}{2}$
$=90^{\circ}+\frac{\angle B A C}{2}=120^{\circ}$.... | \angle B A C=60^{\circ} \text{ or } 100^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,424 |
7. If $n^{2}-19 n+91$ is a perfect square, find the value of the natural number $n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
n^{2}-19 n+91=(n-9)^{2}+(10-n), \\
n^{2}-19 n+91=(n-10)^{2}+(n-9) .
\end{array}
$$
When $n>10$,
$$
(n-10)^{2}<n^{2}-19 n+91<(n-9)^{2} \text {, }
$$
a contradiction. Therefore, $n \leqslant 10$. Direct verification shows that $n=9,10$. | n=9,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,425 |
Three, (50 points) Does there exist a function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for every $n \in \mathbf{N}$, we have
$$
f^{(1989)}(n)=2 n
$$
and $f^{(1900)}(f(n)+1)=2 n+2$?
Prove your conclusion (where, $f^{(1)}(n)=f(n)$,
$$
\left.f^{(k)}(n)=f\left(f^{(k-1)}(n)\right)\right) .
$$ | Three, such a function does not exist. Proof by contradiction. Assume that a function satisfying the given conditions exists.
From $f^{(1989)}(n)=2 n$, we get
$$
f(2 n)=f^{(190)}(n)=f^{(199)}(f(n))=2 f(n) \text {, }
$$
i.e., $f(2 n)=2 f(n)$.
Additionally, from $f^{(1999)}(n)=2 n$, it is easy to see that $f$ is injecti... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,426 |
Let the four-digit number $\overline{a b c d}$ be a perfect square, and its arithmetic square root can be expressed as $\sqrt{\overline{a b c d}}=\overline{a b}+\sqrt{\overline{c d}}$. How many such four-digit numbers are there? | Given $\sqrt{a b c d}=\overline{a b}+\sqrt{\overline{c d}}$, we know
$$
\begin{array}{l}
(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}+\overline{c d} \\
=\overline{a b c d}=100 \overline{a b}+\overline{c d} .
\end{array}
$$
Thus, $(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}=100 \overline{a b}$, whi... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,427 |
Initially 236 Quadrilateral $ABCD$ is inscribed in a circle, the extensions of $AB$ and $DC$ intersect at point $E$, and the extensions of $AD$ and $BC$ intersect at point $F$. If $S_{\triangle ABC}=S_{\triangle ADC}$, prove that line $AC$ bisects $EF$. | Proof: As shown in Figure 3, let the line $AC$ intersect $EF$ at point $M$. It is easy to see that
$$
\begin{array}{l}
\angle ABC + \angle ADC \\
= 180^{\circ}. \\
\text{Therefore, } \sin \angle ABC \\
= \sin \angle ADC. \\
\text{Also, } S_{\triangle ABC} = S_{\triangle ADC}, \text{ then } \\
AB \cdot BC = AD \cdot DC,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,428 |
235 Given real numbers $a, b, c$ satisfy
$$
a+b+c=1, a^{2}+b^{2}+c^{2}=1 \text {. }
$$
Prove: $a^{5}+b^{5}+c^{5} \leqslant 1$. | Prove: Squaring both sides of $a+b+c=1$ yields
$$
\begin{array}{l}
1=(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\
=1+2(a b+b c+c a),
\end{array}
$$
which implies $a b+b c+c a=0$.
Squaring both sides of equation (1) and rearranging, we get
$$
\begin{array}{l}
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\
=-2\left(a b^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,429 |
Given $\triangle ABC$ is an acute triangle, $H$ is the orthocenter. The rays $AH, BH, CH$ intersect the circumcircle of $\triangle ABC$ at points $A', B', C'$ respectively. Prove:
$$
\frac{3}{2} \leqslant \frac{AH}{AA'} + \frac{BH}{BB'} + \frac{CH}{CC'} < 2.
$$ | Proof: As shown in Figure 4, connect \(A^{\prime} B^{\prime}\), \(B^{\prime} C^{\prime}\), \(C^{\prime} A^{\prime}\), and \(A B^{\prime}\). Let \(B^{\prime} C^{\prime}\) intersect \(A A^{\prime}\) at point \(K\). It is easy to prove that
\[
\begin{array}{l}
\angle A A^{\prime} B^{\prime} \\
=\angle A B B^{\prime} \\
=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,430 |
8. Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (Hint: Transform the original equation into
$$
(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}+2 y-y^{2} \text {. }
$$
If $(3 y+1)(y+1)>0$ and $2 y-y^{2}<0$, then it contradicts that $(2 x+1)^{2}$ is a perfect square. Therefore,
$$
(3 y+1)(y+1) \leqslant 0 \text { or } 2 y-y^{2} \geqslant 0 \text {, }
$$
which means $-1 \le... | (x, y)= (0,-1),(-1,-1),(0,0),(-1,0),(5,2),(-6,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,431 |
9. Prove: $\sqrt{\frac{1}{\underbrace{11 \cdots 1}_{n-1} \underbrace{22 \cdots 25}_{n \uparrow}}}$ is a rational number. | (Hint: $\underbrace{11 \cdots 1}_{n-1 \uparrow} \underbrace{22 \cdots 25}_{n \uparrow}=\left(\frac{10^{n}+5}{3}\right)^{2}$.) | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,432 |
10. Given that $a$ is a positive integer. If the roots of the equation $x^{3}+(a+17) x^{2}+(38-a) x-56=0$ are all integers, find the value of $a$ and the integer roots of the equation. | (Tip: The equation has an integer root $x=1$. When $x \neq$ 1, $x^{2}+(a+18) x+56=0$, its discriminant $\Delta=$ $(a+18)^{2}-224$ should be a perfect square, leading to $a=39$ or $a=12$. When $a=39$, the three roots of the original equation are $1,-1$ and -56; when $a=12$, the three roots of the original equation are $... | a= | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,433 |
11. If for all integer values of $x$, the quadratic trinomial $a x^{2}+b x+c$ is a perfect square, prove:
(1) $2a$, $2b$ are integers;
(2) $a$, $b$, $c$ are integers, and $c$ is a perfect square.
Conversely, if (2) holds, is it true that for all integer values of $x$, the quadratic trinomial $a x^{2}+b x+c$ is a perfe... | (提示: Let $f(x)=a x^{2}+b x+c$. Set $x=0$, $\pm 1$, $2$, to get several equations, and then deduce the relevant conclusions; conversely, take $a=b=c=1, x=1$, then $a x^{2}+b x+c=3$ is not a perfect square.) | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,434 |
Theorem 2 For $a_{i j}>0(i=1,2, \cdots, m, j=1$, $2, \cdots, n)$, we have
$$\left(\sum_{i=1}^{m} \prod_{j=1}^{n} a_{i j}\right)^{n} \leqslant \prod_{j=1}^{n} \sum_{i=1}^{m} a_{i j}^{n} .$$ | Prove: Let $\sum_{i=1}^{m} a_{i j}^{n}=A_{j}^{n}$.
