problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 5 Given a positive integer $n$ that satisfies the following condition: In any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$. ${ }^{[2]}$ | Since $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor not exceeding 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17$,
$19,23,29,31,37,41,43$.
On one hand,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,021 |
10. (15 points) For a positive integer $n$, denote
$$
n!=1 \times 2 \times \cdots \times n \text {. }
$$
Find all positive integer tuples $(a, b, c, d, e, f)$ such that
$$
a!=b!+c!+d!+e!+f!,
$$
and $\quad a>b \geqslant c \geqslant d \geqslant e \geqslant f$. | 10. From the problem, we know that $a! \geqslant 5 \cdot f! \geqslant 5$. Therefore, $a \geqslant 3$.
From $a > b$, we get $a! \geqslant a \cdot b!$.
Combining this with the problem statement, we have $a \cdot b! \leqslant a! \leqslant 5 \cdot b!$.
Thus, $a \leqslant 5$. Therefore, $3 \leqslant a \leqslant 5$.
When $a... | (3,2,1,1,1,1),(5,4,4,4,4,4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,022 |
11. (20 points) (1) Prove: There exist integers $x, y$ satisfying $x^{2}+4 x y+y^{2}=2022$.
(2) Question: Do there exist integers $x, y$ satisfying
$$
x^{2}+4 x y+y^{2}=2011 ?
$$
Prove your conclusion. | 11. (1) $(x, y)=(43,1)$ satisfies the equation
$$
x^{2}+4 x y+y^{2}=2022 \text {. }
$$
(2) The answer is certain.
If there exist integers $x, y$ satisfying
$$
x^{2}+4 x y+y^{2}=2011 \text {, }
$$
then $(x+y)^{2}+2 x y=2011$.
Thus, $(x+y)^{2}$ is odd, and therefore, $x+y$ is odd. Hence, $x, y$ are one odd and one even... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,023 |
12. (20 points) For every integer $n$ greater than 1, let its all different prime factors be $p_{1}, p_{2}, \cdots, p_{k}$. For each $p_{i}(1 \leqslant i \leqslant k)$, there exists a positive integer $a_{i}$ such that $p_{i}^{a_{i}} \leqslant n$;
$$
(2) Prove: there exist infinitely many positive integers $n$ such th... | 12. (1) Since $6=2 \times 3$, and $2^{2}6$.
Therefore, $n=6$ is a positive integer that satisfies $p(n)>n$.
(2) Let $n=2^{x} 3^{y} 5^{z} 7^{t}$, where $x, y, z, t$ are any positive integers. Let
$$
\begin{array}{l}
2^{a_{1}} \leqslant n\frac{n}{2}+\frac{n}{3}+\frac{n}{5}+\frac{n}{7} \\
=\left(\frac{1}{2}+\frac{1}{3}+\f... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,024 |
1. The quadratic trinomial
$$
x^{2}+a x+b(a 、 b \in \mathbf{N})
$$
has real roots, and $a b=2^{2011}$. Then the number of such quadratic trinomials is $\qquad$. | $-1.1341$.
It is known that the numbers $a$ and $b$ are powers of 2 with non-negative integer exponents, i.e., $a=2^{k}, b=2^{2011-k}$. Therefore, we have
$$
\begin{array}{l}
\Delta=a^{2}-4 b \geqslant 0 \\
\Rightarrow 2 k \geqslant 2013-k \Rightarrow k \geqslant \frac{2013}{3}=671 .
\end{array}
$$
But $k \leqslant 20... | 1341 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,025 |
2. In the circle $\odot O$ with radius 1, there is an acute $\triangle A B C$ inscribed, $H$ is the orthocenter of $\triangle A B C$, and the angle bisector $A L$ is perpendicular to $O H$. Then $B C=$ $\qquad$ . | 2. $\sqrt{3}$.
As shown in Figure 1, extend $AO$ to intersect $\odot O$ at point $N$, connect $AH$ and extend it to intersect $BC$ at point $H_{1}$, and connect $CN$.
In Rt $\triangle CAN$ and Rt $\triangle AH_{1}B$,
$\angle ANC=\angle ABC$.
Thus, $\angle CAN=\angle BAH_{1}$.
Since $AL$ is the angle bisector of $\tria... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,026 |
3. Given functions defined on $\mathbf{R}$
$$
f(x)=x^{2} \text { and } g(x)=2 x+2 m \text {. }
$$
If the minimum value of $F(x)=f(g(x))-g(f(x))$ is $\frac{1}{4}$, then $m=$ $\qquad$ . | 3. $-\frac{1}{4}$.
Notice,
$$
\begin{array}{l}
F(x)=f(g(x))-g(f(x)) \\
=(2 x+2 m)^{2}-\left(2 x^{2}+2 m\right) \\
=2 x^{2}+8 m x+4 m^{2}-2 m \\
=2(x+2 m)^{2}-4 m^{2}-2 m .
\end{array}
$$
Then $-4 m^{2}-2 m=\frac{1}{4} \Rightarrow m=-\frac{1}{4}$. | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,027 |
4. $\tan 37.5^{\circ}=$ | 4. $\sqrt{6}-2+\sqrt{3}-\sqrt{2}$.
As shown in Figure 2, construct an isosceles right triangle $\triangle ABC$ such that
$$
\begin{array}{l}
\angle C=90^{\circ}, \\
A C=B C=1 .
\end{array}
$$
Then $A B=\sqrt{2}$.
Construct $\angle CAD=30^{\circ}$.
Then $C D=\frac{\sqrt{3}}{3}$,
$$
A D=\frac{2 \sqrt{3}}{3} \text {. }
... | \sqrt{6}-2+\sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,028 |
5. Let $f(x)=\frac{1+x}{1-3 x}$. Define
$$
\begin{array}{l}
f_{1}(x)=f(f(x)), \\
f_{n}(x)=f\left(f_{n-1}(x)\right) \quad(n=2,3, \cdots) .
\end{array}
$$
Then $f_{2011}(2011)=$ | 5. $\frac{1005}{3017}$.
Let $f(x)=f_{0}(x)=\frac{1+x}{1-3 x}$. Then
$f_{1}(x)=f(f(x))=\frac{1+\frac{1+x}{1-3 x}}{1-3 \cdot \frac{1+x}{1-3 x}}=-\frac{1-x}{1+3 x}$,
$f_{2}(x)=f\left(f_{1}(x)\right)=\frac{1-\frac{1-x}{1+3 x}}{1+3 \cdot \frac{1-x}{1+3 x}}=x$,
$f_{3}(x)=f(x)=f_{0}(x)$.
Thus, the expression of $f_{n}(x)$ re... | \frac{1005}{3017} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,029 |
II. (15 points) Given that $D$ is a point on side $BC$ of the equilateral $\triangle ABC$, and the incenter and circumcenter of $\triangle ABD$ and $\triangle ACD$ are $I_{1}, O_{1}$ and $I_{2}, O_{2}$ respectively. Prove:
$$
I_{1} O_{1}^{2}+I_{2} O_{2}^{2}=I_{1} I_{2}^{2} \text {. }
$$ | As shown in Figure 3, perform a transformation $R\left(A, 60^{\circ}\right)$, which is a counterclockwise rotation centered at $A$ by $60^{\circ}$, such that $\triangle A B D$ is transformed to $\triangle A C D_{1}$.
Since $\angle A D C + \angle A D_{1} C$
$$
=\angle A D C + \angle A D B = 180^{\circ} \text{, }
$$
The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,030 |
Three. (15 points) Let $n$ be a positive integer, and denote
$$
n!=1 \times 2 \times \cdots \times n \text {. }
$$
Find all positive integer solutions to the equation
$$
\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{11!}\right]=2011
$$
where $[a]$ denotes the greatest integer not exceeding... | Let $\frac{x}{5!}=6 a+\frac{r_{1}}{5!}=6 a+b+\frac{r_{2}}{5!}$, where $0 \leqslant r$ $\frac{x}{6}-1 .}
\end{array}
$
Then $x+\frac{x-1}{2}+\frac{x}{6}-1<2011$
$
\begin{array}{l}
\Rightarrow \frac{5 x}{3}<2012 \frac{1}{2} \\
\Rightarrow x<1207.5 \\
\Rightarrow\left[\frac{x}{7!}\right]=\left[\frac{x}{8!}\right]=\cdots=\... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,031 |
Example 6 Given six distinct positive integers $a_{1}, a_{2}, \cdots, a_{6}\left(a_{1}<a_{2}<\cdots<a_{6}\right)$, take any three numbers from these six numbers, and denote them as $a_{i} 、 a_{j} 、 a_{k}(i<j<k)$. Let
$$
f(i, j, k)=\frac{1}{a_{i}}+\frac{2}{a_{j}}+\frac{3}{a_{k}} .
$$
Prove: There must exist three diffe... | Prove that among six positive integers, any three numbers can be chosen in $\mathrm{C}_{6}^{3}=20$ ways.
Thus, 20 different triples $(i, j, k)$ are determined.
