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Example 5 Find all positive integers $n$, such that $n+36$ is a perfect square, and apart from 2 or 3, $n$ has no other prime factors.
Find all positive integers $n$, such that $n+36$ is a perfect square, and apart from 2 or 3, $n$ has no other prime factors. | Let $n+36=(x+6)^{2}$. Then $n=x(x+12)$.
According to the problem, $n=2^{\alpha_{1}} \times 3^{\alpha_{2}}\left(\alpha_{1} 、 \alpha_{2} \in \mathrm{N}\right)$.
Then $x=2^{a_{1}} \times 3^{b_{1}}, x+12=2^{a_{2}} \times 3^{b_{2}}$, where $a_{1} 、 b_{1} 、 a_{2} 、 b_{2} \in \mathbf{N}$.
Thus, $2^{a_{2}} \times 3^{b_{2}}-2^{... | 64,108,288,864,1728,10368 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,615 |
Example 6 Let $d$ be a positive integer not equal to $2, 5, 13$. Prove: in the set $\{2, 5, 13, d\}$, there can be found two distinct elements $a, b$ such that $ab-1$ is not a perfect square. | Notice
$$
\begin{array}{l}
2 \times 5-1=3^{2}, 2 \times 13-1=5^{2}, \\
5 \times 13-1=8^{2} .
\end{array}
$$
Therefore, it is only necessary to prove that at least one of $2 d-1$, $5 d-1$, and $13 d-1$ is not a perfect square. Otherwise, assume $x, y, z \in \mathbf{N}_{+}$, such that
$$
\begin{array}{l}
2 d-1=x^{2}, \\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,616 |
2. Let $a_{n}=n^{2}+n+2(n=1,2, \cdots)$. Then in the sequence $\left\{a_{n}\right\}$ ().
(A) there are infinitely many prime numbers
(B) there are infinitely many square numbers
(C) there are only finitely many prime numbers
(D) there are only finitely many square numbers | Given $a_{n}=n(n+1)+2$, we know that $a_{n}$ is a composite number. Also, $a_{1}=4$, and when $n \geqslant 2$, $n^{2}<a_{n}<(n+1)^{2}$.
Therefore, the answer is D. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,617 |
3. After deleting one element $\qquad$ from the set $\{1!, 2!, \cdots, 24!\}$, the product of the remaining elements is exactly a perfect square. | The product of the elements in the set is
$$
\begin{array}{l}
\prod_{i=1}^{12}[(2 i)!\times(2 i-1)!] \\
=\left[\prod_{i=1}^{12}(2 i-1)!\right]^{2} \times 2^{12} \times 12!.
\end{array}
$$
Therefore, the element to be deleted is 12!. | 12! | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,618 |
5. Prove: There does not exist a positive integer $n$, such that $2 n^{2}+1$, $3 n^{2}+1$, and $6 n^{2}+1$ are all perfect squares. | Assume the conclusion does not hold. Then
\[
\begin{array}{l}
36 n^{2}\left(6 n^{2}+1\right)\left(3 n^{2}+1\right)\left(2 n^{2}+1\right) \\
=\left(36 n^{4}+18 n^{2}+1\right)^{2}-1
\end{array}
\]
is a perfect square, which is a contradiction. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,619 |
Example 3 Given $n(n \geqslant 3)$ points in a plane, no three of which are collinear. Prove that among these $n$ points there exist three points $A, B, C$, such that the remaining $n-3$ points are all outside $\triangle A B C$.
untranslated text remains unchanged. | Prove that among $n$ points, if any two points $B$ and $C$ are chosen and a line segment $BC$ is drawn, then the remaining $n-2$ points do not lie on the line containing $BC$. Using $BC$ as the base and the remaining $n-2$ points as vertices, we can form $n-2$ triangles. Choose the triangle with the smallest area, deno... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,620 |
Example 1 Given positive integers $a, b$, such that $(a b+1) \mid\left(a^{2}+b^{2}\right)$.
Prove: $\frac{a^{2}+b^{2}}{a b+1}$ is the square of some positive integer. | 【Analysis and Proof】When $a=b$, there exists an integer $q$ such that $\frac{2 a^{2}}{a^{2}+1}=q \Rightarrow(2-q) a^{2}=q$.
From $2-q>0$, we can get $q=1=1^{2}$.
The conclusion is obviously true.
When $a \neq b$, by symmetry, we may assume $a>b$.
The idea is: if
$$
\frac{a^{2}+b^{2}}{a b+1}=\frac{b^{2}+t^{2}}{b t+1} \t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,621 |
Example 2 Let $p$ be a prime number. Then $p$ can be expressed as the sum of four squares of non-negative integers, i.e., the equation
$$
x^{2}+y^{2}+z^{2}+w^{2}=p
$$
has integer solutions $(x, y, z, w)$. | Prove that when $p=2$, since $2=1^{2}+1^{2}+0^{2}+0^{2}$, the conclusion is correct.
Now consider the case where $p$ is an odd prime.
First, prove: If $p$ is an odd prime, then there exists an
integer $k (k>1)$ which is an odd number, then let integers $a$, $c$, $d$ satisfy the following conditions:
$a \equiv x(\bmod ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,622 |
Example 3 Proof: The indeterminate equation
$$
x^{4}-y^{4}=z^{2}((x, y)=1)
$$
has no positive integer solutions $(x, y, z)$. | Prove that if equation (1) has positive integer solutions, and let $x, y, z$ be the positive integer solutions with the smallest $x$.
If $x$ is even, then by $(x, y)=1$, $y$ is odd. In this case, $x^{4}-y^{4} \equiv 3(\bmod 4)$, so $x^{4}-y^{4}$ cannot be a perfect square. Therefore, $x$ is odd.
If $y$ is odd, then by... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,623 |
Example 4 Proof: There do not exist integers $x, y, z$ satisfying
$$
2 x^{4}+2 x^{2} y^{2}+y^{4}=z^{2}(x \neq 0) \text {. }
$$ | Proof: Let $x, y, z$ be integer solutions of equation (1).
Obviously, from $x \neq 0$, we can deduce that $y \neq 0$.
Without loss of generality, assume $x > 0, y > 0$ and $(x, y) = 1$. Further assume that $x$ is the smallest solution satisfying the above conditions.
Since $z^{2} \equiv 0, 1, 4 \pmod{8}$, it follows t... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,624 |
Example 6 Let $k, n (n>2)$ be positive integers. Prove: the equation $x^{n}-y^{n}=2^{k}$ has no integer solutions.
The text is translated while preserving the original line breaks and format. | Proof by contradiction. Assume the conclusion does not hold, i.e., the equation has integer solutions. Since $n$ is a positive integer, there must be a smallest $n$, let $n_{0}>2$ be the smallest $n$ satisfying
$$
x^{n_{0}}-y^{n_{0}}=2^{m} \quad (m>0)
$$
If $n_{0}$ is even, let $n_{0}=2 l\left(l \in \mathbf{N}_{+}\rig... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,626 |
Example 7 Proof: The equation
$$
2 a^{2}+b^{2}+3 c^{2}=10 n^{2}
$$
has no positive integer solutions $(a, b, c, n)$. | Proof Assume the equation (1) has a set of positive integer solutions $\left(a_{0}, b_{0}, c_{0}, n_{0}\right)$, and it is the smallest $n_{0}$ among all positive integer solutions.
From equation (1), we know that $b_{0}^{2}+3 c_{0}^{2}$ is even, so $b_{0}$ and $c_{0}$ are of the same parity.
When $b_{0}$ and $c_{0}$ ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,627 |
Example 8 Proof: The equation
$$
x^{2}+y^{2}=3\left(z^{2}+u^{2}\right)
$$
has no positive integer solutions $(x, y, z, u)$. | Prove that if equation (1) has a positive integer solution, and $\left(x_{0}, y_{0}, z_{0}, u_{0}\right)$ is the set of positive integer solutions that minimizes $x^{2}+y^{2}$, i.e.,
$$
x_{0}^{2}+y_{0}^{2}=3\left(z_{0}^{2}+u_{0}^{2}\right).
$$
From equation (2), we know that $x_{0}^{2}+y_{0}^{2}$ is a multiple of 3. I... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,628 |
Example 9 Let $a$ be a given positive integer, and $A, B$ be two real numbers. Determine the necessary and sufficient condition for the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=(13 a)^{2}, \\
x^{2}\left(A x^{2}+B y^{2}\right)+y^{2}\left(A y^{2}+B z^{2}\right)+z^{2}\left(A z^{2}+B x^{2}\right) \\... | Solving from (2) $-\frac{B}{2} \times(1)^{2}$, we get
$$
\left(A-\frac{1}{2} B\right)\left(x^{4}+y^{4}+z^{4}\right)=\frac{1}{2}\left(A-\frac{1}{2} B\right)(13 a)^{4} \text {. }
$$
(1) $A \neq \frac{1}{2} B$.
