problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Given $x, y$ are positive real numbers, $n \in \mathrm{N}$, and $n \geqslant 2$. Prove:
$$
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \geqslant 4 \text {. }
$$ | $$
\begin{array}{l}
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \\
\geqslant 2 \sqrt{\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}} \cdot \sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}}} \\
=2 \sqrt[2 n]{\frac{\left(2^{n}-1\right)\left(x^{2}+y^{2}\right)+\left(2^{2 n}-2^{n+1}+2\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false | 726,214 |
In a square $A B C D$ with side length 2, moving points $E$ and $F$ are both on side $A D$, satisfying $A E=D F$. Connecting $C F$ intersects the diagonal $B D$ at point $Q$, and connecting $A Q$ and $B E$ intersect at point $P$. Find the minimum value of $D P$.
---
The translation preserves the original text's line ... | Solve as shown in Figure 3.
From the given conditions,
$$
\begin{array}{l}
A B=D C=D A, \\
\angle B A E \\
=\angle C D F \\
=90^{\circ},
\end{array}
$$
$\angle C D Q=\angle A D Q$.
Since $A E=D F, D Q=D Q$, therefore,
$\triangle A B E \cong \triangle D C F, \triangle C D Q \cong \triangle A D Q$.
Thus, $\angle A E B=\a... | \sqrt{5}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,215 |
Given $a, b, c > 0$, and $abc = 1$, $\lambda (\lambda \geqslant 1)$ is a constant. Prove:
$$
\frac{1}{a+b+\lambda}+\frac{1}{b+c+\lambda}+\frac{1}{c+a+\lambda} \leqslant \frac{3}{2+\lambda},
$$
with equality holding if and only if $a=b=c=1$. | Prove that using “$\sum$” and “$\prod$” to denote cyclic sums and cyclic products, respectively, we have:
$$
\begin{array}{l}
\sum \frac{1}{a+b+\lambda} \leqslant \frac{3}{2+\lambda} \\
\Leftrightarrow (2+\lambda)\left[\sum(a+b)(b+c)+4 \lambda \sum a+3 \lambda^{2}\right] \\
\quad \leqslant 3\left[\prod(a+b)+\lambda \su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,216 |
In $\triangle A B C$, $A B>A C$, points $D_{1}$ and $D_{2}$ are on side $B C$, and $\angle C A D_{1} = \angle B A D_{2}$. $M$ is the midpoint of $B C$, $M E_{1} \parallel A D_{1}$ intersects $A B$ at point $E_{1}$, and $M E_{2} \parallel A D_{2}$ intersects the extension of $B A$ at point $E_{2}$.
Prove:
$$
B E_{1} + B... | Given $\angle C A D_{1}=\angle B A D_{2}$, we know that $A D_{1}$ and $A D_{2}$ are the internal isogonal lines of $\angle B A C$.
By the property theorem of the internal isogonal lines of a triangle, we have
$$
\begin{array}{l}
\frac{A C^{2}}{A B^{2}}=\frac{C D_{1} \cdot C D_{2}}{B D_{1} \cdot B D_{2}} \\
=\frac{\left... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,217 |
Example 3 Let $f(x)$ be a quadratic polynomial, and
$$
|f(-1)| \leqslant 1,|f(0)| \leqslant 1,|f(1)| \leqslant 1 \text {. }
$$
Then for any $x \in[-1,1]$, we have
$$
|f(x)| \leqslant \frac{5}{4} \text {. }
$$ | 【Analysis】Notice the Lagrange interpolation formula
$$
f(x)=\sum_{k=1}^{n+1}\left(\prod_{j \neq k} \frac{x-x_{j}}{x_{k}-x_{j}}\right) f\left(x_{k}\right) \text {. }
$$
From equation (1), we know that $f(-1) 、 f(0) 、 f(1)$ can uniquely determine the quadratic polynomial $f(x)$, i.e.,
$$
\begin{aligned}
f(x)= & \frac{x(... | \frac{5}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 726,218 |
Example 4 Let
$$
\begin{array}{l}
F(x)=x^{3}+a x^{2}+b x+c, \\
G(x)=x^{3}+d x^{2}+e x+f
\end{array}
$$
be two polynomials with real coefficients, and $x_{1}<x_{2}<\cdots<x_{8}$ are exactly the
$$
F G(F-G)=F^{2} G-F G^{2}=0
$$
eight distinct real roots. Prove:
$$
F^{2}\left(x_{1}\right)+F^{2}\left(x_{8}\right) \neq 0 ... | 【Analysis】From the problem, we have
$$
\operatorname{deg} F=\operatorname{deg} G=3, \operatorname{deg}(F-G) \leqslant 2 \text {. }
$$
Thus, $\operatorname{deg}\left(F^{2} G-F G^{2}\right) \leqslant 8$.
Therefore, $x_{1}0 \text {, }
$$
which contradicts $G\left(x_{1}\right)<0<G\left(x_{8}\right)$.
Hence, the assumptio... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,219 |
Example 5 Given an integer $n>1$, let $f(x)$ be an $n$-degree polynomial with integer coefficients, and $F=\{f(x) \mid x \in \mathbf{Z}\}$. Prove: For any $k \in \mathbf{Z}$, $\{k, k+1, \cdots, k+n\} \not \subset F$. ${ }^{[2]}$ | 【Analysis】Proof by contradiction.
Assume there exists $k \in \mathbf{Z}$, such that
$$
\{k, k+1, \cdots, k+n\} \subseteq F \text {, }
$$
i.e., there exists $x_{i} \in \mathbf{Z}$, such that
$$
f\left(x_{i}\right)=k+i(i=0,1, \cdots, n) ;
$$
Clearly, $x_{0}, x_{1}, \cdots, x_{n}$ are all distinct.
Since $f\left(x_{i+1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,220 |
Example 6 Find integers $0<a_{1}<a_{2}<\cdots<a_{n}$, such that
$$
x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)+1
$$
can be factored over the integers. | 【Analysis】Let $a_{0}=0$. Assume
$$
x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)+1=P(x) Q(x) \text {, }
$$
where $P(x)$ and $Q(x)$ are polynomials with integer coefficients, and
$1 \leqslant \operatorname{deg} P(x), \operatorname{deg} Q(x) \leqslant n$.
Let $x=a_{i}(i=0,1, \cdots, n)$. Then
$$
P... | a_1=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,221 |
Example 7 Let $f(x)$ be a polynomial with real coefficients, and when $x \geqslant 0$, $f(x) \geqslant 0$. Prove: There exist polynomials $P(x)$ and $Q(x)$ with real coefficients such that
$$
f(x)=P^{2}(x)+x Q^{2}(x) .
$$ | 【Analysis】When $f(x)=C$ is a constant polynomial, $C \geqslant 0$, the proposition is obviously true.
Assume $\operatorname{deg} f(x)>0$.
Let the real number $a_{i}$ be the $\alpha_{j}$-fold
root of the real-coefficient polynomial $f(x)$, where $1 \leqslant j \leqslant k$, and the complex number $b_{j}+\mathrm{i} c_{j... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,222 |
Example 8 Suppose
$$
x^{n}+a_{2} x^{n-2}+a_{3} x^{n-3}+\cdots+a_{n-1} x+a_{n}=x^{n-1}
$$
has $n$ non-negative real roots. Then
$$
0 \leqslant 2^{2} a_{2}+2^{3} a_{3}+\cdots+2^{n} a_{n} \leqslant 1+\left(\frac{n-2}{n}\right)^{n} .
$$ | 【Analysis】Let the $n$ non-negative real roots be
$$
0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \text {. }
$$
By Vieta's formulas, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=1 \Rightarrow 0 \leqslant x_{i} \leqslant 1 \\
\Rightarrow-1 \leqslant 1-2 x_{i} \leqslant 1 \\
\Rightarrow-1 \leqsl... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,223 |
3. Let $f(x)=x^{2010}-x^{1949}+1, a_{1}=m \in \mathbf{Z}$, and for any $n \in \mathbf{N}_{+}, a_{n+1}=f\left(a_{n}\right)$. Then any two terms in the sequence $\left\{a_{n}\right\}$ are coprime. | When $m=1$, $a_{n}=1$, the proposition is obviously true. When $m>1$, the remainder of $f(m)$ divided by $m$ is 1, i.e.,
$$
\begin{array}{l}
a_{2} \equiv 1(\bmod m), \\
a_{3}=f\left(a_{2}\right)=f(1)(\bmod m) \equiv 1(\bmod m) .