By $0<f(i, j, k)=\frac{1}{a_{i}}+\frac{2}{a_{j}}+\frac{3}{a_{k}}$
$$
\leqslant \frac{1}{3}+\frac{2}{2}+\frac{3}{1}=\frac{13}{3},
$$
the points on the number li... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 725,032 |
Four, (15 points) On a plane, $n$ points are called a "standard $n$-point set" if among any three of these points, there are always two points whose distance is no more than 1. To ensure that a circular paper with a radius of 1 can cover at least 25 points of any standard $n$-point set, find the minimum value of $n$.
| First, prove: $n_{\text {min }}>48$.
Draw a line segment $AB$ of length 5 on the plane, and construct two circles with radii of 0.5 centered at $A$ and $B$, respectively. Take 24 points in each circle. Then there are 48 points on the plane that satisfy the problem's condition (any three points must have at least two po... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,033 |
Five. (15 points) Given a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have
$$
\begin{array}{l}
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4} . \\
\text { Find }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)]
\end{array}
$$
where $[a]$ denotes th... | $$
f(1)=\frac{1}{2} \text {. }
$$
Let $y=z=0$. Then
$$
-\frac{1}{2} f(0)+\frac{1}{2} f(0)-f(x) f(0) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(0)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \leqslant \frac{1}{2} \text {. }
$$
Now let $y=z=1$. Then
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1... | 1011030 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,034 |
2. Let $A$ and $B$ be two sets, and call $(A, B)$ a "pair". When $A \neq B$, consider $(A, B)$ and $(B, A)$ as different pairs. Then the number of different pairs $(A, B)$ that satisfy the condition
$$
A \cup B=\{1,2,3,4\}
$$
is $\qquad$ | 2. 81.
When set $A$ has no elements, i.e., $A=\varnothing$, set $B$ has 4 elements, there is 1 case; when set $A$ contains $k(k=1,2,3,4)$ elements, set $B$ contains the other $4-k$ elements besides these $k$ elements, the elements in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{k} \times 2^{k}$ cas... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,036 |
3. Let the function be
$$
f(x)=x^{2}+x+m\left(m \in \mathbf{R}_{+}\right) \text {. }
$$
If $f(t)<0$, then your judgment on the existence of zeros of the function $y=f(x)$ in the interval $(t, t+1)$ is $\qquad$ . | 3. There exists a zero point.
Because
$$
\begin{array}{l}
f(t)<0, \\
f(t+1)>0 .
\end{array}
$$
Hence,
$$
\begin{array}{l}
-10 \\
\Rightarrow f(t+1)>0 .
\end{array}
$$
Thus, the function $y=f(x)$ has a zero point in the interval $(t, t+1)$. | There\ exists\ a\ zero\ point. | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,037 |
4. Given the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ with two foci $F_{1}$ and $F_{2}$, point $P\left(x_{0}, y_{0}\right)$ satisfies $0<\frac{x_{0}^{2}}{2}+y_{0}^{2} \leqslant 1$. Then the range of $\left|P F_{1}\right|+\left|P F_{2}\right|$ is $\qquad$ | 4. $[2,2 \sqrt{2}]$.
From $0<\frac{x_{0}^{2}}{2}+y_{0}^{2} \leqslant 1$, we know that point $P\left(x_{0}, y_{0}\right)$ is inside the ellipse $C$ (including the boundary).
Therefore, $2 \leqslant\left|P F_{1}\right|+\left|P F_{2}\right| \leqslant 2 \sqrt{2}$. | [2,2 \sqrt{2}] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,038 |
5. Given the complex number $z_{1}$ satisfies
$\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i}$ ( $\mathrm{i}$ is the imaginary unit), and the imaginary part of the complex number $z_{2}$ is 2. Then the condition for $z_{1} z_{2}$ to be a real number is
$$
z_{2}=
$$
$\qquad$ | $5.4+2 \mathrm{i}$
Since $\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i}$, therefore, $z_{1}=2-\mathrm{i}$.
Let $z_{2}=a+2 \mathrm{i}(a \in \mathbf{R})$.
Then $z_{1} z_{2}=(2-\mathrm{i})(a+2 \mathrm{i})$
$$
=(2 a+2)+(4-a) \mathrm{i} \text {. }
$$
Since $z_{1} z_{2}$ is a real number, thus, $a=4$.
Hence $z_{2}=4+2 \ma... | 4+2 \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,039 |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$
a_{n+1}=2 a_{n}+2^{n}-1\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence. Then the value of $\lambda$ is $\qquad$ | 6. -1 .
Notice,
$$
\begin{array}{l}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2^{n}-1-\lambda}{2^{n+1}} . \\
\text { From } \frac{2^{n}-1-\lambda}{2^{n+1}} \text { being a constant, we know } \lambda=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,040 |
7. The function
$$
f(x)=x+\cos x-\sqrt{3} \sin x
$$
passes through a point on its graph where the slope of the tangent line is $k$. Then the range of values for $k$ is $\qquad$ . | $\begin{array}{l}\text { 7. }[-1,3] \text {. } \\ f^{\prime}(x)=1-\sin x-\sqrt{3} \cos x \\ =1-2 \sin \left(x+\frac{\pi}{3}\right) \in[-1,3]\end{array}$ | [-1,3] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,041 |
8. Given three points $A, B, C$ in a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$.
Then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=$ $\qquad$ | 8. -25 .
Given that $\overrightarrow{A B} \perp \overrightarrow{B C}$. Then
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B} \\
=\overrightarrow{C A} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=-\ove... | -25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,042 |
Example 7 Fill in each small square of a $9 \times 9$ grid with a number, with no more than four different numbers in each row and each column. What is the maximum number of different numbers that can be in this grid? ${ }^{[4]}$ | If there are 29 different numbers in this grid, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is the first row).
The remaining 25 numbers are in rows $2 \sim 9$. Similarly, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is t... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,043 |
9. A square $A B C D$ with side length 4 is folded along $B D$ to form a $60^{\circ}$ dihedral angle. Then the distance between the midpoint of $B C$ and $A$ is $\qquad$ . | 9. $2 \sqrt{2}$.
Take the midpoint $O$ of $B D$. It is easy to know that $\triangle A C O$ is an equilateral triangle with side length $2 \sqrt{2}$. Therefore, $A C=2 \sqrt{2}$.
Let the midpoint of $B C$ be $M$.
In $\triangle A C B$, we have
$$
A M=\sqrt{\frac{1}{2}\left(A C^{2}+A B^{2}\right)-\frac{1}{4} B C^{2}}=2 \... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,044 |
10. There are 8 red, 8 white, and 8 yellow chopsticks. Without looking, how many chopsticks must be taken out to ensure that at least two pairs of chopsticks are of different colors? $\qquad$ | 10. 11.
Since among 11 chopsticks there must be a pair of the same color (let's say yellow), the number of black or white chopsticks must be at least 3, among which there must be a pair of the same color, i.e., both black or both white. Therefore, 11 chopsticks ensure success. However, if only 10 chopsticks are taken,... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,045 |
11. The region containing the focus of a parabola is called the interior of the parabola. Can the interiors of 2011 parabolas, allowing for translation and rotation, cover the entire plane? Make a judgment and prove your conclusion. | II. 11. It cannot.
Since each parabola has one axis of symmetry, there are at most 2011 axes of symmetry.
Draw any line $l$ in the plane that is not parallel to any of the 2011 axes of symmetry. Thus, line $l$ intersects the 2011 parabolas at most at $2011 \times 2$ points, dividing line $l$ into a finite number of se... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,046 |
12. Let $a_{k}=\sum_{i=k^{2}}^{(k+1)^{2}-1} \frac{1}{i}$. Prove:
$$
2011 \in\left(\frac{2}{a_{2010}}, \frac{2}{a_{2011}}\right) \text {. }
$$ | 12. It is known that the expression for $a_{k}$ has a total of $2 k+1$ terms.
Considering the sum of the first $k$ terms and the sum of the last $k+1$ terms, we have
$$
\begin{array}{l}
\sum_{i=k^{2}}^{k^{2}+k-1} \frac{1}{i}>\frac{k}{k^{2}+k}=\frac{1}{k+1}, \\
\sum_{i=k^{2}}^{k^{2}+k-1} \frac{1}{i}<\frac{k}{k^{2}}=\fr... | 2011 \in\left(\frac{2}{a_{2010}}, \frac{2}{a_{2011}}\right) | Algebra | proof | Yes | Yes | cn_contest | false | 725,047 |
13. (1) Let $t>0$ be a real number. Prove:
$$
\left(1+\frac{2}{t}\right) \ln (1+t)>2 \text {. }
$$
(2) From 100 cards numbered $1 \sim 100$, each time a card is randomly drawn and then put back. This process is repeated 20 times. Let the probability that the 20 numbers drawn are all different be $P$. Prove:
$$
P<\frac{... | 13. (1) Construct the function
$$
f(x)=\ln (1+x)-\frac{2 x}{x+2} \text {. }
$$
Then \( f^{\prime}(x)=\frac{x^{2}}{(x+1)(x+2)^{2}} \).
When \( x>0 \), \( f^{\prime}(x)>0 \), so \( f(x) \) is an increasing function on \( (0,+\infty) \).
