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 |
|---|---|---|---|---|---|---|---|---|---|
2. If $f\left(\frac{1}{1-x}\right)=\frac{1}{x} f(x)+2$, then $f(3)=$ | 2. $-\frac{1}{2}$.
Let $x=3$, we get $f\left(-\frac{1}{2}\right)=\frac{1}{3} f(3)+2$;
Let $x=-\frac{1}{2}$, we get $f\left(\frac{2}{3}\right)=-2 f\left(-\frac{1}{2}\right)+2$; Let $x=\frac{2}{3}$, we get $f(3)=\frac{3}{2} f\left(\frac{2}{3}\right)+2$.
From the above three equations, we get $f(3)=-\frac{1}{2}$. | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,013 |
3. The sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies
$$
a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}}\left(\left[a_{n}\right]\right. \text { and }
$$
$\left\{a_{n}\right\}$ represent the integer part and the fractional part of $a_{n}$, respectively). Then $a_{2006}=$ $\qquad$ : | $$
3.3009+\sqrt{3} \text {. }
$$
From the given, we have
$$
\begin{array}{l}
a_{0}=1+(\sqrt{3}-1), \\
a_{1}=1+\frac{1}{\sqrt{3}-1}=1+\frac{\sqrt{3}+1}{2}=2+\frac{\sqrt{3}-1}{2}, \\
a_{2}=2+\frac{2}{\sqrt{3}-1}=2+(\sqrt{3}+1)=4+(\sqrt{3}-1), \\
a_{3}=5+\frac{\sqrt{3}-1}{2}, \\
a_{4}=7+(\sqrt{3}-1), \\
\cdots \cdots
\en... | 3009+\sqrt{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,014 |
4. The graphs of the functions $y=x^{2}-x$ and $y=\cos 10 \pi x(x \geqslant 0)$ intersect at $\qquad$ points. | 4.17.
When $x \geqslant 0$, from $\left|x^{2}-x\right| \leqslant 1$, we get
$$
0 \leqslant x \leqslant \frac{1+\sqrt{5}}{2}=1.618 \cdots \text {. }
$$
The period of $y=\cos 10 \pi x$ is $\frac{1}{5}$. Within each period, if $\left|x^{2}-x\right|<1$, then the graphs of $y=x^{2}-x$ and $y=\cos 10 \pi x$ have 2 intersect... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,015 |
5. Let $f(x)=\frac{x^{3}}{1-3 x+3 x^{2}}$, and denote $f_{1}(x)=f(x), f_{n}(x)=f\left(f_{n-1}(x)\right)$. Then $f_{10}(x)=$ $\qquad$ | 5. $\frac{x^{3^{10}}}{x^{3^{10}}-(x-1)^{3^{10}}}$.
Since $f(x)=\frac{x^{3}}{1-3 x+3 x^{2}}=\frac{1}{1-\left(1-\frac{1}{x}\right)^{3}}$, we have
$$
\frac{1}{f(x)}=1-\left(1-\frac{1}{x}\right)^{3},
$$
which means
$$
1-\frac{1}{f(x)}=\left(1-\frac{1}{x}\right)^{3} \text {. }
$$
By substituting $x$ with $f_{n-1}(x)$, we... | \frac{x^{3^{10}}}{x^{3^{10}}-(x-1)^{3^{10}}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,016 |
6. If $n \in\{1,2, \cdots, 100\}$, and $n$ is a multiple of the sum of its digits, then there are $\qquad$ such $n$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ is kept as is in the translation. | 6.33.
Let $n=\overline{a b}=10 a+b,(a+b) \mid (10 a+b)$.
When $a, b$ is 0, it obviously meets the requirement. Therefore, all one-digit numbers and two-digit numbers ending in 0, as well as the three-digit number 100, meet the requirement, making a total of 19 such $n$.
When $a, b$ are both not 0, since $\frac{10 a+b... | 33 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,017 |
4. Let the complex roots of the equation $(3 x)^{2000}+(4 x+15)^{2000}=0$ be $\alpha_{i} 、 \overline{\alpha_{i}}(i=1,2, \cdots, 1000)$. Then $\sum_{i=1}^{1000} \frac{1}{\alpha_{i} a_{i}}$ equals ( ).
(A) $\frac{1000}{3}$
(B) $\frac{2000}{3}$
(C) $\frac{1000}{9}$
(D) $\frac{2000}{9}$ | 4.C.
The original equation is equivalent to $\left(\frac{4 x+15}{3 x}\right)^{2000}=-1$.
Let the 2000th roots of -1 be $\omega_{i} 、 \overline{\omega_{i}}(i=1,2$, $\cdots, 1000)$. Therefore,
$$
\sum_{i=1}^{1000}\left(\omega_{i}+\overline{\omega_{i}}\right)=0 \text {, and } \omega_{i}=\frac{4 \alpha_{i}+15}{3 \alpha_{i}... | \frac{1000}{9} | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,018 |
Three. (20 points) $A$ and $B$ are the endpoints of the major axis of an ellipse. Draw a tangent line $l$ through $A$. Take any point $P$ on $l$, and draw a tangent line $PC$ from $P$ to the ellipse (where $C$ is the point of tangency). Let $CD \perp AB$ at $D$. Prove: the line segment $PB$ passes through the midpoint ... | Three, as shown in Figure 2, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the points $B(a, 0), C\left(x_{0}, y_{0}\right)$. Then the equation of the tangent line $P C$ is
$$
\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1,
$$
and $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$;... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,019 |
Four, (20 points) The inscribed sphere $O$ of tetrahedron $ABCD$ touches the faces $ABD$ and $BCD$ at points $E$ and $F$, respectively. Prove:
$$
\angle AEB = \angle CFD.
$$ | As shown in Figure 3, for the convenience of narration, the points where the inscribed sphere $O$ touches the faces $BCD$, $ACD$, $ABD$, and $ABC$ are respectively denoted as $A_{0}$, $B_{0}$, $C_{0}$, and $D_{0}$. Thus,
$$
\begin{array}{l}
E=C_{0}, \\
F=A_{0}.
\end{array}
$$
Let the radius of sphere $O$ be $r$. It is... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,020 |
Five. (20 points) The sequence $\left\{a_{n}\right\}$ is
$$
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots \text {. }
$$
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, thus, we get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1,1,2$, and add the successor numb... | Five, according to the construction method of $\left\{a_{n}\right\}$, it is easy to know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \text {. }
$$
Generally, there is $a_{2^{n}-1}=n$, that is, the number $n$ first appears at the $2^{n}-1$ term, and if
$m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then ... | 3961 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,021 |
One, (50 points) Through the vertex $A_{0}$ of quadrilateral $A_{0} A_{1} A_{2} A_{3}$, draw perpendicular lines $l_{i}(i=1,2,3)$ to $A_{0} A_{i}$. Let $l_{1} \cap A_{2} A_{3}=\left\{P_{1}\right\}, l_{2} \cap A_{3} A_{1}=\left\{P_{2}\right\}, l_{3} \cap$ $A_{1} A_{2}=\left\{P_{3}\right\}$. Prove: $P_{1} 、 P_{2} 、 P_{3}... | Given the points $P_{1}, P_{2}, P_{3}$ are on the extensions of the sides of $\triangle A_{1} A_{2} A_{3}$, and
$$
\begin{array}{l}
\angle P_{1} A_{0} A_{1} \\
=\angle P_{3} A_{0} A_{3} \\
=90^{\circ},
\end{array}
$$
Therefore, $\angle P_{1} A_{0} A_{3}$
$$
=\angle P_{3} A_{0} A_{1} \text {. }
$$
Since $\angle P_{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,022 |
II. (50 points) Let real numbers $a \geqslant b \geqslant c \geqslant d>0$. Find the minimum value of the function
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right)
\end{array}
$$ | Obviously, $f$ has no upper bound, which is due to when $a=b=1$, $c+d \rightarrow 0$, $f \rightarrow+\infty$. Also note that $f$ is a zero-degree homogeneous function, and when $a=b=c=d$, the value of $f$ is $\left(\frac{3}{2}\right)^{4}$.
The following proof shows that for any positive numbers $a, b, c, d$ satisfying... | \left(\frac{3}{2}\right)^{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,023 |
Three, (50 points) Let $p$ be an odd prime, and $a, b$ be positive integers less than $p$. Prove: $a+b=p$ if and only if, for any positive integer $n$ less than $p$, $\left[\frac{2 a n}{p}\right]+ \left[\frac{2 b n}{p}\right]$ equals a positive odd number. | Three, Necessity.
If $a+b=p, n$ is any positive integer less than $p$. Let
$$
\left[\frac{2 a n}{p}\right]=u,\left[\frac{2 b n}{p}\right]=v \text {. }
$$
Since $p$ is a prime number, $\frac{2 a n}{p}$ and $\frac{2 b n}{p}$ are not integers. Therefore, there exist $\alpha(0<\alpha<1)$ and $\beta(0<\beta<1)$ such that
$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,024 |
2. The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B.
Since $x, y, z$ are all positive integers, we have $x+y \geqslant 2$. Since 23 is a prime number, and $z(x+y)=23$, it follows that $z=1, x+y=23$.
Substituting these into the first equation yields $y^{2}-24 y+63=0$. Solving this, we get $y_{1}=3, y_{2}=21$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,026 |
3. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ is the midpoint of $C D$, and $F$ is the midpoint of $A A_{1}$. The area of the section through points $E$, $F$, and $B_{1}$ is ( ).
(A) $\frac{55 \sqrt{5}}{192} a^{2}$
(B) $\frac{\sqrt{29}}{4} a^{2}$
(C) $\frac{11 \sqrt{29}}{48} a^{2}$
(D) $\f... | 3. C.