By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sum_{i=1}^{m} \prod_{j=1}^{n} \frac{a_{i j}}{A_{j}} \leqslant \sum_{i=1}^{m}\left(\frac{1}{n} \sum_{j=1}^{n} \frac{a_{i j}^{n}}{A_{j}^{n}}\right) \\
=\frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{m} \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,436 |
Example 1 Let the side lengths of $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ be $a, b, c$ and $a_{1}, b_{1}, c_{1}$, and their areas be $S$ and $S_{1}$, respectively. Prove:
$$
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)\left(a_{1}^{2}+b_{1}^{2}+c_{1}^{2}\right)- \\
2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c... | Prove: From the Cauchy-Schwarz inequality and the formula for the area of a triangle,
$$
16 S^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right) \text {, }
$$
we get
$$
\begin{array}{l}
16 S S_{1}+2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c^{2} c_{1}^{2}\right) \\
=4 S \cdot 4 S_{1}+\sqrt{2} a^{2} \cd... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,437 |
Example 2 Prove:
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}>\frac{2 n}{3 n+1} \text {. }
$$ | $$
\begin{array}{l}
\text { Prove: } 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n} \\
=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}+\frac{1}{2 n}\right)- \\
2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) \\
=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot... | \frac{2 n}{3 n+1} | Inequalities | proof | Yes | Yes | cn_contest | false | 720,438 |
Example 11 Prove: 12345678987654321 is a perfect square.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | Proof: Since $\underbrace{11 \cdots 1}_{n \uparrow}=\frac{10^{n}-1}{9}$, and
12345678987654321
is the sum of the following numbers:
11111111111111111
1111111111111110
111111111111100
11111111111000
1111111110000
111111100000
11111000000
1110000000
100000000
Therefore, 12345678987654321
$$
\begin{aligned}
= & \frac{10^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,439 |
Example 1 Let $|a|<1, |b|<1, |c|<1$.
$$
\text { Prove that } (1+a)(1+b)(1+c)>0 \text { and } (1-a)(1-b)(1-c)>0.
$$
Analysis: Treat any one of $a, b, c$ as a variable, and construct a linear function. For example, treat $a$ as a variable, and let $f(x)=(b+c) x+b c+1$. Therefore, it is only necessary to prove that $f(x)... | Proof: Note that
$$
\begin{array}{l}
a b+b c+c a+1=(b+c) a+b c+1, \\
|a|<1, \\
f(1)=b+c+b c+1 \\
=(1+b)(1+c)>0 .
\end{array}
$$
Since $f(x)$ is a linear function, when $x \in(-1,1)$, we always have $f(x)>0$, i.e., when $|a|<1$, we have $a b+b c+c a+1>0$. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,440 |
Example 3 Let $x, y, z$ satisfy
$$
x+y+z=a, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a} \text{. }
$$
Prove: $x, y, z$ must have one equal to $a$. | Proof: Let $x, y, z$ be the roots of the function
$$
f(t)=t^{3}+p t^{2}+q t+r
$$
Then
$$
\left\{\begin{array}{l}
p=-(x+y+z), \\
q=x y+y z+z x, \\
r=-x y z .
\end{array}\right.
$$
Thus, $p=-a$,
$$
q=x y+y z+z x=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) x y z=-\frac{r}{a} \text {. }
$$
Therefore, $f(t)=t^{3}-a ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,442 |
Example 4 If $P(x)$ represents an $n$-degree polynomial, and when $k=0,1, \cdots, n$, $P(k)=\frac{k}{k+1}$, try to find the expression for $P(n+1)$. | Solution: Let
$$
P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} .
$$
Since $P(k)=\frac{k}{k+1}$, we have
$$
(k+1) P(k)-k=0 \quad (k=0,1, \cdots, n) \text {. }
$$
Construct the function $f(x)=(x+1) P(x)-x$.
Thus, when $x=0,1, \cdots, n$, $f(x)=0$, where $f(x)$ is an $(n+1)$-degree polynomial in $x$.
Let $f(x)=... | P(n+1)=\left\{\begin{array}{ll}1, & n=2 k+1 ; \\ \frac{n}{n+2}, & n=2 k .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,443 |
Example 5 Prove:
$$
P=\left[\sum_{k=0}^{n}(-1)^{k}(\sqrt{2008})^{k}\right]\left[\sum_{k=0}^{n}(\sqrt{2008})^{k}\right]
$$
is an integer. | Solution: Construct the function
$$
f(x)=\sum_{k=0}^{n} x^{k} \text { and } F(x)=f(-x) f(x) \text {. }
$$
It is easy to see that $F(-x)=f(-x) f(x)=F(x)$.
Therefore, $F(x)$ is an even function, which means $F(x)$ is a polynomial containing only even powers of $x$.
Taking $x=\sqrt{2008}$, we get
$$
\begin{array}{l}
P=F(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,444 |
Example 8 Given that $f(x)$ is a function defined on $\mathbf{R}$, and for any $a, b \in \mathbf{R}$, it satisfies
$$
f(a b)=a f(b)+b f(a) .
$$
Is $f(x)$ necessarily a constant function? If so, provide a proof; if not, give a counterexample. | Solution: $f(x)$ is not a constant function, a counterexample is
In fact,
$f(x)=\left\{\begin{array}{ll}c x \log _{a}|x|, & x \neq 0 ; \\ 0, & x=0\end{array}(c\right.$ is a non-zero constant) are all counterexamples that satisfy the condition. | not found | Algebra | proof | Yes | Yes | cn_contest | false | 720,447 |
Example 9 Given
$$
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1 .
$$
Prove: $x+y=0$. | Prove: Let
$$
f(x)=x+\sqrt{x^{2}+1}=\tan \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right) .
$$
Then $f(y)=y+\sqrt{y^{2}+1}=\cot \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right)$.
From the above two equations, we get
$$
\begin{array}{l}
x=\frac{\tan ^{2} \theta-1}{2 \tan \theta}, y=\frac{\cot ^{2} \... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,448 |
Question 1 Find all integer pairs $(k, m), m>k \geqslant 0, (m, k)=1$, such that the sequence defined as
$$
x_{0}=\frac{k}{m}, x_{n+1}=\left\{\begin{array}{ll}
\frac{2 x_{n}-1}{1-x_{n}}, & x_{n} \neq 1 ; \\
1, & x_{n}=1
\end{array}\right.
$$
must contain the number 1. | Analysis: Define the Fibonacci sequence $\left\{F_{n}\right\}$:
$$
F_{0}=F_{1}=1, F_{n+2}=F_{n+1}+F_{n}(n=0,1, \cdots) \text {. }
$$
By enumerating a small amount of data, it is not difficult to find that the answer is $(k, m)=\left(F_{2 l}, F_{2 l+1}\right)(l \in \mathbf{N})$.
First, from $x_{n+1}=\frac{2 x_{n}-1}{1-... | (k, m)=\left(F_{2 l}, F_{2 l+1}\right)(l \in \mathbf{N}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,449 |
Example 12 (1) Prove: There exists a pair of non-zero integers $(x, y)$ such that the algebraic expression $11 x^{2}+5 x y+37 y^{2}$ is a perfect square;
(2) Prove: There exist 6 non-zero integers $a_{1} 、 b_{1} 、 c_{1}$ 、 $a_{2} 、 b_{2} 、 c_{2}$, where $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$, such that for any ... | Proof: (1) Start with simple numbers for testing.
Take $x=1, y=2$, then
$$
11 x^{2}+5 x y+37 y^{2}=169=13^{2}
$$
is a perfect square.
(2) From (1), let $x_{0}=t, y_{0}=2 t$. Then
$$
11 x_{0}^{2}+5 x_{0} y_{0}+37 y_{0}^{2}=(13 t)^{2}
$$
is a perfect square.
$$
\begin{array}{l}
\text { Take } t=\frac{n^{2}-53}{185-26 n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,450 |
Question 2 Proof: There are infinitely many pairs of positive integers $a, b$, satisfying: $a\left|\left(b^{2}+1\right), b\right|\left(a^{2}+1\right)$. | Analysis: Starting from the smallest positive integer and enumerating,
$$
\begin{array}{ll}
1^{2}+1=2, & 2^{2}+1=5, \\
5^{2}+1=2 \times 13, & 13^{2}+1=5 \times 34, \\
\cdots \cdots
\end{array}
$$
We find that $(1,2),(2,5),(5,13), \cdots \cdots$ all satisfy the condition, and it can be observed that the required number... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,451 |
Let $3 \overline{F_{n}}$ be the set of all functions satisfying the following recurrence relation: from $\{1,2, \cdots, n\}$ to $\{1,2, \cdots, n\}$, and
(1) $f(k) \leqslant k+1 \ (k=1,2, \cdots, n)$;
(2) $f(k) \neq k \ (k=2,3, \cdots, n)$.
