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Sure, here is the translation:
```
II. (50 points) Let $\mathrm{i}^{2}=-1$. Prove:
$$
\prod_{k=1}^{n-1}\left(2 \cot \frac{\pi}{n}-\cot \frac{k \pi}{n}+\mathrm{i}\right)
$$
is a pure imaginary number.
$$
III. (50 points) Let \( X=\{1,2, \cdots, p\} \), where \( p \)
$$
is a prime number. For a subset \( A \) of \( X ... | When $p=2$, $X=\{1,2\}$, at this time, $X$ has 2 subsets: $\varnothing, \{2\}$, so, $S=2$.
When $p>2$, $p$ is an odd prime, let
$$
\begin{array}{l}
f(x)=(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{p}\right) \\
=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{m} x^{m} .
\end{array}
$$
Consider the number of all subsets of $X=\{1,2,... | S=\left\{\begin{array}{ll}2, & p=2 ; \\ \frac{1}{p}\left(z^{2}+2 p-2\right), & p \text { is an odd prime. }\end{array}\right.} | Number Theory | proof | Yes | Yes | cn_contest | false | 720,183 |
As shown in Figure 2, point $P$ is outside circle $\odot O$, with the radius of $\odot O$ being $r$. Through $P$, any two secants $PAB$ and $PCD$ of $\odot O$ are drawn, and $AD$ and $BC$ intersect at point $Q$. Prove that regardless of the position of point $P$ and the secants $PAB$ and $PCD$, $OP^{2} + OQ^{2} - PQ^{2... | Proof: As shown in Figure 2, let the line $P O$ intersect $\odot O$ at $X$ and $Y$, and the line $Q O$ intersect $\odot O$ at $M$ and $N$. Take a point $K$ on the ray $P Q$ such that $C, D, K, Q$ are concyclic. Therefore,
$$
\begin{array}{l}
P Q \cdot P K = P C \cdot P D = P X \cdot P Y \\
= (O P - r)(O P + r) = O P^{2... | O P^{2} + O Q^{2} - P Q^{2} = 2 r^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 720,184 |
Example $4 P$ is a fixed point inside the acute angle $\angle A$, and a line $l$ passing through $P$ intersects the two sides of $\angle A$ at points $B$ and $C$. Find:
(1) the maximum value of $\frac{1}{P B}+\frac{1}{P C}$ and the minimum value of $P B \cdot P C$;
(2) the line $l$ passing through $P$ such that $S_{\tr... | (1) As shown in Figure 4, take
the constants $A P=a, \angle C A P=\alpha, \angle B A P=\beta$ and the variable $\angle A P C=\theta \in(\beta, \pi-\alpha)$ as the basic quantities.
Notice that
$$
\begin{array}{l}
\frac{1}{P B}+\frac{1}{P C}=\frac{\sin (\theta-\beta)}{a \sin \beta}+\frac{\sin (\theta+\alpha)}{a \sin \al... | \frac{\sin (\alpha+\beta)}{a \sin \alpha \cdot \sin \beta}, \frac{a^{2} \sin \alpha \cdot \sin \beta}{\cos ^{2} \frac{\alpha+\beta}{2}}, PB=PC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,185 |
Find all real numbers $a$ such that in the solution set of the inequality $|x-a|+|1-3 x| \geqslant a+1$ with respect to $x$, there are exactly two distinct solutions whose distances to the point 1 are both $|a|$ (on the real number line).
Translate the text above into English, please keep the original text's line brea... | Solution: The two numbers whose distance to 1 is $|a|$ are $1+|a|$ and $1-|a|$.
(1) The necessary and sufficient condition for the solution of the original inequality to contain the number $x=1+|a|$ is
$$
|1+| a|-a|+|1-3(1+|a|)| \geqslant a+1,
$$
which is equivalent to $||a|-a+1|+|3| a|+2| \geqslant a+1$.
Clearly, $|a... | a<0 \text{ or } 0<a \leqslant \frac{1}{3} \text{ or } a \geqslant 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,186 |
Given $A, B$ are two fixed points on line $l$, and $C, D, E$ are three consecutive moving points on segment $AB$. Squares $C_{1}$ and $C_{2}$ are constructed on the same side of line $l$ with $CD$ and $DE$ as their sides, respectively. $F$ and $G$ are vertices of squares $C_{1}$ and $C_{2}$ (different from points $C, D... | Proof: First, we prove a lemma.
Lemma: As shown in Figure 4, $P$ is a point inside the isosceles right triangle $\triangle A B S$, and $S P$ intersects the hypotenuse $A B$ at point $M$. Let
$$
\begin{array}{l}
\angle P A B=\alpha, \\
\angle P B A=\beta. \\
\text { Then } \frac{A M}{B M} \\
=\frac{\cot \alpha-1}{\cot \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,188 |
Example 5 In a triangle where the sum of two sides is $m$ and their included angle $\alpha$ is a constant, find the minimum value of the perimeter.
In a triangle where the sum of two sides is $m$ and their included angle $\alpha$ is a constant, find the minimum value of the perimeter. | Explanation: Let $m$, $\alpha$, and the two sides $x$, $y$ (variables) be the basic quantities, with the condition that $x+y=m$. Find the minimum value of the third side $c$.
Notice that
$$
\begin{array}{l}
c^{2}=x^{2}+y^{2}-2 x y \cos \alpha \\
=(x+y)^{2}-2 x y(1+\cos \alpha) \\
=m^{2}-2 x y(1+\cos \alpha) .
\end{arr... | m+m \sin \frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,189 |
Example 6 As shown in Figure 5, in $\triangle A B C$, side $B C=a=14$, height $h_{a}$ $=12$, inradius $r$ $=4$. Find $b, c$. | Given
$B$
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \times 14 \times 12 \\
=p \cdot 4=\frac{1}{2} b c \sin \dot{A},
\end{array}
$$
we get
$$
p=21, b+c=2 p-a=28 \text {. }
$$
Notice that
$$
\tan \frac{A}{2}=\frac{O N}{A N}=\frac{r}{p-a}=\frac{4}{7} ,
$$
thus, $\sin A=\frac{2 \tan \frac{\mathrm{A}}{2}}{1+\ta... | b=13, c=15 \text{ or } b=15, c=13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,190 |
Example $7 P$ is a point inside square $A B C D$, and $P A: P B: P C=1: 2: 3$. Find $\angle A P B$. | Explanation: The conditions of this problem only determine the shape of the figure (the ratio of segments and angles). As shown in Figure 6, let the side length of the square be \( a \), and \( PA = m \) be a parameter (the ratio of the two is a constant). Therefore,
\[
\begin{array}{c}
PB = 2m, \, PC = 3m. \\
\text{Th... | 135^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,191 |
Example 8 The three sides of a triangle are consecutive integers. Find the lengths of the three sides according to the following two conditions:
(1) The largest angle is twice the smallest angle;
(2) It is an obtuse or right triangle. | Consider $\triangle A B C$ with side lengths $n-1$, $n$, and $n+1$, and their corresponding angles $\angle A$, $\angle B$, and $\angle C$, which are also arranged in ascending order.
(1) Since $(n-1)+(n+1)=2 n$, by the Law of Sines, we have
$$
\sin A+\sin C=2 \sin B=2 \sin (A+C).
$$
Given $\angle C=2 \angle A$, we get... | 2, 3, 4 \text{ or } 3, 4, 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,192 |
Given positive integer $n$ and real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, $y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}$, satisfying
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i} .
$$
Prove: For any real number $\alpha$, we have
$$
\sum_{i=1}^{n}[i \alpha] x_{i} \geqslant \sum_... | 甲: I want to set
$$
z_{i}=y_{i}-x_{i}(i=1,2, \cdots, n),
$$
then equation (1) becomes $\sum_{i=1}^{n} i z_{i}=0$. Therefore, we only need to prove
$$
\sum_{i=1}^{n}[i \alpha] z_{i} \leqslant 0 .
$$
师: Great idea, it simplifies the problem. Note that
$$
z_{1} \geqslant z_{2} \geqslant \cdots \geqslant z_{n} \text {. }... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,194 |
Example 1 Let $n \in \mathbf{N}_{+}$. Prove:
$$
\frac{1}{2-1}+\frac{1}{2^{2}-1}+\cdots+\frac{1}{2^{n}-1}<\frac{34}{21} .
$$ | Analysis: Suppose the strengthened proposition is
$$
\frac{1}{2-1}+\frac{1}{2^{2}-1}+\cdots+\frac{1}{2^{n}-1} \leqslant \frac{34}{21}-\frac{1}{f(n)},
$$
$f(n)$ should satisfy both
$$
\frac{1}{2-1} \leqslant \frac{34}{21}-\frac{1}{f(1)}
$$
and $\frac{1}{f(k)}-\frac{1}{f(k+1)} \geqslant \frac{1}{2^{k+1}-1}$.
The denomin... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,195 |
Example 2 Given a positive integer $n(n>1)$. Prove:
$$
1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}<\frac{7}{4} .
$$ | Analysis: Suppose the strengthened proposition is
$$
1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!} \leqslant \frac{7}{4}-\frac{1}{f(n)},
$$
$f(n)$ should satisfy both
$$
1+\frac{1}{2} \leqslant \frac{7}{4}-\frac{1}{f(2)}
$$
and $\frac{1}{f(k)}-\frac{1}{f(k+1)} \geqslant \frac{1}{(k+1)!}$.