\end{array}
$$
By induction, when $n \geqslant 2$,
$$
a_{n}^{\cdot}=1(\bmod m) \text {. }
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,224 |
4. Let $f(x)$ be a polynomial with integer coefficients, and
$$
\begin{array}{l}
0 \leqslant f(k) \leqslant 2002 \quad (k=0,1, \cdots, 2003). \\
\text { Then } f(0)=f(1)=\cdots=f(2003).
\end{array}
$$ | Let $n=2002$. Then $(n+1) \mid (f(n+1)-f(0))$.
And $-n \leqslant f(n+1)-f(0) \leqslant n$, so
$$
\begin{array}{l}
f(n+1)-f(0)=0 . \\
\text { Let } F(x)=f(x)-f(0) \text {, then } \\
F(n+1)=F(0)=0 .
\end{array}
$$
Thus, $F(x)=x[x-(n+1)] G(x)$, where $G(x)$ is a polynomial with integer coefficients.
When $2 \leqslant k ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,225 |
Example 3 Mother's Day is coming, and Xiao Hong, Xiao Li, and Xiao Meng went to the flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the three types... | Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $\sqrt{z}$ yuan, respectively. Then,
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16 .
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text{. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text{. }
$$... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,226 |
Example 1 In the tetrahedron $A B C D$, it is known that
$$
A B=A C=A D=D B=5, B C=3, C D=4 \text {. }
$$
Then the volume of the tetrahedron is $\qquad$ . | Given that $\angle B C D=90^{\circ}$.
As shown in Figure 1, take the midpoint $E$ of $B D$ and connect $A E$ and $C E$.
By the properties of a right-angled triangle, we have
$$
B E=C E=D E \text {. }
$$
Since $A B=A C=A D=D B=5$, we have
$\triangle A B E \cong \triangle A C E \cong \triangle A D E$.
Thus, $A E \perp B ... | 5 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,228 |
Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$
and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$ | Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$
Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \a... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,229 |
Example 3 As shown in Figure 3, given that the volume of the regular quadrilateral pyramid $P$ $A B C D$ is 1, $E, F, G, H$ are the midpoints of line segments $A B, C D, P B, P C$ respectively. Then the volume of the polyhedron $B E G-C F H$ is . $\qquad$ | Take the midpoints $M, N$ of $BC, EF$ respectively, and connect $MN$, $GM$, $GN$.
Then the polyhedron $BEG-CFH$ is divided into a quadrilateral pyramid $G-EBMN$ and a triangular prism $HFC-GNM$.
Since $E, F, G, H$ are the midpoints of segments $AB, CD, PB, PC$ respectively, and the volume of the regular quadrilateral... | \frac{5}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,230 |
Example 4 In the tetrahedron $ABCD$, it is known that
$$
\begin{array}{l}
AB=CD=2 \sqrt{13}, BC=AD=\sqrt{41}, \\
AC=DB=\sqrt{61} .
\end{array}
$$
Then the volume of the tetrahedron is $\qquad$ | Notice that,
$$
4^{2}+5^{2}=41,4^{2}+6^{2}=52,5^{2}+6^{2}=61 \text {. }
$$
As shown in Figure 4, the tetrahedron $A B C D$ is extended to the rectangular parallelepiped $A M D N-P C Q B$, where $A N=4, A M=$ $5, A P=6$.
It can be calculated that the volume of the rectangular parallelepiped $A M D N-P C Q B$ is 120.
... | 40 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,231 |
Example 1 Consider using $1 \times m$ and $n \times 1$ rectangles (cannot be rotated) to cover a rectangle without overlap or gaps, where $m, n$ are positive integers. Prove: If a rectangle can be covered in this way, then it can also definitely be covered without rotation, overlap, or gaps using only $1 \times m$ or $... | Proof: Let the rectangle be $a \times b$. Then it can be divided into $a \times b$ unit squares, and the square at the $i$-th row and $j$-th column is labeled as $(i, j)$ $(1 \leqslant i \leqslant a, 1 \leqslant j \leqslant b)$. Take
$$
\varepsilon_{1}=\mathrm{e}^{\frac{2 \pi i n}{n}}, \varepsilon_{2}=\mathrm{e}^{\frac... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,232 |
Example 2: Can a $13 \times 13$ grid be covered without overlap and without gaps using only $1 \times 4$ and $4 \times 1$ rectangles, such that only the center is left uncovered? | Assume it is possible.
Similarly to Example 1, label the small square at $(k, j)$ with $\mathrm{i}^{k+2 j}$.
It can be observed that the sum of the numbers in a $1 \times 4$ or $4 \times 1$ rectangular block is 0.
Thus, if the covering is done as required, the sum of all the numbers in the grid will be the number in ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,233 |
Example 3 A $7 \times 7$ grid is completely covered without overlap by 16 $1 \times 3$ rectangles (which can be rotated $90^{\circ}$). Question: In what positions will the small squares that are not covered be? | Solve: Divide a $7 \times 7$ grid into 49 small squares, and label them with coordinates similar to Example 1.
Assume the square that cannot be covered is $(a, b)$. Label the square $(i, j)$ with $\varepsilon^{i+j}$, where $\varepsilon=\mathrm{e}^{\frac{2 \pi}{3}}$ is a cube root of unity.
Clearly, the sum of the num... | (1,1),(1,4),(1,7),(4,1),(4,4),(4,7),(7,1),(7,4),(7,7) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,234 |
Example 4 Use two types of shapes to cover an $8 \times 9$ grid without rotation or repetition, where one is a $3 \times 1$ rectangle, and the other is a $1 \times 3$ rectangle with the middle square removed. Prove: In the $8 \times 9$ grid, there exists a set of 18 $1 \times 1$ small squares, such that if exactly two ... | Prove: Similar to Example 1, label the $8 \times 9$ grid with coordinates, and label the cell $(k, j)$ with the number $\varepsilon^{k} i^{j}$, where,
$$
\varepsilon=\mathrm{e}^{\frac{2 \pi}{3}} \text {. }
$$
For the two given shapes, if the first one is chosen, then the sum of the numbers in its cells is
$$
\varepsil... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,235 |
Example 4 A group of tourists is traveling by car, with the requirement that the number of passengers in each car is equal. At first, each car carries 22 people, leaving 1 person unable to board. If one car leaves empty, then all the tourists can be evenly distributed among the remaining cars. It is known that each car... | Let's assume there are $k$ cars at the beginning, and after one car leaves empty, the average number of passengers per car is $n$ people.
Obviously, $22<n \leqslant 32$.
From the problem, we have
$22 k+1=n(k-1)$
$\Rightarrow n=\frac{22 k+1}{k-1}=22+\frac{23}{k-1}$.
Since $n$ is a natural number, we have
$k-1=1$ or 23
$... | 24, 529 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,237 |
Example 1 If $a, b, c \in \mathbf{R}_{+}, a+b+c=1$, prove:
$$
\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}>2+\sqrt{5} .
$$ | Prove by introducing an auxiliary function
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,238 |
Example 2 Let $x, y, z$ be real numbers. Prove: For any $\triangle ABC$ we have
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2} \\
\geqslant 2 x y \cos C+2 y z \cos A+2 z x \cos B .
\end{array}
$$ | Prove by constructing an auxiliary function
$$
f(x)=x^{2}+y^{2}+z^{2}-(2 x y \cos C+2 y z \cos A+2 x \cos B) \text {. }
$$
Rearranging, we get
$$
\begin{aligned}
f(x)= & x^{2}-(2 y \cos C+2 z \cos B) x+ \\
& y^{2}+z^{2}-2 y z \cos A .
\end{aligned}
$$
Then, $\Delta=(2 y \cos C+2 z \cos B)^{2}-$
$$
\begin{aligned}
& 4... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,239 |
Example 3 Let $a>0, b>0, c>0$, and $a \neq b, 2c=a+b$.
Prove: $c^{a+b}<a^{a} b^{b}$. | Assume without loss of generality that $a < b$. For $F(x)$, we find the first and second derivatives as follows:
$$
\begin{array}{l}
F^{\prime}(x)=\ln x - \ln (a+x) + \ln 2, \\
F^{\prime \prime}(x)=\frac{1}{x} - \frac{1}{a+x} > 0.