Find the probability that a function $f$ randomly selected from $\overline{F_{n... | This problem involves the application of the Fibonacci sequence in counting problems. By enumeration of smaller values of $n$, it is not difficult to find that the result should be $\frac{F_{n-1}}{F_{n}}$, where $F_{i}$ represents the $i$-th term of the Fibonacci sequence.
Below, we prove this using mathematical induct... | \frac{F_{n-1}}{F_{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,452 |
Question 4 (1) Does there exist 10 integer gram weights (different weights may have the same weight) that can measure any object from $1 \mathrm{~g}$ to $88 \mathrm{~g}$ using a balance, even if any one of the weights is missing?
(2) Does there exist 12 integer gram weights (different weights may have the same weight) ... | Take $n$ weights, and let the weight of the $i$-th weight be $F_{i}$ $(1 \leqslant i \leqslant n)$.
First, prove by induction:
For an object weighing $w\left(1 \leqslant w \leqslant F_{n+2}-1\right)$, its weight can be measured using $n$ weights.
When $n=1$, $F_{3}=F_{2}+F_{1}=2$.
Thus, $F_{3}-1=1, w=1$, which can obvi... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,453 |
1. Given that $n$ is an integer greater than 1. Find all sequences $a_{1}, a_{2}, \cdots, a_{n^{2}+n}$ that satisfy the following conditions.
(1) $a_{i} \in\{0,1\}$, for all $i$ such that $1 \leqslant i \leqslant n^{2}+n$;
$$
\begin{array}{l}
\text { (2) } a_{i+1}+a_{i+2}+\cdots+a_{i+n} \\
<a_{i+n+1}+a_{i+n+2}+\cdots+a... | 1. Such a sequence is unique.
For all $1 \leqslant u \leqslant n, 0 \leqslant v \leqslant n$,
$$
a_{u+m}=\left\{\begin{array}{ll}
0, & u+v \leqslant n ; \\
1, & u+v \geqslant n+1,
\end{array}\right.
$$
It can be written as $n+1$ parts of length $n$:
$$
\begin{array}{l}
(\underbrace{0,0, \cdots, 0}_{n \uparrow})(\unde... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,454 |
2. Divide a unit square into $n(n>1)$ rectangles, with each rectangle's sides parallel to the sides of the unit square. Any line parallel to the sides of the unit square and passing through an internal point of the square also passes through an internal point of some rectangle. Prove: there exists a rectangle whose poi... | 2. The directions of the sides of a unit square are referred to as the horizontal direction and the vertical direction. A horizontal or vertical line passing through a point inside the square but not through any point inside a rectangle is called a "separating line." If a rectangle does not contain any points on the bo... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,455 |
3. Find all positive integers $n$, such that the numbers in the set $S=$ $\{1,2, \cdots, n\}$ are colored red or blue, and satisfy the following condition: the set $S \times S \times S$ contains exactly 2007 ordered triples $(x, y, z)$, such that
(1) $x, y, z$ are of the same color;
(2) $x+y+z$ is divisible by $n$. | 3. $n=69$ and $n=84$.
Assume $1,2, \cdots, n$ are colored red and blue, denoted by $R$ and $B$ for the sets of red and blue numbers, respectively.
Let $|R|=r,|B|=b=n-r$.
If $x, y, z$ are of the same color, then the triplet $(x, y, z) \in S \times S \times S$ is called "monochromatic"; otherwise, it is called "bichroma... | 69 \text{ and } 84 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,456 |
4. Let $A_{0}=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be a sequence of real numbers. For each non-negative integer $k$, construct a new sequence $A_{k+1}$ from the sequence $A_{k}=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ satisfying the following conditions:
(1) Choose a partition $I \cup J$ of $\{1,2, \cdots, n\}$ s... | 4. First, prove a lemma.
Lemma For a sequence $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ where each term satisfies $\left|x_{i}\right| < a (a > 1)$, if $x_{n} \geqslant 0$, then set $I=I^{\prime}, J=J^{\prime} \cup\{n\}$. Otherwise, set $I=I^{\prime} \cup\{n\}, J=J^{\prime}$.
For both cases, since
$$
\sum_{i \in I} x_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,457 |
5. For each integer $n$, define a strip region on the Cartesian coordinate plane as
$$
S_{n}=\{(x, y) \mid n \leqslant x<n+1\} .
$$
Each strip region $S_{n}$ is either colored red or blue. Let $a$ and $b$ be two distinct positive integers. Prove: there exists a rectangle with side lengths $a$ and $b$ such that its ver... | 5. If there exists an integer $n$ such that $S_{n}$ and $S_{n+a}$ have the same color, then select the vertices $(n, 0) \in S_{n}$, $(n, b) \in S_{n}$, $(n+a, 0) \in S_{n+a}$, $(n+a, b) \in S_{n+a}$. These four vertices are the vertices of a rectangle with side lengths $a$ and $b$ and the same color.
For each integer ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,458 |
7. Let $\alpha$ be a positive real number such that in the set $\{1,2, \cdots, n\}$, one can select $2 p$ pairwise distinct subsets
$$
S_{1}, S_{2}, \cdots, S_{p}, T_{1}, T_{2}, \cdots, T_{p} \text {, }
$$
satisfying that for any $1 \leqslant i, j \leqslant p$, $S_{i} \cap T_{j} \neq \varnothing$. | 7. Let $k, m$ be positive integers (conditions will be given later), $n = km$.
Divide the set $\{1,2, \cdots, n\}$ into $k$ pairwise disjoint subsets, each containing $m$ elements, and denote these subsets as $A_{1}, A_{2}, \cdots, A_{k}$. Define the following family of sets:
$\mathscr{E}=\{S \subset\{1,2, \cdots, n\}... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,459 |
8. Given $P$ as a convex $n$-sided polygon on a plane, consider triangles with vertices at the vertices of $P$. If all sides are of unit length, such triangles are called "good". Prove: the number of good triangles does not exceed $\frac{2 n}{3}$. | 8. Consider all good triangles containing a vertex $A$, then the other two vertices of such triangles lie on the circle $\omega_{A}$ with center $A$ and radius 1.
Since $P$ is convex, these vertices lie on an arc with an angle less than $180^{\circ}$.
Let $L_{A} R_{A}$ be the shortest such arc, and the clockwise dire... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,460 |
Example 13 If a perfect square can be written as the sum of a prime number and another perfect square, it is called a "good square number". If a perfect square cannot be written as the sum of a prime number and another perfect square, it is called a "bad square number". Prove that there are infinitely many good square ... | Proof: Let $p$ be an odd prime. Then $\left(\frac{p+1}{2}\right)^{2}$ is a good square number (since $\left(\frac{p+1}{2}\right)^{2}=p+\left(\frac{p-1}{2}\right)^{2}$).
Since there are infinitely many odd primes $p$, it follows that there are infinitely many good square numbers.
Let $n$ be a positive integer.
We will ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,461 |
1. If $a, b$ are real numbers, satisfying $\frac{1}{a}-\frac{1}{b}=\frac{1}{a+b}$, then the value of $\frac{b}{a}-\frac{a}{b}$ is ( ).
(A) -1
(B) 0
(C) $\frac{1}{2}$
(D) 1 | -、1.D.
From the given condition, we know that $b^{2}-a^{2}=a b$. Dividing both sides by $a b$ yields $\frac{b}{a}-\frac{a}{b}=1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,462 |
3. Given a linear function $y=k x+b$, where $k b>$ 0. Then the graph of all linear functions that meet the conditions must pass through ( ).
(A) the first and second quadrants
(B) the second and third quadrants
(C) the third and fourth quadrants
(D) the first and fourth quadrants | 3. B.
From $k b>0$, we know that $k$ and $b$ have the same sign.
When $k$ and $b$ are both positive, the graph of the linear function passes through the first, second, and third quadrants;
When $k$ and $b$ are both negative, the graph of the linear function passes through the second, third, and fourth quadrants.
The... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,464 |
4. In a convex quadrilateral $ABCD$, $E, F, G, H$ are the midpoints of $AB, BC, CD, DA$ respectively, and $EG$ intersects $FH$ at point $O$. Let the areas of quadrilaterals $AEOH, BFOE, CGOF$ be $3, 4, 5$ respectively. Then the area of quadrilateral $DHOG$ is ( ).
(A) $\frac{\sqrt{15}}{2}$
(B) $\frac{15}{4}$
(C) 4
(D) ... | 4.C.
As shown in Figure 3, connect $O A, O B, O C, O D$. Then
$$
\begin{array}{l}
S_{\triangle A E O}=S_{\triangle B E O}, \\
S_{\triangle B F O}=S_{\triangle C F O}, \\
S_{\triangle C O O}=S_{\triangle D C O}, \\
S_{\triangle D H O}=S_{\triangle A H O} .
\end{array}
$$
Therefore, $S_{\text {quadrilateral }}$ EOH $+S... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,465 |
5. Given $x=2007^{2008}$. Then the remainder when $x$ is divided by 10 is ( ).
(A) 1
(B) 3
(C) 7
(D) 9 | 5.A.
$x$ divided by 10 has the same remainder as $7^{2008}$ divided by 10. The remainders of $7,7^{2}, 7^{3}, 7^{4}, 7^{5}, \cdots$ when divided by 10 are $7,9,3,1,7, \cdots$. They repeat with a period of 4.
Since $2008=502 \times 4$, the remainder of $7^{2008}$ divided by 10 is 1, which means the remainder of $x$ div... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,466 |
6. Let $a, b, c$ be distinct rational numbers, satisfying $(b+\sqrt{2})^{2}=(a+\sqrt{2})(c+\sqrt{2})$.
Then the number of sets of $a, b, c$ that satisfy the condition is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 6. A.
Since $(b+\sqrt{2})^{2}=(a+\sqrt{2})(c+\sqrt{2})$, that is, $b^{2}+2+2 \sqrt{2} b=a c+2+(a+c) \sqrt{2}$, then $a c=b^{2}, a+c=2 b$.
Thus, $a^{2}+c^{2}+2 a c=(a+c)^{2}=4 b^{2}=4 a c$.
Therefore, $(a-c)^{2}=0$.
Hence, $a=c$, which contradicts the given condition. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,467 |
1. The sum of all non-negative integer solutions to the inequality $|2 x-1|<6$ with respect to $x$ is $\qquad$ . | 2.1.6.
The original inequality is equivalent to $\left\{\begin{array}{l}2 x-1-6 .\end{array}\right.$
Solving it, we get $-\frac{5}{2}<x<\frac{7}{2}$.
Therefore, all non-negative integer solutions that satisfy the condition are $x=0,1,2,3$.
Thus, the sum of all non-negative integer solutions is 6. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,468 |
2. Given $x=\frac{1}{2+\sqrt{3}}, y=\frac{1}{2-\sqrt{3}}$. Then $x^{3}+12 x y+y^{3}=$ $\qquad$ | 2.64.
It is known that $x=2-\sqrt{3}, y=2+\sqrt{3}$. Therefore, $x+y=4$.
Then, $x^{3}+12 x y+y^{3}=x^{3}+y^{3}+3 x y(x+y)$ $=(x+y)^{3}=64$. | 64 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,469 |
3. Use three regular polygon tiles with equal side lengths to pave the ground, their vertices fit together perfectly to completely cover the ground. Let the number of sides of the regular polygons be $x$, $y$, and $z$. Then the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $\qquad$ . | 3. $\frac{1}{2}$.
According to the problem, we have
$$
\frac{x-2}{x} \times 180^{\circ}+\frac{y-2}{y} \times 180^{\circ}+\frac{z-2}{z} \times 180^{\circ}=360^{\circ} \text {. }
$$
Simplifying, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$. | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,470 |
4. As shown in Figure 1, in trapezoid $A B C D$, $A B / / D C, A B$ $=A C, D A=DB, \angle A D B=90^{\circ}$. Then the degree measure of $\angle A C D$ is $\qquad$. | $4.30^{\circ}$.
As shown in Figure 4, draw $A E \perp C D$ intersecting the extension of $C D$ at $E$, and draw $D F \perp A B$ at $F$.
Since $D A = D B$ and $\angle A D B = 90^{\circ}$, we know that $\triangle A D B$ is an isosceles right triangle. Therefore,
$\angle D B A = \angle D A F = 45^{\circ}$.
Because $A B \... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,471 |
1. Let $a$ and $b$ be positive integers, $1176a=b^{4}$. Find the minimum value of $a$.
untranslated text remains unchanged. | (Note: Observing that $b^{4}=1176 a=2^{3} \times 3 \times$ $7^{2} a$, then $a$ must contain the factor $2 \times 3^{3} \times 7^{2}$. Thus, the minimum value of $a$ is $2 \times 3^{3} \times 7^{2}=2646$.) | 2646 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,472 |
Three, (20 points) As shown in Figure 2, in trapezoid $ABCD$, $AD$ $/ / BC, AC \perp DB, AC=5, \angle DBC=30^{\circ}$.
(1) Find the length of diagonal $BD$;
(2) Find the area of trapezoid $ABCD$. | (1) As shown in Figure 5, draw $A E / /$ $D B$ intersecting the extension of $C B$ at $E$.
Since $A C \perp D B$,
$A E / / D B$
$\Rightarrow A C \perp A E$,
$$
\angle A E C=\angle D B C=30^{\circ}
$$
$\Rightarrow \angle E A C=90^{\circ}$, which means $\triangle E A C$ is a right triangle
$\Rightarrow E C=2 A C=10$
$\Ri... | \frac{25 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,473 |
Four. (25 points) Let the real number $x$ satisfy
$$
\frac{3 x-1}{2}-\frac{4 x-2}{3} \geqslant \frac{6 x-3}{5}-\frac{13}{10} \text {. }
$$
Find the minimum value of $2|x-1|+|x+4|$. | Four, multiplying both sides of the original inequality by 30 gives
$$
15(3 x-1)-10(4 x-2) \geqslant 6(6 x-3)-39 \text {. }
$$
Solving this yields $x \leqslant 2$.
Let $y=2|x-1|+|x+4|$.
(1) When $x \leqslant-4$,
$$
y=-2(x-1)-(x+4)=-3 x-2 \text {. }
$$
Therefore, the minimum value of $y$ is $(-3) \times(-4)-2=$ 10, at... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,474 |
Five. (25 points) Given positive integers $a$, $b$, $c$ satisfying $a < b < c$, and $ab + bc + ca = abc$. Find all sets of $a$, $b$, $c$ that meet the conditions.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $1 \leqslant a<b<c$, we know $a b c=a b+b c+c a<3 b c$.
Therefore, $a<3$. Hence $a=1$ or $a=2$.
(1) When $a=1$, we have $b+b c+c=b c$, which simplifies to $b+c=0$, contradicting the fact that $b$ and $c$ are positive integers.
(2) When $a=2$, we have $2 b+b c+2 c=2 b c$, which simplifies to $b c-2 b-2 c=0$.
Thus,... | a=2, b=3, c=6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,475 |
1. The smallest positive period of the function $f(x)=\cos ^{4} x+\sin ^{2} x(x \in \mathbf{R})$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{2}$
(C) $\pi$
(D) $2 \pi$ | $$
\begin{array}{l}
f\left(x+\frac{\pi}{2}\right)=\sin ^{4} x+\cos ^{2} x \\
=\sin ^{4} x+\cos ^{4} x+\cos ^{2} x \cdot \sin ^{2} x \\
=\cos ^{4} x+\sin ^{2} x=f(x) . \\
\text { Also, } f(0)=1 \text{ and } f\left(\frac{\pi}{4}\right)=\frac{1}{4}+\frac{1}{2} \neq f(0), \text { therefore }
\end{array}
$$
Choose (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,476 |
2. Given the set of points on the plane
$$
\begin{array}{l}
M=\left\{(x, y) \mid y=\sqrt{2 x-x^{2}}\right\}, \\
N=\{(x, y) \mid y=k(x+1)\} .
