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For a positive geometric sequence $\left\{a_{n}\right\}$ with the first term greater than the common ratio, $n$ being a natural number greater than 1. Prove:
$$
a_{1} \sqrt{a_{2}} \cdot \sqrt[3]{a_{3}} \cdots \cdots \sqrt[n]{a_{n}} \leqslant \sqrt{a_{1}^{n} a_{n}} .
$$ | Proof: When $n=2$,
Left side $=a_{1} \sqrt{a_{2}}$,
Right side $=\sqrt{a_{1}^{2} a_{2}}=a_{1} \sqrt{a_{2}}$.
Therefore, the inequality to be proved holds.
When $n>2$, let the common ratio of $\left\{a_{n}\right\}$ be $q$.
Obviously, $q>0$ and $\frac{a_{1}}{q}>1$.
Let $1n$,
$\frac{1}{i}+\frac{1}{n+1-i}=\frac{n+1}{i(n+1-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,074 |
4. As shown in Figure 17, $E F G H$ is an inscribed quadrilateral in square $A B C D$, and $\angle B E G$ and $\angle C F H$ are both acute angles. Given that $E G=3, F H=4$, and the area of quadrilateral $E F G H$ is 5. Find the area of square $A B C D$. | (提示: Auxiliary lines as shown in Figure 17, forming rectangle $P Q R T$. It is easy to know that $S_{\text {square } A B C D}+S_{\text {rectangle } P R R T}=2 S_{\text {quadrilateral } E F G H}$. Let $P Q=$ $x, Q R=y$, and the side length of the square be $a$. Then $x=$ $\sqrt{9-a^{2}}, y=\sqrt{16-a^{2}}$. Substituting... | \frac{44}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,075 |
5. In a $7 \times 7$ unit square grid, there are 64 grid points, and there are many squares with these grid points as vertices. How many different values can the areas of these squares have?
(21st Jiangsu Province Junior High School Mathematics Competition) | (提示: Consider the number of different chordal graphs in a square grid with side length less than or equal to 7 (side length is an integer) (including extreme cases, excluding those with the same area). As shown in Figure 18, the area of the shaded square is $a^{2}+$ $b^{2}$, where $0 \leqslant a+b \leqslant 7$. Without... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,076 |
Example 1 Given that $\alpha^{2008}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$. Find the sum of the coefficients of this polynomial.
(2005, China Western Mathematical Olympiad) | Explanation: Let $S_{n}=\alpha^{n}+\beta^{n}\left(n \in \mathbf{N}_{+}\right)$,
$$
\left(\sigma_{1}, \sigma_{2}\right)=(\alpha+\beta, \alpha \beta) \text {. }
$$
Then $S_{1}=\sigma_{1}$,
$$
S_{2}=\sigma_{1} S_{1}+2(-1)^{3} \sigma_{2}=\sigma_{1}^{2}-2 \sigma_{2}, \cdots \cdots
$$
By Newton's formula, we have
$$
\alpha... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,077 |
Example 2 Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$
10\left(a^{3}+b^{3}+c^{3}\right)-9\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 \text {. }
$$
(2005, China Western Mathematical Olympiad) | Let $S_{k}=a^{k}+b^{k}+c^{k}\left(k \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
\left(\sigma_{1}, \sigma_{2}, \sigma_{3}\right)=(1, a b+b c+c a, a b c), \\
S_{0}=3, S_{1}=1 .
\end{array}
$$
By Newton's formulas, we have
$$
\begin{array}{l}
S_{2}=\sigma_{1} S_{1}+2(-1)^{3} \sigma_{2}=1-2 \sigma_{2}, \\
S_{3}=\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,078 |
Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
x-y+z-w=2, \\
x^{2}-y^{2}+z^{2}-w^{2}=6, \\
x^{3}-y^{3}+z^{3}-w^{3}=20, \\
x^{4}-y^{4}+z^{4}-w^{4}=66^{[2]} .
\end{array}\right.
$$
(2006, National High School Mathematics Competition) | Define the sequences
$$
\begin{array}{l}
\left\{a_{n}\right\}: a_{n}=x^{n}+z^{n}\left(n \in \mathbf{N}_{+}\right), \\
\left\{b_{n}\right\}: b_{n}=y^{n}+w^{n}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Thus, the original system of equations is
$$
\left\{\begin{array}{l}
a_{1}=b_{1}+2, \\
a_{2}=b_{2}+6, \\
a_{3}=... | (x, y, z, w)=(1,2,3,0),(1,0,3,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,079 |
Example 4 Given $x, y, z \in \mathbf{R}$, and $x+y+z=0$.
(1) Prove:
$$
6\left(x^{3}+y^{3}+z^{3}\right)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{3} \text {; }
$$
(2) Find the best constants $\lambda, \mu$, such that
$$
\begin{array}{l}
\lambda\left(x^{6}+y^{6}+z^{6}\right) \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{3}... | (1) If $x=y=z=0$, then the inequality obviously holds.
Assuming $x, y, z$ are not all zero, then at least one is positive and one is negative. Without loss of generality, assume $x y<0$. Thus, the inequality
$$
6\left(x^{3}+y^{3}+z^{3}\right)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{3},
$$
is equivalent to $54 x^... | \lambda_{\text {max }}=\frac{36}{13}, \mu_{\min }=4 | Inequalities | proof | Yes | Yes | cn_contest | false | 720,080 |
Example 5 Find a pair of positive integers $a, b$ that satisfy: (1) $ab(a+b)$ is not divisible by 7; (2) $7^{7} \mid \left[(a+b)^{7}-a^{7}-b^{7}\right]$.
(25th IMO) | Explanation: Substitution
$$
(x, y, z)=(a+b,-a,-b) \text {. }
$$
Then $x+y+z=0$, and
$$
\begin{array}{l}
\left(\sigma_{1}, \sigma_{2}, \sigma_{3}\right)=(0, x y+y z+z x, x y z) . \\
\text { Let } S_{n}=x^{n}+y^{n}+z^{n}(n \in \mathbf{N}) \text {. Then } \\
S_{0}=3, S_{1}=0, S_{2}=-2 \sigma_{2}, \\
S_{n+3}=-\sigma_{2} ... | (a, b)=(18,1),(1,18) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,081 |
Example 6 Find the value of $\cos ^{5} \frac{\pi}{9}+\cos ^{5} \frac{5 \pi}{9}+\cos ^{5} \frac{7 \pi}{9}$. | Let $\left(x_{1}, x_{2}, x_{3}\right)$
$$
=\left(\cos \frac{\pi}{9}, \cos \frac{5 \pi}{9}, \cos \frac{7 \pi}{9}\right) \text {. }
$$
To reduce the degree, first calculate the basic symmetric polynomials of $x_{1}, x_{2}, x_{3}$
$$
\begin{array}{l}
\left(\sigma_{1}, \sigma_{2}, \sigma_{3}\right) \\
=\left(x_{1}+x_{2}+x... | \frac{15}{32} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,082 |
1. Let $a, b, c \geqslant 0$. Prove: $a^{3}+b^{3}+c^{3} \geqslant 3 a b c$. | (Tip: Let $a+b+c=1$. Let $S_{n}=a^{n}+$ $b^{n}+c^{n}$. By Newton's power sum formula, we get $S_{3}=\sigma_{1}^{3}-$ $3 \sigma_{1} \sigma_{2}+3 \sigma_{3}$. Therefore, the original inequality is equivalent to $\sigma_{1}^{3}-$ $3 \sigma_{1} \sigma_{2} \geqslant 0 \Leftrightarrow \sigma_{1}^{2}-3 \sigma_{2} \geqslant 0$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,083 |
Example 3 As shown in Figure 5, quadrilateral $ABCD$ is a square, $\odot O$ passes through vertex $A$ and the intersection point $P$ of the diagonals, and intersects $AB, AD$ at points $F, E$ respectively.
(1) Prove: $DE = AF$;
(2) If the radius of $\odot O$ is $\frac{\sqrt{3}}{2}$, and $AB = \sqrt{2} + 1$, find the va... | Analysis: (1) It is easy to think of connecting $E P$ and $F P$ to prove $\triangle D E P \cong \triangle A F P$. This is because $A B C D$ is a square, so $D P=A P, \angle A D P=\angle B A P=45^{\circ}, \angle E P F=$ $180^{\circ}-\angle B A D=90^{\circ}=\angle A P D$, hence $\angle E P D=$ $\angle A P F$ is proved.