The right side of equation ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,196 |
Example 3 A uniform cube with the numbers $1,1,2,3,3,5$ marked on its six faces is shown in Figure 3. When this cube is thrown once, the number on the top face is recorded as the x-coordinate of a point in the Cartesian coordinate system, and the number on the bottom face is recorded as the y-coordinate of that point. ... | Explanation: Each roll may result in six points with coordinates (two of which are overlapping):
$$
(1,1),(1,1),(2,3),(3,2),(3,5),(5,3) \text {. }
$$
By plotting the points and calculations, it can be found that the line determined by any two of the points $(1,1)$, $(2,3)$, and $(3,5)$ all pass through the point $P(4,... | \frac{2}{3} | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,197 |
Question 1 In a non-obtuse $\triangle A B C$, prove:
$$
\begin{array}{l}
\frac{(1-\cos 2 A)(1-\cos 2 B)}{1-\cos 2 C}+ \\
\frac{(1-\cos 2 C)(1-\cos 2 A)}{1-\cos 2 B}+ \\
\frac{(1-\cos 2 B)(1-\cos 2 C)}{1-\cos 2 A} \geqslant \frac{9}{2} .
\end{array}
$$ | Prove that equation (1) is
$$
\sum \frac{\sin ^{2} B \cdot \sin ^{2} A}{\sin ^{2} C} \geqslant \frac{9}{4} \text {. }
$$
The left side of equation (2)
$$
\begin{array}{l}
=\frac{[\cos (A-B)-\cos (A+B)]^{2}}{4 \sin ^{2} C}+ \\
\sin ^{2} C \cdot \frac{\sin ^{4} A+\sin ^{4} B}{\sin ^{2} A \cdot \sin ^{2} B} \\
=\frac{[\c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,198 |
Example 2 Figure 3 is a Pythagorean tree grown by squares and isosceles right triangles according to a certain pattern. Observe the figure and answer the questions.
(1) At the far right end of each layer of the tree, there is a square marked with a number indicating the layer, starting from this square and counting cou... | Solution: (1) From the figure, we observe that the number of squares in each upper layer is twice the number of squares in the previous layer. Therefore, the fifth layer has 16 squares, and the $n$-th layer has $2^{n-1}$ squares.
(2) The sum of the areas of the two smaller squares on each branch equals the area of the ... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,199 |
Example 3 Figure 4 is a "sheep's head" pattern, the method of which is: starting from square (1), using one of its sides as the hypotenuse, an isosceles right triangle is constructed outward, and then using its legs as sides, squares (2) and (2)' are constructed outward, ... and so on. If the side length of square (1) ... | Solution: By the Pythagorean theorem, the side length of square (2) is $64 \times \frac{\sqrt{2}}{2}=32 \sqrt{2}(\mathrm{~cm})$,
the side length of square (3) is
$$
32 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{2}=32(\mathrm{~cm}) \text {, }
$$
the side length of square (4) is
$$
32 \... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,200 |
Example 4 As shown in Figure 5, connect isosceles right triangles on a square in an infinite repetitive process. The side length of the first square is 1. The sum of the area of the first square and the first isosceles right triangle is $S_{1}$, the sum of the area of the second square and the second isosceles right tr... | Given that the sum of the area of the first square and the first isosceles right triangle is
$$
S_{1}=1+\frac{1}{4}=\frac{5}{4} \text {. }
$$
The sum of the area of the second square and the second isosceles right triangle is
$$
S_{2}=\frac{1}{4} \times 2+\frac{1}{4} \times 2 \times \frac{1}{4}=\frac{5}{4} \times \fra... | S_{n}=\frac{5}{4} \times \frac{1}{2^{n-1}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,201 |
1. Given that $H$ is the orthocenter of acute $\triangle A B C$, the circle with the midpoint of side $B C$ as its center and passing through point $H$ intersects line $B C$ at points $A_{1}$ and $A_{2}$; the circle with the midpoint of side $C A$ as its center and passing through point $H$ intersects line $C A$ at poi... | 1. As shown in Figure $1, B_{0}, C_{0}$ are the midpoints of sides $C A, A B$ respectively. Let the circle with center $B_{0}$ passing through point $H$ and the circle with center $C_{0}$ passing through point $H$ intersect at another point $A^{\prime}$. Then $A^{\prime} H \perp C_{0} B_{0}$.
Since $B_{0}, C_{0}$ are ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,202 |
2. (1) Let real numbers $x, y, z$ be not equal to 1, and satisfy $xyz=1$. Prove:
$$
\frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1 \text {; }
$$
(2) Prove: There exist infinitely many triples of rational numbers $(x, y, z)$, where $x, y, z$ are not equal to 1, and $xyz=1$, such that... | 2. (1) Let $\frac{x}{x-1}=a, \frac{y}{y-1}=b, \frac{z}{z-1}=c$. Then $x=\frac{a}{a-1}, y=\frac{b}{b-1}, z=\frac{c}{c-1}$.
From $x y z=1$
$$
\begin{aligned}
\Rightarrow & a b c=(a-1)(b-1)(c-1) \\
\Rightarrow & a+b+c-1=a b+b c+c a \\
\Rightarrow & a^{2}+b^{2}+c^{2} \\
& =(a+b+c)^{2}-2(a b+b c+c a) \\
& =(a+b+c)^{2}-2(a+b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,203 |
4. Find all functions
$$
f:(0,+\infty) \rightarrow(0,+\infty),
$$
satisfying for all positive real numbers $w, x, y, z, w x=y z$, we have
$$
\frac{(f(w))^{2}+(f(x))^{2}}{f\left(y^{2}\right)+f\left(z^{2}\right)}=\frac{w^{2}+x^{2}}{y^{2}+z^{2}} .
$$ | 4. Let $w=x=y=z=1$. Then $(f(1))^{2}=f(1)$.
So, $f(1)=1$.
For any $t>0$, let $w=t, x=1, y=z=\sqrt{t}$. Then $\frac{(f(t))^{2}+1}{2 f(t)}=\frac{t^{2}+1}{2 t}$.
Clearing the denominator and rearranging gives
$$
(t f(t)-1)(f(t)-t)=0 \text {. }
$$
Therefore, for each $t>0$,
$$
f(t)=t \text { or } f(t)=\frac{1}{t} \text {.... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,205 |
5. Let $n$ and $k$ be positive integers, $k \geqslant n$, and $k-n$ is an even number. $2 n$ lamps are sequentially numbered as $1,2, \cdots, 2 n$, and each lamp can be either "on" or "off". Initially, all lamps are off. An operation can be performed on these lamps: each operation changes the state of one lamp (i.e., o... | 5. (Zhang Ruixiang's solution) The required ratio is $2^{k-n}$.
Lemma: Let $t$ be a positive integer. If a $t$-element 0,1 array $\left(a_{1}, a_{2}, \cdots, a_{t}\right)\left(a_{1}, a_{2}, \cdots, a_{t} \in\{0,1\}\right)$ contains an odd number of 0s, then it is called "good". The number of good arrays is $2^{t-1}$.
... | 2^{k-n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,206 |
6. In a convex quadrilateral $ABCD$, $BA \neq BC$. Circles $\omega_{1}$ and $\omega_{2}$ are the incircles of $\triangle ABC$ and $\triangle ADC$, respectively. Suppose there exists a circle $\omega$ that is tangent to ray $BA$ (the tangency point is not on segment $BA$), tangent to ray $BC$ (the tangency point is not ... | 6. (Mou Xiaosheng's solution) First, prove two lemmas.
Lemma 1 Let quadrilateral $ABCD$ be a convex quadrilateral, and circle $\omega$ be tangent to ray $BA$ (excluding segment $BA$), ray $BC$ (excluding segment $BC$), and lines $AD$ and $CD$. Then
$$
AB + AD = CB + CD.
$$
Proof of Lemma 1: As shown in Figure 2, let ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,207 |
Example 4: A fair cube die, with faces numbered $1, 2, 3, 4, 5, 6$, is thrown twice. If the numbers on the top faces are $m$ and $n$ respectively, then the probability that the graph of the quadratic function $y=x^{2}+m x+n$ intersects the $x$-axis at two distinct points is ( ).
(A) $\frac{5}{12}$
(B) $\frac{4}{9}$
(C)... | Explanation: The total number of basic events is $6 \times 6=36$, meaning we can get 36 quadratic functions. Using the discriminant of a quadratic equation $\Delta=b^{2}-4 a c=m^{2}-4 n>0$, we know $m^{2}$ $>4 n$. By enumeration, we find that there are 17 pairs of $m, n$ that satisfy the condition, so $P=\frac{17}{36}$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,208 |
2. For all $m, n \in \mathbf{N}_{+}$, the function $f: \mathbf{N}_{+} \rightarrow$ $\mathbf{N}_{+}$ satisfies
$$
f(m+n) \geqslant f(m)+f(f(n))-1 .
$$
Find all possible values of $f(2007)$, where $\mathbf{N}_{+}$ is the set of positive integers. | 2. All possible values of $f(2007)$ are $1,2, \cdots$, 2008.
For any positive integers $m, n$, if $m>n$, then
$$
f(m)=f(n+(m-n)) \geqslant f(n)+f(f(m-n))-1 \geqslant f(n).
$$
Therefore, $f$ is non-decreasing.