\end{array}
$$
Thus, the function $F^{\prime}(x)$ is monotonically increasing.
Since $F^... | c^{a+b} < a^{a} b^{b} | Inequalities | proof | Yes | Yes | cn_contest | false | 726,240 |
Example 4 Let $n$ be a positive integer. Prove:
$$
\frac{(1+\sqrt{n})^{2010}-(1-\sqrt{n})^{2010}}{\sqrt{n}}
$$
is an integer. | Prove by constructing an auxiliary function
$$
G(x)=(1+x)^{2010}-(1-x)^{2010} \text {. }
$$
It is easy to see that $G(-x)=-G(x)$, which means $G(x)$ is an integer polynomial containing only odd powers.
Thus, $\frac{G(x)}{x}$ is an integer polynomial containing only even powers.
$$
\text { Hence } \frac{G(\sqrt{n})}{\s... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,241 |
Example 5 If the equation $x^{3}-27 x+m=0$ has three distinct real roots, find the range of real number $m$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Define the auxiliary function $F(x)=27 x-x^{3}$.
Differentiate the function $F(x)$ to get
$$
F^{\prime}(x)=27-3 x^{2} \text {. }
$$
Let $F^{\prime}(x)=0$. Then $x= \pm 3$.
Therefore, $F(x)_{\text {max }}=F(3)=54$,
$$
F(x)_{\text {min }}=F(-3)=-54 \text {. }
$$
From the graph, we know that when $-54<m<54$,
$$
F(x)=m \... | -54<m<54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,242 |
Example 6 Let the functions $f(x)=\ln x, g(x)=\frac{1}{2} x^{2}$. If $x_{1}>x_{2}>0$, for what value of $m(m \in \mathbf{Z}, m \leqslant 1)$ is it always true that
$$
m\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)>x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)
$$
holds. | Introduce an auxiliary function
$$
t(x)=m g(x)-x f(x)=\frac{m}{2} x^{2}-x \ln x \quad (x>0) \text {. }
$$
By the problem, $x_{1}>x_{2}>0$.
Therefore, if the original inequality always holds for $x>0$, i.e., the function $t(x)$ is monotonically increasing, then
$$
t^{\prime}(x)=m x-\ln x-1 \geqslant 0
$$
always holds.... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,243 |
Example 1 Let $f(x)$ be a periodic function defined on the set of real numbers with a period of 2, and it is an even function. When $x \in[2,3]$, $f(x)=-x$. Then when $x \in[-2,0]$, $f(x)=$ $\qquad$ | When $x \in[-2,-1]$, $x+4 \in[2,3]$. By the periodicity of $f(x)$ with period 2, we have
$$
f(x)=f(x+4)=-x-4 \text {. }
$$
When $x \in[-1,0]$, $-x+2 \in[2,3]$.
By the fact that $f(x)$ is an even function with period 2, we have
$$
f(x)=f(-x)=f(-x+2)=x-2 \text {. }
$$
In summary, when $x \in[-2,0]$,
$$
f(x)=-3+|x+1| .
... | f(x)=-3+|x+1| | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,244 |
Example 3 If $f(1+\cos x)=\sin ^{2} x+\cos x-1$, then $f(x)=$ $\qquad$ . | Notice that,
$$
\begin{array}{l}
f(1+\cos x)=\sin ^{2} x+\cos x-1 \\
=-\cos ^{2} x+\cos x .
\end{array}
$$
Let $1+\cos x=m$. Then $\cos x=m-1$.
So $f(m)=-(m-1)^{2}+m-1$ $=-m^{2}+3 m-2$.
Therefore, $f(x)=-x^{2}+3 x-2$. | -x^{2}+3 x-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,246 |
Example 4 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x_{1}, x_{2}, \cdots, x_{2000}$, we have
$$
\begin{array}{l}
\sum_{i=1}^{2005} f\left(x_{i}+x_{i+1}\right)+f\left(\sum_{i=1}^{2006} x_{i}\right) \\
\leqslant \sum_{i=1}^{2006} f\left(2 x_{i}\right) .
\end{array}
$$ | Let $x_{1}=x_{2}=x$,
$$
x_{3}=x_{4}=\cdots=x_{2006}=0 \text {. }
$$
Substitute into equation (1) to get
$$
\begin{array}{l}
f(2 x)+f(x)+2003 f(0)+f(2 x) \\
\leqslant 2 f(2 x)+2004 f(0) .
\end{array}
$$
Thus, $f(x) \leqslant f(0)$.
Now let $x_{1}=x_{3}=\cdots=x_{2005}=\frac{x}{2}$,
$$
x_{2}=x_{4}=\cdots=x_{2006}=-\fra... | f(x)=c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,247 |
Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had ... | Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$
Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$
which simplifies to ... | 131 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,248 |
Example 5 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x+y) f(x-y) \\
=(f(x)+f(y))^{2}-4 x^{2} f(y) .
\end{array}
$$ | Let $x=y=0$, we get
$$
f^{2}(0)=4 f^{2}(0) \text {. }
$$
Thus, $f(0)=0$.
Let $x=y \neq 0$. Then
$$
0=4 f^{2}(x)-4 x^{2} f(x) \text {. }
$$
Hence, $f(x)=0$ or $f(x)=x^{2}$.
Upon verification, $f(x)=0$ or $f(x)=x^{2}$ satisfy the original equation. | f(x)=0 \text{ or } f(x)=x^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,249 |
Example 6 Given that $f(x)$ is monotonically increasing on $[1,+\infty)$, and for any $x, y \in[1,+\infty)$, we have
$$
f(x+y)=f(x)+f(y)
$$
Prove: There exists a real number $k$, such that $f(x)=k x$ holds for $x \in [1,+\infty)$. | Prove (1) From $f(x+y)=f(x)+f(y)$, we know
$$
\begin{array}{l}
f\left(x_{1}+x_{2}+\cdots+x_{n}\right) \\
=f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right) .
\end{array}
$$
Taking $x_{1}=x_{2}=\cdots=x_{n}=1$, then
$$
f(n)=n f(1)\left(n \geqslant 1, n \in \mathbf{N}_{+}\right) \text {. }
$$
(2) Let $x... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,250 |
1. Given that the function $f(x)$ is a monotonically increasing function defined on $[-4,+\infty)$, to ensure that for all real numbers $x$,
$$
f\left(\cos x-b^{2}\right) \geqslant f\left(\sin ^{2} x-b-3\right)
$$
holds true. Find the range of real numbers $b$. | Answer $: b \in\left[\frac{1}{2}-\sqrt{2}, 1\right]$. | b \in\left[\frac{1}{2}-\sqrt{2}, 1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,251 |
4. If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+xy(x+y) \\
f^{\prime}(0)=1 .
\end{array}
$$
Find the analytical expression of the function $f(x)$. | 提示:由原式得
$$
\begin{array}{l}
f(x+y) \\
=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] .
\end{array}
$$
上式可化为柯西方程
$$
g(x+y)=g(x)+g(y) \text {, }
$$
其中, $g(x)=f(x)-\frac{1}{3} x^{3}$.
则 $f(x)-\frac{1}{3} x^{3}=a x$.
再由 $f^{\prime}(0)=1$, 得 $a=1$.
于是, $f(x)=\frac{1}{3} x^{3}+x$.
Hint: From the original equatio... | f(x)=\frac{1}{3} x^{3}+x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,253 |
1. Let $1 \leqslant x \leqslant 3$. Then the sum of the maximum and minimum values of $|x-1|-|x-3|$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 1. A.
From the given, we know
$$
|x-1|-|x-3|=(x-1)-(3-x)=2 x-4 \text{, }
$$
its maximum and minimum values are $2$ and $-2$, respectively. Therefore, the required sum is 0. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,254 |
2. Let $x=\sqrt{5}, y$ be the greatest integer not exceeding the real number $x$. Then $\frac{1}{x-y}=(\quad$.
(A) $\sqrt{5}-2$
(B) $\sqrt{5}+2$
(C) $\sqrt{5}-1$
(D) $\sqrt{5}+1$ | 2. B.
From the given, we know $y=2$.