(... | \sqrt{2} \text{ or } \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,084 |
2. Solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=n \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n \\
\cdots \cdots \\
x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}=n
\end{array}\right.
$$ | ( Hint: Using Newton's formulas and the initial conditions $S_{1}=$ $S_{2}=\cdots=S_{n}=n$, we get $\sigma_{1}-\sigma_{2}+\cdots+(-1)^{n-2} \sigma_{n-1}+(-1)^{n-1} \sigma_{n}=1$. Therefore, among $x_{1}, x_{2}, \cdots, x_{n}$, there must be one that is 1. By symmetry, assume $x_{n}=1$, which reduces the original system... | x_{1}=x_{2}=\cdots=x_{n}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,085 |
4. Find the integer solutions $(x, y, z)$ of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=0, \\
x^{3}+y^{3}+z^{3}=-18
\end{array}\right.
$$ | (Tip: Note that $x y z \neq 0$. Using Newton's formula, we derive $x y z = -6$. Thus, the system of equations has 6 sets of integer solutions: $(x, y, z) = (1,2,-3), (2,1,-3), (1,-3,2), (2,-3,1), (-3,1,2), (-3,2,1)$. | (1,2,-3), (2,1,-3), (1,-3,2), (2,-3,1), (-3,1,2), (-3,2,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,087 |
5. Definition
$f_{n}=x^{n} \sin n A+y^{n} \sin n B+z^{n} \sin n C\left(n \in \mathbf{N}_{+}\right)$, where $\angle A, \angle B, \angle C \in \mathbf{R}$, and $\angle A+\angle B+\angle C=k \pi(k \in \mathbf{Z})$. If $f_{1}=f_{2}=0$, prove that $f_{n}=0$ for all $n \in \mathbf{N}_{+}$.
(Ninth United States of America Mat... | (Hint: Let $(\alpha, \beta, \gamma)=(x(\cos A+\mathrm{i} \sin A)$, $y(\cos B+\mathrm{i} \sin B), z(\cos C+\mathrm{i} \sin C)), S_{n}=$ $\alpha^{n}+\beta^{n}+\gamma^{n}(n \in \mathbf{N})$. Then $\operatorname{Im}\left(S_{n}\right)=f_{n}$. Supplement $S_{0}$ $=3$, combined with $\sigma_{1} 、 \sigma_{2} 、 \sigma_{3} 、 S_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,088 |
Example 1 Given a positive integer $n(n>3)$, let real numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2} .
\end{array}
$$
Find the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$.
(28th Uni... | The above equation holds for $a \geqslant 2$.
Try taking $a=2$, at this point, $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ $=2$. From this, we can conjecture that the minimum value of $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is 2.
Solution: First, we prove that for any real numbers $a_{1}, a_{2}, \cdots... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,089 |
Example 2 Let $m, n$ be positive integers, $m<2001, n<$ 2002, and there are $2001 \times 2002$ different real numbers. Fill these numbers into the squares of a $2001 \times 2002$ chessboard so that each square contains exactly one number. If a number in a square is less than at least $m$ numbers in its column and also ... | Analysis: To facilitate the calculation of the number of bad cells $S$, we can examine a scenario arranged according to specific rules:
Fill the numbers $1, 2, \cdots, 2001 \times 2002$ in natural order from left to right, row by row, into the $2001 \times 2002$ chessboard (as shown below).
\begin{tabular}{cccc}
1 & 2... | (2001-m)(2002-n) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,090 |
Example 3 Find the smallest positive number $c$, such that for all infinite positive sequences $\left\{x_{n}\right\}$ satisfying $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, the inequality
$$
\begin{array}{l}
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \\
\leqslant c \sqrt{x_{1}+x_{2}+\cdots+x_{n}}(n=1,2, \cdots) .
... | Analysis: To find the smallest positive number $c$, it is necessary to find a constant $A$ such that $c \geqslant A$. This can be achieved by selecting an appropriate infinite sequence of positive numbers $\left\{x_{n}\right\}$ that satisfies $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, and substituting it into inequa... | \sqrt{2}+1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,091 |
Example 4 Given a positive integer $n(n \geqslant 3)$, arrange several distinct positive integers in a circle so that the product of any two adjacent numbers is less than $n$. How many numbers can there be at most on the circle? | Analysis: Let the maximum number of numbers on a circle when the product of any two adjacent numbers is less than $n$ be $f(n)$.
Starting with special cases, it is easy to know
$$
f(3)=f(4)=f(5)=f(6)=2 \text {. }
$$
The key to solving the problem is to realize that any two of the numbers $2,3,4,5,6$ cannot be adjacent... | f_{\text {max }}=2 k-1 \text{ when } k(k+1)<n \leqslant k(k+2); \text{ } f_{\text {max }}=2 k \text{ when } k(k+2)<n \leqslant(k+1)(k+2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,092 |
Example 5 A $98 \times 98$ chessboard is displayed on a computer screen, with its squares colored like a chessboard. You are allowed to select any rectangle with the mouse (the sides of the rectangle must lie on the grid lines), and then click the mouse button, which will change the color of each square in the rectangl... | Analysis: If we consider "all cells of the same color," it would be very difficult (because too many objects can obscure the essential relationships), leading to the idea of adopting a local thinking strategy.
First local thinking: Transform "all cells of the same color" into "any two adjacent cells of the same color.... | 98 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,093 |
Example 4 As shown in Figure 6, point $A$ is on the line segment $B G$, quadrilaterals $A B C D$ and $D E F G$ are both squares, with areas of $7 \mathrm{~cm}^{2}$ and $11 \mathrm{~cm}^{2}$, respectively. Find the area of $\triangle C D E$. | Analysis: Since $CD = \sqrt{7}$, we only need to find the height $EH$ on side $CD$ (as shown in Figure 6).
From the given conditions, it is easy to see that $\triangle EDH$ is obtained by rotating $\triangle DAG$ around point $D$ by $90^{\circ}$, so,
$$
EH = AG = \sqrt{DG^2 - AD^2} = 2.
$$
Therefore, $S_{\triangle EC... | \sqrt{7} \, \text{cm}^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,095 |
A frog starts from point $(1,1)$ on a Cartesian coordinate system and can jump in the following two ways:
(1) It can jump from any point $(a, b)$ to point $(2a, b)$ or $(a, 2b)$.
(2) For point $(a, b)$, if $a > b$, it can jump from $(a, b)$ to $(a-b, b)$; if $a < b$, it can jump from $(a, b)$ to $(a, b-a)$.
Please con... | The answer is that it can reach points $(3,5)$ and $(200,6)$.
In [1], it is explored that the necessary and sufficient condition for jumping to point $(a, b)$ according to the two specified ways is that the greatest common divisor of $a$ and $b$ is $2^{k}$. It also provides a construction method for the jumping method ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,096 |
Let $a, b, n>0, 0<x<\frac{\pi}{2}$. Find the minimum value of the function
$$
y=\frac{a}{\cos ^{n} x}+\frac{b}{\sin ^{n} x}
$$ | Solution: Notice that
$$
\begin{array}{l}
n+2 \\
=\frac{2 a}{y \cos ^{n} x}+n \cos ^{2} x+\frac{2 b}{y \sin ^{n} x}+n \sin ^{2} x \\
\geqslant(n+2) \sqrt[n+2]{\left(\frac{a}{y \cos ^{n} x}\right)^{2}\left(\cos ^{2} x\right)^{n}}+ \\
(n+2) \sqrt[n+2]{\left(\frac{b}{y \sin ^{n} x}\right)^{2}\left(\sin ^{2} x\right)^{n}} ... | \left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,097 |
1. Find all pairs of positive integers $(k, n)$ such that $\left(7^{k}-3^{n}\right) \mid\left(k^{4}+n^{2}\right)$ | 1. $(2,4)$.
Assume the positive integer pair $(k, n)$ satisfies the condition.
Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even. Therefore, $k$ and $n$ have the same parity.
If $k$ and $n$ are both odd, then
$$
k^{4}+n^{2} \equiv 1+1=2(\bmod 4).
$$
However, $7^{k}-3^{n} \equiv 7-3 \equiv 0(\bmod 4)$, which is ... | (2,4) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,098 |
2. Let $b$ and $n$ be integers greater than 1. If for every positive integer $k$ greater than 1, there exists an integer $a_{k}$ such that $k!\left(b-a_{k}^{n}\right)$, prove: there exists an integer $A$ such that $b=A^{n}$. | 2. Let the prime factorization of $b$ be
$$
b=p_{1}^{a_{1}} p_{2}^{\alpha_{2}} \cdots p_{s}^{\alpha_{s}},
$$
where $p_{1}, p_{2}, \cdots, p_{s}$ are distinct prime numbers, and $\alpha_{1}$, $\alpha_{2}, \cdots, \alpha_{s}$ are positive integers.
We will prove that all exponents $\alpha_{i}$ are divisible by $n$.
Let ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,099 |
3. Let $X$ be a set consisting of 10000 integers, and none of the numbers in the set is a multiple of 47. Prove: There exists a 2007-element subset $Y$ of $X$, such that for any $a, b, c, d, e \in Y$, we have
$$
47 \times (a - b + c - d + e).
$$ | 3. If for any $a, b, c, d, e \in M$, there is $47 \times (a-b+c-d+e)$, then the integer set $M$ is called "good".
Consider the set
$$
J=\{-9,-7,-5,-3,-1,1,3,5,7,9\} \text {. }
$$
The proof that $J$ is good is as follows:
Actually, for any $a, b, c, d, e \in J$, the number $a-b+c-d+e$ is odd, and
$$
\begin{array}{l}
-4... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,100 |
4. For each integer $k(k \geqslant 2)$, prove:
$$
2^{3 k} \|\left(C_{2^{k+1}}^{\mathrm{C}^{k}}-\mathrm{C}_{2^{k}}^{2^{k-1}}\right) \text {. }
$$ | 4. Let $(2 n-1)!!=1 \cdot 3 \cdot 5 \cdots \cdots(2 n-1)$,
$$
(2 n)!!=2 \cdot 4 \cdot 6 \cdots \cdots(2 n)=2^{n} n!,
$$
where $n$ is any positive integer. Then
$$
\begin{array}{l}
(2 n)!=(2 n)!!(2 n-1)!! \\
=2^{n} n!(2 n-1)!!.