Therefore, \( f(t)>f(0) \), which means
$$
\begin{array}{l}
\ln (1+t)-\frac{2 t}{t+2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,048 |
14. Given that two rays $A X$ and $A Y$ drawn from vertex $A$ of $\triangle A B C$ intersect $B C$ at points $X$ and $Y$ respectively. Prove: $A B^{2} \cdot C Y \cdot C X=A C^{2} \cdot B X \cdot B Y$ holds if and only if $\angle B A X=\angle C A Y$. | 14. Sufficiency.
If $\angle B A X=\angle C A Y$, let
$$
\begin{array}{l}
\angle B A X=\angle C A Y=\alpha . \\
\text { Then } \frac{S_{\triangle A B X}}{S_{\triangle A C Y}}=\frac{A B \cdot A X \sin \alpha}{A C \cdot A Y \sin \alpha}=\frac{B X}{C Y} \\
\Rightarrow \frac{A B \cdot A X}{A C \cdot A Y}=\frac{B X}{C Y} .
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,049 |
1. The left and right directrices $l_{1} 、 l_{2}$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ trisect the line segment $F_{1} F_{2}$ ($F_{1} 、 F_{2}$ are the left and right foci of the hyperbola, respectively). Then the eccentricity $e$ of the hyperbola is ( ).
(A) $\frac{\sqrt{6}}{2}$
(B) $\sqrt{3}$
(... | 1. B.
From $2 c=3 \times \frac{2 a^{2}}{c} \Rightarrow e=\sqrt{3}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,050 |
2. Given the cubic function
$$
f(x)=a x^{3}+b x^{2}+c x+d(a, b, c, d \in \mathbf{R}) \text {. }
$$
Proposition $p: y=f(x)$ is a monotonic function on $\mathbf{R}$;
Proposition $q: y=f(x)$ intersects the $x$-axis at exactly one point.
Then $p$ is a ( ) condition for $q$.
(A) Sufficient but not necessary
(B) Necessary b... | 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 | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,051 |
3. Three people, A, B, and C, are playing the game "Scissors, Rock, Paper." In each round, A, B, and C simultaneously show one of the gestures: scissors, rock, or paper, and they do so independently. Let the number of people who beat A in a round be $\xi$. Then the expected value $E \xi$ of the random variable $\xi$ is... | 3. C.
Notice,
$$
\begin{array}{l}
P(\xi=0)=\frac{3 \times 4}{27}=\frac{4}{9}, \\
P(\xi=1)=\frac{3 \times 4}{27}=\frac{4}{9}, \\
P(\xi=2)=\frac{3 \times 1}{27}=\frac{1}{9} .
\end{array}
$$
Then $E \xi=\frac{4}{9} \times 0+\frac{4}{9} \times 1+\frac{1}{9} \times 2=\frac{2}{3}$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 725,052 |
4. The function
$$
f(x)=\sqrt{x-5}+\sqrt{24-3 x}
$$
has a maximum value of ( ).
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) $3 \sqrt{3}$ | 4. C.
The domain of $f(x)$ is $5 \leqslant x \leqslant 8$.
From $f^{\prime}(x)=\frac{1}{2 \sqrt{x-5}}+\frac{-3}{2 \sqrt{24-3 x}}=0$ $\Rightarrow x=\frac{23}{4}$.
Since $f(5)=3, f\left(\frac{23}{4}\right)=2 \sqrt{3}, f(8)=\sqrt{3}$,
thus, $f(x)_{\max }=f\left(\frac{23}{4}\right)=2 \sqrt{3}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,053 |
Example 1 Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ (1) | Solve As shown in Figure 1, the figure determined by inequality (1) is the quadrilateral $\square A B C D$ and its interior, enclosed by four straight lines, where,
$$
\begin{array}{l}
A(-1,4), B(1,1), \\
C(-1,-2), D(-3,1) .
\end{array}
$$
Consider the family of lines $2 x-3 y=k$, i.e.,
$$
y=\frac{2}{3} x-\frac{k}{3} ... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,054 |
5. Given the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, and the sequence $\left\{b_{n}\right\}$ satisfies
$$
b_{1}=a_{1}, b_{2}=a_{2}+a_{3}, b_{3}=a_{4}+a_{5}+a_{6}, \cdots .
$$
If $\lim _{n \rightarrow \infty} \frac{b_{n}}{n^{3}}=2$, then the common difference $d$ of the sequence $\left\{a_{n}\right\}... | 5. D.
Notice,
$$
\begin{array}{l}
b_{n}=a \frac{n(n-1)}{2}+1+a \frac{n(n-1)}{2}+2+\cdots+a \frac{n(n-1)}{2}+n \\
=\frac{n}{2}\left[a_{\frac{n(n-1)}{2}}^{2}+1+a_{\frac{n(n-1)}{2}}^{2}+n\right. \\
=\frac{n}{2}\left\{a_{1}+\frac{n(n-1)}{2} d+a_{1}+\left[\frac{n(n-1)}{2}+n-1\right] d\right\} \\
=\frac{n}{2}\left(2 a_{1}-d... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,055 |
6. As shown in Figure 1, the squares $A B C D$ and $A B E F$ with side length 2 are in planes that form a $60^{\circ}$ angle, and points $M$ and $N$ are on segments $A C$ and $B F$ respectively, with $A M=F N$. The range of the length of segment $M N$ is ( ).
(A) $\left[\frac{1}{2}, 2\right]$
(B) $[1,2]$
(C) $[\sqrt{2}... | 6. B.
As shown in Figure 2, draw $M H / / B C$ intersecting $A B$ at point $H$. Then
$$
\frac{A M}{A C}=\frac{A H}{A B} .
$$
Since $A M=F N$,
$$
A C=F B,
$$
thus, $\frac{F N}{F B}=\frac{A H}{A B} \Rightarrow N H / / A F$.
Therefore, $N H \perp A B, M H \perp A B$.
Hence, $\angle M H N=60^{\circ}$.
Let $A H=x(0 \leqs... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,056 |
7. Given that the real number $x$ satisfies
$$
|2 x+1|+|2 x-5|=6 \text {. }
$$
Then the range of values for $x$ is $\qquad$ | $$
\text { II.7. }\left[-\frac{1}{2}, \frac{5}{2}\right] \text {. }
$$
Notice that,
$$
\begin{array}{l}
|2 x+1|+|2 x-5| \\
\geqslant|(2 x+1)+(5-2 x)|=6 .
\end{array}
$$
The equality holds if and only if $(2 x+1)(2 x-5) \leqslant 0$, that is,
$$
-\frac{1}{2} \leqslant x \leqslant \frac{5}{2}
$$ | \left[-\frac{1}{2}, \frac{5}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,057 |
8. Let two non-zero vectors $a$ and $b$ in the plane be perpendicular to each other, and $|b|=1$. Then the sum of all real numbers $m$ such that the vectors $a+m b$ and $a+(1-m) b$ are perpendicular is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output th... | 8. 1.
Notice,
$$
\begin{aligned}
0 & =(a+m b) \cdot[a+(1-m) b] \\
& =a^{2}+a \cdot b+m(1-m) b^{2} \\
& =|a|^{2}+m(1-m),
\end{aligned}
$$
i.e., $m^{2}-m-|a|^{2}=0$.
By the relationship between roots and coefficients, the sum of all real $m$ that satisfy the condition is 1. | null | Other | math-word-problem | Yes | Yes | cn_contest | false | 725,058 |
9. Let the sum of the first $n$ terms of the real geometric sequence $\left\{a_{n}\right\}$ be
$S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}=$ $\qquad$ . | 9. 150 .
$$
\begin{array}{l}
\text { Let } b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, \\
b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30} .
\end{array}
$$
Let $q$ be the common ratio of $\left\{a_{n}\right\}$. Then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with common ratio $r=q^{10}$. Therefore,
$$
\begin{array}{l}
70=S... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,059 |
10. Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than the real number $x$ (for example, $\lceil 3.2 \rceil = 4, \lceil -\pi \rceil = -3$). Then, the sum of all real roots of the equation
$$
\lceil 3 x+1\rceil=2 x-\frac{1}{2}
$$
is equal to | 10. -4 .
Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$. Then
$$
x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4} \text {. }
$$
Thus, the original equation is equivalent to
$$
\begin{array}{l}
{\left[\frac{2 k+3}{4}\right]=-1 \Rightarrow-2<\frac{2 k+3}{4} \leqslant-1} \\
\Rightarrow-\frac{11}{2}<k \leqslant-\frac{7}{2} \\
\Right... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,060 |
11. Given
$$
(1+\sqrt{3})^{n}=a_{n}+b_{n} \sqrt{3} \text {, }
$$
where $a_{n}$ and $b_{n}$ are integers. Then $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}=$ $\qquad$ . | 11. $\sqrt{3}$.
From the given, we know
$$
(1-\sqrt{3})^{n}=a_{n}-b_{n} \sqrt{3} \text {. }
$$
Then $a_{n}=\frac{1}{2}\left[(1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}\right]$,
$$
b_{n}=\frac{1}{2 \sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right] \text {. }
$$
Therefore, $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,061 |
12. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, and $SA=SB=SC=AB=2$. Suppose points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$ | 12. $\frac{\sqrt{3}}{3}$.
