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3. Given condition $p: a, b, c, d$ form a geometric sequence; condition $q: ad=bc$. Then condition $p$ is a (.) condition of condition $q$.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary | 3. A.
Sufficiency obviously holds, necessity does not hold.
For example, $a=1, b=2, c=5, d=10$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,797 |
5. Given $x, y \in \mathbf{R}_{+}$. Then the maximum value of $\frac{x}{2 x+y}+\frac{y}{x+2 y}$ is ( ).
| 5. B.
Let $s=2x+y, t=x+2y$. Then
$$
\begin{array}{l}
x=\frac{1}{3}(2s-t), \\
y=\frac{1}{3}(2t-s) .
\end{array}
$$
$$
\text{Thus } \frac{x}{2x+y}+\frac{y}{x+2y}=\frac{4}{3}-\frac{1}{3}\left(\frac{t}{s}+\frac{s}{t}\right) \leqslant \frac{2}{3} \text{. }
$$ | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,799 |
7. Six students consulted about their grades. The teacher said: A is not the best among the six, B is not the worst among the six, and the six students have different grades. Then the number of different possible rankings of their grades is ( ) .
(A) 120
(B) 216
(C) 384
(D) 504 | 7. D.
Let $A$ be the set of all possible rankings where A ranks first, and $B$ be the set of all possible rankings where B ranks last. Then
$$
|A|=|B|=5!,|A \cap B|=4!.
$$
The number of all possible rankings where A ranks first or B ranks last is
$$
|A \cup B|=|A|+|B|-|A \cap B|=216.
$$
According to the teacher, the... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,801 |
Example 8 Equation
$$
x^{3}-\sqrt{3} x^{2}-(2 \sqrt{3}+1) x+3+\sqrt{3}=0
$$
The three roots of the equation are $\qquad$ | Let $\sqrt{3}=a$.
Transform the main variable to get
$$
\begin{array}{l}
a^{2}-\left(x^{2}+2 x-1\right) a+x^{3}-x=0 \\
\Rightarrow[a-(x-1)]\left[a-\left(x^{2}+x\right)\right]=0 \\
\Rightarrow x=a+1=\sqrt{3}+1 \text { or } x^{2}+x-\sqrt{3}=0 .
\end{array}
$$
Therefore, the solutions to the original equation are
$$
\sqr... | \sqrt{3}+1, \frac{-1 \pm \sqrt{1+4 \sqrt{3}}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,802 |
8. If point $P$ is on the curve $y=-x^{2}-1$, and point $Q$ is on the curve $x=1+y^{2}$, then the minimum value of $|P Q|$ is ( ).
(A) $3 \sqrt{2}$
(B) $\frac{3 \sqrt{2}}{2}$
(C) $\frac{3 \sqrt{2}}{4}$
(D) $\frac{3 \sqrt{2}}{8}$ | 8. C.
Given that the two parabolas are symmetric about the line $y=-x$, the minimum value of $|P Q|$ is twice the minimum distance from point $P\left(x,-x^{2}-1\right)$ to the line $y=-x$. Therefore,
$$
\begin{array}{l}
d=\frac{\left|x-x^{2}-1\right|}{\sqrt{2}}=\frac{x^{2}-x+1}{\sqrt{2}}, \\
d_{\min }=\frac{3 \sqrt{2}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,803 |
9. Given the function
$$
f(x)=\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right) x^{2}+b x+6,
$$
where $a$ and $b$ are constants, $a>1$, and
$$
f\left(\lg \log _{8} 1000\right)=8.
$$
Then the value of $f(\lg \lg 2)$ is ( ).
(A) 8
(B) 4
(C) -4
(D) -8 | 9. B.
From the given information,
$$
\begin{array}{l}
f\left(\lg \log _{8} 1000\right)=f\left(\lg \log _{2} 10\right) \\
=f(-\lg \lg 2)=8 . \\
\text { Also, } \frac{1}{a^{-x}-1}+\frac{1}{2}=\frac{a^{x}}{1-a^{x}}+\frac{1}{2} \\
=-1+\frac{1}{1-a^{x}}+\frac{1}{2} \\
=-\frac{1}{a^{x}-1}-\frac{1}{2},
\end{array}
$$
Let $F... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,804 |
10. In the arithmetic sequence $\left\{a_{n}\right\}$, if $\frac{a_{11}}{a_{10}}<-1$, and the sum of the first $n$ terms $S_{n}$ has a maximum value. Then when $S_{n}$ takes the smallest positive value, $n=(\quad)$.
(A) 1
(B) 10
(C) 19
(D) 20 | 10. C.
Let the common difference of the arithmetic sequence be $d$. Clearly, $d0, a_{11}0$.
Also, $S_{19}-S_{1}=9\left(a_{10}+a_{11}\right)<0$, so $n=19$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,805 |
11. Given
$$
f(x)=\cos 2x + p|\cos x| + p \quad (x \in \mathbf{R}) \text{.}
$$
Let the maximum value of $f(x)$ be $h(p)$. Then the expression for $h(p)$ is $\qquad$ . | 11. $\left\{\begin{array}{ll}p-1, & p\frac{1}{2}\end{array}\right.$
Therefore, $h(\dot{p})=\left\{\begin{array}{ll}p-1, & p<-2 ; \\ 2 p+1, & p \geqslant-2 .\end{array}\right.$ | h(p)=\left\{\begin{array}{ll}p-1, & p<-2 ; \\ 2 p+1, & p \geqslant-2 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,806 |
12. Given
$$
\sin (x+\sin x)=\cos (x-\cos x),
$$
where, $x \in[0, \pi]$. Then $x=$ $\qquad$ | 12. $\frac{\pi}{4}$.
The original equation is equivalent to
$$
\cos \left(\frac{\pi}{2}-x-\sin x\right)=\cos (x-\cos x).
$$
Then $x-\cos x=2 k \pi+\frac{\pi}{2}-x-\sin x(k \in \mathbf{Z})$,
or $x-\cos x=2 k \pi-\left(\frac{\pi}{2}-x-\sin x\right)(k \in \mathbf{Z})$.
From equation (1), we get
$$
2 x+\sin x-\cos x=2 k ... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,807 |
13. Let $A$ and $B$ be two distinct points on the parabola
$$
y^{2}=2 p x(p>0)
$$
Then the minimum value of $|\overrightarrow{O A}+\overrightarrow{O B}|^{2}-|\overrightarrow{A B}|^{2}$ is $\qquad$. | 13. $-4 p^{2}$.
Let $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$. Then
$$
\begin{array}{l}
|\overrightarrow{O A}+\overrightarrow{O B}|^{2}=\left(x_{A}+x_{B}\right)^{2}+\left(y_{A}+y_{B}\right)^{2}, \\
|\overrightarrow{A B}|^{2}=\left(x_{A}-x_{B}\right)^{2}+\left(y_{A}-y_{B}\right)^{2}, \\
|\overrightarrow{... | -4 p^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,808 |
14. In $\triangle A B C$, it is known that $G$ is the centroid, the opposite sides of the three interior angles $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, and
56a $\overrightarrow{G A}+40 b \overrightarrow{G B}+35 c \overrightarrow{G C}=0$.
Then $\angle B=$ $\qquad$ | 14. $60^{\circ}$.
Since $\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0$, therefore, $40 b \overrightarrow{G A}+40 b \overrightarrow{G B}+40 b \overrightarrow{G C}=0$.
Thus, $(56 a-40 b) \overrightarrow{G A}+(35 c-40 b) \overrightarrow{G C}=0$. Since $\overrightarrow{G A}$ and $\overrightarrow{G C}$... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,809 |
15. (12 points) Inequality
$$
\begin{array}{l}
\sin 2 \theta-(2 \sqrt{2}+\sqrt{2} a) \sin \left(\theta+\frac{\pi}{4}\right)-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)} \\
>-3-2 a
\end{array}
$$
holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$. Find the range of real number $a$. | Three, 15. Let $x=\sin \theta+\cos \theta \in[1, \sqrt{2}]$. Then
$$
\begin{array}{l}
\sin 2 \theta=x^{2}-1, \\
\sin \left(\theta+\frac{\pi}{4}\right)=\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x .