The above equation simplifies to
$$
2\left(x^{4}+y^{4}+z^{4}\right)=(13 a)^{4} \text {. }
$$
Assume $x, y, z$ a... | A=\frac{1}{2} B | Algebra | proof | Yes | Yes | cn_contest | false | 723,629 |
Example 1 Given $a, b, c > 0, a^{2}+b^{2}+c^{2}=14$. Prove: $a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant 14$. | Prove the construction of the $3 \times 5$ matrix
$$
\left(\begin{array}{ccccc}
a^{5} & a^{5} & 1 & 1 & 1 \\
\frac{b^{5}}{8} & \frac{b^{5}}{8} & 4 & 4 & 4 \\
\frac{c^{5}}{27} & \frac{c^{5}}{27} & 9 & 9 & 9
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\begin{array}{l}
{\left[\left(a^{5}+\frac{b^{5}}{8... | 14 | Inequalities | proof | Yes | Yes | cn_contest | false | 723,630 |
Example 4 On the plane, there are $n$ points, any three of which can form a triangle, and the area of each triangle does not exceed 1. Prove: there exists a triangle with an area not exceeding 4 that can cover all $n$ points. | Prove that the number of triangles formed by $n$ points on a plane is finite, and among them, there must be a triangle with the largest area (let it be $\triangle A B C$).
As shown in Figure 1, draw lines through each vertex parallel to the opposite sides, resulting in a new $\triangle A^{\prime} B^{\prime} C^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,631 |
Example 2 Let $a, b, c, d \geqslant 0, ab + bc + cd + da = 1$. Prove: $\sum \frac{a^{3}}{b+c+d} \geqslant \frac{1}{3}$.
where, “ $\sum$ ” denotes the cyclic sum. | Prove the construction of the $4 \times 2$ matrix
$$
\left(\begin{array}{cc}
\frac{a^{3}}{b+c+d} & b+c+d \\
\frac{b^{3}}{c+d+a} & c+d+a \\
\frac{c^{3}}{d+a+b} & d+a+b \\
\frac{d^{3}}{a+b+c} & a+b+c
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\left[\sum \frac{a^{3}}{b+c+d} \cdot 3(a+b+c+d)\right]^{\f... | \sum \frac{a^{3}}{b+c+d} \geqslant \frac{1}{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 723,632 |
Example 3 Let $x_{i}>0(i=1,2, \cdots, n), m \in \mathbf{R}_{+}$, $a \geqslant 0, \sum_{i=1}^{n} x_{i}=s \leqslant n$. Prove:
$$
\prod_{i=1}^{n}\left(x_{i}^{m}+\frac{1}{x_{i}^{m}}+a\right) \geqslant\left[\left(\frac{s}{n}\right)^{m}+\left(\frac{n}{s}\right)^{m}+a\right]^{n} \text {. }
$$ | Prove the construction of a $3 \times n$ matrix
$$
\left(\begin{array}{cccc}
x_{1}^{m} & x_{2}^{m} & \cdots & x_{n}^{m} \\
\frac{1}{x_{1}^{m}} & \frac{1}{x_{2}^{m}} & \cdots & \frac{1}{x_{n}^{m}} \\
a & a & \cdots & a
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\begin{array}{l}
{\left[\prod_{i=1}^{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,633 |
Consider an interval $(a, b)$. Prove: There must exist an irrational number $x \in (a, b)$, such that the sequence $\left\{\cos \left(6^{n} \pi x\right)\right\}$ has infinitely many terms greater than $\frac{1}{2}$. | To prove that there are infinitely many terms in the sequence $\left\{\cos \left(6^{n} \pi x\right)\right\}$ greater than $\frac{1}{2}$, it suffices to prove that there are infinitely many positive integers $n$ such that
$$
2 k+\frac{1}{3}>6^{n} x>2 k-\frac{1}{3}(k \in \mathbf{Z}) \text {. }
$$
Consider the fractional... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,634 |
Question As shown in Figure 1, in $\triangle A B C$, $A B=A C$, and $D$ is the midpoint of side $B C$. $E$ is a point outside $\triangle A B C$ such that $C E \perp A B$ and $B E=$ $B D$. Through the midpoint $M$ of line segment $B E$, draw a line $M F \perp B E$, intersecting the minor arc $\overparen{A D}$ of the cir... | Prove that from $\triangle A B D \backsim \triangle C B N$, we know
$$
B D: B N=A B: B C=B O: B D \text {. }
$$
Then $B E: B N=B D: B N$
$$
=B O: B D=B O: B E \text {. }
$$
Since $\angle N B E=\angle E B O$, we have
$$
\begin{array}{l}
\triangle E B N \backsim \triangle O B E \\
\Rightarrow \angle O E B=\angle B N E=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,635 |
2. All real number pairs $(x, y)$ that satisfy the equation
$$
(x+3)^{2}+y^{2}+(x-y)^{2}=3
$$
are $\qquad$ . | 2. $(-2,-1)$.
Expanding and rearranging the given equation yields $x^{2}+3 x+y^{2}-x y+3=0$. Completing the square gives $\left(y-\frac{x}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2} x+\sqrt{3}\right)^{2}=0$.
From this, we get $x=-2, y=-1$. | (-2, -1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,637 |
3. In Rt $\triangle A B C$, it is known that $\angle C=90^{\circ}, B C$ $=6, C A=3, C D$ is the angle bisector of $\angle C$. Then $C D=$ $\qquad$ . | $3.2 \sqrt{2}$.
Let $CD=t$. Then, by the relationship of the areas of triangles, we have
$$
\begin{array}{l}
\frac{1}{2} \times 3 t \sin 45^{\circ}+\frac{1}{2} \times 6 t \sin 45^{\circ} \\
=S_{\triangle M C B}=\frac{1}{2} \times 3 \times 6 .
\end{array}
$$
Solving this, we get $t=2 \sqrt{2}$. | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,638 |
4. If the product of the first 2011 positive integers
$$
1 \times 2 \times \cdots \times 2011
$$
can be divided by $2010^{k}$, then the maximum value of the positive integer $k$ is | 4.30.
$$
2010=2 \times 3 \times 5 \times 67 \text {. }
$$
In $1 \times 2 \times \cdots \times 2011$, the exponent of 67 is
$$
\left[\frac{2011}{67}\right]+\left[\frac{2011}{67^{2}}\right]+\cdots=30 \text {. }
$$
Obviously, the exponents of $2,3,5$ in $1 \times 2 \times \cdots \times 2011$ are all greater than 30.
The... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,639 |
5. As shown in Figure 1, in the Cartesian coordinate system, the vertices of the equilateral triangle $\triangle ABC$ are $B(1,0)$ and $C(3,0)$. A line passing through the origin $O$ intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. If $OM=MN$, then the coordinates of point $M$ are $\qquad$ | 5. $M\left(\frac{5}{4}, \frac{\sqrt{3}}{4}\right)$.
Draw a line parallel to the $x$-axis through point $N$ intersecting $A B$ at point $L$. Since $O M=M N$, we have $\triangle O B M \cong \triangle N L M \Rightarrow M L=O B=1$.
Given that $\triangle A B C$ is an equilateral triangle with side length 2, we know that $\... | \left(\frac{5}{4}, \frac{\sqrt{3}}{4}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,640 |
6. As shown in Figure 2, in rectangle $A B C D$, $A B=5, B C=$ 8, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=2, B F=5, D G=3, A H=3$. Point $O$ is on line segment $H F$ such that the area of quadrilateral $A E O H$ is 9. Then the area of quadrilateral $O F C G$ is | 6.6.5.
As shown in Figure 5, connect $E F$,
$F G$, $G H$, $H E$.
It is easy to see that quadrilateral $E F G H$ is a parallelogram, and
$$
\begin{array}{l}
S_{\text {OEFCH }} \\
=40-\left(S_{\triangle A E H}+S_{\triangle E B F}+S_{\triangle F C G}+S_{\triangle G D H}\right) \\
=19 .