As shown in Figure 3.
Since $S A=S B=S C$, the projection of point $S$ on the plane $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $H$ of $A B$.
Similarly, the projection of point $O$ on the plane $A B C$ is also $H$. Therefore, we only need to find the length of $O ... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,062 |
13. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$. | Three, 13. Let $t=\sqrt{100-m x}$. Then $x=\frac{100-t^{2}}{m}$.
Hence $y=\frac{100-t^{2}}{m}+t$
$$
=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} \text {. }
$$
When $t=\frac{m}{2}$, $y$ has a maximum value $\frac{100}{m}+\frac{m}{4}$, that is,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,063 |
14. Given the function
$$
f(x)=2\left(\sin ^{4} x+\cos ^{4} x\right)+m(\sin x+\cos x)
$$
has a maximum value of 5 for $x \in\left[0, \frac{\pi}{2}\right]$. Find the value of the real number $m$. | 14. Notice,
$$
f(x)=2-(2 \sin x \cdot \cos x)^{2}+m(\sin x+\cos x)^{4} \text {. }
$$
Let $t=\sin x+\cos x$
$$
=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in[1, \sqrt{2}] \text {. }
$$
Thus, $2 \sin x \cdot \cos x=t^{2}-1$.
Therefore, $f(x)=2-\left(t^{2}-1\right)^{2}+m t^{4}$
$=(m-1) t^{4}+2 t^{2}+1$.
Let $u=t^{2} \i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,064 |
Example 2 Bivariate function
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}-2 x+10}+\sqrt{y^{2}-4 y+5}(x \leqslant 1)
\end{array}
$$
The minimum value of the function is $\qquad$ | Solve As shown in Figure 2, let $A(x, 0)(x \leqslant-1), B(0, y)$, $C(1,-3), D(1,2)$. Then
$$
\begin{array}{l}
f(x, y)=|A B|+|A C|+|B D| \\
=(|A B|+|B D|)+|A C| \\
\geqslant|A D|+|A C|,
\end{array}
$$
Equality holds if and only if points $A$, $B$, and $D$ are collinear, i.e.,
$$
\frac{y}{0-x}=\frac{2}{1-x} \Rightarrow... | 2 \sqrt{2}+\sqrt{13} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,065 |
15. Given the parabola $y=x^{2}$ and the line $l$ passing through the point $P(-1,-1)$ intersects the parabola at points $P_{1}$ and $P_{2}$. Find:
(1) The range of the slope $k$ of the line $l$;
(2) The locus of point $Q$ on the segment $P_{1} P_{2}$ that satisfies the condition
$$
\frac{1}{P P_{1}}+\frac{1}{P P_{2}}=... | 15. (1) The line $l: y+1=k(x+1)$ intersects with the parabola equation $y=x^{2}$. By eliminating $y$ and rearranging, we get
$$
x^{2}-k x-(k-1)=0 \text {. }
$$
From $\Delta=(-k)^{2}+4(k-1)>0$, solving gives
$k>-2+2 \sqrt{2}$ or $k<-2-2 \sqrt{2}$.
\end{array}
$$
Thus, $x_{1}+1, x_{2}+1, x+1$ have the same sign.
$$
\be... | 2 x-y+1=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,066 |
16. Given that $m$ is a real number, the sequence $\left\{a_{n}\right\}$ has the sum of its first $n$ terms as $S_{n}$, satisfying
$$
S_{n}=\frac{9}{8} a_{n}-\frac{4}{3} \times 3^{n}+m \text {, and } a_{n} \geqslant \frac{64}{3}
$$
for any positive integer $n$. Prove: when $m$ takes its maximum value, for any positive... | 16. When $n=1$, from $a_{1}=\frac{9}{8} a_{1}-4+m$, we get $a_{1}=8(4-m)$.
When $n \geqslant 1$,
$$
\begin{array}{l}
S_{n}=\frac{9}{8} a_{n}-\frac{4}{3} \times 3^{n}+m, \\
S_{n+1}=\frac{9}{8} a_{n+1}-\frac{4}{3} \times 3^{n+1}+m .
\end{array}
$$
Then $a_{n+1}=\frac{9}{8} a_{n+1}-\frac{9}{8} a_{n}-\frac{8}{3} \times 3^... | \frac{3}{16} | Algebra | proof | Yes | Yes | cn_contest | false | 725,067 |
1. Equation
$$
\begin{array}{l}
\left(x^{2}+2011 x-2012\right)^{2}+\left(2 x^{2}-2015 x+2014\right)^{2} \\
=\left(3 x^{2}-4 x+2\right)^{2}
\end{array}
$$
The sum of all real roots of the equation is ( ).
(A) $\frac{2007}{2}$
(B) $-\frac{2007}{2}$
(C) 4
(D) -4 | -、1. B.
Notice,
$$
\begin{array}{l}
\left(x^{2}+2011 x-2012\right)+\left(2 x^{2}-2015 x+2014\right) \\
=3 x^{2}-4 x+2 \text {. } \\
\text { Let } x^{2}+2011 x-2012=y, \\
2 x^{2}-2015 x+2014=z .
\end{array}
$$
Then $3 x^{2}-4 x+2=y+z$.
So $y^{2}+z^{2}=(y+z)^{2}$
$$
\begin{array}{l}
\Rightarrow y=0 \text { or } z=0 \\
\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,068 |
2. Let $\alpha$ be an acute angle. If
$$
\begin{array}{l}
M=\frac{\sin ^{4} \alpha+\cos ^{4} \alpha}{\sin ^{6} \alpha+\cos ^{6} \alpha}, \\
N=\frac{\sin ^{4} \alpha+\cos ^{4} \alpha}{\sqrt{\sin ^{4} \alpha+\cos ^{4} \alpha}}, \\
P=\frac{\sqrt{\sin ^{4} \alpha+\cos ^{4} \alpha}}{\sin ^{6} \alpha+\cos ^{6} \alpha},
\end{... | 2. D.
Let $\alpha=45^{\circ}$. Then
$$
M=2, N=\frac{\sqrt{2}}{2}, P=2 \sqrt{2} \text {. }
$$
Thus $N<M<P$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,069 |
3. As shown in Figure 1, points $B, C, D$ are on the same straight line, and $A B=B C=C A, C D=D E=E C, A D$ intersects $B E$ at point $F$. If $B C: C D=1: 2$, then $B F: D F=$ ( ).
(A) $1: 2$
(B) $2: 3$
(C) $3: 4$
(D) None of the above | $$
\begin{aligned}
\triangle A C D \cong \triangle B C E \\
\Rightarrow A D=B E \\
\Rightarrow C G=C H \\
\Rightarrow F C \text { bisects } \angle D F B \\
\Rightarrow B F: D F=B C: C D=1: 2 . \\
\quad \text { 3. A. }
\end{aligned}
$$
As shown in Figure 6, connect $C F$, draw $C G \perp A D$ at point $G$, $C H \perp B... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,070 |
4. As shown in Figure 2, in the right trapezoid $M N P Q$, it is known that $\angle M N P=90^{\circ}, P M \perp N Q$. If $\frac{N Q}{P M}=\frac{\sqrt{2}}{2}$, then $\frac{M Q}{N P}$ $=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{1}{4}$
(D) $\frac{2}{3}$ | 4. A.
It is easy to prove $\triangle M Q N \backsim \triangle N M P$
$$
\begin{array}{l}
\Rightarrow \frac{M Q}{N M}=\frac{N Q}{P M}=\frac{M N}{N P} \\
\Rightarrow \frac{M Q}{N P}=\left(\frac{N Q}{P M}\right)^{2}=\frac{1}{2}
\end{array}
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,071 |
5. Given that $D$ is a point on the hypotenuse $AB$ of right $\triangle ABC$, $DE \perp BC$ at point $E$, and $BE = AC$. If $BD = 1$, $DE + BC = 2$, then $\tan B = (\quad)$.
(A) $\frac{\sqrt{3}}{3}$
(B) 1
(C) $\sqrt{3}$
(D) None of the above | 5. A.
As shown in Figure 7, let $D E=x$.
Then $B C=2-x$,
$$
B E=A C=\sqrt{1-x^{2}} \text {. }
$$
By the given condition,
$$
\frac{D E}{A C}=\frac{B E}{B C} \text {, }
$$
which means $\frac{x}{\sqrt{1-x^{2}}}=\frac{\sqrt{1-x^{2}}}{2-x}$.
Solving this, we get $x=\frac{1}{2}$. Therefore, $D E=\frac{1}{2} B D \Rightarro... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,072 |
6. As shown in Figure 3, given that $AB$ and $CD$ are two perpendicular diameters of $\odot O$, $E$ is a point on segment $OB$, and $BE=2OE$. The extension of segment $CE$ intersects $\odot O$ at point $F$, and segment $AF$ intersects $DO$ at point $G$. Then
$$
DG: GC=(\quad) .
$$
(A) $1: 2$
(B) $1: 3$
(C) $2: 3$
(D) $... | 6. B.