\end{array}
$$
Thus, the original inequality becomes
$$
\begin{array}{l}
x^{2}-1-\frac{\sqrt{2}}{2}(2 \sqr... | a>3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,810 |
16. (12 points) As shown in Figure 1, in the cube $A B C D$ $-A_{1} B_{1} C_{1} D_{1}$, $O$, $E$, $F$, $G$ are the midpoints of $B D$, $B B_{1}$, $A_{1} D_{1}$, $D_{1} C_{1}$ respectively, and $A B=1$. Find the volume of the tetrahedron $O E F G$. | 16. Connect $B_{1} D_{1}$ and $F G$ intersecting at point $H$. Clearly, points $F$ and $G$ are symmetric with respect to the plane $B B_{1} D_{1} D$. Then
$$
\begin{array}{l}
V_{O-E F G}=2 V_{F-O E H}=\frac{2}{3} S_{\triangle O E H} \cdot F H \text {. } \\
\text { Also, } S_{\triangle O E H}=S_{\text {trapezoid } O B B... | \frac{5}{48} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,811 |
17. (12 points) In the Cartesian coordinate system, it is known that $\odot C_{1}$ and $\odot C_{2}$ intersect at points $P$ and $Q$, with the coordinates of $P$ being $(3,2)$. The product of the radii of the two circles is $\frac{13}{2}$. If $\odot C_{1}$ and $\odot C_{2}$ are both tangent to the line $l: y=k x$ and t... | 17. From the problem, we know that points $O, C_{1}, C_{2}$ are collinear.
Let the radii of $\odot C_{1}$ and $\odot C_{2}$ be $r_{1}$ and $r_{2}$, respectively, and the slope of line $C_{1} C_{2}$ be $\tan \alpha \neq 0$.
Let $m=\cot \alpha$. Then $C_{1}\left(m r_{1}, r_{1}\right), C_{2}\left(m r_{2}, r_{2}\right)$,
... | y=2\sqrt{2}x | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,812 |
Example 9 Let
$$
x^{3}-3 \sqrt{2} x^{2}+6 x-2 \sqrt{2}-8=0 \text {. }
$$
Then the value of $x^{5}-41 x^{2}+1$ is ( ).
(A) $13-\sqrt{2}$
(B) $-13+\sqrt{2}$
(C) -13
(D) 13 | Solve the given conditions:
$$
\begin{array}{l}
(x-\sqrt{2})^{3}=8 \Rightarrow x-\sqrt{2}=2 \\
\Rightarrow(x-2)^{2}=2 \Rightarrow x^{2}-4 x+2=0 .
\end{array}
$$
Therefore, $x^{5}-41 x^{2}+1$
$$
\begin{array}{l}
=\left(x^{2}-4 x+2\right)\left(x^{3}+4 x^{2}+14 x+7\right)-13 \\
=-13 .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,813 |
18. (15 points) Two players, A and B, are playing a certain game with the following rules: In each round, the winner gets 1 point, and the loser gets 0 points; the game ends when one player's score is 2 points higher than the other's. Additionally, the game is limited to a maximum of 20 rounds; if after 20 rounds, the ... | 18. Let the probability that the game ends after $k$ rounds be $p(\xi=k)$.
If $k$ is odd, then the difference in scores between A and B is also odd, and the game cannot end.
Consider when $k$ is even, the results of the first two rounds:
(1) A wins both or B wins both, referred to as a decisive two rounds, with a prob... | 2<E \xi \leqslant 4-\left(\frac{1}{2}\right)^{8} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,814 |
1. Given that $P$ is a point on the plane of $\triangle A B C$, satisfying
$$
\overrightarrow{P A}+\overrightarrow{P B}+2 \overrightarrow{P C}=3 \overrightarrow{A B} \text {. }
$$
Then the ratio of the area of $\triangle A B P$ to the area of $\triangle A B C$ is
$\qquad$ | 1. $1: 2$ | 1: 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,816 |
6. For all real numbers $x, y$ satisfying
$$
|x|+|y| \leqslant 1
$$
the inequality
$$
\left|2 x-3 y+\frac{3}{2}\right|+|y-1|+|2 y-x-3| \leqslant a
$$
always holds. Then the minimum value of the real number $a$ is . $\qquad$ | 6. $\frac{23}{2}$ | \frac{23}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,821 |
8. Given the equation about $x$
$$
|x-k|=\frac{\sqrt{2}}{2} k \sqrt{x}
$$
has two distinct real roots in the interval $[k-1, k+1]$. Then the range of the real number $k$ is $\qquad$. | 8. $0<k \leqslant 1$ | 0<k \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,823 |
Example 10 If the equation
$$
\left(x^{2}-1\right)\left(x^{2}-4\right)=k
$$
has four non-zero real roots, and the four points corresponding to them on the number line are equally spaced, find the value of $k$. | Let $x^{2}=y$. Then the original equation becomes
$$
y^{2}-5 y+4-k=0 \text {. }
$$
By the given conditions, this equation has two distinct positive roots $\alpha, \beta$, without loss of generality, let $\alpha>\beta$.
Thus, the four real roots of the original equation are $\pm \sqrt{\alpha}, \pm \sqrt{\beta}$.
Since ... | \frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,824 |
9. (16 points) Given the quadratic function
$$
y=f(x)=x^{2}+b x+c \text {. }
$$
the graph passes through the point $(1,13)$, and
$$
y=f\left(x-\frac{1}{2}\right)
$$
is an even function.
(1) Find the analytical expression of $f(x)$.
(2) Does the graph of the function $y=f(x)$ have any points where the x-coordinate is ... | (1) Since the function
$$
y=f\left(x-\frac{1}{2}\right)
$$
is an even function, the axis of symmetry of the quadratic function $f(x)$ is
$$
x=-\frac{1}{2} .
$$
Thus, $b=1$.
Given that the graph of the quadratic function $f(x)$ passes through the point $(1,13)$, we have $1+b+c=13$.
Therefore, $c=11$.
Hence, $f(x)=x^{2... | (10,121) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,825 |
11. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, a moving line $l$ passing through the point $P\left(\frac{\sqrt{2}}{3},-\frac{1}{3}\right)$ but not through the point $Q(\sqrt{2}, 1)$ intersects the ellipse $C$ at points $A$ and $B$.
(1) Find $\angle A Q B$;
(2) Let the area of $\triangle Q A B... | 11. (1) If the slope of line $l$ exists, let its equation be $y=k x+b$.
Since point $P$ is on line $l$, we have,
$$
\begin{array}{l}
-\frac{1}{3}=\frac{\sqrt{2}}{3} k+b \\
\Rightarrow b=-\frac{1}{3}(\sqrt{2} k+1) .
\end{array}
$$
Combining the equation of line $l$ and the equation of ellipse $C$ and eliminating $y$ gi... | S<3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,827 |
1. Let the sets be
$$
\begin{array}{l}
M=\left\{x \left\lvert\, x=\frac{7 \pi}{6}+2 k \pi\right. \text { or } x=\frac{5 \pi}{6}+2 k \pi, k \in \mathbf{Z}\right\}, \\
N=\left\{x \left\lvert\, x= \pm \frac{7 \pi}{6}+2 k \pi\right., k \in \mathbf{Z}\right\} .
\end{array}
$$
Then the relationship between $M$ and $N$ is ( ... | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,828 |
2. In the regular pentagonal prism $A B C D E-A_{1} B_{1} C_{1} D_{1} E_{1}$, there is a point $P$ on the lateral edge $C C_{1}$. If the section $P A E_{1}$ is perpendicular to the lateral face $A E E_{1} A_{1}$, then such a point $P$ ( ).
(A) has and only has one
(B) sometimes has two, sometimes has one
(C) has exac... | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,829 |
3. Let the function $y=f(x)(x \in \mathbf{R})$ satisfy
$$
f(x+2)=f(x) \text {, }
$$
and when $x \in(-1,1]$, $f(x)=1-x^{2}$. The function
$$
g(x)=\left\{\begin{array}{ll}
\lg |x|, & x \neq 0 ; \\
1, & x=0 .
\end{array}\right.
$$
Then the number of zeros of the function $t(x)=f(x)-g(x)$ in the interval $[-5,10]$ is ( )... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,830 |
4. In $\triangle A B C$, it is known that
$6 \overrightarrow{A C} \cdot \overrightarrow{A B}=2 \overrightarrow{A B} \cdot \overrightarrow{B C}=3 \overrightarrow{B C} \cdot \overrightarrow{C A}$.
Then $\angle A=$ ( $\quad$.
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $135^{\circ}$ | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,831 |
5. Point $P(x, y)$ satisfies the planar region
$$
\left\{\begin{array}{l}
\cos \theta \leqslant x \leqslant 3 \cos \theta, \\
\sin \theta \leqslant y \leqslant 3 \sin \theta
\end{array}(\theta \in \mathbf{R}),\right.
$$
Point $M(x, y)$ satisfies
$$
(x+5)^{2}+(y+5)^{2}=1 \text {. }
$$
Then the minimum value of $|\over... | 5. D.
Notice that, point $P$ satisfies the planar region which is part of an annulus centered at the origin:
$$
1 \leqslant x^{2}+y^{2} \leqslant 9(x \geqslant 0, y \geqslant 0) .
$$
Therefore, the minimum value of $|\overrightarrow{P M}|$ is
$$
\sqrt{(1+5)^{2}+(0+5)^{2}}-1=\sqrt{61}-1 \text {. }
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,832 |
6. Given $a_{1}, a_{2}, \cdots, a_{2011}$ is a sequence of distinct positive integers. If the order of these 2011 numbers is arbitrarily changed and denoted as $b_{1}, b_{2}, \cdots, b_{2011}$. Then the value of the number
$$
M=\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \cdots\left(a_{2011}-b_{2011}\right)
$$
mus... | 6. D.
