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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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
14. Given the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{0}=0$, $a_{1}=1$, for all positive integers $n$, we have
$$
a_{n+1}=2 a_{n}+2007 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2008 \mid a_{n}$. | 14. Solution 1: Let $m=2008, a_{n+1}=2 a_{n}+2007 a_{n-1}$, the characteristic equation is $\lambda^{2}-2 \lambda-2007=0$, with characteristic roots $1 \pm \sqrt{m}$. Combining $a_{0}=0, a_{1}=1$, we get
$$
a_{n}=\frac{1}{2 \sqrt{m}}\left[(1+\sqrt{m})^{n}-(1-\sqrt{m})^{n}\right] \text {. }
$$
By the binomial theorem,
... | 2008 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,426 |
15. There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are... | 15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.
When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, ... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,427 |
1. Given that $a$ and $b$ are the roots of the equation $\log _{3 x} 3+\log _{77}(3 x)=$ $-\frac{4}{3}$. Then $a+b=(\quad)$.
(A) $\frac{10}{27}$
(B) $\frac{4}{81}$
(C) $\frac{10}{81}$
(D) $\frac{28}{81}$ | $-1 . C$.
The original equation is transformed into
$$
\frac{1}{1+\log _{3} x}+\frac{1+\log _{3} x}{3}=-\frac{4}{3} \text {. }
$$
Let $1+\log _{3} x=t$, then $\frac{1}{t}+\frac{t}{3}=-\frac{4}{3}$.
Solving for $t$ yields $t_{1}=-1, t_{2}=-3$.
Therefore, $1+\log _{3} x=-1$ or $1+\log _{3} x=-3$.
Thus, the two roots of ... | \frac{10}{81} | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,428 |
2. Let $D$ be a point on side $A B$ of $\triangle A B C$, and $P$ be a point inside $\triangle A B C$ such that
$$
\begin{array}{l}
\boldsymbol{A D}=\frac{3}{4} \boldsymbol{A B}, \boldsymbol{A P}=\boldsymbol{A} D+\frac{2}{5} \boldsymbol{B} C . \\
\text { Then } \frac{S_{\triangle P D}}{S_{\triangle M B C}}=(\quad) .
\e... | 2. A.
Connect $P D$, then $D P=\frac{2}{5} B C$.
Therefore, $D P / / B C, \angle A D P=\angle B$.
$$
\text { Hence } \frac{S_{\triangle A P D}}{S_{\triangle A B C}}=\frac{\frac{1}{2}|A D| \cdot|D P| \sin \angle A D P}{\frac{1}{2}|A B| \cdot|B C| \sin B}=\frac{3}{10} \text {. }
$$ | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,429 |
3. The function $f(x)$ defined on $\mathbf{R}$ is both an odd function and a periodic function. If the smallest positive period of $f(x)$ is $\pi$, and when $x \in\left[0, \frac{\pi}{2}\right)$, $f(x)=\sin x$, then the value of $f\left(\frac{8 \pi}{3}\right)$ is $(\quad)$.
(A) $\frac{\sqrt{3}}{2}$
(B) $-\frac{\sqrt{3}}... | 3. B.
According to the given conditions, we have
$$
\begin{array}{l}
f\left(\frac{8 \pi}{3}\right)=f\left(-\frac{\pi}{3}+3 \pi\right)=f\left(-\frac{\pi}{3}\right) \\
=-f\left(\frac{\pi}{3}\right)=-\sin \frac{\pi}{3}=-\frac{\sqrt{3}}{2}
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,430 |
4. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube with edge length 1, $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$, and $M$ is a point on the edge $B B_{1}$ such that $S_{\triangle D B M}: S_{\triangle O_{1} B_{1} M}=$ 2:3. Then the volume of the tetrahedron $O_{1} A D M$ is ( ).
(A) $\frac{7}{24... | 4.C.
As shown in Figure 1, it is easy to see that $A C \perp$ plane $D_{1} B_{1} B D$. Let $O$ be the center of the base $A B C D$, then $A O \perp$ plane $D O_{1} M$.
$$
\begin{aligned}
& \text { Since } \frac{S_{\triangle D B M}}{S_{\triangle O_{1} B_{1} M}} \\
= & \frac{B D \cdot B M}{O_{1} B_{1} \cdot B_{1} M}=2 \... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,431 |
5. There are 5 red balls and 5 black balls, numbered $1,2,3,4,5$ respectively, and 4 balls are drawn from them. The probability that the numbers on the drawn balls are all different is ( ).