\end{array}
$$
From $S_{\triangle... | 6.5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,641 |
Example 5 In a convex pentagon $A B C D E$, there are no parallel segments among its sides and diagonals. Extend side $B C$ to intersect diagonal $A D$ at some point, and then mark an arrow on side $B C$ pointing in the direction of the intersection point. Follow this method to mark arrows on all five sides. Prove: The... | Proof Consider the five triangles formed by every three consecutive vertices of a convex pentagon. Among these, there must be one with the smallest area. Let $\triangle ABC$ have the smallest area.
Below we prove: the arrow on side $BC$ must point to point $B$.
Since $DA$ is not parallel to $BC$ (as shown in Figure 2),... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,642 |
7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ . | 7. -2278 .
Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
... | -2278 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,643 |
8. Given real numbers $a, b, c$ satisfy
$$
a \geqslant b \geqslant c, a+b+c=0 \text {, and } a \neq 0 \text {. }
$$
Let $x_{1}, x_{2}$ be the two real roots of the equation $a x^{2}+b x+c=0$. Then the maximum distance between two points $A\left(x_{1}, x_{2}\right)$ and $B\left(x_{2}, x_{1}\right)$ in the Cartesian coo... | 8. $3 \sqrt{2}$.
From the problem, we know $a>0, c<0$.
Thus, from $a \geqslant b \geqslant c$, we get $\frac{c}{a} \leqslant \frac{b}{c} \leqslant 1$.
Also, from $a+b+c=0$, we get $\frac{b}{a}=-1-\frac{c}{a}$.
Therefore, $\frac{c}{a} \leqslant-1-\frac{c}{a} \leqslant 1, -2 \leqslant \frac{c}{a} \leqslant-\frac{1}{2}$.... | 3 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,644 |
9. As shown in Figure 3, given that pentagon $A B C D E$ is a regular pentagon, and the area of the pentagram $A C E B D$ (shaded part) is 1. Let $A C$ intersect $B E$ at point $P$, and $B D$ intersect $C E$ at point $Q$. Then the area of quadrilateral $A P Q D$ is $\qquad$ | 9. $\frac{1}{2}$.
As shown in Figure 6, connect
$Q R$.
It is easy to see that quadrilateral
$A P Q R$ is a rhombus,
$\triangle Q P R \cong \triangle A R P$,
$\triangle S P Q \cong \triangle T R Q$.
Let $S_{\triangle A P R}=a$,
$S_{\triangle P S Q}=b$.
Then $1=S_{\text {figure}}=6a+2b$,
$S_{\text {quadrilateral } A P Q... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,645 |
10. Let $a, b, c (1 \leqslant a < b < c \leqslant 9)$ be integers, and $\overline{a b c} \cdot \overline{b c a} \cdot \overline{c a b} + 1$ is divisible by 9. Then the minimum value of $a + b + c$ is $\qquad$, and the maximum value is $\qquad$. | 10.8, 23.
Let $a+b+c \equiv r(\bmod 9)$. Then the remainders of $\overline{a b c} 、 \overline{b c a} 、 \overline{c a b}$ when divided by 9 are all $r$. Therefore,
$$
\begin{array}{l}
\overline{a b c} \cdot \overline{b c a} \cdot \overline{c a b}+1 \equiv r^{3}+1 \equiv 0(\bmod 9), \\
r \equiv 2,5,8(\bmod 9) .
\end{arr... | 8, 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,646 |
11. Given that the area of $\triangle A B C$ is 4, with side lengths $B C=a, C A=b, A B=c(c>b)$, and $A D$ is the angle bisector of $\angle A$, $C^{\prime}$ is the reflection of point $C$ about the line $A D$. If $\triangle C^{\prime} B D$ is similar to $\triangle A B C$, find the minimum perimeter of $\triangle A B C$... | II. 11. As shown in Figure 7, it is clear that point $C^{\prime}$ lies on side $A B$, and
$B C^{\prime}=c-b$.
By the property of internal angle bisectors, we have $B D=\frac{c a}{b+c}$.
Since $\triangle C^{\prime} B D \backsim \triangle A B C$, and $\angle B$ is a common angle, we have
$$
\frac{c-b}{c}=\frac{\frac{c a... | 4 \sqrt{2}+4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,647 |
12. Place the nine digits $1,2, \cdots, 9$ into the nine small squares in Figure 4, so that the seven three-digit numbers $\overline{a b c} 、 \overline{d e f} 、 \overline{g h i} 、 \overline{a d g} 、 \overline{b e h} 、 \overline{c f i}$ and $\overline{a e i}$ are all divisible by 11. Find the maximum value of the three-... | 12. According to the problem, for modulo 11 we have
$$
\begin{array}{l}
a+c \equiv b, d+f \equiv e, g+i \equiv h, a+g \equiv d, \\
b+h \equiv e, c+i \equiv f, a+i \equiv e . \\
\text { Then }(a+c)+(d+f)+(g+i)+(b+h)+e \\
\equiv b+h+3 e \equiv 4 e(\bmod 11) .
\end{array}
$$
The left side of the above equation is
$$
1+2+... | 734 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,648 |
13. Let real numbers $x, y, z$ satisfy $x+y+z=0$, and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \leqslant 2$. Find the maximum and minimum values of $x$. | 13. From the problem, we know
$$
\begin{aligned}
2 & \geqslant(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \\
= & 2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x \\
= & 2 x^{2}+2(y+z)^{2}-4 y z-2(y+z)-2 y z \\
& 6 x^{2}-6 y z \\
& =6 x^{2}-6 \cdot \frac{(y+z)^{2}-(y-z)^{2}}{4} \\
& =\frac{9}{2} x^{2}+\frac{3}{2}(y-z)^{2} \geqslant \frac{9}... | -\frac{2}{3} \leqslant x \leqslant \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,649 |
14. Numbers of the form $a^{2}+161 b^{2}$, where $a$ and $b$ are integers, are called "good numbers". Prove:
(1) $100$ and $2010$ are good numbers;
(2) There exist positive integers $x$ and $y$ such that $x^{161}+y^{161}$ is a good number, but $x+y$ is not a good number. | 14. (1) From $100=10^{2}+161 \times 0^{2}$,
$$
2010=43^{2}+161 \times 1^{2} \text {, }
$$
we know that 100 and 2010 are good numbers.
(2) Let $x=2^{162}+2=2\left(2^{161}+1\right), y=2^{161}+1$.
Then $x^{161}+y^{161}=2^{161}\left(2^{161}+1\right)^{161}+\left(2^{161}+1\right)^{161}$
$$
=\left(2^{161}+1\right)^{162}=\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,650 |
1. Given that the function $y=f(x)$ is a periodic function with a period of 3 defined on $\mathbf{R}$, Figure 1 shows the graph of this function in the interval $[-2,1]$. Then the value of $\frac{f(2010)}{f(5) f(16)}$ is equal to | $-\sqrt{1}-\frac{1}{2}$.
Note that
$$
\begin{array}{l}
f(5)=f(-1)=-1, \\
f(16)=f(1)=2, \\
f(2010)=f(0)=1 . \\
\text { Then } \frac{f(2010)}{f(5) f(16)}=-\frac{1}{2} .
\end{array}
$$ | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,651 |
2. The sum of the cubes of all roots of the equation $\left|x^{2}-x+\frac{1}{2010}\right|=\frac{1}{2010}$ is equal to | 2. $\frac{669}{335}$.
The original equation is equivalent to
$x^{2}-x=0$ or $\frac{-1}{1005}$.
Its roots are $x_{1}=1, x_{2}=0, x_{3}, x_{4}$ satisfying $x_{3}+x_{4}=1, x_{3} x_{4}=\frac{1}{1005}$.
Then $x_{3}^{3}+x_{4}^{3}$
$$
\begin{array}{l}
=\left(x_{3}+x_{4}\right)\left[\left(x_{3}+x_{4}\right)^{2}-3 x_{3} x_{4}\... | \frac{669}{335} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,652 |
Example 6 At a party, $n(n \geqslant 2)$ pairs of young men and women dance together. Suppose no man has danced with all the women, and each woman has danced with at least one man. Prove that there must be two men $b_{1} , b_{2}$ and two women $g_{1} , g_{2}$, such that $b_{1}$ has danced with $g_{1}$, $b_{2}$ has danc... | Proof 1: Let one of the male youths who danced with the most female youths be $b_{1}$.
Since $b_{1}$ has not danced with all the female youths, there exists a female youth $g_{2}$ who has not danced with $b_{1}$.