\end{array}
$$
For each positive integer $n$, we have
$$
\begin{array}{l}
\mathrm{C}_{4 n}^... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,101 |
5. Find all surjective functions $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ such that for any $m, n \in \mathbf{N}_{+}$ and any prime $p$, $f(m+n)$ is divisible by $p$ if and only if $f(m) + f(n)$ is divisible by $p$. | 5. $f(n)=n$.
Assume the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ satisfies the condition.
Lemma For any prime $p$ and any $x, y \in \mathbf{N}_{+}, x \equiv y(\bmod p)$ if and only if $f(x) \equiv f(y)(\bmod p)$. Additionally, $p \mid f(x)$ if and only if $p \mid x$.
Proof of the lemma: For any prime $... | f(n)=n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,102 |
6. Given $k \in \mathbf{N}_{+}$. Prove: $\left(4 k^{2}-1\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even. | 6. Lemma For any positive integers $x, y$,
$$
(4 x y-1) \mid\left(4 x^{2}-1\right)^{2}
$$
if and only if $x=y$.
This lemma is the 5th problem of this year's IMO.
When $x=k, y=2 n$, by the lemma we have
$$
(8 k n-1) \mid\left(4 k^{2}-1\right)^{2}
$$
if and only if $k=2 n$.
Therefore, when $k$ is odd, there is no such ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,103 |
1. Let $a^{2}+1=3 a, b^{2}+1=3 b$, and $a \neq b$, then the value of the algebraic expression $\frac{1}{a^{2}}+\frac{1}{b^{2}}$ is ( ).
(A) 5
(B) 7
(C) 9
(D) 11 | -、1.B.
From the given conditions, we have
$$
a^{2}-3 a+1=0, b^{2}-3 b+1=0,
$$
and $a \neq b$, so $a$ and $b$ are the two roots of the quadratic equation $x^{2}-3 x+$ $1=0$. Therefore,
$$
\begin{array}{l}
a+b=3, a b=1 . \\
\text { Hence } \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{(a+b)^{2}-2... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,105 |
Example 5 As shown in Figure 7, in trapezoid $A B C D$, $A B / / C D$,
with $A D$ and $B C$ as sides, construct squares $A D E F$ and square $B C G H$ outward, respectively. Let $I$ be the midpoint of line segment $E G$, and connect $I D$ and $I C$. Prove: $I D=I C$.
(2004, National Junior High School Mathematics Compe... | Analysis: At first glance, it seems difficult to effectively establish the connection between the conditions and the conclusion. However, upon careful consideration, from $A B / / C D$, we know that the distance between these two parallel lines is equal everywhere. Therefore, draw the heights $C K$ and $D J$ of the tra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,106 |
2. As shown in Figure 1, let $A D$, $B E$, and $C F$ be the three altitudes of $\triangle A B C$. If $A B=6, B C=$ $5, E F=3$, then the length of line segment $B E$ is ( ).
(A) $\frac{18}{5}$
(B) 4
(C) $\frac{21}{5}$
(D) $\frac{24}{5}$ | 2.D.
Since $A D, B E, C F$ are the three altitudes of $\triangle A B C$, it is easy to see that $B, C, E, F$ are concyclic.
Thus, $\triangle A E F \backsim \triangle A B C$.
Therefore, $\frac{A F}{A C}=\frac{E F}{B C}=\frac{3}{5}$, which means $\cos \angle B A C=\frac{3}{5}$.
Hence, $\sin \angle B A C=\frac{4}{5}$.
In... | \frac{24}{5} | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,107 |
3. From 5 cards each written with the numbers $1, 2, 3, 4, 5$, any 2 cards are taken out. The number on the first card is used as the tens digit, and the number on the second card is used as the units digit to form a two-digit number. Then the probability that the number formed is a multiple of 3 is ( ).
(A) $\frac{1}{... | 3. C.
The two-digit numbers that can be formed are
$$
\begin{array}{l}
12,13,14,15,21,23,24,25,31,32, \\
34,35,41,42,43,45,51,52,53,54,
\end{array}
$$
a total of 20. Among these, the numbers that are multiples of 3 are
$$
12,15,21,24,42,45,51,54 \text {, }
$$
a total of 8. Therefore, the probability that a number fo... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,108 |
4. In $\triangle A B C$, $\angle A B C=12^{\circ}, \angle A C B=$ $132^{\circ}, B M$ and $C N$ are the external angle bisectors of these angles, and points $M, N$ are on lines $A C, A B$ respectively. Then ( ).
(A) $B M>C N$
(B) $B M=C N$
(C) $B M<C N$
(D) The size relationship between $B M$ and $C N$ is uncertain | 4. B.
Since $\angle A B C=12^{\circ}$ and $B M$ is the external angle bisector of $\angle A B C$, we have
$$
\angle M B C=\frac{1}{2}\left(180^{\circ}-12^{\circ}\right)=84^{\circ} .
$$
Also, $\angle B C M=180^{\circ}-\angle A C B=48^{\circ}$, thus
$$
\angle B M C=180^{\circ}-84^{\circ}-48^{\circ}=48^{\circ} \text {. ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,109 |
5. There are five different goods with the same price. Starting from today, each will be discounted by $10\%$ or $20\%$ every day. After several days, the prices of these five goods are all different. Let the ratio of the highest price to the lowest price be $r$. Then the minimum value of $r$ is ( ).
(A) $\left(\frac{9... | 5.B.
It is easy to know that after 4 days, the prices of the five goods can be all different.
Let the price of the five goods before the discount be $a$, and after $n$ days. The price of each good after $n$ days can certainly be expressed as
$$
\begin{array}{l}
a(1-10 \%)^{k}(1-20 \%)^{n-k} \\
=a\left(\frac{9}{10}\ri... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,110 |
6. Given real numbers $x, y$ satisfy $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$. Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ( ).
(A) -2008
(B) 2008
(C) -1
(D) 1 | 6.D.
From $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$,
we get
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008} \\
=\frac{2008}{y-\sqrt{y^{2}-2008}}=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$
From the above two equations, we ... | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,111 |
1. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ | $$
\begin{array}{l}
\text { Given } a^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{3-\sqrt{5}}{2}=1-a \\
\Rightarrow a^{2}+a=1 . \\
\text { Therefore, } \frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a} \\
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a \cdot a^{2}-a} \\
=\frac{a^{3}-2 a^{3}-1+2}{a(1-... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,112 |
3. Given the quadratic function $y=x^{2}+a x+b$ whose graph intersects the $x$-axis at two points with the x-coordinates $m$ and $n$, and $|m|+|n| \leqslant 1$. Let the maximum and minimum values of $b$ that satisfy the above conditions be $p$ and $q$, respectively. Then $|p|+|q|$ | 3. $\frac{1}{2}$.
According to the problem, $m$ and $n$ are the roots of the quadratic equation $x^{2} + a x + b = 0$, then
$$
\begin{array}{l}
m+n=-a, \quad m n=b. \\
\text { Also, } |m|+|n| \leqslant 1, \text { then } \\
|m+n| \leqslant |m|+|n| \leqslant 1, \\
|m-n| \leqslant |m|+|n| \leqslant 1.
\end{array}
$$
Sin... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,114 |
4. Arrange the squares of positive integers $1,2, \cdots$ in a sequence: $149162536496481100121144 \cdots$, the digit at the 1st position is 1, the digit at the 5th position is 6, the digit at the 10th position is 4, the digit at the 2008th position is $\qquad$. | 4.1.
$1^{2}$ to $3^{2}$, each result occupies 1 digit, totaling $1 \times 3$ $=3$ digits;
$4^{2}$ to $9^{2}$, each result occupies 2 digits, totaling $2 \times 6$ $=12$ digits;
$10^{2}$ to $31^{2}$, each result occupies 3 digits, totaling $3 \times$ $22=66$ digits;
$32^{2}$ to $99^{2}$, each result occupies 4 digits, t... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,115 |
One, (20 points) Given $a^{2}+b^{2}=1$. For all real numbers $x$ satisfying $0 \leqslant x \leqslant 1$, the inequality
$$
\begin{array}{l}
a(1-x)(1-x-a x)-b x(b-x-b x) \\
\geqslant 0
\end{array}
$$
always holds. When the product $a b$ takes the minimum value, find the values of $a$ and $b$. | I. Organize equation (1) and substitute $a^{2}+b^{2}=1$ to get
$$
(1+a+b) x^{2}-(2 a+1) x+a \geqslant 0 \text {. }
$$
In equation (2), let $x=0$, we get $a \geqslant 0$; let $x=1$, we get $b \geqslant 0$.