For any positive integer $n, f(n) \equiv 1$, clearly satisfies the condition.
Assume $f(n) \neq \equiv 1$, th... | 1,2, \cdots, 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,209 |
3. Given $n \in \mathbf{N}_{+}, x, y$ are positive real numbers, and satisfy $x^{n}+y^{n}=1$. Prove:
$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)} .
$$ | 3. For any real number $t \in(0,1)$, we have
$$
\frac{1+t^{2}}{1+t^{4}}=\frac{1}{t}-\frac{(1-t)\left(1-t^{3}\right)}{t\left(1+t^{4}\right)}<\frac{1}{t} \text {. }
$$
Let $t=x^{k}$ and $t=y^{k}$, then
$$
\begin{array}{l}
0<\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}<\sum_{k=1}^{n} \frac{1}{x^{k}} \\
=\frac{1-x^{n}}{x^{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,210 |
4. Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that for all $x, y \in \mathbf{R}_{+}$,
$$
f(x+f(y))=f(x+y)+f(y),
$$
where $\mathbf{R}_{+}$ is the set of positive real numbers. | 4. $f(x)=2 x$.
First, we prove: for all $y \in \mathbf{R}_{+}$, we have $f(y) \rightarrow y$.
In fact, from the condition, we get
$$
f(x+f(y))>f(x+y) \text {. }
$$
Therefore, $f(y) \neq y$.
If there exists a positive real number $y$ such that $f(y) < y$,
$$
\begin{array}{l}
f(y) < y,
\end{array}
$$
which is a contra... | f(x)=2x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,211 |
5. Let $c>2$, and a sequence of non-negative real numbers $a(1), a(2), \cdots$ satisfies:
(1) For any positive integers $m, n$,
$$
a(m+n) \leqslant 2 a(m)+2 a(n) \text {; }
$$
(2) For any non-negative integer $k$,
$$
a\left(2^{k}\right) \leqslant \frac{1}{(k+1)^{c}} \text {. }
$$
Prove: The sequence $\{a(n)\}$ is boun... | 5. Define $a(0)=0$. Then condition (1) holds for all non-negative integers $m, n$.
Lemma For any non-negative integers $n_{1}, n_{2}, \cdots, n_{k}$, we have:
(i) $a\left(\sum_{i=1}^{k} n_{i}\right) \leqslant \sum_{i=1}^{k} 2^{i} a\left(n_{i}\right)$;
(ii) $a\left(\sum_{i=1}^{k} n_{i}\right) \leqslant 2 k \sum_{i=1}^{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,212 |
6. Given non-negative real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}=1$. Prove:
$$
a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\cdots+a_{100}^{2} a_{1}<\frac{12}{25} .
$$ | 6. Let $S=\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$, where $a_{101} = a_{1}$ and $a_{102} = a_{2}$. By the Cauchy-Schwarz inequality and the AM-GM inequality, we have
$$
\begin{array}{l}
(3 S)^{2}=\left[\sum_{k=1}^{100} a_{k+1}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)\right]^{2} \\
\leqslant\left(\sum_{k=1}^{100} a_{k+1}^{2}\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,213 |
1. From the set of fractions $\left\{\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \frac{1}{12}\right\}$, remove two fractions so that the sum of the remaining numbers is 1. The two numbers to be removed are ( ).
(A) $\frac{1}{4}$ and $\frac{1}{8}$
(B) $\frac{1}{4}$ and $\frac{1}{10}$
(C) $\frac{1}{... | $-1 . C$
From $\frac{1}{4}+\frac{1}{12}=\frac{1}{3}$, and $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, thus by removing $\frac{1}{8}$ and $\frac{1}{10}$, the sum of the remaining numbers can be 1. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,214 |
2. Simplify $\frac{\sqrt[3]{2+\sqrt{5}}}{1+\sqrt{5}}$ to get ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{5}}{4}$
(C) $\frac{3}{8}$
(D) $\frac{1+\sqrt{5}}{7}$ | 2. A.
$$
\begin{array}{l}
\sqrt[3]{2+\sqrt{5}}=\sqrt[3]{\frac{8(2+\sqrt{5})}{8}}=\frac{1}{2} \sqrt[3]{16+8 \sqrt{5}} \\
\quad=\frac{1}{2} \sqrt[3]{(1+\sqrt{5})^{3}}=\frac{1+\sqrt{5}}{2} \\
\Rightarrow \frac{\sqrt[3]{2+\sqrt{5}}}{1+\sqrt{5}}=\frac{1}{2} .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,215 |
$3.5^{55}$ The last three digits are ( ).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. A.
Notice that $5^{55}=5 \times 5^{54}$.
Since $5^{2}$ leaves a remainder of 1 when divided by 8, $5^{54}$ also leaves a remainder of 1 when divided by 8. Therefore, $5^{55}$ leaves a remainder of 5 when divided by 8. Among the four numbers 125, 375, 625, and 875, only 125 leaves a remainder of 5 when divided by 8. | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,216 |
4. If the real numbers $x, y, z$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
\frac{x y}{x+2 y}=1, \\
\frac{y z}{y+2 z}=2, \\
\frac{z x}{z+2 x}=3,
\end{array}\right.
$$
then ( ).
(A) $x+2 y+3 z=0$
(B) $7 x+5 y+2 z=0$
(C) $9 x+6 y+3 z=0$
(D) $10 x+7 y+z=0$ | 4. D.
From equations (1) and (3), we get $y=\frac{x}{x-2}, z=\frac{6 x}{x-3}$, hence $x \neq 0$.
Substituting into equation (2), we solve to get $x=\frac{27}{10}$.
Therefore, $y=\frac{27}{7}, z=-54$.
Verification shows that this set of solutions satisfies the original system of equations.
Thus, $10 x+7 y+z=0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,217 |
5. Divide each side of an equilateral triangle into four equal parts, and then draw lines through these points parallel to the other two sides. Then the number of rhombuses formed by the line segments in Figure 1 is ( ).
(A) 15
(B) 18
(C) 21
(D) 24 | 5.C.
In Figure 1, there are only two types of rhombi with side lengths of 1 or 2. Each rhombus has exactly one diagonal equal to its side length. Inside the original equilateral triangle, each line segment of length 1 is exactly the diagonal of a rhombus with side length 1; there are 18 such line segments, correspondi... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,218 |
Example 5 As shown in Figure 4, place 3 identical coins sequentially into a $4 \times 4$ square grid (each square can only hold 1 coin). Find the probability that any two of the 3 coins are neither in the same row nor in the same column. | Explanation: (1) Calculate the total number of ways to place the coins $n$: The first coin can be placed in 16 different cells, which gives 16 ways; the second coin can be placed in the remaining 15 cells, which gives 15 ways; the third coin can be placed in the remaining 14 cells, which gives 14 ways.
So, the total nu... | \frac{6}{35} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,219 |
6. Someone viewed 2008 as a number-filling game: $2 \square \square 8$. Thus, he filled each box with a two-digit number $\overline{a b}$ and $\overline{c d}$, and found that the resulting six-digit number $\overline{2 a b c d 8}$ is exactly a perfect cube. Then $\overline{a b}+\overline{c d}=$ (. ).
(A) 40
(B) 50
(C) ... | 6.D.
Let $\overline{2 a b c d 8}=(\overline{x y})^{3}$.
According to the characteristic of the last digit, we get $y=2$, and then determine $\overline{x y}$.
Since $60^{3}=216000, 70^{3}=343000$, it follows that $60<\overline{x y}<70$.
Therefore, $\overline{x y}=62$ only. And $62^{3}=238328$, so $\overline{a b}=38, \o... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,220 |
1. Let $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+4}\right)=9$. Then $x \sqrt{y^{2}+4}+y \sqrt{x^{2}+1}=$ $\qquad$ . | II. $1 . \frac{77}{18}$.
According to the given equation, we have
$$
\begin{array}{l}
x y+y \sqrt{x^{2}+1}+x \sqrt{y^{2}+4}+ \\
\sqrt{\left(x^{2}+1\right)\left(y^{2}+4\right)}=9 . \\
\text { Let } x \sqrt{y^{2}+4}+y \sqrt{x^{2}+1}=z .
\end{array}
$$
Let $x \sqrt{y^{2}+4}+y \sqrt{x^{2}+1}=z$.
Then equation (1) becomes
... | \frac{77}{18} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,221 |
3. A book has a total of 61 pages, sequentially numbered as 1, 2, ..., 61. Someone, while adding these numbers, mistakenly reversed the digits of two two-digit page numbers (a two-digit number of the form $\overline{a b}$ was treated as $\overline{b a}$), resulting in a total sum of 2008. Therefore, the maximum sum of ... | 3.68.
Notice that $1+2+\cdots+61=1891$,
$$
2008-1891=117 \text {. }
$$
Since the page numbers in the form of $\overline{a b}$ are read as $\overline{b a}$, the difference in the sum will be $9|a-b|$, because $a$ and $b$ can only take values from 1,2, $\cdots, 9$, $|a-b| \leqslant 8$, so,
$9|a-b| \leqslant 72$.
Since ... | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,223 |
4. The largest integer not exceeding $(\sqrt{5}+\sqrt{3})^{6}$ is
$\qquad$ | 4.3903.