Assume $M$ is odd. Then
$$
a_{i}-b_{i}(i=1,2, \cdots, 2011)
$$
must all be odd, and their sum is also odd;
$$
\begin{array}{l}
\text { but }\left(a_{1}-b_{1}\right)+\left(a_{2}-b_{2}\right)+\cdots+\left(a_{2011}-b_{2011}\right) \\
=\left(a_{1}+a_{2}+\cdots+a_{2011}\right)- \\
\left(b_{1}+b_{2}+\cdots+b_{2011}\... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,833 |
7. There are 5 pairs of shoes with different sizes. If 4 shoes are taken out, the probability of getting exactly one pair is $\qquad$ . | $\begin{array}{l}\text { II.7. } \frac{4}{7} . \\ P=\frac{\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}}{\mathrm{C}_{10}^{4}}=\frac{120}{210}=\frac{4}{7} .\end{array}$ | \frac{4}{7} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,834 |
Example 1 Given that $a$ and $b$ are the roots of the equation
$$
x^{2}-4 x+m=0
$$
and $b$ and $c$ are the roots of the equation
$$
x^{2}-8 x+5 m=0
$$
then $m=$ $\qquad$ | Given that $b$ is a common root of the two given equations, we have
$$
\begin{array}{l}
b^{2}-4 b+m=0, \\
b^{2}-8 b+5 m=0 .
\end{array}
$$
From equation (1), we get $m=4 b-b^{2}$.
Substituting this into equation (2) and simplifying, we get $b^{2}-3 b=0$.
Solving this, we find $b=0$ or 3.
Therefore, $m=4 b-b^{2}=0$ or ... | 0 \text{ or } 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,835 |
Example 2 If the two quadratic equations
$$
x^{2}+x+m=0 \text { and } m x^{2}+x+1=0
$$
each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$. | Given that the common real root of the two equations is $\alpha$, and $m \neq 1$. Then
$$
\begin{array}{l}
\alpha^{2}+\alpha+m=0, \\
m \alpha^{2}+\alpha+1=0 .
\end{array}
$$
From equation (1), we get $m=-\alpha^{2}-\alpha$.
Substituting into equation (2), we get
$$
\begin{array}{l}
\alpha^{4}+\alpha^{3}-\alpha-1=0 \\
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,836 |
17. Let $x, y, z$ be integers, and
$$
x+y+z=3, x^{3}+y^{3}+z^{3}=3 \text {. }
$$
Then $x^{2}+y^{2}+z^{2}=$ $\qquad$ . | 17.3 or 57.
Substitute $z=3-x-y$ into $x^{3}+y^{3}+z^{3}=3$, we get $x y=3(x+y)-9+\frac{8}{x+y}$.
Since $x, y$ are integers, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 1 , } \\
{ x y = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
x+y=4, \\
x y=5 ;
\end{array}\right.\right. \\
\left\{\begin{ar... | 3 \text{ or } 57 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,838 |
18. Let $a \leqslant 2$. Find the maximum and minimum values of $y=(x-2)|x|$ on $[a, 2]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, 18. When $x \leqslant 0$, $y=-(x-1)^{2}+1$; when $x>0$, $y=(x-1)^{2}-1$.
From this, we know $y_{\max }=0$.
When $1 \leqslant a \leqslant 2$, $y_{\text {min }}=a^{2}-2 a$;
When $1-\sqrt{2} \leqslant a<1$, $y_{\text {min }}=-1$;
When $a<1-\sqrt{2}$, $y_{\min }=-a^{2}+2 a$. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,839 |
19. Given two sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ satisfying
$$
\begin{array}{l}
x_{0}=y_{0}=1, \\
x_{n}=\frac{x_{n-1}}{2+x_{n-1}}(n \geqslant 1), y_{n}=\frac{y_{n-1}^{2}}{1+2 y_{n-1}}(n \geqslant 1) .
\end{array}
$$
Prove: For any natural number $n$, there exists a natural number $j_{n}$, such... | 19. From the given, we have
$$
\begin{array}{l}
\frac{1}{x_{n}}=1+\frac{2}{x_{n-1}} \\
\Rightarrow \frac{1}{x_{n}}+1=2\left(1+\frac{1}{x_{n-1}}\right)=\cdots=2^{n+1} \\
\Rightarrow x_{n}=\frac{1}{2^{n+1}-1} .
\end{array}
$$
From the given, we also have
$$
\begin{array}{l}
y_{n}+1=\frac{\left(y_{n-1}+1\right)^{2}}{1+2 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,840 |
20. Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line is drawn through its left focus $F_{1}$ intersecting the ellipse at points $A$ and $B$. Point $D(a, 0)$ is a point to the right of $F_{1}$. Connecting $A D$ and $B D$ intersects the left directrix of the ellipse at points $M$ and $N$. If the circ... | 20. It is known that $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$, and $l_{A B}: y=k(x+3)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k(x+3), \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array}\right.$
$$
\Rightarrow\left(16+25 k^{2}\right) x^{2... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,841 |
21. In an acute triangle $\triangle ABC$, $\angle A=\frac{\pi}{3}$, let the set of points $P$ inside it that simultaneously satisfy $P k \leqslant P B$ and $P A \leqslant P C$ form a region $G$ whose area is $\frac{1}{3}$ of the area of $\triangle ABC$. Prove: $\triangle ABC$ is an equilateral triangle. | $$
\begin{array}{l}
\text { Also, } S_{\triangle A O B}+S_{\triangle M O C}=2 S_{\triangle A E O}+2 S_{\triangle A O F} \\
=2 S_{\text {quadrilateral } A E O F}=\frac{2}{3} S_{\triangle A B C},
\end{array}
$$
Then $S_{\triangle O B C}=\frac{1}{3} S_{\triangle A B C}$.
Given $\angle B O C=120^{\circ}$, then
$O M=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,842 |
22. Let $a, b, c \in \mathbf{R}_{+}$, and $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$.
Prove:
$$
\frac{a+b}{2+a+b}+\frac{b+c}{2+b+c}+\frac{c+a}{2+c+a} \geqslant \frac{3}{2},
$$
and specify the condition for equality. | 22. From the Cauchy-Schwarz inequality $\sum_{i=1}^{n} \frac{a_{i}^{2}}{b_{i}} \geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{\sum_{i=1}^{n} b_{i}}$, we get
$$
\begin{array}{l}
\frac{a+b}{2+a+b}+\frac{b+c}{2+b+c}+\frac{c+a}{2+c+a} \\
\geqslant \frac{(\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c})^{2}}{6+2(a+b+c)} .
\end{arr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,843 |
1. Let real numbers $x, y$ satisfy the equation
$$
2 x^{2}+3 y^{2}=4 x \text {. }
$$
Then the minimum value of $x+y$ is ( ).
(A) $1+\frac{\sqrt{15}}{3}$
(B) $1-\frac{\sqrt{15}}{3}$
(C) 0
(D) None of the above | - 1. B.
Equation (1) can be transformed into
$$
(x-1)^{2}+\frac{y^{2}}{\frac{2}{3}}=1 \text {. }
$$
Let $x+y=t$.
When the line (3) is tangent to the ellipse (2), $t$ takes its extreme values.
At this point, $2 x^{2}+3(x-t)^{2}=4 x$, which simplifies to
$$
5 x^{2}-(6 t+4) x+3 t^{2}=0
$$
It has equal roots.
Thus, $\De... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,844 |
2. Given that the lengths of two medians of $\triangle A B C$ are 5 and 10, respectively. If the length of the third median is also an integer, then the maximum length of the third median is ( ).
(A) 7
(B) 8
(C) 14
(D) 15 | 2. C.
Let the length of the third median be $x$.
It is well known that the three medians of a triangle can form a triangle, and the original triangle can be determined by the lengths of the three medians. Therefore,
$$
10-5<x<10+5 \text {. }
$$
Since $x$ is an integer, the maximum value of $x$ is 14. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,845 |
3. Given that the units digit of $2^{n}+7^{n}(n \in \mathbf{N}, n>2010)$ is 3. Then the smallest $n$ is ( ).
(A) 2011
(B) 2012
(C) 2013
(D) 2014 | 3. D.
Notice that the units digit of $2^{k}$ cycles through $2, 4, 8, 6$, while the units digit of $7^{k}$ cycles through $7, 9, 3, 1$.
Since the units digit of $2^{n}+7^{n}$ is 3, it follows that $n$ is of the form $4k+2$.
