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 |
|---|---|---|---|---|---|---|---|---|---|
II. (25 points) As shown in Figure 3, from a point $P$ outside $\odot O$, two tangents to $\odot O$ are drawn, touching the circle at points $A$ and $B$. On the minor arc $\overparen{A B}$, take any point $C$. First, draw a tangent from point $C$ to $\odot O$, intersecting $P A$ and $P B$ at points $D$ and $E$, respect... | As shown in Figure 6, draw $DM \perp AB$ and $EN \perp AB$ through points $D$ and $E$, with the feet of the perpendiculars being $M$ and $N$ respectively.
From $PA = DB$ we know
$\angle DAM = \angle EBN$.
Also, $\angle AMD = \angle BNE = 90^\circ$, so,
Rt $\triangle ADM \sim$ Rt $\triangle BEN$.
$$
\text{Therefore, } \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,260 |
Three piles of stones have the numbers 21, 10, and 11, respectively. Now, the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. The question is:
(1) Can the number of stones in the three piles be 4, 14, and 24 after several such opera... | (1) It can be achieved.
The minimum number of operations required is 6, for example:
$$
\begin{array}{l}
(21,10,11) \rightarrow(23,9,10) \rightarrow(22,8,12) \\
\rightarrow(24,7,11) \rightarrow(23,6,13) \rightarrow(25,5,12) \\
\rightarrow(24,4,14) .
\end{array}
$$
Since the minimum number of stones in one pile decreas... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,261 |
1. Prove the proposition by contradiction: If $p$ then $q$, the first step is to assume the proposition is not true. The correct assumption is ( ).
(A) If $p$ then not $q$
(B) If not $p$ then $q$
(C) If not $p$ then not $q$
(D) $p$ and not $q$ | -、1.D.
The negation of “if $p$ then $q$” is no longer any of the “four types of propositions,” but rather represents a “conjunction” proposition: $p$ and not $q$. | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 715,262 |
2. "Real numbers $a=b=c$" is the ( ) condition for the inequality $a^{3}+b^{3}+c^{3} \geqslant 3 a b c$ to hold with equality.
(A) Sufficient but not necessary
(B) Sufficient and necessary
(C) Necessary but not sufficient
(D) Neither sufficient nor necessary | 2.A.
$$
\begin{array}{l}
\text { Given } a^{3}+b^{3}+c^{3}-3 a b c \\
=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]
\end{array}
$$
we know that $a=b=c$ is a sufficient condition for $a^{3}+b^{3}+c^{3}=3 a b c$, but it is not necessary, because $a+b+c=0$ can also make the equality hold. | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 715,263 |
3. The point on the circle $x^{2}+y^{2}=4$ that is closest to the line $4 x+3 y-12$ $=0$ has coordinates ( ).
(A) $\left(\frac{6}{5}, \frac{8}{5}\right)$
(B) $\left(\frac{8}{5}, \frac{6}{5}\right)$
(C) $\left(\frac{12}{5}, \frac{16}{5}\right)$
(D) $\left(\frac{16}{5}, \frac{12}{5}\right)$ | 3. B.
It is known that $A(3,0), B(0,4)$. As shown in Figure 1, the point with the minimum distance should lie on the tangent line parallel to the given line. Connecting the radius $O P$, we have $O P \perp A B$, thus
$\triangle P Q O \backsim \triangle A O B$.
Therefore, $\frac{x}{O P}=\frac{O B}{A B} \Rightarrow x=\f... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,264 |
4. If $\triangle A B C$ can be folded along its three midlines to form a triangular pyramid, then the shape of $\triangle A B C$ is ( ).
(A) acute triangle
(B) obtuse triangle
(C) right triangle
(D) cannot be determined, any of the above is possible | 4.A.
From the problem, we know that the median line divides the triangle into 4 congruent triangles. After assembling into a triangular pyramid, the three dihedral angles sharing the same vertex are exactly the three interior angles of the original triangle, denoted as $\alpha, \beta, \gamma$, and assume $\gamma \geqs... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,265 |
Example 4 Let $x_{i}>0(i=1,2, \cdots, n), \mu>0$, $\lambda-\mu>0$, and $\sum_{i=1}^{n} x_{i}=1$. For a fixed $n\left(n \in \mathbf{N}_{+}\right)$, find the minimum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n} \frac{x_{i}}{\lambda-\mu x_{i}}$. | $$
\begin{array}{l}
\text { Explanation: } f\left(x_{1}, x_{2}, \cdots, x_{n}\right)+\frac{n}{\mu} \\
=\sum_{i=1}^{n}\left(\frac{x_{i}}{\lambda-\mu x_{i}}+\frac{1}{\mu}\right)=\frac{\lambda}{\mu}\left(\sum_{i=1}^{n} \frac{1}{\lambda-\mu x_{i}}\right) \\
=\frac{\lambda}{\mu(n \lambda-\mu)}\left[\sum_{i=1}^{n}\left(\lamb... | \frac{n}{n \lambda-\mu} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,266 |
5. In $\triangle A B C$, it is known that $\tan ^{2} B=\tan A \cdot \tan C$. Then the range of $\angle B$ is ( ).
(A) $\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$
(B) $\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right]$
(C) $\left(0, \frac{\pi}{3}\right] \cup\left(\frac{2 \pi}{3}, \pi\right)$
(D) $\left[\frac{\pi}{3}, \frac{\p... | 5.A.
From $\tan A \cdot \tan C=\tan ^{2} B$,
we get $\tan B=\frac{\tan A+\tan C}{\tan A \cdot \tan C-1}=\frac{\tan A+\tan C}{\tan ^{2} B-1}$.
Thus, $\tan A+\tan C=\tan B\left(\tan ^{2} B-1\right)$.
(2) ${ }^{2}-4 \times$ (1) gives
$$
\begin{array}{l}
0 \leqslant(\tan A-\tan C)^{2} \\
=\tan ^{2} B\left(\tan ^{2} B-1\ri... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,267 |
6. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} .
\end{array}
$$
Then $\lim _{n \rightarrow \infty}\left(\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}\right)=$ ( ).
(A) 2
(B) 3
(C)... | 6.C.
$$
\begin{array}{l}
\text { Because } a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1} \\
=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}, \\
\text { so } \frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\fr... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,268 |
1. In the square $A B C D$, construct the inscribed circle $\odot 0$, which is tangent to $A B$ at point $Q$ and to $C D$ at point $P$. Connect $P A$ and $P B$ to form the isosceles $\triangle P A B$. Rotating these three figures (the square, the circle, and the isosceles triangle) around their common axis of symmetry ... | ニ、1.3:2:1.
Let the side length of the square be $2a$, then
$$
\begin{array}{l}
V_{\text {rounded }}=S h=\pi a^{2} \cdot 2 a=2 \pi a^{3} \text {, } \\
V_{\text {** }}=\frac{4}{3} \pi a^{3}, V_{\text {base }}=\frac{1}{3} S h=\frac{2}{3} \pi a^{3} \text {. } \\
\end{array}
$$
Thus, $V_{\text {rounded }}: V_{\text {** }}:... | 3:2:1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,269 |
2. Given $\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta=\frac{3}{4}$, $\sin (\alpha-\beta) \neq \frac{1}{2}$. Then $\sin (\alpha+\beta)=$ $\qquad$ . | $$
\begin{array}{l}
\text { 2. } \frac{1}{2} \text {. } \\
0=\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta-\frac{3}{4} \\
=\frac{1-\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}- \\
\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]-\frac{3}{4} \\
=\frac{1}{4}-\frac{1}{2}(\cos 2 \alpha-\cos 2 \beta)- \\
\... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,270 |
$$
\begin{aligned}
& 3 . \mathrm{C}_{2004}^{0}-\mathrm{C}_{2004}^{2}+\mathrm{C}_{2004}^{4}-\mathrm{C}_{2004}^{6}+\cdots-\mathrm{C}_{2004}^{2002} \\
+ & \mathrm{C}_{2004}^{200}=
\end{aligned}
$$ | 3. $-2^{1002}$
In the binomial expansion
$$
(1+x)^{n}=\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{1} x+\mathrm{C}_{n}^{2} x^{2}+\cdots+\mathrm{C}_{n}^{n} x^{n}
$$
if we take $n=2004, x=\mathrm{i}$, we get
$$
\begin{array}{l}
(1+\mathrm{i})^{2004} \\
=\mathrm{C}_{2004}^{0}-\mathrm{C}_{2004}^{2}+\mathrm{C}_{2004}^{4}-\mathrm{C}... | -2^{1002} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,271 |
4. Let set $A=\left\{(x, y) \mid y=x^{2}\right\}, B=$ $\left\{(x, y) \mid x^{2}+(y-m)^{2}=1\right\}$. If $A \cap B \neq \varnothing$, then the range of values for $m$ is $\qquad$. | 4. $\left[-1, \frac{5}{4}\right]$.