Therefore, $\frac{1}{x-y}=\frac{1}{\sqrt{5}-2}=\sqrt{5}+2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,255 |
3. As in Figure 1, in quadrilateral $A B C D$, it is known that
$$
\begin{array}{l}
\text { } \angle A C B=\angle B A D \\
=105^{\circ}, \angle A B C= \\
\angle A D C=45^{\circ} \text {. Then } \\
\angle C A D=(\quad) .
\end{array}
$$
(A) $65^{\circ}$
(B) $70^{\circ}$
(C) $75^{\circ}$
(D) $80^{\circ}$ | 3. C.
It is easy to know that, $\angle C A D=105^{\circ}-\left(180^{\circ}-105^{\circ}-45^{\circ}\right)=75^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,256 |
5. Given that $x, y, z$ are non-negative real numbers, and satisfy
$$
\left\{\begin{array}{l}
3 x+2 y+z=5 . \\
2 x+y-3 z=1 .
\end{array}\right.
$$
Let $s=3 x+y-7 z$. Then the maximum value of $s$ is ( ).
(A) $-\frac{1}{11}$
(B) $\frac{1}{11}$
(C) $-\frac{5}{7}$
(D) $-\frac{7}{5}$ | 5. A.
From the problem, we have
$$
\left\{\begin{array}{l}
x=7 z-3 \geqslant 0, \\
y=7-11 z \geqslant 0
\end{array} \Rightarrow \frac{3}{7} \leqslant z \leqslant \frac{7}{11}\right. \text {. }
$$
Then $s=3 x+y-7 z$
$$
\begin{array}{l}
=3(7 z-3)+(7-11 z)-7 z \\
=3 z-2 .
\end{array}
$$
Therefore, $s_{\max }=3 \times \... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,258 |
Example 6 In the donation activity of a school to the "Hope Project", the total donation amount of $m$ male students and 11 female students in Class A is equal to the total donation amount of 9 male students and $n$ female students in Class B, which is $mn + 9m + 11n + 145$ yuan. It is known that each person's donation... | Notice,
$$
\begin{array}{l}
m n+9 m+11 n+145 \\
=(m+11)(n+9)+46 .
\end{array}
$$
From the problem, we know that $m+11$ and $n+9$ are both divisors of $m n+9 m+11 n+145$. Since each person's donation amount is the same, therefore,
$$
m+11=n+9 \text {, }
$$
and $m+11$ and $n+9$ are both divisors of 46.
Given that $m, n... | 47 \text{ yuan or } 25 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,259 |
6. As shown in Figure 2, given
$$
\begin{array}{l}
\angle D A P=\angle P B C= \\
\angle C D P=90^{\circ}, A P= \\
P B=4, A D=3 . \text { Then } \\
B C=(\quad) .
\end{array}
$$
(A) $\frac{32}{3}$
(B) 16
(C) $\frac{41}{3}$
(D) $\frac{41}{2}$ | 6. C.
Connect $P C$, and draw $D E \perp B C$ at point $E$.
It is easy to know that $P D=5, D E=8$.
Let $E C=x$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
P C^{2}=P D^{2}+D C^{2}=25+64+x^{2}, \\
P C^{2}=P B^{2}+B C^{2}=16+(3+x)^{2} \\
\Rightarrow 89=25+6 x \Rightarrow x=\frac{32}{3} .
\end{array}
$$
Thu... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,260 |
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$
the solution is $1 \leqslant x \leqslant 3$. Then $a=$ | $=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$. | -12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,261 |
2. If $p$ and $8p^{2}+1$ are both prime numbers, then $p=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. 3.
Notice that, if $p=3k$, then $8p^2+1=72k^2+1;$
if $p=3k+1$, then
$$
\begin{array}{l}
8p^2+1=72k^2+48k+9 \\
=3\left(24k^2+16k+3\right);
\end{array}
$$
if $p=3k+2$, then
$$
\begin{array}{l}
8p^2+1=72k^2+96k+33 \\
=3\left(24k^2+32k+11\right).
\end{array}
$$
Also, since $p$ and $8p^2+1$ are both prime numbers, i... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 726,262 |
3. Let $x$ and $y$ be two distinct non-negative integers, and satisfy $x y + 2x + y = 13$. Then the minimum value of $x + y$ is $\qquad$ | 3. 5 .
From the problem, we know that $(x+1)(y+2)=15$.
$$
\begin{array}{l}
\text { Then }(x+1, y+2) \\
=(15,1),(5,3),(3,5),(1,15) \\
\Rightarrow(x, y)=(14,-1),(4,1),(2,3),(0,13) .
\end{array}
$$
Therefore, the minimum value of $x+y$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,263 |
4. As shown in Figure 3, it is known that quadrilateral $A B C D$ is a square, $\triangle A E P$ is an isosceles right triangle, $\angle E \Lambda P$ $=90^{\circ}$, and $D 、 P 、 E=$ are collinear points. If $E A=P A$ $=1, P B=\sqrt{5}$, then $D P$ $=$ . $\qquad$ | 4. $\sqrt{3}$.
Connect $B E$. It is easy to know, $\triangle A E B \cong \triangle A P D$.
Then $P D=E B$, and $P E \perp E B$.
So $B E=\sqrt{B P^{2}-E P^{2}}=\sqrt{5-2}=\sqrt{3}$. | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,264 |
-、(20 points) Let the real number $k$ satisfy $0<k<1$. Solve the fractional equation about $x$
$$
\frac{2 k}{x-1}-\frac{1}{x^{2}-x}=\frac{k+1}{x} \text {. }
$$ | $$
\begin{array}{l}
\frac{2 k x}{(x-1) x}-\frac{1}{(x-1) x}=\frac{(k+1)(x-1)}{(x-1) x} \\
\Rightarrow 2 k x-1=(k+1)(x-1) \\
\Rightarrow(k-1) x=-k . \\
\text { Also, } 0<k<1 \text {, so } x=-\frac{k}{k-1} .
\end{array}
$$
Upon inspection, when $k=\frac{1}{2}$, $x=1$ is an extraneous root, and the original equation has ... | x=-\frac{k}{k-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,265 |
Four. (25 points) Let the linear function $y=k x+b(k \neq 0)$ intersect the positive half-axes of the $x$-axis and $y$-axis at points $E$ and $F$, respectively, and intersect the linear function $y=2 x-1$ at point $A(m, 2)$, with $A$ being the midpoint of $E F$.
(1) Find the expression of the linear function;
(2) If th... | (1) Since point $A(m, 2)$ lies on the graph of $y=2 x-1$, we have,
$$
2=2 m-1 \Rightarrow m=\frac{3}{2} \Rightarrow A\left(\frac{3}{2}, 2\right) \text {. }
$$
Since $A$ is the midpoint of $E F$, the points are $E(3,0), F(0,4)$.
Thus, the required linear function expression is
$$
y=-\frac{4}{3} x+4 \text {. }
$$
(2) Ac... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,266 |
Five. (25 points) As shown in Figure 4, given $A B=A C$, $\angle B A C=\angle C D E=$ $90^{\circ}$, $D C=D E$, and $F$ is the midpoint of $B E$. Prove:
$$
F A=F D,
$$
and $F A \perp F D$. | Extend $A F$ to point $G$, such that $A F = F G$. Connect $G E, G D$.
In $\triangle A F B, \triangle G F E$, we have
$$
\begin{array}{l}
A F = G F, \angle A F B = \angle G F E, F B = F E \\
\Rightarrow \triangle A F B \cong \triangle G F E \\
\Rightarrow A B = G E, \angle B = \angle 2 . \\
\text { Also } \angle C = 18... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,267 |
1. Given an integer $n \geqslant 2$, for any pairwise coprime positive integers $a_{1}, a_{2}, \cdots, a_{n}$, let
$A=a_{1}+a_{2}+\cdots+a_{n}$.
For $i=1,2, \cdots, n$, let the greatest common divisor of $A$ and $a_{i}$ be $d_{i}$; the greatest common divisor of the remaining $n-1$ numbers after removing $a_{i}$ from $... | 1. Consider $D_{1}=\left(a_{2}, a_{3}, \cdots, a_{n}\right)$ and $d_{2}=\left(a_{2}, \dot{A}\right)=\left(a_{2}, a_{1}+a_{2}+\cdots+a_{n}\right)$.
Let $\left(D_{1}, d_{2}\right)=d$. Then $d\left|a_{2}, d\right| a_{3}, \cdots, d\left|a_{n}, d\right| A$.