Given that $n>2010$, the smallest $n$ is 2014. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,846 |
4. Let $a=\sqrt{5}-2$. Then the value of the algebraic expression
$$
3 a^{3}+10 a^{2}-12 a+6
$$
is ( ).
(A) 4
(B) 6
(C) $6-\sqrt{5}$
(D) $6+\sqrt{5}$ | $$
\begin{array}{l}
\text { Given } a=\sqrt{5}-2 \Rightarrow a^{2}+4 a-1=0 \text {. } \\
\text { Then } 3 a^{3}+10 a^{2}-12 a+6 \\
=\left(a^{2}+4 a-1\right)(3 a-2)-a+4 \\
=4-a=6-\sqrt{5} \text {. }
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,847 |
5. Satisfy
$$
y=\sqrt[3]{x+31}+\sqrt[3]{x+2012}
$$
the integer pairs $(x, y)$ ().
(A) Only one pair
(B) Exactly two pairs
(C) At least three pairs
(D) Do not exist | 5. B.
Let $a^{3}=x+31, b^{3}=x+2012(a, b \in \mathbf{Z})$.
Then $\boldsymbol{y}=a+b$,
$$
\begin{array}{l}
b^{3}-a^{3}=(b-a)\left[(b-a)^{2}+3 a b\right] \\
=1981=7 \times 283 .
\end{array}
$$
Since $b>a$, we have
$$
\begin{array}{l}
\left(b-a,(b-a)^{2}+3 a b\right) \\
=(1,1981),(7,283),(283,7),(1981,1) .
\end{array}
$... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,849 |
6. Let $a, b, c$ be the lengths of the three sides of $\triangle ABC$, and $t_{a}, t_{b}, t_{c}$ be the lengths of the corresponding angle bisectors of $\triangle ABC$. Given the following conclusions:
(1) $t_{a} t_{b} t_{c} \geqslant a b c$,
(2) $t_{a} t_{b} t_{c}<\frac{\sqrt{3}}{2} a b c$,
(3) $t_{a}^{2}+t_{b}^{2}+t_... | 6. D.
Notice,
$$
\begin{aligned}
t_{a} & =\frac{b c \sin A}{(b+c) \sin \frac{A}{2}}=\frac{2 b c \cos \frac{A}{2}}{b+c}, \\
t_{b} & =\frac{2 a c \cos \frac{B}{2}}{a+c}, \\
t_{c} & =\frac{2 a b \cos \frac{C}{2}}{a+b} .
\end{aligned}
$$
Assume without loss of generality that $a \geqslant b \geqslant c$. Then
$$
\angle A... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,850 |
1. As shown in Figure 1, in square $A B C D$, points $M$ and $N$ are on sides $A D$ and $C D$ respectively, and $M D=2 A M, N D=3 C N$. Then $\sin \angle M B N=$ | 2.1. $\frac{11 \sqrt{170}}{170}$.
Connect $M N$. Without loss of generality, let $A D=12$. Then
$$
A M=4, D M=8, C N=3, D N=9 \text {. }
$$
Thus, $B M=\sqrt{A B^{2}+A M^{2}}=4 \sqrt{10}$,
$$
\begin{array}{l}
B N=\sqrt{B C^{2}+C N^{2}}=3 \sqrt{17} . \\
\text { Also, } S_{\triangle B B M}=24, S_{\triangle B C N}=18, S_{... | \frac{11 \sqrt{170}}{170} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,851 |
2. Let positive real numbers $a, b, c, d, e$ satisfy $a<b<c<d$ $<e$, and the smallest three of the 10 products of any two numbers are $28, 32, 56$, and the largest two are 128, 240. Then $e=$ $\qquad$ | 2.16.
From the problem, we know
$$
a b=28, a c=32, c e=128, d e=240 \text {. }
$$
Then $c=\frac{8}{7} b, d=\frac{15}{8} c=\frac{15}{7} b$.
Thus, $a d=\frac{15}{7} a b=60>56$.
Therefore, $b c=56 \Rightarrow \frac{8}{7} b^{2}=56 \Rightarrow b=7$.
Hence, $a=4, c=8, d=15, e=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,852 |
3. A fair cube die has the numbers $1, 2, 3, 4, 5, 6$ on its six faces. If the die is rolled three times, the probability that the sum of the numbers on the top faces is a multiple of 3 is | 3. $\frac{1}{3}$.
Let the numbers on the top faces of the dice after three rolls be $x$, $y$, and $z$.
Notice that the multiples of 3 are
$$
\begin{array}{l}
3=1+1+1, \\
6=1+1+4=1+2+3=2+2+2, \\
9=1+2+6=1+3+5=1+4+4 \\
=2+2+5=2+3+4=3+3+3, \\
12=1+5+6=2+4+6=2+5+5 \\
=3+3+6=3+4+5=4+4+4, \\
15=3+6+6=4+5+6=5+5+5, \\
18=6+6+... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,853 |
4. A positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that the sum of two of these divisors is six times the sum of the other two. Then $n=$ $\qquad$ . | 4.287 or 143.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p<q$).
When $n=p^{3}$, all its positive divisors are $1, p, p^{2}, p^{3}$. According to the problem, we have
$$
\{i, j, k\}=\{1,2,3\},\{s, t\}=\{1,6\} \text {, }
$$
sat... | 287 \text{ or } 143 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,854 |
一、(20 points) Given that the radii of $\odot A$, $\odot B$, and $\odot C$ are all 3, and they are mutually externally tangent. On the plane where these three circles lie, take all points that are a distance of $2 \sqrt{3} - 3$ from at least one point on these circles. Find the area of the region formed by all such poin... | Given that $\triangle ABC$ is an equilateral triangle with a side length of 6.
Let the center of $\triangle ABC$ be $O$. Then
$$
O A=O B=O C=2 \sqrt{3} \text{. }
$$
As shown in Figure 2, construct the reflections of point $O$ about the sides $AB$, $BC$, and $CA$ as points $O_{1}$, $O_{2}$, and $O_{3}$, respectively. C... | 18 \sqrt{3}+\pi(72 \sqrt{3}-120) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,855 |
II. (25 points) Given the quadratic function
$$
y=x^{2}+b x-c
$$
the graph passes through three points
$$
P(1, a), Q(2,3 a)(a \geqslant 3), R\left(x_{0}, y_{0}\right) .
$$
If the centroid of $\triangle P Q R$ is on the $y$-axis, find the minimum perimeter of $\triangle P Q R$. | Given points $P$ and $Q$ are on the graph of the quadratic function
$$
y=x^{2}+b x-c
$$
we have,
$$
\left\{\begin{array} { l }
{ 1 + b - c = a , } \\
{ 4 + 2 b - c = 3 a }
\end{array} \Rightarrow \left\{\begin{array}{l}
b=2 a-3, \\
c=a-2 .
\end{array}\right.\right.
$$
Since the centroid of $\triangle P Q R$ lies on ... | 4 \sqrt{2}+5 \sqrt{5}+\sqrt{37} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,856 |
Three. (25 points) If placing the positive integer $N$ to the left of the positive integer $n$ results in a new number that is divisible by 7, then $N$ is called a "magic number" of $n$. $M$ is a set of positive integers such that for any positive integer $n$, there exists a positive integer in set $M$ that is a magic ... | For $n=1,2, \cdots, 7$.
If $|M| \leqslant 6$, then by the pigeonhole principle, there must be a positive integer $N$ in set $M$ that is a common magic number of $i, j(1 \leqslant i<j \leqslant 7)$, i.e.,
$71(10 N+i), 7 I(10 N+j)$.
Then $7!(j-i)$, but $0<j-i \leqslant 6$, a contradiction.
Therefore, $|M| \geqslant 7$.
T... | 28 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,857 |
1. Let the function $f(x)=x^{2}(x \in D, D$ be the domain) have the range $\left\{1^{2}, 2^{2}, \cdots, 2012^{2}\right\}$. Then the number of functions $f(x)$ that satisfy this condition is $\qquad$ . | $-、 1.3^{2012}$.
Since 1 is in the range of $f(x)=x^{2}(x \in D)$, therefore, $1 \in D$, or $-1 \in D$, or $-1 、 1 \in D$, there are three possibilities,
Similarly, for any number $k(k=2,3, \cdots, 2012)$, there are also three possibilities.
Therefore, $D$ has $3^{2012}$ possibilities, which means the number of funct... | 3^{2012} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,858 |
Property 3 A cyclic quadrilateral $ABCD$ is a harmonic quadrilateral if and only if $AC$ is the symmedian of $\triangle ABD$. | Proof As shown in Figure 6, let $M$ be the midpoint of $BD$.
Then $AC$ is the A-symmedian of $\triangle ABD$
$$
\begin{array}{l}
\Leftrightarrow \angle BAC = \angle MAD \\
\Leftrightarrow \frac{\sin \angle BAC}{\sin \angle CAD} = \frac{\sin \angle DAM}{\sin \angle BAM}.