As shown in Figure 8, let the radius of $\odot O$ be 3. Then
$$
O E=1, C E=\sqrt{10} .
$$
Connect $D F$. Then
$$
\begin{array}{l}
\angle C F D=90^{\circ} \\
\Rightarrow \triangle E O C \sim \triangle D F C \\
\Rightarrow \frac{E C}{D C}=\frac{O C}{C F} \\
\Rightarrow C F=\frac{9 \sqrt{10}}{5} \\
\Rightarrow E F... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 725,073 |
$$
\begin{aligned}
M= & |2012 x-1|+|2012 x-2|+\cdots+ \\
& |2012 x-2012|
\end{aligned}
$$
The minimum value of the algebraic expression is . $\qquad$ | 2. 1012036 .
By the geometric meaning of absolute value, we know that when
$$
1006 \leqslant 2012 x \leqslant 1007
$$
$M$ has the minimum value.
$$
\begin{array}{l}
\text { Then } M_{\text {min }}=(-1-2-\cdots-1006)+ \\
(1007+1008+\cdots+2012) \\
=1006 \times 1006=1012036 .
\end{array}
$$ | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,075 |
3. If real numbers $m, n, p, q$ satisfy the conditions
$$
\begin{array}{l}
m+n+p+q=22, \\
m p=n q=100,
\end{array}
$$
then the value of $\sqrt{(m+n)(n+p)(p+q)(q+m)}$ is
$\qquad$ | 3. 220 .
From the given, we have
$$
\begin{aligned}
& (m+n)(n+p)(p+q)(q+m) \\
= & {[(m+n)(p+q)][(n+p)(q+m)] } \\
= & (200+m q+n p)(200+m n+p q) \\
= & 200^{2}+100\left(m^{2}+n^{2}+p^{2}+q^{2}+2 m n+\right. \\
& 2 m q+2 n p+2 p q) \\
= & 200^{2}+100\left[(m+n+p+q)^{2}-400\right] \\
= & {[10(m+n+p+q)]^{2} . }
\end{align... | 220 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,077 |
4. Let the function be
$$
f(x)=a x^{2}+4 x+b \text{, }
$$
the equation $f(x)=x$ has two real roots $\beta_{1}$ and $\beta_{2}$. If $a$ and $b$ are both negative integers, and $\left|\beta_{1}-\beta_{2}\right|=1$, then the coordinates of the vertex of the graph of the function $f(x)$ are $\qquad$ | 4. $(2,2)$.
$f(x)=x$ can be transformed into $a x^{2}+3 x+b=0$.
From the given conditions, we have
$$
\begin{array}{l}
\beta_{1}+\beta_{2}=-\frac{3}{a}, \beta_{1} \beta_{2}=\frac{b}{a} \\
\Rightarrow\left|\beta_{1}-\beta_{2}\right|=\sqrt{\left(\beta_{1}+\beta_{2}\right)^{2}-4 \beta_{1} \beta_{2}} \\
\quad=\sqrt{\frac{9... | (2,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,078 |
One, (20 points) If the two sides of a right triangle, $x, y$, are both prime numbers, and make the algebraic expressions $\frac{2 x-1}{y}$ and $\frac{2 y+3}{x}$ both positive integers, find the inradius $r$ of this right triangle. | (1) If $x>y$, then
$$
1 \leqslant \frac{2 y+3}{x}<\frac{2 x+3}{x}<4 \text {. }
$$
It is easy to see that $\frac{2 y+3}{x}=1$ or 2.
$$
\begin{array}{l}
\text { (i) From } \frac{2 y+3}{x}=1 \\
\Rightarrow x=2 y+3 \\
\Rightarrow \frac{2 x-1}{y}=\frac{2(2 y+3)-1}{y}=4+\frac{5}{y} \\
\Rightarrow y=5, x=13 ;
\end{array}
$$
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,079 |
II. (25 points) As shown in Figure 5, the circumcenter of $\triangle ABC$ is $O$, and the orthocenter is $H$, with $\angle ACB = 60^{\circ}$. Does there exist a point $M$ on the arc $\overparen{AB}$ that does not contain point $C$, such that $CM \perp OH$? If it exists, please specify the position of $M$ and prove that... | II. Existence.
$M$ is the midpoint of arc $\overparen{A B}$, such that $C M \perp O H$.
The reason is as follows:
As shown in Figure 9, let the radius of $\odot O$ be $R$. Connect $O A$ and $O M$. By the corollary of the perpendicular diameter theorem, we get $A B \perp O M$.
Extend $A O$ to intersect $\odot O$ at poin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,080 |
Three. (25 points) There are five numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that satisfy the following conditions:
(1) One of the numbers is $\frac{1}{2}$;
(2) From these five numbers, any two numbers taken, there must exist one number among the remaining three such that the sum of this number and the two taken numbe... | Let's assume $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$. From condition (2), taking $a_{1} 、 a_{2}$, there must exist $a_{i}$ ( $3,4,5$) such that $a_{1}+a_{2}+a_{i}=1$; taking $a_{4} 、 a_{5}$, there must exist $a_{j}(j=1,2,3)$ such that $a_{j}+a_{4}+a_{5}=1$.
$$
\begin{array}{l}
\text { He... | \frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 725,081 |
1. Given the function
$$
f(x)=\frac{1}{4}\left(x^{2}+\ln x\right) \text {. }
$$
When $x \in\left(0, \frac{\pi}{4}\right)$, the size relationship between $\mathrm{e}^{\cos 2 x}$ and $\tan x$ is | $$
\text { - 1. } \mathrm{e}^{\cos 2 x}>\tan x \text {. }
$$
Obviously, $f(x)=x^{2}+\ln x$ is an increasing function on $(0,+\infty)$.
Since when $x \in\left(0, \frac{\pi}{4}\right)$, $\cos x>\sin x$, therefore,
$$
\begin{array}{l}
f(\cos x)>f(\sin x) \\
\Rightarrow \cos ^{2} x+\ln \cos x>\sin ^{2} x+\ln \sin x \\
\Ri... | \mathrm{e}^{\cos 2 x}>\tan x | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,082 |
3. The equation of the line passing through the intersection points of the parabolas
$$
y=2 x^{2}-2 x-1 \text { and } y=-5 x^{2}+2 x+3
$$
is $\qquad$ . | 3. $6 x+7 y-1=0$.
Solve the system of equations
$$
\left\{\begin{array}{l}
y=2 x^{2}-2 x-1, \\
y=-5 x^{2}+2 x+3 .
\end{array}\right.
$$
$5 \times$ (1) $+2 \times$ (2) and simplify to get
$$
6 x+7 y-1=0
$$ | 6 x+7 y-1=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,084 |
4. If the function
$$
f(x)=a x+\sin x
$$
has perpendicular tangents on its graph, then the real number $a$ is
$\qquad$ . | 4.0.
Notice that, $f^{\prime}(x)=a+\cos x$.
If the function $f(x)$ has two perpendicular tangents, then there exist $x_{1}, x_{2} \in \mathbf{R}$, such that
$$
\begin{array}{l}
f^{\prime}\left(x_{1}\right) f^{\prime}\left(x_{2}\right)=-1 \\
\Leftrightarrow\left(a+\cos x_{1}\right)\left(a+\cos x_{2}\right)=-1 \\
\Leftr... | 0 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 725,085 |
5. The terms of the sequence $\left\{a_{n}\right\}$ are all positive, and the sum of the first $n$ terms $S_{n}$ satisfies
$$
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) .
$$
Then $a_{n}=$ | 5. $\sqrt{n}-\sqrt{n-1}$.
From $a_{1}=S_{1}=\frac{1}{2}\left(a_{1}+\frac{1}{a_{1}}\right)$, we get $a_{1}=S_{1}=1$.
When $n>1$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) \\
\Rightarrow S_{n-1}+a_{n}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) \\
\Rightarrow S_{n-1}=\frac{1}{2}\l... | a_{n}=\sqrt{n}-\sqrt{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,086 |
Example 1 The number of integer solutions to the equation $\frac{x+3}{x+1}-y=0$ is ( ) groups.
(A) 1
(B) 2
(C) 3
(D) 4
(2004, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | From $\frac{x+3}{x+1}-y=0$, we know
$$
y=\frac{x+3}{x+1}=1+\frac{2}{x+1} \text{.}
$$
Since $x, y$ are both integers, therefore, $x+1= \pm 1$ or $\pm 2$.
Thus, $(x, y)$
$$
=(-2,-1),(0,3),(1,2),(-3,0) \text{.}
$$
Therefore, the original equation has 4 integer solutions.
Hence, the answer is D. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 725,087 |
Example 2 The equation $2 x^{2}+5 x y+2 y^{2}=2007$ has $\qquad$ different integer solutions.
(2007, National Junior High School Mathematics League Sichuan Preliminary Competition) | The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2007 \text {. }
$$
Since $x$ and $y$ are integers, without loss of generality, assume $x \leqslant y$, so,
$$
2 x+y \leqslant x+2 y \text {. }
$$
Notice that, $31[2 x+y)+(x+2 y)]$.
Thus, from equation (1), we get the system of equations
$$
\left\{\begin{... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,088 |
Example 3 Let positive integers $x, y, p, n, k$ satisfy
$$
x^{n}+y^{n}=p^{k} \text {. }
$$
Prove: If $n$ is an odd number greater than 1, and $p$ is an odd prime, then $n$ can be expressed as a power of $p$ with a natural number as the exponent.