Since $g_{2}$ has danced with at least one male youth, there exists $b_{2} \left(\neq b_{1}\right)$ who h... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,653 |
3. As shown in Figure $2, A B$ is tangent to $\odot O$ at point $A$, connecting point $B$ with a point $D$ inside $\odot O$ intersects $\odot O$ at point $C, A B$ $=6, D C=C B=3$, $O D=2$. Then the radius of $\odot O$ is . $\qquad$ | 3. $\sqrt{22}$.
As shown in Figure 5, extend $B D$ to intersect $\odot O$ at point $E$, and extend $O D$ to intersect $\odot O$ at points $F$ and $G$, where $F G$ is the diameter of $\odot O$. Let the radius of $\odot O$ be $r$.
By the secant-tangent theorem, we have
$$
B C \cdot B E = B A^{2} \text {, }
$$
which mea... | \sqrt{22} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,654 |
4. The function $f(x)$ that satisfies the equation
$$
f(x)+(x-2) f(1)+3 f(0)=x^{3}+2 \quad (x \in \mathbf{R})
$$
is $f(x)=$ . $\qquad$ | 4. $x^{3}-x+1$.
Substitute $x=1$ and $x=0$ into the equation, we get
$$
\begin{array}{l}
f(0)=1, f(1)=1 \text {. } \\
\text { Therefore, } f(x)=x^{3}+2-(x-2) f(1)-3 f(0) \\
=x^{3}-x+1 \text {. }
\end{array}
$$ | x^{3}-x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,655 |
5. A natural number is called a "good number" if it is exactly 2007 more than the sum of its digits. Then the sum of all good numbers is $\qquad$ . | 5.20145.
Let $f(n)=n-S(n)$ ($S(n)$ is the sum of the digits of the natural number $n$). Then the function $f(n)$ is a non-strictly increasing function, and
$$
\begin{array}{l}
f(2009)<f(2010) \\
=f(2011)=\cdots=f(2019)=2007 \\
<f(2020) .
\end{array}
$$
Therefore, there are only 10 natural numbers that satisfy the con... | 20145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,656 |
Three, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, determine all possible values of $k$, and provide an example for each corresponding $k$ value; if not, briefly explain the reason.
---
Translate the above text into English, please retain the original text's line breaks... | Three, (1) If 2010 can be written as the sum of squares of $k$ prime numbers, and the sum of the squares of the smallest 10 distinct prime numbers is
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010,
\end{array}
$$
Therefore, $k \leqslant 9$.
It is known that there is only one even prime number ... | k=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,658 |
Four, (15 points) Given nine points on a plane, the distance between any two points is no less than 1. Prove: there exist at least two points whose distance is no less than $\sqrt{3}$.
| Given a set of nine points $S$. Then a line $l$ can be drawn in the plane such that all points in $S$ are on the same side of $l$. Translate the line $l$ until it encounters a point $O$ in $S$. At this point, all points in $S$ are on the line $l$ or on the same side of $l$.
As shown in Figure 7, with $O$ as the center... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,659 |
Five. (15 points) As shown in Figure 4, given that the diagonals of the convex quadrilateral $ABCD$ intersect at point
$$
O, O_{1}, O_{2}, O_{3}, O_{4}
$$
are the incenters of $\triangle AOB$,
$\triangle BOC$, $\triangle COD$,
$\triangle DOA$, respectively, and the corresponding inradii $r_{1}, r_{2},$
$r_{3}, r_{4... | $$
\begin{array}{l}
\text { (1) Let } A B=a, B C=b, C D=c, \\
D A=d, A O=x, B O=y, C O=u, \\
D O=v, \cos \angle A O B=k .
\end{array}
$$
Then $\cos \angle B O C=-k$.
From the given conditions, we have
$$
\begin{array}{l}
\frac{a+x+y}{x y}+\frac{c+u+v}{u v}=\frac{b+y+u}{y u}+\frac{d+x+v}{x v} \\
\Rightarrow a u v+a x y... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,660 |
1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$ | $-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqsla... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,661 |
2. Given the function $f(x)=a x^{2}-\sqrt{2}$. If $f(f(\sqrt{2}))=-\sqrt{2}$,
then $a=$ $\qquad$ | 2.0 or $\frac{\sqrt{2}}{2}$.
From $a\left[a(\sqrt{2})^{2}-\sqrt{2}\right]^{2}-\sqrt{2}=-\sqrt{2}$, we get $a(2 a-\sqrt{2})^{2}=0$.
Solving this, we get $a=0$ or $\frac{\sqrt{2}}{2}$. | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,662 |
3. Let $a_{n}=\sqrt{1 \times 2}+\sqrt{2 \times 3}+\cdots+\sqrt{n \times(n+1)}$. Then $\left[\frac{2 a_{n}}{n}\right]=$ $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$). | 3. $n+1$.
Since $k<\sqrt{k(k+1)}<k+\frac{1}{2}$, we have
$$
\sum_{k=1}^{n} k<a_{n}<\sum_{k=1}^{n}\left(k+\frac{1}{2}\right) \text {, }
$$
which means $\frac{n(n+1)}{2}<a_{n}<\frac{n(n+2)}{2}$.
Therefore, $\left[\frac{2 a_{n}}{n}\right]=n+1$. | n+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,663 |
Example $7 A B$ is a fixed chord in the fixed circle $\odot O$, construct chords $C_{1} D_{1}, C_{2} D_{2}, \cdots, C_{1988} D_{1988}$ in $\odot O$. For each chord $C_{i} D_{i}(i=1,2, \cdots, 1988)$, it is bisected by chord $A B$ at point $M_{i}$; draw tangents to $\odot O$ at $C_{i}$ and $D_{i}$, and let the two tange... | 【Analysis】As shown in Figure 3, let the tangents to $\odot O$ through points $A$ and $B$ intersect at point $P$. Suppose the tangents to $\odot O$ through points $C_{i}$ and $D_{i}$ intersect at a point infinitely close to $P$, and the endpoints of chord $C_{i} D_{i}$ are infinitely close to $A$. Then the intersection ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,664 |
4. Let $A(0, b)$ be the endpoint of the minor axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $B$ be a point on the ellipse, and $C(0,-1)$ be the projection of point $B$ on the $y$-axis. If $A B=3 \sqrt{2}, A C=$ $B C$, then the focal distance of the ellipse is | 4. $4 \sqrt{2}$.
It is known that $AC=BC=3, OC=1, OA=2$, i.e., $b=2$, and point $B( \pm 3,-1)$. Substituting into the ellipse equation yields $a^{2}=12$.
Then $c=\sqrt{a^{2}-b^{2}}=2 \sqrt{2}$.
Thus, the focal distance $2 c=4 \sqrt{2}$. | 4 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,665 |
5. If $x \in [0, \pi]$, then the range of the function
$$
y=\frac{\sin x \cdot \cos x}{1+\sin x+\cos x}
$$
is | 5. $\left[-1, \frac{\sqrt{2}-1}{2}\right]$.
Let $t=\sin x+\cos x$. Then $\sin x \cdot \cos x=\frac{t^{2}-1}{2}$.
Thus, $y=\frac{t-1}{2}$.
Also, $t=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$, and $\frac{\pi}{4} \leqslant x+\frac{\pi}{4}<\frac{5 \pi}{4}$, so $\sin \left(x+\frac{\pi}{4}\right) \in\left[-\frac{\sqrt{2}}{... | \left[-1, \frac{\sqrt{2}-1}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,666 |
6. Let the side length of the lateral edges of a regular quadrilateral pyramid be 1. Then the maximum value of its volume is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6. $\frac{4 \sqrt{3}}{27}$.
As shown in Figure 2, let the center of the base of the regular quadrilateral pyramid $P-ABCD$ be $H$, and $\angle PAH = \theta$. Then
$$
\begin{array}{l}
AH = \cos \theta, \\
PH = \sin \theta, \\
V = \frac{2}{3} \cos^2 \theta \cdot \sin \theta.
\end{array}
$$
Let $\sin \theta = x$. Then $... | \frac{4 \sqrt{3}}{27} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,667 |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}+a_{n+1}=-n^{2} \text {. }
$$
then $a_{15}=$ $\qquad$ | 7. -104 .
Rewrite $a_{n}+a_{n+1}=-n^{2}$ as
$$
\left(a_{n}+\frac{n^{2}}{2}-\frac{n}{2}\right)+\left[a_{n+1}+\frac{(n+1)^{2}}{2}-\frac{n+1}{2}\right]=0 \text {. }
$$
Let $b_{n}=a_{n}+\frac{n^{2}}{2}-\frac{n}{2}$. Then
$$
b_{1}=1 \text {, and } b_{n+1}=-b_{n} \text {. }
$$
Therefore, $b_{2 k-1}=1, b_{2 k}=-1(k=1,2, \c... | -104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,668 |
8. If a four-digit number $n$ contains at most two different digits among its four digits, then $n$ is called a "simple four-digit number" (such as 5555 and 3313). Then, the number of simple four-digit numbers is | 8. 576.