\end{array}
$$
Since $M$ is the midpoint of $BD$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,859 |
2. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=4, A A_{1}=A D=2$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, B C$ respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$ | 2.3.
Take point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{2}$. Then $H E / / B_{1} G$. Therefore, $H E / /$ plane $B_{1} F G$.
Thus, $V_{B_{1}-E F G}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H}$.
And $S_{\triangle B_{1} F H}=\frac{9}{2}$, the distance from point $G$ to plane $B_{1} F H$ is ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,860 |
3. In the geometric sequence $\left\{a_{n}\right\}$ with all terms being positive, if $a_{3}+a_{4}-a_{2}-a_{1}=8$, then the minimum value of $a_{5}+a_{6}+a_{7}+a_{8}$ is $\qquad$ | $3.44+20 \sqrt{5}$.
Let the common ratio be $q$. From the problem, we know $q>1$.
From $a_{3}+a_{4}-a_{2}-a_{1}=8$, we get
$a_{1}(q+1)\left(q^{2}-1\right)=8$.
And $a_{5}+a_{6}+a_{7}+a_{8}=a_{1}(q+1) q^{4}\left(q^{2}+1\right)$
$$
=\frac{8 q^{4}\left(q^{2}+1\right)}{q^{2}-1} \text {. }
$$
Let $t=q^{2}-1$. Then $t>0$, an... | 44+20 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,861 |
4. Let $M=\frac{a^{2}+b^{2}+c^{2}}{a b c+1}\left(a, b, c \in \mathbf{R}_{+}\right)$. If $a+b \leqslant c+1, b+c \leqslant a+1, c+a \leqslant b+1$, then the maximum value of $M$ is | 4. $\frac{3}{2}$.
Adding $a+b \leqslant c+1, b+c \leqslant a+1$ yields $2 b \leqslant 2 \Rightarrow b \leqslant 1$.
Similarly, $a \leqslant 1, c \leqslant 1$.
When $a=b=c=1$, $M=\frac{3}{2}$.
Next, we prove: $M \leqslant \frac{3}{2}$.
Let $x=1-a, y=1-b, z=1-c$. Then $0 \leqslant x, y, z \leqslant 1$, and $x+y \geqslan... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,862 |
5. Let $M$ and $N$ be the points of intersection of the left and right directrices of the ellipse $G$ with the major axis. If there exists a point $P$ on the ellipse $G$ such that $\angle M P N=120^{\circ}$, then the range of the eccentricity of the ellipse $G$ is $\qquad$ | 5. $\left(0, \frac{\sqrt{3}}{3}\right]$.
Let the ellipse $G: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Then $M\left(-\frac{a^{2}}{c}, 0\right), N\left(\frac{a^{2}}{c}, 0\right)$.
Also, $P\left(x_{0}, y_{0}\right)\left(y_{0}>0\right), \angle P M N=\alpha$, $\angle P N M=\beta$, then
$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{... | \left(0, \frac{\sqrt{3}}{3}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,863 |
6. Given the three sides of $\triangle A B C$ are $a, b, c$. If $a+b+c=16$, then
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
$=$ | 6. 64 .
Let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,864 |
7. Let $p(p>2)$ be a prime number, and the set $A=\{1,2$, $\cdots, p-1\}$. If a non-empty subset $B \subseteq A$ is such that $p$ divides the sum of the elements in set $B$, then the number of subsets $B$ that satisfy this condition is $\qquad$ . | 7. $\frac{2^{p-1}-1}{p}$.
$$
\text { Let } A_{1}=A \cup\{0\} \text {. }
$$
Let $B=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ be a non-empty subset of $A_{1}$, and $B_{i}=\left\{b_{i 1}, b_{i 2}, \cdots, b_{i k}\right\}(i=0,1, \cdots, p-1)$, where $b_{i j} \equiv i+a_{j}(\bmod p)(j=1, \cdots, k)$. Then the sums of all... | \frac{2^{p-1}-1}{p} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,865 |
8. An $8 \times 8$ chessboard is colored in the usual way, with 32 black squares and 32 white squares. A "path" consists of 8 white squares, one in each row, and adjacent white squares share a common vertex. The number of such paths is $\qquad$. | 8. 296
We can use the number labeling method:
\begin{tabular}{cccccccc}
1 & Black & 1 & Black & 1 & Black & 1 & Black \\
Black & 2 & Black & 2 & Black & 2 & Black & 1 \\
2 & Black & 4 & Black & 4 & Black & 3 & Black \\
Black & 6 & Black & 8 & Black & 7 & Black & 3 \\
6 & Black & 14 & Black & 15 & Black & 10 & Black \\... | 296 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,866 |
9. (16 points) In the geometric sequence $\left\{a_{n}\right\}$ where all terms are positive, what is the maximum number of terms that can be integers between $100 \sim 1000$? | 9. Let the geometric sequence $\left\{a q^{n-1}\right\}$ satisfy
$$
100 \leqslant a q^{n-1} \leqslant 1000 \quad (a, q > 1) \text{ are integers.}
$$
Clearly, $q$ must be a rational number.
Let $q=\frac{t}{s}(t>s \geqslant 1, (t, s)=1)$.
Since $a q^{n-1}=a\left(\frac{t}{s}\right)^{n-1}$ is an integer, $a$ must be a mult... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,867 |
10. (20 points) Let $A$ and $B$ be two points on the hyperbola $x^{2}-\frac{y^{2}}{2}=1$.
$O$ is the origin, and it satisfies $\overrightarrow{O A} \cdot \overrightarrow{O B}=0, \overrightarrow{O P}=\alpha \overrightarrow{O A}+(1-\alpha) \overrightarrow{O B}$.
(1) When $\overrightarrow{O P} \cdot \overrightarrow{A B}=... | 10. (1) From the problem, we know $O P \perp A B, O A \perp O B$.
Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\left\{\begin{array}{l}
2 x_{1}^{2}-y_{1}^{2}=2, \\
2 x_{2}^{2}-y_{2}^{2}=2, \\
x_{1} x_{2}+y_{1} y_{2}=0 .
\end{array}\right.
$$
$$
\begin{array}{l}
\text { and }|O P|^{2}=\fra... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,868 |
11. (20 points) The rules of a coin-tossing game are: toss a coin, and get 1 point for heads, and 2 points for tails.
(1) Let the score after 5 tosses be $\xi$, find the distribution of $\xi$ and the mathematical expectation $E \xi$;
(2) Find the probability of getting exactly $n\left(n \in \mathbf{N}_{+}\right)$ point... | 11. (1) The probability of scoring $\xi$ from 5 throws is
$$
P(\xi=i)=\mathrm{C}_{5}^{i-5}\left(\frac{1}{2}\right)^{5}(i=5,6, \cdots, 10) \text {. }
$$
Its distribution is shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$\xi$ & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$P$ & $\frac{1}{32}$ & $\frac{5}{32}$ & $\... | p_{n}=\frac{1}{3}\left[2+\left(-\frac{1}{2}\right)^{n}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,869 |
One, (40 points) Let $H_{\triangle A B C}$ denote the sum of the three altitudes of $\triangle A B C$. Given that $A_{1}, B_{1}, C_{1}$ are the excenters of $\triangle A B C$. Prove: $H_{\triangle A_{1} B_{1} C_{1}} \geqslant 2 H_{\triangle A B C}$.
| Given the figure 1, draw perpendiculars from points $A_{1}$, $B_{1}$, and $C_{1}$ to $BC$, $AC$, and $AB$ respectively, with the feet of the perpendiculars being $H_{1}$, $H_{2}$, and $H_{3}$. Let the lengths of the three altitudes of $\triangle ABC$ be $h_{A}$, $h_{B}$, and $h_{C}$; and the radii of the three excircle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,871 |
II. (40 points) Let integers $x, y, z$ satisfy
$$
(x-y)(y-z)(z-x)=x+y+z \text {. }
$$
Prove: 27 divides $(x+y+z)$. | Second, from the known conditions, we deduce that $x, y \backslash z$ must be pairwise congruent modulo 3.
Proof by contradiction.
Assume $x \equiv y(\bmod 3), x \neq z(\bmod 3)$.
Then 3 I $(x-y) \Rightarrow 3$ । $(x+y+z)$
$\Rightarrow 3$ । $(2 x+z) \Rightarrow 3$ । $(z-x)$,
Contradiction.
Next, assume $x, y, z$ have ... | 27 \text{ divides } (x+y+z) | Number Theory | proof | Yes | Yes | cn_contest | false | 724,872 |
Three. (50 points) Find the largest positive real number $\lambda$ such that the inequality
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant \lambda \sum_{k=1}^{n}\left(1+\sum_{i=1}^{k} a_{i}^{2}\right)^{-\frac{1}{2}}
$$
holds for all $n \in \mathbf{N}_{+}$ and $a_{i} \in \mathbf{R}$. | Three, first prove a lemma.