As shown in Figure 1, draw $H G / / F B_{1}$, intersecting $D D_{1}$ and $C C_{1}$ at points $H$ and $G$ respectively, and $H F$ intersects $A D$ at point $I$. Then the pentagon $E I F B_{1} G$ is the desired cross-section.
It is easy to see that $C_{1} G=\frac{3}{4} a$,
$$
D H=\frac{1}{4} a \text {. }
$$
After... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,027 |
4. The number of solutions to the equation $\sin ^{x} a+\cos ^{x} \alpha=1\left(0<\alpha<\frac{\pi}{2}\right)$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 4. B.
When $x>2$, from $\sin ^{x} \alpha\sin ^{2} \alpha, \cos ^{x} \alpha>\cos ^{2} \alpha$, we get $\sin ^{x} \alpha+\cos ^{x} \alpha>1$, in this case there is also no solution. Therefore, the only solution is $x=2$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,028 |
5. Let $x \in \mathbf{R}_{+}$. Then the maximum value of $y=\sqrt{\frac{1}{1+x^{2}}}+2 \sqrt{\frac{x}{1+x}}$ is ( ).
(A) $\frac{\sqrt{2}}{3}$
(B) $\frac{2 \sqrt{2}}{3}$
(C) $\frac{\sqrt{2}}{2}$
(D) $\frac{3 \sqrt{2}}{2}$ | 5.D.
Let $x=\frac{1}{t}$, then,
$$
\begin{array}{l}
y=\sqrt{\frac{t^{2}}{1+t^{2}}}+2 \sqrt{\frac{1}{1+t}}=\frac{t}{\sqrt{1+t^{2}}}+\frac{2}{\sqrt{1+t}} \\
\leqslant \frac{t}{\sqrt{\frac{(1+t)^{2}}{2}}}+\frac{2}{\sqrt{1+t}}=\frac{\sqrt{2} t}{1+t}+\frac{2}{\sqrt{1+t}} \\
=\sqrt{2}-\frac{\sqrt{2}}{1+t}+\frac{2}{\sqrt{1+t... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,029 |
5. Given that $P$ is a point on the line $y=x+1$, and $M, N$ are points on the circles $C_{1}:(x-4)^{2}+(y-1)^{2}=4$ and $C_{2}: x^{2}+(y-2)^{2}=1$ respectively. Then the maximum value of $|P M| - |P N|$ is $(\quad)$.
(A) 4
(B) 5
(C) 6
(D) 7 | 5.C.
As shown in Figure 2, it is easy to see that the circle $C_{3}: x^{2}+(y-5)^{2}=4$ is symmetric to the circle $C_{1}$:
$$
\begin{array}{l}
(x-4)^{2}+ \\
(y-1)^{2}=4 \text { with respect to the line } y=x+1.
\end{array}
$$
Therefore, for any point $M$ on $C_{1}$, there exists a point $M^{\prime}$ on circle $C_{3}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,030 |
6. If $x_{i}=\frac{i}{101}$, then $T=\sum_{i=0}^{101} \frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}$ is ( ).
(A) 51
(B) 52
(C) 53
(D) 54 | 6.A.
$$
\begin{array}{l}
\frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{\left(1-x_{i}\right)^{3}}{3\left(1-x_{i}\right)^{2}-3\left(1-x_{i}\right)+1} \\
=\frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}}{3-6 x_{i}+3 x_{i}^{2}-3+3 x_{i}+1} \\
=\frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{1-3 ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,031 |
1. The maximum value $M(a)$ of the function $f(x)=\left|x^{2}-a\right|$ in the interval $[-1,1]$ has its minimum value as $\qquad$ . | $$
\text { 2.1. } \frac{1}{2} \text {. }
$$
(1) When $a \leqslant 0$, it is easy to see that $M(a)=1-a$.
(2) When $a>0$, $M(a)=\max \{a,|1-a|\}$.
When $0<a \leq \frac{1}{2}$, $M(a)=1-a$; when $a>\frac{1}{2}$, $M(a)=a$.
Therefore, $(M(a))_{\min }=\frac{1}{2}$. | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,032 |
2. Let the volume of tetrahedron $ABCD$ be $V$, $E$ be the midpoint of edge $AD$, and point $F$ be on the extension of $AB$ such that $BF = AB$. The plane through points $C$, $E$, and $F$ intersects $BD$ at point $G$. Then the volume of tetrahedron $CDGE$ is | 2. $\frac{1}{3} V$.
As shown in Figure 3, it is easy to see that
$V_{\text {WWibuncis }}$
$=V_{\text {m flower position }}$.
By Menelaus' Theorem, we know
$$
\frac{B G}{G D} \cdot \frac{D E}{E A} \cdot \frac{A F}{F B}=1 .
$$
Therefore, $\frac{B G}{G D}=\frac{1}{2}$.
Then, $S_{\text {. } \triangle A R C}=\frac{1}{2} D... | \frac{1}{3} V | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,033 |
3. For an integer $m$, its unit digit is denoted by $f(m)$, and let $a_{n}=f\left(2^{n+1}-1\right)(n=1,2, \cdots)$. Then $a_{2006}$ $=$ . $\qquad$ | 3.7.
It is known that $f\left(2^{m}\right)=f\left(2^{m+4}\right)$. Therefore,
$$
a_{2 \omega 0}=f\left(2^{2 \omega 1}-1\right)=f\left(2^{2 \omega 07}\right)-1=f(8)-1=7 .
$$ | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,034 |
4. Given $n(n \geqslant 3)$ lines where exactly $m(m \geqslant 2)$ lines are parallel, and no three lines intersect at the same point. The maximum number of regions these $n$ lines can divide the plane into is $\qquad$ | 4. $\frac{1}{2}\left(n^{2}+n-m^{2}+m\right)+1$.
Consider these $m$ parallel lines, which divide the plane into $m+1$ parts, denoted as $a_{m}=m+1$. Adding one more line, it is divided into $m+1$ segments by the $m$ lines, at this point, the number of parts the plane is divided into increases by $m+1$ parts, and the pl... | \frac{1}{2}\left(n^{2}+n-m^{2}+m\right)+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,035 |
5. Connecting the centers of the faces of a regular polyhedron, we get a new regular polyhedron, which we call the positive sub-polyhedron of the original polyhedron. A cube $T_{1}$ has a surface area of $a_{1}=16$, its positive sub-polyhedron is $T_{2}$, with a surface area of $a_{2}$, the positive sub-polyhedron of $... | 5. $18+3 \sqrt{3}$.
Given that $T_{1}, T_{3}, \cdots$ are cubes, and $T_{2}, T_{4}, \cdots$ are regular octahedra. Let the side length of $T_{i}$ be $b_{i}$. From Figure 4, it is easy to see that $b_{2}=\frac{\sqrt{2}}{2} b_{1}$. From Figure 5, we calculate that $C H=\frac{1}{2} A C=\frac{\sqrt{2}}{2} b_{2}$,
$$
b_{3}... | 18+3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,036 |
6. Let $f(x)=x^{2}+a x+b-2 (|x| \geqslant 2)$. If the graph of the function $y=f(x)$ intersects the $x$-axis, then the minimum value of $a^{2}+b^{2}$ is $\qquad$ . | 6. $\frac{4}{5}$.
The function intersects the $x$-axis in five scenarios as shown in Figure 6.
For Figures 6(a), (b), and (c), it is easy to see that at least one of $f(-2) \leqslant 0$ and $f(2) \leqslant 0$ holds, i.e.,
$$
-2 a+b+2 \leqslant 0 \text { or } 2 a+b+2 \leqslant 0 \text {. }
$$
Thus, the region in the ... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,037 |
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$, where
$$
a_{1}=1, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{4}{a_{n}}\right)\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: For $n \geqslant 2$, we have $a_{n}>2+4\left(\frac{1}{3}\right)^{2^{n-1}}$. | $$
\begin{array}{l}
a_{n+1}+2=\frac{a_{n}^{2}+4 a_{n}+4}{2 a_{n}}=\frac{\left(a_{n}+2\right)^{2}}{2 a_{n}} . \\
a_{n+1}-2=\frac{a_{n}^{2}-4 a_{n}+4}{2 a_{n}}=\frac{\left(a_{n}-2\right)^{2}}{2 a_{n}} .
\end{array}
$$
Thus, $\frac{a_{n+1}+2}{a_{n+1}-2}=\left(\frac{a_{n}+2}{a_{n}-2}\right)^{2}$.
Therefore, when $n \geqsl... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,038 |
Four. (20 points) Given the function $f(x)=\left|\log _{2}(x-1)\right|$. If real numbers $a, b(1<a<b)$ satisfy
$$
f(a)=f\left(\frac{b}{b-1}\right), f(b)=2 f\left(\frac{a+b}{2}\right),
$$
Prove: $4<b<5$. | From $f(a)=f\left(\frac{b}{b-1}\right)$, we can get
$$
\left|\log _{2}(a-1)\right|=\left|\log _{2}(b-1)\right| .
$$
Then $a-1=b-1$ or $a-1=\frac{1}{b-1}$, the former does not meet the problem's requirements, so it is discarded.
Also, $1=(a-1)(b-1)2$.
Thus, $f(b)=\left|\log _{2}(b-1)\right|=\log _{2}(b-1)$.
And $f\left... | 4<b<5 | Algebra | proof | Yes | Yes | cn_contest | false | 717,039 |
6. Let $n>1, f(x)$ be a monotonically increasing function defined on the finite set $A=$ $\{1,2, \cdots, n\}$, and for any $x$ 、 $y \in A$, we have $\frac{f(x)}{f(y)}=f(x) f(y)$. Then, ( ) .
(A) $n=2$
(B) $n=3$
(C) $n=4$
(D) $n \geqslant 5$ | 6. A.
For any $x \neq y, x, y \in A$, we have
$$
\frac{f(x)}{f(y)}=f(x) f(y), \frac{f(y)}{f(x)}=f(y) f(x) \text{. }
$$
Therefore, $\frac{f(x)}{f(y)}=\frac{f(y)}{f(x)}$, which means $f^{2}(x)=f^{2}(y)$.