If the four digits of a four-digit number are all the same, then there are nine such four-digit numbers.
If the four digits of a four-digit number have two different values, the first digit \( a \in \{1,2, \cdots, 9\} \) has 9 possible choices. After choosing \( a \), select \( b \in \{0,1, \cdots, 9\} \) and... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,669 |
II. (16 points) As shown in Figure $1, M$ is the center of the equilateral $\triangle A_{1} A_{2} A_{3}$, and $N$ is any point on the plane. The circle with diameter $M N$ intersects the lines $M A_{i}$
$$
\text { ( } i=1,2,3) \text { at }
$$
points $B_{i}$. Prove:
$$
M B_{1}^{2}+M B_{2}^{2}+M B_{3}^{2}=N B_{1}^{2}+N ... | Given that points $M, B_{1}, B_{2}, N, B_{3}$ are concyclic, we have
$$
\angle B_{1} M B_{3}=120^{\circ},
$$
which implies
$$
\begin{array}{l}
\angle B_{1} B_{2} B_{3}=60^{\circ}, \\
\angle B_{1} B_{3} B_{2}=\angle B_{1} M B_{2}=\angle A_{1} M B_{2}=60^{\circ}.
\end{array}
$$
Therefore, $\triangle B_{1} B_{2} B_{3}$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,670 |
Three. (20 points) Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and agree that $x_{n+1}=x_{1}$. Prove:
$$
\sum_{k=1}^{n} \sqrt{\frac{1}{\left(x_{k}+1\right)^{2}}+\frac{x_{k+1}^{2}}{\left(x_{k+1}+1\right)^{2}}} \geqslant \frac{n}{\sqrt{2}} .
$$ | Let $x_{k}=\tan ^{2} \theta_{k}$, where $\theta_{k} \in\left[0, \frac{\pi}{2}\right)$ $(k=1,2, \cdots, n)$, and it is agreed that $\theta_{n+1}=\theta_{1}$. Then
$$
\begin{array}{l}
\sqrt{\frac{1}{\left(x_{k}+1\right)^{2}}+\frac{x_{k+1}^{2}}{\left(x_{k+1}+1\right)^{2}}} \\
=\sqrt{\cos ^{4} \theta_{k}+\sin ^{4} \theta_{... | \frac{n}{\sqrt{2}} | Inequalities | proof | Yes | Yes | cn_contest | false | 723,671 |
Four. (20 points) In a football invitational tournament, a total of $n$ teams are scheduled to participate, with each team's scheduled number of matches being $m_{1}, m_{2}, \cdots, m_{n}$. If any two teams are scheduled to play at most one match against each other, then $\left(m_{1}, m_{2}, \cdots, m_{n}\right)$ is ca... | Let the team that is scheduled to play $m_{i}(i=1,2, \cdots, n)$ matches be $A_{i}$.
(1) If the $m_{1}$ matches of $A_{1}$ are exactly against $A_{2}, A_{3}, \cdots, A_{m_{1}+1}$, then $A_{1}$ can be directly removed (of course, all the matches involving $A_{1}$ are also canceled). Thus, the matches between the remaini... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,672 |
1. Given the condition $p: \sqrt{1+\sin 2 \alpha}=\frac{4}{3}$ and the condition $q:|\sin \alpha+\cos \alpha|=\frac{4}{3}$. Then $p$ is ( ) of $q$.
(A) a sufficient but not necessary condition
(B) a necessary but not sufficient condition
(C) a necessary and sufficient condition
(D) neither a sufficient nor a necessary ... | $-1 . \mathrm{C}$.
$$
\begin{array}{l}
\text { Since } \sqrt{1+\sin 2 \alpha}=\sqrt{(\sin \alpha+\cos \alpha)^{2}} \\
=|\sin \alpha+\cos \alpha|,
\end{array}
$$
Therefore, $p$ is a necessary and sufficient condition for $q$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,673 |
2. Among 5 products, there are 4 genuine items and 1 defective item. If two items are randomly selected, let the number of genuine items among them be the random variable $\xi$. Then the mathematical expectation $E \xi$ is ( ).
(A) $\frac{6}{5}$
(B) $\frac{7}{5}$
(C) $\frac{8}{5}$
(D) $\frac{9}{5}$ | 2. C.
The mathematical expectation is $1 \times \frac{\mathrm{C}_{4}^{1}}{\mathrm{C}_{5}^{2}}+2 \times \frac{\mathrm{C}_{4}^{2}}{\mathrm{C}_{5}^{2}}=\frac{8}{5}$. | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,674 |
2. Prove: In any convex pentagon, three diagonals can be found that can form a triangle. | In a convex pentagon $A B C D E$, the longest diagonal is $B E$. It is not difficult to prove: a triangle can be formed with $B E$, $C E$, and $B D$ as sides. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,675 |
3. Let the base edge length of the regular tetrahedron $S-ABC$ be 3, and the side edge length be 2. Then the angle formed by the side edge $SA$ and the base $ABC$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $\arctan \sqrt{2}$ | 3. A.
Let the projection of vertex $S$ on the base $\triangle A B C$ be $H$. Then $H$ is the circumcenter of $\triangle A B C$.
Thus, $A H=\frac{2}{3} \times 3 \times \frac{\sqrt{3}}{2}=\sqrt{3}$.
Therefore, $\angle S A H=30^{\circ}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,676 |
4. Given the function
$$
f(x)=\frac{x^{4}+\left(k^{2}+2 k-4\right) x^{2}+4}{x^{4}+2 x^{2}+4}
$$
has a minimum value of 0. Then the value of the non-zero real number $k$ is ( ).
(A) -4
(B) -2
(C) 2
(D) 4 | 4. B.
Notice that
$$
f(x)=1+\left(k^{2}+2 k-6\right) \frac{x^{2}}{x^{4}+2 x^{2}+4} \text {. }
$$
Since $x^{4}+4 \geqslant 4 x^{2}$, we have
$$
0 \leqslant \frac{x^{2}}{x^{4}+2 x^{2}+4} \leqslant \frac{1}{6} \text {. }
$$
When $k^{2}+2 k-6 \geqslant 0$, $f_{\text {min }}=1>0$;
When $k^{2}+2 k-6<0$,
$$
f_{\min }=1+\fr... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,677 |
5. The eight vertices of the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ are all on the surface of sphere $O$, where,
$$
A A_{1}=1, A B=2 \sqrt{2}, A D=3 \sqrt{3} .
$$
Then the spherical distance between points $B$ and $C$ is ( ).
(A) $\frac{2 \pi}{3}$
(B) $\frac{4 \pi}{3}$
(C) $2 \pi$
(D) $4 \pi$ | 5. C.
The radius of sphere $O$ is
$$
R=\frac{1}{2} \sqrt{1+(2 \sqrt{2})^{2}+(3 \sqrt{3})^{2}}=3 \text {. }
$$
In $\triangle O B C$,
$$
O B=O C=3, B C=A D=3 \sqrt{3} \text {, }
$$
then $\cos \angle B O C=-\frac{1}{2}$.
Thus, $\angle B O C=\frac{2 \pi}{3}$.
Therefore, the spherical distance between points $B$ and $C$ ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,678 |
6. For any real number $m$, the tangent line to the graph of the function
$$
f(x)=x^{2}+m x+1
$$
at the point $(2, f(2))$ always passes through a fixed point $P$. Then the coordinates of $P$ are ( ).
(A) $(0,3)$
(B) $(0,-3)$
(C) $\left(\frac{3}{2}, 0\right)$
(D) $\left(-\frac{3}{2}, 0\right)$ | 6. B.
Since $f^{\prime}(x)=2 x+m$, we have
$$
f^{\prime}(2)=4+m \text{. }
$$
Therefore, the equation of the tangent line at $(2, f(2))$ is
$$
y-(5+2 m)=(4+m)(x-2),
$$
which simplifies to $y=(m+4) x-3$.
Thus, the tangent line always passes through $(0,-3)$. | B | Calculus | MCQ | Yes | Yes | cn_contest | false | 723,679 |
7. Let $A_{1}$ and $A_{2}$ be the left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. If there exists a point $P$ on the ellipse, different from $A_{1}$ and $A_{2}$, such that $\overrightarrow{P O} \cdot \overrightarrow{P A_{2}}=0$, where $O$ is the origin, then the range of the e... | 7. D.