From $A \cap B \neq \varnothing$, we know that the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}, \\
x^{2}+(y-m)^{2}=1
\end{array}\right.
$$
has real solutions.
Solution 1: Substitute (1) into (2) to eliminate $x^{2}$, we get
$$
y+(y-m)^{2}=1 \text {, }
$$
which simplifies... | \left[-1, \frac{5}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,272 |
5. Given real numbers $a, b, c$ satisfy $a+b+c=0$, $a^{2}+b^{2}+c^{2}=1$. Then the maximum value of $a b c$ is $\qquad$ | 5. $\frac{\sqrt{6}}{18}$.
Solution 1: From $a^{2}+b^{2}+c^{2}=1$, we know that $a^{2}, b^{2}, c^{2}$ must have one that is no greater than $\frac{2}{3}$. Without loss of generality, let $c^{2} \leqslant \frac{2}{3}$. Let $c=\sqrt{\frac{2}{3}} \cos \theta$, then
$$
\begin{array}{l}
a b c=\frac{1}{2}\left[(a+b)^{2}-\lef... | \frac{\sqrt{6}}{18} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,273 |
6. Given that the odd number $n$ is a three-digit number, and the sum of the last digits of all its factors (including 1 and $n$) is 33. Then $n$ $=$ . $\qquad$ | 6.729 (or $27^{2}$).
From $n$ being an odd number, we know that each of its factors is also odd, and of course, the unit digit of each factor is odd. Since the sum of the unit digits of all factors is 33, which is an odd number, $n$ must have an odd number of factors. Therefore, $n$ is a perfect square.
Since a prime... | 729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,274 |
Three. (20 points) Draw two perpendicular rays through the origin intersecting the ellipse $\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$. Let the midpoint of segment $AB$ be $P$. Find the equation of the locus of point $P$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is a note and not part o... | Let $P(x, y), A\left(x_{A}, y_{A}\right)$. Since $P$ is the midpoint of $A B$, we have $B\left(2 x-x_{A}, 2 y-y_{A}\right)$. Given that $A$ and $B$ lie on the ellipse and $O A \perp O B$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{A}=\sqrt{2} \cos \alpha, \\
y_{A}=\sin \alpha ;
\end{array}\right. \\
\left\{... | 3\left(x^{2}+2 y^{2}\right)^{2}-2 x^{2}-8 y^{2}=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,275 |
Four. (20 points) There is a circle $C$ with a radius of 1. Inside circle $C$, construct an inscribed equilateral triangle and its incircle $C_{3}$, with radius denoted as $r_{3}$; inside incircle $C_{3}$, construct an inscribed square and its incircle $C_{4}$, with radius denoted as $r_{4}$; and so on, inside incircle... | $$
r_{n}=r_{n-1} \cos \frac{\pi}{n} \text {. }
$$
Thus, for $n \geqslant 3$, we have
$$
r_{n}=\cos \frac{\pi}{3} \cdot \cos \frac{\pi}{4} \cdots \cdots \cos \frac{\pi}{n} \text {. }
$$
When $01-x^{2} .
\end{array}
$$
Therefore, we have
$$
\begin{array}{l}
r_{n}^{2}=\cos ^{2} \frac{\pi}{3} \cdot \cos ^{2} \frac{\pi}{... | r_{n}>\frac{1}{24} | Geometry | proof | Yes | Yes | cn_contest | false | 715,276 |
Example 5 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and
$$
\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1 \text {. }
$$
Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.
(2001, National High School Mathematics Competition) | Explanation: First, find the minimum value of $\sum_{i=1}^{n} x_{i}$. From $k<j$ we know
$$
\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j} \leqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2} \text {. }
$$
Equality holds if and only if at most one of $x_{1}, x_{2}, \cdots, x_{n}... | \left(\sum_{i=1}^{n} x_{i}\right)_{\min }=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,277 |
Five. (20 points) Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ $(n \geqslant 3)$ that have the following property: each subset contains at least 2 elements, and the difference (absolute value) between any 2 elements in each subset is greater than 1. Find $a_{10}$. | Solution 1: Consider the recurrence relation of $a_{n}$, dividing the subsets that satisfy the condition into two categories:
The first category contains $n$. This type of subset, apart from the $a_{n-2}$ subsets that satisfy the condition in $\{1,2, \cdots, n-2\}$ combined with the element $n$, also includes $n-2$ tw... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,278 |
One, (50 points) In quadrilateral $A B C D$, $\angle A +$ $\angle C=120^{\circ}$. Prove:
$$
\begin{aligned}
(A C \cdot B D)^{2}= & (A B \cdot C D)^{2}+(B C \cdot D A)^{2} \\
& +A B \cdot B C \cdot C D \cdot D A .
\end{aligned}
$$ | As shown in Figure 5, construct $\angle E A D = \angle B C D, \angle E D A = \angle B D C$, and connect $E B$. Then we have
$$
\angle E A B = \angle D A B +
$$
$\angle B C D = 120^{\circ}$,
and $\triangle B C D \backsim \triangle E A D$.
Thus, $\frac{B C}{E A} = \frac{C D}{D A}$.
In $\triangle B D E$ and $\triangle C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,279 |
II. (50 points) An old man divides his savings of $m$ gold coins among his $n$ children ($m, n$ are positive integers greater than 1). First, he gives the eldest child 1 gold coin and $\frac{1}{7}$ of the remainder; then, from the remaining coins, he gives the second child 2 gold coins and $\frac{1}{7}$ of the remainde... | Let the number of gold coins left after giving to the $k$-th child be $a_{k}$, then $a_{0}=m, a_{n-1}=n$,
$$
a_{k}=a_{k-1}-\left(k+\frac{a_{k-1}-k}{7}\right)=\frac{6}{7}\left(a_{k-1}-k\right) \text {. }
$$
Thus, $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,280 |
Three. (50 points) For the bivariate polynomial $f(x, y)$ in terms of $x$ and $y$, it satisfies
$$
\begin{array}{l}
f(x, 0)=1, \\
f(f(x, y), z)=f(z, x y)+z .
\end{array}
$$
Find the expression for $f(x, y)$. | $$
\begin{array}{l}
f(1, y)=f(f(x, 0), y)=f(y, 0)+y=1+y . \\
\text { Then, } f(x, 1)=f(1+(x-1), 1) \\
=f(f(1, x-1), 1) \text { (by (1) } \\
=f(1,1 \cdot(x-1))+1 \text { (by the given) } \\
=[1+(x-1)]+1 \text { (by (1) } \\
=x+1 .
\end{array}
$$
On one hand, by (2) we have
$$
f(f(x, y), 1)=f(x, y)+1 \text {. }
$$
On t... | f(x, y) = xy + 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,281 |
149 Given the quadratic function $f(x)=x^{2}-2 m x+1$. Does there exist a real number $x$, such that for any real numbers $a, b, c$ satisfying $0 \leqslant x \leqslant 1$, $f(a), f(b), f(c)$ can form the lengths of the three sides of a triangle. | Solution: This problem is equivalent to:
For $0 \leqslant x \leqslant 1, f(x)_{\text {min }} \geqslant 0$ and $2 f(x)_{\text {min }}>f(x)_{\text {max }}$.
(1) When $m 0 , } \\
{ 2 f ( 0 ) > f ( 1 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
1>0, \\
2>2-2 m
\end{array} \Rightarrow m>0 .\right.\right.
$$
When $m0... | 0<m<\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,282 |
Find all integers $n$ such that $\frac{70 n+200}{n^{2}+1}$ is an integer.
Initial text: 150 | Solution: Since $\left(n^{2}+1\right) \mid 10(7 n+20)$, then,
$$
\begin{array}{l}
\left(n^{2}+1\right)\left\{\left[10 \cdot 7^{2}\left(n^{2}+1\right)-10(7 n+20)(7 n-20)\right] .\right. \\
\text { Also, } 10 \cdot 7^{2}\left(n^{2}+1\right)-10(7 n+20)(7 n-20) \\
=4490=2 \times 5 \times 449,
\end{array}
$$
Since 449 is a... | n=0, \pm 1, \pm 2, \pm 3, -67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,283 |
Given $x, y$ are positive numbers satisfying $x+y=1$. Prove: $\frac{x}{x^{2}+y^{3}}+\frac{y}{x^{3}+y^{2}} \leqslant 2\left(\frac{x}{x+y^{2}}+\frac{y}{x^{2}+y}\right)$. | Prove: Let $t=xy$, then
$$
\begin{aligned}
x^{2}+y^{2} & =(x+y)^{2}-2xy=1-2t, \\
x^{3}+y^{3} & =(x+y)^{3}-3xy(x+y)=1-3t, \\
x^{4}+y^{4} & =(x^{2}+y^{2})^{2}-2x^{2}y^{2}=1-4t+2t^{2}, \\
x^{5}+y^{5} & =(x^{2}+y^{2})(x^{3}+y^{3})-x^{2}y^{2}(x+y) \\
& =1-5t+5t^{2}.