Thus, $f^{2}(x)=c_{1}\left(c_{1}>0\right)$.
Hence, $f(x) \in\{c,-c\}(c>0)$.
If $n>2$, by the pigeonhole principle, t... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,040 |
Five. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Draw any line $l$ that is not perpendicular to the $x$-axis, intersecting the ellipse $C$ at points $P$ and $Q$. Point $S$ is the reflection of point $P$ across the $x$-axis. Find the maximum value of the area of $\triangle O S Q$.
---... | When $P Q // x$-axis, points $S, O, Q$ are collinear, not forming a triangle. As shown in Figure 8, let the equation of line $l$ be
$$
x=m+k y .
$$
We will prove that $SQ$ passes through point $B\left(\frac{a^{2}}{m}, 0\right)$.
To do this, we need to prove $k_{\text {BQ }}=k_{S B}$.
Substituting equation (1) into th... | \frac{1}{2} a b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,041 |
One, (50 points) In a convex quadrilateral $ABCD$, $AB$ and $DC$ intersect at point $E$, and $AD$ and $BC$ intersect at point $F$. There is a point $P$ inside the quadrilateral $ABCD$ such that $\angle APB + \angle CPD = \pi$. Prove:
$$
\angle FPD = \angle BPE.
$$ | Figure 9
I. Proof 1: As shown in Figure 9, take a point $Q$ on $F P$ such that $P, C, Q$, and $D$ are concyclic. Let $Q D$ intersect $A P$ at point $L$, and $Q C$ intersect $P B$ at point $M$. Using Menelaus' theorem:
The line $E C D$ intersects $\triangle F A B$, giving
$$
\frac{F D}{D A} \cdot \frac{A E}{E B} \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,042 |
Three. (50 points) Let $S$ be a set of $2n+1$ points in the plane, where no three points are collinear and no four points are concyclic. A circle is called a "good circle" if there are three points of $S$ on the circle, $n-1$ points inside the circle, and $n-1$ points outside the circle. Prove that the number of good c... | Three, consider $\mathrm{C}_{2 n+1}^{2}$ pairs of points $A_{1}, A_{2}, \cdots, A_{C_{2 n+1}}^{2}$. Let the number of good circles containing the point pair $A_{i}$ be $a_{i}$. Then the total number of good circles should be $f=\frac{1}{3} \sum_{i=1}^{C_{2 n+1}^{2}} a_{i}$ (since each circle contains three point pairs)... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,044 |
1. If an $n$-faced polyhedron has $m$ faces that are right triangles, we say that the polyhedron's straightness is $\frac{m}{n}$. If an $n(n \geqslant 4)$-faced polyhedron has a straightness of 1, and the number of edges is $k$, then $n$ and $k$ should satisfy ( ).
(A) $k=3 n$
(B) $k=\frac{3}{2} n$
(C) $k=\frac{4}{n} n... | -.1.B.
Since the straightness of an $n$-faced polyhedron is 1, all its $n$ faces are right-angled triangles. Also, because each face of the $n$-faced polyhedron contains three edges, and each edge belongs to two faces, we have $2k = 3n$, which means $k = \frac{3}{2} n$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,045 |
2. As shown in Figure 1, $P$ is a point inside $\triangle A B C$, and satisfies $\boldsymbol{A P}=$ $\frac{2}{5} A B+\frac{1}{5} A C$. Then the ratio of the area of $\triangle P B C$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{5}$
(D) $\frac{2}{5}$ | 2.D.
As shown in Figure 6, let $A M=\frac{2}{5} A B$ and $A N=\frac{1}{5} A C$, then $A P=A M + A N$. By the parallelogram rule, we have
$$
M P \parallel A C, N P \parallel A B \text{. }
$$
Draw a line through point $P$ parallel to $B C$, intersecting $A B$ and $A C$ at points $B^{\prime}$ and $C^{\prime}$, respectiv... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,046 |
3. Let the function be
$$
f(x)=\frac{\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)+2 x^{2}+x}{2 x^{2}+\cos x}
$$
with the maximum value $M$ and the minimum value $m$. Then $M$ and $m$ satisfy ( ).
(A) $M+m=2$
(B) $M+m=4$
(C) $M-m=2$
(D) $M-m=4$ | 3.A.
$$
f(x)=\frac{\sin x+\cos x+2 x^{2}+x}{2 x^{2}+\cos x}=\frac{\sin x+x}{2 x^{2}+\cos x}+1 \text {. }
$$
Since $g(x)=\frac{\sin x+x}{2 x^{2}+\cos x}$ is an odd function, and $f(x)$ has a maximum value and a minimum value, therefore, $g(x)$ also has a maximum value $M'$ and a minimum value $m'$, and $M'+m'=0$.
Thus,... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,047 |
4. As shown in Figure 2, a tangent line is drawn from the left focus $F$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to the circle $x^{2}+y^{2}=a^{2}$. The point of tangency is $T$, and the extension of $F T$ intersects the right branch of the hyperbola at point $P$. If the midpoint of segmen... | 4.C.
Let the right focus of the hyperbola be $\boldsymbol{F}^{\prime}$, then
$$
|P F|=\left|P F^{2}\right|+2 a \text{. }
$$
Since $|P F|=2|M F|,\left|P F^{\prime}\right|=2|O M|$, we have
$$
|M F|=|O M|+a \text{. }
$$
In the right triangle $\triangle O T F$, since
$$
|T F|=\sqrt{|O F|^{2}-|O T|^{2}}=b,
$$
we get $|M... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,048 |
5. In the acute triangle $\triangle A B C$, the side lengths $a, b, c$ and the area $S$ satisfy $S=\frac{c^{2}-(a-b)^{2}}{k}$, and $\angle C$ is neither the largest nor the smallest interior angle of $\triangle A B C$. Then the range of the real number $k$ is ( ).
(A) $(0,4)$
(B) $(4(\sqrt{2}-1), 4)$
(C) $(0,4(\sqrt{2}... | 5. B.
Assume $0<\angle A \leqslant \angle C \leqslant \angle B<\frac{\pi}{2}$, then
$$
\frac{\pi}{2}<\angle A+\angle C \leqslant 2 \angle C \leqslant \angle C+\angle B<\pi \text {. }
$$
Thus, $\frac{\pi}{4}<\angle C<\frac{\pi}{2}$.
Also, $k=\frac{c^{2}-(a-b)^{2}}{S}=\frac{2 a b-\left(a^{2}+b^{2}-c^{2}\right)}{\frac{1... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,049 |
6. Given that $S$ is a set composed of $n(n \geqslant 3)$ positive numbers. If there exist three distinct elements in $S$ that can form the three sides of a triangle, then $S$ is called a "triangle number set". Consider the set of consecutive positive integers $\{4,5, \cdots, m\}$, where all its 10-element subsets are ... | 6. C.
Let there be three positive numbers $a_{1}, a_{2}, a_{3} \left(a_{1}<a_{2}<a_{3}\right)$.
If $a_{1}+a_{2} \leqslant a_{3}$, then these three numbers cannot form the sides of a triangle. To make $a_{3}$ the smallest, take $a_{1}+a_{2}=a_{3}$.
$$
\begin{array}{l}
\text { Let } a_{1}=4, a_{2}=5, a_{3}=a_{1}+a_{2}=9... | 253 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,050 |
1. Let $A$ be a finite set, for any $x, y \in A$, if $x \neq y$, then $x+y \in A$. Then, the maximum number of elements in $A$ is $\qquad$ . | Let the number of elements in $A$ be $n$.
If $n>3$, then there must be two elements with the same sign. Without loss of generality, assume there are two positive numbers. Let the largest two positive numbers be $a, b$ (with $a < b$). Then, $a+b \notin A$, which is a contradiction.
Therefore, $n \leqslant 3$.
Also, $A=\... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,051 |
1. Positive integers $a, b, c$ satisfy
$$
\log _{6} a+\log _{6} b+\log _{6} c=6 \text {, }
$$
$a, b, c$ form an increasing geometric sequence, and $b-a$ is a perfect square. Then the value of $a+b+c$ is $\qquad$ | $=.1 .111$.
From the given, we have $abc=6$, and $b^2=ac$. Therefore, $b=36$.
Also, $b-a$ is a perfect square, so $a$ can take the values $11, 20, 27, 32$.
Upon verification, only $a=27$ meets the conditions. In this case, $c=48$.
Thus, $a+b+c=27+36+48=111$. | 111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,052 |
2. Given point $A(5 \sqrt{3}, 5)$, line $l: x=m y+n$ $(n>0)$ passes through point $A$. If the feasible region
$$
\left\{\begin{array}{l}
x \leqslant m y+n, \\
x-\sqrt{3} y \geqslant 0, \\
y \geqslant 0
\end{array}\right.
$$
has an circumcircle with a diameter of 20, then the value of $n$ is $\qquad$ | 2. 10 $\sqrt{3}$.
Notice that the line $l^{\prime}: x-\sqrt{3} y=0$ also passes through point $A$, so $A$ is the intersection point of lines $l$ and $l^{\prime}$.
The feasible region is the shaded area (including the boundary) of $\triangle A O B$ as shown in Figure 7.
Let the inclination angle of line $l$ be $\alph... | 10 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,053 |
3. Two spheres $O_{1}$ and $O_{2}$, both with a radius of 1, are tangent to each other, and they are also tangent to the two half-planes of a dihedral angle $\alpha-l-\beta$ of $60^{\circ}$. There is another larger sphere $O$ that is tangent to both half-planes of the dihedral angle and is externally tangent to spheres... | 3. $\frac{5+\sqrt{13}}{3}$.