From the given, we know $\angle O P A_{2}=90^{\circ}$.
Let $P(x, y)(x>0)$.
The equation of the circle with $\mathrm{OA}_{2}$ as its diameter is
$$
\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4} \text {, }
$$
Combining this with the ellipse equation, we get
$$
\left(1-\frac{b^{2}}{a^{2}}\right) x^{2}-a x+b... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,680 |
8. Let $F(x, y)=(x-y)^{2}+\left(\frac{x}{3}+\frac{3}{y}\right)^{2}(y \neq 0)$. Then the minimum value of $F(x, y)$ is ( ).
(A) $\frac{12}{5}$
(B) $\frac{16}{5}$
(C) $\frac{18}{5}$
(D) 4 | 8. C.
Let the moving points be $P\left(x,-\frac{x}{3}\right)$ and $Q\left(y, \frac{3}{y}\right)$. Then $F(x, y)=|P Q|^{2}$.
Thus, the trajectory of point $P$ is the line $y=-\frac{x}{3}$, and the trajectory of point $Q$ is the hyperbola $y=\frac{3}{x}$. The distance from any point $\left(x_{0}, \frac{3}{x_{0}}\right)... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,681 |
9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$
Then $f(2010)=$ $\qquad$ . | $$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$
From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,682 |
10. 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 $\qquad$ . | 10.4.
The figure determined by $3|x+1|+2|y-1| \leqslant 6$ is the quadrilateral $ABCD$ and its interior, where,
$$
A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) \text {. }
$$
By the knowledge of linear programming, the maximum value of $2x-3y$ is 4.
The maximum value can be achieved when $x=-1, y=-2$. | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,683 |
11. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, when $n \geqslant 2$, $a_{n} 、 S_{n} 、 S_{n}-\frac{1}{2}$ form a geometric sequence. Then $\lim _{n \rightarrow \infty} n^{2} a_{n}=$ $\qquad$ . | 11. $-\frac{1}{2}$.
From the condition, when $n \geqslant 2$,
$$
S_{n}^{2}=a_{n}\left(S_{n}-\frac{1}{2}\right)=\left(S_{n}-S_{n-1}\right)\left(S_{n}-\frac{1}{2}\right) \text {. }
$$
Thus, $\frac{1}{S_{n}}-\frac{1}{S_{n-1}}=2$.
$$
\text { Then } \frac{1}{S_{n}}=\frac{1}{S_{1}}+2(n-1)=2 n-1 \text {. }
$$
Therefore, $S... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,684 |
12. The capacity of a set refers to the sum of its elements. Then the total capacity of all non-empty sets $A$ that satisfy the condition “ $A \subseteq\{1,2, \cdots, 7\}$, and if $a \in A$ then $8-a \in A$ ” is
(Answer with a specific number).
| 12.224.
First, find the single-element and two-element sets that satisfy the conditions:
$$
A_{1}=\{4\}, A_{2}=\{1,7\}, A_{3}=\{2,6\}, A_{4}=\{3,5\} .
$$
Then, any combination of elements from these four sets also meets the requirements.
Therefore, the total sum of elements in all sets $A$ that satisfy the condition... | 224 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,685 |
3: There is an $8 \times 8$ chessboard, and at the start, each of the 64 small squares contains a "castle" chess piece. If a castle chess piece can attack an odd number of other castle chess pieces still on the board, it is removed. Question: What is the maximum number of castle chess pieces that can be removed (a cast... | First, prove that the pieces on the four corners of the chessboard will not be taken away. Since a castle on a corner can attack at most two other castles, if this castle is taken away, it means that the castle can only attack one. Without loss of generality, we can assume that the castle in the upper left corner is th... | 59 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,686 |
13. Given the function
$$
\begin{aligned}
f(x)= & \sin \left(x+\frac{\pi}{4}\right)+2 \sin \left(x-\frac{\pi}{4}\right)- \\
& 4 \cos 2 x+3 \cos \left(x+\frac{3 \pi}{4}\right) .
\end{aligned}
$$
(1) Determine whether the function $f(x)$ is odd, even, or neither, and provide a proof;
(2) Find the minimum and maximum valu... | Three, 13. (1) Notice that
$$
\begin{aligned}
f(x)= & \frac{\sqrt{2}}{2}(\sin x+\cos x)+\sqrt{2}(\sin x-\cos x)- \\
& 4 \cos 2 x-\frac{3 \sqrt{2}}{2}(\cos x+\sin x) \\
= & -2 \sqrt{2} \cos x-4 \cos 2 x .
\end{aligned}
$$
Since $\cos x$ and $\cos 2 x$ are both even functions, $f(x)$ is an even function.
(2) Notice that... | (f(x))_{\min }=2 \sqrt{2}-4, (f(x))_{\max }=\frac{17}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,687 |
14. Given that $F$ is the focus of the parabola $y^{2}=4 x$, point $M(4,0)$, a line with slope $k_{1}$ is drawn through point $F$ intersecting the parabola at points $A$ and $B$. Extend $A M$ and $B M$ to intersect the parabola again at points $C$ and $D$, respectively. Let the slope of line $C D$ be $k_{2}$.
(1) Find ... | 14. (1) From the given conditions, we know $F(1,0)$. Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right)$, and $D\left(x_{4}, y_{4}\right)$.
Without loss of generality, assume $y_{1}>0$.
By combining $l_{A B}: y=k_{1}(x-1)$ with $y^{2}=4 x$, we get $y^{2}-\frac{4}{k_{1}} y-4=0$.
... | \frac{k_{1}}{k_{2}}=4, \left(0, \arctan \frac{3}{4}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,688 |
15. Given the function
$$
f(x)=x^{3}-m x^{2}-x+1(m \in \mathbf{R}) \text {. }
$$
(1) Find the monotonic intervals of the function $f(x)$;
(2) If for all real numbers $x$, we have
$$
f^{\prime}(x) \geqslant|x|-\frac{7}{4}
$$
holds, find the range of the real number $m$. | 15. (1) Since $f^{\prime}(x)=3 x^{2}-2 m x-1$, and
$$
\Delta=4 m^{2}+12>0 \text {, }
$$
therefore, $f^{\prime}(x)=0$ has two distinct real roots
$$
x_{1}=\frac{m-\sqrt{m^{2}+3}}{3}, x_{2}=\frac{m+\sqrt{m^{2}+3}}{3} .
$$
Clearly, $x_{1}x_{2}$ or $x0$, i.e., $f(x)$ is monotonically increasing.
In summary, the monotonic... | [-1,1] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,689 |
16. Given that $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, for any positive integer $n$, we have
$$
(1-b) S_{n}=-b a_{n}+4^{n}(b>0)
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $c_{n}=\frac{a_{n}}{4^{n}}\left(n \in \mathbf{N}_{+}\right)$, if... | 16. (1) When $n=1$, we have
$$
(1-b) a_{1}=-b a_{1}+4 \text {. }
$$
Thus, $a_{1}=4$.
When $n \geqslant 2$,
$$
(1-b) S_{n}=-b a_{n}+4^{n}
$$
and $(1-b) S_{n-1}=-b a_{n-1}+4^{n-1}$.
Then $(1-b) a_{n}=-b\left(a_{n}-a_{n-1}\right)+3 \times 4^{n-1}$,
which simplifies to $a_{n}=b a_{n-1}+3 \times 4^{n-1}$.