Hence, $d \mid a_{1}$.
Thus, $d \mid\left(a_{1}, a_{2}, \cdots, a_{... | (n-1)^{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,268 |
2. As shown in Figure 1, let $\triangle ABC$ be inscribed in $\odot O$, $P$ be the midpoint of arc $\overparen{BAC}$, $Q$ be the antipode of $P$, $I$ be the incenter of $\triangle ABC$, and $PI$ intersect side $BC$ at point $D$. The circumcircle of $\triangle AID$ intersects the extension of $PA$ at point $F$. Point $E... | 2. From $\angle A E F=\angle A P E$, we know
$$
A F \cdot P F=E F^{2} \text {. }
$$
Also, from the fact that points $A, I, D, F$ are concyclic, we have
$$
\begin{array}{l}
P A \cdot P F=P I \cdot P D . \\
\text { Therefore, } P F^{2}=A F \cdot P F+P A \cdot P F \\
=E F^{2}+P I \cdot P D .
\end{array}
$$
As shown in F... | \sin ^{2} \angle BAC = \frac{2r}{R} | Geometry | proof | Yes | Yes | cn_contest | false | 726,269 |
1. Movie tickets have three prices: 10 yuan, 15 yuan, and 20 yuan. The class leader bought 30 movie tickets with 500 yuan, among which the 20 yuan tickets are ( ) more than the 10 yuan tickets.
(A) 20
(B) 15
(C) 10
(D) 5 | Let the number of movie tickets bought at 10 yuan, 15 yuan, and 20 yuan be $x$, $y$, and $z$ respectively.
According to the problem, we can set up the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=30, \\
10 x+15 y+20 z=500 .
\end{array}\right.
$$
Subtracting (1) $\times 15$ from (2) gives
$$
5(z-x)=50 \Rightarr... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,270 |
4. Given that $p$ is a prime number, $a, k$ are positive integers, satisfying $p^{a}<k<2 p^{a}$. Prove: There exists a positive integer $n$, such that
$$
n<p^{2 a} \text {, and } \mathrm{C}_{n}^{k} \equiv n \equiv k\left(\bmod p^{a}\right) \text {. }
$$ | 4. Prove: $\mathrm{C}_{k+1 p^{0}}^{k}\left(t=-1,0,1, \cdots, p^{a}-2\right)$ these $p^{a}$ numbers form a complete residue system modulo $p^{a}$.
Let $\nu_{p}(m)$ be the exponent of the prime $p$ in the prime factorization of $m$. Let $r_{p}(m)=\frac{m}{p^{\nu_{p}(m)}}$.
Notice that; $p^{\circ}<k<2 p^{\circ}$.
Thus, fo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,272 |
5. Let the integer $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots$, $b_{n}$ be non-negative real numbers. Prove:
$$
\begin{array}{l}
\left(\frac{n}{n-1}\right)^{n-1} \frac{1}{n} \sum_{i=1}^{n} a_{i}^{2}+\left(\frac{1}{n} \sum_{i=1}^{n} b_{i}\right)^{2} \\
\geqslant \prod_{i=1}^{n}\left(a_{i}^{2}+b_{i... | 5. Let $\lambda=\left(\frac{n}{n-1}\right)^{n-1}$. Clearly, $\lambda>1$.
For $i \in\{1,2, \cdots, n\}$, fix $a_{j}, b_{j}(j \neq i)$, and simultaneously fix $p=a_{i}^{2}+b_{i}^{2}$. Thus, the right side of the inequality remains fixed, and at this point, the left side of the inequality is
$$
\frac{\lambda}{n}\left(p-b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,273 |
6. In the Cartesian coordinate plane, let point sets $P, Q$ be convex polygonal regions (including interiors and boundaries) with vertices all being integer points, $T=P \cap Q$. Prove: if the point set $T$ is non-empty and contains no integer points, then $T$ is a non-degenerate convex quadrilateral region. | 6. Given that $P$ and $Q$ are both closed convex polygons, we know that $T$ is a (possibly degenerate) closed convex polygon.
If $T$ is a point, then $T$ must be a vertex of $P$ or $Q$, which is an integer point, leading to a contradiction.
If $T$ is a line segment, then $T$ must be the intersection of an edge of $P$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,274 |
1. Given the sets
$$
A=\{1, b, a+b\}, B=\{a-b, a b\} \text {, }
$$
and $A \cap B=\{-1,0\}$.
Then the values of $a$ and $b$ are ( ).
(A) $-1,0$
(B) $0,-1$
(C) $-1,1$
(D) $1,-1$ | -,1. A.
From $1 \notin A \cap B$, we know that one of $b$ or $a+b$ must be 0, and the other must be -1.
Since $a \neq 0$ (otherwise, the elements in set $A$ would contradict the principle of distinctness), therefore, $a=-1, b=0$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,275 |
2. In the Cartesian coordinate system $x O y$, the area of the figure enclosed by the curve
$$
2|x|+3|y|=5
$$
is ( ).
(A) $\frac{5}{3}$
(B) 5
(C) $\frac{20}{3}$
(D) $\frac{25}{3}$ | 2. D.
As shown in Figure 1, it is clear that the figure enclosed by the curve $2|x|+3|y|=5$ is symmetric about the $x$-axis and $y$-axis. Therefore, we only need to consider the part of the graph in the first quadrant. In this case, $x>0, y>0$, so the curve is $2 x+3 y=5$.
Thus, the figure enclosed by the curve $2|x|... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,276 |
3. Given that in the expansion of $\left(x^{2}-\frac{1}{x}\right)^{n}$, the constant term is 15. Then the value of $n$ is ( ).
(A) 3
(B) 6
(C) 9
(D) 12 | 3. B.
From the given, we have
$$
\mathrm{C}_{n}^{k}\left(x^{2}\right)^{k}\left(-\frac{1}{x}\right)^{n-k}=15 \text {. }
$$
Notice that,
$$
2 k+k-n=0 \Rightarrow k=\frac{n}{3} \Rightarrow C_{n}^{\frac{n}{3}}=15 \text {. }
$$
When $n=3$, $\mathrm{C}_{3}^{1}=3 \neq 15$;
When $n=6$, $C_{6}^{2}=15$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,277 |
4. Let $A, B, C, D$ be four points on a sphere with center $O$, such that $AB, AC, AD$ are mutually perpendicular, and $AB=3$ cm, $AC=4$ cm, $AD=\sqrt{11}$ cm. Then the radius $r$ of the sphere is $(\quad)$ cm.
(A) 3
(B) 4
(C) 5
(D) 6 | 4. A.
Taking $A$ as the origin, and the rays $AB$, $AC$, and $AD$ as the $x$-axis, $y$-axis, and $z$-axis respectively to establish a rectangular coordinate system. Then $B(3,0,0), C(0,4,0), D(0,0, \sqrt{11})$. Clearly, the center of the sphere should be the intersection point of the following three planes:
$$
\alpha:... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,278 |
5. Given three points $A, B, C$ in a plane satisfying
$$
|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=5,|\overrightarrow{C A}|=6 \text {. }
$$
Then the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is ( ). | 5. C.
From the cosine theorem, we know
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-\overrightarrow{B A} \cdot \overrightarrow{B C} \\
=-\frac{|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}-|\overrightarrow{A C}|^{2}}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\overrightarrow{B ... | -35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,279 |
6. Let $z_{1}, z_{2}$ be a pair of unequal conjugate complex numbers, and $\left|z_{1}\right|=\sqrt{3}, \frac{z_{1}^{2}}{z_{2}}$ is a real number. Then $\left|z_{1}-z_{2}\right|=(\quad)$.
(A) $\sqrt{3}$
(B) $\sqrt{6}$
(C) 3
(D) $2 \sqrt{3}$ | 6. C.