Lemma $\frac{1}{a_{k}}+\frac{1}{\sqrt{1+\sum_{i=1}^{k-1} a_{i}^{2}}} \geqslant \frac{2 \sqrt{2}}{\sqrt{1+\sum_{i=1}^{k} a_{i}^{2}}}$.
Proof Let $1+\sum_{i=1}^{k-1} a_{i}^{2}=a_{k}^{2} t_{k}\left(t_{k}>0\right)$.
Then (1) $\Leftrightarrow 1+\frac{1}{\sqrt{t_{k}}} \geqslant \frac{2 \sqrt{2}}{\... | 2 \sqrt{2}-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,873 |
Four. (50 points) There are 12 points on a plane, and no three points are collinear. Using any one of these points as the starting point and another point as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of zero t... | Let the 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $C_{12}^{3}$. Let the number of vectors starting from $P_{i}(i=1,2, \cdots, 12)$ be $x_{i}$ $\left(0 \leqslant x_{i} \leqslant 11\right)$.
If a triangle with three points as vertices is a non-zero triangle, th... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,874 |
Given the equation
\[
\begin{array}{l}
a^{3} x^{4}+2 a^{2} b x^{3}+\left(a b^{2}+2 a^{2} c+a b\right) x^{2}+ \\
\left(2 a b c+b^{2}\right) x+a c^{2}+b c+c=0(a>0)
\end{array}
\]
has real roots. Prove:
\[
\left(\sqrt{b^{2}-4 a c}+1\right)^{2} \geqslant 2 b+1 .
\] | Prove that the original equation can be transformed into
$$
a\left(a x^{2}+b x+c\right)^{2}+b\left(a x^{2}+b x+c\right)+c=0 \text {. }
$$
Let $f(x)=a x^{2}+b x+c$. Then the original equation can be transformed into
$$
f(f(x))=0 .
$$
Suppose the real roots of $f(x)=0$ are $x_{1}$ and $x_{2}\left(x_{1} \leqslant x_{2}\... | \left(\sqrt{b^{2}-4 a c}+1\right)^{2} \geqslant 2 b+1 | Algebra | proof | Yes | Yes | cn_contest | false | 724,875 |
As shown in Figure 2, points $P$ and $Q$ are both outside the square $ABCD$, satisfying
$$
\begin{array}{l}
\angle ABP \\
=\angle ADQ \\
=\angle PCQ \\
=90^{\circ} .
\end{array}
$$
Prove: $S_{\triangle APQ}=S_{\triangle BPC}+S_{\triangle DCQ}$. | Proof: Let $\angle ABC = \angle ADC = \alpha$,
$$
\angle BCD = \angle BAD = \beta.
$$
Clearly, $\alpha + \beta = 180^{\circ}$.
Let $\angle PBC = \gamma$.
Since $\alpha + \gamma = \angle ABP = 90^{\circ}$, we have
$$
\begin{array}{l}
\angle CDQ = \angle ADQ - \angle ADC = 90^{\circ} - \alpha = \gamma. \\
\text{And } \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,876 |
Given 315, find
$$
S=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots+\sqrt[2011]{2011}}}}
$$
the integer part of \( S \). | Let $n=2011$, and set
$$
\begin{array}{l}
a_{1}=\sqrt[n]{n}, \\
a_{k}=\sqrt[n-k+1]{n-k+1+a_{k-1}}(2 \leqslant k \leqslant n-1) .
\end{array}
$$
Then $S=a_{n-1}$.
We will prove by induction that:
$$
1<a_{m}<2(m=1,2, \cdots, n-1) \text {. }
$$
(1) When $m=1$, since $1<n<2^{n}$, we have
$$
1<\sqrt[n]{n}<2 \text {. }
$$
(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,877 |
316 Given that the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ is a monotonically increasing function. If $f(f(n))=3 n$, find $f(2011)$. | Let $f(1)=m$. Then $f(m)=f(f(1))=3$.
Thus, $m>1 \Rightarrow f(m)>f(1) \Rightarrow 3>m$.
Therefore, $f(1)=m=2 \Rightarrow f(2)=f(f(1))=3$.
Similarly, $f(3)=6, f(6)=9$.
From $f(3)<f(4)<f(5)<f(6)$, we get
$6<f(4)<f(5)<9$.
Hence, $f(4)=7, f(5)=8$.
Furthermore, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)... | 3846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,878 |
Example 4 Given three fixed points on a line in sequence as $A, B$, and $C$, $\Gamma$ is a circle passing through $A$ and $C$ with its center not on $AC$. The tangents to circle $\Gamma$ at points $A$ and $C$ intersect at point $P$, and $PB$ intersects circle $\Gamma$ at point $Q$. Prove: The bisector of $\angle AQC$ i... | Proof As shown in Figure 9, let $P Q$ intersect the circle $\Gamma$ at point $M$, and connect $A M, C M$.
Let $Q L$ be the angle bisector of $\angle A Q C$ and intersect $A C$ at point $L$.
By Property 2, we know that
quadrilateral $A M C Q$ is a harmonic quadrilateral
$$
\begin{array}{l}
\Rightarrow A Q \cdot C M \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,881 |
For example, $5 \angle A P B$ contains an inscribed circle that is tangent to $A P, P B$ at points $A, B$. $P C D$ is any secant line intersecting the circle at points $C, D$. Point $Q$ is on side $C D$, and $\angle Q A D = \angle P B C$. Prove: $\angle P A C = \angle Q B D$.
---
The translation maintains the origina... | Prove as shown in Figure 10, connect $A B$.
By property 2, we know that quadrilateral $A D B C$ is a harmonic quadrilateral, and
$$
\begin{array}{l}
\angle B A C \\
=\angle P B C \\
=\angle Q A D .
\end{array}
$$
By property 3, we know that
$A B$ is the $A$-symmedian of $\triangle A C D$.
Therefore, $A Q$ is the medi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,882 |
Example 6 Two circles intersect at points $A$ and $B$. Let $PQ$ be a common tangent, with $P$ and $Q$ being the points of tangency. $S$ is the intersection of the tangents to the circumcircle of $\triangle APQ$ at points $P$ and $Q$. $H$ is the reflection of point $B$ over $PQ$. Prove that $A$, $S$, and $H$ are colline... | Prove as shown in Figure 11, connect $QB$, $BP$, extend $AB$ to intersect $PQ$ at point $M$, connect $SA$ to intersect the circumcircle of $\triangle APQ$ at point $H'$.
Then $MP^2 = MB \cdot MA = MQ^2$
$\Rightarrow M$ is the midpoint of side $PQ$.
$$
\begin{array}{l}
\text{Also, } \angle PHQ = \angle PBQ \\
= 180^{\ci... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,883 |
Example 7 Let $A, B$ be two points inside $\odot O$, and $O$ is the midpoint of line segment $AB$. $P$ is a point on $\odot O$, and the lines $PA, PB$ intersect $\odot O$ again at points $C, D$. The tangents to $\odot O$ at points $C, D$ intersect at point $Q$, and $M$ is the midpoint of $PQ$. Prove:
$$
OM \perp AB.
$$ | Prove as shown in Figure 12, draw $P E / / A B$, connect $P O$ and extend it to intersect $\odot O$ at point $F$.
From $P E, P A, P O$,
$P B$ forming a harmonic pencil, we know quadrilateral $E C F D$ is a harmonic quadrilateral.
By property 2, we know points $E, F, Q$ are collinear.
$$
\begin{array}{l}
\text { Also }... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,884 |
Example 3 Given three different real numbers $a, b, c$, satisfying $a-b+c=3$, the equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+b x+c=0
$$
have one common real root, and the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$
also have one common real root. Find the values of $a, b, c$. | Let $x_{1}$ be a common real root of the equations
$x^{2}+a x+1=0$ and $x^{2}+b x+c=0$.
Then
$$
\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}\right.
$$
Subtracting the two equations, we get
$$
x_{1}=\frac{c-1}{a-b} \text {. }
$$
Let $x_{2}$ be a common real root of the equation... | a=-2, b=-3, c=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,885 |
1. In a non-isosceles $\triangle ABC$, the incenter is $I$, and the incircle $\odot I$ touches $BC$, $CA$, and $AB$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. $AA_{1}$ and $BB_{1}$ intersect $\odot I$ again at points $A_{2}$ and $B_{2}$. In $\triangle A_{1}B_{1}C_{1}$, the angle bisectors of $\angle A_{1}$ ... | From property 2, we know that quadrilateral $A_{2} C_{1} A_{1} B_{1}$ is a harmonic quadrilateral. Therefore,
$$
\frac{A_{2} C_{1}}{B_{1} A_{2}}=\frac{A_{1} C_{1}}{A_{1} B_{1}}=\frac{A_{3} C_{1}}{A_{3} B_{1}} .
$$
Thus, $A_{2} A_{3}$ is the angle bisector of $\angle B_{1} A_{2} C_{1}$.