\end{aligned}
$$
Since $x^{2}+y=x+y^{2}$, the original in... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,284 |
Given $a_{1}>0, i=1,2, \cdots, n, \sum_{i=1}^{n} a_{i}=1, k \in$
$N$. Prove:
$$
\begin{array}{l}
\left(a_{1}^{k}+\frac{1}{a_{1}^{k}}\right)\left(a_{2}^{k}+\frac{1}{a_{2}^{k}}\right) \cdots\left(a_{n}^{k}+\frac{1}{a_{n}^{k}}\right) \\
\geqslant\left(n^{k}+\frac{1}{n^{k}}\right)^{n} .
\end{array}
$$ | Prove: $a_{1}^{k}+\frac{1}{a_{1}^{k}}=a_{1}^{k}+\underbrace{\frac{1}{n^{2 k} a_{1}^{k}}+\frac{1}{n^{2 k} a_{1}^{k}}+\cdots+\frac{1}{n^{2 k} a_{1}^{k}}}_{n^{2 k} \text { terms }}$
$$
\begin{array}{l}
\geqslant\left(n^{2 k}+1\right)\left[\frac{a_{1}^{k}}{\left(n^{2 k} a_{1}^{k}\right)^{n^{2 k}}}\right]^{-\frac{1}{n^{2 k}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,285 |
Example 6 Let $n$ be a fixed integer, $n \geqslant 2$. Determine the smallest positive number $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$
holds for all non-negative real numbers $x_{i}(i=1,2, \cdots, ... | Explanation: When $x_{i}=0(i=1,2, \cdots, n)$, it holds for all $c$.
When $x_{i}$ are not all 0, the problem transforms into
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)}{\left(\sum_{i=1}^{n} x_{i}\right)^{4}} \leqslant c
$$
always ho... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,286 |
Example 8 Let $F=\max _{1 \leqslant x \leqslant 3}\left|x^{3}-a x^{2}-b x-c\right|$. Find the minimum value of $F$ when $a, b, c$ take all real numbers. (2001, IMO China National Training Team Selection Exam) | Let $f(x)=(x-2)^{3}-\frac{3}{4}(x-2)$, $x \in[1,3]$. Set $x-2=\cos \theta, \theta \in[0, \pi]$, then $|f(x)|=\left|\cos ^{3} \theta-\frac{3}{4} \cos \theta\right|=\frac{1}{4}|\cos 3 \theta| \leqslant \frac{1}{4}$. When $\theta=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi$, i.e., $x=3, \frac{5}{2}, \frac{3}{2}, 1$, $|f(x)|=\f... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,287 |
1. Given $\left|x_{i}\right|<1(i=1,2, \cdots, n)$, let $\sum_{i=1}^{n}\left|x_{1}\right|=19+\left|\sum_{k=1}^{n} x_{k}\right|$. Find the minimum value of $n$. | (Given: $\left|\sum_{i=1}^{n} x_{i}\right| \geqslant 0$, then $\sum_{i=1}^{n}\left|x_{i}\right| \geqslant 19$, since $\left|x_{i}\right|<1$, then $n \geqslant 20$. Taking $x_{i}=\frac{19}{20}(i=1,2, \cdots, 10), x_{j}=-\frac{19}{20}(j=11,12, \cdots, 20)$, we have $n=20$, hence $n_{\min }=20$.) | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,288 |
2. Given positive numbers $x_{i}(i=1,2, \cdots, n)$, such that $\sum_{i=1}^{n} x_{1}=1$. Let
$$
S=\max \left\{\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \cdots, \frac{x_{n}}{1+x_{1}+x_{2}+\cdots+x_{n}}\right\} \text {. }
$$
Find the minimum value of $S$. | (Tip: $1-S \leqslant \frac{1+x_{1}+\cdots+x_{k-1}}{1+x_{1}+\cdots+x_{k}}(k=1,2, \cdots$, $n)$, then $(1-S)^{n} \leqslant \frac{1}{2} \Rightarrow S \geqslant 1-\sqrt[n]{\frac{1}{2}}$.)
3. Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}=1(n \geqslant 2)$. Find the maximum value of $\sum_{1 \leqslant ... | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,289 |
4. In space, there are 2004 points, no three of which are collinear. Divide them into 30 groups with different numbers of points. From any 3 different groups, take 1 point from each to form a triangle. To maximize the total number of such triangles, how many points should be in each group? | (Let the number of points in each group be $x_{1}, x_{2}, \cdots, x_{30}$, and $x_{1}<x_{2}<\cdots<x_{30}$. First, prove $1 \leqslant x_{t+1}-x_{i} \leqslant 2$. Then, prove that between 1 to 30, there is at most one $i$ such that $x_{t+1}-x_{t}=2$. Finally, find $52,53, \cdots, 72,74, \cdots, 82$.) | 52,53, \cdots, 72,74, \cdots, 82 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,290 |
Example 3 Find all rational numbers $r$, such that all roots of the equation $r x^{2}+(r+1) x+r-1=0$ in $x$ are integers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Analysis: When $r=0$, the original equation is a linear equation in $x$; when $r \neq 0$, the original equation is a quadratic equation in $x$. Since $r$ is a rational number, solving it directly or using the discriminant is difficult, so we can consider using Vieta's formulas to eliminate $r$ first.
Solution: When $r... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,291 |
Example 1 Let $S=\{1,2, \cdots, 1000000\}, A$ be a subset of $S$ containing exactly 101 elements. Prove that there exist numbers $t_{1}, t_{2}, \cdots, t_{100}$ in $S$ such that the following sets
$$
A_{j}=\left\{x+t_{j} \mid x \in A\right\}, j=1,2, \cdots, 100
$$
are pairwise disjoint.
(44th IMO) | This problem can be generalized as follows:
Let $n (n \geqslant 2)$ be a positive integer, $S=\left\{1,2, \cdots, n^{3}\right\}$, and $A$ be a subset of $S$ containing $n+1$ elements. Then for any positive integer $m \leqslant n$, there exists an $m$-element subset $T=\left\{t_{1}, t_{2}, \cdots, t_{m}\right\}$ of $S$ ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,293 |
Example 2: In a city, there are $n$ high schools, and the $i$-th high school sends $c_{t}$ students $\left(1 \leqslant c_{i} \leqslant 39,1 \leqslant i \leqslant n\right)$ to the stadium to watch a ball game, with the total number of students $\sum_{i=1}^{n} c_{i}=1899$. Each row in the stands has 199 seats, and studen... | Proof: First, number the schools in descending order of the number of students, i.e., the numbered $\{c_i\}$ forms a non-increasing sequence. Then, seat the students in order: starting from the first row, the 1st school, the 2nd school, etc. If the students of the last school in the first row cannot all sit in the firs... | 11 | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,294 |
Example 3 If $a_{1}=1, a_{k}=k+a_{k-1}(2 \leqslant k \leqslant$ $n)$, then $a_{1}, a_{2}, \cdots, a_{n}$ is called a regular number. Question: How many numbers in the set $\{1,2, \cdots, 2001\}$ are either a regular number itself or the sum of several different regular numbers? | Solution: It is easy to know that $a_{k}=\frac{k(k+1)}{2}=C_{k+1}^{2}$, these regular numbers form the sequence $1,3,6,10,15,21,28,36,45,55$, ... First, we check the numbers that can be represented by regular numbers (i.e., can be written as the sum of several different regular numbers):
Among the positive integers le... | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,295 |
For example, 42000 people are divided into $n$ groups, satisfying:
(i) No one in each group knows everyone in the same group;
(ii) In any three people in each group, at least two of them do not know each other;
(iii) For any two people who do not know each other in each group, there is exactly one person in the same gr... | Proof: Representing people as points, connect a line between points representing people who do not know each other within the same group, and do not connect a line between points representing people who know each other. The original conditions can be restated as:
(i) There are no isolated points in any group;
(ii) Amon... | 400 | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,296 |
Example 5 Let $n$ be a positive integer, and let $A_{1}, A_{2}, \cdots, A_{n+1}$ be $n+1$ non-empty subsets of the set $\{1,2, \cdots, n\}$. Prove: there exist two non-empty disjoint subsets $\left\{i_{1}, i_{2}, \cdots, i_{k}\right\}$ and $\left\{j_{1}, j_{2}, \cdots, j_{m}\right\}$ of $\{1,2, \cdots, n+1\}$, such tha... | Proof: By mathematical induction.