Solution 1 From the problem, we have
$$
\frac{z_{1}^{2}}{z_{2}}=\overline{\left(\frac{z_{1}^{2}}{z_{2}}\right)} \Rightarrow z_{1}^{2} \overline{z_{2}}=z_{2} \overline{z_{1}^{2}} \text {. }
$$
Multiplying both sides by $z_{1} z_{2}$, we get
$$
z_{1}^{3}\left|z_{2}\right|^{2} \doteq \overline{z_{1}} z_{2}^{2}\lef... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,280 |
2. A forest farm has arranged seven days of tree planting work. Starting from the second day, 5 more tree planters are added each day compared to the previous day, but starting from the second day, each person plants 5 fewer trees per day than the previous day, and all planters on the same day plant the same number of ... | Prompt: Example 6. Answer: $1521 ; 54$ or 24.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Example 6. Answer: $1521 ; 54$ or 24. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,281 |
7. Given $S$ is a parallelepiped. Proposition $p: S$ is a cube; Proposition $q: S$'s any body diagonal is perpendicular to any face diagonal that does not intersect it. Then proposition $p$ is a ( ) condition for proposition $q$.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and neces... | 7. C.
Sufficiency is obvious.
Necessity.
As shown in Figure 3, the top and bottom faces are parallelograms, and each side face is a rectangle. $\square$
Draw the projections of the body diagonals $A_{1} C$ and $B_{1} D$ on the plane $A A_{1} B_{1} B$.
Since the plane $A A_{1} B_{1} B \perp$ plane $A B C D$, the feet... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,282 |
8. A fraction whose numerator and denominator have a greatest common divisor of 1 is called a reduced fraction. Then the sum of all positive reduced proper fractions with a denominator of 100 is ( ).
(A) 20
(B) 30
(C) 35
(D) 45 | 8. A.
From $100=2^{2} \times 5^{2}$, let
$$
\begin{array}{l}
A=\left\{n|2| n, n \leqslant 100, n \in \mathbf{N}_{+}\right\}, \\
B=\left\{n|5| n, n \leqslant 100, n \in \mathbf{N}_{+}\right\} .
\end{array}
$$
Then $|A \cup B|=|A|+|B|-|A \cap B|$
$$
=50+20-10=60 \text {. }
$$
Therefore, there are 40 proper fractions t... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 726,283 |
9. A department arranges the duty schedule for four people, Jia, Yi, Bing, and Ding, during the eight-day National Day holiday. It is known that Jia and Yi each work four days, Jia cannot work on the first day and Jia and Yi do not work on the same day; Bing needs to work three days and cannot work consecutively; Ding ... | 9. B.
First, consider A and B. A does not work on the first day, so A can choose any four days from the remaining seven days, thus there are $\mathrm{C}_{7}^{4}$ ways to arrange, and the remaining days are assigned to B, so B has $\mathrm{C}_{4}^{4}$ ways to arrange.
, Suppose C's working days are arranged on
$B_{2}$ ... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 726,284 |
10. Given
$$
f(x)=\frac{x^{4}+k x^{2}+1}{x^{4}+x^{2}+1}(k, x \in \mathbf{R}) \text {. }
$$
Then the product of the maximum and minimum values of $f(x)$ is
(A) $\frac{4 k-1}{3}$
(B) $k$
(C) $\frac{k+1}{2}$
(D) $\frac{k+2}{3}$ | 10. D.
Notice that, $f(x)=1+\frac{(k-1) x^{2}}{x^{4}+x^{2}+1}$.
And $x^{4}+1 \geqslant 2 x^{2}$, then $0 \leqslant \frac{x^{2}}{x^{4}+x^{2}+1} \leqslant \frac{1}{3}$.
When $k \geqslant 1$,
$$
f(x)_{\max }=\frac{k+2}{3}, f(x)_{\min }=1 \text {; }
$$
When $k<1$,
$$
f(x)_{\min }=\frac{k+2}{3}, f(x)_{\max }=1 \text {. }
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,285 |
11. The function
$$
y=\cos 2x + 2 \sin x \quad (x \in (0, 2\pi))
$$
is monotonically decreasing in the interval $\qquad$ . | $=$ 11. $\left(\frac{\pi}{6}, \frac{\pi}{2}\right),\left(\frac{5 \pi}{6}, \frac{3 \pi}{2}\right)$.
Differentiating the given function, we get
$$
y^{\prime}=-2 \sin 2 x+2 \cos x=-2 \cos x(2 \sin x-1) \text {. }
$$
Let $y^{\prime} 0 , } \\
{ \cos x > 0 }
\end{array} \text { or } \left\{\begin{array}{l}
2 \sin x-1<0 , \\... | \left(\frac{\pi}{6}, \frac{\pi}{2}\right),\left(\frac{5 \pi}{6}, \frac{3 \pi}{2}\right) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 726,286 |
12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$
Then $1992 a c-1811 b c-181 a b=$ | 12. 0 .
Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$
Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,287 |
13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$ | 13. 1 .
Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$
we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$
Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,288 |
14. Let $O$ be a point inside $\triangle A B C$, and
$$
\overrightarrow{A O}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \text {. }
$$
Then $S_{\triangle O A B}: S_{\triangle O B C}=$ | 14. $\frac{3}{5}$.
From the given, we have $5 \overrightarrow{O A}+4 \overrightarrow{O B}+3 \overrightarrow{O C}=0$.
As shown in Figure 6, extend $O A, O B, O C$ to $O A^{\prime}, O B^{\prime}, O C^{\prime}$, such that
$$
\begin{array}{l}
\left|O A^{\prime}\right|=5|O A|,\left|O B^{\prime}\right|=4|O B|, \\
\left|O C^... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,289 |
15. (12 points) Given a positive integer $n$, the arithmetic sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}^{2}+a_{n+1}^{2} \leqslant \frac{2}{5} .
$$
Prove: $\sum_{i=n+1}^{2 n+1} a_{i} \leqslant n+1$. | $$
\begin{array}{l}
\sum_{i=n+1}^{2 n+1} a_{i}=\frac{(n+1)\left(a_{n+1}+a_{2 n+1}\right)}{2} \\
=\frac{(n+1)\left(3 a_{n+1}-a_{1}\right)}{2} \\
\leqslant \frac{n+1}{2} \sqrt{a_{1}^{2}+9 a_{n+1}^{2}-6 a_{1} a_{n+1}} \\
=\frac{n+1}{2} \sqrt{a_{1}^{2}+9 a_{n+1}^{2}+2 \times 3 a_{1}\left(-a_{n+1}\right)} \\
\leqslant \frac... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,290 |
16. (12 points) In the tetrahedron $A-BCD$, it is known that $\triangle BCD$ and $\triangle ACD$ are both equilateral triangles with side lengths of 2, and $\triangle BCD$ lies in plane $\alpha$. The side length $AB=\sqrt{3}$. Now, four of the eight labels numbered $1 \sim 8$ are randomly attached to the four vertices,... | 16. (1) Let $M_{1}$ denote the event that “$f(C)$ and $f(D)$ are both odd numbers,” and $M_{2}$ denote the event that “$f(C)$ and $f(D)$ are both even numbers.”
From the problem, we have
$$
\begin{array}{l}
P\left(M_{1}\right)=\frac{\mathrm{A}_{4}^{2}}{\mathrm{~A}_{8}^{2}}=\frac{4 \times 3}{8 \times 7}=\frac{3}{14}, \\... | \frac{3}{7}, \frac{9}{56} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,291 |
3. Place several parts in at least 10 boxes, requiring that each box contains the same number of parts. If each box is filled with 12 parts, one part is left over; if three more boxes are added, all the parts can be evenly distributed among the boxes. How many boxes were there originally? How many parts are there? | Prompt: Example 5. Answer: $32 ; 385$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 32 ; 385 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,292 |
17. (12 points) Given the ellipse $C$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0),
$$
Let $A_{d}$ be the set of all lines that are at a distance $d$ from the origin. Does there exist a constant $d(0<d<b)$, such that for any line $l \in A_{d}$, there exist $l_{1} 、 l_{2} \in A_{d}$, where $l_{1} 、 l_{2}$ pass ... | 17. Suppose there exists a constant $d$ that satisfies the conditions of the problem.
Take $l \in A_{d}$ as a special line: $x=d$, and let it intersect the ellipse $C$ at points $P$ and $Q$.
Construct a circle $O_{d}$ with the origin $O$ as the center and $d$ as the radius, intersecting the positive x-axis at point $... | d=\frac{a b}{\sqrt{a^{2}+b^{2}}} | Geometry | proof | Yes | Yes | cn_contest | false | 726,293 |
18. (12 points) Let the function
$$
f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)
$$
satisfy that when $|x| \leqslant 1$, we have $|f(x)| \leqslant 1$. Let the maximum value of $\left|f^{\prime}(x)\right|$ for $|x| \leqslant 1$ be $K$. Find all functions $f(x)$ such that there exists $x_{0} \in[-1,1]$, satisfyi... | 18. Note that, $|f(0)|=|c| \leqslant 1$.