As shown in Figure 8, it is easy to see that the centers of the three spheres $O$, $O_{1}$, and $O_{2}$ are all on the bisector plane of the dihedral angle $\alpha-l-\beta$. Let the sphere $O_{1}$ be tangent to the planes $\alpha$ and $\beta$ at points $A$ and $B$ respectively, and the sphe... | \frac{5 + \sqrt{13}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,054 |
4. If $x, y \in[-2006,2006]$, and satisfy
$$
\begin{array}{l}
1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(1+x)(1-y)}{x-y+1},
\end{array}
$$
then the minimum value of $xy$ is | 4. $\frac{1}{25}$.
Since $1+\cos ^{2}(2 x+3 y-1)>0$, we have
$$
\begin{array}{l}
\frac{x^{2}+y^{2}+2(1+x)(1-y)}{x-y+1} \\
=\frac{(x-y+1)^{2}+1}{x-y+1} \\
=x-y+1+\frac{1}{x-y+1}>0 .
\end{array}
$$
Thus, $x-y+1+\frac{1}{x-y+1} \geqslant 2$.
Also, $1 \leqslant 1+\cos ^{2}(2 x+3 y-1) \leqslant 2$, so, $1+\cos ^{2}(2 x+3 ... | \frac{1}{25} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,055 |
5. In the Cartesian coordinate system, circle $C_{1}$ intersects with circle $C_{2}$ at points $P$ and $Q$, where the coordinates of point $P$ are $(3,2)$, and the product of the radii of the two circles is $\frac{13}{2}$. If the line $y=k x(k>0)$ and the $x$-axis are both tangent to circles $C_{1}$ and $C_{2}$, then $... | 5. $2 \sqrt{2}$.
Let the coordinates of the centers of the two circles be $C_{i}\left(x_{i}, y_{i}\right)(i=1,2)$. Then $\left|P C_{i}\right|=y_{i}$, that is, $\sqrt{\left(x_{i}-3\right)^{2}+\left(y_{i}-2\right)^{2}}=y_{i}$, which also means $x_{i}^{2}-6 x_{i}-4 y_{i}+13=0(i=1,2)$.
The line connecting the centers of t... | 2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,056 |
6. Given the function $f(x)=\frac{x}{\sqrt{1-x^{2}}}$. Define the function
$$
f_{n}(x)=\underbrace{f(f(f \cdots f(x)) \cdots)}_{n \uparrow},
$$
where the inverse function of $f_{n}(x)$ is $f_{n}^{-1}(x)$. Then
$$
f_{40}\left(\frac{1}{\sqrt{41}}\right) f_{* 0}^{-1}\left(\frac{1}{\sqrt{41}}\right)=
$$ | 6. $\frac{1}{9}$.
$$
\begin{array}{l}
f_{1}(x)=f(x)=\frac{x}{\sqrt{1-x^{2}}}, \\
f_{2}(x)=f(f(x)) \\
=\frac{\frac{x}{\sqrt{1-x^{2}}}}{\sqrt{1-\left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2}}}=\frac{x}{\sqrt{1-2 x^{2}}}, \\
f_{3}(x)=f(f(f(x))) \\
=\frac{\frac{x}{\sqrt{1-2 x^{2}}}}{\sqrt{1-\left(\frac{x}{\sqrt{1-2 x^{2}}}\righ... | \frac{1}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,057 |
Three. (20 points) A bag contains $m$ red balls and $n$ white balls $(m > n \geqslant 4)$, which are identical except for their colors. Now, two balls are drawn at random.
(1) If the probability of drawing two red balls is an integer multiple of the probability of drawing one red and one white ball, prove that $m$ must... | (1) Let "drawing two red balls" be event $A$, and "drawing one red and one white ball" be event $B$. Then,
$$
P(A)=\frac{\mathrm{C}_{m}^{2}}{\mathrm{C}_{m+n}^{2}}, P(B)=\frac{\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}}{\mathrm{C}_{m+n}^{2}}.
$$
According to the problem, $P(A)=k P(B)\left(k \in \mathbf{N}_{+}\right)$. There... | (10,6),(15,10),(21,15) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,058 |
Four. (20 points) As shown in Figure 3, $A_{1}$ and $A_{2}$ are the left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a > b > 0)$. $M(m, 0) (m > a)$ is a fixed point on the $x$-axis. A line passing through point $M$ intersects the ellipse at two different points $A$ and $B$. The lines $A... | Let $l_{A B}: x=k y+m$, substituting into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
\left(a^{2}+k^{2} b^{2}\right) y^{2}+2 k m b^{2} y+b^{2}\left(m^{2}-a^{2}\right)=0 \text {. }
$$
Thus, $y_{1}+y_{2}=-\frac{2 k m b^{2}}{a^{2}+k^{2} b^{2}}, y_{1} y_{2}=\frac{b^{2}\left(m^{2}-a^{2}\right)}{a^{2}+k^{2} b^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,059 |
Five. (20 points) Let 2006 real numbers $a_{1}, a_{2}, \cdots, a_{2006}$ satisfy
$$
\begin{array}{l}
\frac{a_{1}}{2}+\frac{a_{2}}{3}+\cdots+\frac{a_{2006}}{2007}=\frac{4}{3}, \\
\frac{a_{1}}{3}+\frac{a_{2}}{4}+\cdots+\frac{a_{2006}}{2008}=\frac{4}{5}, \\
\frac{a_{1}}{4}+\frac{a_{2}}{5}+\cdots+\frac{a_{2006}}{2009}=\fra... | Let $n=2006$, and define
$$
R(x)=\frac{a_{1}}{x+1}+\frac{a_{2}}{x+2}+\cdots+\frac{a_{n}}{x+n} .
$$
Then the required algebraic expression is
$$
\frac{a_{1}}{3}+\frac{a_{2}}{5}+\frac{a_{3}}{7}+\cdots+\frac{a_{2006}}{4013}=\frac{1}{2} R\left(\frac{1}{2}\right) \text {. }
$$
Let $R(x)=\frac{q(x)}{p(x)}$,
where $p(x)=(x+... | 1-\frac{1}{4013^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,060 |
One, (50 points) As shown in Figure $4, G$ and $H$ are two points inside $\triangle ABC$, and they satisfy $\angle ACG = \angle BCH, \angle CAG = \angle BAH$. Through point $G$, perpendiculars $GD \perp BC, GE \perp CA, GF \perp AB$ are drawn, with feet of the perpendiculars being $D$, $E$, and $F$ respectively. If $\a... | 一、If 9, connect $D H$.
Since $G D \perp B C, G E \perp$ $C A$, therefore, points $C, D, G, E$ are concyclic.
Thus $\angle G D E=\angle A C G$
$=\angle B C H$.
Hence, $D E \perp C H$.
Similarly, $E F \perp A H$.
Also, $\angle D E F=90^{\circ}$, so, $\angle A H C=90^{\circ}$.
Since $\frac{C D}{C H}=\frac{G C \cos \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,061 |
2. If $|\omega|<6$, and the line $x=\frac{\pi}{6}$ is an axis of symmetry for the function $f(x)$ $=\sin \left(\omega x+\frac{\pi}{3}\right)$, then $\omega=$ $\qquad$ | $$
2.0,1,-5 \text {. }
$$
First, $f(x)$ is not necessarily a trigonometric function; when $\omega=0$, it satisfies the condition.
Second, let $\sin \left(\omega \cdot \frac{\pi}{6}+\frac{\pi}{3}\right)= \pm 1$.
Solving this, we get $\omega=1,-5$. | 2.0, 1, -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,062 |
II. (50 points) Prove that there exists a unique sequence of positive integers $a_{1}, a_{2}, \cdots$, such that
$$
\begin{aligned}
& a_{1}=1, a_{2}>1, \\
& a_{n+1}\left(a_{n+1}-1\right)=\frac{a_{n} a_{n+2}}{\sqrt[3]{a_{n} a_{n+2}-1}+1}-1 \\
(n= & 1,2, \cdots) .
\end{aligned}
$$ | $$
\begin{array}{l}
=\sqrt{n+1}\left(a_{n+1}-1\right)=\frac{a_{n} a_{n+2}}{\sqrt[3]{a_{n} a_{n+2}-1}+1}-1 \\
=\frac{\left(\sqrt[3]{a_{n} a_{n+2}-1}\right)^{3}+1}{\sqrt[3]{a_{n} a_{n+2}-1}+1}-1 \\
=\left(\sqrt[3]{a_{n} a_{n+2}-1}\right)^{2}-\sqrt[3]{a_{n} a_{n+2}-1},
\end{array}
$$
That is, $\left(a_{n+1}-\sqrt[3]{a_{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,063 |
Three. (50 points) For a row consisting of $n$ P's, $n$ Q's, and $n$ R's, redefine a new row below it (one letter shorter). If the two letters above it are different, write the third letter in that position; if the two letters above it are the same, write that letter in that position. Repeat the operation on the newly ... | When $n=1$, there are only 6 cases as shown in Figure 10:
PQR PRQ QPR QRP RPQ RQP
$R P \quad Q P \quad R Q \quad P Q \quad Q R \quad P R$
$\begin{array}{llllll}\mathbf{Q} & \mathbf{R} & \mathbf{P} & \mathbf{R} & \mathbf{P} & \mathbf{P}\end{array}$
Figure 10
All of them meet the requirements.
For the case of $n \geqsla... | n=1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,064 |
1. If $\{x\}=x-[x]([x]$ represents the greatest integer not exceeding $x$), then the number of real solutions to the equation $2005 x+\{x\}=\frac{1}{2006}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | - 1.C.
Since $x=[x]+\{x\}$, we have
$$
2005[x]+2006\{x\}=\frac{1}{2006} \text {. }
$$
Also, $0 \leqslant 2006\{x\}<2006$, so, $[x]=0$ or -1.