It is easy to see that $1+a+b>0,0<\frac{2 a+1}{2(1+a+b)}<1$. Therefore, the graph (parabola) of the quadratic fun... | a=\frac{\sqrt{6}-\sqrt{2}}{4}, b=\frac{\sqrt{6}+\sqrt{2}}{4} \text{ or } a=\frac{\sqrt{6}+\sqrt{2}}{4}, b=\frac{\sqrt{6}-\sqrt{2}}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,116 |
Example 6 As shown in Figure $9, A, B, C$ are three villages located on an east-west oriented highway, $A B=$ $2 \text{ km}, B C=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle A D C$ $=45^{\circ}$. Now the $\triangle A D C$ area is planned as a development zone, except fo... | Analysis: The problem is clearly to find the length of $BD$, but there is no effective connection between the problem and the given conditions. Since $\angle ADC = 45^{\circ}$, considering its relationship with the right angle, we think of performing an axial symmetry transformation on $\triangle DAB$ and $\triangle DC... | 11 \text{ km}^2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,117 |
II. (25 points) As shown in Figure 3, circle $\odot O$ intersects with circle $\odot D$ at points $A$ and $B$. $BC$ is a tangent to $\odot D$, and point $C$ lies on $\odot O$, with $AB = BC$.
(1) Prove that point $O$ lies on the circumference of $\odot D$;
(2) Let the area of $\triangle ABC$ be $S$. Find the minimum va... | (1) As shown in Figure 4, connect $O A$, $O B$, and $O C$.
Since $O$ is the center of the circle and $A B = B C$, we have
$\triangle O B A \backsim \triangle O B C$.
Thus, $\angle O B A$
$$
= \angle O B C.
$$
Also, since $O D \perp A B$ and $D B \perp B C$, we have
$$
\begin{array}{l}
\angle D O B = 90^{\circ} - \ang... | \frac{\sqrt{2 S}}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 720,118 |
Three. (25 points) Let $a$ be a prime number, $b$ be a positive integer, and
$$
9(2 a+b)^{2}=509(4 a+511 b) \text {. }
$$
Find the values of $a$ and $b$. | Three, Equation (1) is $\left(\frac{6 a+3 b}{509}\right)^{2}=\frac{4 a+511 b}{509}$.
Let $m=\frac{6 a+3 b}{509}, n=\frac{4 a+511 b}{509}$. Then
$$
b=\frac{509 m-6 a}{3}=\frac{509 n-4 a}{511} \text {. }
$$
Thus, $3 n-511 m+6 a=0$.
Since $n=m^{2}$, we have
$$
3 m^{2}-511 m+6 a=0 .
$$
From Equation (1), we know that $(2... | a=251, b=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,119 |
One, (20 points) Given $a^{2}+b^{2}=1$. For all real number pairs $(x, y)$ satisfying $x+y=1, x y \geqslant 0$, the inequality
$$
a y^{2}-x y+b x^{2} \geqslant 0
$$
always holds. When the product $a b$ takes the minimum value, find the values of $a$ and $b$. | Given $x+y=1, x y \geqslant 0$, we know that
$0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1$.
In equation (1), let $x=0, y=1$, we get $a \geqslant 0$; let $x=1, y=0$, we get $b \geqslant 0$.
Substituting $y=1-x$ into equation (1) yields
$$
a(1-x)^{2}-x(1-x)+b x^{2} \geqslant 0,
$$
which simplifies to $(1+a+b) x^... | a=\frac{\sqrt{6}-\sqrt{2}}{4}, b=\frac{\sqrt{6}+\sqrt{2}}{4} \text{ or } a=\frac{\sqrt{6}+\sqrt{2}}{4}, b=\frac{\sqrt{6}-\sqrt{2}}{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,120 |
Three. (25 points) Let $a$ be a prime number, $b$ and $c$ be positive integers, and satisfy
$$
\left\{\begin{array}{l}
9(2 a+2 b-c)^{2}=509(4 a+1022 b-511 c), \\
b-c=2 .
\end{array}\right.
$$
Find the value of $a(b+c)$. | Three, Equation (1) is
$$
\left(\frac{6 a+6 b-3 c}{509}\right)^{2}=\frac{4 a+1022 b-511 c}{509} \text {. }
$$
Let $m=\frac{6 a+6 b-3 c}{509}$,
$$
n=\frac{4 a+1022 b-511 c}{509} \text {. }
$$
Then $2 b-c=\frac{509 m-6 a}{3}=\frac{509 n-4 a}{511}$.
Thus, $3 n-511 m+6 a=0$.
Since $n=m^{2}$, we have
$$
3 m^{2}-511 m+6 a=... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,121 |
1. In $\triangle A B C$, $\lg \sin A-\lg \cos B-\lg \sin C=\lg 2$. Then $\triangle A B C$ must be ( ).
(A) isosceles triangle
(B) equilateral triangle
(C) right triangle
(D) shape uncertain | - 1.A.
From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sin A=2 \cos B \cdot \sin C, \\
\angle A=\pi-(\angle B+\angle C)
\end{array}\right. \\
\Rightarrow \sin (B+C)=2 \cos B \cdot \sin C .
\end{array}
$$
Rearranging, we get $\sin (B-C)=0$.
Thus, $\angle B=\angle C$.
Therefore, $\trian... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,122 |
2. In tetrahedron $ABCD$, if $AB=CD=5$, $AC=BD=4$, $AD=BC=x$, then the range of values for $x$ is ( ).
(A) $1<x<9$
(B) $1<x<\sqrt{41}$
(C) $3<x<9$
(D) $3<x<\sqrt{41}$ | 2.D.
Completing the shape to make the edges of tetrahedron $ABCD$ the face diagonals of a rectangular prism. It is easy to see that $\triangle ABC$ is an acute triangle, hence, $x^{2}+4^{2}>5^{2}$ and $x^{2}<4^{2}+5^{2}$, i.e., $3<x<\sqrt{41}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,123 |
3. Let $E_{i} F_{i}(i=1,2,3,4)$ be line segments parallel to the upper and lower bases of a trapezoid and whose endpoints lie on the legs of the trapezoid. If $E_{1} F_{1}$ divides the trapezoid into two equal areas, $E_{2} F_{2}$ is the midline of the trapezoid, $E_{3} F_{3}$ divides the trapezoid into two similar fig... | 3. B.
Let the upper and lower bases of the trapezoid be $a_{1}$ and $a_{2}$. Then
$$
\begin{array}{l}
E_{4} F_{4}=\frac{2 a_{1} a_{2}}{a_{1}+a_{2}}, E_{3} F_{3}=\sqrt{a_{1} a_{2}}, \\
E_{2} F_{2}=\frac{a_{1}+a_{2}}{2}, E_{1} F_{1}=\sqrt{\frac{a_{1}^{2}+a_{2}^{2}}{2}} .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,124 |
4. Given $x, y \geqslant 1$ and $\sqrt[y]{x-1}+\sqrt[y]{y-1} \leqslant 2$. Then ( ).
(A) $x \geqslant y$
(B) $x \leqslant y$
(C) $x+y \geqslant x y$
(D) $x+y \leqslant x y$ | 4.C.
Due to the symmetry of $x$ and $y$, it is clear that options (A) and (B) are both incorrect.
Let $a=\sqrt[n]{x-1}, b=\sqrt[n]{y-1}$. Then $a^{n}+1=x, b^{n}+1=y$.
Given that $a b \leqslant \frac{a+b}{2}=\frac{\sqrt[n]{x-1}+\sqrt[n]{y-1}}{2} \leqslant 1$,
thus, $a^{n} b^{n} \leqslant 1$. Therefore,
$$
\begin{array... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 720,125 |
5. Let $A=\{0,1,2,3\}, B=\{2,3,4,5,6\}$, and the mapping $f: A \rightarrow B$ satisfies that for any $x \in A, x + f(x) + x f^{2}(x)$ is an odd number. Then the number of such mappings is ( ).
(A) 80
(B) 100
(C) 250
(D) 625 | 5. B.
Notice that when $x=0,2$, $f(x)$ can take the values 3,5; when $x=1,3$, $f(x)$ can take any value in $B$. Therefore, the number of $f(x)$ is $2 \times 2 \times 5 \times 5=100$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,126 |
6. Let quadrilateral $ABCD$ be a square on the $xy$ plane, with vertices $A(0,0)$, $B(1,0)$, $C(1,1)$, and $D(0,1)$. Given the transformation $u=x^{2}-y^{2}$, $v=2xy$ from the $xy$ plane to the $uv$ plane, the image set of the square $ABCD$ is ( ).
Figure 1 is one of the options. | 6. C.
$$
\begin{array}{l}
\text { Line segment } A B:\left\{\begin{array}{l}
0 \leqslant x \leqslant 1, \\
y=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array} { l }
{ u = x ^ { 2 } ( 0 \leqslant u \leqslant 1 ) , } \\
{ v = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
0 \leqslant u \leqslant 1, \\
v=0 .
\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,127 |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=1, a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0 \text {. }
$$
Then $a_{2008}=$ $\qquad$ | 7.2007!.
Dividing both sides of the recurrence relation by $a_{n+1} a_{n}$, we get
$$
\frac{a_{n+2}}{a_{n+1}}-\frac{a_{n+1}}{a_{n}}=1 \text {. }
$$
Thus, the sequence $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. Therefore, $\frac{a_{n+1}}{a_{n}}=n$.