From the fact that one of $b$ and $-b$ must have the same sign as $a+c$, we have
$$
\begin{array}{l}
|f(1)|=|a+b+c| \leqslant 1, \\
|f(-1)|=|a-b+c| \leqslant 1 \\
\Rightarrow|a+c|+|b| \leqslant 1 \\
\Rightarrow|a+c| \leqslant 1-|b| \\
\Rightarrow|a|-|c| \leqslant|a+c| \leqslant... | f(x)= \pm\left(2 x^{2}-1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,294 |
19. (18 points) Let $A \subseteq \mathbf{R}, \mathscr{B}$ be the set of all subsets of $A$. If the set $A$ satisfies that for any mapping $f: \mathscr{B} \rightarrow \mathscr{B}$, there always exists $X \in \mathscr{B}$, such that
$$
\underbrace{f(f(\cdots}_{2 n \uparrow}(f(X)) \cdots)) \neq A-X
$$
holds, where $A-X$ ... | 19. Let $f^{k}(X)=\underbrace{f(f(\cdots(f(X)) \cdots))}_{k \uparrow}$.
If there exists a finite subset $B \subseteq A$, satisfying
$$
|B|=n+1 \text {. }
$$
First, prove: There exists a mapping $f: \mathscr{A} \rightarrow \mathscr{B}$, such that for any set $Y \in \mathscr{B}$, we have $f^{2^{n}}(Y)=A-Y$.
Let the set... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,295 |
1. Let $a$ be a real number, and $a^{3}+a^{2}-a+2=0$. Then
$$
\begin{array}{l}
(a+1)^{2011}+(a+1)^{2012}+(a+1)^{2013} \\
=(\quad) .
\end{array}
$$
(A) 3
(B) -3
(C) 1
(D) -1 | $\begin{array}{l}\text { I. 1. D. } \\ \text { Given } 0=a^{3}+a^{2}-a+2=(a+2)\left(a^{2}-a+1\right) \\ \Rightarrow a+2=0 \Rightarrow a+1=-1 \text {. } \\ \text { Therefore, }(a+1)^{2011}+(a+1)^{2012}+(a+1)^{2013} \\ =(-1)^{2011}+(-1)^{2012}+(-1)^{2013}=-1 \text {. }\end{array}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,296 |
$$
\begin{array}{l}
\frac{1+\sqrt{2013}(\sqrt{2012}-\sqrt{2011})}{\sqrt{2011}+\sqrt{2012}+\sqrt{2013}}+\sqrt{2011} \\
=(\quad) .
\end{array}
$$
(A) $\sqrt{2010}$
(B) $\sqrt{2011}$
(C) $\sqrt{2012}$
(D) $\sqrt{2013}$ | 2. C.
$$
\begin{array}{l}
\text { Original expression }=\frac{2012+\sqrt{2013} \cdot \sqrt{2012}+\sqrt{2011} \cdot \sqrt{2012}}{\sqrt{2011}+\sqrt{2012}+\sqrt{2013}} \\
=\sqrt{2012} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,297 |
3. As shown in Figure 1, given the parabola $C: y=x^{2}-2 x+4$ and the line $l: y=-2 x + 8$, the line $y=k x$ $(k>0)$ intersects the parabola $C$ at two different points $A$ and $B$, and intersects the line $l$ at point $P$. Perpendiculars are drawn from $A, B, P$ to the $x$-axis, with the feet of the perpendiculars be... | 3. B.
$$
\begin{array}{l}
\text { Let } A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) . \\
\text { From }\left\{\begin{array}{l}
y=k x, \\
y=x^{2}-2 x+4
\end{array}\right. \\
\Rightarrow x^{2}-(2+k) x+4=0 \\
\Rightarrow x_{1}+x_{2}=2+k, x_{1} x_{2}=4 \\
\Rightarrow \frac{1}{O A_{1}}+\frac{1}{O B_{1}}=\frac{1}{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,298 |
4. Given that point $A$ in the first quadrant lies on the graph of the inverse proportion function $y=\frac{1}{x}$, and point $B$ in the second quadrant lies on the graph of the inverse proportion function $y=\frac{k}{x}$. If $O A \perp O B, \tan A=$ $\frac{\sqrt{2}}{2}$, then $k=(\quad)$.
(A) $\frac{1}{4}$
(B) $-\frac... | 4. D.
Draw $A M \perp x$-axis through point $A$, and draw $B N \perp x$-axis through point $B$, with the feet of the perpendiculars being $M$ and $N$ respectively.
$$
\begin{array}{l}
\text { Since } O A \perp O B \Rightarrow \angle A O B=90^{\circ} . \\
\Rightarrow \angle M A O=\angle B O N \\
\Rightarrow \triangle A... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,299 |
5. If $a b c=4, a+b+c=2$, and $a \geqslant b \geqslant c$, then the minimum value of $|a|+|b|+|c|$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 5. C.
If $a>0$.
Also, from the problem, we know $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Then, $\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \Rightarrow a \geqslant 4$.
Therefore, $b, c$ are both negative.
$$
\begin{array}{l}
\text... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,300 |
6. As shown in Figure 2, in rectangle $A B C D$, it is known that $A B=4$, $B C=2$. Points $F$ and $E$ are on $A B$ and $C D$ respectively. Connecting $D F$, $C F$, $A E$, and $B E$ intersect at points $Q$ and $P$. If the area of quadrilateral $P E Q F$ is $S$, then ( ).
(A) $S \geqslant 2$
(B) $S \leqslant 2$
(C) $S \... | 6. B.
As shown in Figure 5(a), in trapezoid $ABCD$, $AC \parallel BD$, and $AD$ intersects $BC$ at point $E$. Let
\[
\begin{array}{l}
S_{\triangle AEC}=a, S_{\triangle AEB}=c, \\
S_{\triangle BED}=b, S_{\triangle DEC}=d .
\end{array}
\]
Then $ab=cd$, and $c=d$.
Therefore, $c=d=\sqrt{ab}$.
Also, $(\sqrt{a}-\sqrt{b})^{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 726,301 |
1. Every day at 5 PM when school is over, Xiao Ming's father always drives from home to pick him up on time and take him back. One day, the school dismissed an hour early, and Xiao Ming walked home by himself. On the way, he met his father who was coming to pick him up, and as a result, they arrived home 20 minutes ear... | 二、1.50 minutes.
As shown in Figure 6, Xiao Ming
starts walking home from point $A$
and meets the car coming to pick him up
at point $C$. As a result, the car returns from $C$ to $B$ 20 minutes earlier than usual. This indicates that the car takes 20 minutes to travel from $C$ to $A$ and back to $C$.
Therefore, the car ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 726,302 |
4. Use square tiles to cover a piece of land without overlapping or leaving gaps. If tiles with a side length of $x$ cm are used, exactly $n$ tiles are needed; if tiles with a side length of $y$ cm are used, exactly 124 more tiles are needed. Given that $x, y, n$ are integers, and $x, y$ are coprime. Question: How many... | From the problem, we know
$$
n x^{2}=(n+124) y^{2} \text {. }
$$
Given that $x$ and $y$ are coprime, we have
$$
\begin{array}{l}
n+124=k x^{2}, n=k y^{2}\left(k \in \mathbf{N}_{+}\right) \\
\Rightarrow 124=k(x-y)(x+y) \\
\Rightarrow\left\{\begin{array} { l }
{ x + y = 3 1 , } \\
{ x - y = 1 }
\end{array} \text { or }... | 23.04 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,303 |
2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$ | 2. 45 .
From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$
When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,304 |
3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$
has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$
has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$
has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$ | 3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$
Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,305 |
4. As shown in Figure 3, in Rt $\triangle A B C$, $C D$ and $C E$ are the altitude and median to the hypotenuse $A B$, respectively, $B C = a, A C = b (b > a)$. If $\tan \angle D C E = \frac{1}{3}$, then $\frac{a}{b}=$ $\qquad$ | 4. $\frac{\sqrt{10}-1}{3}$.
It is known that, $C D=\frac{a b}{\sqrt{a^{2}+b^{2}}}$.