When $n=2$, the conclusion is obviously true.
Assume the conclusion holds for a positive integer $n-1 (n \geqslant 3)$, we need to prove that the conclusion also holds for the positive integer $n$.
(1) Obviously, when some element in the set $\{1,2, \cdots, n\}$ does not belong to any ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,297 |
Given $\triangle A B C$ is an acute triangle, and the circle $\odot K$ with diameter $A B$ intersects $A C$ and $B C$ at points $P$ and $Q$ respectively. The tangents to $\odot K$ from $A$ and $Q$ intersect at point $R$, and the tangents to $\odot K$ from $B$ and $P$ intersect at point $S$. Prove that point $C$ lies on... | Proof: As shown in Figure 1, let $R Q$ intersect $P S$ and $A C$ intersect $R K$, and $B C$ intersect $S K$ at points $W$, $Y$, and $N$ respectively. Connect $P K$, $W K$, $Q K$, $W N$, $W Y$, and $B P$. Then we have
$$
\begin{array}{l}
\angle Y K W \\
=\angle Y K Q-\angle W K Q \\
=\frac{\angle A K Q-\angle P K Q}{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,298 |
Two circles are externally tangent at point $A$, and internally tangent to another circle $\odot O$, with points of tangency $B$ and $C$. Let $D$ be the midpoint of the chord of $\odot O$ cut by the common internal tangent of the smaller circles. Prove: When $B$, $C$, and $D$ are not collinear, $A$ is the incenter of $... | Proof: As shown in Figure 2, let the radical center of the three circles be $T$.
Then $\angle O B T = \angle O C T = \angle O D T = 90^{\circ}$.
Therefore, $B, O, D, C, T$ are concyclic.
Hence $\angle C D T = \angle O D C - 90^{\circ} = 90^{\circ} - \angle O B C$
$= 90^{\circ} - \angle O C B = 90^{\circ} - \angle O D B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,299 |
Given $a, b, c$ are positive real numbers. Prove:
$$
a^{2}+b^{2}+c^{2}+a b c=4 \Rightarrow a+b+c \leqslant 3 \text {. }
$$
(20th Iranian Mathematical Olympiad) | Proof: Assume $a+b+c=t>3$, and $a^{2}+b^{2}+c^{2}+a b c=4$, then we have $c \in(0,2)$. Since
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+a b c \\
=(a+b)^{2}+c^{2}-(2-c) a b,
\end{array}
$$
fixing $c$ and $a+b$, we know that the above expression achieves its minimum value when $a=b$.
Similarly, when $a=b=c=\frac{t}{3}$, this... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 715,300 |
Question 4 Let $S$ be a finite non-empty set of positive integers greater than 1, and has the following property: there exists a number $s \in S$, such that for any positive integer $n$, either $(s, n)=1$, or $(s, n)=s$. Prove: there must exist two numbers $s, t \in S$ ($s, t$ not necessarily distinct), such that $(s, ... | Proof: For an integer $t > 1$, if for any $n \in \mathbf{N}_{+}$ we have $(t, n)'=1$ or $t$, then $t$ is a prime number. Otherwise, taking $n$ as any prime factor of the composite number $t$ would lead to a contradiction. Therefore, the set $S$ contains a prime number $t$, and $(t, t)=t$ is still a prime number. Hence,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,301 |
Example 4 Given that $a$ is a positive integer, and makes the quadratic equation in $x$
$$
a x^{2}+2(2 a-1) x+4(a-3)=0
$$
have at least one integer root. Find the value of $a$.
| Solution: Transform the original equation into
$$
(x+2)^{2} a=2(x+6) \text {. }
$$
Obviously, $x+2 \neq 0$, thus,
$$
a=\frac{2(x+6)}{(x+2)^{2}} .
$$
Since $a$ is a positive integer, then $a \geqslant 1$, i.e.,
$$
\begin{array}{l}
\frac{2(x+6)}{(x+2)^{2}} \geqslant 1 \Rightarrow x^{2}+2 x-8 \leqslant 0 \\
\Rightarrow(... | 1,3,6,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,302 |
Question 5: Two circles $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P$ and $Q$, and the common tangent line closer to point $P$ touches $\odot O_{1}$ at point $A$ and $\odot O_{2}$ at point $B$. A line tangent to $\odot O_{1}$ at point $P$ intersects $\odot O_{2}$ again at point $C$, and lines $AP$ and $BC$ in... | Prove: As shown in Figure 3, connect $A Q, B Q, P Q, R Q$, and extend $C P$ to intersect $A B$ at $X$. Then
$$
\begin{array}{l}
\angle B P R=\angle P B A+\angle B A P \\
=\angle B C X+\angle A P X \\
=\angle B C X+\angle C P R=\angle B R P .
\end{array}
$$
Therefore, $B P=B R$.
By the converse of the tangent-chord ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,303 |
In $\triangle ABC$, the circumcenter is $O$, and the feet of the altitudes $AH$, $BK$, and $CL$ are $H$, $K$, and $L$, respectively. $A_0$, $B_0$, and $C_0$ are the midpoints of $AH$, $BK$, and $CL$, respectively. The incircle with center $I$ touches the sides $BC$, $CA$, and $AB$ of $\triangle ABC$ at points $D$, $E$,... | Proof: As shown in Figure 4, let $M$ be the midpoint of $BC$, $T$ be the intersection of $OI$ and $A_0D$, and extend $A_0D$ to intersect $OM$ at $G$. Let $AB = c$, $BC = a$, $CA = b$, and $r$, $R$ be the inradius and circumradius of $\triangle ABC$, respectively. Then,
\[
\begin{array}{l}
OG = OM + MG = OM + A_0H \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,304 |
Question 7 Let $n$ be a positive integer. Prove that $2^{n}+1$ has no prime factor that is congruent to -1 modulo 8.
(44th IMO Vietnam National Team Selection Exam) | Proof: Let $p=8k-1, k \in \mathbf{N}_{+}, p$ be a prime number, and the symbol
$\left(\frac{-1}{p}\right)=1 \Leftrightarrow$ odd prime $p \equiv 1(\bmod 4)$;
$\left(\frac{2}{p}\right)=1 \Leftrightarrow$ odd prime $p \equiv \pm 1(\bmod 8)$.
From the given, we know $p=7(\bmod 8)$, hence
$$
\left(\frac{-1}{p}\right)=-1 \t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,305 |
Find all positive integers $a$, $b$, $c$ such that $a$, $b$, $c$ satisfy $(a!)(b!)=a!+b!+c!$.
(2002-2003 British Mathematical Olympiad) | Let $\alpha(n)$ be the exponent of 2 in the prime factorization of $n$.
If $a \neq b$, assume without loss of generality that $a > b$.
Clearly, $b \neq 1$ or 2.
Thus, $c! = (a!)(b!) - a! - b!$
$> (a!)(b! - 2) > a!$.
If $a = b + 1$, then
$(b + 1)! = b + 2 + \prod_{t=b+1}^{c} t$.
The left side of the above equation is a ... | (3, 3, 4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,306 |
Question 9 Set $S$ is a set of $n$ points in the plane, and the distance between any two points in $S$ is at least 1 unit. Prove: there exists a subset $T$ of $S$, $T$ contains at least $\frac{n}{7}$ points, and the distance between any two points in $T$ is at least $\sqrt{3}$ units. (2003, Canadian Mathematical Olympi... | Proof: As shown in Figure 5, establish an arbitrary Cartesian coordinate system on the plane, and select the point W with the largest y-coordinate (if there are more than one, choose any one). Remove the points that are less than $\sqrt{3}$ units away from W. If more than 6 points are removed, then at least two of thes... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 715,307 |
For a given prime $p$, determine whether the equation
$$
x^{2}+y^{2}+p z=2003
$$
always has integer solutions $x, y, z$. Prove your conclusion. (2003, Singapore Mathematical Olympiad) | Solution: For $p=2$, take $(x, y, z)=(1,0,1001)$; for $p=2003$, take $(x, y, z)=(0,0,1)$.
Now, assume $p \neq 2$ and $p \neq 2003$. We need to prove that for a given $p$, there exist $x, y \in \mathbf{Z}$ such that
$$
x^{2} + y^{2} \equiv 2003 \pmod{p}.
$$
Introduce the Legendre symbol.
