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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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6. Given that $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, and a perpendicular line $D E$ is drawn from point $D$ to the leg $B C$. If the segments $A C, C D, D E$ can form a triangle (i.e., they are the three sides of a triangle), then the range of $\frac{B C}{A B}$ is ( ).
(A... | 6.C.
In Rt $\triangle A B C$,
$$
\begin{array}{l}
0A C \\
\Rightarrow A C \sin A + A C \sin ^{2} A > A C \\
\Rightarrow \sin ^{2} A + \sin A - 1 > 0 \\
\Rightarrow \left(\sin A + \frac{\sqrt{5} + 1}{2}\right)\left(\sin A - \frac{\sqrt{5} - 1}{2}\right) > 0 .
\end{array}
$$
Since $\sin A + \frac{\sqrt{5} + 1}{2} > 0$,... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,292 |
1. Given that $a$, $b$, and $c$ are all integers, and it always holds that
$$
(x-a)(x-10)+1=(x+b)(x+c) \text {. }
$$
then the integer $a=$ $\qquad$ | II. 1.8 or 12.
From the given, we have
$$
(x-a)(x-10)+1=(x+b)(x+c) \text {. }
$$
Then $x^{2}-(a+10) x+10 a+1$
$$
=x^{2}+(b+c) x+b c \text {. }
$$
Thus, $\left\{\begin{array}{l}-(a+10)=b+c, \\ 10 a+1=b c .\end{array}\right.$
Therefore, $-99=10 b+10 c+b c$, which means
$$
c=\frac{-10 b-100+1}{b+10}=-10+\frac{1}{b+10} ... | 8 \text{ or } 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,293 |
2. In the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CD$ bisects $\angle ACB$. If $BC=a$, $CA=b(a<b)$, $CD=t$, and $t=b-a$, then $\tan A=$ $\qquad$ (express in exact numerical form). | 2. $\frac{\sqrt{6}-\sqrt{2}}{2}$.
As shown in Figure 4, draw $D E \perp A C, D F \perp$ $B C$, with the feet of the perpendiculars being $E$ and $F$ respectively. It is easy to see that quadrilateral $C E D F$ is a square. Therefore,
$$
C E=D E=\frac{t}{\sqrt{2}} .
$$
From $\frac{A E}{A C}=\frac{D E}{B C}$, we have $... | \frac{\sqrt{6}-\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,294 |
$\begin{array}{l}\text { 3. Given real numbers } a, b, c. \text { If } \\ \frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 c a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1, \\ \text { then }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2008}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2008}+ \\ \left(\frac{a^{2}+b^{... | 3.3.
Notice
$$
\begin{array}{l}
\frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 a a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+1+\frac{(c-a)^{2}-b^{2}}{2 c a}+1+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}-1=1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,295 |
4. Let positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=49 \\
y^{2}+y z+z^{2}=36 \\
z^{2}+z x+x^{2}=25
\end{array}\right.
$$
Then $x+y+z=$ $\qquad$ | 4. $\sqrt{55+36 \sqrt{2}}$.
(1) + (2) + (3) gives
$$
2(x+y+z)^{2}-3(x y+y z+z x)=110 \text {. }
$$
Also, (1) - (2) gives
$$
(x-z)(x+y+z)=13 \text {, }
$$
(2) - (3) gives
$$
(y-x)(x+y+z)=11 \text {, }
$$
(1) - (3) gives
$$
(y-z)(x+y+z)=24 \text {. }
$$
From $\left.\frac{1}{2}(\text { (5) })^{2}+(6)^{2}+(7)^{2}\right)$... | \sqrt{55+36 \sqrt{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,296 |
One, (20 points) Given that $x$, $a$, and $b$ are real numbers, and satisfy $y=\frac{a x^{2}+8 x+b}{x^{2}+1}$ with a maximum value of 9 and a minimum value of 1. Find the values of $a$ and $b$.
| Given that $x^{2}+1>0$, we can simplify and rearrange the given equation by eliminating the denominator, resulting in
$$
(y-a) x^{2}-8 x+(y-b)=0 .
$$
Since $x$ is a real number, we have
$$
\Delta=(-8)^{2}-4(y-a)(y-b) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-(a+b) y+(a b-16) \leqslant 0$.
Given that $1 \le... | a=b=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,297 |
Theorem 2 Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{n+1}=\frac{a a_{n}+b}{c a_{n}+d}(c \neq 0, a d-b c \neq 0),
$$
and the function $f(x)=\frac{a x+b}{c x+d}$, with the initial term $a_{1} \neq f\left(a_{1}\right)$.
(1) If the function $f(x)$ has two distinct fixed points $x_{1}$ and $x_{2}$, then
$$
\frac... | Proof: (1) Since the function $f(x)$ has two distinct fixed points $x_{1}$ and $x_{2}$, from $x=\frac{a x+b}{c x+d}$ we get
$$
c x^{2}+(d-a) x-b=0 \text {. }
$$
Rearranging gives $c x^{2}-a x=b-d x$, i.e., $b-d x=(c x-a) x$.
Thus, $b-d x_{1}=\left(c x_{1}-a\right) x_{1}$.
Since $a_{n+1}=\frac{a a_{n}+b}{c a_{n}+d}$, w... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,298 |
II. (25 points) As shown in Figure 2, in $\triangle ABC$, $AB = AC$, point $D$ is on segment $AB$, and point $E$ is on the extension of segment $AC$, such that $DE = AC$. On the circumcircle of $\triangle ABC$ at arc $\overparen{AC}$ (excluding point $B$), take any point $F$ (different from points $A$ and $C$), and con... | In $\triangle A B C$ and $\triangle C E E_{1}$, we have
$$
\begin{array}{l}
A B=A C, \\
G E=G E_{1}.
\end{array}
$$
Therefore, $\triangle A B C$ and $\triangle G E E_{1}$ are both isosceles triangles.
$$
\begin{array}{l}
\text { Hence } \angle A B C=\angle A C B, \angle G E E_{1}=\angle G E_{1} E. \\
\text { Also, } \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,299 |
Three. (25 points) Given the system of equations about $x$ and $y$:
$$
\left\{\begin{array}{l}
k x^{2} + y + (k - a) = 0, \\
y = -(k + a) x + b c
\end{array}\right.
$$
It has only one integer solution, where $k, a, b, c$ are all integers, and $a > 0, a, b, c$ satisfy $a^{2} - a + b c = -1, b + c = 2$. Find:
(1) The va... | (1) From $b+c=2, bc=a^{2}-a+1$, it is easy to know that $b, c$ are the roots of the equation $x^{2}-2x+a^{2}-a+1=0$.
Since $b, c$ are integers, we have
$\Delta=(-2)^{2}-4\left(a^{2}-a+1\right) \geqslant 0$.
Thus, $a^{2}-a \leqslant 0$.
Also, $a>0$, so $a \leqslant 1$.
Since $a$ is an integer, hence $a \leqslant 1$.
(2)... | a=1, (k=0, x=0, y=1); (k=1, x=1, y=-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,300 |
1. If the points $(x, y)$ in a plane region satisfy the inequality
$$
\sqrt{\frac{x^{2}}{25}}+\sqrt{\frac{y^{2}}{16}} \leqslant 1 \text {, }
$$
then the area of the plane region is ( ).
(A) 30
(B) 40
(C) 50
(D) 60 | -,1.B.
The original inequality is $\frac{|x|}{5}+\frac{|y|}{4} \leqslant 1$. It represents the boundary and interior of a rhombus with vertices at $(5,0)$, $(0,-4)$, $(-5,0)$, and $(0,4)$. Its area is
$$
S=\frac{1}{2} \times 10 \times 8=40 \text{. }
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,301 |
2. Given $x+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)$ $=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$,
and $a_{0}+a_{2}+a_{3}+\cdots+a_{n-1}=60-\frac{n(n+1)}{2}$. Then the possible value of the positive integer $n$ is ( ).
(A) 7
(B) 6
(C) 5
(D) 4 | 2.C.
Let $x=1$, we get
$$
\begin{array}{l}
1+2^{2}+2^{3}+\cdots+2^{n} \\
=a_{0}+a_{1}+a_{2}+\cdots+a_{n}=2^{n+1}-3
\end{array}
$$
From $a_{1}$ being the coefficient of $x$, we know
$$
a_{1}=1+2+3+\cdots+n=\frac{n(n+1)}{2} \text {. }
$$
From $a_{n}$ being the coefficient of $x^{n}$, we know $a_{n}=\mathrm{C}_{n}^{n}=... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,302 |
3. If the equation about $x$
$$
x^{2}-\left(a^{2}+b^{2}-6 b\right) x+a^{2}+b^{2}+2 a-4 b+1=0
$$
has two real roots $x_{1} 、 x_{2}$ satisfying $x_{1} \leqslant 0 \leqslant x_{2} \leqslant 1$, then the range of $a^{2}+b^{2}+4 a+4$ is $(\quad)$.