\end{array}
$$
When $M \cap N \neq \varnothing$, the range of values for $k$ is ( ).
(A) $\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$
(B) $\left[0, \frac{\sqrt{3}}{3}\r... | 2. B.
From the geometric meaning of point sets $M$ and $N$, we can easily know. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,477 |
4. Given the equation in $x$, $x^{2}-2 a x+a^{2}-4 a=0$ has at least one complex root with a modulus of 3. Then the possible values of the real number $a$ are ( ).
(A) 1,9
(B) $-1,9,2-\sqrt{13}$
(C) $1,9,2+\sqrt{13}$
(D) $1,9,2-\sqrt{13}$ | 4.D.
Rewrite the equation as $(x-a)^{2}=4 a$.
When $a \geqslant 0$, the equation has real roots, giving $x= \pm 3$, so one root is 3 or -3.
$$
\text { From }(a \pm 3)^{2}=4 a \Rightarrow a=1 \text { or } 9 \text {. }
$$
When $a<0$, we get $x=a \pm 2 \sqrt{|a|} \mathrm{i}$, so $|x|^{2}=a^{2}-4 a=9$, giving $a=2-\sqrt{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,479 |
6. Given a geometric sequence $\left\{a_{n}\right\}$ with common ratio $q$, compare $a_{9} S_{8}$ and $a_{8} S_{9}$
(B) $a_{9} S_{8}<a_{8} S_{9}$
(C) $a_{9} S_{8}=a_{8} S_{9}$
(D) It depends on the value of $a_{1}$ | $\begin{array}{l}\text { 6. A. } \\ \begin{array}{l}a_{9} S_{8}-a_{8} S_{9} \\ =\frac{a_{1}^{2}}{1-q}\left[q^{8}\left(1-q^{8}\right)-q^{7}\left(1-q^{9}\right)\right] \\ =-a_{1}^{2} q^{7}>0 .\end{array}\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 6. A. } \\ \begin{array}{l}a_{9} S_{8}-a_{8} S_{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,481 |
7. Set
$$
A=\left\{x \left\lvert\, x=\left[\frac{5 k}{6}\right]\right., k \in \mathbf{Z}, 100 \leqslant k \leqslant 999\right\},
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then the number of elements in set $A$ is $\qquad$ | When $k=100$, $\left[\frac{5 k}{6}\right]=83$;
When $k=999$, $\left[\frac{5 k}{6}\right]=832$.
It is also easy to see that for $100 \leqslant k \leqslant 999$, we have
$$
0 \leqslant\left[\frac{5(k+1)}{6}\right]-\left[\frac{5 k}{6}\right] \leqslant 1 \text {. }
$$
Therefore, the elements in $A$ can cover all int... | 750 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,482 |
2. Let $\overline{x x y y}$ be a perfect square. Find $x, y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | (Tip: $\overline{x x y y}=\overline{x x 00}+\overline{y y}=11 x \cdot 100+11 y$ $=11(100 x+y)$, thus, 11 divides $(x+y)$. From this, we get $x=7, y=4$. At this point, $\overline{x x y y}=7$744.) | null | Number Theory | proof | Yes | Yes | cn_contest | false | 720,483 |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=5, a_{n}=\frac{2 a_{n-1}-1}{a_{n-1}-2}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the sum of its first 100 terms is $\qquad$ | 8.400.
Since $a_{1}=5$, then $a_{2}=3, a_{3}=5, a_{4}=3$, so the sequence has a period of 2. Therefore, the sum of the first 100 terms is 400. | 400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,484 |
9. Among the eight vertices of a regular octagon, three vertices are chosen at random. The probability that these three points form the vertices of a right triangle is $\qquad$ .
| 9. $\frac{3}{7}$.
Connecting three vertices of a regular octagon yields 56 triangles, of which 24 are right triangles, hence, $P=\frac{3}{7}$. | \frac{3}{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,485 |
10. The equation $x^{2}+a|x|+a^{2}-3=0$ $(a \in \mathbf{R})$ has a unique real solution. Then $a=$ | 10. $\sqrt{3}$.
Since $f(x)=x^{2}+a|x|+a^{2}-3$ is an even function, the only real solution must be 0, so $a^{2}-3=0$ and $a>0$. Therefore, $a=\sqrt{3}$. | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,486 |
11. The line $l:(2 m+1) x+(m+1) y-7 m-$ $4=0$ is intersected by the circle $C:(x-1)^{2}+(y-2)^{2}=25$ to form the shortest chord length of $\qquad$ . | $11.4 \sqrt{5}$
Line $l$ passes through point $D(3,1)$, with the center of the circle being $C(1,2)$. The shortest chord is perpendicular to $CD$, and $CD^{2}=5$.
The half-chord length of the circle is $2 \sqrt{5}$, so the chord length is $4 \sqrt{5}$. | 4 \sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,487 |
12. Let the foci of an ellipse be $F_{1}(-1,0)$ and $F_{2}(1,0)$ with eccentricity $e$. A parabola with vertex at $F_{1}$ and focus at $F_{2}$ intersects the ellipse at a point $P$. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=e$, then the value of $e$ is $\qquad$. | 12. $\frac{\sqrt{3}}{3}$.
In the parabola, the directrix $x=-3$, $|PF_{2}|$ is the distance from point $P$ to the directrix;
In the ellipse, $\frac{|PF_{1}|}{|PF_{2}|}=e$, $|PF_{2}|$ is also the distance from point $P$ to the left directrix.
Therefore, the directrix of the parabola coincides with the left directrix o... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,488 |
13. Let the function $f(x)=x-k \sqrt{x^{2}-1}(x \geqslant 1$, $k$ is a given real number, $0<k<1)$. Try to find the range of $f(x)$. | Three, 13. When $x>1$, the derivative of $f(x)$ is
$f^{\prime}(x)=1-\frac{k x}{\sqrt{x^{2}-1}}$.
Let $f^{\prime}(t)=0$.
When $t>1$, the solution is $t=\frac{1}{\sqrt{1-k^{2}}}$.
Thus, $f(t)=f\left(\frac{1}{\sqrt{1-k^{2}}}\right)=\sqrt{1-k^{2}}$, $f(1)=1$.
Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & 1 & $(... | \left[\sqrt{1-k^{2}},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,489 |
14. A tangent line is drawn from the left focus $F$ of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ to the circle $x^{2}+y^{2}=9$, with the point of tangency being $T$. Extend $F T$ to intersect the right branch of the hyperbola at point $P$. If $M$ is the midpoint of segment $F P$, and $O$ is the origin, find th... | 14. Without loss of generality, place point $P$ in the first quadrant. As shown in Figure 2, let $F^{\prime}$ be the right focus of the hyperbola, and connect $P F^{\prime}$. Since $M$ and $O$ are the midpoints of $F P$ and $F F^{\prime}$ respectively, we have $|M O|=\frac{1}{2}\left|P F^{\prime}\right|$. By the defini... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,490 |
15. Given that the diameter of the circumcircle of $\triangle A B C$ is 25, the lengths of the three sides are all integers, and the distances from the circumcenter $O$ to the sides $A B$ and $B C$ are also integers, with $A B > B C$. Find the lengths of the three sides of $\triangle A B C$. | 15. As shown in Figure 3, draw perpendiculars from the center $O$ to $AB$ and $BC$, with the feet of the perpendiculars being $D$ and $E$ respectively. Let $AB = x, OD = y$ (where $x, y$ are positive integers). Then
$BD = \frac{x}{2}$.
Since $DB^2 + OD^2 = OB^2$, we have $x^2 + 4y^2 = 25^2$.