(1) If $\left(\frac{2003}{p}\r... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,308 |
2. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a^{2}+b^{2}=t c^{2}$, and $\cot C=$ $2004(\cot A+\cot B)$, then the value of the constant $t$ is $\qquad$ | $2.4009$ | 2.4009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,310 |
3. The three sides are three consecutive positive integers, and its perimeter is less than or equal to 100 of the acute triangle has $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note is not part of the translation but is provided to clarify the instruction. The actual translation is above. | 3. 29 | 29 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,311 |
Example 5 Given the quadratic equation in $x$
$$
\left(k^{2}-8 k+15\right) x^{2}-2(13-3 k) x+8=0
$$
both roots are integers. Find the value of the real number $k$.
Analysis: Since $k$ is a real number, we cannot solve it using the discriminant. We can first find the two roots of the equation $x_{1}=$ $\frac{2}{5-k}, x... | Solution: The original equation is
$$
(k-3)(k-5) x^{2}-2(13-3 k)+8=0 \text {, }
$$
which is $[(k-5) x+2][(k-3) x+4]=0$.
Solving, we get $x_{1}=\frac{2}{5-k}, x_{2}=\frac{4}{3-k}$.
Therefore, $k=5-\frac{2}{x_{1}}, k=3-\frac{4}{x_{2}}$.
Eliminating $k$, we get $\frac{2}{x_{1}}-\frac{4}{x_{2}}=2$, which is $\left(x_{1}-1... | k=4,7, \frac{13}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,313 |
$5 . n$ is a known positive integer, $0 \leqslant r \leqslant n, r \in \mathbf{Z}$. Then when $r!(n-r)!$ takes the minimum value, $r=$ $\qquad$ . | 5. $\left\{\begin{array}{cl}\frac{n}{2}, & n \text { is even; } \\ \frac{n-1}{2} \text { or } \frac{n+1}{2}, & n \text { is odd }\end{array}\right.$ | \left\{\begin{array}{cl}\frac{n}{2}, & n \text{ is even; } \\ \frac{n-1}{2} \text{ or } \frac{n+1}{2}, & n \text{ is odd }\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,314 |
7. Given the set $M=\left\{(a, b) \mid\left(y^{2}+4\right) a^{2}-\right.$ $2(x y+b y+8) a+x^{2}+2 b x+2 b^{2}+12$ is a square of a linear expression in $x$ and $y$ $\}$. When $(a, b)$ takes all elements in the set $M$, the maximum distance from the point $(a, b)$ to the origin $O$ is $\qquad$ | $7 . \frac{2 \sqrt{21}}{3}$ | \frac{2 \sqrt{21}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,316 |
8. As shown in Figure 1, three small balls with a radius of $10 \mathrm{~cm}$ each are placed in a hemispherical bowl, with the tops of the balls exactly at the same level as the rim of the bowl. The radius of the bowl is $\qquad$ $\mathrm{cm}$. | $8.10\left(1+\frac{\sqrt{21}}{3}\right)$ | 10\left(1+\frac{\sqrt{21}}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,317 |
9. As shown in Figure $2, A B$ is a chord of the parabola passing through the focus $F$, and it forms a $45^{\circ}$ angle with the axis of symmetry of the parabola, $O$ is the vertex of the parabola. Then the size of $\angle A O B$ is $\qquad$ (expressed in terms of an inverse trigonometric function).
Translate the a... | 9. $\pi-\arccos \frac{3 \sqrt{41}}{41}\left(\right.$ or $\left.\pi-\arctan \frac{4 \sqrt{2}}{3}\right)$ | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,318 |
Given $f(x)=x^{3}+m x^{2}+n x+5, m 、 n$ $\in \mathbf{Z}$. Find:
(1) All pairs $(m, n)$ such that $f(x)=0$ has 3 integer roots (including repeated roots);
(2) All pairs $(m, n)$ such that $f(x)=0$ has at least 1 integer root, and $0 \leqslant m \leqslant 5,0 \leqslant n \leqslant 5$. | (1) If $\alpha$ is an integer root of $f(x)=0$, then from $m, n \in \mathbf{Z}$ and $\alpha\left(\alpha^{2}+m \alpha+n\right)=-5$ we know $\alpha \mid 5$.
Let $\alpha, \beta, \gamma$ be the 3 integer roots of $f(x)=0$, then $\alpha \beta \gamma = -5$. Since 5 is a prime number, $\alpha, \beta, \gamma$ can only be -1, ... | (1,5), (0,4), (5,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,320 |
Three, the sequence $\left\{a_{n}\right\}$ satisfies
$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1) \text {, }
$$
$n=1,2, \cdots$, and $a_{100}=10098$. Find the general term formula of the sequence $\left\{a_{n}\right\}$. | Three, Solution 1: The given equation can be transformed into
$$
(n-1)\left(a_{n+1}-2 n\right)=(n+1)\left[a_{n}-2(n-1)\right] \text {. }
$$
Let $b_{n}=a_{n}-2(n-1)$, then we have
$(n-1) b_{n+1}=(n+1) b_{n}$.
When $n \geqslant 2$, $b_{n+1}=\frac{n+1}{n-1} \cdot b_{n}$;
When $n \geqslant 3$, $b_{n}=\frac{n}{n-2} \cdot b... | a_{n}=(n-1)(n+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,321 |
Four, among the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion.
In the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion. | Let $n=\overline{a b c d e f g h i j}$ satisfy the conditions, then $11111 \mid n$, and $a, b, \cdots, j$ are a permutation of $0,1,2, \cdots, 9$ $(a \neq 0)$.
Since $a+b+\cdots+j=0+1+\cdots+9=45$,
it follows that $9 \mid(a+b+\cdots+j)$. Therefore, $9 \mid n$.
Since 11111 and 9 are coprime, then $99999 \mid n$.
Let $x=... | 3456 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,322 |
1. Given that the function $f(x)$ is an odd function on $\mathbf{R}$, and $g(x)$ is an even function on $\mathbf{R}$. If $f(x)-g(x)=$ $x^{2}+9 x+12$, then $f(x)+g(x)=(\quad)$.
(A) $-x^{2}+9 x-12$
(B) $x^{2}+9 x-12$
(C) $-x^{2}-9 x+12$
(D) $x^{2}-9 x+12$ | $-1 . A$
From $f(x) 、 g(x)$ being odd and even functions respectively, we know
$$
f(-x)=-f(x), g(-x)=g(x) .
$$
Substituting $-x$ for $f(x)-g(x)=x^{2}+9 x+12$, we get $f(-x)-g(-x)=x^{2}-9 x+12$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,323 |
Example 6 Find all positive integer triples $(a, b, c)$ such that the roots of the following three quadratic equations in $x$
$$
\begin{array}{l}
x^{2}-3 a x+2 b=0, \\
x^{2}-3 b x+2 c=0, \\
x^{2}-3 c x+2 a=0
\end{array}
$$
are all positive integers. | Let the roots of the equation $x^{2}-3 a x+2 b=0$ be $x_{1}, x_{2}$, with $x_{1} \leqslant x_{2}$; the roots of $x^{2}-3 b x+2 c=0$ be $x_{3}, x_{4}$, with $x_{3} \leqslant x_{4}$; and the roots of $x^{2}-3 c x+2 a=0$ be $x_{5}, x_{6}$, with $x_{5} \leqslant x_{6}$. By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1... | (1,1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,324 |
3. The solution set of the equation $x^{2}+x-1=x \pi^{x^{2}-1}+\left(x^{2}-1\right) \pi^{x}$ is $A$ (where $\pi$ is an irrational number, $\pi=3.141 \cdots$, and $x$ is a real number). Then the sum of the squares of all elements in $A$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 3.C.
Let $y=x^{2}-1$, then the original equation becomes
$$
x\left(\pi^{y}-1\right)+y\left(\pi^{x}-1\right)=0.
$$
When $x>0$, we have $\pi^{x}-1>0$.
Thus, $x$ and $\pi^{x}-1$ have the same sign.
Similarly, $y$ and $\pi^{y}-1$ have the same sign.
Therefore, when $x y \neq 0$, we have $x\left(\pi^{x}-1\right) y\left(\p... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,326 |
4. Given point $P(x, y)$ satisfies $(x-4 \cos \theta)^{2}+$ $(y-4 \sin \theta)^{2}=4(\theta \in \mathbf{R})$. Then the area of the region where point $P(x, y)$ is located is ( ).
(A) $36 \pi$
(B) $32 \pi$
(C) $20 \pi$
(D) $16 \pi$ | 4.B.
The points that satisfy the condition lie within the annulus centered at $(0,0)$ with radii 2 and 6, and its area is $36 \pi-4 \pi=32 \pi$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,327 |
5. Put 10 identical balls into 3 boxes numbered $1, 2, 3$ (all 10 balls must be placed each time), with the requirement that the number of balls in each box is no less than the box's number. The number of such ways to place the balls is ( .
(A) 9
(B) 12
(C) 15
(D) 18 | 5.C.