(A) $\left[\frac{\sqrt{2}}{2}, \sqrt{5}+2\right]$
(B) $[1,9-4 \sqrt{5}]$
(C)... | 3. D.
$$
\text { Let } \begin{aligned}
f(x)= & x^{2}-\left(a^{2}+b^{2}-6 b\right) x+ \\
& a^{2}+b^{2}+2 a-4 b+1 .
\end{aligned}
$$
Given that the two roots $x_{1}, x_{2}$ of $f(x)=0$ satisfy $x_{1} \leqslant 0 \leqslant x_{2} \leqslant 1$, we know that $f(0) \leqslant 0, f(1) \geqslant 0$, which means
$$
(a+1)^{2}+(b-... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,303 |
4. Two spheres with radius $r$ are tangent to each other, and both are tangent to the two faces of a dihedral angle. A third sphere is also tangent to the two faces of the dihedral angle and is tangent to the two spheres with radius $r$. Given that the plane angle of the dihedral angle is $60^{\circ}$, and the radius o... | 4.C.
Let the radius of the large sphere be $R$. Denote the centers of the two small spheres as $O_{1}$ and $O_{2}$, their tangency point as $\mathrm{A}$, the center of the large sphere as $O_{3}$, and one of the half-planes of the dihedral angle as $\alpha$. The tangency point of the large sphere with the plane $\alph... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,304 |
5. In the Cartesian coordinate system $x O y$, the coordinates of the vertices of a triangle are $\left(x_{i}, y_{i}\right)(i=1,2,3)$, where $x_{i} 、 y_{i}$ are integers and satisfy $1 \leqslant x_{i} 、 y_{i} \leqslant n$ ( $n$ is an integer). If there are 516 such triangles, then the value of $n$ is ( ).
(A) 3
(B) 4
(... | 5.B.
It is known that when $n \geqslant 5$, the number of triangles exceeds 516. Therefore, we only need to consider the case where $n \leqslant 4$. In this case, all sets of three points total $\mathrm{C}_{n}^{3_{2}}$.
Among these, the sets of three points that cannot form a triangle include: sets of three collinear... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,305 |
6. A line $l$ passing through the origin intersects the hyperbola $x y=-2 \sqrt{2}$ at points $P$ and $Q$, where point $P$ is in the second quadrant. Now, fold the upper and lower half-planes to form a right dihedral angle, at which point $P$ falls to point $P^{\prime}$. Then the minimum length of the line segment $P^{... | 6. D.
Let the equation of line $l$ be $y=k x(k<0)$.
From $x y=-2 \sqrt{2}, y=k x$, we get
$$
\begin{array}{l}
P\left(-\sqrt{\frac{-2 \sqrt{2}}{k}}, \sqrt{-2 \sqrt{2} k}\right), \\
Q\left(\sqrt{\frac{-2 \sqrt{2}}{k}},-\sqrt{-2 \sqrt{2} k}\right) .
\end{array}
$$
Draw $P^{\prime} M \perp x$-axis, and connect $M Q$.
The... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,306 |
2. Given the function $y=x^{2}+6 a x-a$ intersects the $x$-axis at two distinct points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$, and
$$
\begin{array}{l}
\frac{a}{\left(1+x_{1}\right)\left(1+x_{2}\right)}-\frac{3}{\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)} \\
=8 a-3 .
\end{array}
$$
Then the value of ... | 2. $\frac{1}{2}$.
From $\Delta=36 a^{2}+4 a>0$, we get $a>0$ or $a<-\frac{1}{9}$. According to the problem,
$$
\begin{array}{l}
y=x^{2}+6 a x-a=\left(x-x_{1}\right)\left(x-x_{2}\right) . \\
\text { Then }\left(1+x_{1}\right)\left(1+x_{2}\right)=f(-1)=1-7 a, \\
\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right) \\
=f(1-6... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,308 |
Example 3 Given $a_{1}=\frac{1}{2}, a_{n+1}=\frac{a_{n}+3}{2 a_{n}-4}(n \in$ $\mathbf{N}_{+}$). Find the general term formula $a_{n}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x=\frac{x+3}{2 x-4}$. Then
$$
2 x^{2}-5 x-3=0 \text {. }
$$
Therefore, $x_{1}=-\frac{1}{2}, x_{2}=3$ are the two fixed points of $f(x)=\frac{x+3}{2 x-4}$. Hence
$$
\begin{array}{l}
a_{n+1}+\frac{1}{2}=\frac{a_{n}+3}{2 a_{n}-4}+\frac{1}{2}=\frac{2\left(a_{n}+\frac{1}{2}\right)}{2 a_{n}-4}, \\
a_{n+1}-3=\frac{a_{n}... | a_{n}=\frac{(-5)^{n}+3 \times 2^{n+1}}{2^{n+1}-2(-5)^{n}}\left(n \in \mathbf{N}_{+}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,309 |
3. In tetrahedron $ABCD$,
\[
\begin{array}{l}
\angle ABC=\angle BAD=60^{\circ}, \\
BC=AD=3, AB=5, AD \perp BC,
\end{array}
\]
$M, N$ are the midpoints of $BD, AC$ respectively. Then the size of the acute angle formed by lines $AM$ and $BN$ is \qquad (express in radians or inverse trigonometric functions). | 3. $\arccos \frac{40}{49}$.
From $A M=\frac{1}{2}(A B+A D)$, we get
$|A M|^{2}=\frac{1}{4}\left(|A B|^{2}+|A D|^{2}+2 A B \cdot A D\right)=\frac{49}{4}$.
Thus, $|A M|=\frac{7}{2}$.
Similarly, $|B N|=\frac{7}{2}$.
Also, $A M \cdot B N=\frac{1}{4}(A B+A D) \cdot(B A+B C)$
$=\frac{1}{4}\left(-|A B|^{2}-A B \cdot A D-B A ... | \arccos \frac{40}{49} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,310 |
4. Let $a_{1}, a_{2}, \cdots, a_{100}$ be
$$
\{1901,1902, \cdots, 2000\}
$$
an arbitrary permutation, and the partial sum sequence
$$
\begin{array}{l}
S_{1}=a_{1}, S_{2}=a_{1}+a_{2}, \\
S_{3}=a_{1}+a_{2}+a_{3}, \\
\cdots \cdots \\
S_{100}=a_{1}+a_{2}+\cdots+a_{100} .
\end{array}
$$
If every term in the sequence $\lef... | 4. $C_{9}^{33} \times 33!\times 33!\times 34!=\frac{99!\times 33!\times 34!}{66!}$.
Let $\{1901,1902, \cdots, 2000\}=A_{0} \cup A_{1} \cup A_{2}$,
where the elements in $A_{i}$ are congruent to $i \pmod{3}$ for $i=0,1,2$.
Then $\left|A_{0}\right|=\left|A_{1}\right|=33,\left|A_{2}\right|=34$.
Let $a_{i}$ modulo 3 be $a... | \frac{99!\times 33!\times 34!}{66!} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,311 |
5. The solution set of the equation about $x$
$$
|1-2 \sin x+\cos x|+2 \sin x+1=\cos 2 x
$$
is $\qquad$ . | 5. $\{x \mid x=2 k \pi+\pi, k \in \mathbf{Z}\}$.
From the original equation, we get
$$
\begin{array}{l}
|1-2 \sin x+\cos x| \\
=\cos 2 x-2 \sin x-1 \\
=-2 \sin ^{2} x-2 \sin x \\
=-2 \sin x(\sin x+1) \geqslant 0 \\
\Rightarrow-1 \leqslant \sin x \leqslant 0 \\
\Rightarrow 1-2 \sin x+\cos x \geqslant 0 \\
\Rightarrow 1... | \{x \mid x=2 k \pi+\pi, k \in \mathbf{Z}\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,312 |
6. Given positive real numbers $a, b$ satisfy
$$
a^{2}+b^{2}=25 \ (b>3),
$$
complex numbers $u, v, w$ satisfy
$$
w=a+b \mathrm{i}, \ u-w=3 v .
$$
If $|v|=1$, then when the principal value of the argument of $u$ is the smallest, the value of $\frac{u}{w}$ is $\qquad$ | 6. $\frac{16}{25}-\frac{12}{25}$ i.
Given $|v|=1$, we know $|u-w|=3|v|=3$.
Thus, in the complex plane, the point $P$ corresponding to $u$ lies on the circle $C$ with center at the point $M$ corresponding to $w$ and radius 3.