He... | 2007! | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,129 |
8 . The positive integer solutions $(x, y)$ of the equation $1!+2!+\cdots+x!=y^{2}$ are $\qquad$ . | 8. $(1,1),(3,3)$.
When $x=1$, $y=1$, the equation is satisfied;
When $x=2$, $1!+2!=3$, the equation has no solution;
When $x=3$, $1!+2!+3!=9, y=3$, the equation is satisfied;
When $x=4$, $1!+2!+3!+4!=33$, the equation has no solution;
When $x \geqslant 5$, $1!+2!+\cdots+x!$ ends in 3, while the last digit of a squar... | (1,1),(3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,130 |
$$
\begin{array}{l}
\text { 9. Let }\left(1+x+x^{2}\right)^{n} \\
=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {. }
\end{array}
$$
Then the value of $a_{0}+a_{3}+a_{6}+\cdots+a_{3\left[\frac{2 n}{3}\right]}$ is (where, $[x]$ is the greatest integer function of $x$). | $9.3^{n-1}$.
Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$. Then
$$
\begin{array}{l}
\omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}, \\
\omega^{3}=1, \\
\cdots \cdots \\
\omega^{3 k}=1, \\
\omega^{3 k+1}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \\
\omega^{3 k+2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} \math... | 3^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,131 |
10. Given $2<\frac{A_{n+1}^{5}}{A_{n-1}^{3}} \leqslant 42$. Then the value of $n$ is | $10.4,5,6$.
Since $\frac{A_{n+1}^{5}}{\mathrm{~A}_{n-1}^{3}}=\frac{(n+1) n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)}$ $=n(n+1)$,
thus, $22, \\ n^{2}+n \leqslant 42,\end{array}\right.$ we get
$-7 \leqslant n<-2$ or $1<n \leqslant 6$.
Based on the range of $n$, we have $4 \leqslant n \leqslant 6$ and $n \in$ $\mathbf{N}$, hence $... | n=4,5,6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,132 |
11. The number of tetrahedra that can be formed with the vertices of a regular hexagonal prism is $\qquad$.
untranslated part: $\qquad$ | 11.426.
Among 12 vertices, choosing any 4, there are $\mathrm{C}_{12}^{4}$ different ways, but when the 4 chosen points are coplanar, they cannot form a tetrahedron and should be excluded.
The 4 chosen points being coplanar can be divided into 4 cases:
(1) 4 points are on the same side, with 6 sides, and the side is a ... | 426 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,133 |
12. Let the quadratic function $f(x)$ have a positive leading coefficient, and satisfy the inequality
$$
f\left(\frac{3 \pi}{2}-x\right)=f\left(\frac{\pi}{2}+x\right)(x \in \mathbf{R}) \text {. }
$$
Then the solution set of the inequality $f(\arcsin |x|)<f(\arccos |x|)$ is | 12. $\left[-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right]$.
From the problem, we have
$$
f\left(\pi-\left(x-\frac{\pi}{2}\right)\right)=f\left(\pi+\left(x-\frac{\pi}{2}\right)\right) \text {. }
$$
Thus, the parabola $f(x)$, which opens upwards, has its axis of symmetry at $x=\pi$. Therefore, $f(... | \left[-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,134 |
13. Given
$$
\begin{array}{l}
\tan (\alpha+\beta-\gamma) \cdot \tan \beta=\tan (\alpha-\beta+\gamma) \cdot \tan \gamma . \\
\text { Prove: } \frac{\sin (\beta+\gamma)}{\sin \alpha} \cdot \frac{\cos (\beta-\gamma)}{\cos \alpha}=-1 .
\end{array}
$$ | $$
\begin{array}{l}
\frac{\sin (\alpha+\beta-\gamma)}{\cos (\alpha+\beta-\gamma)} \cdot \frac{\cos (\alpha-\beta+\gamma)}{\sin (\alpha-\beta+\gamma)} \\
=\frac{\sin \gamma}{\cos \gamma} \cdot \frac{\cos \beta}{\sin \beta} .
\end{array}
$$
Both sides of the equation above are transformed using the product-to-sum identi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,135 |
14. Let $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, and the equation $\cos 2 x-4 a \cos x-a+2=0$
has two distinct solutions. Find the range of values for $a$. | 14. From the given,
$2 \cos ^{2} x-1-4 a \cos x-a+2=0$,
which is $2 \cos ^{2} x-4 a \cos x-a+1=0$.
Let $t=\cos x$. From $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$ we have
$0 \leqslant t \leqslant 1$.
Thus, the original equation having two distinct solutions is equivalent to the equation $2 t^{2}-4 a t-a+1=0$ ... | \frac{3}{5} \leqslant a<1 \text{ or } a=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,136 |
15. There is a 6-row $n$-column matrix composed of 0s and 1s, where each row contains exactly 5 ones, and the number of columns in which any two rows both have a 1 is at most 2. Find the minimum value of $n$. | 15. First, the total number of 1s in the matrix is $5 \times 6=$ 30.
Let the number of 1s in the $k$-th column be $l_{k}$. Then
$$
\sum_{k=1}^{n} l_{k}=30 \text {. }
$$
For any $1 \leqslant i<j \leqslant 6$ and $1 \leqslant k \leqslant n$, consider the triplet $\{i, j, k\}$: such that the $i$-th row and the $j$-th row... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,137 |
1. The integer part of the real number $\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}$ is ( ).
(A) 969
(B) 970
(C) 971
(D) None of the above conclusions is correct | $-1 . \mathrm{A}$.
$$
\left(\tan 60^{\circ}+2 \cos 45^{\circ}\right)^{6}=(\sqrt{3}+\sqrt{2})^{6} \text {. }
$$
Let $\sqrt{3}+\sqrt{2}=a, \sqrt{3}-\sqrt{2}=b$. Then
$$
a+b=2 \sqrt{3}, a b=1 \text {. }
$$
Thus, $a^{2}+b^{2}=(a+b)^{2}-2 a b=10$,
$$
a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)=18 \sqrt{3} \text {. }
$$
... | 969 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,138 |
Example 8 As shown in Figure 12, a square $B C E F$ is constructed on the hypotenuse $B C$ of $\mathrm{Rt} \triangle A B C$ on the same side as $\triangle A B C$. Let the center of the square be $O$, and connect $A O$. If $A B=4, A O=6 \sqrt{2}$, then the length of $A C$ is ( )
(A) 12
(B) 16
(C) $4 \sqrt{3}$
(D) $8 \sq... | Analysis 1: Take a point $G$ on $AC$ such that $CG = AB = 4$.
Since $\angle BAC = \angle BOC = 90^{\circ}$, points $A$, $O$, $C$, and $B$ are concyclic. Therefore, $\angle ABO = \angle ACO$.
Because $BO = CO$, $\triangle BAO \cong \triangle CGO$.
From $\angle AOB = \angle GOC$, we know $\angle AOG = 90^{\circ}$.
Since ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,139 |
2. Given a rhombus $A B C D$ with side $A B=10$, diagonal $B D=12$, and 100 different points $P_{1}, P_{2}$, $\cdots, P_{100}$ on $B D$. Draw $P_{i} E_{i} \perp A B$ at $E_{i}$, and $P_{i} F_{i} \perp A D$ at $F_{i}$. Then
$$
P_{1} E_{1}+P_{1} F_{1}+P_{2} E_{2}+P_{2} F_{2}+\cdots+P_{100} E_{100}+P_{100} F_{100}
$$
is $... | 2. C.
As shown in Figure 5, extend $E_{i} P_{i}$ to intersect $C D$ at point $G_{i}$. Then,
$$
P_{i} G_{i} \perp C D \text {. }
$$
Thus, $P_{i} F_{i}=P_{i} G_{i}$. Connect $A C$ to intersect $B D$ at point $O$. Therefore,
$$
\begin{array}{l}
A C \perp B D, \\
B O=\frac{1}{2} B D=6 .
\end{array}
$$
Then $A O=\sqrt{A ... | 960 | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,140 |
3. Given the function
$y=\left(k^{2}+k\right) x^{2}-\sqrt{5}(k+1) x+1(k$ is a constant $)$, the graph of the function intersects the coordinate axes at two points. Which of the following conclusions is correct? ( ).
(A) $k=-5$
(B) $k=-1$ or -5
(C) $k=0$ or -5
(D) $k-1$ | 3. C.
From the problem, we know that the given function is either a linear function or a quadratic function that intersects the $x$-axis at one point. Therefore,
$$
\left\{\begin{array}{l}
k^{2}+k=0, \\
-\sqrt{5}(k+1) \neq 0
\end{array}\right.
$$
or $\left\{\begin{array}{l}k^{2}+k \neq 0, \\ \Delta=5(k+1)^{2}-4\left(... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,141 |
4. As shown in Figure 1, in $\triangle A B C$ with perimeter $l$,
$$
\begin{array}{l}
\angle A B C=\alpha, \\
\angle A C B=2 \alpha .
\end{array}
$$
A circle $\odot O$ with radius $\frac{l}{2008}$ rolls without slipping along the sides of $\triangle A B C$ in a counterclockwise direction until it returns to its origin... | 4. B.