Let the number of balls in the boxes numbered $1, 2, 3$ be $x_{1}, x_{2}, x_{3}$ respectively, then
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=10\left(x_{1} \geqslant 1, x_{2} \geqslant 2, x_{3} \geqslant 3\right) . \\
\text { Let } y_{1}=x_{1}, y_{2}=x_{2}-1, y_{3}=x_{3}-2, \text { then } \\
y_{1}+y_{2}+y_{3}=7\left(... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,328 |
6. Given the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, and $S_{5}=$ $28, S_{10}=36$. Then $S_{15}$ equals ( ).
(A) 80
(B) 40
(C) 24
(D) -48 | 6.C.
It is known that $S_{5}, S_{10}-S_{5}, S_{15}-S_{10}$ form an arithmetic sequence. It is easy to get $S_{15}=24$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,329 |
7. Given the curve $C: y=\sqrt{-x^{2}-2 x}$ and the line $l: x+y-m=0$ intersect at two points. Then the range of $m$ is ( ).
(A) $(-\sqrt{2}-1, \sqrt{2})$
(B) $(-2, \sqrt{2}-1)$
(C) $[0, \sqrt{2}-1)$
(D) $(0, \sqrt{2}-1)$ | 7.C.
The curve $(x+1)^{2}+y^{2}=1(y \geqslant 0)$ represents the upper half (including endpoints) of a circle with center at $(-1,0)$ and radius 1. The line $l$ has a slope of -1 and an intercept of $m$. Given that the line $l$ intersects the curve $C$ at two points, we have
$\frac{|-1-m|}{\sqrt{2}}<1$, and $m \geqsl... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,330 |
8. The cross-sectional area of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ through the diagonal $B D_{1}$ is $S$, and $S_{1}$ and $S_{2}$ are the maximum and minimum values of $S$, respectively. Then $\frac{S_{1}}{S_{2}}$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{2 \sqrt{3}}{3}$
(D) $\frac{2 \... | 8.C.
As shown in Figure 1, let the section through $B D_{1}$ be $\square B E D_{1} E_{1}$,
then $S=2 S_{\triangle B D_{1} E_{1}}=h$.
$B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. It is easy to see that $h$ is minimized when it is the distance $h^{\prime}$ between the skew lines $B D_{1}$ and $B_{1} C_... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,331 |
9. Let $x=0.82^{0.5}, y=\sin 1, z=\log _{3} \sqrt{7}$. Then the size relationship of $x, y, z$ is ( ).
(A) $x<y<z$
(B) $y<z<x$
(C) $z<x<y$
(D) $z<y<x$ | 9. B.
From $x=0.82^{0.5}>0.81^{0.5}=0.9, y=\sin 17^{5}$ we get $9>5 \log _{3} 7$, which means $0.9>\log _{3} \sqrt{7}$, so $x>z$.
Also, from $3^{7}<7^{4}$, we can get $3^{\frac{7}{4}}<7$.
And $3^{\sqrt{3}}<7, \sqrt{3}<\log _{3} 7, \frac{\sqrt{3}}{2}<\log _{3} \sqrt{7}$, which means $y<z$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,332 |
10. If in the quadratic equation $x^{2}-2(a-3) x-$ $b^{2}+9=0$, $a$ and $b$ are the numbers obtained by rolling a die, then the probability $P=$ ( ) that the quadratic equation has two positive roots.
(A) $\frac{1}{18}$
(B) $\frac{1}{9}$
(C) $\frac{1}{6}$
(D) $\frac{13}{18}$ | 10.A.
From the quadratic equation having two positive roots, we know $\left\{\begin{array}{l}\Delta \geqslant 0, \\ a-3>0, \\ -b^{2}+9>0 \text {. }\end{array}\right.$
Solving, we get $a=6, b=1$ or $a=6, b=2$.
Thus, $P=\frac{2}{36}=\frac{1}{18}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,333 |
$1 . m$ is what integer when the equation
$$
\left(m^{2}-1\right) x^{2}-6(3 m-1) x+72=0
$$
has two distinct positive integer roots? | (Tip: $m^{2}-1 \neq 0, m \neq \pm 1$. Since $\Delta=36(m-3)^{2}>$ 0, hence $m \neq 3$. Using the quadratic formula, we get $x_{1}=\frac{6}{m-1}, x_{2}=$ $\frac{12}{m+1}$. Therefore, $(m-1)|6,(m+1)| 12$. So $m=2$, at this point, $x_{1}=6, x_{2}=4$.) | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,335 |
12. Given in $\triangle A B C$, $\boldsymbol{A} \boldsymbol{B}=\boldsymbol{a}, \boldsymbol{A} \boldsymbol{C}=\boldsymbol{b}$. Try to express $S_{\triangle A B C}=$ $\qquad$ using vector operations of $\boldsymbol{a}$ and $\boldsymbol{b}$. | $\begin{array}{l}\text { 12. } \frac{1}{2} \sqrt{(|\boldsymbol{a}| \cdot|\boldsymbol{b}|)^{2}-(\boldsymbol{a} \cdot \boldsymbol{b})^{2}} . \\ \text { Given } S_{\triangle A B C}=\frac{1}{2}|A \boldsymbol{B}| \cdot|A C| \sin A \\ =\frac{1}{2}|\boldsymbol{a}| \cdot|\boldsymbol{b}| \sin A, \\ \text { we have } S_{\triangl... | \frac{1}{2} \sqrt{(|\boldsymbol{a}| \cdot|\boldsymbol{b}|)^{2}-(\boldsymbol{a} \cdot \boldsymbol{b})^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,336 |
13. From 3 boys and $n$ girls, any 3 people are selected to participate in a competition, given that the probability of having at least 1 girl among the 3 people is $\frac{34}{35}$. Then $n=$ $\qquad$ . | 13.4. From the condition, $1-\frac{\mathrm{C}_{3}^{3}}{\mathrm{C}_{n+3}^{3}}=\frac{34}{35}$, solving for $n$ yields $n=4$. | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,337 |
14. There are 10 table tennis players participating in a round-robin tournament. The match results show that there are no draws, and among any 5 players, there is 1 player who wins against the other 4, and 1 player who loses to the other 4. Then the number of players who won exactly two matches is $\qquad$. | 14.1.
It can be proven that under the given conditions, no two players have the same number of wins. Therefore, the number of wins for 10 players are 10 different numbers: $0,1, \cdots, 9$. Hence, the number of players who win exactly two games is 1.
If not, suppose there exist $A$ and $B$ with the same number of win... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,338 |
15. (12 points) For the function $f(x)$, if $f(x)=x$, then $x$ is called a "fixed point" of $f(x)$; if $f(f(x))=x$, then $x$ is called a "stable point" of $f(x)$. The sets of "fixed points" and "stable points" of the function $f(x)$ are denoted as $A$ and $B$, respectively, i.e., $A=\{x \mid f(x)=x\}, B=\{x \mid f(f(x)... | (1) If $A=\varnothing$, then $A \subseteq B$ is obviously true.
If $A \neq \varnothing$, let $t \in A$, then
$$
f(t)=t, f(f(t))=f(t)=t \text {. }
$$
Thus, $t \in B$, hence $A \subseteq B$.
(2) The elements of $A$ are the real roots of the equation $f(x)=x$, i.e., $a x^{2}-1=x$. Since $A \neq \varnothing$, we have
$$
a... | \left[-\frac{1}{4}, \frac{3}{4}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,339 |
16. (12 points) A clothing workshop has 4 teams, A, B, C, and D, with their daily production capacities for shirts and pants as shown in Table 1. Now, it is required to produce matching sets of shirts and pants (one shirt and one pair of pants make a set). How many sets can these 4 teams produce in 7 days? | 16. The ratio of shirts to skirts produced daily by groups $A$, $B$, $C$, and $D$ are $\frac{8}{10}$, $\frac{9}{12}$, $\frac{7}{11}$, and $\frac{6}{7}$, respectively, and
$$
\frac{6}{7}>\frac{8}{10}>\frac{9}{12}>\frac{7}{11} \text {. }
$$
Only by having the group with the highest efficiency in producing shirts make sh... | 125 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 715,340 |
17. (12 points) Let the sequence $\left\{a_{n}\right\}$ satisfy the conditions: $a_{1}=$ $1, a_{2}=2$, and $a_{n+2}=a_{n+1}+a_{n}, n=1,2, \cdots$. Prove that for any positive integer $n$, we have
$$
\sqrt[n]{a_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{a_{n}}} .
$$ | 17. Let $a_{0}=1$, then we have $a_{k+1}=a_{k}+a_{k-1}$, and
$$
1=\frac{a_{k}}{a_{k+1}}+\frac{a_{k-1}}{a_{k+1}}, k=1,2, \cdots \text {. }
$$
Thus, $n=\sum_{k=1}^{n} \frac{a_{k}}{a_{k+1}}+\sum_{k=1}^{n} \frac{a_{k-1}}{a_{k+1}}$.