When the principal value of the argument of $u$ is the smallest, $OP$ is tangent to the circl... | \frac{16}{25}-\frac{12}{25} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,313 |
Three. (20 points) In the Cartesian coordinate system $x O y$, a line segment $A B$ of fixed length $m$ has its endpoints $A$ and $B$ sliding on the $x$-axis and $y$-axis, respectively. Let point $M$ satisfy
$$
A M=\lambda A B(\lambda>0, \lambda \neq 1) .
$$
Do there exist two distinct fixed points $E$ and $F$ such th... | Three, let $A(a, 0)$ and $B(0, b)$. Then $a^{2}+b^{2}=m^{2}$. From $A M=\lambda A M$, we know that $M$ divides $A B$ in the ratio $\frac{\lambda}{1-\lambda}$.
Let $M(x, y)$. Then
$$
\left\{\begin{array}{l}
x=\frac{a}{1+\frac{\lambda}{1-\lambda}}=(1-\lambda) a, \\
y=\frac{\frac{\lambda}{1-\lambda} b}{1+\frac{\lambda}{1-... | E(-\sqrt{1-2 \lambda} m, 0), F(\sqrt{1-2 \lambda} m, 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,314 |
Four, (20 points) Given positive real numbers $a$ and $b$, it is known that real numbers $x$ and $y$ satisfy $a x^{2} - b x y + a y^{2} = 1$. Try to find the range of the bivariate function $f(x, y) = x^{2} + y^{2}$.
Translate the above text into English, please retain the line breaks and format of the source text, an... | Let $x=\frac{m+n}{\sqrt{a}}, y=\frac{m-n}{\sqrt{a}}$. Then $a \cdot \frac{(m+n)^{2}}{a}-b \cdot \frac{m^{2}-n^{2}}{a}+a \cdot \frac{(m-n)^{2}}{a}=1$,
which means
$(2 a-b) m^{2}+(2 a+b) n^{2}=a$.
Thus, $f(x, y)=x^{2}+y^{2}$
$=\frac{(m+n)^{2}+(m-n)^{2}}{a}=\frac{2}{a}\left(m^{2}+n^{2}\right)$.
When $2 a-b>0$, i.e., $b<2... | \left[\frac{2}{2 a+b}, \frac{2}{2 a-b}\right] \text{ when } b < 2a; \left[\frac{2}{2 a+b},+\infty\right) \text{ when } b \geqslant 2a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,315 |
Five. (20 points) Determine whether there exist infinitely many $\triangle ABC$, satisfying that the lengths of $AB$, $BC$, and $CA$ ($AB < BC < CA$) form an arithmetic sequence of coprime positive integers, and that the altitude from $B$ to $BC$ and the area of $\triangle ABC$ are both positive integers. Provide a pro... | There exist infinitely many triangles $\triangle A B C$ that satisfy the conditions.
Let $A B=a-d, B C=a, C A=a+d(a, d \in \mathbf{N}_{+}, a>d)$, and the area of $\triangle A B C$ be $S$, with the height from side $B C$ being $h_{a}$.
According to Heron's formula, we have
$$
\begin{array}{l}
S=\sqrt{\frac{3 a}{2}\left(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,316 |
一、(50 Points) As shown in Figure 1, in $\triangle A B C$, the excircle opposite to $\angle B A C$ touches the lines $B C$ and $A B$ at points $K$ and $L$, respectively. The excircle opposite to $\angle A B C$ touches the lines $A B$ and $B C$ at points $M$ and $N$, respectively. Let the intersection of lines $K L$ and ... | Let $B C=a, C A=b, A B=c$. Then $A L=B N=B M$
$$
=\frac{1}{2}(a+b+c)
$$
Construct the angle bisector of $\angle A C N$, intersecting $A N, M N$, and the extension of $B A$ at points $Y$, $X^{\prime}$, and $Z$ respectively.
When $a \neq b$, assume without loss of generality that $a>b$. Then
$$
\frac{A Y}{Y N}=\frac{A C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,317 |
(50 points) (1) Prove that there exist infinitely many positive integers $n$, such that $3^{n}+2$ and $5^{n}+2$ are both composite;
(2) Determine whether there exist positive integers $p, q$, such that for any positive integer $n (n \geqslant 2008)$, at least one of $3^{n} p+2$ and $5^{n} q+2$ is a prime number? Prove ... | II. (1) By Fermat's Little Theorem, we know that
$$
3^{4 k} \equiv 1(\bmod 5)\left(k \in \mathbf{N}_{+}\right) \text {. }
$$
And $3^{n}+2=3\left(3^{n-1}-1\right)+5$,
thus, it suffices to let $n-1=4 k$.
Then $5 \mid \left(3^{n}+2\right)$, i.e., when $n=4 k+1$, $3^{n}+2$ is composite.
By Fermat's Little Theorem, we know... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,318 |
Three. (50 points) Given that $n$ is a positive integer greater than 10, and set $A$ contains $2n$ elements. If the family of sets
$$
\left\{A_{i} \subseteq A \mid i=1,2, \cdots, m\right\}
$$
satisfies the following two conditions, it is called "suitable":
(1) For any $i=1,2, \cdots, m, \operatorname{Card}\left(A_{i}\... | The maximum positive integer $m=4$.
Take two non-complementary $n$-element subsets $A_{1}, A_{2}$ of $A$, and let $A_{3}, A_{4}$ be the complements of $A_{1}, A_{2}$, respectively. From these four sets, any three sets will always have two sets that are complementary, hence the intersection of these three sets is the em... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,319 |
$$
\begin{array}{l}
\frac{1}{2222}+\frac{1}{3333}+\frac{1}{4444}+\frac{1}{5555}+ \\
\frac{1}{6666}+\frac{1}{7777}+\frac{1}{8888}+\frac{1}{9999} .
\end{array}
$$
The calculation is as follows:
$$
\begin{array}{l}
\frac{1}{2222}+\frac{1}{3333}+\frac{1}{4444}+\frac{1}{5555}+ \\
\frac{1}{6666}+\frac{1}{7777}+\frac{1}{8888... | $$
\begin{array}{l}
\frac{1}{2222}+\frac{1}{9999}=\frac{1111(9+2)}{1111^{2} \times 2 \times 9} \\
=\frac{11}{1111 \times 18}=\frac{1}{101 \times 18}. \\
\text { Similarly, } \frac{1}{3333}+\frac{1}{8888}=\frac{1}{101 \times 24}, \\
\frac{1}{4444}+\frac{1}{7777}=\frac{1}{101 \times 28}, \\
\frac{1}{5555}+\frac{1}{6666}=... | \frac{419}{254520} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,320 |
In $\triangle A B C$, $\angle A=90^{\circ}, A B=3, A C=4$. $P$ is a point inside $\triangle A B C$, $P D \perp B C$ at $D, P E \perp A C$ at $E, P F \perp A B$ at $F$. Line $E F$ intersects lines $B P$ and $C P$ at points $M$ and $N$. If
$$
\frac{A B}{P F}+\frac{A C}{P E}+\frac{B C}{P D}=12,
$$
prove: $\triangle A B C... | Proof: As shown in Figure 2, connect $AP$ and $BN$. For the sake of convenience, let $PD=x$, $PE=y$, and $PF=z$. It is easy to see that $BC=5$. Therefore, the given equation becomes
$$
\frac{3}{z}+\frac{4}{y}+\frac{5}{x}=12.
$$
Also, $S_{\triangle PBC} + S_{\triangle PCA} + S_{\triangle PAB} = S_{\triangle ABC}$, whic... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,321 |
The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a_{2}=1$, and for $n \geqslant 3$, we always have $a_{n}=3 a_{n-1}+a_{n-2}$. It is known that the tangent of the acute angle $\alpha_{2 i+1}$ is $\frac{3 \sqrt{3}}{a_{2 i+1}}(i=1,2, \cdots, n)$, and the tangent of the acute angle $\alpha_{2 n+2}$ is $\frac{\sqrt{3}}{... | Proof: First, we prove a lemma.
Lemma $a_{2 n} a_{2 n+3}=a_{2 n+1} a_{2 n+2}-9\left(n \in \mathbf{N}_{+}\right)$.
Proof of the lemma: Use mathematical induction on $n$.
When $n=1$, $a_{2}=1, a_{3}=4, a_{4}=13, a_{5}=$
43, we have $a_{2} a_{5}=a_{3} a_{4}-9$. The lemma holds.
Assume that when $n=k$, the lemma holds, i.e... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,322 |
234 Let $R$ be the set of all real numbers.
(1) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x+y) \\
=-2006 x+2007 f(x)-2008 f(y)+2009 y ;
\end{array}
$$
(2) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x... | Solution: (1) From $f(x+y)=f(y+x)$ and the given conditions, we have
$$
\begin{array}{l}
-2006 x+2007 f(x)-2008 f(y)+2009 y \\
=-2006 y+2007 f(y)-2008 f(x)+2009 x,
\end{array}
$$
which simplifies to $f(x)-x=f(y)-y$.