As shown in Figure 6, the path taken by the center $O$ is $O \rightarrow B_{1} \rightarrow$
$$
\begin{array}{l}
B_{2} \rightarrow C_{1} \rightarrow C_{2} \rightarrow \\
A_{1} \rightarrow A_{2} \rightarrow O .
\end{array}
$$
Let $\angle A_{1} A A_{2}$
$$
=x^{\circ}, \angle B_{1} B B_{2}=y^{\circ}, \angle C_{1} C... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,142 |
5. There are two uniform small cubes $A$ and $B$. The six faces of cube $A$ are labeled with the numbers $1, 2, 3, 4, 5, 6$; the six faces of cube $B$ are labeled with the numbers $1, 1, 2, 3, 4, 6$. Xiao Guang rolls cube $A$, and the number facing up is $x$; Xiao Ming rolls cube $B$, and the number facing up is $y$. T... | 5.D.
List 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$x$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline$y$ & $(1,1)$ & $(2,1)$ & $(3,1)$ & $(4,1)$ & $(5,1)$ & $(6,1)$ \\
\hline 1 & $(1,1)$ & $(2,1)$ & $(3,1)$ & $(4,1)$ & $(5,1)$ & $(6,1)$ \\
\hline 2 & $(1,2)$ & $(2,2)$ & $(3,2)$ & $(4,2)$ & $(5,2)$ & $(6,2)$ \\
\hline 3 & ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,143 |
6. Given that $x$ and $y$ are both negative integers, and satisfy $y=$ $\frac{10 x}{10-x}$. Then the maximum value of $y$ is ( ).
(A) 90
(B) -9
(C) -11
(D) None of the above conclusions are correct | 6. D.
From the given conditions, we have
$$
x=\frac{10 y}{y+10}=10-\frac{100}{y+10} \text{. }
$$
For \( x \) to be a negative integer, \( y+10 \) must be a divisor of 100, and \( -10 < y < 0 \).
Therefore, \( y+10=1,2,4,5 \).
Thus, \( y=-9,-8,-6,-5 \).
Substituting and verifying, we find that when \( y=-5 \), \( x=-1... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,144 |
1. Calculate: $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\cdots+\frac{10}{3^{9}}+$ $\frac{23}{3^{9}+3^{10}}=$ $\qquad$ | $=1 . \frac{9}{4}$.
Let $S=1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\cdots+\frac{10}{3^{9}}$.
Then $\frac{S}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+\cdots+\frac{10}{3^{10}}$.
(1) - (2) gives
$\frac{2}{3} S=1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\cdots+\frac{1}{3^{9}}-\frac{10}{3^{10}... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,145 |
3. Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=a^{2}+b^{2}+c^{2}+d^{2}=3 \text {. }
$$
Then the range of values for $d$ is $\qquad$ . | 3.0 $\leqslant d \leqslant \frac{3}{2}$.
Let $y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \geqslant 0$.
Then $y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \geqslant 0$.
Since the coefficient of $x^{2}$ is $3>0$, we have,
$$
\begin{array}{l}
\Delta=[-2(a+b+c)]^{2}-4 \times 3\left(a^{2}+b^{2}+c^{2}\right) \leqslant 0 \\
\Rig... | 0 \leqslant d \leqslant \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,147 |
One. (20 points) Given the one-variable quartic equation $x^{4}+4 x^{3}+(k+7) x^{2}+2(k+3) x+3 k=0$ has real roots.
(1) Find the range of values for $k$;
(2) If the product of all real roots of the equation is -4, and $m, n$ are two of the real roots, find the value of $-m^{3}-\frac{32}{n^{2}}-4 n+\frac{4}{n}$. | (1) A quartic equation can be transformed into
$$
\begin{array}{l}
x^{2}\left(x^{2}+2 x+3\right)+2 x\left(x^{2}+2 x+3\right)+ \\
k\left(x^{2}+2 x+3\right)=0,
\end{array}
$$
which is $\left(x^{2}+2 x+3\right)\left(x^{2}+2 x+k\right)=0$.
Therefore, the quartic equation can be reduced to two quadratic equations:
$$
x^{2}... | 9 \mp \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,149 |
1. As shown in Figure 14, quadrilateral $A B C D$ is a square, and an equilateral triangle $\triangle A B E$ is constructed outward from side $A B$. $C E$ intersects $B D$ at point $F$. Then $\angle A F D=$ $\qquad$ degrees. | (Given that $\angle E C B=\angle C E B=$ $15^{\circ}$, and by the axial symmetry of the square, we get $\angle F A B=$ $\angle E C B=15^{\circ}$. Therefore, $\angle A F D=60^{\circ}$.) | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,150 |
II. (25 points) As shown in Figure 4, in $\triangle ABC$, $\angle ACB = 90^{\circ}$, $AC = BC$. Point $P$ is on $AB$, and $2AP < PB < 3AP$. The circle $\odot O$ passing through points $C$ and $P$ intersects $AC$, $BC$, and $AB$ at points $D$, $E$, and $G$ respectively. $AF$ is tangent to $\odot O$ at point $F$, and $DE... | II. As shown in Figure 9, connect $P E$.
Since $\angle A C B = 90^{\circ}$, therefore, $D E$ is the diameter of $\odot O$.
Thus, $\angle D P E = 90^{\circ}$.
Therefore, $\tan \angle P D E = \frac{P E}{P D}$.
Draw $P H \perp B C$ at $H$, and $P I \perp A C$ at $I$.
Since quadrilateral $E C D P$ is a cyclic quadrilatera... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,151 |
Three. (25 points) Given $n(n>1)$ integers (which can be the same) $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
x_{1}+x_{2}+\cdots+x_{n}=x_{1} x_{2} \cdots x_{n}=9111 \text {. }
$$
Find the maximum value of $x_{1}, x_{2}, \cdots, x_{n}$ when $n$ takes the minimum value. | Three, from $x_{1} x_{2} \cdots x_{n}=9111$, we know that $x_{1}, x_{2}, \cdots, x_{n}$ are all odd numbers.
Since $x_{1}+x_{2}+\cdots+x_{n}=9111$ is odd, $n (n>1)$ must be odd.
If $n=3$, then
$$
x_{1}+x_{2}+x_{3}=x_{1} x_{2} x_{3}=9111 \text {. }
$$
By the symmetry of the condition equations, without loss of general... | 9111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,152 |
1. Let $y=f(x)$ be the exponential function $y=a^{x}$. Among the four points $P(1,1)$, $Q(1,2)$, $M(2,3)$, $N\left(\frac{1}{2}, \frac{1}{4}\right)$, the common points of the graph of the function $y=f(x)$ and its inverse function $y=f^{-1}(x)$ can only be point ( ).
(A) $P$
(B) $Q$
(C) $M$
(D) $N$ | -1.D.
Since the inverse function of the exponential function $y=a^{x}(a>0, a \neq 1)$ is the logarithmic function $y=\log _{a} x$, and the logarithmic function passes through the point $(1,0)$, it cannot pass through points $P$ and $Q$.
If it passes through point $M$, then $3=a^{2}, a=\sqrt{3}$.
But $y=\log _{\sqrt{3}}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,153 |
2. Given real numbers $a, b, c$ satisfy
$$
a+b+c=2, a b c=4 \text {. }
$$
Then the minimum value of $|a|+|b|+|c|$ is ( ).
(A) 2
(B) 3
(C) $3 \sqrt[3]{4}$
(D) 6 | 2.D.
Obviously, at least one of $a$, $b$, and $c$ is positive. Without loss of generality, let it be $a$.
From $b+c=2-a$ and $bc=\frac{4}{a}$, we get the quadratic equation in $x$
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
which has two real roots.
Thus, $(2-a)^{2}-\frac{16}{a} \geqslant 0$.
Combining this with $a>0$, we get ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,154 |
3. Arrange 4 identical red balls and 4 identical blue balls in a row, with each ball corresponding to a sequence number from $1,2, \cdots, 8$ from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the sequence numbers corresponding to the 4 red balls is less th... | 3. A.
1 to 8, taking any four different numbers to find their sum, we can get $\mathrm{C}_{8}^{4}=70$ answers (which can be the same).
Among them, the sum being 18 has 8 cases:
$$
\begin{array}{l}
(8,7,2,1),(8,6,3,1),(8,5,4,1), \\
(8,5,3,2),(7,6,4,1),(7,6,3,2), \\
(7,5,4,2),(6,5,4,3) .
\end{array}
$$
Therefore, the nu... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,155 |
4. Let $O$ be the center of the base $\triangle ABC$ of the regular tetrahedron $P-ABC$. A moving plane through $O$ intersects $PC$ at point $S$, and the extensions of $PA$ and $PB$ at points $Q$ and $R$, respectively. Then the sum $\frac{1}{PQ} + \frac{1}{PR} + \frac{1}{PS}(\quad)$
(A) has a maximum value but no minim... | 4.D.
Let the regular tetrahedron $P-ABC$ have dihedral angles between each pair of lateral edges as $\alpha$, and the angle between $PC$ and the plane $PAB$ as $\beta$. Then
$$
\begin{array}{l}
V_{S-PQR}=\frac{1}{3} S_{\triangle PQR} h \\
=\frac{1}{3}\left(\frac{1}{2} PQ \cdot PR \sin \alpha\right) PS \sin \beta .