If $[x]=0$, then $\{x\}=\frac{1}{2006^{2}}$, i.e., $x=\frac{1}{2006^{2}}$;
If $[x]=-1$, then $2006\{x\}=2005+\frac{1}{2006}$, i.e.,
$$
\begin{array}{l}
\{x\}=\frac{2005}{2006}+\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,065 |
2. $\cos \left(\frac{\arccos 0.25-\arccos 0.875}{2}\right)$ equals
(A) $\frac{\sqrt{6}}{3}$
(B) $\frac{\sqrt{6}}{4}$
(C) $\frac{1}{4}$
(D) $\frac{3 \sqrt{6}}{8}$ | 2.D.
$$
\begin{array}{l}
\text { Original expression }=\cos \left(\frac{\arccos \frac{1}{4}-\arccos \frac{7}{8}}{2}\right) \\
=\sqrt{\frac{1+\cos \left(\arccos \frac{1}{4}-\arccos \frac{7}{8}\right)}{2}} \\
=\sqrt{\frac{1+\frac{1}{4} \times \frac{7}{8}+\sqrt{1-\left(\frac{1}{4}\right)^{2}} \times \sqrt{1-\left(\frac{7}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,066 |
3. If $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a unit cube, and $M$ is the midpoint of edge $B B^{\prime}$, then the distance from point $M$ to the plane $A^{\prime} C^{\prime} D$ is ( ).
(A) 1
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\sqrt{3}-1$ | 3. C.
As shown in Figure 2, connect $B D$, $B A^{\prime}$, $B C^{\prime}$, and $B^{\prime} D$.
Since $M$ is the midpoint of $B B^{\prime}$, the distance from point $M$ to the plane $A^{\prime} C^{\prime} D$ is half the sum of the distances from point $B$ to the plane $A^{\prime} C^{\prime} D$ and from point $B^{\prim... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,067 |
4. Let point $P$ be on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0$, $c=\sqrt{a^{2}-b^{2}}$ ), and the equation of line $l$ be $x=-\frac{a^{2}}{c}$. The coordinates of point $F$ are $(-c, 0)$. Draw $P Q \perp l$ at point $Q$. If points $P$, $F$, and $Q$ form an isosceles right triangle, then the eccent... | 4. A.
As shown in Figure 3, let $T$ be the intersection of line $l$ with the $x$-axis, and construct $P R \perp x$-axis at point $R$. From the problem, we know
$$
\begin{array}{l}
\angle P F Q=90^{\circ}, \\
P F=Q F, P Q \parallel R T .
\end{array}
$$
Then $T F=Q T=P R$
$$
=F R \text {. }
$$
Thus, we have $y=x+c=-c+... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,068 |
5. The number of ordered pairs of integers $(a, b)$ that make the three numbers $a+5$, $b-2$, and $a+b$ the lengths of the sides of a right triangle is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 5.A.
If $a+5$ is the hypotenuse, then
$$
(a+5)^{2}=(a+b)^{2}+(b-2)^{2} \text {. }
$$
Rearranging gives $2 a(5-b)=2 b(b-2)-21$.
The left side is even, while the right side is odd, leading to a contradiction.
If $a+b$ is the hypotenuse, then
$$
(a+b)^{2}=(a+5)^{2}+(b-2)^{2} .
$$
Rearranging gives $2(a b-5 a+2 b)=29$, ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,069 |
6. Let $1 \leqslant a \leqslant b \leqslant 100$, and $57 \mid a b$. The number of positive integer pairs $(a, b)$ is ( ).
(A) 225
(B) 228
(C) $\overline{231}$
(D) 234 | 6.B.
Obviously, $57=3 \times 19$.
We proceed with the classification and counting.
(1) If $57 \mid a$, then $a=57, b=57,58, \cdots, 100$, i.e., there are $100-56=44$ pairs $(a, b)$.
(2) If $57 \mid b$, and $57 \nmid a$, then $b=57, a=1,2, \cdots, 56$, i.e., there are 56 pairs $(a, b)$.
(3) If $3 \mid a, 19 \mid b$, an... | 228 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,070 |
1. Let $a$ be a real number. If for any real number $x$, the inequality $x^{2} \log _{\frac{1}{2}}\left(a^{2}-2 a-3\right)-2 x+1 \geqslant 0$ always holds, then the range of values for $a$ is $\qquad$ | \[
\begin{array}{l}
0.1 .1-\frac{3 \sqrt{2}}{2} \leqslant a0, \\
\log _{\frac{1}{2}}\left(a^{2}-2 a-3\right)>0, \\
\Delta=4-4 \log _{\frac{1}{2}}\left(a^{2}-2 a-3\right) \leqslant 0
\end{array}
\]
\[
\Leftrightarrow\left\{\begin{array}{l}
0 \text{I} \text{ when } 2<a-1 \leqslant \frac{3}{\sqrt{2}}, \text{ i.e., } 3<a \... | 1-\frac{3}{\sqrt{2}} \leqslant a \leqslant 1+\frac{3}{\sqrt{2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,071 |
2. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a unit cube, $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$. $M$ is the midpoint of edge $B B_{1}$. Then the volume of the tetrahedron $O_{1} A D M$ is $\qquad$ . | 2. $\frac{1}{8}$.
As shown in Figure 4, it is easy to know that $A C \perp$ plane $D_{1} B_{1} B D$. Let $O$ be the center of plane $A B C D$, then $A O \perp$ plane $D O, M$. Therefore,
$$
\begin{aligned}
& V_{4-\omega_{1}} M I \\
= & \frac{1}{3} S_{\Delta X_{1} M} \cdot A O \\
= & \frac{1}{3}\left(1 \times \sqrt{2}-... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,072 |
3. If the function $y=\log _{a}\left(x^{2}+a x+1\right)$ has no minimum value, then the set of all possible values of $a$ is $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. $\{a \mid 01$ when, $y=\log _{a}\left(x^{2}+a x+1\right)$ does not have a minimum value equivalent to $t=x^{2}+a x+1$ not having a minimum value under the constraint $x^{2}+a x+1>0$, which is also equivalent to the minimum value of $t=x^{2}+a x+1$ being no greater than 0, i.e., $\Delta=a^{2}-4 \geqslant 0$. Combinin... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,073 |
3. The volume of the solid of revolution obtained by rotating a triangle with side lengths $3, 4, 5$ around the angle bisector of its largest interior angle is $\qquad$ . | 3. $\frac{32 \sqrt{2} \pi}{7}$.
As shown in Figure 5, $CD$ is the bisector of $\angle ACB=90^{\circ}$, point $D$ is on $AB$, and the symmetric point of $A$ about $CD$ is $A_{1}$. $AA_{1}$ intersects the extension of $CD$ at point $E$. The volume of the solid of revolution is equal to the volume obtained by rotating $\... | \frac{32 \sqrt{2} \pi}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,074 |
4. Let $a$ be a positive number. Then the function
$$
f(x)=a \sin x+\sqrt{a} \cos x\left(-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}\right)
$$
the sum of the maximum and minimum values $g(a)(a>0)$ is in the range $\qquad$ . | 4. $\left(0, \frac{1}{2}\right)$.
First, we have
$$
f(x)=\sqrt{a^{2}+a} \sin (x+\varphi)\left(-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}\right) .
$$
where, $\varphi=\arcsin \frac{\sqrt{a}}{\sqrt{a^{2}+a}}=\arcsin \frac{1}{\sqrt{a+1}}$.
At this point, we have
$$
\begin{array}{l}
-\frac{\pi}{2}+\arcsin \frac{1}... | \left(0, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,075 |
5. For any $x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$, the inequality
$$
\sin ^{2} x+a \sin x+a+3 \geqslant 0
$$
always holds. Then the range of the real number $a$ is $\qquad$ | 5. $a \geqslant-2$.
Let $f(x)=\sin ^{2} x+a \sin x+a+3$, then
$$
f(x)=\left(\sin x+\frac{a}{2}\right)^{2}+a+3-\frac{a^{2}}{4} \geqslant 0 .
$$
Since $x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$, therefore, $\sin x \in\left[-\frac{1}{2}, 1\right]$.
Let $t=\sin x$, then $-\frac{1}{2} \leqslant t \leqslant 1$.
$$
\... | a \geqslant -2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,076 |
6. The number of isosceles triangles (triangles that are the same in shape and size but with different vertices are considered different triangles) with vertices from the set of grid points $T=\{(x, y) \mid x, y=0,1,2\}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks a... | 6.36.
In Figure 6, there are 6 different grid squares, and each grid square has 4 isosceles right triangles, so there are a total of $6 \times 4 = 24$ isosceles triangles.
In addition, there are 4 isosceles triangles of the same shape and size as $\triangle A O D$, 4 isosceles triangles of the same shape and size as ... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,077 |
$$
\begin{array}{l}
\text { Three. (20 points) Let } x_{1}=3, \\
x_{n+1}=\left(\frac{2}{n^{2}}+\frac{3}{n}+1\right) x_{n}+n+1
\end{array}
$$
$(n=1,2, \cdots)$. Find the general formula for $x_{n}$. | From the problem, we have
$$
\begin{array}{l}
x_{n+1}=\frac{n^{2}+3 n+2}{n^{2}} \cdot x_{n}+n+1 \\
=\frac{(n+1)(n+2)}{n^{2}} \cdot x_{n}+n+1 \\
=(n+1)\left(\frac{n+2}{n^{2}} \cdot x_{n}+1\right) .
\end{array}
$$
Let $x_{n}=n a_{n}$, then
$$
(n+1) a_{n+1}=(n+1)\left(\frac{n+2}{n^{2}} \cdot n a_{n}+1\right) \text {, }
$... | x_{n}=n^{2}(2 n+1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,078 |
Four. (20 points) In the right triangle $\triangle ABC$, $\angle ACB = 90^{\circ}$, $BC = a$, $CA = b$ ($a > 0$, $b > 0$), and point $P$ is on side $AB$. The right triangle $\triangle ABC$ is folded along the line $PC$ to form a tetrahedron $PABC$. Find the maximum possible volume of this tetrahedron. | Let the volume of tetrahedron $PABC$ be $V$.