Notice that $(\sqrt{5}+\sqrt{3})^{6}=(8+2 \sqrt{15})^{3}$.
Let $8+2 \sqrt{15}=a, 8-2 \sqrt{15}=b$. Then we have
$$
a+b=16, ab=4 \text{. }
$$
We know that $a$ and $b$ are the roots of the equation $x^{2}-16 x+4=0$, so we have
$$
\begin{array}{l}
a^{2}=16 a-4, b^{2}=16 b-4 ; \\
a^{3}=16 a^{2}-4 a, b^{3}=16 b^{2... | 3903 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,224 |
One. (20 points) Let $a$ be an integer such that the equation $a x^{2}-(a+5) x+a+7=0$ has at least one rational root. Find all possible rational roots of the equation.
| When $a=0$, the rational root of the equation is $x=\frac{7}{5}$.
The following considers the case where $a \neq 0$. In this case, the original equation is a quadratic equation, and by the discriminant
$$
(a+5)^{2}-4 a(a+7) \geqslant 0,
$$
which simplifies to $3 a^{2}+18 a-25 \leqslant 0$.
Solving this, we get $-\frac... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,225 |
II. (25 points) As shown in Figure 3, in quadrilateral $ABCD$, $E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively, $P$ is any point on the extension of diagonal $AC$, $PF$ intersects $AD$ at point $M$, $PE$ intersects $BC$ at point $N$, and $EF$ intersects $MN$ at point $K$. Prove that $K$ is the midpoin... | Proof 1: As shown in Figure 3, $EF$ intersects $\triangle PMN$, then
\[
\frac{NK}{KM} \cdot \frac{MF}{FP} \cdot \frac{PE}{EN} = 1.
\]
(1)
$BC$ intersects $\triangle PAE$, then
\[
\frac{EB}{BA} \cdot \frac{AC}{CP} \cdot \frac{PN}{NE} = 1.
\]
Thus, $\frac{PN}{NE} = \frac{2CP}{AC}$.
Therefore, $\frac{PE}{EN} = \frac{2CP +... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,226 |
Three, (25 points) 120 people participated in a math competition, with the test consisting of 5 major questions. It is known that questions $1, 2, 3, 4, 5$ were correctly solved by 96, $83, 74, 66, 35$ people respectively. If at least 3 questions must be answered correctly to win an award, how many people won awards at... | Three, number these 120 people as $P_{1}, P_{2}, \cdots, P_{120}$, and consider them as 120 points on a number line. Let $A_{k}(k=1, 2,3,4,5)$ represent the group of people among these 120 who did not answer the $k$-th question correctly, and $\left|A_{k}\right|$ be the number of people in that group. Then,
$\left|A_{1... | 40 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,227 |
1. To obtain the graph of the function $y=\sin \left(2 x-\frac{\pi}{6}\right)$, the graph of the function $y=\cos 2 x$ can be ( ).
(A) shifted to the right by $\frac{\pi}{6}$ units
(B) shifted to the right by $\frac{\pi}{3}$ units
(C) shifted to the left by $\frac{\pi}{6}$ units
(D) shifted to the left by $\frac{\pi}{3... | $$
\begin{array}{l}
\text {-1. B. } \\
y=\sin \left(2 x-\frac{\pi}{6}\right)=\cos \left[\frac{\pi}{2}-\left(2 x-\frac{\pi}{6}\right)\right] \\
=\cos \left(\frac{2 \pi}{3}-2 x\right)=\cos 2\left(x-\frac{\pi}{3}\right)
\end{array}
$$
It is easy to know that the graph of $y=\cos 2 x$ shifts to the right by $\frac{\pi}{3}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,228 |
2. The mascots of the 2008 Beijing Olympic Games in China consist of 5 "Fuwa", named Beibei, Jingjing, Huanhuan, Yingying, and Nini. There are two sets of Fuwa of different sizes (a total of 10 Fuwa). From the two sets of Fuwa, any 5 Fuwa are selected, and the number of ways to select them such that exactly one is miss... | 2.A.
Since $\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1} \times 2 \times 2 \times 2=160$, the answer is $(\mathrm{A})$. | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,229 |
Example 6 A square ground is paved with regular hexagonal tiles with a side length of $36 \mathrm{~cm}$. Now, a circular disc with a radius of $6 \sqrt{3} \mathrm{~cm}$ is thrown upwards. The probability that the disc, after landing, does not intersect with the gaps between the tiles is approximately $\qquad$ . | Explanation: As shown in Figure 5, to ensure that the circular disc does not overlap with the gaps between the tiles, the center of the circular disc must fall within a smaller regular hexagon that is concentric with the tiles, with sides parallel to the tile edges and a distance of \(6 \sqrt{3} \mathrm{~cm}\) from the... | \frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,230 |
3. Given that $A, B, C$ are three non-collinear points on a plane, $O$ is the centroid of $\triangle A B C$, and a moving point $P$ satisfies
$$
O P=\frac{1}{3}\left(\frac{1}{2} O A+\frac{1}{2} O B+2 O C\right) .
$$
Then $P$ must be the ( ) of $\triangle A B C$.
(A) midpoint of the median from side $A B$
(B) one of th... | 3. B.
As shown in Figure 2, since $O$ is the centroid of $\triangle A B C$, we have:
$$
\begin{aligned}
& O A+O B+O C \\
= & 0 . \\
& \text { Therefore, } O P \\
= & \frac{1}{3}\left(\frac{1}{2} O A+\frac{1}{2} O B+2 O C\right) \\
= & \frac{1}{6}(O A+O B+4 O C) \\
= & \frac{1}{6}(-O C+4 O C)=\frac{1}{2} O C .
\end{ali... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,231 |
4. If there exists an obtuse angle $\alpha$, such that
$$
\sin \alpha-\sqrt{3} \cos \alpha=\log _{2}\left(x^{2}-x+2\right)
$$
holds, then the range of real number $x$ is ( ).
(A) $\{x \mid-1 \leqslant x<0$ or $1<x \leqslant 2\}$
(B) $\{x \mid-1<x<0$ or $1<x<2\}$
(C) $\{x \mid 0 \leqslant x \leqslant 1\}$
(D) $\{x \mid-... | 4.A.
Let $\frac{\pi}{2}<\alpha<\pi$. Then
$$
\sin \alpha-\sqrt{3} \cos \alpha=2 \sin \left(\alpha-\frac{\pi}{3}\right) \in(1,2] \text {. }
$$
Therefore, $1<\log _{2}\left(x^{2}+x+2\right) \leqslant 2$
Thus $2<x^{2}-x+2 \leqslant 4$.
Solving this, we get $-1 \leqslant x<0$ or $1<x \leqslant 2$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,232 |
For sets $M$ and $N$, define
$$
\begin{array}{l}
M-N=\{x \mid x \in M \text { and } x \notin N\}, \\
M \oplus N=(M-N) \cup(N-M) . \\
\text { Let } A=\left\{y \mid y=x^{2}-3 x, x \in \mathbf{R}\right\}, \\
B=\left\{y \mid y=-2^{x}, x \in \mathbf{R}\right\} .
\end{array}
$$
Then $A \oplus B=(\quad$.
(A) $\left(-\frac{9}... | 5.C.
Let $A=\left[-\frac{9}{4},+\infty\right), B=(-\infty, 0)$, then $A \oplus B=\left(-\infty,-\frac{9}{4}\right) \cup[0,+\infty)$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,233 |
6. Given the sequence $\left\{a_{n}\right\}$ with the general term $a_{n}=\left(\frac{2}{3}\right)^{n-1}\left[\left(\frac{2}{3}\right)^{n-1}-1\right]$. Which of the following statements is correct? (A) The maximum term is $a_{1}$, the minimum term is $a_{4}$ (B) The maximum term is $a_{1}$, the minimum term does not ex... | 6.D.
$$
a_{1}=0, a_{2}=-\frac{2}{9}, a_{3}=-\frac{20}{81} \text {. }
$$
When $n>3$, $t=\left(\frac{2}{3}\right)^{n-1}$ is a decreasing function, and
$$
0>a_{n}>a_{n-1}>\cdots>a_{4}>a_{3}$.
Therefore, $a_{1}$ is the largest, and $a_{3}$ is the smallest. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,234 |
7. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+b_{2} x^{2}+\cdots+b_{n} x^{n},
\end{array}
$$
and it satisfies $b_{1}+b_{2}+\cdots+b_{n}=26$. Then a possible value of the positive integer $n$ is $\qquad$ | Take $x=0$, we get $b_{0}=n$.
Take $x=1$, we get
$$
2+2^{2}+2^{3}+\cdots+2^{n}=b_{0}+b_{1}+b_{2}+\cdots+b_{n} \text {, }
$$
i.e., $\frac{2\left(1-2^{n}\right)}{1-2}=n+26$.
Thus, $2\left(2^{n}-1\right)=n+26$.
Therefore, $n=4$ is a possible value. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,235 |
8. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a unit cube, two ants, one black and one white, start from point $A$ and crawl along the edges. Each time they complete an edge, it is called “completing a segment”. The white ant’s crawling route is $A A_{1} \rightarrow A_{1} D_{1} \rightarrow \cdots$, and the black a... | 8. $\sqrt{2}$.