\en... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,156 |
5. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$. Then $\sum_{n=1}^{2009} x_{n}=(\quad)$.
(A) 1
(B) -1
(C) $-2+\sqrt{3}$
(D) $2+\sqrt{3}$ | 5. C.
From the problem, we know
$$
\begin{array}{c}
x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}} . \\
\text { Let } x_{n}=\tan \alpha_{n} \text {. Then } \\
x_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) .
\end{array}
$$
Thus, $x_{n+6}=x_{n}$.
It is easy to see that $x_{1}=1, x_{2}=2+\sqrt{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,157 |
6. For any point $P$ on the ellipse $C: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw a perpendicular line $P H$ from $P$ to the right directrix of the ellipse $C$ (where $H$ is the foot of the perpendicular), and extend $P H$ to point $Q$ such that $|H Q|=\lambda|P H|(\lambda \geqslant 1)$. As point $P$ moves on the ellips... | 6.C.
Let $P\left(x_{1}, y_{1}\right), Q(x, y)$.
Since the equation of the right directrix is $x=3$, we have $H(3, y)$.
Also, $H Q=\lambda P H$, then
$$
\frac{H P}{P Q}=\frac{-1}{1+\lambda} \text {. }
$$
By the section formula, we get
$$
\left\{\begin{array}{l}
x_{1}=\frac{3(1+\lambda)-x}{\lambda}, \\
y_{1}=y .
\end{a... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,158 |
1. $[x]$ represents the greatest integer not exceeding the real number $x$. Suppose the real number $x$ is not an integer, and $x+\frac{99}{x}=[x]+\frac{99}{[x]}$. Then, the value of $x$ is | Ni.1. -9.9 .
Let $x=[x]+t, 0<t<1$.
Substitute into the equation (note $t \neq 0$ )
$$
t=\frac{99-[x]^{2}}{[x]}, [x] \in \mathbf{Z} \text {. }
$$
From $0<t<1$ we know $[x]=-10$.
Thus, $t=0.1, x=-9.9$. | -9.9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,159 |
2. Positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=\sqrt{14}, \\
x^{2}+z^{2}-\sqrt{2+\sqrt{3}} x z=7-4 \sqrt{3}, \\
y^{2}+z^{2}+\sqrt{2+\sqrt{3}} y z=7+4 \sqrt{3} .
\end{array}\right.
$$
Then $z=$ | 2. $\frac{\sqrt{3}+1}{\sqrt{7}}$.
According to the conditions of the system of equations, we can draw Figure 2.
Notice that
$$
\cos 15^{\circ}=\frac{1}{2} \sqrt{2+\sqrt{3}}, 7 \pm 4 \sqrt{3}=(2 \pm \sqrt{3})^{2},
$$
Therefore, equation (2) represents the third side $AC = 2 - \sqrt{3}$ of $\triangle ADC$ with sides $D... | \frac{\sqrt{3}+1}{\sqrt{7}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,160 |
2. As shown in Figure 15, in trapezoid $A B C D$, $A D / / B C (B C > A D), \angle D=90^{\circ}, B C=$ $C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$ .
(2004, National Junior High School Mathematics Competition) | (Hint: First, complete the trapezoid $ABCD$ into a square $CBFD$, then rotate $\triangle ABF$ $90^{\circ}$ around point $B$ to the position of $\triangle BCG$. It is easy to see that $\triangle ABE \cong \triangle GBE$. Therefore, $AE = EG = EC + AF = 10$. Let $EC = x$. Then $AF = 10 - x$, $DE = 12 - x$, $AD = 2 + x$. ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,161 |
3. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$, and $Q, R, S, P$ are the midpoints of segments $PA, QB, RC, SD$ respectively. Then $V_{P-ABC}: V_{P-BCD}: V_{P-CDA}: V_{P-DAB}=$ $\qquad$ | $3.8: 1: 2: 4$.
Let $H_{P, B C D}$ be the distance from point $P$ to plane $B C D$, and so on.
As shown in Figure 3.
Given $V_{P-B C D}=1$.
$$
\begin{array}{l}
\text { By } \frac{H_{S, B C D}}{H_{P, B C D}}=\frac{S D}{P D} \\
=2 \\
\Rightarrow V_{S-B C D}=2 ; \\
\text { By } \frac{H_{R, B C D}}{H_{S, B C D}}=\frac{R C... | 8: 1: 2: 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,162 |
4. Let $n \in \mathbf{N}_{+}$, and the set $M=\{1,2, \cdots, 2 n\}$. There exists a positive integer $k$, such that for any $k$-element subset of $M$, there must be four distinct elements whose sum equals $4 n+1$, then $k_{\text {min }}=$ $\qquad$. | 4. $n+3$.
Consider the $(n+2)$-element subset of $M$
$$
P=\{n-1, n, n+1, \cdots, 2 n\} \text {. }
$$
By the problem's condition, the sum of any four distinct elements in $P$ is not less than
$$
(n-1)+n+(n+1)+(n+2)=4 n+2 \text {. }
$$
Therefore, $k \geqslant n+3$.
Pair the elements of $M$ into $n$ pairs:
$$
B_{i}=(i,... | n+3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,163 |
5. Notice that $6!=8 \times 9 \times 10$. Then, the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers is $\qquad$ . | 5.23.
According to the requirements of the problem, write $n!$ as
$$
n!=n \times(n-1) \times \cdots \times 5 \times 24 \text {. }
$$
Therefore, $n+1 \leqslant 24, n \leqslant 23$.
Hence, the maximum value of $n$ is 23. At this point,
$$
23!=24 \times 23 \times \cdots \times 5 \text {. }
$$
The right side of the abov... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,164 |
6. Given the quadratic function $f(x)=a x^{2}+b x+c, a$ $\in \mathbf{N}_{+}, c \geqslant 1, a+b+c \geqslant 1$, the equation $a x^{2}+b x+c$ $=0$ has two distinct positive roots less than 1. Then the minimum value of $a$ is | 6.5.
Let the two roots of the equation $a x^{2}+b x+c=0$ be $x_{1}$ and $x_{2}$, and $0<x_{1}<x_{2}<1, a>4$.
Therefore, $a \geqslant 5$.
Take $f(x)=5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)$, which meets the conditions. Hence, the minimum value of $a$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,165 |
Three. (20 points) Let $n(n \geqslant 2)$ be a given positive integer. Find all integer tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, satisfying the conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$;
(2) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$. | Three, by the mean inequality we have
$$
\begin{array}{l}
n=\frac{n^{2}}{n} \leqslant \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \\
\leqslant \sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}} \leqslant \sqrt{\frac{n^{3}+1}{n}} \\
<n+\frac{1}{2 n^{2}},
\end{array}
$$
(Note: $\left(n+\frac{1}{2 n^{2}}\right)^{2}=\frac{n^{3}+... | (n, n, \cdots, n) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,166 |
Four. (20 points) Let the hyperbola \( S: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \), \( M\left(x_{0}, y_{0}\right) \notin S \), and \( x_{0} y_{0} \neq 0, N\left(\lambda x_{0}, \lambda y_{0}\right) \), where \( \frac{1}{\lambda}=\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}} \). A line \( l \) passing through point... | Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $C\left(x_{3}, y_{3}\right)$.
As shown in Figure 4, draw a line $m$ through point $M$ with a slope of $\frac{b^{2} x_{0}}{a^{2} y_{0}}$. The equation of line $m$ is
$y-y_{0}=$ $\frac{b^{2} x_{0}}{a^{2} y_{0}}\left(x-x_{0}\right)$.
Let line $m$ interse... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,167 |
One, (50 points) As shown in Figure $1, \triangle A B C$ has an area of 1. Points $E, F, G$ are on sides $B C, C A, A B$ respectively, and $A E$ is bisected by $B F$ at point $R$, $B F$ is bisected by $C G$ at point $S$, and $C G$ is bisected by $A E$ at point $T$. Find the area of $\triangle R S T$. | $$
\text { I. Let } \frac{C F}{F A}=x, \frac{A G}{G B}=y, \frac{B E}{E C}=z(x, y, z>0) \text {. }
$$
By $B F$ intersecting $\triangle C A G$, according to Menelaus' theorem, we get
$$
\frac{C F}{F A} \cdot \frac{A B}{B G}=1 \text {. }
$$
Similarly, by $C G$ and $A E$ intersecting $\triangle A B E$ and $\triangle B C ... | \frac{7-3 \sqrt{5}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,168 |
Given a sequence of non-negative numbers $a_{1}, a_{2}, \cdots, a_{n}$, for any integer $k$ from 1 to $n$, let $m_{k}$ be the value
$$
m_{k}=\max _{l=1,2, \cdots, k} \frac{a_{k-l+1}+a_{k-l+2}+\cdots+a_{k}}{l} .