As shown in Figure 7, construct $PD \perp AC$ at point $D$, and $BE \perp PC$ at point $E$. Then,
$$
\begin{array}{l}
V \leqslant \frac{1}{3}\left(\frac{1}{2} PD \cdot AC\right) \cdot BE \\
=\frac{b}{6} PD \cdot a \sin \theta=\frac{ab}{6} PD \sin \theta \\
=\frac{ab}{6} PD \... | \frac{1}{6} \cdot \frac{a^2 b^2}{\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,079 |
Five. (20 points) Let the binary function be
$$
z=f(x, y)=2 x^{2}+3 y^{2}-6 y
$$
with the domain
$$
D=\left\{(x, y) \mid 3 x^{2}+2 y^{2} \leqslant 7 x y, x, y \in \mathbf{R}\right\} \text {. }
$$
(1) Find the range of values of $z=f(x, y)$ (for $(x, y) \in D$);
(2) Find all real numbers $a$ such that in the Cartesian ... | (1) When $x=0$, $2 y_{-}^{2} \leqslant 0, y=0$,
$$
f(x, y)=f(0,0)=0 \text {; }
$$
When $x \neq 0$, $2\left(\frac{y}{x}\right)^{2}-7 \times \frac{y}{x}+3 \leqslant 0$, that is
$$
\left(\frac{y}{x}-\frac{1}{2}\right)\left(\frac{y}{x}-3\right) \leqslant 0 \text {. }
$$
Solving, we get $\frac{1}{2} \leqslant \frac{y}{x} ... | -\frac{81}{26} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,080 |
One. (50 points) As shown in Figure 1, in $\triangle ABC$, $AB = AC = 3, BC = 2, AD \perp BC$ at point $D, BE \perp AC$ at point $E, EF \perp BC$ at point $F, AD$ intersects $BE$ at point $H$. A circle $\odot O$ is drawn with $HF$ as its diameter, intersecting $HE$ and $EF$ at points $P$ and $Q$ respectively. Extend $Q... | $$
\begin{array}{c}
B D=D C=D E \\
=\frac{1}{2} B C=1 . \\
\text { Let } \angle C E F=\angle E B C \\
=\angle B E D=\angle B A D= \\
\angle C A D=\alpha \text {. Then } \\
\sin \alpha=\frac{C D}{A C}=\frac{1}{3}, \\
\cos 2 \alpha=1-2 \sin ^{2} \alpha \\
=\frac{7}{9}, \\
D F=D E \sin \angle D E F=1 \times \sin \left(90^... | P Q=\frac{\sqrt{946}}{108}, K H=\frac{329 \sqrt{2}}{576} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,081 |
II. (50 points) Let $k$ be a non-zero real number, and $\alpha, \beta$ be the two real roots of the quadratic equation $x^{2}-7 x+8 k=0$. Does there exist such a $k$ that the relation
$$
\frac{2}{\alpha}+3 \beta^{2}=\frac{7+\sqrt{49-32 k}}{8 k}
$$
holds? Please explain your reasoning. | II. Such a non-zero real number $k$ must exist. The reason is as follows:
Assume that a non-zero real number $k$ satisfying the problem's conditions exists. By Vieta's formulas, we have $\alpha+\beta=7, \alpha \beta=8 k$. Then
$$
\begin{array}{l}
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=49-16 k, \\
(\alph... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,082 |
Three. (50 points) Let $n$ be a positive integer. In the decimal digits of an $n$-digit number, if it contains the digit 8, then the digit immediately before each 8 cannot be the digit 3 (i.e., the substring "38" cannot appear). Find the number of all such $n$-digit numbers.
Translate the above text into English, pres... | Three, consider the number of $n+1(n \geqslant 1)$-digit numbers $a_{n+1}$, and divide into the following two cases.
(1) When the unit digit is not 8, the first $n$ digits have $a_{n}$ ways of selection, and the unit digit has 9 ways of selection, thus, the $n+1$-digit number has $9 a_{n}$ ways of selection.
(2) When t... | a_{n}=\frac{22+9 \sqrt{6}}{2 \sqrt{6}}(5+2 \sqrt{6})^{n-1}+\frac{-22+9 \sqrt{6}}{2 \sqrt{6}}(5-2 \sqrt{6})^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,083 |
4. In a cube $A C^{\prime}$ with an edge length of 1, $E$ and $F$ are the midpoints of edges $B C$ and $D D^{\prime}$, respectively. The volume of the tetrahedron $A B^{\prime} E F$ is $\qquad$ . | 4. $\frac{5}{24}$.
Although the areas of the tetrahedron's faces can be calculated, it is difficult to find any height, so it is not advisable to directly calculate the volume. Instead, consider an equivalent volume transformation, making one face of the tetrahedron $A B^{\prime} E F$ parallel to the base of the cube.... | \frac{5}{24} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,084 |
1. Let $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}$. Find the value of $a_{2}+a_{4}+\cdots+a_{2 n}$ is ( ).
(A) $3^{n}$
(B) $3^{n}-2$
(C) $\frac{3^{n}-1}{2}$
(D) $\frac{3^{n}+1}{2}$ | $-1 . \mathrm{C}$.
Let $x=0$, we get $a_{0}=1$;
Let $x=-1$, we get
$$
a_{0}-a_{1}+a_{2}-a_{3}+\cdots+a_{2 n}=1;
$$
Let $x=1$, we get
$$
a_{0}+a_{1}+a_{2}+a_{3}+\cdots+a_{2 n}=3^{n} \text{. }
$$
(2) + (3) gives
$$
2\left(a_{0}+a_{2}+a_{4}+\cdots+a_{2 n}\right)=3^{n}+1 \text{. }
$$
Thus, $a_{0}+a_{2}+a_{4}+\cdots+a_{2 ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,085 |
2. If $\sin x+\sin y=1$, then the range of $\cos x+\cos y$ is ( ).
(A) $[-2,2]$
(B) $[-1,1]$
(C) $[0, \sqrt{3}]$
(D) $[-\sqrt{3}, \sqrt{3}]$ | 2.D.
Let $\cos x+\cos y=t$, then
$$
\begin{array}{l}
\cos ^{2} x+2 \cos x \cdot \cos y+\cos ^{2} y=t^{2} . \\
\text { Also } \sin x+\sin y=1 \text {, so } \\
\sin ^{2} x+2 \sin x \cdot \sin y+\sin ^{2} y=1 .
\end{array}
$$
Therefore, $2(\cos x \cdot \cos y+\sin x \cdot \sin y)=t^{2}-1$, that is
$$
2 \cos (x-y)=t^{2}-... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,086 |
3. Let $f_{1}(x)=\sqrt{2}, f_{2}(x)=\sin x+\cos \sqrt{2} x$, $f_{3}(x)=\sin \frac{x}{\sqrt{2}}+\cos \sqrt{2} x, f_{4}(x)=\sin x^{2}$. Among the above functions, the number of periodic functions is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. B.
$f_{1}(x)=\sqrt{2}$ is a periodic function with any positive real number as its period.
Since $\sin x$ is a periodic function with period $T_{1}=2 \pi$, and $\cos \sqrt{2} x$ is a periodic function with period $T_{2}=\frac{2 \pi}{\sqrt{2}}$, and the ratio of $T_{1}$ to $T_{2}$ is not a rational number, hence $f_{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,087 |
5. Given that $\boldsymbol{a}, \boldsymbol{b}$ are two mutually perpendicular unit vectors, and $|c|=13, \boldsymbol{c} \cdot \boldsymbol{a}=3, \boldsymbol{c} \cdot \boldsymbol{b}=4$. Then for any real numbers $t_{1}, t_{2}, \mid \boldsymbol{c}-t_{1} a-t_{2} b$ | the minimum value is ( ).
(A) 5
(B) 7
(C) 12
(D) 13 | 5.C.
From the given conditions, we have
$$
\begin{array}{l}
\left|c-t_{1} a-t_{2} b\right|^{2}=|c|^{2}-6 t_{1}-8 t_{2}+t_{1}^{2}+t_{2}^{2} \\
=169+\left(t_{1}-3\right)^{2}+\left(t_{2}-4\right)^{2}-25 \\
=144+\left(t_{1}-3\right)^{2}+\left(t_{2}-4\right)^{2} \geqslant 144 .
\end{array}
$$
When $t_{1}=3, t_{2}=4$, $\le... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,089 |
6. Let the function $y=f(x)$ satisfy $f(x+1)=f(x)+1$. Then the number of roots of the equation $f(x)=x$ could be ( ).
(A) infinitely many
(B) none or a finite number
(C) a finite number
(D) none or infinitely many | 6.D.
$f(x+1)=f(x)+1$ obviously has a solution $f(x)=x+c$, where $c$ is any real number.
When $c \neq 0$, $f(x)=x$ has no solution;
When $c=0$, $f(x)=x$ has infinitely many solutions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,090 |
$$
\begin{array}{l}
\text { 7. Let } M=\left\{x \left\lvert\, \frac{x-2}{3}+\frac{x-3}{2}=\frac{3}{x-2}+\frac{2}{x-3}\right.\right\}, \\
N=\left\{x \left\lvert\, \frac{x-6}{5}+\frac{x-5}{6}=\frac{5}{x-6}+\frac{6}{x-5}\right.\right\} .
\end{array}
$$
Find $M \cap N=$ | $$
N i, 7. M \cap N=\{0\} \text{. }
$$
From the given information, we can solve for
$$
M=\left\{0,5, \frac{13}{5}\right\}, N=\left\{0,11, \frac{61}{11}\right\} .
$$ | M \cap N=\{0\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,091 |
8. Given the sequence $\left\{x_{n}\right\}$, satisfying $(n+1) x_{n+1}=x_{n}$ $+n$, and $x_{1}=2$. Then $x_{200 s}=$ $\qquad$ . | 8. $\frac{2005!+1}{2005!}$.