(22nd Russian Mathematical Olympiad (9th grade)) | Prove that since $n$ is odd, then
$$
(x+y) \mid\left(x^{n}+y^{n}\right) \text {. }
$$
From the equation, we know $x+y=p^{r}$ (positive integer $r \leqslant k$). Let $p^{s} \| n$. By the lemma, we have
$$
p^{s+r} \|\left[x^{n}-(-y)^{n}\right]=x^{n}+y^{n}=p^{k} \text {. }
$$
Therefore, $s=k-r$.
If $n=p^{s} q(q>1,(p, q)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,089 |
4. As shown in Figure 3, in $\triangle A B C$, it is known that $\angle B A C=60^{\circ}, A B$ $=2 A C$, point $P$ is inside $\triangle A B C$, and $P A=$ $\sqrt{3}, P B=5, P C=2$. Then the degree of $\angle A P C$ is $\qquad$ and the area of $\triangle A B C$ is . $\qquad$ | 4. $120^{\circ}, 3+\frac{7 \sqrt{3}}{2}$.
As shown in Figure 5, take the midpoint $D$ of $AB$, and the midpoint $F$ of $AP$. Rotate $\triangle APC$ clockwise by $60^{\circ}$ to $\triangle AED$, and connect $EF$, $DF$, and $PE$. Then $\triangle AEP$ is an equilateral triangle, $DE = PC = 2$, $EF = 1.5$, and $DF = \frac... | 120^{\circ}, 3+\frac{7 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,090 |
One. (20 points) As shown in Figure 4, in $\triangle ABC$, $\angle C = 90^{\circ}, AC = BC, O$ is a point inside $\triangle ABC$, and the distances from point $O$ to the sides of $\triangle ABC$ are all equal to 1. $\triangle ABC$ is rotated $45^{\circ}$ clockwise around point $O$ to get $\triangle A_{1} B_{1} C_{1}$, ... | (1) Connect $O C, O C_{1}$, intersecting $Q P, P N$ at points $D, E$ respectively.
According to the problem, $\angle C O C_{1}=45^{\circ}$.
Since the distances from point $O$ to $A C, B C$ are both equal to 1, we know that $O C$ is the angle bisector of $\angle A C B$.
Since $\angle A C B=90^{\circ}$, therefore,
$\angl... | 4 \sqrt{2} - 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,091 |
II. (25 points) Let the quadratic function
$$
y=x^{2}+p x+q
$$
pass through the point $(2,-1)$, and intersect the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Let $M$ be the vertex of the quadratic function. Find the analytical expression of the quadratic function that minimizes th... | II. Since $-1=2^{2}+2 p+q$, we have
$$
2 p+q=-5 \text {. }
$$
Given that $x_{1}$ and $x_{2}$ are the roots of $x^{2}+p x+q=0$, we get
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=q \text {. }
$$
Then $|A B|=\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{p^{2}-4 q} .
\end{... | y=x^{2}-4 x+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,092 |
Three. (25 points) Select $k$ numbers from 1 to 2012, such that among the selected $k$ numbers, there are definitely three numbers that can form the lengths of the sides of a triangle (the lengths of the three sides of the triangle must be distinct). What is the minimum value of $k$ that satisfies the condition? | Three, the problem is equivalent to:
Select $k-1$ numbers from $1,2, \cdots, 2012$, such that no three of these numbers can form the sides of a triangle with unequal sides. What is the maximum value of $k$ that satisfies this condition?
Now consider the arrays that meet the above conditions.
When $k=4$, the smallest th... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,093 |
1. Given a real number $a>0$, and $a \neq 1$, the area of square $A B C D$ is 36, where $A B / / x$-axis, and points $A$, $B$, and $C$ are on the graphs of the functions
$$
y=\log _{a} x, y=2 \log _{a} x, y=3 \log _{a} x
$$
respectively. Then $a=$ $\qquad$ . | $-1.3^{ \pm^{\frac{1}{6}}}$.
When $a>1$, let $B\left(x_{0}, 2 \log _{a} x_{0}\right)\left(x_{0}>0\right)$. Then $A\left(x_{0}+6, \log _{a}\left(x_{0}+6\right)\right), C\left(x_{0}, 3 \log _{a} x_{0}\right)$. It is known that the side length of the square $A B C D$ is 6.
From $A B / / x$-axis, we have
$$
\left\{\begin{a... | 3^{\pm\frac{1}{6}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,094 |
2. In $\triangle A B C$, it is known that $A B=2, B C=4$, $\angle A B C=120^{\circ}$, and a point $P$ outside the plane satisfies $P A=P B$ $=P C=4$. Then the volume of the tetrahedron $P-A B C$ is equal to | 2. $\frac{4 \sqrt{5}}{3}$.
In $\triangle A B C$, by the cosine rule, we get $A C=2 \sqrt{7}$.
Let the circumradius of $\triangle A B C$ be $R$. Then, by the sine rule, we have
$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{2 \sqrt{21}}{3} .
$$
Draw $P O \perp$ plane $A B C$, with the foot of the perpendicular being $O$.
... | \frac{4 \sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,095 |
3. Given sets
$$
\begin{array}{c}
A=\left\{x \mid 2 x^{3}+5 x^{2}+x-2>0\right\}, \\
B=\left\{x \mid x^{2}-a x+b \leqslant 0\right\} .
\end{array}
$$
If sets $A$ and $B$ satisfy
$$
A \cap B=\left\{x \left\lvert\, \frac{1}{2}<x \leqslant 3\right.\right\},
$$
then the trajectory equation of point $(a, b)$ in the rectang... | 3. $b=3 a-9\left(2 \leqslant a \leqslant \frac{7}{2}\right)$.
From $2 x^{3}+5 x^{2}+x-2>0$
$\Rightarrow(x+1)(x+2)(2 x-1)>0$
$\Rightarrow-2<x<-1$ or $x>\frac{1}{2}$
$\Rightarrow A=(-2,1) \cup\left(\frac{1}{2},+\infty\right)$.
Let $B=\{x \mid \alpha \leqslant x \leqslant \beta\}$.
Given $A \cap B=\left\{x \left\lvert\, ... | b=3 a-9\left(2 \leqslant a \leqslant \frac{7}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,096 |
4. Given a positive integer $n$ less than 2011. Then
$$
\sum_{i=1}^{2010}\left[\frac{2 i n}{2011}\right]=
$$
$\qquad$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 4. $1005(2 n-1)$
Notice,
$$
\begin{array}{l}
\sum_{i=1}^{2010}\left[\frac{2 i n}{2011}\right] \\
=\frac{1}{2} \sum_{i=1}^{2010}\left(\left[\frac{2 i n}{2011}\right]+\left[\frac{2(2011-i) n}{2011}\right]\right) \\
=\frac{1}{2} \times 2010 \times(2 n-1) \\
=1005(2 n-1) .
\end{array}
$$ | 1005(2 n-1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,097 |
5. In the Cartesian coordinate system, a circle with center at $(1,0)$ and radius $r$ intersects the parabola $y^{2}=x$ at four points $A, B, C, D$. If the intersection point $F$ of $A C$ and $B D$ is exactly the focus of the parabola, then $r=$ $\qquad$ | 5. $\frac{\sqrt{15}}{4}$.
Combining the equations of the circle and the parabola, we get
$$
\left\{\begin{array}{l}
y^{2}=x, \\
(x-1)^{2}+y^{2}=r^{2} .
\end{array}\right.
$$
Eliminating $y$ yields $x^{2}-x+1-r^{2}=0$.
From $\Delta=1-4\left(1-r^{2}\right)>0$, we solve to get $r>\frac{\sqrt{3}}{2}$.
According to the pr... | \frac{\sqrt{15}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,098 |
6. A five-digit number $\overline{a b c d e}$ satisfies
$$
ac>d, dd, b>e
$$
(for example, 37201, 45412), if its digits change with the position in a manner similar to the monotonicity of a sine function over one period, then this five-digit number is said to conform to the "sine rule". Therefore, there are $\qquad$ fiv... | 6.2892 .
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers among $a, b, c, d, e$. It is easy to see that $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$. In this case, there are $10-k$ ways to choose $(b, d)$, and $a, c, e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ g... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,099 |
7. Given that the circumcenter of $\triangle A B C$ is $O$, and
$$
\overrightarrow{A O} \cdot \overrightarrow{B C}+2 \overrightarrow{B O} \cdot \overrightarrow{C A}+3 \overrightarrow{C O} \cdot \overrightarrow{A B}=0
$$
then the minimum value of $\cot A+\cot C$ is $\qquad$ | 7. $\frac{2 \sqrt{3}}{3}$.
Let $B C=a, C A=b, A B=c$. Draw $O D \perp B C$, with the foot of the perpendicular being $D$. It is easy to see that $D$ is the midpoint of side $B C$, so we have
$$
\overrightarrow{A D}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \text {. }
$$
Then $\overrightarrow{A O} \cdot \... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,101 |
8. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{2}=1, \\
a_{n+2}=\frac{n(n+1) a_{n+1}+n^{2} a_{n}+5}{n+2}-2\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then the general term formula of $\left\{a_{n}\right\}$ is $a_{n}=$ | 8. $\frac{(n-1)!+1}{n}$.