It is easy to know that the routes taken by the two ants have a cycle of 6. After the black and white ants have completed 2008 segments, they stop at vertices $D_{1}$ and $C$ of the cube, respectively, with a distance of $D_{1} C=\sqrt{2}$. | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,236 |
9. Given that the function $f(x)$ is defined on $\mathbf{R}$, and satisfies:
(1) $f(x)$ is an even function;
(2) For any $x \in \mathbf{R}$, $f(x+4) = f(x)$, and when $x \in [0,2]$, $f(x) = x + 2$.
Then the distance between the two closest points of intersection between the line $y=4$ and the graph of the function $f(... | 9.4.
Draw the graph of the function $f(x)$ (as shown in Figure 3).
From Figure 3, it is easy to see that the distance between the two closest intersection points of the line $y=4$ and the graph of the function $f(x)$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,237 |
10. There are six thin sticks, among which the lengths of the two longer ones are $\sqrt{3} a$ and $\sqrt{2} a$, and the lengths of the other four are all $a$. Using them to form a triangular pyramid. Then the cosine value of the angle between the two longer edges is $\qquad$ | $10 \cdot \frac{\sqrt{6}}{3}$.
As shown in Figure 4.
At first glance, it seems that two diagrams, Figure 4(a) and Figure 4(b), can be drawn, but if we reconsider the existence of the two diagrams, we immediately find that the figure is actually a trap, and Figure 4(a) does not exist at all. Because by taking the midpoi... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,238 |
$$
\begin{array}{l}
\text { 11. Let } f(x)=x^{2}+a x, \\
\{x \mid f(x)=0, x \in \mathbf{R}\} \\
=\{x \mid f(f(x))=0, x \in \mathbf{R}\} \neq \varnothing .
\end{array}
$$
Then the range of all real numbers $a$ that satisfy the condition is | 11. $\{a \mid 0 \leqslant a<4\}$.
Since the roots of $f(x)=0$ are $x=0$ or $x=-a$, we can transform $f(f(x))=0$ into
$$
f(x)=0 \text { or } f(x)+a=0 \text {. }
$$
From the problem, $f(x)+a=0$ has no solution or the roots of $f(x)+a=0$ are $x=0$ or $x=-a$.
And $f(x)+a=0$, which is $x^{2}+a x+a=0$.
Thus, $\Delta=a^{2}-... | 0 \leqslant a<4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,239 |
12. If sets $A_{1}$ and $A_{2}$ satisfy $A_{1} \cup A_{2}=A$, then we denote $\left[A_{1}, A_{2}\right]$ as a pair of subset partitions of $A$. It is stipulated that: $\left[A_{1}, A_{2}\right]$ and $\left[A_{2}, A_{1}\right]$ are considered the same pair of subset partitions of $A$. Given the set $A=\{1,2,3\}$. Then, ... | 12. 14.
Divide into three categories.
First category: When at least one of $A_{1}$ and $A_{2}$ is $A$ (let's assume $A_{1}=A$), then $A_{2}$ can be any subset of $A$, for a total of 8 different subset decompositions.
Second category: $A_{1}$ and $A_{2}$ are one-element and two-element subsets, respectively, with the ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,240 |
Example 7: From 5 cards each written with a number $1,2,3,4,5$, any two cards are taken out. The number on the first card is used as the tens digit, and the number on the second card is used as the units digit to form a two-digit number. Then the probability that the number formed is a multiple of 3 is ( ).
(A) $\frac{... | Explanation: The two-digit numbers that can be formed are
$$
\begin{array}{l}
12,13,14,15,21,23,24,25,31,32,34, \\
35,41,42,43,45,51,52,53,54,
\end{array}
$$
a total of 20. Among them, the numbers that are multiples of 3 are $12,15,21$, $24,42,45,51,54$, a total of 8.
Therefore, the probability that a number formed is... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,241 |
13. As shown in Figure 1, the basic unit of a certain building can be approximately constructed as follows: First, construct a rhombus $ABCD$ on the ground plane $\alpha$, with side length $1$ and $\angle BAD=60^{\circ}$. Then, above $\alpha$, install two identical regular pyramids $P-ABD$ and $O-CBD$ with $\triangle A... | Three, 13. (1) As shown in Figure 5, draw $P O_{1} \perp$ plane $A B C D$ at point $O_{1}$, and $Q O_{2} \perp$ plane $A B C D$ at point $O_{2}$.
Since $P-A B D$ and $Q-C B D$ are both regular tetrahedrons, $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ are the centers of the equilateral triangles $\triangle A B D$ and $\triang... | \frac{1}{3}, \frac{\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,242 |
14. Given a line segment $C D$ of length 6 with midpoint $M$. Two triangles $\triangle A C D$ and $\triangle B C D$ are constructed on the same side with $C D$ as one side, and both have a perimeter of 16, satisfying $\angle A M B = 90^{\circ}$. Find the minimum area of $\triangle A M B$. | 14. As shown in Figure 7, points $A$ and $B$ move on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ (with $C$ and $D$ being the left and right foci, respectively).
Let $l_{M A}: y=k x$. Then,
$$
\left\{\begin{array}{l}
y = k x, \\
\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1
\end{array} \Rightarrow \left\{\begin{array}{l... | \frac{400}{41} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,243 |
15. Let the function $f(x)=x^{2}+a x+b(a, b$ be real constants). It is known that the inequality $|f(x)| \leqslant 12 x^{2}+4 x-30|$ holds for any real number $x$. Define the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ as:
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, 2 a_{n}=f\left(a_{n-1}\right)+15(n=2,3, \c... | 15. (1) Let the two real roots of the equation $2 x^{2}+4 x-30=0$ be $\alpha, \beta$. Then $\alpha+\beta=-2, \alpha \beta=-15$.
Taking $x=\alpha$ in $|f(x)| \leqslant\left|2 x^{2}+4 x-30\right|$, we get $|f(x)| \leqslant 0$, hence $f(\alpha)=0$.
Similarly, $f(\beta)=0$.
Therefore, $f(x)=(x-\alpha)(x-\beta)$
$$
=x^{2}-... | 2 | Algebra | proof | Yes | Yes | cn_contest | false | 720,244 |
16. In four-dimensional space, the distance between point $A\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ and point $B\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ is defined as
$$
A B=\sqrt{\sum_{i=1}^{4}\left(a_{i}-b_{i}\right)^{2}} .
$$
Consider the set of points
$I=\left\{P\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \mid c_{i}=... | 16. Construct the following 8 points:
$$
\begin{array}{l}
P_{1}(0,0,0,0), P_{2}(0,1,0,0), P_{3}(0,0,0,1), \\
P_{4}(0,0,1,1), P_{5}(1,1,0,0), P_{6}(1,1,1,0), \\
P_{7}(1,1,1,1), P_{8}(1,0,1,1).
\end{array}
$$
By calculation, it is known that no three of these points can form an equilateral triangle, so, $n_{\min } \geqs... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,245 |
17. Given positive numbers $a, b, c$ satisfying $2a + 4b + 7c \leqslant 2abc$.
Find the minimum value of $a + b + c$. | 17. When $a=3, b=\frac{5}{2}, c=2$, $a+b+c$ takes the minimum value $\frac{15}{2}$.
When $a=3, b=\frac{5}{2}, c=2$, $a+b+c=3+$ $\frac{5}{2}+2=\frac{15}{2}$.
The following is the proof: $\frac{15}{2}$ is the required minimum value.
Let $a=3 x, b=\frac{5}{2} y, c=2 z$.
From the given, we have
$$
\begin{array}{l}
2 \time... | \frac{15}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,246 |
1. If the real numbers $m, n, x, y$ satisfy
$$
m^{2}+n^{2}=a, x^{2}+y^{2}=b \text {, }
$$
where $a, b$ are constants, then the maximum value of $m x+n y$ is $(\quad)$.
(A) $\frac{a+b}{2}$
(B) $\sqrt{a b}$
(C) $\sqrt{\frac{a^{2}+b^{2}}{2}}$
(D) $\frac{\sqrt{a^{2}-b^{2}}}{2}$ | -、1.B.
Solution 1: By Cauchy-Schwarz inequality,
$$
\begin{array}{l}
(m x+n y)^{2} \leqslant\left(m^{2}+n^{2}\right)\left(x^{2}+y^{2}\right)=a b \\
\Rightarrow m x+n y \leqslant \sqrt{a b} .
\end{array}
$$
Equality holds if and only if $m=n=\sqrt{\frac{a}{2}}$, i.e., $x=y=\sqrt{\frac{b}{2}}$,
$$
m x+n y=\sqrt{a b} \te... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,247 |
2. Let $y=f(x)$ be the exponential function $y=a^{x}$. Among the four points $P(1,1)$, $Q(1,2)$, $M(2,3)$, $N\left(\frac{1}{2}, \frac{1}{4}\right)$, the common points of the graph of the function $y=f(x)$ and its inverse function $y=f^{-1}(x)$ can only be point ( ).
(A) $P$
(B) $Q$
(C) $M$
(D) $N$ | 2.D.
Since the inverse function of the exponential function $y=a^{x}(a>0, a \neq 1)$ is the logarithmic function $y=\log _{a} x$, and the logarithmic function passes through the point $(1,0)$, it cannot pass through points $P$ and $Q$.
If it passes through point $M$, then
$3=a^{2} \Rightarrow a=\sqrt{3}$.