Thus, we can write $f(x)-x=c$ (where $c$ is a constant to be determined).
Therefore, $f(x)=x+c$.
Substi... | f(x)=x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,323 |
Example 7 Given $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, and satisfies the conditions:
(1) For any $x, y \in \mathbf{R}_{+}$, $f(x f(y))=y f(x)$;
(2) As $x \rightarrow+\infty$, $f(x) \rightarrow 0$.
Try to find the function $f(x)$. | Solution: Let $x=y$, then we have $f(x f(x))=x f(x)$.
Obviously, $x f(x)$ is a fixed point of $f(x)$, and the set of fixed points is $\left\{x f(x) \mid x \in \mathbf{R}_{+}\right\}$.
Now let $x=y=1$, substituting into condition (1) we get
$f(f(1))=f(1)$.
Next, let $x=1, y=f(1)$, substituting into the original equation... | f(x)=\frac{1}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,324 |
Example 3 Let $a$ be a prime number, $b$ be a positive integer, and $9(2 a+b)^{2}=509(4 a+511 b)$. Find the values of $a$ and $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Since $9(2 a+b)^{2}=3^{2}(2 a+b)^{2}$ is a perfect square, therefore, $509(4 a+511 b)$ is a perfect square. Since 509 is a prime number, we can let
$$
4 a+511 b=509 \times 3^{2} k^{2} \text {. }
$$
Thus, the original equation becomes
$$
9(2 a+b)^{2}=509^{2} \times 3^{2} k^{2} \text {, }
$$
which means $2 a+... | a=251, b=7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,325 |
4. Given the sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfies the relation $\left(3-a_{n+1}\right)\left(6+a_{n}\right)=18$, and $a_{0}=3$. Then $\sum_{i=0}^{n} \frac{1}{a_{i}}=$ $\qquad$ . | (Let $b_{n}=\frac{1}{a_{n}}(n=0,1, \cdots)$. Then $\left(3-\frac{1}{b_{n+1}}\right)\left(6+\frac{1}{b_{n}}\right)=18$, i.e., $3 b_{n+1}-6 b_{n}-1=0$. Therefore, $b_{n+1}=2 b_{n}+\frac{1}{3}$. Using the fixed point method, we can find $b_{n}$. Hence $\sum_{i=0}^{n} \frac{1}{a_{i}}=\frac{1}{3}\left(2^{n+2}-n-3\right)$.) | \frac{1}{3}\left(2^{n+2}-n-3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,326 |
Example 1 Let $\triangle ABC$ be an acute triangle with unequal side lengths, and let $O$ be its circumcenter. Point $A_{0}$ lies on the extension of segment $AO$ such that $\angle B A_{0} A=\angle C A_{0} A$. Through point $A_{0}$, draw $A_{0} A_{1} \perp AC$ and $A_{0} A_{2} \perp AB$, with the feet of the perpendicu... | Proof: First, it is easy to see that points $A_{0}, B, O, C$ are concyclic.
In fact, as shown in Figure 2, construct the circumcircle of $\triangle A_{0} B C$, and let the intersection point of this circle with line $A O$ be $P$ (different from point $O$). By the property of equal angles subtending equal chords, we hav... | \frac{1}{R_{A}} + \frac{1}{R_{B}} + \frac{1}{R_{C}} = \frac{2}{R} | Geometry | proof | Yes | Yes | cn_contest | false | 720,327 |
Example 2 As shown in Figure 3, $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$, respectively. $AD$ is the altitude from $A$ to side $BC$, and $I$ lies on segment $OD$. Prove that the circumradius of $\triangle ABC$ is equal to the exradius of the excircle opposite to side $BC$. | Proof: As shown in Figure 3, let $I_{1}$ be the excenter, $A I_{1}$ intersects $B C$ at point $E$ and intersects $\odot O$ at point $M$. Then $M$ is the midpoint of $\overparen{B C}$.
Connecting $O M$, then $O M \perp B C$.
Draw $I_{1} F \perp B C$ at point $F$. Then
$\triangle A D I \backsim \triangle M O I, \triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,328 |
Example 3 As shown in Figure 4, in $\triangle A B C$, let $A B>A C$, and draw the tangent line $l$ to the circumcircle of $\triangle A B C$ at point $A$. Also, construct a circle with center $A$ and radius $A C$ that intersects line segment $A B$ at point $D$, and intersects line $l$ at points $E$ and $F$. Prove: lines... | Proof: (1) As shown in Figure 5, let the angle bisector of $\angle BAC$ intersect $E_1F_1$ at point $I$.
Since $AE_1 = AC$, point $C$ and $E_1$ are symmetric with respect to line $AI$. Also, $E_1$, $C$, and $F_1$ lie on $\odot A$, thus
\[
\angle ACI = \angle AE_1I = \angle AF_1I.
\]
Therefore, points $A$, $I$, $C$, an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,329 |
Example 1 Given 20 distinct positive integers, all not exceeding 100. Prove: among the differences obtained by subtracting the smaller number from the larger number for each pair, at least 3 are equal.
| Proof: By making an ordered assumption, let these 20 distinct positive integers be arranged in ascending order as $a_{1}, a_{2}, \cdots, a_{20}$.
If the proposition does not hold, then
$$
a_{20}-a_{19}, a_{19}-a_{18}, \cdots, a_{2}-a_{1}
$$
these 19 positive integers also do not have 3 that are the same.
Thus, 1,2,3,4... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,330 |
Example 2 Place $3 n$ chess pieces on a $2 n \times 2 n$ chessboard. Prove: there must exist $n$ rows and $n$ columns that include all $3 n$ chess pieces. | Proof: Let $d_{1}, d_{2}, \cdots, d_{2 n}$ represent the number of chess pieces in each of the $2 n$ rows (where $d_{i}$ is not necessarily the number of pieces in the $i$-th row), and make the following ordered assumption:
$$
\begin{array}{l}
d_{1} \geqslant d_{2} \geqslant \cdots \geqslant d_{n} \geqslant d_{n+1} \ge... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,331 |
Example 3 Suppose there are 5 line segments, any three of which can form a triangle. Prove: at least one of the triangles is an acute triangle.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: Let the 5 line segments be $a_{1}, a_{2}, a_{3}$, $a_{4}, a_{5}$, and $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$.
Assume that all triangles formed are not acute triangles, then by the cosine rule we have
$$
\begin{array}{l}
a_{3}^{2}=a_{1}^{2}+a_{2}^{2}-2 a_{1} a_{2} \cos \alpha \\
... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,332 |
Example 4 Distribute 100 apples to 7 people, with each person receiving a different number of apples. Prove: There must exist 3 people whose total number of apples is less than 50. | Proof: Let's denote the number of apples each of the 7 people have, in ascending order, as $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<a_{6}<a_{7}$, and
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=100 .
$$
We will consider two cases:
(1) If $a_{4} \geqslant 15$, then
$$
a_{5}+a_{6}+a_{7} \geqslant 16+17+18=51 \text {, }
$$
the p... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,333 |
Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-negative real numbers, not all zero, $m$ be a given rational number greater than 1, and $r_{1}, r_{2}, \cdots, r_{n}$ be non-negative real numbers such that the inequality
$$
\sum_{k=1}^{n} r_{k}\left(x_{k}-a_{k}\right) \leqslant\left(\sum_{k=1}^{n} x_{k}^{m}\right)^{\frac... | Solution: Let $x_{1}=t \neq a_{1}, x_{2}=a_{2}, x_{3}=a_{3}, \cdots \cdots$ $x_{n}=a_{n}$.
From equation (1) we get
$$
\begin{array}{l}
r_{1}\left(t-a_{1}\right) \\
\leqslant\left(t^{m}+\sum_{k=2}^{n} a_{k}^{m}\right)^{\frac{1}{m}}-\left(a_{1}^{m}+\sum_{k=2}^{n} a_{k}^{m}\right)^{\frac{1}{m}} . \\
\text { Let } b=\sum_... | r_{i}=a_{i}^{m-1}\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{1-m}{m}}(i=1,2, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,334 |
2. Given isosceles $\triangle A B C, A B=A C, M$ is the midpoint of side $B C$, $X$ is a moving point on the minor arc $\overparen{M A}$ of the circumcircle of $\triangle A B M$, $T$ is a point inside $\angle B M A$, and satisfies $\angle T M X=90^{\circ}, T X=B X$. Prove: $\angle M T B - \angle C T M$ is independent o... | Proof 1: As shown in Figure 1, let $N$ be the midpoint of line segment $B T$. Then line $X N$ is the axis of symmetry of isosceles $\triangle B X T$. Therefore,
$$
\begin{array}{l}
\angle T N X \\
=90^{\circ}, \angle B X N= \\
\angle N X T .