Let $b_{n}=n a_{n}$. Then
$$
\begin{array}{l}
b_{1}=b_{2}=2, \\
b_{n+2}=n b_{n+1}+n b_{n}-2 n+1 .
\end{array}
$$
Let $c_{n}=b_{n}-1$. Thus,
$$
c_{1}=c_{2}=1, c_{n+2}=n c_{n+1}+n c_{n} \text {. }
$$
Then $c_{n+1}-n c_{n}=-\left[c_{n}-(n-1) c_{n-1}\right]$
$$
=(-1)^{n-1}\left(c_{2}-c_{1}\right... | \frac{(n-1)!+1}{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,102 |
9. (16 points) $F$ is the left focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ centered at the origin $O$. A line $l$ with a non-zero slope passing through $F$ intersects the ellipse at points $A$ and $B$. The line segment $AO$ is extended to intersect the ellipse at point $C$. Find the maximum ... | When the semi-focal distance of the ellipse is $c$, and the inclination angle of the line $l$ is $\theta(0c$ is considered, the function
$$
f(x)=\frac{b^{2}}{x}+c^{2} x(0c$ is considered, the maximum area of $\triangle A B C$ is $\frac{2 b^{2} c}{a}$. | \frac{2 b^{2} c}{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,103 |
10. (20 points) From the set
$$
S=\{1,2, \cdots, n\}\left(n \in \mathbf{N}_{+}, n \geqslant 2\right)
$$
two different subsets $P$ and $Q$ are taken successively. Find the probability of the following events:
(1) $P \nsubseteq Q$, and $Q \nsubseteq P$;
(2) $\operatorname{Card}(P \cap Q)=k(0 \leqslant k \leqslant n-1)$. | 10. Given that the set $S$ has $2^{n}$ subsets, the number of ordered subset pairs $(P, Q)$ is $\mathrm{A}_{2^{n}}^{2}=2^{n}\left(2^{n}-1\right)$.
(1) Consider the complementary event of “$P \nsubseteq Q$, and $Q \nsubseteq P$”:
“$P \varsubsetneqq Q$ or $Q \varsubsetneqq P$.
If $P \varsubsetneqq Q$, let $\operatorname{... | \frac{3^{k}}{2^{n}\left(2^{n}-1\right)} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,104 |
11. (20 points) Let positive real numbers \(a, b, c, d \in \left[\frac{1}{2}, 2\right]\), and satisfy \(a b c d = 1\). Try to find
\[
\begin{array}{l}
f(a, b, c, d) \\
=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{d}\right)\left(d+\frac{1}{a}\right)
\end{array}
\]
the maximum and minimum values... | 11. Given $a b c d=1$, we have
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{d}\right)\left(d+\frac{1}{a}\right) \\
=(a b+1)(b c+1)(c d+1)(d a+1) .
\end{array}
$$
By the AM-GM inequality, we get
$$
\begin{array}{l}
(a b+1)(c d+1)=2+a b+c d \\
\geqslant 2+2 \... | 16 \text{ and } 25 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 725,105 |
One, (40 points) As shown in Figure $1, \triangle ABC (AB > AC)$ has an incircle $\odot I$ that touches sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Through a point $P$ on the extension of $BC$, draw another tangent to $\odot I$, which touches $\odot I$ at point $G$ and intersects $AB$ and $AC$ ... | First, prove that $P, Q, F$ and $P, E, R$ are collinear respectively.
By the tangent length theorem, we have
$$
\begin{array}{l}
P G=P D, M G=M F, B D=B F . \\
\text { Then } \frac{P G}{G M} \cdot \frac{M F}{F B} \cdot \frac{B D}{D P}=1 .
\end{array}
$$
By the converse of Ceva's theorem, we know that $B G, M D, P F$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,106 |
II. (40 points) Let $p$ be a prime number, and the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and for any non-negative integer $n$,
$$
a_{n+2}=2 a_{n+1}-p a_{n} \text {. }
$$
If -1 is a term in the sequence $\left\{a_{n}\right\}$, find all possible values of $p$. | The only $p$ that satisfies the condition is $p=5$.
It is easy to see that when $p=5$, $a_{3}=-1$.
Next, we prove that it is the only solution.
Assume $a_{m}=-1\left(m \in \mathbf{N}_{+}\right)$.
Clearly, $p \neq 2$, otherwise, from $a_{n+2}=2 a_{n+1}-2 a_{n}$, we know that for $n \geqslant 2$, $a_{n}$ is always even.
... | p=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,107 |
Three. (50 points) Given positive real numbers $a, b, c$ satisfying $a+b+c=3$. Prove:
$$
\sum \frac{a}{1+(b+c)^{2}} \leqslant \frac{3\left(a^{2}+b^{2}+c^{2}\right)}{a^{2}+b^{2}+c^{2}+12 a b c},
$$
where, “ ”” denotes the cyclic sum. | The original inequality is equivalent to
$$
\sum\left[a-\frac{a}{1+(b+c)^{2}}\right] \geqslant 3-\frac{3\left(a^{2}+b^{2}+c^{2}\right)}{a^{2}+b^{2}+c^{2}+12 a b c} \text {, }
$$
which means $\sum \frac{a(b+c)^{2}}{1+(b+c)^{2}} \geqslant \frac{36 a b c}{a^{2}+b^{2}+c^{2}+12 a b c}$.
By the Cauchy-Schwarz inequality, we... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,108 |
Four, (50 points) Given a positive integer $n(n \geqslant 3)$. Three types of fruits are distributed into $n$ boxes. Try to find the smallest positive integer $k$, such that no matter how the fruits are distributed, it is always possible to select $k$ boxes, in which the three types of fruits are each at least half of ... | Let's denote three types of fruits as $x, y, z$, and $A_{i}(i=1,2, \cdots, n)$ represents the amount of $x$ fruit in the $i$-th box.
Consider such a distribution: one box contains all the $x$ fruits, another box contains all the $y$ fruits, and the remaining $n-2$ boxes evenly distribute all the $z$ fruits. Clearly,
$... | \left[\frac{n+3}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 725,109 |
Given positive real numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{a}, \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{b}, \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{c} .
\end{array}\right.
$$
Prove: (1) The real numbers $a, b, c$ can be the lengths of the sides of a t... | Prove (1) Since
$$
\begin{array}{l}
b+c-a=\frac{2 y z}{x+y+z}>0, \\
c+a-b=\frac{2 z x}{x+y+z}>0, \\
a+b-c=\frac{2 x y}{x+y+z}>0,
\end{array}
$$
Therefore, \(a\), \(b\), and \(c\) can be the lengths of the three sides of a triangle.
(2) Note that,
$$
\begin{array}{l}
p-a=\frac{y z}{x+y+z}, p-b=\frac{z x}{x+y+z}, \\
p-c... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 725,110 |
Initially 320, a line that divides the perimeter of a triangle into two equal parts is called the "perimeter bisector" of the triangle. Let $P$ be any point on the side of $\triangle A B C$, can a perimeter bisector of $\triangle A B C$ be drawn through this point $P$? If so, please write down the method; if not, pleas... | Solution as follows:
If $P$ is one of the three vertices of $\triangle ABC$, let's assume it is $A$.
As shown in Figure 2, construct the incircle of $\triangle ABC$, which touches sides $BC$, $CA$, and $AB$ at points $D'$, $E'$, and $F'$, respectively.
According to the properties of tangents to a circle, we have
$AF' ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 725,112 |
In acute $\triangle A B C$, $A D 、 B E 、 C F$ are three altitudes, $H$ is the orthocenter, $O$ is the circumcenter, $M$ is the midpoint of side $A C$, the extension of $B O$ intersects $A C$ at point $P$, and $D F$ intersects $B E$ at point $Q$. Prove: $P Q / / M H$. | Proof As shown in Figure 3, let $a, b, c, \angle A, \angle B, \angle C$ represent the three sides and three interior angles of $\triangle ABC$.
From the fact that $A, E, H, F$ and $B, F, H, D$ and $C, D, H, E$ are each sets of four concyclic points, it is easy to deduce that $H$ is the incenter of $\triangle DEF$.
By ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 725,113 |
320 Suppose there are $2 k+1$ consecutive natural numbers,
where the sum of the squares of the $k$ larger natural numbers is equal to the sum of the squares of the remaining $k+1$ smaller natural numbers (such as $5^{2}=4^{2}+3^{2}$ $\left.(k=1), 14^{2}+13^{2}=12^{2}+11^{2}+10^{2}(k=2)\right)$. For convenience, such an... | Let
\[
\begin{array}{l}
(a+k)^{2}+(a+k-1)^{2}+\cdots+(a+1)^{2} \\
=a^{2}+(a-1)^{2}+\cdots+(a-k)^{2}
\end{array}
\]
be the $k$-th transformation method,
\[
w_{k}=(a+k, a-k)
\]
indicates that the largest number is $a+k$ and the smallest number is $a-k$.