$$
Prove: For any $\alpha>0$, the number of $k$ such that $m_{k}>\alpha$ is less than $\frac{a_{1}+a_{2}+\cd... | II. An integer $k$ is called a "good number" if it satisfies $m_{k}>\alpha$. Let $[i, j]$ be an integer satisfying $1 \leqslant i \leqslant j \leqslant n$, and let
\[ S(i, j)=\frac{a_{i}+a_{i+1}+\cdots+a_{j}}{j-i+1}. \]
Consider selecting $\left[p_{i}, q_{i}\right]$, where $p_{1}$ satisfies
\[ a_{p_{1}-m} \leqslant \al... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,169 |
Three. (50 points) For which positive integers $n$, do there exist positive integers $m$ and positive integers $a_{1}, a_{2}, \cdots, a_{m-1}$, such that
$$
n=\sum_{k=1}^{m-1} a_{k}\left(m-a_{k}\right) \text { ? }
$$
where $a_{1}, a_{2}, \cdots, a_{m-1}$ can be the same, and $1 \leqslant a_{i} \leqslant m-1(i=1,2, \cd... | Three, let $n \geqslant 12$ and $n$ does not meet the requirement. Let $m$ be an even number,
$$
m=2 l, b_{i}=\left|l-a_{i}\right|(i=1,2, \cdots, m-1) .
$$
Then $0 \leqslant b_{i} \leqslant l-1(1 \leqslant i \leqslant m-1)$, and
$$
\sum_{k=1}^{m-1} a_{k}\left(m-a_{k}\right)=l^{2}(2 l-1)-\sum_{i=1}^{2 l-1} b_{i}^{2} \t... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,170 |
Five. (20 points) The sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=2 a_{n}-n^{2}+3 n\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Do there exist constants $\lambda, \mu$ such that the sequence $\left\{a_{n}+\lambda n^{2}+\mu n\right\}$ is a geometric sequence? If they exist, find the values of $... | (1) Let $a_{n+1}=2 a_{n}-n^{2}+3 n$ be transformed into
$$
\begin{array}{l}
a_{n+1}+\lambda(n+1)^{2}+\mu(n+1) \\
=2\left(a_{n}+\lambda n^{2}+\mu n\right),
\end{array}
$$
which is
$$
a_{n+1}=2 a_{n}+\lambda n^{2}+(\mu-2 \lambda) n-\lambda-\mu \text {. }
$$
Thus, $\left\{\begin{array}{l}\lambda=-1, \\ \mu-2 \lambda=3, ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,171 |
Example 1: In a "level-passing game", the rule is: on the $n$-th level, a die must be rolled $n$ times. If the sum of the points that appear in these $n$ rolls is greater than $\frac{3^{n}}{4}$, then it is considered passing the level; otherwise, it is not considered passing. There are the following statements:
(1) Pas... | Explanation: To pass the first level, the score needs to be greater than $\frac{3}{4}$. Clearly, rolling a die once will yield at least one point, so (1) is correct.
To pass the second level, the sum of the scores needs to be greater than $\frac{9}{4}$, which means the sum of the scores must be at least 3. Since the s... | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 720,172 |
Example 3 In a regular $\triangle A B C$ with side length 8, take a moving point $P$ on side $B C$. Find the maximum and minimum values of the segment $M N$ intercepted by the perpendicular bisector of $A P$ on $\triangle A B C$. | As shown in Figure 3, take the side length 8 and the variable $\angle B A P = \alpha \in \left[0, \frac{\pi}{3}\right]$ as the basic quantities. Then
$$
\begin{array}{l}
A D = \frac{1}{2} A P \\
= \frac{1}{2} \cdot \frac{8 \sin 60^{\circ}}{\sin \left(60^{\circ} + \alpha\right)} \\
= \frac{2 \sqrt{3}}{\sin \left(60^{\ci... | 4 \sqrt{3} \text{ and } 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,174 |
3. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the first term $a$ and common difference $d$. Define
$$
\begin{array}{l}
S_{n}^{(0)}=a_{n}, \\
S_{n}^{(k)}=S_{1}^{(k-1)}+S_{2}^{(k-1)}+\cdots+S_{n}^{(k-1)}(k>0) .
\end{array}
$$
Then, $S_{n}^{(m)}=$ $\qquad$ (express in terms of $a$, $d$, and $n$). | 3. $\mathrm{C}_{n+m-1}^{m} a_{1}+\mathrm{C}_{n+m-1}^{m+1} d$.
Notice that
$$
\begin{array}{l}
S_{n}^{(1)}=a_{1}+a_{2}+\cdots+a_{n}=\mathrm{C}_{n}^{1} a_{1}+\mathrm{C}_{n}^{2} d, \\
S_{n}^{(2)}=S_{1}^{(1)}+S_{2}^{(1)}+\cdots+S_{n}^{(1)} \\
=\left(\mathrm{C}_{1}^{1}+\mathrm{C}_{2}^{1}+\cdots+\mathrm{C}_{n}^{1}\right) a_... | S_{n}^{(m)}=\mathrm{C}_{n+m-1}^{m} a_{1}+\mathrm{C}_{n+m-1}^{m+1} d | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,175 |
5. A polygon is cut with one cut (the cut does not pass through the vertices of the polygon) to divide it into 2 polygons, then one of the polygons is cut with another cut (the cut does not pass through the vertices of the polygon) to divide it into another polygon, ... and so on. If starting from a square, to cut out ... | 5.2011014.
Let a total of $k$ cuts be made. Since each cut increases the number of polygons by one, there are a total of $k+1$ polygons. Apart from one triangle, one quadrilateral, one pentagon, ... one 2008-sided polygon, there are
$$
(k+1)-2006=k-2005
$$
polygons.
Considering the total sum $S$ of the number of side... | 2011014 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,177 |
6. Let $a_{i} \in\{1,2, \cdots, 9\}(1 \leqslant i \leqslant n)$, and $a_{1}>a_{2}>\cdots>a_{n}$. If $A$ is a $p$-digit number, $B$ is a $q$-digit number $(p<q, p+q=n)$, and the union of the sets of digits of $A$ and $B$ is exactly $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, then the maximum value of the product $A B$... | 6. $a_{2} a_{3} \cdots a_{2 p-1} a_{2 p+1} a_{2 p+2} \cdots a_{n} \cdot a_{1} a_{4} a_{6} \cdots a_{2 p}$.
First, arrange the first digits of the two numbers, one of which must be $a_{1}$ and the other $a_{2}$.
Next, arrange the second digits (if any) of the two numbers, one of which must be $a_{3}$ and the other $a_... | \overline{a_{2} a_{3} \cdots a_{2 p-1} a_{2 p+1} a_{2 p+2} \cdots a_{n}} \cdot \overline{a_{1} a_{4} a_{6} \cdots a_{2 p}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,178 |
Three. (20 points) Let the edge length of the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ be 1. Find a point $P$ on the diagonal $B D_{1}$ and a point $Q$ on the base quadrilateral $A B C D$ such that the value of $P Q+P C_{1}$ is maximized.
---
The translation retains the original text's line breaks and formatting. | $$
\begin{array}{l}
\text { Three, as shown in Figure 6, let } \\
P B=x, \angle D_{1} B C_{1} \\
=\alpha \text {. Then } \\
\cos \alpha=\frac{\sqrt{2}}{\sqrt{3}} \\
=\frac{\sqrt{6}}{3} \text {, } \\
P C_{1}^{2} \\
=x^{2}+2-2 \sqrt{2} x \cos \alpha \\
=x^{2}+2-\frac{4 \sqrt{3}}{3} x \text {. } \\
\end{array}
$$
Let $A ... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,179 |
Four. (20 points) As shown in Figure 3, let $F$ be one of the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $A$ be the vertex of the ellipse that is farthest from $F$. Take 2008 distinct points $P_{i}(i=1,2, \cdots, 2008)$ on the minor axis $B C$ of the ellipse. Let $F P_{i}$ intersect $A B$ or $A C$ a... | For any point $P_{i}(i=1,2, \cdots, 2008)$, by symmetry, we can assume $P_{i}$ is on the positive half of the $y$-axis, as shown in Figure 7.
In $\triangle A F C$, since $A N_{i}$, $C O$, and $F M_{i}$ are concurrent, by Ceva's Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F O}{... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,180 |
Five, (20 points) Does there exist periodic functions $f(x), g(x): \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x \in \mathbf{R}$, we have $f(x) + g(x) = x$? | Five, Existence.
Partition the set of real numbers into several classes, where two numbers $x, y$ belong to the same class if and only if there exist $m, n \in \mathbf{Z}$ such that $x - y = m + n \sqrt{2}$. In each class, select one element, and these elements form a set $M$.
For any $x \in \mathbf{R}$, let $x - x_{0... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,181 |
一、(50 points) As shown in Figure 4, in quadrilateral $ABCD$,
$AC, BD$
intersect at point
$O, AO=$
$OC, \quad BO$
$>DO$. Let
points $M, N$
be the midpoints of
$DO, BO$ respectively, draw $OP \parallel BC$, intersecting $AD$ at point $P$, and connect $PM, PN$. Try to determine: when $\angle ACB$ is what value,
$$
\angle ... | One, as shown in Figure 8, take the midpoints $S, T$ of $AO, OC$ respectively, and connect $SM, NT$.
Then $SM \parallel AD, NT \parallel BC$.
Also, since $M, N$ are the midpoints of $DO, BO$ respectively, $S, T$ are the midpoints of $AO, CO$ respectively.
Let the intersection of lines $SM, NT$ be point $Q$, and $SQ$ i... | \angle ACB \leqslant 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,182 |
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