Solving this, we get $(x, y... | 15, 7, 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,491 |
One, (50 points) Given point $O$ is a point inside a convex quadrilateral $A B C D$, $A O=O B, C O=O D, \angle A O B=$ $\angle C O D=120^{\circ}, E, F, G$ are the midpoints of segments $A B, B C,$ and $C D$ respectively. Prove: $\triangle E F G$ is an equilateral triangle. | Connect $A C$ and $B D$.
Then
$$
\begin{array}{l}
E F \Perp \frac{1}{2} A C, \\
F G \Perp \frac{1}{2} B D .
\end{array}
$$
Since $O A=O B, O C=O D$, and
$$
\angle A O B=\angle C O D=120^{\circ},
$$
therefore, with $O$ as the center, rotating $\triangle A O C$ counterclockwise by $120^{\circ}$ makes it $\triangle B O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,492 |
II. (50 points) Given that $a, b, c, d$ are positive real numbers, and $a+b+c+d=4$. Prove:
$$
a^{2} b c+b^{2} d a+c^{2} d a+d^{2} b c \leqslant 4 \text {. }
$$ | $\begin{array}{l}\text { II. } a^{2} b c+b^{2} d a+c^{2} d a+d^{2} b c \\ =(a b+c d)(a c+b d) \\ \leqslant\left(\frac{a b+c d+a c+b d}{2}\right)^{2} \\ =\frac{[(a+d)(b+c)]^{2}}{4} \\ \leqslant \frac{1}{4}\left(\frac{a+d+b+c}{2}\right)^{4}=4 .\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,493 |
3. Is $A=20082008 \cdots 2008$ a perfect square? | (It is not a perfect square, because the unit digit of a perfect square cannot be 8.) | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,494 |
Three, (50 points) Find the smallest positive integer $n$ such that there exists an $(n+1)$-term sequence $a_{0}, a_{1}, \cdots, a_{n}$, satisfying $a_{0}=0, a_{n}=2008$, and
$$
\left|a_{i}-a_{i-1}\right|=i^{2}(i=1,2, \cdots, n) .
$$ | If $n \leqslant 17$, then
$$
\begin{array}{l}
a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
\leqslant \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right|=\frac{1}{6} n(n+1)(2 n+1) \\
\leqslant \frac{1}{6} \times 17 \times 18 \times 35<2008,
\end{array}
$$
a contradiction.
If $n=18$, then
$$
\begin{array}{l}
a_{n}=\sum_... | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,495 |
9.1. Does there exist 14 positive integers such that their product is equal to $\frac{1}{2008}$ of the product of the 14 numbers obtained by adding 1 to each of them? | 9.1. Existence.
$\underbrace{1,1, \cdots, 1,4,4,4,250}_{10 \uparrow}$ is a sequence of 14 positive integers that satisfies the condition.
In fact,
$$
\begin{array}{l}
(1+1)^{10}(4+1)^{3}(250+1)=2^{10} \times 5^{3} \times 251 \\
=2008 \times 1^{10} \times 4^{3} \times 250
\end{array}
$$ | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,496 |
9.2. Given the real-coefficient equation
$$
x^{3}+a x^{2}+b x+c=0
$$
has three real roots. If $-2 \leqslant a+b+c \leqslant 0$, prove: this cubic equation has at least one root in the interval $[0,2]$. | 9.2. Let $P(x)=x^{3}+a x^{2}+b x+c$.
Suppose $x_{1}, x_{2}, x_{3}$ are the three real roots of $P(x)$. Then $P(x)=(x-x_{1})(x-x_{2})(x-x_{3})$.
Notice that $P(1)=1+a+b+c$, so $-1 \leqslant(1-x_{1})(1-x_{2})(1-x_{3}) \leqslant 1$,
which means $|1-x_{1}||1-x_{2}||1-x_{3}| \leqslant 1$.
Thus, at least one of $|1-x_{1}|, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,497 |
9.3. Let $H$ and $M$ be the orthocenter and centroid of a non-isosceles $\triangle ABC$, respectively. Prove that the centroid of the triangle formed by the lines passing through points $A$, $B$, and $C$ and perpendicular to $AM$, $BM$, and $CM$, respectively, lies on the line $MH$. | 9.3. Let $\triangle A^{\prime} B^{\prime} C^{\prime}$ be the triangle formed by the three lines passing through points $A$, $B$, and $C$ and perpendicular to $A M$, $B M$, and $C M$, respectively. Let $G$ be its centroid.
Construct $\square B M C A_{1}$, then segment $M A_{1}$ bisects side $B C$. Therefore, point $A_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,498 |
9.4. A group of scientists work at a research institute. During an 8-hour workday, each scientist visits the cafeteria at least once. It is known that for any two scientists, the total time that at least one of them is in the cafeteria is at least $x(x>4)$ hours. Find the maximum possible number of scientists working a... | 9.4.2 $\left[\frac{x}{2 x-8}\right]$.
Suppose there are $n$ scientists in the institute, $t_{ij}$ represents the time when exactly one of the $i$-th and $j$-th scientists is in the coffee shop, and let $S=\sum_{1 \leqslant i<j \leqslant n} t_{ij}$. Then $S \geqslant \mathrm{C}_{n}^{2} x$.
On the other hand, divide th... | 2\left[\frac{x}{2 x-8}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,499 |
9.5. On an infinitely large chessboard, the distance between any two squares is defined as the minimum number of steps a King needs to move from one square to another. Given three squares that are pairwise 100 steps apart, find the number of squares that are 50 steps away from each of these three squares. | 9.5. Let the side length of a small square be 1.
Represent a small square by the coordinates of its center. Let the absolute difference of the x-coordinates of two small squares be $x$, and the absolute difference of the y-coordinates be $y$.
It is not hard to see that a king can reach from one square to another in $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,500 |
9.6. Given that the incircle of $\triangle ABC$ touches sides $AB, AC$ at points $X, Y$, and $K$ is the midpoint of the arc $\overparen{AB}$ (not containing point $C$) of the circumcircle of $\triangle ABC$, the line $XY$ bisects the segment $AK$. Find the size of $\angle BAC$. | 9.6. Let $I$ be the incenter of $\triangle ABC$, and $XY$ intersects $AI$ at point $S$, $T$ is the midpoint of $AI$.
From the collinearity of $C$, $I$, and $K$, we get
$$
\begin{array}{l}
\angle AIK = \angle CAI + \angle ACI \\
= \angle BAI + \angle KAB = \angle KAI.
\end{array}
$$
Thus, $KT \perp AI$.
Since $XY \perp... | 120^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,501 |
9.7. On a blackboard that has a positive integer written on it, perform the following operations: if the number $x$ is written on the blackboard, then you can write the number $2x+1$ or $\frac{x}{x+2}$. It is known that at some point the number 2008 is written on the blackboard. Prove that the original number on the bl... | 9.7. Clearly, only positive rational numbers can appear on the blackboard.
Suppose at a certain moment the rational number $x=\frac{p}{q}$ $((p, q)=1)$ is on the blackboard. Then from $x$ we can get
$$
2 z+1=\frac{2 p+q}{q} \text { or } \frac{x}{x+2}=\frac{p}{p+2 q} \text {. }
$$
Notice that
$$
\begin{array}{l}
(2 p+... | 2.008 | Number Theory | proof | Yes | Yes | cn_contest | false | 720,502 |
9.8. Among $3^{2 k}$ coins that look identical, there is exactly 1 counterfeit coin. It is known that the counterfeit coin is lighter than the genuine ones. There are three pan balances without weights, two of which are good and one is bad. When using the bad balance, it may give the correct result or it may give a wro... | 9.8. First, prove a lemma.
Lemma: Through $3 k$ weighings, one can either find the counterfeit coin among $3^{2 k}$ coins, or find three coins containing the counterfeit coin while also identifying a good balance.
Proof of the lemma: Use induction on $k$.