\end{array}
$$
Since $M N$ is the midline of $\triangle B C T$ parallel to $C... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,335 |
Example $4^{\circ}$ If a perfect square number has the last 3 digits the same and not 0, find the minimum value of this number.
保留源文本的换行和格式，直接输出翻译结果。 | Solution: Since the last two digits of a perfect square can only be even 0, even 1, even 4, even 9, $\overline{25}$, or odd 6, the last three digits of a perfect square can only be 444. Since 444 is not a perfect square, we have $n \geqslant 1444$. And $1444=38^{2}$ is a perfect square, so the smallest perfect square i... | 1444 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,336 |
3. Given trapezoid $A B C D$ with diagonals $A C$ and $B D$ intersecting at point $P$, point $Q$ is between the parallel lines $B C$ and $A D$, satisfying $\angle A Q D=\angle C Q B$, and $P$ and $Q$ are on opposite sides of line $C D$. Prove: $\angle B Q P=\angle D A Q$.
| 3. Let $t=\frac{A D}{B C}$. The homothety $h$ with center $P$ and ratio $-t$ transforms $\triangle P B C$ to $\triangle P D A$. As shown in Figure 3, let $Q^{\prime}=h(Q)$. Then $Q, P, Q^{\prime}$ are collinear.
Since points $P, Q$ are on the same side of side $A D$ and also on the same side of side $B C$, thus, point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,337 |
5. Given that $\triangle A B C$ is a determined triangle, $A_{1}, B_{1}, C_{1}$ are the midpoints of sides $B C, C A, A B$ respectively. $P$ is a moving point on the circumcircle of $\triangle A B C$, and $P A_{1}, P B_{1}, P C_{1}$ intersect the circumcircle of $\triangle A B C$ at another points $A^{\prime}, B^{\prim... | 5. Proof 1: As shown in Figure 4, let $A_{0}, B_{0}, C_{0}$ be the three vertices of the triangle formed by the intersections of lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$.
We will prove: $S_{\triangle A_{0} B_{0} C_{0}}=\frac{1}{2} S_{\triangle A B C}$.
Consider the cyclic hexagon $A B C C^{\prime} P A^{\prime}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,338 |
7. Let $\triangle A B C$ be an acute-angled triangle with angles $\angle A, \angle B, \angle C$ being $\alpha, \beta, \gamma$ respectively, and $\beta>\gamma, I$ being its incenter, and $R$ being its circumradius. $A D$ is the altitude from $A$ to side $B C$, and $K$ is a point on line $A D$ such that $A K=2 R$, with p... | 7. First prove: $\angle K I D=\frac{\beta-\gamma}{2}$.
We can even prove that this conclusion is correct without the condition $I E=I F$.
As shown in Figure 11, let the circumcenter of $\triangle A B C$ be $O$, and connect $A O$ with $\odot O$ intersecting at another point $P$, and the angle bisector of $\angle A$, $... | \beta \leqslant 3 \gamma | Geometry | proof | Yes | Yes | cn_contest | false | 720,340 |
8. Given that point $P$ is a point on side $AB$ of convex quadrilateral $ABCD$, $\omega$ is the incircle of $\triangle CPD$, and $I$ is its center. If $\omega$ is tangent to the incircles of $\triangle APD$ and $\triangle BPC$ at points $K$ and $L$, respectively, and $AC$ intersects $BD$ at point $E$, and $AK$ and $BL$... | 8. Let $\Omega$ be a circle tangent to segment $AB$ and rays $AD$ and $BC$, with center $J$.
Below is the proof: Points $E$ and $F$ lie on line $IJ$.
Let the incircles of $\triangle APD$ and $\triangle BPC$ be $\omega_A$ and $\omega_B$, respectively, and let $h_1$ be the homothety that transforms $\omega$ into $\Omega$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,341 |
2. Person A and Person B start from point $A$ at the same time and travel along the same route to point $B$. If Person A walks at a speed of $a \mathrm{~km} / \mathrm{h}$ for half the time and at $b \mathrm{~km} / \mathrm{h}$ for the other half; and Person B walks at a speed of $a \mathrm{~km} / \mathrm{h}$ for half th... | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,343 |
4. As shown in Figure 2, in $\triangle A B C$, it is known that $\angle B A C=$ $45^{\circ}$. If $A D \perp B C$ at point $D$, and $B D=2, C D=3$, then the area of $\triangle A B C$ is ( ).
(A) $\frac{5}{2}$
(B) 5
(C) $\frac{15}{2}$
(D) 15 | 4. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,345 |
5. A "level-passing game" stipulates: On the $n$-th level, a die must be rolled $n$ times. If the sum of the points from these $n$ rolls is greater than $\frac{3^{n}}{4}$, it counts as passing the level; otherwise, it does not count as passing. Among the following statements:
(1) Passing the first level is a certain ev... | 5. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 720,346 |
Example 5 There is a four-digit number
$$
N=\overline{(a+1) a(a+2)(a+3)} \text {, }
$$
which is a perfect square. Find $a$. | Solution: Note that the last digit of a perfect square can only be $0, 1, 4, 5, 6, 9$, thus
$$
a+3 \equiv 0,1,4,5,6,9(\bmod 10) .
$$
Also, $a \geqslant 0, a+3 \leqslant 9$, so $a=1,2,3,6$.
Therefore, the last two digits of $N$ are
$$
\overline{(a+2)(a+3)}=34,45,56,89 \text {. }
$$
Since the last two digits of a perfe... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,347 |
6. If the function of $x$
$$
y=(a-3) x^{2}-(4 a-1) x+4 a
$$
intersects the coordinate axes at two points, then the value of $a$ is
$\qquad$ . | ii. $6.3,0$ or $-\frac{1}{40}$.
When $a-3=0$, the original function becomes $y=-11 x+12$, whose graph intersects the coordinate axes at two points.
When $a-3 \neq 0$, the original function is a quadratic function, and its graph must intersect the $y$-axis at one point $(0,4a)$.
If this intersection point is not the o... | 3, 0, -\frac{1}{40} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,348 |
7. As shown in Figure 3, in $\triangle A B C$, it is known that $\angle B=$ $40^{\circ}, \angle B A D=30^{\circ}$. If $A B$ $=C D$, then the size of $\angle A C D$ is $\qquad$ (degrees). | $7.40^{\circ}$.
As shown in Figure 6, $\triangle A B D$ is folded along the line $A D$, making point $B$ land at point $E$, resulting in $\triangle A E D$, and $A E$ intersects $C D$ at point $O$.
Since $\triangle A E D \cong \triangle A B D$, we have
$\angle 1=\angle B A D=30^{\circ}, \angle 2=\angle B=40^{\circ}$.
... | 40^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,349 |
8. As shown in Figure 4, there are five circles that are externally tangent to each other in sequence, and all are tangent to lines $a$ and $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, then the diameter of $\odot \mathrm{O}_{3}$ is $\qquad$ | 8.24.
As shown in Figure 7, let the radii of the five circles $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O_{5}$ be $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$, respectively. Draw perpendiculars from points $O_{1}, O_{2}, O_{3}$ to line $a$, with the feet of the perpendiculars being $A_{1}, A_{2}, A_{3}$, respe... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,350 |
9. Given real numbers $a, b, c, d$, and $a \neq b, c \neq d$. If the equations: $a^{2}+a c=2, b^{2}+b c=2, c^{2}+a c=$ $4, d^{2}+a d=4$ all hold, then the value of $6 a+2 b+3 c+2 d$ is $\qquad$. | 9.0 .
$$
\begin{array}{l}
\text { Given }\left(a^{2}+a c\right)-\left(b^{2}+b c\right)=2-2=0, \\
\left(c^{2}+a c\right)-\left(d^{2}+a d\right)=4-4=0,
\end{array}
$$
we get
$$
\begin{array}{l}
(a-b)(a+b+c)=0, \\
(c-d)(a+c+d)=0 .
\end{array}
$$
Since $a \neq b, c \neq d$, we have
$$
a+b+c=0, a+c+d=0 .
$$
Thus, $b=d=-(... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,351 |
10. Given that $m$ and $n$ are positive integers. If $1 \leqslant m \leqslant n$ $\leqslant 30$, and $mn$ is divisible by 21, then the number of pairs $(m, n)$ that satisfy the condition is $\qquad$ . | 10.57.
Given that positive integers $m, n$ satisfy $mn$ is divisible by 21, and $1 \leqslant m \leqslant n \leqslant 30$, therefore,
(1) If $m=21$, then $n=21,22, \cdots, 30$. Hence, there are 10 pairs $(m, n)$ that satisfy the condition.