(i) If $b=4$, t... | \left(0, \frac{5}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,690 |
2. Let the sequence $\left\{8 \times\left(-\frac{1}{3}\right)^{n-1}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the smallest integer $n$ that satisfies the inequality
$$
\left|S_{n}-6\right|<\frac{1}{125}
$$
is | 2. 7 | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,692 |
3. Given $n(n \in \mathbf{N}, n \geqslant 2)$ is a constant, and $x_{1}$, $x_{2}, \cdots, x_{n}$ are any real numbers in the interval $\left[0, \frac{\pi}{2}\right]$. Then the function
$$
\begin{array}{l}
f\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
=\sin x_{1} \cdot \cos x_{2}+\sin x_{2} \cdot \cos x_{3}+\cdots+ \\
\q... | 3. $\frac{n}{2}$ | \frac{n}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,693 |
5. A bug crawls along a triangular iron ring, and at each vertex it has an equal chance of crawling to one of the other two vertices. Then the probability that it is back at the starting point after $n$ crawls is
保留源文本的换行和格式，这里的翻译已经按照要求完成。如果需要进一步的帮助或调整，请告知。 | 5.
$$
\frac{2^{n}+2 \times(-1)^{n}}{3 \times 2^{n}}
$$ | \frac{2^{n}+2 \times(-1)^{n}}{3 \times 2^{n}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,695 |
6. Let $O$ be a fixed point on a plane, and $A, B, C$ be three non-collinear points on the plane. A moving point $P$ satisfies
$$
\overrightarrow{O P}-\lambda \frac{\overrightarrow{A C}}{|\overrightarrow{A C}|}=\overrightarrow{O A}+\lambda \frac{\overrightarrow{A B}}{|\overrightarrow{A B}|}(\lambda \in[0,+\infty)) \tex... | 6. The angle bisector of $\angle B A C$ | \text{The angle bisector of } \angle B A C | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,696 |
4. Let $S$ be a finite set of points in the plane (the number of points is greater than or equal to 5), some of which are colored red, and the rest are colored blue. Suppose that no three or more points of the same color are collinear. Prove that there exists a triangle such that
(1) its three vertices are of the same ... | For any five points in $S$, coloring them red or blue must result in three points of the same color (pigeonhole principle), so conclusion (1) holds. There are finitely many triangles with three vertices of the same color, and among them, there must be one with the smallest area (let it be $\triangle A B C$). Then $\tri... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,697 |
1. (16 points) Does there exist a real number $a$, such that the line $y=a x+1$ intersects the hyperbola $3 x^{2}-y^{2}=1$ at two points $A$ and $B$, and the circle with diameter $AB$ passes exactly through the origin of the coordinate system? | 1. Let points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
Substitute $y=a x+1$ into $3 x^{2}-y^{2}=1$, and eliminate $y$ to get
$$
\left(3-a^{2}\right) x^{2}-2 a x-2=0 \text {. }
$$
Then $x_{1}$ and $x_{2}$ are the roots of the equation
$$
\left(3-a^{2}\right) x^{2}-2 a x-2=0
$$
By Vieta's formulas... | a= \pm 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,700 |
2. (20 points) Prove: There does not exist a function $f: \mathbf{Z} \rightarrow\{1,2,3\}$, such that for any integers $x, y$, if $|x-y| \in\{2,3,5\}$, then $f(x) \neq f(y)$.
| 2. Suppose there exists such a function $f$. Then for any integer $n$, let
$$
f(n)=a, f(n+5)=b,
$$
where, $a, b \in\{1,2,3\}$.
By the condition, $a \neq b$.
$$
\begin{array}{l}
\text { Since }|(n+5)-(n+2)|=3, \\
|n-(n+2)|=2,
\end{array}
$$
thus $f(n+2) \neq a$ or $b$, meaning $f(n+2)$ is the remaining number in $\{1,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,701 |
3. (20 points) Given non-negative real numbers $a, b, c$ satisfying $a+b+c=1$. Prove:
$$
9 a b c \leqslant a b+b c+c a \leqslant \frac{1}{4}(1+9 a b c) .
$$ | 3. Proof 1 First, we prove the left inequality.
Using the AM-GM inequality, we get
$$
\begin{array}{l}
a b + b c + c a \\
= (a b + b c + c a)(a + b + c) \\
\geqslant 3 \sqrt[3]{a^{2} b^{2} c^{2}} \cdot 3 \sqrt[3]{a b c} = 9 a b c .
\end{array}
$$
Next, we prove the right inequality.
Assume without loss of generality ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,702 |
1. Given real numbers $a, b$ satisfy $a^{2}+b^{2}=a+b$. Then the range of values for $a+b$ is ().
(A) $a+b \geqslant 0$
(B) $0 \leqslant a+b \leqslant 2$
(C) $0 \leqslant a+b \leqslant 2 \sqrt{2}$
(D) $0 \leqslant a+b \leqslant 3$ | -.1. B.
Let $a+b=t$. Then $ab=\frac{t^{2}-t}{2}$. Therefore, $0 \leqslant(a-b)^{2}=(a+b)^{2}-4ab=2t-t^{2}$. Solving this, we get $0 \leqslant t \leqslant 2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,703 |
2. As shown in Figure 1, in a $3 \times 4$ table, one $1 \times 1$ small square in the top-left corner is colored red. Then, the number of rectangles in this table that include the red small square is ( ) .
(A) 11
(B) 12
(C) 13
(D) 14 | 2. B.
In this table, the rectangle can be determined by the two endpoints of its diagonal. Since it needs to include the red small square, one endpoint is determined, and the other endpoint of the diagonal has $3 \times 4=12$ choices. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,704 |
3. Given that the three sides of the acute triangle $\triangle A B C$ are three consecutive positive integers, and $A B>B C>C A$. If the altitude from $A$ to side $B C$ is $A D$, then $B D-D C=(\quad)$.
(A) 3
(B) 4
(C) 5
(D) 6 | 3. B.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
A B^{2}-B D^{2}=A C^{2}-C D^{2} . \\
\text { Then }(B D+C D)(B D-C D) \\
=(A B+A C)(A B-A C) .
\end{array}
$$
Let $A B=n+2, B C=n+1, A C=n$. Then
$$
B D+D C=n+1 \text {, }
$$
Substituting into the above equation will yield the solution. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,705 |
4. Given rational numbers $a, b, c$ satisfy
$$
a^{2}+b^{2}+1=2\left(c^{2}+a b+b-a\right) .
$$
Then the value of $a-b-c$ is ( ).
(A) 0
(B) -1
(C) $\pm 1$
(D) uncertain | 4. B.
The original equation can be transformed into $(a-b+1)^{2}=2 c^{2}$.
Since $a$, $b$, and $c$ are rational numbers, it follows that,
$$
c=0, a-b+1=0 \text {. }
$$
Therefore, $a-b-c=-1$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,706 |
5. In trapezoid $A B C D$, $A D / / B C, B C=B D$, $A D=C D$, and $\angle C=80^{\circ}$. Then the degree measure of $\angle A$ is ( ).
(A) $120^{\circ}$
(B) $130^{\circ}$
(C) $140^{\circ}$
(D) $150^{\circ}$ | 5. D.
It is known that $\angle B D A=20^{\circ}$.
Construct a regular $\triangle C D O$ inward with $C D$ as a side, and connect $O B$.
Then $\triangle A B D \cong \triangle O B D \cong \triangle O B C$.
Thus, $\angle A=\angle D O B=150^{\circ}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,707 |
Example 1 Given that $AB$ is a chord of $\odot O$ with radius 1, and $AB=a<1$. Using $AB$ as one side, construct a regular $\triangle ABC$ inside $\odot O$, and let $D$ be a point on $\odot O$ different from point $A$, such that $DB=AB=a$, and the extension of $DC$ intersects $\odot O$ at point $E$. Then the length of ... | Solve As shown in Figure 2, connect $A D, O E, O A$.
Notice that
$$
D B=A B=B C \text {. }
$$
By Judgment 1, $B$ is the circumcenter of $\triangle A C D$.
Then, by Property 2(1), $\angle A D C=\frac{1}{2} \angle A B C$.
Thus, $O$ is the circumcenter of $\triangle A D E$.
Therefore, $\angle A D E=\frac{1}{2} \angle A ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,708 |
$$
\begin{array}{l}
\text { 3. Given } E F=C E \\
=C F, E A=B F \\
=2 A B, A B=B D \\
=D A, \text { and } A P= \\
C P=B Q=C Q \\
=P D=D Q=1 \text {. Then the line segment } B D=\text {. }
\end{array}
$$ | From the problem, we know that the conditions determine the symmetry of the entire figure.
As shown in Figure 4, connect $C A, C B, C D$.
It is easy to prove
$$
\begin{array}{l}
\triangle A E C \cong \\
\triangle B F C \\
\Rightarrow C A=C B .
\end{array}
$$
Since $D A=D B, C D=C D$, therefore, $\triangle A D C \cong... | B D=\frac{\sqrt{19}}{19} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,709 |
5. In $\triangle A B C$, it is known that $A B=A C$, point $D$ is on side $A C$, $\angle A D B=60^{\circ}$, point $E$ is on side $B D$, $\angle E C B=30^{\circ}$. Find the degree measure of $\angle A E C$. | ```
Let point $F$ be the reflection of point $E$ over $BC$.
\[
\begin{array}{l}
\text { Then } \angle B F C=\angle B E C \\
=180^{\circ}-\frac{1}{2} \angle B A C .