When $k=1$, arrange 9 coins in a 3x3 grid, and use $ij$ to den... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,503 |
1. Several people participated in a test, required to answer five questions, and it was stipulated that those who answered 3 or more questions correctly would be qualified. The test results are: the number of people who answered the first to fifth questions correctly are $81, 91, 85, 79, 74$, respectively, and 70 of th... | - 1.C.
Let the number of people participating in the test be $n$.
Among these $n$ people, the total number of questions answered incorrectly is:
$$
\begin{array}{l}
5 n-(81+91+85+79+74) \\
=5 n-410 \text{ (questions). }
\end{array}
$$
Since those who answer 3 or more questions incorrectly are considered不合格 (unqualifi... | 100 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,504 |
4. Let $a, b$ be positive integers, $24a^{2}+1=b^{2}$. Prove: one and only one of $a, b$ is a multiple of 5. | (提示: If $a$ and $b$ are both multiples of 5, then $1 \equiv$ $24 a^{2}+1=b^{2} \equiv 0(\bmod 5)$, a contradiction; If $a$ and $b$ are not multiples of 5, then $a^{2} \equiv 1,4(\bmod 5), b^{2} \equiv 1$, $4(\bmod 5)$. Thus, $24 a^{2}+1 \equiv 0,2(\bmod 5)$, which contradicts $24 a^{2}+1=b^{2}$.)
| proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,505 |
3. Given real numbers $a, b, c$ satisfy $2^{a}=5, 2^{b}=10, 2^{c}=80$. Then the value of the algebraic expression $2006 a-3344 b+1338 c$ is ( ).
(A) 2007
(B) 2008
(C) 2009
(D) 2010 | 3. B.
From $2^{b-a}=2^{b} \div 2^{a}=2^{1}$, we get
$$
b=a+1 \text {; }
$$
From $2^{c-a}=2^{c} \div 2^{a}=80 \div 5=2^{4}$, we get $c=a+4$.
Thus, $2006 a-3344 b+1338 c$
$$
\begin{array}{l}
=2006 a-3344(a+1)+1338(a+4) \\
=2008 .
\end{array}
$$ | 2008 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,507 |
4. Among the 1001 radicals $\sqrt{1008}, \sqrt{1009}, \cdots, \sqrt{2008}$, the number of radicals that are of the same type as the quadratic radical $\sqrt{2016}$ is ( ) .
(A) 3
(B) 4
(C) 5
(D) 6 | 4. A.
Notice that
$$
\sqrt{2016}=\sqrt{2^{5} \times 3^{2} \times 7}=12 \sqrt{14} \text {. }
$$
Let the similar radical be $m \sqrt{14}$ (where $m$ is a positive integer). Then $\sqrt{1008} \leqslant m \sqrt{14} \leqslant \sqrt{2008}$, which means $\sqrt{72} \leqslant m \leqslant \sqrt{143 \frac{3}{7}}$.
And $8<\sqrt... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,508 |
5. As shown in Figure 2, points $E$ and $F$ are on the sides $BC$ and $AB$ of rectangle $ABCD$, respectively, with $BF=3$, $BE=4$, $CE=3$, and $AE$ intersects $CF$ at point $P$. Given that $\angle APC = \angle AEB + \angle CFB$, the area of rectangle $ABCD$ is ( ).
(A) 70
(B) 63
(C) 77
(D) 65.8 | 5. A.
As shown in Figure 6, draw $E Q / / C F$, and draw $F Q / / B C$. Connect $A Q$ and $B Q$.
Notice that
$\angle A P C$
$=\angle A E B +$
$\angle C F B$,
$\angle A P C = \angle E P F$,
$\angle A B C = 90^{\circ}$,
then $\angle E P F = 135^{\circ}$,
$\angle A P F = 45^{\circ}$.
Thus, $\angle Q E A = \angle A P F ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,509 |
6. Given that $P$ is a point inside the circle $\odot O$ with a radius of $15 \mathrm{~cm}$, and among all the chords passing through $P$, there are 25 chords whose lengths are integer centimeters. Then the range of $O P$ is ( ).
(A) $\left(12, \frac{\sqrt{611}}{2}\right)$
(B) $\left(12, \frac{\sqrt{613}}{2}\right)$
(C... | 6.A.
In circle $\odot O$ with a radius of $15 \mathrm{~cm}$, the diameter is the longest chord. Among the chords passing through point $P$, the longest chord (diameter) is unique, and there are 24 other chords with integer lengths of $29 \mathrm{~cm}, 28 \mathrm{~cm}, \cdots, 18 \mathrm{~cm}$, each occurring twice. In... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,510 |
1. The number of simplest proper fractions with 2088 as the denominator is
$\qquad$.
| II.1.672.
Since $2088=2^{3} \times 3^{2} \times 29$, the number of integers from $1,2, \cdots$, 2088 that contain at least one of the factors $2,3,29$ is
$$
\begin{array}{l}
\left(\left[\frac{2088}{2}\right]+\left[\frac{2088}{3}\right]+\left[\frac{2008}{29}\right]\right)- \\
\left(\left[\frac{2088}{2 \times 3}\right]+\... | 672 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,511 |
3. Given that $x$, $m$, and $n$ are positive integers, $m+n=5$, and $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers. Then the number of possible values of $x$ is $\qquad$ . | 3.2.
From the problem, $m$ can take the values $1,2,3,4$, and accordingly, $n$ can be $4,3,2,1$, with $m$ and $n$ being one odd and one even.
Thus, $x^{2}+m$ and $\left|x^{2}-n\right|$ are one odd and one even.
Since $x^{2}+m$ and $\left|x^{2}-n\right|$ are both prime numbers, it follows that $x^{2}+m=2$ or $\left|x^{... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,513 |
4. Given that $P$ is a point inside a square $A B C D$ with center $O$, $A P \perp B P, O P=\sqrt{2}, P A=6$. Then the side length of the square $A B C D$ is $\qquad$ | $4.2 \sqrt{13}$ or 10.
As shown in Figure 8, when point $P$ is inside $\triangle O A D$, it is clear that $P A > P B$. On $P A$, intercept $A Q = P B$, and connect $O Q$.
It is easy to prove that
$\triangle O A Q \cong \triangle O B P$,
$$
O Q = O P = \sqrt{2} \text{. }
$$
It is evident that $\angle P O Q = 90^{\circ}... | 2 \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,514 |
One, (20 points) A toy factory has a total of 273 labor hours and 243 units of raw materials for producing a batch of teddy bears and kittens. Producing a teddy bear requires 9 labor hours and 12 units of raw materials, with a profit of 144 yuan; producing a kitten requires 6 labor hours and 3 units of raw materials, w... | Let the number of bears and cats produced be $x$ and $y$, respectively, with the total profit being $z$ yuan, requiring $m$ labor hours and $n$ units of raw materials. Then,
$$
\left\{\begin{array}{l}
m \leqslant 273, n \leqslant 243, \\
9 x+6 y=m, \\
12 x+3 y=n .
\end{array}\right.
$$
From equations (1) and (2), we g... | 3978 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,515 |
5. Let $m, n$ be positive integers. Prove: $A=3^{m}+3^{n}+1$ is not a perfect square. | (Tip: Since $3^{m}+3^{n} \equiv 2,4,6(\bmod 8)$, then $3^{m}+$ $3^{n}+1 \equiv 3,5,7(\bmod 8)$. And $x^{2} \equiv 0,1,4(\bmod 8)$, contradiction.) | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,516 |
II. (25 points) As shown in Figure 4, the incircle $\odot I$ of $\triangle ABC$ touches $AB$ and $AC$ at points $D$ and $E$, respectively. Extend $DI$ to $M$ and $EI$ to $N$ such that $IM = IN = 9IE$. Points $B$, $I$, $C$, $M$, and $N$ are concyclic. The ratio of the perimeter of $\triangle ABC$ to side $BC$ is the sim... | As shown in the figure,
10, let $A I$ intersect the circumcircle of $\triangle A B C$ at point $P, Q$ be the midpoint of side $B C$, and draw $P F \perp D M$ at point $F$, connecting $P C$. Then
$P Q \perp B C, C Q=\frac{1}{2} B C$,
Rt $\triangle A D I \sim$ Rt $\triangle P F I$,
Rt $\triangle C Q P \sim \mathrm{Rt} \t... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,517 |
Three. (25 points) Solve the system of equations
$$
\left\{\begin{array}{l}
a+b+c+d=2 \\
a^{2}-b^{2}+c^{2}-d^{2}=0 \\
a^{3}+b^{3}+c^{3}+d^{3}=7 \\
a^{4}-b^{4}+c^{4}-d^{4}=0
\end{array}\right.
$$ | $$
\begin{array}{l}
c^{2}-d^{2}=-\left(a^{2}-b^{2}\right), \\
\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)+\left(c^{2}-d^{2}\right)\left(c^{2}+d^{2}\right)=0 .