(2) If $m \neq 21$,
(i) When $n=21$, $m=1,2, \cdots, 20$. There are 20 pairs $(m... | 57 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,352 |
11. Given that $a$ and $b$ are real numbers, and $a^{2}+a b+b^{2}=3$. If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. | Three, 11. Let $a^{2}-a b+b^{2}=k$. From
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=3, \\
a^{2}-a b+b^{2}=k
\end{array} \Rightarrow a b=\frac{3-k}{2} .\right.
$$
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$$
=3+\frac{3-k}{2}=\frac{9-k}{2} \text {. }
$$
Since $(a+b)^{2} \geqslant 0$, i.e., $\frac{9-k}{2} \geqs... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,353 |
12. As shown in Figure 5, in $\triangle A B C$, it is given that $A C=B C, \angle C=$ $20^{\circ}, D$ and $E$ are points on sides $B C$ and $A C$ respectively. If $\angle C A D=20^{\circ}$, $\angle C B E=30^{\circ}$, find the size of $\angle A D E$. | 12. As shown in Figure 8, draw $D G / / B A$, intersecting $A C$ at point $G$, and connect $B G$ with $A D$ intersecting at point $H$. Then
$$
\begin{array}{l}
A D=B G, \\
A H=B H, \\
D H=G H .
\end{array}
$$
In $\triangle A B C$, since $A C=$ $B C, \angle C=20^{\circ}$, therefore,
$$
\angle C A B=\angle C B A=80^{\ci... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,354 |
13. Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1}+x_{2}+\cdots+x_{n}=2008 .
$$
Find the maximum value of the product $x_{1} x_{2} \cdots x_{n}$ of these $n$ positive integers. | 13. Let the maximum value of $x_{1} x_{2} \cdots x_{n}$ be $M$.
Since $x_{1}+x_{2}+\cdots+x_{n}=2008$, it is clear that each $x_{i} (i=1,2, \cdots, n)$ in $M$ is greater than 1.
If there is an $x_{i} \geqslant 4$, we can split $x_{i}$ into $x_{i}-2$ and 2, and examine their product:
$$
\begin{array}{l}
\left(x_{i}-2\... | 2^{2} \times 3^{668} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,355 |
Example 6 Proof: For any positive integer $n$, the algebraic expression
$$
n^{6}+3 n^{5}-9 n^{4}-11 n^{3}+8 n^{2}+12 n+3
$$
is never a perfect square. | Proof: Modulo 4 we have
$$
\begin{array}{l}
n^{6}+3 n^{5}-9 n^{4}-11 n^{3}+8 n^{2}+12 n+3 \\
\equiv n^{6}-n^{5}-n^{4}+n^{3}+3 \\
\equiv n^{3}\left(n^{3}-n^{2}-n+1\right)+3 \\
\equiv n^{3}\left[(n+1)\left(n^{2}-n+1\right)-n(n+1)\right]+3 \\
\equiv n^{3}(n+1)\left(n^{2}-2 n+1\right)+3 \\
\equiv n^{3}(n+1)(n-1)^{2}+3(\bmo... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,358 |
3. Given $(a+b)^{2}=8,(a-b)^{2}=12$. Then the value of $a^{2}+$ $b^{2}$ is ( ).
(A) 10
(B) 8
(C) 20
(D) 4 | 3.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,359 |
4. The equation about $x$
$$
2 k x^{2}+(8 k+1) x=-8 k
$$
has two distinct real roots. Then the range of values for $k$ is ( ).
(A) $k>-\frac{1}{16}$
(B) $k \geqslant-\frac{1}{16}$ and $k \neq 0$
(C) $k=-\frac{1}{16}$
(D) $k>-\frac{1}{16}$ and $k \neq 0$ | 4.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,360 |
5. Given that $a$, $b$, and $c$ are the lengths of the three sides of $\triangle ABC$, and the quadratic equation in $x$
$$
(c-b) x^{2}+2(b-a) x+(a-b)=0
$$
has two equal real roots. Then this triangle is ( ).
(A) Equilateral triangle
(B) Right triangle
(C) Isosceles triangle
(D) Scalene triangle | 5. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,361 |
7. If real numbers $a, b$ satisfy the equation
$$
a^{2}=7-3 a, b^{2}=7-3 b \text {, }
$$
then the value of the algebraic expression $\frac{b}{a}+\frac{a}{b}$ is ( ).
(A) $-\frac{23}{7}$
(B) $\frac{23}{7}$
(C) 2 or $-\frac{23}{7}$
(D) 2 or $\frac{23}{7}$ | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,363 |
8. If $\alpha$ is an acute angle, and $\cos \alpha=0.6$, then $(\quad)$.
(A) $0^{\circ}<\alpha<30^{\circ}$
(B) $30^{\circ}<\alpha<45^{\circ}$
(C) $45^{\circ}<\alpha<60^{\circ}$
(D) $60^{\circ}<\alpha<90^{\circ}$ | 8.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,364 |
9. If $|x-a|=a-|x|(x \neq 0, x \neq a)$, then
$$
\begin{array}{l}
\sqrt{a^{2}-2 a x+x^{2}}-\sqrt{a^{2}+2 a x+x^{2}} \\
=(\quad) .
\end{array}
$$
(A) $2 a$
(B) $2 x$
(C) $-2 a$
(D) $-2 x$ | 9. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,365 |
11. Simplify: $(a+1)^{2}-(a-1)^{2}=(\quad)$.
(A) 2
(B) 4
(C) $4 a$
(D) $2 a^{2}+2$ | 11. C | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,367 |
12. As shown in Figure 2, point $A$
is on the diagonal
of the parallelogram. Determine the relationship between $S_{1}$ and $S_{2}$ ( ).
(A) $S_{1}=S_{2}$
(B) $S_{1}>S_{2}$
(C) $S_{1}<S_{2}$
(D) Cannot be determined | 12. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,368 |
Example 7: Prove that 30000 cannot be expressed as the sum of the squares of two positive integers. | Proof 1: Assume $x^{2}+y^{2}=30000$.
Since $x^{2}, y^{2} \equiv 0,1(\bmod 4)$, therefore, $x^{2}+y^{2} \equiv 0,1,2(\bmod 4)$.
But $30000 \equiv 0(\bmod 4)$, hence
$$
x^{2} \equiv y^{2} \equiv 0(\bmod 4).
$$
Therefore, $x \lesssim y$ are even.
Let $x=2 x_{1}, y=2 y_{1}$. Substituting into the original equation gives $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,369 |
20. Given that the average of $x_{1}, x_{2}, x_{3}$ is $5$, and the average of $y_{1}, y_{2}, y_{3}$ is $7$. Then the average of $2 x_{1}+3 y_{1}, 2 x_{2}+3 y_{2}, 2 x_{3}+3 y_{3}$ is ( ).
(A) 31
(B) $\frac{31}{3}$
(C) $\frac{93}{5}$
(D) 17 | 20.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,377 |
Example 8 Find the positive integer solutions of the equation
$$
3 x^{2}-8 x y+7 y^{2}-4 x+2 y=109
$$ | Solution: The original equation can be transformed to
$$
(3 x-4 y-2)^{2}+5(y-1)^{2}=336 \text {. }
$$
From $5(y-1)^{2}=336-(3 x-4 y-2)^{2} \leqslant 336$,
we get $1 \leqslant y \leqslant 9$.
Moreover, since a perfect square is congruent to 0 or 1 modulo 4, we have that both sides of equation (1) modulo 4 are 0 or
$$
1... | (x, y)=(2,5),(14,9) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,380 |
30. Observe the triangular number array in Figure 14. Then the last number in the 50th row is ( ).
(A) 125
(B) 1260
(C) 1270
(D) 1275 | 30.D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,388 |
1. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the sum of the first $n$ terms denoted as $S_{n}$. If $S_{2}=10, S_{5}=55$, then a direction vector of the line passing through points $P\left(n, a_{n}\right)$ and $Q(n+2$, $a_{n+2})$ can be ( ).
(A) $\left(2, \frac{1}{2}\right)$
(B) $\left(-\frac{1}{2},-2\ri... | -1.B.
Let $\left\{a_{n}\right\}$ be an arithmetic sequence with a common difference of $d$. Then $\left\{\begin{array}{l}S_{2}=2 a_{1}+d=10, \\ S_{5}=5 a_{1}+10 d=55\end{array} \Rightarrow\left\{\begin{array}{l}a_{1}=3, \\ d=4 .\end{array}\right.\right.$
Therefore, $k_{P Q}=\frac{a_{n+2}-a_{n}}{(n+2)-n}=d=4$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,389 |
2. As shown in Figure 1, in the dihedral angle $\alpha-l-\beta$ of $120^{\circ}$, $\odot O_{1}$ and $\odot O_{2}$ are respectively
in the half-planes $\alpha, \beta$,
and are tangent to the edge $l$ at the same
point $P$. Then the sphere with
$\odot O_{1}$ and $\odot O_{2}$ as cross-sections ( ).
(A) has only 1
(B) has... | 2. A
It is known that $\angle O_{1} P O_{2}$ is the plane angle of the dihedral angle $\alpha-l-\beta$.