By the arithmetic-geometric mean inequality, we get
$1 \geqslant \sqrt[n]{\frac{a_{1}}{a_{2... | \sqrt[n]{a_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{a_{n}}} | Inequalities | proof | Yes | Yes | cn_contest | false | 715,341 |
18. (16 points) In $\triangle A B C$ with a fixed perimeter, it is known that $|A B|=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ that in... | 18. (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $|CA| + |CB| = 2a (a > 3)$ be a constant, then the locus of point $C$ is an ellipse with foci at $A$ and $B$. Therefore, the focal distance is $2c = |AB| = 6$.
Since $\cos C = \... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,342 |
19. (16 points) Given a tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ that are mutually perpendicular, $P$ is any point within the base $\triangle ABC$, and $OP$ forms angles $\alpha, \beta, \gamma$ with the three lateral faces. Prove:
$$
\frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}... | 19. From the problem, we have
$$
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1,
$$
and $a 、 \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$. Then | proof | Geometry | proof | Yes | Yes | cn_contest | false | 715,343 |
1. Given that the point $(x, y)$ moves on the line $x+2 y=3$. When $2^{x}+4^{y}$ takes the minimum value, the distance from the point $(x, y)$ to the origin is ( ).
(A) $\frac{3 \sqrt{5}}{4}$
(B) $\frac{45}{16}$
(C) $\frac{3 \sqrt{2}}{4}$
(D) $\frac{9}{8}$ | $$
\begin{array}{l}
-1 . \text { A. } \\
2^{x}+4^{y} \geqslant 2 \sqrt{2^{x+2 y}}=4 \sqrt{2} .
\end{array}
$$
When $x=\frac{3}{2}, y=\frac{3}{4}$, the above expression achieves its minimum value. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,344 |
2. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with eccentricity $e \in$ $\left[\frac{2 \sqrt{3}}{3}, 2\right]$. The range of the angle $\alpha$ between the two asymptotes of the hyperbola is ( ).
(A) $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$
(B) $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$
(C) $\lef... | 2.C.
Let the inclination angle of the asymptote $y=\frac{b}{a} x$ be $\beta$. Note that
$$
\begin{array}{l}
1+\frac{b^{2}}{a^{2}}=e^{2} \in\left[\frac{4}{3}, 4\right], \\
\tan \beta=\frac{b}{a} \in\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right], \beta \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] .
\end{array}
$$
Thus, $a=\... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 715,345 |
2. Let $m$ be a non-zero integer, and the quadratic equation in $x$, $m x^{2}-(m-1) x+1=0$, has rational roots. Find the value of $m$.
| (Given: Let $\Delta=(m-1)^{2}-4 m=n^{2}, n$ be a non-negative integer, then $(m-3)^{2}-n^{2}=8$, i.e., $(m-3-n)(m-3+n)=$ 8. Following Example 2, we get $\left\{\begin{array}{l}m=6, \\ n=1\end{array}\right.$ or $\left\{\begin{array}{l}m=0, \\ n=1\end{array}\right.$ (discard). Therefore, $m=6$, and the two roots of the e... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,346 |
3. On the 4 faces of a regular tetrahedron, the numbers $1, 2, 3,$ and $4$ are written. Four such uniform regular tetrahedra are thrown onto a table, and the product of the 4 numbers on the 4 faces that touch the table is divisible by 4. The probability of this happening is ( ).
(A) $\frac{1}{8}$
(B) $\frac{9}{64}$
(C)... | 3.D.
The probability of the event "all 4 numbers are odd" is
$$
P_{1}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \text {; }
$$
The probability of the event "3 numbers are odd, 1 number is 2" is
$$
\begin{array}{l}
P_{2}=\mathrm{C}_{4}^{1} \cdot \frac{1}{4} \cdot\left(\frac{1}{2}\right)^{3}=\frac{1}{8} . \\
\text { The... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 715,347 |
4. Three people, A, B, and C, are training in a broadcast tournament format. In each round, 2 people play a singles match, and the other 1 person is the referee. The loser of each round becomes the referee for the next round, and the previous referee challenges the winner. At the end of the half-day training, it was fo... | 4.A.
A total of $12+21-8=25$ games were played, with Jia serving as the referee for $25-12=13$ games. Since the same person cannot serve as the referee for two consecutive games, Jia must have been the referee for the 1st, 3rd, ..., 11th, ..., 25th games. Therefore, the loser of the 10th game was Jia. | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 715,348 |
5. The curve $x^{2}+y^{2}-a y=0$ intersects with $a x^{2}+b x y+x$ $=0$ at exactly 3 distinct points. Therefore, it must be that ( ).
(A) $\left(a^{4}+4 a b+4\right)(a b+1)=0$
(B) $\left(a^{4}-4 a b-4\right)(a b+1)=0$
(C) $\left(a^{4}+4 a b+4\right)(a b-1)=0$
(D) $\left(a^{4}-4 a b-4\right)(a b-1)=0$ | 5. B.
It is clear that $a \neq 0$. The curve $a x^{2}+b x y+x=0$ consists of the two lines $x=0$ and $a x+b y+1=0$. The line $x=0$ intersects the circle $x^{2}+y^{2}-a y=0$ at two distinct points $(0,0)$ and $(0, a)$. According to the problem, there are two possibilities:
(1) The line $a x+b y+1=0$ is tangent to the c... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,349 |
6. Two periodic functions $y_{1}$ and $y_{2}$ have the smallest positive periods $a$ and $b$ respectively, and $b=n a(n \geqslant 2, n$ is an integer $)$. If the function $y_{3}=y_{1}+y_{2}$ has the smallest positive period $t$, then among the following 5 scenarios:
(1) $tb$
the number of scenarios that cannot occur i... | 6. B
$b$ is a period of $y_{3}$; hence $t \leqslant b$. If $t=a$, then from $y_{2}=y_{3}-y_{1}$ we get $b \leqslant a$, which is a contradiction. Therefore, (2) and (5) are impossible.
The following examples show that the other three cases can occur:
Take $y_{2}=\sin x+\sin \frac{2 x}{3}$, then $b=6 \pi$.
(1) Let $y_{1... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,350 |
7. Given $\log _{a} x=24, \log _{b} x=40, \log _{\text {abc }} x=$
12. Then, $\log _{c} x=$ $\qquad$ . | \begin{array}{l}\text { II.7.60. } \\ \log _{x} c=\log _{x} a b c-\log _{x} a-\log _{x} b \\ =\frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60} .\end{array} | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,351 |
8. Let $f(x)=\prod_{i=1}^{4}\left(x^{2}-8 x+c_{i}\right), M=|x|$ $f(x)=0\}$. It is known that $M=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right.$, $\left.x_{7}, x_{8}\right\} \subseteq \mathbf{N}$. Then, $\max \left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}-$ $\min \left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}=$ $\qquad$ | 8.15.
Let the two roots of $x^{2}-8 x+c=0$ be $\alpha, \beta$, then $\alpha+\beta=8$. The unequal non-negative integer values of $(\alpha, \beta)$ are only $(0,8),(1,7),(2,6)$, $(3,5)$. Therefore, $\left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}=\{0,7,12,15\}$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,352 |
9. If real numbers $x, y$ satisfy $3 x+2 y-1 \geqslant 0$, then the minimum value of $u=x^{2}+y^{2}+6 x-2 y$ is $\qquad$ . | 9. $-\frac{66}{13}$.
$u=(x+3)^{2}+(y-1)^{2}-10$. The minimum distance from a point in the half-plane $3 x+2 y-1$ $\geqslant 0$ to the fixed point $(-3,1)$ is $\frac{|-9+2-1|}{\sqrt{13}}=\frac{8}{\sqrt{13}}$. Therefore,
$$
u_{\text {min }}=\left(\frac{8}{\sqrt{13}}\right)^{2}-10=-\frac{66}{13} .
$$ | -\frac{66}{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,353 |
10. The solution set of the inequality system $\sin x>\cos x>\tan x>\cot x$ in $(0,2 \pi)$ (expressed in intervals) is $\qquad$ | 10. $\left(\frac{3 \pi}{4}, \pi-\arcsin \frac{\sqrt{5}-1}{2}\right)$.
As shown in Figure 1, solve using the interval method on the quadrant diagram.