Similarly, $B_{2} B_{3}$ is the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,886 |
2. $M, N$ are the midpoints of the arcs $\overparen{BC}$ and $\overparen{AC}$ of the circumcircle $\Gamma$ of acute $\triangle ABC (\angle A < \angle B)$, respectively. A line through point $C$ parallel to $MN$ intersects the circle $\Gamma$ at point $P$, and $I$ is the incenter of $\triangle ABC$. Connecting $PI$ and ... | (1) Quadrilateral MPNI is a parallelogram, $TC$ is the $T$-median of $\triangle N T M$, so quadrilateral NCMT is a harmonic quadrilateral, hence
$$
M C=N P, N C=M P \text {. }
$$
(2) First prove $N C=N I_{1}, M C=M I_{2}, \triangle I_{1} N T \backsim$ $\triangle I_{2} M T$, then prove $Q 、 I_{1} 、 I_{2} 、 T$ are concyc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,887 |
3. Let $P$ be a point outside $\odot O$, and draw two tangents from $P$ to $\odot O$, touching at points $A$ and $B$. Let $PO$ intersect $AB$ at point $Q$, and draw any chord $CD$ of $\odot O$ through $Q$. Prove: $\triangle PAB$ and $\triangle PCD$ have the same incenter. | Let $P D$, $P C$, $P O$ intersect $\odot O$ at points $C^{\prime}$, $D^{\prime}$, $I$ respectively. Quadrilateral $D A C^{\prime} B$ is a harmonic quadrilateral. Points $C$ and $C^{\prime}$, $D$ and $D^{\prime}$ are symmetric with respect to $P O$. Line $C I$ bisects $\angle D C P$. Similarly, $D I$ bisects $\angle C D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,888 |
4. From a point outside a circle $\odot O$, draw tangents $P A$ and $P B$, and a secant $P C D$. Through point $C$, draw a line parallel to $P A$, intersecting $A B$ and $A D$ at points $E$ and $F$ respectively. Prove that $C E = E F$. | Connect $A C$.
Let $A B$ and $C D$ intersect at point $K$. It is easy to prove
$$
\triangle A C F \backsim \triangle A D C \Rightarrow A C^{2}=A D \cdot A F \text {. }
$$
Applying Menelaus' theorem to $\triangle F D C$ and the transversal $A E K$ yields
$$
\frac{C K}{K D} \cdot \frac{D A}{A F} \cdot \frac{F E}{E C}=1 ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,889 |
Example 1 Simplify:
$$
\frac{1}{\sqrt{8}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{14}}+\cdots+\frac{1}{\sqrt{6038}+\sqrt{6041}} .
$$ | Solve: By observation and induction, we get the general term
$$
f(n)=\frac{1}{\sqrt{3 n+5}+\sqrt{3 n+8}} .
$$
Decomposing the general term, we get
$$
\begin{array}{l}
f(n)=\frac{\sqrt{3 n+8}-\sqrt{3 n+5}}{(\sqrt{3 n+5}+\sqrt{3 n+8})(\sqrt{3 n+8}-\sqrt{3 n+5})} \\
=\frac{1}{3}(\sqrt{3 n+8}-\sqrt{3 n+5}) .
\end{array}
$... | \frac{1}{3}(\sqrt{6041}-2\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,890 |
Example 2 Note
$$
\begin{array}{l}
S=\frac{2}{1 \times 3}+\frac{2^{2}}{3 \times 5}+\cdots+\frac{2^{49}}{97 \times 99}, \\
T=\frac{1}{3}+\frac{2}{5}+\frac{2^{2}}{7}+\cdots+\frac{2^{48}}{99} .
\end{array}
$$
Then the value of $S-T$ is ( ).
(A) $\frac{2^{49}}{99}$
(B) $1-\frac{2^{49}}{99}$
(C) $\frac{2^{49}}{99}-1$
(D) $... | Solve: By observation and induction, we get the general term
$$
f(n)=\frac{2^{n}}{(2 n-1)(2 n+1)}-\frac{2^{n-1}}{2 n+1} \text {. }
$$
Decomposing the general term, we get
$$
\begin{array}{l}
f(n)=2^{n-1} \cdot \frac{(2 n+1)-(2 n-1)}{(2 n-1)(2 n+1)}-\frac{2^{n-1}}{2 n+1} \\
=2^{n-1}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,891 |
Example 3 For any positive integer $n$, prove:
$$
\frac{1}{1 \times 2+2^{2}}+\frac{1}{2 \times 3+3^{2}}+\cdots+\frac{1}{n(n+1)+(n+1)^{2}}<\frac{5}{12} .
$$ | $$
\begin{array}{l}
\frac{1}{k(k+1)+(k+1)^{2}}=\frac{1}{(k+1)(2 k+1)} \\
<\frac{1}{2 k(k+1)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}\right) . \\
\text { Therefore, } \frac{1}{1 \times 2+2^{2}}+\frac{1}{2 \times 3+3^{2}}+\cdots+\frac{1}{n(n+1)+(n+1)^{2}} \\
=\frac{1}{6}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{... | \frac{5}{12} | Inequalities | proof | Yes | Yes | cn_contest | false | 724,892 |
Example 4 For any positive integer $n$, prove:
$$
\frac{1}{1 \times 3}+\frac{1}{2 \times 4}+\frac{1}{3 \times 5}+\cdots+\frac{1}{n(n+2)}<\frac{3}{4}
$$ | Notice that,
$$
\begin{array}{l}
\frac{1}{k(k+2)}=\frac{1}{2} \cdot \frac{(k+2)-k}{k(k+2)} \\
=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right) .
\end{array}
$$
Substituting $1,2, \cdots, n$ into the above equation and summing them up, we get
$$
\begin{array}{l}
\frac{1}{1 \times 3}+\frac{1}{2 \times 4}+\cdots+\frac{... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 724,893 |
Example 5 Calculate:
$$
\frac{\left(2^{4}+2^{2}+1\right)\left(4^{4}+4^{2}+1\right) \cdots\left(2012^{4}+2012^{2}+1\right)}{\left(3^{4}+3^{2}+1\right)\left(5^{4}+5^{2}+1\right) \cdots\left(2013^{4}+2013^{2}+1\right)} \text {. }
$$ | Notice that,
$$
\begin{aligned}
& \frac{(2 n)^{4}+(2 n)^{2}+1}{(2 n+1)^{4}+(2 n+1)^{2}+1} \\
= & \frac{\left[(2 n)^{2}+1\right]^{2}-(2 n)^{2}}{\left[(2 n+1)^{2}+1\right]^{2}-(2 n+1)^{2}} \\
= & \frac{\left(4 n^{2}+2 n+1\right)\left[(2 n)^{2}-2 n+1\right]}{\left[(2 n+1)^{2}+(2 n+1)+1\right]\left(4 n^{2}+2 n+1\right)} \\... | \frac{3}{4054183} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,894 |
Example 6 For any positive integer $n$, prove:
$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{2 n-1}\right)>\sqrt{2 n+1} .
$$ | Prove the general term:
$$
\begin{array}{l}
1+\frac{1}{2 k-1}=\frac{2 k}{2 k-1}=\sqrt{\frac{(2 k)^{2}}{(2 k-1)^{2}}} \\
>\sqrt{\frac{(2 k)^{2}-1}{(2 k-1)^{2}}}=\frac{\sqrt{2 k+1}}{\sqrt{2 k-1}} .
\end{array}
$$
Substitute $k=1,2, \cdots, n$ into the above expression and multiply them:
$$
\begin{array}{l}
\left(1+\frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,895 |
Example 4 Given the equations
$$
x^{2}+a_{1} x+a_{2} a_{3}=0 \text { and } x^{2}+a_{2} x+a_{1} a_{3}=0
$$
have exactly one non-zero common root. Prove: The other two roots (excluding the common root) of these two equations are the roots of the equation
$$
x^{2}+a_{3} x+a_{1} a_{2}=0
$$ | Let the equation
$$
x^{2}+a_{1} x+a_{2} a_{3}=0
$$
have two roots $\alpha, \beta$, and the equation
$$
x^{2}+a_{2} x+a_{1} a_{3}=0
$$
have two roots $\alpha, \gamma$, where $\alpha$ is the common root of these two equations.