In the plane $O_{1} \mathrm{PO}_{2}$, draw perpendiculars from points $O_{1}$ and $O_{2}$ to $P O_{1}$ and $\mathrm{PO}_{2}$, respectively, and let the intersection of these two perpendiculars be $O$. Since $\angle... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,390 |
Theorem 1 If the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=p a_{n}+q$ $(p \notin\{0,1\})$, and $x_{0}$ is a fixed point of the function $f(x)=p x+q$, then we have $a_{n+1}-x_{0}=p\left(a_{n}-x_{0}\right)$. | Proof: Since $x_{0}$ is a fixed point of the function $f(x)=p x+q$, we have,
$$
x_{0}=p x_{0}+q .
$$
Also, $a_{n+1}=p a_{n}+q$.
Subtracting (1) from (2) gives
$$
a_{n+1}-x_{0}=p\left(a_{n}-x_{0}\right) \text {. }
$$
Note: Setting $x=p x+q$, the fixed point of the function $f(x)=p x$ $+q$ is found to be $x_{0}=\frac{q... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,391 |
3. Given the function
$$
f(x)=\left\{\begin{array}{ll}
(3-a) x-3, & x \leqslant 7 ; \\
a^{x-6}, & x>7,
\end{array}\right.
$$
The sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=f(n)\left(n \in \mathbf{N}_{+}\right)$, and $\left\{a_{n}\right\}$ is an increasing sequence. Then the range of the real number $a$ is ( ).
(... | 3. D.
From $\left\{a_{n}\right\}$ being an increasing sequence, we get
$$
\left\{\begin{array}{l}
3-a>0, \\
a>1
\end{array} \Rightarrow 12$. Therefore, the range of real number $a$ is $(2,3)$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,392 |
4. Arrange all the simplest proper fractions with a denominator of 24 in ascending order, which are $a_{1}, a_{2}, \cdots, a_{n}$. Then the value of $\sum_{i=1}^{n} \cos a_{i} \pi$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) 0
(D) $-\frac{1}{2}$ | 4.C.
All the simplest proper fractions with a denominator of 24 are
$$
\frac{1}{24}, \frac{5}{24}, \frac{7}{24}, \frac{11}{24}, \frac{13}{24}, \frac{17}{24}, \frac{19}{24}, \frac{23}{24} \text {. }
$$
Notice that
$$
\begin{array}{l}
\frac{1}{24}+\frac{23}{24}=\frac{5}{24}+\frac{19}{24}=\frac{7}{24}+\frac{17}{24}=\fra... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,393 |
5. Set
$$
M=\left\{(x, y) \| \log _{4} x+\log _{4} y \leqslant 1, x, y \in \mathbf{N}_{+}\right.
$$
has ( ) subsets.
(A) 4
(B) 16
(C) 64
(D) 256 | 5.D.
From $\log _{4} x+\log _{4} y \leqslant 1$, we get $1 \leqslant x y \leqslant 4$.
Thus, $(x, y)=(1,1),(1,2),(1,3),(1,4)$,
$$
(2,1),(2,2),(3,1),(4,1) \text {. }
$$
Therefore, the set $M$ has $2^{8}=256$ subsets. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,394 |
6. Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$
a_{n}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and its partial sum $S_{n}$. Then, in the sequence $S_{1}, S_{2}, \cdots, S_{2008}$, the number of rational terms is ( ).
(A) 43
(B) 44
(C) 45
(D) 46 | 6.A.
$$
\begin{array}{l}
\text { Given } a_{k}=\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots),
\end{array}
$$
then $S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)$
$$
=1-\fr... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,395 |
7. The maximum value of the function $y=\tan x-\frac{2}{|\cos x|}$ is ( ).
(A) $1-2 \sqrt{2}$
(B) $1+2 \sqrt{2}$
(C) $-\sqrt{3}$
(D) $\sqrt{3}$ | 7.C.
To maximize $y$, we should have $\tan x > 0$.
Assume $0 < x < \frac{\pi}{2}$. Then
$$
y = \tan x - \frac{2}{\cos x} = \frac{\sin x - 2}{\cos x} < 0,
$$
which implies $\sin x - y \cos x = 2$.
Thus, $\sin (x - \theta) = \frac{2}{\sqrt{1 + y^2}} (\tan \theta = y)$.
From $\sin (x - \theta) \leq 1$, we get $\frac{2}{... | C | Calculus | MCQ | Yes | Yes | cn_contest | false | 720,396 |
8. Given the function
$$
f(x)=x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right) \text {. }
$$
For any real numbers $a, b (a+b \neq 0)$, the value of $\frac{f(a)+f(b)}{a^{3}+b^{3}}$ is ( ).
(A) always greater than zero
(B) always equal to zero
(C) always less than zero
(D) the sign is uncertain | 8. A.
$$
\begin{array}{l}
\text { Since } f(-x)+f(x) \\
=-x^{3}-\log _{2}\left(\sqrt{x^{2}+1}+x\right)+x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right) \\
=0,
\end{array}
$$
Therefore, $f(-x)=-f(x)$, which means $f(x)$ is an odd function.
$$
\begin{array}{l}
\text { Also, } f(x)=x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,397 |
10. Let $k$, $m$, $n$ be integers. From a point $P\left(m^{3}-m, n^{3}-n\right)$ outside the circle $x^{2}+y^{2}=(3 k+1)^{2}$, two tangents are drawn to the circle, touching it at points $A$ and $B$. The number of integer points (points with both coordinates as integers) on the line $AB$ is ( ).
(A) 2
(B) 1
(C) 0
(D) i... | 10. C.
It is known that the equation of the line $AB$ on which the chord of contact lies is
$$
\left(m^{3}-m\right) x+\left(n^{3}-n\right) y=(3 k+1)^{2} \text {. }
$$
If the line $AB$ contains an integer point $\left(x_{0}, y_{0}\right)$, then
$$
\begin{array}{l}
(m-1) m(m+1) x_{0}+(n-1) n(n+1) y_{0} \\
=(3 k+1)^{2} ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,398 |
11. Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$. Then the value of $\frac{\tan \alpha}{\tan \beta}$ is | $$
\begin{array}{l}
\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\
= 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\
\sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\
\text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \c... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,399 |
12. If $x^{5}+3 x^{3}+1=a_{0}+a_{1}(x-1)+$ $a_{2}(x-1)^{2}+\cdots+a_{5}(x-1)^{5}$ holds for any real number $x$, then the value of $a_{3}$ is $\qquad$ (answer with a number). | 12.13.
$$
\begin{array}{l}
\text { In } x^{5}+3 x^{3}+1 \\
=[(x-1)+1]^{5}+3[(x-1)+1]^{3}+1
\end{array}
$$
the coefficient of the $(x-1)^{3}$ term in the expanded form is $\mathrm{C}_{5}^{2}+3=$ 13, so, $a_{3}=13$. | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,400 |
Example 1 Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1, a_{n}=$ $2 a_{n-1}+1(n>1)$. Then the general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=
$$ | Let $x=2x+1$, we get $x=-1$. Then $a_{n}+1=2\left(a_{n-1}+1\right)$.
Therefore, the sequence $\left\{a_{n}+1\right\}$ is a geometric sequence with the first term $a_{1}+1=2$ and common ratio 2.
Hence $a_{n}+1=2 \times 2^{n-1}=2^{n}$, which means $a_{n}=2^{n}-1$. | a_{n}=2^{n}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,401 |
13. If real numbers $x, y$ satisfy $x^{2}+y^{2}=1$, then the minimum value of $\frac{2 x y}{x+y-1}$ is $\qquad$ | $13.1-\sqrt{2}$.
Notice
$$
\begin{array}{l}
\frac{2 x y}{x+y-1}=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{x+y-1} \\
=\frac{(x+y)^{2}-1}{x+y-1}=x+y+1 .
\end{array}
$$
Since $\left(\frac{x+y}{2}\right)^{2} \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$, therefore, $-\sqrt{2} \leqslant x+y \leqslant \sqrt{2}$.
Thus, $\frac... | 1-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,402 |
14. There are 20 cards, each with a number from 1, 2, $\cdots$, 20, placed in a box. Four people each draw a card without replacement. The two people who draw the two smaller numbers are in one group, and the two people who draw the two larger numbers are in another group. If two of them draw $5$ and $14$, the probabil... | 14. $\frac{7}{51}$.
Since two people have already drawn the cards 5 and 14, the other two only need to draw from the remaining 18 cards, which gives $\mathrm{A}_{18}^{2}$ possible situations.