From $(n+1) x_{n+1}=x_{n}+n$, we derive $x_{n+1}-1=\frac{x_{n}-1}{n+1}$
Therefore, $x_{n+1}-1=\frac{x_{n}-1}{n+1}=\frac{x_{n-1}-1}{(n+1) n}$
$$
\begin{array}{l}
=\frac{x_{n-2}-1}{(n+1) n(n-1)}=\cdots \\
=\frac{x_{1}-1}{(n+1) n(n-1) \cdots 2}=\frac{1}{(n+1)!} .
\end{array}
$$
Hence $x_{n+1}... | \frac{2005!+1}{2005!} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,092 |
9. Let the function be
$$
2 f(x)+x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}-x^{2}+4 x+3}{x+1} \text {. }
$$
Then $f(x)=$ $\qquad$ . | 9. $x^{2}-3 x+6-\frac{5}{x+1}$.
Let $x=\frac{1}{y}$, we get
$$
f(y)+2 y^{2} f\left(\frac{1}{y}\right)=\frac{3 y^{3}+4 y^{2}-y+3}{y+1} .
$$
Substitute $y$ with $x$ to get
$$
f(x)+2 x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}+4 x^{2}-x+3}{x+1} \text {. }
$$
Also, $2 f(x)+x^{2} f\left(\frac{1}{x}\right)=\frac{3 x^{3}... | x^{2}-3 x+6-\frac{5}{x+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,093 |
Example 1 On a distant island, there live two tribes, one is the "Honest Tribe," and the other is the "Lying Tribe." As the names suggest, the people of the Lying Tribe always lie when speaking or answering questions, while the people of the Honest Tribe always tell the truth.
A journalist encountered four islanders o... | Explanation: From the first person's answer, we can determine:
(1) The first person is from the Lying Tribe (if the first person were from the Truthful Tribe, they would not claim to be from the Lying Tribe);
(2) Among the four people, there must be someone from the Truthful Tribe (otherwise, the first person would be ... | The fourth person is from the Truthful Tribe. | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,095 |
Example 2 A woman and her brother, son, and daughter, four people in total, are all chess players. The twin of the weakest player (also one of the four chess players) and the strongest player are of opposite sexes, and the weakest player and the strongest player are of the same age. Who is the weakest player? | Explanation: The ages of the woman, brother, son, and daughter have the following possibilities:
Below, we start from the worst chess player and make inferences.
(1) If the woman is the worst chess player $\rightarrow$ the brother is a twin $\rightarrow$ the daughter is the best chess player $\rightarrow$ the woman and... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,096 |
Example 4 Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0$, and
$$
\begin{array}{l}
\sum_{i=1}^{k} a_{i} \leqslant \sum_{i=1}^{k} b_{i}(1 \leqslant k \leqslant n) . \\
\text { Then } a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2} .
\end{array}
$$ | Let $S_{k}=\sum_{i=1}^{1} a_{i}, S_{k}^{\prime}=\sum_{i=1}^{1} b_{i}$.
By the Abel transformation formula, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i}^{2}=a_{n} S_{n}+\sum_{k=1}^{n-1} S_{k}\left(a_{k}-a_{k+1}\right) \\
\leqslant a_{n} S_{n}^{\prime}+\sum_{k=1}^{n-1} S_{k}^{\prime}\left(a_{k}-a_{k+1}\right)=\sum_{i... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,097 |
3. The integer sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+2}=a_{n+1}^{2}+a_{n}^{2}(n \geqslant 1) .
$$
If the positive integer $m$ satisfies $a_{m}=2005$, then the set of all possible $m$ is ( ).
(A) $\{1,2\}$
(B) $\{1,2,3\}$
(C) $\{1,2,3,4\}$
(D) $\{1,2,3,4,5\}$ | 3. B.
Let $a_{1}=2005$, then $m=1$ is possible;
Let $a_{2}=2005$, then $m=2$ is possible;
Let $a_{1}=18, a_{2}=41$. Then
$$
a_{3}=a_{1}^{2}+a_{2}^{2}=324+1681=2005,
$$
so $m=3$ is also possible.
If $m=4$ is possible, then we have
$$
2005=a_{2}^{2}+\left(a_{1}^{2}+a_{2}^{2}\right)^{2}\left(a_{1}, a_{2} \in \mathbf{N}_... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,098 |
4. Let the solution set of the equation $x^{2}-x-1=\left(x^{2}-1\right) \pi^{x}-$ $x \pi^{x^{2}-1}$ be $M$. Then the sum of the cubes of all elements in $M$ is ( ).
(A) 0
(B) 2
(C) 4
(D) 5 | 4.C.
Obviously, $x=0, \pm 1$ are all solutions to the original equation.
When $x \neq 0, \pm 1$, from
$$
\left(x^{2}-1\right)\left(\pi^{x}-1\right)=x\left(\pi^{x^{2}-1}-1\right)
$$
and $\pi$ being a transcendental number, we know $\pi^{x}-1=\pi^{x^{2}-1}-1$, which means $x=x^{2}-1$. At this point, there are two real ... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,099 |
5. Given the equation about the real number $x$: $(1+\mathrm{i} x)^{n}=$ (1-ix)"A (where $A \in \mathbf{C}, n \in \mathbf{N}_{+}$) has exactly one solution. Then $\Lambda$ $n$ satisfies ( ).
(A) $|A|=1, n \in \mathbf{N}_{+}$
(B) $|A|=1, n=1$ or 2
(C) $A=1, n \in \mathbf{N}_{+}$
(D) $|A|=1, A \neq-1, n=1$ | 5.D.
Let $x=\tan \alpha, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
A=\left(\frac{1+\mathrm{i} x}{1-\mathrm{i} x}\right)^{n}=\left(\frac{\cos \alpha+\mathrm{i} \sin \alpha}{\cos \alpha-\mathrm{i} \sin \alpha}\right)^{n} \\
=(\cos 2 \alpha+\mathrm{i} \sin 2 \alpha)^{n}=\cos 2 n \alp... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,100 |
6. Given $x, y, z \in \mathbf{R}_{+}$, H.
$$
\begin{array}{l}
s=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}, \\
t=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} .
\end{array}
$$
Then the minimum value of $s^{2}-t^{2}$ is ( ).
(A) $3 \sqrt{5}$
(B) $4 \sqrt{10}$
(C) 36
(D) 45 | 6.C.
$$
\begin{array}{l}
\text { Given } s+t=(\sqrt{x+2}+\sqrt{x+1})+(\sqrt{y+5}+ \\
\sqrt{y+1})+(\sqrt{z+10}+\sqrt{z+1}) \text {, } \\
s-t=\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{4}{\sqrt{y+1}+\sqrt{y+5}}+ \\
\frac{9}{\sqrt{z+1}+\sqrt{z+10}} \text {, } \\
\end{array}
$$
we know $s^{2}-t^{2}=(s+t)(s-t) \geqslant(1+2+3)^... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,101 |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=u, a_{n+1}=\frac{v}{a_{n}} + 1\left(u \in \mathbf{R}, v \geqslant-\frac{1}{4}\right)$. If the set formed by the sequence $\left\{a_{n}\right\}$ is a finite set, then $u, v$ should satisfy $\qquad$ . | 1. $u^{2}=u+v$.
Since the sequence $\left\{a_{n}\right\}$ forms a finite set, there exist $m, n \in \mathbf{N}_{+}$ such that $a_{m}=a_{n}$. Also, $a_{n+1}$ is uniquely determined by $a_{n}$, so from the $n$-th or $m$-th term onward, the sequence $\left\{a_{n}\right\}$ is periodic.
Since $a_{n+1}-\lambda=\frac{(1-\lam... | u^{2}=u+v | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,102 |
2. If real numbers $a, b (a \neq b)$ satisfy $f(x)=$ $-\frac{x+a}{x+b}$, and the inverse function $F(x)$ has a center of symmetry $M$, then the coordinates of point $M$ are | 2. $(-1,-b)$.
First, $F(x)=\frac{-a-b x}{1+x}=-b+\frac{b-a}{1+x}$.
Let $M\left(x_{0}, y_{0}\right)$, then for any $x_{1}, x_{2} \in \mathbf{R}$, if $x_{1}+x_{2}=2 x_{0}$, we have $F\left(x_{1}\right)+F\left(x_{2}\right)=2 y_{0}$, that is
$$
\begin{array}{l}
(b-a)\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}\right)=2\left(y... | (-1,-b) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,103 |
3. Given $P$ as a regular $n$-sided polygon inscribed in a unit circle, $M(n)$ as the maximum number of points within $P$ or on its boundary such that the distance between any two points is not less than 1. Then the set of positive integers $n$ that satisfy $M(n)=n+1$ is
| 3. $\{3,4,5,6\}$.
As shown. When $n=3,4,5,6$, the requirements are met.
When $n \geqslant 7$, the center $O$ and $n$ vertices $A_{1}, A_{2}, \cdots, A_{n}$ divide $P$ into $n$ triangles. By the pigeonhole principle, there must be a triangle with 2 points inside (as shown in Figure 2), suppose it is in $\triangle O A_{... | \{3,4,5,6\} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,104 |
4. Given that $\theta$ is an acute angle, define the function
$$
f(\theta)=\frac{\sin ^{2} \theta}{\cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta} \text {. }
$$
Then the range of the function $f(\theta)$ is $\qquad$ | $$
\begin{array}{l}
\text { 4. }[\sqrt{2},+\infty) \text {. } \\
\text { Let } \theta \rightarrow 0 \text {, then } f(\theta) \rightarrow+\infty \text {. } \\
\text { Also, } f(\theta)=\frac{1}{\cos \theta}-\cos \theta+\frac{1}{\sin \theta}-\sin \theta \\
=\frac{\sin \theta+\cos \theta}{\sin \theta \cdot \cos \theta}-(... | [\sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,105 |
5. Given $p$ is a prime number, $S=\{1,2, \cdots, p-1\}$, $a_{1}, a_{2}, \cdots, a_{p-1} \in \mathbf{N}_{+}$.
$A$ : For any $i$ 、 $j \in S$, we have $a_{i} \equiv a_{j} \equiv \equiv(\bmod p)$.