But $y=\log ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,248 |
3. In the table shown in Figure 1, if each cell is filled with a number such that each row forms an arithmetic sequence and each column forms a geometric sequence, then the value of $x + y + z$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. A.
The last two numbers in the first and second rows are $2.5, 3$ and $1.25, 1.5$; in the third, fourth, and fifth columns, $x=0.5, y=\frac{5}{16}, z=\frac{3}{16}$. Thus, $x+y+z=1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,249 |
4. If the cosines of the three interior angles of $\triangle A_{1} B_{1} C_{1}$ are the sines of the three interior angles of $\triangle A_{2} B_{2} C_{2}$, then, ( . . .
(A) $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ are both acute triangles
(B) $\triangle A_{1} B_{1} C_{1}$ is an acute triangle, ... | 4.B.
Since the interior angles of the two triangles cannot be right angles, and the cosine values of the interior angles of $\triangle A_{1} B_{1} C_{1}$ are all greater than zero, $\triangle A_{1} B_{1} C_{1}$ is an acute triangle.
If $\triangle A_{2} B_{2} C_{2}$ is an acute triangle, without loss of generality, we ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,250 |
5. Let $a$ and $b$ be skew lines with an angle of $30^{\circ}$ between them. Then the planes $\alpha$ and $\beta$ that satisfy the condition “$a \subseteq \alpha, b \subseteq \beta$, and $\alpha \perp \beta$” ( ).
(A) do not exist
(B) exist and are unique
(C) exist and there are exactly two pairs
(D) exist and there ar... | 5.D.
Any plane $\alpha$ can be made through $a$, and there can be infinitely many.
Take any point $M$ on $b$, and draw a perpendicular line to plane $\alpha$ through $M$. The plane $\beta$ determined by $b$ and the perpendicular line is perpendicular to $\alpha$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,251 |
Example 8 A math game is played between two very smart students, A and B. The referee first writes the integers $2,3, \cdots, 2006$ on the blackboard, then randomly erases one number. Next, A and B take turns erasing one number (i.e., B erases one of the numbers first, then A erases one number, and so on). If the last ... | Explanation: The key to winning is to see which number the referee erases. Note that $2,3, \cdots, 2006$, contains 1002 odd numbers and 1003 even numbers.
(1) If the referee erases an odd number, at this point, player B will definitely win.
No matter what number player A takes, as long as there are still odd numbers, ... | \frac{1003}{2005} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,252 |
6. Let set $A=\left\{x \mid x^{2}-[x]=2\right\}$, $B=\{x|| x \mid<2\}$,
where, $[x]$ denotes the greatest integer less than or equal to $x$.
Then $A \cap B=$ $\qquad$ | $Two \mathbf{6} . A \cap B=\{-1, \sqrt{3}\}$.
Since $|x|<2$, therefore, the value of $[x]$ can be -2, $-1,0,1$.
When $[x]=-2$, $x^{2}=0$, no solution;
When $[x]=-1$, $x^{2}=1 \Rightarrow x=-1$;
When $[x]=0$, $x^{2}=2$ no solution;
When $[x]=1$, $x^{2}=3 \Rightarrow x=\sqrt{3}$.
Therefore, $x=-1$ or $\sqrt{3}$. | A \cap B=\{-1, \sqrt{3}\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,253 |
7. When rolling three dice simultaneously, the probability of at least one die showing a 6 is $\qquad$ (the result should be written as a reduced fraction). | $7 \cdot \frac{91}{216}$.
Considering the complementary event, $P=1-\left(\frac{5}{6}\right)^{3}=\frac{91}{216}$. | \frac{91}{216} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,254 |
8. Given point $O$ inside $\triangle A B C$, $O A+2 O B+$ $2 O C=0$. Then the ratio of the area of $\triangle A B C$ to the area of $\triangle O C B$ is $\qquad$ | 8.5:1.
As shown in Figure 3, note that the base of $\triangle A B C$ and $\triangle O C B$ is the same, and the ratio of their heights is $5: 1$. Therefore, the ratio of their areas is $5: 1$. | 5:1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,255 |
11. (15 points) Given the function $f(x)=-2 x^{2}+b x+c$ has a maximum value of 1 at $x=1$, and $0<m<n$. When $x \in [m, n]$, the range of $f(x)$ is $\left[\frac{1}{n}, \frac{1}{m}\right]$. Find the values of $m$ and $n$. | Three, 11. From the problem, we have
$$
f(x)=-2(x-1)^{2}+1 \text {. }
$$
Then $f(x) \leqslant 1 \Rightarrow \frac{1}{m} \leqslant 1 \Rightarrow m \geqslant 1$.
Thus, $f(x)$ is monotonically decreasing on $[m, n]$. Therefore,
$$
f(m)=-2(m-1)^{2}+1=\frac{1}{m} \text {, }
$$
and $f(n)=-2(n-1)^{2}+1=\frac{1}{n}$.
Hence, ... | m=1, n=\frac{1+\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,258 |
12. (15 points) $A, B$ are two moving points on the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$, satisfying $O A \cdot O B=0$. Prove:
(1) $\frac{1}{|\boldsymbol{O A}|^{2}}+\frac{1}{|\boldsymbol{O B}|^{2}}$ is a constant;
(2) A moving point $P$ lies on the line segment $A B$, satisfying $\boldsymbol{O P} \cdot \boldsy... | 12. (1) Let point $A(r \cos \theta, r \sin \theta)$, $B\left(r^{\prime} \cos \theta^{\prime}, r^{\prime} \sin \theta^{\prime}\right)$.
Then $r=|O A|, r^{\prime}=|O B|$.
Point $A$ is on the hyperbola, so $r^{2}\left(\frac{\cos ^{2} \theta}{4}-\frac{\sin ^{2} \theta}{4}\right)=1$.
Therefore, $\frac{1}{r^{2}}=\frac{\cos ^... | \frac{6 \sqrt{5}}{5} | Geometry | proof | Yes | Yes | cn_contest | false | 720,259 |
13. (20 points) As shown in Figure 2, planes $M$ and $N$ intersect at line $l$. Points $A$ and $D$ are on line $l$, ray $DB$ is in plane $M$, and ray $DC$ is in plane $N$. Given that $\angle BDC = \alpha$, $\angle BDA = \beta$, $\angle CDA = \gamma$, and $\alpha$, $\beta$, $\gamma$ are all acute angles. Find the cosine... | 13. In planes $M$ and $N$, draw a perpendicular from point $A$ to $DA$, intersecting rays $DB$ and $DC$ at points $B$ and $C$ respectively.
Given $DA=1$. Then
\[
\begin{array}{l}
AB=\tan \beta, DB=\frac{1}{\cos \beta}, \\
AC=\tan \gamma, DC=\frac{1}{\cos \gamma},
\end{array}
\]
and $\angle BAC=\varphi$ is the plane an... | \frac{\cos \alpha - \cos \beta \cdot \cos \gamma}{\sin \beta \cdot \sin \gamma} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,260 |
14. (20 points) Can the numbers in the following arrays be filled into a $3 \times 3$ grid, with each small square containing one number, such that the product of the 3 numbers in each row, each column, and the two diagonals are all equal? If so, provide one way to fill the grid; if not, provide a proof.
(1) $2,4,6,8,1... | Because if the product of each row is equal, then the product of the 9 numbers is a cube number. But
$$
\begin{array}{l}
2 \times 4 \times 6 \times 8 \times 12 \times 18 \times 24 \times 36 \times 48 \\
=2^{1+2+1+3+2+1+3+2+4} \times 3^{1+1+2+1+2+1} \\
=2^{19} \times 3^{8}
\end{array}
$$
is not a cube number, hence it ... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,261 |
1. Given $A=(2-x)-\sqrt{2}(\sqrt{2} x-5)$. The natural number $x$ that makes $A$ positive has ( ).
(A) 1
(B) 2
(C) more than 2 but a finite number
(D) an infinite number | $-1 . C$.
Let $A=(2-x)-(\sqrt{2} x-5) \sqrt{2}>0$.
Solving, we get $x<\frac{2+5 \sqrt{2}}{3}$.
Also, $3<\frac{2+5 \sqrt{2}}{3}<4$, so the natural numbers $x=0,1,2,3$ satisfy the condition, for a total of 4. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,262 |
Example 1 As shown in Figure 1, on the extension of the diameter $AB$ of a semicircle, take a fixed point $L$, draw a line $l \perp AB$ through $L$, and draw any secant line through $L$ intersecting the semicircle at points $C$ and $D$. The lines $BC$ and $BD$ intersect $l$ at points $N$ and $M$ respectively. Prove tha... | Explanation: Take the radius of the semicircle $r$, $BL = a$, and the variable $\angle ALD = \theta$ as the basic quantities. It is easy to see that
$$
\begin{array}{l}
LC + LD = 2(a + r) \cos \theta, \\
LC \cdot LD = a(a + 2r),
\end{array}
$$
thus $LC$ and $LD$ can be determined (solved). Noting that
$$
\tan \angle L... | a^2 + 2ar | Geometry | proof | Yes | Yes | cn_contest | false | 720,263 |
2. Given that $a$ is an integer, $14 a^{2}-12 a-27 \mid$ is a prime number. Then the sum of all possible values of $a$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. D.
From the problem, we know
$$
\left|4 a^{2}-12 a-27\right|=|(2 a+3)(2 a-9)|
$$
is a prime number.