By the projection theorem, we have
$$
\begin{array}{l}
B C^{2}=B D \cdot B A \\
\Rightarrow B D=\frac{B C^{2}}{B A}=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} \\
\Rightarrow D E=B E-B D=\frac{b^{2}-a^{2}}{2 \sqrt{a^{2}+b^{2}}} \\
\Rightarrow \ta... | \frac{\sqrt{10}-1}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,306 |
One, (20 points) In a dormitory of a school, there are several students. On New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator, who in turn gives a card back to each student. In this way, a total of 51 g... | Let there be $x$ students and $y$ administrators living in the dormitory $\left(x, y \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
x(x-1)+x y+y=51 \\
\Rightarrow y=\frac{51+x-x^{2}}{x+1}=\frac{49}{x+1}-x+2 .
\end{array}
$$
Since $x, y$ are positive integers, we know that $(x, y)=(6,3)$.
Therefore, there are 6 s... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,307 |
II. (25 points) As shown in Figure 4, with the side $BC$ of the acute triangle $\triangle ABC$ as the diameter, construct circle $\odot O$. $AD$ is the tangent to $\odot O$, and from a point $E$ on $AB$, draw a perpendicular to $AB$ intersecting the extension of $AC$ at point $F$. If $\frac{AB}{AF}=\frac{AE}{AC}$, prov... | Let $\odot O$ intersect $AB$ at point $G$, and connect $CG$. Since $BC$ is the diameter of $\odot O$, we have $CG \perp AB$.
$$
\begin{array}{l}
\text { Also, } EF \perp AB \Rightarrow CG \parallel EF \\
\Rightarrow \frac{AG}{AE}=\frac{AC}{AF}=\frac{AE}{AB} \\
\Rightarrow AE^2=AB \cdot AG .
\end{array}
$$
From the fac... | AD=AE | Geometry | proof | Yes | Yes | cn_contest | false | 726,308 |
Three. (25 points) Given the equation
$$
\left(m^{2}-1\right) x^{2}-3(3 m-1) x+18=0
$$
has two positive integer roots, where $m$ is an integer.
(1) Find the value of $m$;
(2) The sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $c=2 \sqrt{3}$, $m^{2}+a^{2} m-8 a=0$, $m^{2}+b^{2} m-8 b=0$, find the area of $\tria... | Three, (1) It is easy to know that the two roots of the equation are
$$
x_{1}=\frac{6}{m+1}, x_{2}=\frac{3}{m-1} \text {. }
$$
From the problem, we have
$$
\left\{\begin{array}{l}
m+1=1,2,3,6, \\
m-1=1,3 .
\end{array}\right.
$$
Thus, $m=2$.
(2) According to the problem, $a$ and $b$ are both roots of the equation
$$
x... | 1 \text{ or } \sqrt{9+12 \sqrt{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,309 |
2. Let $0 \leqslant \alpha, \beta < 2 \pi, \alpha, \beta \neq \frac{\pi}{3}, \frac{5 \pi}{3}$, and
$$
\frac{2 \sin \alpha - \sqrt{3}}{2 \cos \alpha - 1} + \frac{2 \sin \beta - \sqrt{3}}{2 \cos \beta - 1} = 0 \text{. }
$$
Then $\cot \frac{\alpha + \beta}{2} = $ $\qquad$ | 2. $-\frac{\sqrt{3}}{3}$.
Notice that,
$$
\frac{2 \sin \alpha-\sqrt{3}}{2 \cos \alpha-1}+\frac{2 \sin \beta-\sqrt{3}}{2 \cos \beta-1}=0
$$
has a geometric meaning that the slopes of the lines connecting point $P\left(1, \frac{3}{2}\right)$ on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ to points $M(2 \cos \alpha,... | -\frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,311 |
3. In $\triangle A B C$, it is known that $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=$ $\qquad$ | 3. 0 .
It is known that, $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O H}$.
Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{M N}$
$$
\begin{aligned}
= & \overrightarrow{O A} \cdot(\overright... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,312 |
4. If three dice of different sizes are rolled randomly, the probability that the sum of the numbers shown on two or three of the dice is 7 is $\qquad$ .
| 4. $\frac{35}{72}$.
Let $A_{i}=\{$ three dice do not show $i\}(i=1,2$, $\cdots, 6$.
By the principle of inclusion-exclusion, we have
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{A_{6}}\right|=\left|\overline{A_{1} \cup A_{6}}\right| \\
=6^{3}-\left(\left|A_{1}\right|+\left|A_{6}\right|-\left|A_{1} \cap A_... | \frac{35}{72} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,313 |
Example 1 There are more than $n^{2}$ stones on the table. Betja and Watt play a game: Betja starts, and the two take turns to remove stones from the table. Each time, the number of stones removed is either a prime number less than $n$, a positive integer multiple of $n$, or 1. The player who takes the last stone wins.... | Prove that the set of positive integers $\mathbf{N}_{+}$ can be partitioned into two disjoint subsets $I$ and $J$, where:
$I=\left\{i \in \mathbf{N}_{+}\right\}$ when there are $i$ stones on the table, the first player wins\},
$J=\left\{j \in \mathbf{N}_{+}\right.$ when there are $j$ stones on the table, the second pla... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,314 |
Example 7 Minkowski's Theorem.
A convex region \( R \) in the plane, symmetric with respect to the origin, if its area is greater than 4, then it must contain an integer point other than the origin. | Prove that for each even integer point $(2n, 2m)$ as the center, a square $S_{2n, 2m}$ with a side length of 2.
If $S_{2n, 2m}$ contains a part of $R$, then use the transformation
$$
x-2n=x', y-2m=y',
$$
to move this part into the square $S_{0,0}$.
In this way, all parts of the convex region $R$ are concentrated into ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,316 |
6. Let the parabola $y=x^{2}+x-1$ intersect the $x$-axis at points $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$. Then $10 x_{2}^{3}+4 x_{1}^{5}-58=(\quad)$.
(A) 0
(B) -100
(C) -2
(D) 82 | 6. B.
From the given conditions, we have
$$
\begin{array}{l}
x_{1}^{2}=1-x_{1} \\
\Rightarrow x_{1}^{5}=\left(1-x_{1}\right)^{2} x_{1}=\left(1-2 x_{1}+1-x_{1}\right) x_{1} \\
\quad=2 x_{1}-3 x_{1}^{2}=5 x_{1}-3 .
\end{array}
$$
Similarly, $x_{2}^{3}=2 x_{2}-1$.
Then $10 x_{2}^{3}+4 x_{1}^{5}-58$
$$
\begin{array}{l}
=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,317 |
1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$ | $\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{arra... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,318 |
2. After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of fi... | 2. 60.
Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. S... | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,319 |
3. As shown in Figure 2, in
rhombus $A B C D$, it is
known that $\angle A B C=60^{\circ}$, line
$E F$ passes through point $D$, and
intersects the extensions of
$B A$ and $B C$ at points $E$ and $F$, respectively. $M$
is the intersection of $C E$ and $A F$. If $C M=4, E M=5$, then $C A=$ $\qquad$ | 3. 6 .
It is easy to prove $\triangle E A D \backsim \triangle D C F$.
$$
\begin{array}{l}
\text { Therefore, } \frac{E A}{A D}=\frac{D C}{C F} \Rightarrow \frac{E A}{A C}=\frac{A C}{C F} \\
\Rightarrow \triangle E A C \backsim \triangle A C F \\
\Rightarrow \angle A E C=\angle C A F \\
\Rightarrow C A^{2}=C E \cdot C... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,320 |
4. Given an equilateral $\triangle A B C$ with side length $6 \sqrt{3}, \odot O$ has a radius of $r$. When the center $O$ moves from point $A$ along the path $A B \rightarrow B C \rightarrow C A$ back to point $A$, $\odot O$ moves with point $O$. Let the area of the part of $\triangle A B C$ that $\odot O$ does not pas... | 4. $S=3 \sqrt{3}(3-r)^{2}(0<r<3)$ When $r>0$, inside $\triangle ABC$, the part not covered by $\odot O$ forms a regular $\triangle A_{0} B_{0} C_{0}$. The sides of this triangle are parallel to the sides of the original equilateral triangle, and the distance between the parallel lines is $r$.
Connecting $A A_{0}$ and... | S=3 \sqrt{3}(3-r)^{2}(0<r<3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,321 |
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