$B$ : For any non-empty subset $M$ of $S$, we have
$$
\sum_{i \in M}^{1} a_{i} \not \equiv 0(\bmod p) \text {. }
$$
Do you th... | 5. Sufficient and Necessary.
When $B$ is satisfied, consider $p$ numbers
$$
a_{1}, a_{2}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, a_{1}+a_{2}+\cdots+a_{p-1} .
$$
Since none of them are divisible by $p$, there must be two numbers that are congruent modulo $p$.
$$
\begin{array}{l}
\text { Also } a_{1} \not\equiv a_{1}+... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,106 |
Example 5 Given $x, y, z \in \mathbf{R}$, and $x+y+z=0$. Prove:
$$
6\left(x^{3}+y^{3}+z^{3}\right)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right)^{3} \text {. }
$$ | Explanation: Bowman's Triangle Substitution
$$
x=r \cos \theta, y=r \sin \theta,
$$
then $z=-r(\cos \theta+\sin \theta)$.
Assume $r \neq 0$. Thus, the original inequality is equivalent to
$$
\begin{array}{l}
6\left[\cos ^{3} \theta+\sin ^{3} \theta-(\cos \theta+\sin \theta)^{3}\right]^{2} \\
\leqslant\left[\cos ^{2} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,108 |
Three. (20 points) Given $0 \leqslant x, y, z \leqslant 1$, and $|x-y|$, $|y-z|$, $|z-x|$ are all no greater than $\frac{1}{2}$. Try to find the maximum and minimum values of $S=x+y+z - xy - yz - zx$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the t... | $$
\text { Three, } S_{\text {min }}=0, S_{\text {max }}=\frac{5}{6} \text {. }
$$
From the problem, we know that $x+y+z \geqslant x y+y z+z x$. That is, $S \geqslant 0$.
When $x=y=z=0$ or 1, $S$ reaches its minimum value 0.
$$
\begin{array}{l}
\text { Also, } S=x+y+z-x y-y z-z x \\
=x(1-y)+y(1-z)+z(1-x),
\end{array}
... | S_{\text {min }}=0, S_{\text {max }}=\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,109 |
Four. (20 points) Given the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ with the left vertex $C$ and the right focus $F$. A line $l$ passing through $F$ intersects the ellipse at points $A$ and $B$. The lines $C A$ and $C B$ intersect the line $x=10$ at points $M$ and $N$, respectively. Try to find the expression for... | Four, the problem can be transformed into finding the expression and minimum value of the length of the segment $M^{\prime} N^{\prime}$ intercepted by the lines $C A, C B$ on the directrix.
As shown in Figure 4, draw $A P$ perpendicular to the left directrix $l_{1}$ at $P$, $B Q \perp l_{1}$ at $Q$, $T$ is the interse... | |M N|=\frac{12}{\sin \theta}, \text{ and the minimum value of } |M N| \text{ is } 12. | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,110 |
Five. (20 points) Given the sequence of positive integers $\left\{T_{m, n}\right\}$ satisfies:
(1) $T_{m, m}=2^{m}$;
(2) $T_{m, n}=T_{n, m}$;
(3) $\left(T_{m, n}\right)^{m+n}=\left(T_{m, m+n}\right)^{n}$.
Here $m, n$ are any positive integers. Try to find the general term formula for the sequence $\left\{T_{m, n}\righ... | Five, $[m, n]$ is the least common multiple of $m$ and $n$.
Assume $n \geqslant m$, let $n=q m+r, 0 \leqslant r<m$. From (3) we know $\left(T_{m, q n+r}\right)^{(\varphi+1) m+r}=\left(T_{m,(q+1) m+r}\right)^{m m+r}$,
i.e., $\left(T_{m \cdot q m+r}\right)^{\frac{1}{m+r}}=\left(T_{m \cdot(q+1) m+1)^{(4+1) m+r}}\right.$.
... | 2^{(m \cdot n)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,111 |
One, (50 points) Given an equilateral $\triangle ABC$, and $P$ is a point on its plane. Points $E$ and $F$ lie on segments $PB$ and $PC$, respectively. Point $D$ satisfies that $A$ and $D$ are on the same side of line $EF$ and $\triangle DEF$ is an equilateral triangle. Prove: $BE + CF \geq AD$, and find the locus of p... | (1) Points $P, A$ are on the same side of line $BC$, as shown in Figure 5. Given $DI \perp BE, DJ \parallel FC$, extending $BI, CJ$ to intersect at point $G$. Thus, quadrilaterals BIDE and FDJC are both parallelograms. Therefore,
$$
\begin{array}{l}
\angle BGC = \angle EDF \\
= 60^{\circ}.
\end{array}
$$
Also, $\angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,112 |
II. (50 points) Given $x_{i} \in \mathbf{R}_{+}, \prod_{i=1}^{6} x_{i}=1$, and $\prod_{1 \leqslant i<j \leqslant 6}\left(x_{i} x_{j}-1\right) \geqslant 0$. Prove:
$$
\sum_{i=1}^{6} \frac{1}{\left(1+x_{i}\right)^{2}} \geqslant \frac{3}{2} \text {. }
$$ | $$
\begin{array}{l}
\text { II. Let } a_{1}=\left(x_{1} x_{2}-1\right)\left(x_{3} x_{4}-1\right)\left(x_{5} x_{6}-1\right), \\
a_{2}=\left(x_{1} x_{3}-1\right)\left(x_{2} x_{5}-1\right)\left(x_{4} x_{6}-1\right), \\
a_{3}=\left(x_{1} x_{4}-1\right)\left(x_{2} x_{6}-1\right)\left(x_{3} x_{5}-1\right), \\
a_{4}=\left(x_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,113 |
Three. (50 points) Given $r, m, n \in \mathbf{N}_{+}(m>1)$, the set $M=\{1,2, \cdots, r\}$, and $A_{1}, A_{2}, \cdots, A_{m n}$ are $m n$ different $m$-element subsets of $M$. Prove: There exist sets $A$ and $B$ satisfying the following conditions:
(1) $A \cap B=\varnothing, A \cup B=M$;
(2) Among the $2 m n$ sets $A \... | Three, we call the pair $(A, B)$ that satisfies condition $(1)$ a binary partition of $M$. Thus, by taking $A$ through all $2^{r}-2$ proper subsets of $M$, we know that there are $2^{\prime}-2$ such binary partitions $\left(A^{(i)}, B^{(i)}\right)$. For each binary partition $\left(A^{(i)}, B^{(i)}\right)$, define:
$$
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,114 |
As shown in Figure 2, in the square $ABCD$, with the midpoint $O_1$ of side $AB$ as the center and $\frac{AB}{2}$ as the radius, draw the semicircle $\odot O_1$. The center $O_2$ of the semicircle $\odot O_2$ is on side $BC$, and it is tangent to side $CD$, and externally tangent to the semicircle $\odot O_1$ at point ... | Prove: Extend $BC$ to $E$, such that $CE=AO_{1}$, and connect $DE$, $DO_{1}$, $DO_{2}$, $O_{1}O_{2}$. Thus, we have $O_{2}E=O_{1}O_{2}$.
It is easy to prove that $\mathrm{Rt} \triangle DCE \cong \mathrm{Rt} \triangle DAO_{1}$.
Then $DE=DO_{1}$, $\angle EDC=\angle O_{1}DA$.
Therefore, $\triangle EDO_{2} \cong \triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,115 |
Let $n$ be a positive integer, $S_{n}=\sin 1-\sin 4+\cdots+(-1)^{n-1} \sin (3 n-2)$. Prove: $S_{n} \neq 0$. | Proof: Notice
$$
\begin{array}{l}
(-1)^{n-1} \sin (3 n-2)=\sin [(n-1) \pi+3 n-2] \\
=\sin [(n-1)(3+\pi)+1],
\end{array}
$$
Then \( S_{n}=\sum_{i=1}^{n}(-1)^{i-1} \sin (3 i-2) \)
$$
\begin{array}{l}
=\sum_{i=1}^{n} \sin [(i-1)(3+\pi)+1] \\
=\frac{1}{\sin \frac{3+\pi}{2}} \cdot \sum_{i=1}^{n} \sin [(i-1)(3+\pi)+1] \cdot... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,117 |
In $\triangle A B C$, prove:
$$
\begin{array}{l}
\left(\tan \frac{B}{2} \cdot \tan \frac{C}{2}+1\right) \cos A+ \\
\left(\tan \frac{C}{2} \cdot \tan \frac{A}{2}+1\right) \cos B+ \\
\left(\tan \frac{A}{2} \cdot \tan \frac{B}{2}+1\right) \cos C=2 .
\end{array}
$$ | Proof: Note that
$$
\begin{array}{l}
\sin 2 A+\sin 2 B+\sin 2 C \\
=\sin 2 A+2 \sin (B+C) \cdot \cos (B-C) \\
=2 \sin A \cdot[\cos A+\cos (B-C)] \\
=-2 \sin A \cdot[\cos (B+C)-\cos (B-C)] \\
=4 \sin A \cdot \sin B \cdot \sin C, \\
\cos A+\cos B+\cos C-1 \\
=-2 \sin ^{2} \frac{A}{2}+2 \cos \frac{B+C}{2} \cdot \cos \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,118 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.