Therefore, $2 a+3= \pm 1$ or $2 a-9= \pm 1$, which means $a=-1,-2$ or $a=5,4$.
Thus, the sum of all possible values of $a$ is 6. | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,264 |
3. As shown in Figure 1, extend the sides $AB$ and $DC$ of quadrilateral $ABCD$ to meet at point $E$, and extend sides $BC$ and $AD$ to meet at point $F$. $AC$ intersects $EF$ at $X$, and $BD$ intersects $EF$ at $Y$. The relationship between $\frac{E X}{X F}$ and $\frac{E Y}{Y F}$ is ( ).
(A) $\frac{E X}{X F}>\frac{E Y... | 3. B.
Notice that $\frac{E X}{X F}=\frac{S_{\triangle A B C}}{S_{\triangle A F C}}=\frac{S_{\triangle E C}}{S_{\triangle E C F}} \cdot \frac{S_{\triangle B C F}}{S_{\triangle A F C}}$ $=\frac{A D}{D F} \cdot \frac{E B}{B A}=\frac{S_{\triangle A B D}}{S_{\triangle B B F}} \cdot \frac{S_{\triangle B D E}}{S_{\triangle A... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,265 |
4. Right trapezoid $A B C D$ is shown in Figure 2, and a moving point $P$ starts from $B$, moving along the sides of the trapezoid in the order $B \rightarrow C \rightarrow D \rightarrow A$. Let the distance traveled by point $P$ be $x$, and the area of $\triangle A B P$ be $f(x)$.
If the graph of the function $y=f(x)... | 4. B.
According to the image, we get $B C=4, C D=5, D A=5$. Therefore, $A B=8$.
Thus, $S_{\triangle A B C}=\frac{1}{2} \times 8 \times 4=16$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,266 |
5. Given $f(x)=1-(x-a)(x-b)$, and $m, n$ are the roots of the equation $f(x)=0$. Then the possible order of the real numbers $a, b, m, n$ is $(\quad)$.
(A) $m<a<b<n$
(B) $a<m<n<b$
(C) $a<m<b<n$
(D) $m<a<n<b$ | 5.A.
Notice that $f(a)=$ $f(b)=1$, the parabola $y=$ $-x^{2}+(a+b) x+1-$ $a b$ opens downwards, therefore, the graph can only be as shown in Figure 6 (assuming $m<$ $n$). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,267 |
6. The areas of two right trapezoids are $S_{1}$ and $S_{2}$, respectively. Their oblique sides are equal, and their acute angles are complementary. The bases of one trapezoid are $4$ and $8$, and the bases of the other trapezoid are $5$ and $13$. Then $S_{1}+S_{2}$ $=(\quad)$.
(A) 96
(B) 92
(C) 88
(D) 84 | 6. D.
As shown in Figure 7, two right trapezoids can be combined into a large rectangle minus a small rectangle, that is,
$$
S_{1}+S_{2}=8 \times 13-4 \times 5=84 .
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,268 |
1. Given $a+\frac{1}{a+1}=b+\frac{1}{b-1}-2$, and $a-$ $b+2 \neq 0$. Then the value of $a b-a+b$ is $\qquad$ . | $=、 1.2$.
Let $a+1=x$,
$$
b-1=y \text {. }
$$
Then $x-y$
$$
\begin{array}{l}
=a-b+2 \\
\neq 0 .
\end{array}
$$
Substitute into the given equation,
we get $x+\frac{1}{x}=y+\frac{1}{y}$,
which means $(x-y)(x y-1)=0$.
Since $x-y \neq 0$, it follows that $x y=1$.
Therefore, $(a+1)(b-1)=1$, which means $a b-a+b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,269 |
2. Given that when $0 \leqslant \theta \leqslant \frac{\pi}{2}$, it always holds that $\cos ^{2} \theta+2 m \sin \theta-2 m-2<0$. Then the range of real number $m$ is $\qquad$. | 2. $m>-\frac{1}{2}$.
Let $x=\sin \theta$. Then the original inequality becomes $x^{2}+1>2 m x-2 m(0 \leqslant x \leqslant 1)$. Let $y_{1}=x^{2}+1, y_{2}=2 m x-2 m$.
In the Cartesian coordinate system of Figure 8, draw the graphs of $y_{1}$ and $y_{2}$, where $y_{2}$ is a family of lines passing through the point $(1,... | m>-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,270 |
3. The integer solutions $(x, y)$ of the equation $2 x^{2}+x y+y^{2}-x+2 y+1=0$ are $\qquad$ _. | 3. $(0,-1),(1,-1),(1,-2)$.
The original equation can be transformed into
$$
2 x^{2}+(y-1) x+y^{2}+2 y+1=0 \text {. }
$$
Considering $\Delta=(y-1)^{2}-8(y+1)^{2} \geqslant 0$.
Solving yields $\frac{-9-4 \sqrt{2}}{7} \leqslant y \leqslant \frac{-9+4 \sqrt{2}}{7}$.
Thus, $y=-1,-2$.
Therefore, when $y=-1$, $x=0,1$;
when ... | (0,-1),(1,-1),(1,-2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,271 |
One, (20 points) Given real numbers $a, b, c, d$ are all distinct and satisfy
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{a}=x \text {. }
$$
Find the value of $x$. | $$
\begin{array}{l}
d=x-\frac{1}{a}=\frac{a x-1}{a}, \\
c=x-\frac{1}{d}=x-\frac{a}{a x-1}=\frac{a x^{2}-x-a}{a x-1}, \\
b=x-\frac{1}{c}=x-\frac{a x-1}{a x^{2}-x-a} \\
=\frac{a x^{3}-x^{2}-2 a x+1}{a x^{2}-x-a}, \\
a+\frac{a x^{2}-x-2 a}{a x^{3}-x^{2}-2 a x+1}=x .
\end{array}
$$
Therefore, $a x^{4}-a^{2} x^{3}-x^{3}-2 ... | x=\sqrt{2} \text{ or } -\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,273 |
Example 2 Given that $A$ is a fixed point inside $\odot O$, and the two sides of the moving right angle $\angle A$ intersect $\odot O$ at points $B$ and $C$. Prove: The product of the distances from points $A$ and $O$ to $BC$, $AX \cdot OY$, is a constant. | Explanation: As shown in Figure 2, take
the radius $r$ of $\odot O$, $OA = a$, and the variables $\angle OCB = \beta$, $\angle ACB = \alpha$ as the basic quantities. Then,
$$
\begin{array}{l}
AC = BC \cos \alpha \\
= 2r \cos \beta \cdot \cos \alpha,
\end{array}
$$
where $r$, $a$, $\beta$, and $\alpha$ satisfy the cond... | 2r^{2} \cdot \frac{1 - \frac{a^{2}}{r^{2}}}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 720,274 |
II. (25 points) As shown in Figure 5, the quadrilateral $ABCD$ is tangent to $\odot O$ at points $E, G, F, H$. Given that $EF \perp HG$. Prove: the area of quadrilateral $ABCD$
$$
S=\sqrt{AB \cdot BC \cdot CD \cdot DA} .
$$ | As shown in Figure 10, let the radius of $\odot O$ be $r$, $A E = A H = a$, $B E = B G = b$, $C G = C F = c$, $D F = D H = d$. Connect $O A$, $O C$, $O E$, and $O G$. Then $O E = O G = r$.
Notice that
$$
\begin{array}{l}
\angle B A D = \frac{1}{2}(\overparen{H F} + \overparen{F G} + \overparen{G E} - \overparen{E H}), ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,275 |
$$
\text { Three. (25 points) In } \triangle A B C \text {, } A B=8+2 \sqrt{6} \text {, }
$$
$B C=7+2 \sqrt{6}, C A=3+2 \sqrt{6}, O, I$ are the circumcenter and incenter of $\triangle A B C$, respectively. Is the length of segment $O I$ a rational number or an irrational number? Make a judgment and explain your reasoni... | Three, OI=2.5 is a rational number.
Let $a, b, c, p, S, R, r$ represent the three sides, the semi-perimeter, the area, the circumradius, and the inradius of $\triangle ABC$, respectively. Thus,
$$
\begin{array}{l}
p=\frac{1}{2}(a+b+c)=9+3 \sqrt{6}, \\
p-a=2+\sqrt{6}, p-b=6+\sqrt{6}, \\
p-c=1+\sqrt{6}.
\end{array}
$$
T... | 2.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,276 |
1. Given the following two propositions:
Proposition $P$ : There exist functions $f(x)$, $g(x)$, and an interval $I$, such that $f(x)$ is an increasing function on $I$, $g(x)$ is also an increasing function on $I$, but $f(g(x))$ is a decreasing function on $I$;
Proposition $Q$ : There exist an odd function $f(x)(x \i... | $-1 . A$
Take $f(x)=1-x^{2}, g(x)=\left(\frac{1}{2}\right)^{\frac{1}{x}}$, interval $I=(-\infty, 0)$.
Then $f(x)$ is an increasing function on the interval $(-\infty, 0)$, and $g(x)$ is also an increasing function on the interval $(-\infty, 0)$.
But $f(g(x))=1-\left(\frac{1}{2}\right)^{\frac{2}{x}}$ is a decreasing f... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,277 |
2. $\triangle A B C$ satisfies $\boldsymbol{A B} \cdot \boldsymbol{B C}=\boldsymbol{B} \boldsymbol{C} \cdot \boldsymbol{C A}$. Then $\triangle A B C$ is ( ).