Clearly, $a^{2}=4 \times(1+2+\cdots+k) a$, or $a=4 \times(1+2+\cdo... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,114 |
Example 6 For any integer $a(a>3)$, prove: there exists a positive integer $N$, which is the product of exactly 2011 different prime numbers, such that $N \mid\left(a^{N}-1\right)$.
untranslated text remains the same as the source, only the example statement is translated. | First, prove that there exists a prime factor $p$ in $a-1$, such that $a^{p}-1$ has a prime factor $q$ that is not $p$.
If $a$ is even, by the lemma, the exponent of $p$ in $a^{p}-1$ is exactly one more than the exponent of $p$ in $a-1$.
Since $\sum_{i=0}^{p-1} a^{i}>p$, $a^{p}-1$ must have a prime factor $q$ that is... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,115 |
Example 7 Given that $N$ is a positive integer whose last two digits are 25, and $m$ is a positive integer. Prove: there exists a positive integer $n$, such that the last $m$ digits of $5^{n}$ have the same parity as the last $m$ digits of $N$, i.e., for $1 \leqslant k \leqslant m$, the $k$-th digit from the right of $... | Prove by induction on $m$.
Obviously, when $m=1,2$, taking $n=2$ suffices.
Assume that the last $m$ digits of $5^{n}$ have the same parity pattern as the last $m$ digits of $N$.
For $m+1$, construct $5^{n}$ and $5^{n+2^{m-2}}$.
Since $2^{2} \|(5-1)$, by the lemma we know that
$2^{m} \|\left(5^{2 m-2}-1\right)$.
And cle... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,116 |
For odd integer $n$ and prime $p$, if
$$
(n, p)=1,(2012, n)=1 \text {, }
$$
then there exist infinitely many multiples $m$ of $p$, such that for positive integers $s \backslash t$, whenever $m^{\prime} \equiv 2012^{s}\left(\bmod n^{t}\right)$, it follows that $n^{t-1} \mid s$. | Proof Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$, where $p_{1}, p_{2}$, $\cdots, p_{k}$ are odd primes.
$$
\text { Let } P=p_{1} p_{2} \cdots p_{k} \text {. }
$$
Consider the system of congruences
$$
x \equiv p_{i}+2012\left(\bmod p_{i}^{2}\right) \text {. }
$$
By the Chinese Remainder Theorem, there mu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 725,117 |
Example 1 Let $a, b, c$ be non-negative real numbers, satisfying
$$
S=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1 \text{. }
$$
Find the maximum and minimum values of the elementary symmetric polynomials. | Solve: From the inequality of sum and reciprocal sum, we get
$$
\begin{array}{l}
S(1+a+1+b+1+c) \geqslant 3^{2} \\
\Rightarrow a+b+c \geqslant 6 . \\
\quad \text { By } \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \geqslant \frac{a+b+c}{3} \geqslant \frac{6}{3}=2 \\
\quad \Rightarrow a^{2}+b^{2}+c^{2} \geqslant 12 .
\end{array}
... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,118 |
Example 2 Non-negative real numbers $a$, $b$, $c$ satisfy
$$
S=\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+a c}=1 \text{. }
$$
Find the extremum of the elementary symmetric polynomial. | $$
\begin{array}{l}
\text { Given } \\
S(1+a b+1+b c+1+c a) \geqslant 3^{2} \\
\Rightarrow a b+b c+a c \geqslant 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Since } a^{2}+b^{2}+c^{2} \geqslant a b+b c+a c \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Since }(a+b+c)^{2}=a^{... | \frac{3 \sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,119 |
Example 3 Let $a, b, c$ be non-negative real numbers, satisfying
$$
\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}=6 \text{. }
$$
Prove: $a b c \leqslant \frac{27}{64}$. | Proof: Let $\sqrt{4 a+1}=x, \sqrt{4 b+1}=y$,
$$
\sqrt{4 c+1}=z \text{. }
$$
$$
\begin{array}{l}
\text{Then } x+y+z=6, a=\frac{x^{2}-1}{4}, \\
b=\frac{y^{2}-1}{4}, c=\frac{z^{2}-1}{4} .
\end{array}
$$
Therefore, $a b c$
$$
\begin{array}{l}
=\frac{1}{64}(x-1)(y-1)(z-1)(x+1)(y+1)(z+1) \\
\leqslant \frac{1}{64}\left(\frac... | \frac{27}{64} | Inequalities | proof | Yes | Yes | cn_contest | false | 725,120 |
Example 4 Let non-negative real numbers $a, b, c$ satisfy
$$
\sqrt{a^{2}+a}+\sqrt{b^{2}+b}+\sqrt{c^{2}+c}=2 \text {. }
$$
Find the maximum and minimum values of $abc$, $a^{2}+b^{2}+c^{2}$, and $a+b+c$. | (1) From Example 3, we know $0 \leqslant a b c \leqslant \frac{1}{27}$.
(2) From $a+b+c \leqslant \sqrt{3} \sqrt{a^{2}+b^{2}+c^{2}}$, we get
$$
\begin{array}{l}
\frac{2}{3}=\frac{\sqrt{a^{2}+a}+\sqrt{b^{2}+b}+\sqrt{c^{2}+c}}{3} \\
\leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}+a+b+c}{3}} \\
\Rightarrow a^{2}+b^{2}+c^{2}+a+b+c... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,121 |
Example 3 Suppose the lengths of the two legs of a right triangle are $a$ and $b$, and the length of the hypotenuse is $c$. If $a$, $b$, and $c$ are all positive integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
(2010, National Junior High School Mathematics Competi... | Solve: By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Also, $c=\frac{1}{3} a b-(a+b)$, so
$$
c^{2}=\left[\frac{1}{3} a b-(a+b)\right]^{2} \text {. }
$$
Rearranging gives $a b-6(a+b)+18=0$.
Thus, $(a-6)(b-6)=18$.
Since $a$ and $b$ are both positive integers, and without loss of generality, let $a<b$, then,
$$... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 725,122 |
Example 5 Let non-negative real numbers $x, y, z$ (all not equal to 1) satisfy
$$
\left(\frac{x}{x-1}\right)^{2}+\left(\frac{y}{y-1}\right)^{2}+\left(\frac{z}{z-1}\right)^{2}=1 \text {. }
$$
Prove: $\frac{1}{2} \leqslant x+y+z \leqslant \frac{3}{2}(\sqrt{3}-1)$. | Proof: Let $\left(\frac{x}{x-1}\right)^{2}=m,\left(\frac{y}{y-1}\right)^{2}=n$,
$$
\left(\frac{z}{z-1}\right)^{2}=p \text {. }
$$
Then $m+n+p=1$.
Since $x-11$, the condition does not hold), so,
$$
\frac{x}{x-1}=-\sqrt{m} \Rightarrow x=\frac{\sqrt{m}}{1+\sqrt{m}} \text {. }
$$
Similarly, $y=\frac{\sqrt{n}}{1+\sqrt{n}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,123 |
Example 1 Find the value of $\sin ^{4} 10^{\circ}+\sin ^{4} 50^{\circ}+\sin ^{4} 70^{\circ}$. (2010, Tsinghua Characteristic Examination)
【Analysis】When encountering higher powers, generally reduce the power first. | Notice,
$$
\begin{array}{l}
\sin ^{4} 10^{\circ}+\sin ^{4} 50^{\circ}+\sin ^{4} 70^{\circ} \\
=\left(\frac{1-\cos 20^{\circ}}{2}\right)^{2}+\left(\frac{1-\cos 100^{\circ}}{2}\right)^{2}+ \\
\left(\frac{1-\cos 140^{\circ}}{2}\right)^{2} \\
= \frac{1}{4}\left[3-2\left(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\ri... | \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,124 |
Example 2 Given that $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$. Prove:
$$
\cos B + \cos C + \frac{2a}{b+c} \geqslant 4 \sin \frac{A}{2} \text{. }
$$
(2008, Zhejiang University Independent Admission Examination) | $$
\begin{array}{l}
\cos B+\cos C+\frac{2 a}{b+c} \\
=2 \cos \frac{B+C}{2} \cdot \cos \frac{B-C}{2}+\frac{4 \sin \frac{A}{2} \cdot \cos \frac{A}{2}}{2 \sin \frac{B+C}{2} \cdot \cos \frac{B-C}{2}} \\
=2 \sin \frac{A}{2} \cdot\left(\cos \frac{B-C}{2}+\frac{1}{\cos \frac{B-C}{2}}\right) \\
\geqslant 4 \sin \frac{A}{2} .
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 725,125 |
Example 3: Does there exist $0<x<\frac{\pi}{2}$, such that $\sin x$, $\cos x$, $\tan x$, $\cot x$ can be arranged in an arithmetic sequence?
(2010, Peking University Independent Admission Examination) | (1) If $\sin x+\cos x=\tan x+\cot x$, then $\tan x=\frac{\sin x}{\cos x}>\sin x$, $\cot x=\frac{\cos x}{\sin x}>\cos x$. Therefore, $\sin x+\cos x>1$, so, $\sin x=\cos x\left(x=\frac{\pi}{4}\right)$.
But at this point, the four numbers are $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$, $1$, $1$, which cannot form an arith... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 725,126 |
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