For the two people who drew 5 and 14 to be in the same group, there are two scenarios:
(1) 5 and 14 are the smaller two number... | \frac{7}{51} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,403 |
15. The three lines $y=\sqrt{2} x, y=-\sqrt{2} x$ and $x=m$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into several parts. Now, using 6 different colors to color these parts, each part is colored with one color, and any two parts are of different colors, there are 720 different coloring methods. T... | 15. $|m|-2<m \leqslant-\frac{2}{3}$ or $m=0$ or $\left.\frac{2}{3} \leqslant m<2\right\}$.
Obviously, the two fixed lines $y= \pm \sqrt{2} x$ divide the elliptical region $\frac{x^{2}}{4}+y^{2} \leqslant 1$ into 4
blocks (as shown in Figure 5). When $|m|$
$\geqslant 2$, using 6 colors to color these 4 blocks, the numb... | \left\{m \left\lvert\,-2<m \leqslant-\frac{2}{3}\right. \text { or } m=0 \text { or } \frac{2}{3} \leqslant m<2\right\} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,404 |
One, (15 points) Find the maximum value of the function
$$
y=\left[\sin \left(\frac{\pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)\right] \sin \left(\frac{\pi}{3}+x\right)
$$
and the set of $x$ values at which the maximum value is attained. | $$
\begin{array}{l}
y=\left[\sin \left(\frac{\pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)\right] \sin \left(\frac{\pi}{3}+x\right) \\
=\sqrt{2} \sin x\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right) \\
=\frac{\sqrt{6}}{2} \sin x \cdot \cos x+\frac{\sqrt{2}}{2} \sin ^{2} x \\
=\frac{\sqrt{6}}{4} \sin 2 x+... | \frac{3 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,405 |
II. (15 points) Let the quadratic function $f(x)$ have a maximum value of 12 on the interval $[-1,4]$, and the solution set of the inequality $f(x)<0$ with respect to $x$ is the interval $(0,5)$.
(1) Find the analytical expression of the function $f(x)$;
(2) If for any $x \in \mathbf{R}$, the inequality
$$
f(2-2 \cos x... | (1) According to the problem, let
$$
f(x)=a x(x-5)(a>0),
$$
i.e., $f(x)=a\left(x-\frac{5}{2}\right)^{2}-\frac{25}{4} a$.
Since $f(x)$ achieves its maximum value of 12 on the interval $[-1,4]$, and $a>0$, when $x=-1$, $f(x)$ reaches its maximum value, i.e., $f(-1)=6 a=12$, thus $a=2$.
Therefore, $f(x)=2 x^{2}-10 x$.
(2... | (-\infty,-5) \cup(1,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,406 |
Three. (20 points) Let $x>0, y>0, n \in \mathbf{N}_{+}$. Prove: $\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leqslant \frac{x^{n}+y^{n}}{1+x y}$. | Three, because $x>0, y>0, n \in \mathbf{N}_{+}$, so,
$$
\begin{array}{l}
\left(x^{n-1} y+x y^{n-1}\right)-\left(x^{n}+y^{n}\right) \\
=-(x-y)\left(x^{n-1}-y^{n-1}\right) \leqslant 0,
\end{array}
$$
that is
$$
\begin{array}{l}
x^{n-1} y+x y^{n-1} \leqslant x^{n}+y^{n} . \\
\text { Also }\left(1+x^{2}\right)\left(1+y^{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,407 |
In the right-angled coordinate plane $x O y$, $\triangle A_{i} B_{i} A_{i+1}(i=1,2, \cdots)$ are equilateral triangles, and satisfy
$$
\boldsymbol{O A}_{1}=\left(-\frac{1}{4}, 0\right), \boldsymbol{A}_{i} \boldsymbol{A}_{i+1}=(2 i-1,0) \text {. }
$$
(1) Prove that the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on... | Let $B_{n}(x, y)$. Then
$$
\left\{\begin{array}{l}
x=-\frac{1}{4}+1+3+\cdots+(2 n-3)+\frac{2 n-1}{2}=\left(n-\frac{1}{2}\right)^{2}, \\
|y|=\frac{\sqrt{3}}{2}(2 n-1) .
\end{array}\right.
$$
Eliminating $n$ gives $y^{2}=3 x$.
Thus, the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma: y^{2}... | 18 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,408 |
Five. (20 points) As shown in Figure 3, $AB$ is the diameter of the semicircle $\odot O$, $C$ is the midpoint of $\overparen{AB}$, $M$ is the midpoint of chord $AC$, and $CH \perp BM$, with the foot of the perpendicular being $H$. Prove:
$$
CH^{2}=AH \cdot OH .
$$ | Five, as shown in Figure 3, connect $O C$ and $B C$. Then
$$
\angle B O C = \angle B H C = 90^{\circ} \text{. }
$$
Therefore, $O$, $B$, $C$, and $H$ are concyclic.
Hence $\angle O H B = \angle O C B = 45^{\circ}$.
Also, $\angle B C M = 90^{\circ}$, $C H \perp B M$, and $M$ is the midpoint of $A C$, then
$$
\begin{arra... | CH^{2}=AH \cdot OH | Geometry | proof | Yes | Yes | cn_contest | false | 720,409 |
Six. (30 points) A student has 40 days to prepare for the National High School Mathematics Winter Camp. During this period, he can arrange a maximum of 60 hours for review, and plans to review at least 1 hour each day (assuming the number of hours reviewed each day is an integer). Prove: no matter how he arranges his r... | Six, let the review time for each of the 40 days be $b_{1}, b_{2}, \cdots, b_{40}$, and consider the partial sums:
$$
\begin{array}{l}
a_{1}=b_{1}, a_{2}=b_{1}+b_{2}, \cdots \cdots \\
a_{40}=b_{1}+b_{2}+\cdots+b_{40} .
\end{array}
$$
According to the problem,
$$
\begin{array}{l}
b_{i} \geqslant 1(1 \leqslant i \leqsla... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,410 |
1. Given $a b c \neq 0$, and
$$
a+b+c=a^{2}+b^{2}+c^{2}=2 \text {. }
$$
Then the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$ is ( ).
(A) 3
(B) -3
(C) 1
(D) -1 | - 1.A.
Since $a+b+c=a^{2}+b^{2}+c^{2}=2$, we have
$$
\begin{array}{l}
a b+b c+c a \\
=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]=1
\end{array}
$$
Notice that
$$
\begin{array}{l}
a(1-a)^{2}=a\left(1-2 a+a^{2}\right) \\
=a\left[(a b+b c+c a)-(a+b+c) a+a^{2}\right] \\
=a b c .
\end{array}
$$
Sim... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,411 |
Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect square.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect sq... | Solution: When $n>27$,
$$
A=4^{27}+4^{500}+4^{n}=4^{27}\left(1+4^{473}+4^{n-27}\right) \text {. }
$$
Since $4^{27}$ is a perfect square, for $A$ to be a perfect square, $1+4^{473}+4^{n-27}$ must be a perfect square.
Notice that
$$
\begin{array}{l}
1+4^{473}+4^{n-27}=1+2^{2 \times 473}+\left(2^{n-27}\right)^{2} \\
=1+2... | 972 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,412 |
Example 10 Proof: The product of any 5 consecutive positive integers is not a perfect square. | Proof: Let 5 consecutive positive integers be $a-2, a-1, a, a+1, a+2 (a>2)$. Then
$$
\begin{array}{l}
N=(a-2)(a-1) a(a+1)(a+2) \\
=a\left(a^{2}-1\right)\left(a^{2}-4\right)=a\left(a^{4}-5 a^{2}+4\right) .
\end{array}
$$
Assume $N$ is a perfect square. For any odd prime factor $p$ of $a$, if $p \mid (a-2)$, then
$$
p \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,413 |
6. Let $n$ be a positive integer. Prove that $A=n^{4}+2 n^{3}+$ $2 n^{2}+2 n+1$ is not a perfect square. | (Tip: $\left.\left(n^{2}+n\right)^{2}<A<\left(n^{2}+n+1\right)^{2}.\right)$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,414 |
3. Real numbers $x, y, z$ satisfy $x^{2}+y^{2}-x+y=1$. Then the range of the function
$$
f(x, y, z)=(x+1) \sin z+(y-1) \cos z
$$
is . $\qquad$ | 3. $\left[-\frac{3 \sqrt{2}+\sqrt{6}}{2}, \frac{3 \sqrt{2}+\sqrt{6}}{2}\right]$.
$$
f(x, y, z)=\sqrt{x^{2}+y^{2}+2 x-2 y+2} \sin (z+\theta),
$$
where, $\cos \theta=\frac{x+1}{\sqrt{(x+1)^{2}+(y-1)^{2}}}$,
$$
\sin \theta=\frac{y-1}{\sqrt{(x+1)^{2}+(y-1)^{2}}} \text {. }
$$
Since $z$ can take any value in $\mathbf{R}$,... | \left[-\frac{3 \sqrt{2}+\sqrt{6}}{2}, \frac{3 \sqrt{2}+\sqrt{6}}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,416 |
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