The boundary lines are $\frac{k \pi}{4}(0 \leqslant k \leqslant 7)$ and the solutions to the equation $\cos x = \tan x$ are
$$
\begin{array}{l}
\arcsin \frac{\sqrt{5}-1}{... | \left(\frac{3 \pi}{4}, \pi-\arcsin \frac{\sqrt{5}-1}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,354 |
11. In tetrahedron $ABCD$, $AB=CD=a, BC=AD=b, CA=BD=c$. If the angle between the skew lines $AB$ and $CD$ is $\theta$, then, $\cos \theta=$ $\qquad$ . | 11. $\frac{\left|b^{2}-c^{2}\right|}{a^{2}}$.
Solution 1: Let $D C=a, D A=b, D B=c$. From the given conditions, we have
$$
|\boldsymbol{a}-\boldsymbol{b}|=|\boldsymbol{c}|, a^{2}+b^{2}-2 \boldsymbol{a} \cdot \boldsymbol{b}=c^{2} .
$$
Thus, $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{a^{2}+b^{2}-c^{2}}{2}$.
Similarly, ... | \frac{\left|b^{2}-c^{2}\right|}{a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,355 |
12. Let $a, b, x \in \mathbf{N}_{+}, a \leqslant b, X$ be the solution set of the inequality $\lg b - \lg a < \lg x < \lg b + \lg a$. Given that $\operatorname{card}(X) = 50$. When $ab$ takes its maximum possible value, $\sqrt{a+b}=$ $\qquad$ . | 12.6.
$$
\begin{array}{l}
\frac{b}{a}<x<a b, a \geqslant 2, \\
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 .
\end{array}
$$
Therefore, $a b \leqslant 68$.
Equality holds if and only if $a=2, b=34$. | 12.6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 715,356 |
3. Given that the equation $x^{2}+(a-6) x+a=0$ has two integer roots with respect to $x$. Find the value of $a$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
(As this is a note or instruction, it is not part of the translation task and thus not translated.) | (Tip: Let the two integer roots be $x_{1} \geqslant x_{2}$, by Vieta's formulas we get $x_{1}+x_{2}=6-a, x_{1} x_{2}=a$, eliminating $a$, we get $x_{1} x_{2}+x_{1}+x_{2}$ $=6,\left(x_{1}+1\right)\left(x_{2}+1\right)=7$. Solving, we get $x_{1}=6, x_{2}=0$ or $x_{1}=$ $-2, x_{2}=-8$. Therefore, $a=x_{1} x_{2}=0$ or 16.) | 0 \text{ or } 16 | Algebra | MCQ | Yes | Yes | cn_contest | false | 715,357 |
13. (20 points) Find the minimum value of the function
$$
\begin{aligned}
f(x)= & \mid \sin x+\cos x+\tan x+\cot x+ \\
& \sec x+\csc x \mid
\end{aligned}
$$
where $\sec x=\frac{1}{\cos x}, \csc x=\frac{1}{\sin x}$. | Three, 13. Let $u=\sin x+\cos x$, then $\sin x \cdot \cos x=\frac{1}{2}\left(u^{2}-1\right)$.
Thus, $\sin x+\cos x+\tan x+\cot x+\sec x+\csc x$ $=u+\frac{2}{u-1}$.
When $u>1$, we have
$$
f(x)=1+u-1+\frac{2}{u-1} \geqslant 1+2 \sqrt{2} .
$$
When $u<1$, we have
$$
f(x)=-1+1-u+\frac{2}{1-u} \geqslant 2 \sqrt{2}-1 \text {... | 2 \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,358 |
14. (20 points) In the ellipse $x^{2}+4 y^{2}=8$, $AB$ is a moving chord of length $\frac{5}{2}$, and $O$ is the origin. Find the range of the area of $\triangle AOB$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 14. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the equation of line $AB$ be $y=k x+b$. Substituting into the ellipse equation and rearranging, we get
$$
\left(4 k^{2}+1\right) x^{2}+8 k b x+4\left(b^{2}-2\right)=0 .
$$
Thus, $x_{1}+x_{2}=-\frac{8 k b}{4 k^{2}+1}, x_{1} x_{2}=\frac{4\left(b^{2}-2... | \left[\frac{5 \sqrt{103}}{32}, 2\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 715,359 |
15. (20 points) In the infinite sequence $\left\{x_{n}\right\}(n \geqslant 1)$, for each odd $n$, $x_{n}, x_{n+1}, x_{n+2}$ form a geometric sequence, and for each even $n$, $x_{n}, x_{n+1}, x_{n+2}$ form an arithmetic sequence. Given $x_{1}=a, x_{2}=b$.
(1) Find the general term formula of the sequence, and what neces... | 15. (1) Observe the first few terms: $a, b, \frac{b^{2}}{a}, \frac{b(2 b-a)}{a}$, $\frac{(2 b-a)^{2}}{a}, \frac{(2 b-a)(3 b-2 a)}{a}, \frac{(3 b-2 a)^{2}}{a}, \cdots$, and guess
$$
\begin{array}{l}
x_{2 k-1}=\frac{[(k-1) b-(k-2) a]^{2}}{a}, \\
x_{2 k}=\frac{[(k-1) b-(k-2) a][k b-(k-1) a]}{a}(k \geqslant 1) .
\end{array... | n b-(n-1) a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,360 |
16. (30 points) (1) Given a positive integer $n(n \geqslant 5)$, the set $A_{n}=\{1,2, \cdots, n\}$, does there exist a one-to-one mapping $\varphi$ : $A_{n} \rightarrow A_{n}$ that satisfies the condition: for all $k(1 \leqslant k \leqslant n-1)$, we have $k \mid(\varphi(1)+\varphi(2)+\cdots+\varphi(k))$?
(2) $\mathbf... | 16. (1) Does not exist.
Let $S_{k}=\sum_{i=1}^{k} \varphi(i)$.
When $n=2 m+1(m \geqslant 2)$, by $2 m \mid S_{2 m}$ and
$$
S_{2 m}=\frac{(2 m+1)(2 m+2)}{2}-\varphi(2 m+1)
$$
we get $\varphi(2 m+1) \equiv m+1(\bmod 2 m)$.
But $\varphi(2 m+1) \in A_{2 m+1}$, hence $\varphi(2 m+1)=m+1$.
Again, by $(2 m-1) \mid S_{2 m-1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 715,361 |
6.1. In the fields of Dreamland, there grows a kind of tree that bears gold coins, and the number of gold coins on each tree may differ. Every night, each tree bears one new gold coin. On March 1st, there were a total of 1000 gold coins. On a certain day in March, someone planted another such tree, and by March 31st, t... | 6. 1. A total of 993 gold coins were produced in 30 days. Examining the remainder of 993 divided by 30, we have $993=33 \times 30+3$. The remainder 3 is the number of days the newly planted tree participated in producing gold coins. Therefore, the person planted the tree on March 27. | March 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,362 |
6.3. Among the positive integers $1,2, \cdots, 1992$, which of the following two categories contains more integers?
(1) Integers divisible by 8 but not by 9;
(2) Integers divisible by 9 but not by 8. | 6.3. Add to both sets integers that can be divided by both 8 and 9, so the first set consists of all integers divisible by 8 among the positive integers 1, 2, $\cdots, 1992$, and the second set consists of all integers divisible by 9 among the positive integers $1, 2, \cdots, 1992$. Obviously, the first set contains mo... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,364 |
6.4. There is a 36-digit number, in which the digits $1,2, \cdots, 9$ each appear 4 times, and except for 9, all other digits are less than the digit that follows them. It is known that the first digit of the number is 9. What is the last digit of the number? Please provide all possible answers and prove that there are... | 6.4. The last digit of this number is 8.
According to the problem, only 9 can follow 8. If all four 8s are located within the number, then each of them is followed by one 9, and adding the one 9 at the beginning, there are a total of 5 nines. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 715,365 |
7.2. Person A and Person B perform division with remainder on the same number. A divides it by 8, and B divides it by 9. It is known that the sum of the quotient obtained by A and the remainder obtained by B equals 13. Try to find the remainder obtained by A.
| 7.2. Let $a$ and $b$ represent the quotient and remainder obtained by Jia, and let $c$ and $d$ represent the quotient and remainder obtained by Yi. Thus, we have
$$
8 a+b=9 c+d, a+d=13 \text{. }
$$
Substituting $d=13-a$ into the first equation, we get $9(a-c)=$ $13-b$, so $13-b$ is divisible by 9. Since $b$ can only b... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 715,366 |
4. The equation $a x^{2}+2(a-3) x+(a-2)=0$ about $x$ has at least one integer solution, and $a$ is an integer. Find the value of $a$. | (提示: When $a=0$, there is no integer solution. When $a \neq 0$, the equation is a quadratic equation with integer coefficients, so, $\Delta=4(a-3)^{2}-4 a(a$ $-2)=4(9-4 a)$ is a perfect square. Therefore, $9-4 a$ is a square | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 715,368 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.