\end{array}
\]
By criterion 3, $A$ is the circumcenter of $\triangle B C F$.
It is also easy to prove that $\triangle C E F$ is an isosceles triangle.
Thus... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,710 |
4. Let the line $l$ passing through the point $M(2,0)$ intersect the parabola $y^{2}=4 x$ at points $A$ and $B$, and intersect the circle $(x-4.5)^{2}+y^{2}=16$ at points $C$ and $D$. If $|A C|=|B D|$ and $|A B| \neq|C D|$, then the equation of the line $l$ is . $\qquad$ | 4. $x=2, y-2 x+4=0, y+2 x-4=0$.
Among the 24 permutations of $A, B, C, D$, the 8 that satisfy $|A C|=|B D|$ and $|A B| \neq|C D|$ are as follows:
$A, C, D, B ; \quad A, D, C, B ; \quad B, C, D, A$;
$B, D, C, A ; \quad C, A, B, D ; \quad D, A, B, C$;
$C, B, A, D ; \quad D, B, A, C$.
Their common feature is that the mid... | x=2, y-2 x+4=0, y+2 x-4=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,711 |
5. Given the six edges of a tetrahedron are $\sqrt{5}$, $\sqrt{3}$, $\sqrt{2}$, $\sqrt{2}$, $\sqrt{2}$, and $\sqrt{2}$. Then the cosine of the angle between the two longer edges is $\qquad$. | 5. $\frac{\sqrt{15}}{5}$.
In Figure 2, in the tetrahedron $A-BCD$,
$$
AB=\sqrt{5} \text {. }
$$
If the two longer edges are skew lines, then
$$
CD=\sqrt{3} \text {. }
$$
Take the midpoint $E$ of $CD$, and connect $AE$ and $BE$.
In the isosceles triangle, we have
$$
AE=BE=\sqrt{(\sqrt{2})^{2}-\left(\frac{\sqrt{3}}{2}... | \frac{\sqrt{15}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,712 |
6. For $0<x<1$, if the complex number
$$
z=\sqrt{x}+\mathrm{i} \sqrt{\sin x}
$$
corresponds to a point, then the number of such points inside the unit circle is $n=$ | 6. 1 .
From the point on the unit circle, we have
$$
x+\sin x=1(00(x \in(0,1))$, which means $\varphi(x)$ is a strictly increasing function.
Also, $\varphi(0)=-10$, so the equation $x+\sin x=1$ has exactly one real root in $(0,1)$.
Therefore, $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,713 |
7. From $S=\{0,1, \cdots, 7\}$, randomly select a subset $A$ $=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$, and the remaining elements form the complement set $B=$ $\left\{b_{1}, b_{2}, b_{3}, b_{4}\right\}$. Then the probability that the selected subset $A$ satisfies
$$
a_{1}+a_{2}+a_{3}+a_{4}>b_{1}+b_{2}+b_{3}+b_{4}
$... | 7. $\frac{31}{70}$.
The number of ways to choose a 4-element subset is $\mathrm{C}_{8}^{4}=70$, among which, the number of ways that satisfy
$$
a_{1}+a_{2}+a_{3}+a_{4}=b_{1}+b_{2}+b_{3}+b_{4}=14
$$
is 8:
$$
\begin{array}{l}
(7,6,1,0),(7,5,2,0),(7,4,3,0), \\
(7,4,2,1),(6,5,3,0),(6,5,2,1), \\
(6,4,3,1),(5,4,3,2) .
\end... | \frac{31}{70} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,714 |
8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.
Then $m=$ $\qquad$ $(\bmod 2011)$. | 8. 0 .
Notice
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,715 |
9. (16 points) In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $A B$. Find the distance between the skew lines $D E$ and $B_{1} C$.
In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $A B$. Find the distance between the skew lines $D E$ and $B_{1} C$. | 9. Solution 1 As shown in Figure 3, connect $A_{1} C$, $A_{1} D$, and $A_{1} E$.
Since $B_{1} C / / A_{1} D$, we know
$B_{1} C / /$ plane $A_{1} D E$.
Let the distance between the skew lines $D E$ and $B_{1} C$ be $d$. Then the distance from point $C$ to plane $A_{1} D E$ is $d$.
Notice that
$$
\begin{array}{l}
S_{\tri... | \frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,716 |
10. (20 points) Given the equation of curve $C$ as
$$
\frac{\left(16 \lambda^{2}-9\right) x^{2}}{112}+\frac{\lambda^{2} y^{2}}{7}=1(|x| \leqslant 4) \text {. }
$$
(1) For $\lambda>0$, discuss the shape of curve $C$;
(2) Let $M$ be a moving point on curve $C$, and $P$ be a point on the line passing through $M$ and perpe... | 10. (1) Discuss in four cases (see Fig. 4).
(i) When $0<\lambda<\frac{3}{4}$, the curve
$C$ is
$$
\frac{x^{2}}{\frac{112}{16 \lambda^{2}-9}}+\frac{\lambda^{2} y^{2}}{7}=1(|x| \leqslant 4),
$$
This is the part of an ellipse centered at the origin with the major axis on the $x$-axis, satisfying $|x| \leqslant 4$.
(ii) W... | \frac{x^{2}}{16}+\frac{y^{2}}{7}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,717 |
11. (20 points) Let the function
$$
f(x)=\frac{a}{2} x^{2}+x-a
$$
have its maximum value $g(a)$ on the domain $[\sqrt{2}, 2]$.
(1) Find the analytical expression for $g(a)$;
(2) When $g(a)>g\left(\frac{1}{a}\right)$, find the range of real numbers $a$.
| 11. (1) When $a \geqslant 0$, $f^{\prime}(x)=a x+1>0$, so $f(x)$ is monotonically increasing on $[\sqrt{2}, 2]$. Therefore, the maximum value is obtained at the right endpoint, and
$$
g(a)=f(2)=a+2 .
$$
When $a<0$,
$$
f^{\prime}(x)=a x+1=0 \Rightarrow x=-\frac{1}{a}>0 .
$$
We discuss the following three cases.
(i) Wh... | \left(-\frac{\sqrt{2}}{2}, 0\right) \cup(1,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,718 |
一、(40 points) As shown in Figure 1, in the cyclic pentagon $A B C D E$, $\overparen{A E D} = \overparen{A B}$, the diagonals $A C$ and $B D$ intersect at point $P$, and a point $Q$ is taken on the extension of $B D$. Prove that the necessary and sufficient condition for points $C, P, E,$ and $Q$ to be concyclic is that... | Connect $A D$. From $\overparen{A E D}=\overparen{A B}$, we know
$$
\angle A B D=\angle A D B,
$$
and the extension of $B D$ intersects the extension of $A E$.
When $A$, $E$, and $Q$ are collinear, the extension of $B D$ intersects the extension of $A E$ at point $Q$. Then
$\angle D E Q=\angle A B D=\angle A D B$.
Thu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,719 |
For the function $f(x)$, if $f\left(x_{0}\right)=$ $x_{0}$, then $x_{0}$ is called a fixed point of $f(x)$. Given the function
$$
f(x)=\frac{x}{a x+b}(b>0), f(2)=1,
$$
and it has a unique fixed point in its domain.
(1) Find the expression for $f(x)$;
(2) If the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\pi, a... | (1) Given $f(2)=1$, we have
$$
2 a+b=2 \text {. }
$$
Also, since $f(x)$ has a unique fixed point, the equation
$$
x(a x+b-1)=0
$$
has a unique root (obviously 0).
We will discuss this in two cases.
(i) When $a=0$, from equation (1) we get $b=2$, hence
$$
f(x)=\frac{x}{2} \text {. }
$$
(ii) When $a \neq 0$, from equat... | \frac{2011(2+1005 \pi)}{2 \pi} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,720 |
Example 1 Given in the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and $a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$.
Find $a_{n}$. 1$]$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $b_{n}=\sqrt{1+24 a_{n}} \geqslant 0$. Then $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$.
Thus, $4 b_{n+1}^{2}=\left(b_{n}+3\right)^{2}$.
Hence $2 b_{n+1}=b_{n}+3 \Rightarrow \frac{b_{n+1}-3}{b_{n}-3}=\frac{1}{2}$.
It is easy to get $b_{n}=3+\left(\frac{1}{2}\right)^{n-2}$.
Therefore, $a_{n}=\frac{1}{24}\left[\left... | a_{n}=\frac{1}{24}\left[\left(\frac{1}{2}\right)^{n-2}+3\right]^{2}-\frac{1}{24} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,721 |
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