(A) right triangle
(B) acute triangle
(C) obtuse triangle
(D) isosceles triangle | 2.D.
Let the midpoint of $B C$ be $D$. Then
$$
\begin{array}{l}
A \boldsymbol{B} \cdot \boldsymbol{B C}=\boldsymbol{B C} \cdot \boldsymbol{C A} \\
\Rightarrow(\boldsymbol{A B}-\boldsymbol{C A}) \cdot B C=0 \\
\Rightarrow(A B+A C) \cdot B C=0 \Rightarrow 2 A D \cdot B C=0 \\
\Rightarrow A D \perp B C \Rightarrow A B=A ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,278 |
3. Let the curve $f(x)=a \cos x+b \sin x$ have a line of symmetry at $x=\frac{\pi}{5}$. Then a point of symmetry for the curve $y=f\left(\frac{\pi}{10}-x\right)$ is ( ).
(A) $\left(\frac{\pi}{5}, 0\right)$
(B) $\left(\frac{2 \pi}{5}, 0\right)$
(C) $\left(\frac{3 \pi}{5}, 0\right)$
(D) $\left(\frac{4 \pi}{5}, 0\right)$ | 3. B.
Since $f(x)=a \cos x+b \sin x$
$$
=\sqrt{a^{2}+b^{2}} \sin (x+\theta)
$$
has a period of $2 \pi$, therefore, a point of symmetry for the curve
$$
f(x)=a \cos x+b \sin x
$$
is $\left(\frac{\pi}{5}+\frac{\pi}{2}, 0\right)$, which is $\left(\frac{7 \pi}{10}, 0\right)$.
Thus, a point of symmetry for the curve $y=f... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,279 |
4. Let the function $f(x)$ satisfy: for any real number $x \geqslant 0$, $f(2 x+1)=\sqrt{x}$. Then such a function $f(x)$ ( ).
(A) does not exist
(B) exists uniquely
(C) exists exactly two
(D) exists infinitely many
Translate the above text into English, please retain the original text's line breaks and format, and ou... | 4.D.
Let $M$ be any non-empty subset of the set $\{x \mid x<1\}$, and $t(x)$ be any function defined on $M$. Then the function
$$
f_{M}(x)=\left\{\begin{array}{ll}
\sqrt{\frac{x-1}{2}}, & x \geqslant 1 ; \\
t(x), & x \in M
\end{array}\right.
$$
all satisfy the condition. | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 720,280 |
6. There is a special train, with a total of 20 stations along the way (including the starting and ending stations). For safety reasons, it is stipulated that passengers who board at the same station cannot disembark at the same station. To ensure that all boarding passengers have seats (one seat per passenger), the tr... | 6. B.
Consider the number of seats required when the train departs from station $j$.
For a fixed $i(1 \leqslant i \leqslant j)$, among the passengers boarding at station $i$, at most $20-j$ people will still be on the train when it passes through station $j$. This is because the passengers boarding at the same statio... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,282 |
1. The solution set of the inequality $x^{3}+\left(1-x^{2}\right)^{\frac{3}{2}} \geqslant 1$ is | 1. $\{0,1\}$.
Let $\sqrt{1-x^{2}}=y$, then the inequality becomes
$$
x^{2}+y^{2}=1, x^{3}+y^{3} \geqslant 1 \text {. }
$$
Thus, $0 \leqslant x^{3}+y^{3}-x^{2}-y^{2}$
$$
\begin{array}{l}
=x^{2}(x-1)+y^{2}(y-1) \\
=\left(1-y^{2}\right)(x-1)+\left(1-x^{2}\right)(y-1) \\
=-\left(y^{2}-1\right)(x-1)-\left(x^{2}-1\right)(y-... | \{0,1\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,283 |
Example 1 Let $N=23x+92y$ be a perfect square, and $N$ does not exceed 2392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs. | Solution: Since $N=23(x+4 y)$, and 23 is a prime number, there exists a positive integer $k$, such that $x+4 y=23 k^{2}$.
Also, because $23(x+4 y)=N \leqslant 2392$, we have $x+4 y \leqslant 104$.
Therefore, $23 k^{2}=x+4 y \leqslant 104$, which means $k^{2} \leqslant 4$.
Thus, $k^{2}=1,4$.
When $k^{2}=1$, $x+4 y=23$, ... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,285 |
Example 2 Let the perfect square $y^{2}$ be the sum of the squares of 11 consecutive integers. Then the minimum value of $|y|$ is $\qquad$ . | Solution: Let the middle number of 11 consecutive integers be $x$. Then
$$
\begin{aligned}
y^{2}= & (x-5)^{2}+(x-4)^{2}+\cdots+x^{2}+\cdots+ \\
& (x+4)^{2}+(x+5)^{2} \\
= & x^{2}+2\left(x^{2}+1^{2}\right)+2\left(x^{2}+2^{2}\right)+\cdots+2\left(x^{2}+5^{2}\right) \\
= & 11 x^{2}+2\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\rig... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,286 |
Example 2 In the positive term sequence $\left\{a_{n}\right\}$,
$$
a_{1}=10, a_{k+1}=10 \sqrt{a_{k}}(k=1,2, \cdots) \text {. }
$$
Find the general term formula $a_{n}$. | Solution: Let $b_{k}=\lg a_{k}$. Then $b_{k+1}=\frac{1}{2} b_{k}+1$.
Let $x=\frac{1}{2} x+1$, we get $x=2$. Then $b_{k+1}-2=\frac{1}{2}\left(b_{k}-2\right)$.
Therefore, the sequence $\left\{b_{n}-2\right\}$ is a geometric sequence with the first term $b_{1}-2=-1$ and the common ratio $\frac{1}{2}$.
Thus, $b_{n}-2=-1 \t... | a_{n}=10^{2-\left(\frac{1}{2}\right)^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,287 |
2. Let the three sides of $\triangle A B C$ be $a, b, c$, and the radius of its circumcircle be $R$. If $R=\frac{a \sqrt{b c}}{b+c}$, then the largest angle of $\triangle A B C$ is ( ).
(A) $75^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $150^{\circ}$ | 2.C.
From the given, we know that $(b-c)^{2} \geqslant 0$, which means $b+c \geqslant 2 \sqrt{b c}$.
Therefore, $\frac{\sqrt{b c}}{b+c} \leqslant \frac{1}{2}$.
From the known condition, $\frac{R}{a}=\frac{\sqrt{b c}}{b+c} \leqslant \frac{1}{2}$.
Thus, $a \geqslant 2 R$.
Also, $a \leqslant 2 R$, so $a=2 R$.
Hence, $\fr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,288 |
3. Given the quadratic function $y=a x^{2}+b x+c(a, b, c$ are real numbers) whose graph intersects the $x$-axis at points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $a>0, a>2 c>b$, when $x=1$, $y=-\frac{a}{2}$. Then the range of $| x_{1}-x_{2} |$ is ( ).
(A) $(0,1)$
(B) $(1,2)$
(C) $\left(\frac{1}{2}, \... | 3. D.
Since when $x=1$, $y=-\frac{a}{2}$, we have
$$
3 a+2 b+2 c=0,
$$
which means $2 c=-3 a-2 b$.
From $a>2 c>b$, we know $a>-3 a-2 b>b$.
Since $a>0$, then $1>-3-2 \times \frac{b}{a}>\frac{b}{a}$, which means
$$
-2<\frac{b}{a}<-1 \text {. }
$$
Given that the graph of the quadratic function intersects the $x$-axis a... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,289 |
4. As shown in Figure 1, in trapezoid $A B C D$, $A B / / C D$, $A B=a, C D=b$. If $\angle A D C=\angle B F E$, and the area of quadrilateral $A B F E$ is equal to the area of quadrilateral $C D E F$, then the length of $E F$ is ( ).
(A) $\frac{a+b}{2}$
(B) $\sqrt{a b}$
(C) $\frac{2 a b}{a+b}$
(D) $\sqrt{\frac{a^{2}+b^... | 4.D.
As shown in Figure 3, extend $DA$ and $CB$ to intersect at point $P$.
From the given conditions, we have
$$
\begin{array}{l}
\angle PAB = \angle ADC \\
= \angle BFE.
\end{array}
$$
Therefore,
$\triangle PAB \backsim \triangle PFE$
$\backsim \triangle PDC$.
Let $EF = x, S_{\triangle PAB} = S_0, S_{\text{quadrilat... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,290 |
5. Let $\sqrt{x}+\sqrt{y}=\sqrt{2009}$, and $0<x<y$. Then the number of different integer pairs $(x, y)$ that satisfy this equation is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5.C.
From $2009=49 \times 41$, we get $\sqrt{2009}=7 \sqrt{41}$.
Also, $0<x<y$, so $\sqrt{2009}$ can be rewritten as
$$
\begin{array}{l}
\sqrt{2009}=\sqrt{41}+6 \sqrt{41}=2 \sqrt{41}+5 \sqrt{41} \\
=3 \sqrt{41}+4 \sqrt{41},
\end{array}
$$
which means
$$
\begin{array}{l}
\sqrt{2009}=\sqrt{41}+\sqrt{1476} \\
=\sqrt{164... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,291 |
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