\end{array}
$$
Substituting, we get
$$
\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}-c^{2}-d^{2}\right)=0 \text {. }
$$
Thus, $a^{2}-b^{2}=0$ or $a^{2}+b^{2}-... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,518 |
1. Let $\left\{a_{n}\right\}_{n \geqslant 1}$ be an arithmetic sequence with the first term 2 and common difference 1, and $\left\{b_{n}\right\}_{n \geqslant 1}$ be a geometric sequence with the first term 1 and common ratio 2. Let
$$
M_{n}=a_{b_{1}}+a_{b_{2}}+\cdots+a_{b_{n}}(n=1,2, \cdots) \text {. }
$$
Then the num... | -、1.C.
Notice
$$
\begin{array}{l}
M_{n}=a_{b_{1}}+a_{b_{2}}+\cdots+a_{b_{n}} \\
=\left[2+\left(b_{1}-1\right)\right]+\cdots+\left[2+\left(b_{n}-1\right)\right] \\
=n+\left(1+2^{1}+2^{2}+\cdots+2^{n-1}\right) \\
=n+2^{n}-1 .
\end{array}
$$
From $n+2^{n}-1 \leqslant 2008$, we get $n \leqslant 10$. Therefore, the maximum... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,520 |
2. Let $x, y \in \mathbf{R}$, the complex numbers $z_{1}=x+1+y \mathrm{i}$ and $z_{2}$ $=x+(y+1) \mathrm{i}$ have a sum of their magnitudes equal to 2. Then the range of $x^{2}+y^{2}$ is ( ).
(A) $[0, \sqrt{2}]$
(B) $[0,2]$
(C) $[0, \sqrt{3}]$
(D) $[0,3]$ | 2. B.
From the conditions of the problem, the locus of point $(x, y)$ is an ellipse with foci at $F_{1}(-1,0)$ and $F_{2}(0,-1)$, and this ellipse passes through the origin $O$. The problem requires finding the maximum distance from a moving point $(x, y)$ on the ellipse to the origin $O$ (the minimum distance is 0), ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,521 |
3. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$, and when $x<0$, $f(x)<0$. If the graph of $y=f(x)$ intersects or is tangent to the line $2007 x + 2008 y = 2009$, then the number of intersection or tangent points is ( ).
(A) 1
(B) 2
(C) 3
(D) Uncerta... | 3. A.
From the problem, for any $x, y \in \mathbf{R}$, we have
$$
f(x+y)=f(x)+f(y) \text {. }
$$
Taking $x=y=0$ in equation (1), we get
$$
f(0)=0 \text {. }
$$
Taking $y=-x$ in equation (1) and using equation (2), we get
$$
f(-x)=-f(x) \text {. }
$$
According to the problem and equation (3), when $x>0$,
$$
f(x)=-f(... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,522 |
4. As shown in Figure 1, in the tetrahedron $P-ABC$, the three lateral edges $PA$, $PB$, and $PC$ are mutually perpendicular. The dihedral angles formed by the lateral faces $PAB$, $PBC$, and $PCA$ with the base $ABC$ have plane angles of $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$, respectively. The area of the base ... | 4. C.
As shown in Figure 5, let's assume
$$
\begin{array}{l}
P A=a, P B=b, \\
P C=c. \text{ Draw } P N \perp \\
A B \text{ at } N \text{, and connect } C N. \\
\text{ It is easy to see that } \angle P N C=\theta_{1} \\
\Rightarrow \tan \theta_{1} \\
=\tan \angle P N C \\
=\frac{P C}{P N} \\
=\frac{P C}{\frac{2 S_{\tri... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,523 |
5. Regarding the set
$$
S=\left\{n \mid n \in \mathbf{Z}, \text { and } \frac{n^{3}+n^{2}+n+1}{3 n^{2}+7 n-1} \in \mathbf{Z}\right\}
$$
the correct statement about the number of elements $|S|$ is ( ).
(A) $|S|=2$
(B) $|S|=3$
(C) $|S|=4$
(D) $S$ is a finite set, and $|S| \geqslant 5$ | 5.B.
Consider the following three cases.
(1) Clearly, $n=0$ satisfies the condition.
(2) When $n>0$,
$$
\begin{array}{l}
\frac{n^{3}+n^{2}+n+1}{3 n^{2}+7 n-1} \in \mathbf{Z} \\
\Leftrightarrow \frac{n^{3}+n^{2}+n+1}{3 n^{2}+7 n-1}+1 \in \mathbf{Z} \\
\quad \Leftrightarrow \frac{n\left(n^{2}+4 n+8\right)}{3 n^{2}+7 n-1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,524 |
6. Among the 8 vertices, 12 midpoints of edges, 6 centers of faces, and 1 center of the cube, these 27 points, the number of "6-point coplanar groups" is ( ).
(A) 1320
(B) 1326
(C) 1332
(D) 1336 | 6.C.
As shown in Figure 6, let the plane containing the 6-point set be $\alpha$. We will count the number of 6-point sets in the same plane $\alpha$ as follows:
(1) If $\alpha$ is a side face (such as $ABCD$), there are 6 side faces, and each side face has 9 points, forming a total of $6 C_{9}^{6}$ 6-point sets.
(2) I... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,525 |
1. If $S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B$, where $b$ is the side opposite to $\angle B$, then $\frac{\cot B}{\cot A+\cot C}=$ $\qquad$ . | $\begin{array}{l}\text { II. 1. } \frac{1}{2} . \\ S_{\triangle A B C}=\frac{1}{4} b^{2} \tan B \Leftrightarrow \frac{1}{2} a c \sin B=\frac{1}{4} b^{2} \tan B \\ \Leftrightarrow 2 \cot B=\frac{\sin (A+C)}{\sin A \cdot \sin C}=\cot A+\cot C \\ \Leftrightarrow \frac{\cot B}{\cot A+\cot C}=\frac{1}{2} .\end{array}$ | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,526 |
Example 1 Let $n$ be a positive integer. Prove: 57 divides
$$
8^{2 n+1}+7^{n+2}
$$ | Proof: Notice
$$
\begin{array}{l}
8^{2 n+1}+7^{n+2}=8 \times 8^{2 n}+49 \times 7^{n} \\
=8 \times 64^{n}+(57-8) \times 7^{n} \\
=57 \times 7^{n}+8\left(64^{n}-7^{n}\right) .
\end{array}
$$
Since $64^{n}-7^{n}$ contains the factor $64-7=57$, and $57 \times 7^{n}$ is clearly a multiple of 57, therefore, 57 divides $\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,527 |
Example 2 Let $n$ be a positive integer. Prove:
$$
n\left(n^{2}-1\right)\left(n^{2}-5 n+26\right)
$$
is divisible by 120. | Analysis: Since $120=5 \times 3 \times 2^{3}$, and $3, 5, 8$ are pairwise coprime, it follows from Property 6 that we only need to prove
$$
n\left(n^{2}-1\right)\left(n^{2}-5 n+26\right)
$$
is divisible by $3$, $5$, and $8$.
Now consider the target expression (1), where $n\left(n^{2}-1\right)$ can be factored into the... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,528 |
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