According to the problem, we have
$$
\begin{array}{l}
\alpha^{2}+a_{1} \alpha+a_{2} a_{3}=0, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,896 |
Example 1 Given $a, b \in \mathbf{R}$, the equation about $x$
$$
x^{4}+a x^{3}+2 x^{2}+b x+1=0
$$
has one real root. Find the minimum value of $a^{2}+b^{2}$. | 【Analysis】This is a quartic equation, and parameters $a, b$ are not easy to handle. We might as well regard $a, b$ as the main variables and $x$ as a parameter. Since $a^{2}+b^{2}$ represents the square of the distance from the moving point $P(a, b)$ to the origin, and $P(a, b)$ lies on the line
$$
x^{3} a + x b + x^{4... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,897 |
Example 2 Let $k \geqslant 9$. Solve the equation
$$
x^{3}+2 k x^{2}+k^{2} x+9 k+27=0 \text {. }
$$ | 【Analysis】Since the equation is a cubic equation, it is difficult to factorize, so this approach is hard. However, by changing the main variable, viewing $x$ as a parameter and $k$ as the main variable, it can be seen as a quadratic equation in $k$
$$
\begin{array}{l}
x k^{2}+\left(2 x^{2}+9\right) k+\left(x^{3}+27\rig... | x=\frac{(3-k) \pm \sqrt{(k-9)(k+3)}}{2} \text{ or } x=k+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,898 |
Example 3 Given the equation in $x$
$$
x^{3}-a x^{2}-2 a x+a^{2}-1=0
$$
has exactly one real root. Find the range of real values for $a$.
| 【Analysis】Regarding the cubic equation in $x$ as a quadratic equation in $a$.
The original equation is transformed into
$$
a^{2}-\left(x^{2}+2 x\right) a+x^{3}-1=0 \text {. }
$$
Solving for $a$ yields $a=x-1$ or $a=x^{2}+x+1$.
Since the original equation has only one real root, the equation $a=x^{2}+x+1$ in $x$ has no... | a<\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,899 |
Example 4 Let the inequality
$$
x^{2}-(a+1) x-a^{2}>0
$$
hold for all $a \in(1,2)$. Find the range of $x$. | 【Analysis】In this inequality, the parameter $a$ has both quadratic and linear terms, which makes it difficult to handle. Instead of treating $x$ as the main variable, we can view the original inequality as a quadratic inequality in $a$ with $x$ as a parameter, and then find the range of $x$.
The original inequality can... | x \geqslant 4 \text{ or } x \leqslant -1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,900 |
3. Prove: The equations
$$
x^{2}-x-1=0 \text { and } x^{4}-3 x-2=0
$$
have a common root. | Prompt: By polynomial division, we know
$$
x^{4}-3 x-2=\left(x^{2}-x-1\right)\left(x^{2}+x+2\right) \text {. }
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,901 |
4. Let the equations
$$
a x^{2}+b x+c=0 \text { and } c x^{2}+b x+a=0
$$
have only one common positive real root. Try to find the relationship between $a$, $b$, and $c$. | Given that $a \neq c$, let the common positive real root be $t$. Then
$$
\begin{array}{l}
a t^{2}+b t+c=0, \\
c t^{2}+b t+a=0 .
\end{array}
$$
Subtracting the two equations and rearranging gives
$$
(a-c) t^{2}=(a-c) \Rightarrow t=1(t>0) .
$$
Therefore, $a+b+c=0$. | a+b+c=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,902 |
5. If the quadratic equation with positive integer coefficients
$$
4 x^{2}+m x+n=0
$$
has two distinct rational roots $p$ and $q$ ($p<q$), and the equations
$$
x^{2}-p x+2 q=0 \text { and } x^{2}-q x+2 p=0
$$
have a common root, find the other root of the equation
$$
x^{2}-p x+2 q=0
$$ | Let the common root be $x_{0}$. Then
$$
\begin{array}{l}
x_{0}^{2}-p x_{0}+2 q=0, x_{0}^{2}-q x_{0}+2 p=0 \\
\Rightarrow x_{0}=-2 \Rightarrow p+q=-2 .
\end{array}
$$
Given $p+q=-\frac{m}{4}$, then $m=-8$.
From the equation $4 x^{2}+8 x+n=0$, we know
$$
\Delta=64-16 n>0 \text {. }
$$
Thus, $n<4$, and $n$ is a positive... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,903 |
Example 5 Let the function
$$
f(x)=a x^{2}+x-a
$$
be defined on the interval $[-1,1]$. If $|a| \leqslant 1$, prove:
$$
|f(x)| \leqslant \frac{5}{4} \text {. }
$$ | Let $g(a)=\left(x^{2}-1\right) a+x$.
(1) When $x^{2}-1=0$, i.e., $x= \pm 1$, we have $g(a)= \pm 1$.
Obviously, $|f(x)|=|g(a)| \leqslant \frac{5}{4}$ holds.
(2) When $x^{2}-1<0$, i.e., $-1<x<1$, the linear function $g(a)$ is a decreasing function on the closed interval $[-1,1]$, then
$$
\begin{array}{l}
g(-1)_{\max }=-x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,904 |
Question In the sequence $\left\{a_{n}\right\}$, it is known that $a_{1} \in(1,2)$, $a_{n+1}=a_{n}^{3}-3 a_{n}^{2}+3 a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Prove: $\sum_{k=1}^{n}\left(a_{k}-a_{k+1}\right)\left(a_{k+2}-1\right)<\frac{1}{4}$. | Let $f(x)=x^{3}-3 x^{2}+3 x$.
It is easy to see that $f^{\prime}(x)=3 x^{2}-6 x+3 \geqslant 0$.
Therefore, the function $f(x)$ is monotonically increasing on the interval $(1,2)$.
Also, $f(1)=1, f(2)=2, a_{1} \in(1,2)$, and $a_{n+1}=f\left(a_{n}\right)$. By mathematical induction, we get $a_{n} \in(1,2)\left(n \in \mat... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,905 |
In $\triangle A B C$, let the angle bisectors of $\angle A$ and $\angle C$ intersect at point $I$, and intersect $C B$ and $A B$ at points $A_{1}$ and $C_{1}$, respectively, and intersect the circumcircle of $\triangle A B C$ at points $A_{2}$ and $C_{2}$. Let $K$ be the intersection of $A_{1} C_{2}$ and $A_{2} C_{1}$,... | Extend $I K$ to intersect $B C$ at point $N$. Connect $A_{2} N$ to intersect the circumcircle of $\triangle A B C$ at point $P$. Connect $P C_{2}$ and $P B$. Connect $P I$ and $C_{2} A_{2}$ to intersect at point $Q$.
By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{A_{1} K}{K C_{2}} \cdot \frac{C_{2} I}{I C} \cd... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,906 |
Example 5 If the two roots of the equation
$$
x^{2}-3 x-1=0
$$
are also roots of the equation
$$
x^{4}+a x^{2}+b x+c=0
$$
then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0 | $$
\begin{array}{l}
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow... | -13 | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,907 |
1. As shown in Figure 1, in the inscribed $\triangle ABC$ within a circle, $\angle A$ is the largest angle, and points $D$ and $E$ on the arc $\overparen{BC}$ not containing point $A$ are the midpoints of arcs $\overparen{ABC}$ and $\overparen{ACB}$, respectively. Let the circle passing through points $A$ and $B$ and t... | 1. As shown in Figure 2, connect $E P, A E, B E, B P, C D$.
Let $\angle B A C, \angle A B C, \angle A C B$ be $\angle A, \angle B, \angle C$ respectively, and let $X, Y$ be any points on the extensions of $C A$ and $D A$ respectively.
From the given conditions, it is easy to see that
$$
A D = D C, A E = E B.
$$
Combi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,908 |
2. Given a prime number $p$. Let $A=\left(a_{i j}\right)$ be a $p \times p$ matrix, satisfying
$$
\left\{a_{i j} \mid 1 \leqslant i, j \leqslant p\right\}=\left\{1,2, \cdots, p^{2}\right\} .
$$
We are allowed to perform the following operation on a matrix: select a row or a column, and simultaneously add 1 or subtract... | 2. By the commutative and associative laws of addition and subtraction, operations on the same row or column can be combined and performed without considering the order of the operations.
Assume that the final result of all operations is to subtract $x_{i}$ from each number in the $i$-th row and to subtract $y_{j}$ fr... | 2(p!)^2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,909 |
3. Prove: For any real number $M>2$, there exists a strictly increasing sequence of positive integers $a_{1}, a_{2}$, $\cdots$ such that:
(1) For each positive integer $i$, $a_{i}>M^{i}$;
(2) If and only if the integer $n \neq 0$, there exist a positive integer $m$ and $b_{1}, b_{2}, \cdots, b_{m} \in\{-1,1\}$, such th... | 3. Construct the sequence of positive integers $\left\{a_{n}\right\}$ recursively as follows: take an integer $a_{1}>M^{2}$, and $a_{2}=a_{1}+1$. For $k \geqslant 2$, take the integer $a_{2 k-1}>M^{2 k}+\sum_{i=1}^{2 k-2} a_{i}, a_{2 k}=k+\sum_{i=1}^{2 k-1} a_{i}$.
We will prove that this sequence satisfies the conditi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,910 |
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