(A) $\frac{5}{21}$
(B) $\frac{2}{7}$
(C) $\frac{1}{3}$
(D) $\frac{8}{21}$ | 5.D.
From 10 balls, choosing 4, there are $\mathrm{C}_{10}^{4}=210$ different ways. If it is required that the numbers of the chosen balls are all different, one can first select 4 numbers from 5 numbers, which has $\mathrm{C}_{5}^{4}$ ways. For each number, there are two colors to choose from, so the number of ways t... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,432 |
Example 5 Convert the mixed periodic continued fraction
into a quadratic irrational.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Explanation: Let $y=[4,5, \overline{2,3}]$, with the repeating part being $x=[\overline{2,3}]$. Then,
$$
y=4+\frac{1}{5}+\frac{1}{x}=[4,5, x] \text {. }
$$
Thus, we have Table 5.
Table 5
\begin{tabular}{|c|c|c|c|c|c|}
\hline$a_{n}$ & & & 4 & 5 & $x$ \\
\hline$p_{n}$ & 0 & 1 & 4 & 21 & $21 x+4$ \\
\hline$q_{n}$ & 1 & 0... | \frac{75-\sqrt{15}}{17} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,433 |
6. The number of positive integers $n$ that make $3^{n}+81$ a perfect square is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 6.B.
When $n \leqslant 4$, it is easy to see that $3^{n}+81$ is not a perfect square. Therefore, let $n=k+4\left(k \in \mathbf{N}_{+}\right)$. Then
$$
3^{n}+81=81\left(3^{k}+1\right) \text {. }
$$
Since $3^{n}+81$ is a perfect square and 81 is a square number, there must exist a positive integer $x$ such that
$$
3^{k... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,434 |
7. Let $[x]$ denote the greatest integer not greater than $x$, and the sets
$$
\begin{array}{l}
A=\left\{x \mid x^{2}-2[x]=3\right\}, \\
B=\left\{x \left\lvert\, \frac{1}{8}<2^{x}<8\right.\right\} .
\end{array}
$$
Then $A \cap B=$ $\qquad$ | $=7 . \mid-1, \sqrt{7}\}$.
Since the solution to the inequality $\frac{1}{8}<2^{x}<8$ is $-3<x<3$, therefore, $B=(-3,3)$.
When $x \in A \cap B$, $\left\{\begin{array}{l}x^{2}-2[x]=3, \\ -3<x<3 .\end{array}\right.$
Thus, $[x]$ can only take the values $-3,-2,-1,0,1,2$.
If $[x] \leqslant-2$, then $x^{2}=3+2[x]<0$, which ... | \{-1, \sqrt{7}\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,435 |
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)} \text {. }
$$
Then $a_{2007}=$ | 8.1343352 .
Given $a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, squaring both sides yields $3\left(a_{n+1}-a_{n}\right)^{2}=2\left(a_{n+1}+a_{n}\right)$.
Also, $3\left(a_{n}-a_{n-1}\right)^{2}=2\left(a_{n}+a_{n-1}\right)$, subtracting the two equations gives
$$
\begin{array}{l}
3\left(a_{n+1}-a_{n-1}\... | 1343352 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,436 |
9. Let the complex number $z_{1}=(6-a)+(4-b) \mathrm{i}$,
$$
\begin{array}{l}
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value, $3 a+4 b=$ $\qquad$ | 9.12.
It is easy to find that $z_{1}+z_{2}+z_{3}=12+9 \mathrm{i}$. Therefore,
$$
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=15 \text {. }
$$
The equality holds if and only if $\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9}$. Solving this, we ge... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,437 |
10. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the function
$$
y=\frac{225}{4 \sin ^{2} x}+\frac{2}{\cos x}
$$
has a minimum value of | 10.68
Since $x \in\left(0, \frac{\pi}{2}\right)$, we have $\sin x>0, \cos x>0$. Let $k>0$, then
$$
\begin{array}{l}
y=\frac{225}{4 \sin ^{2} x}+k \sin ^{2} x+\frac{1}{\cos x}+\frac{1}{\cos x}+ \\
\quad k \cos ^{2} x-k \\
\geqslant 15 \sqrt{k}+3 \sqrt[3]{k}-k .
\end{array}
$$
Equality in (1) holds if and only if
$$
\le... | 68 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,438 |
11. For the function $f(x)=\sqrt{a x^{2}+b x}$, there exists a positive number $b$, such that the domain and range of $f(x)$ are the same. Then the value of the non-zero real number $a$ is $\qquad$. | 11. -4 .
If $a>0$, for the positive number $b$, the domain of $f(x)$ is
$$
D=\left(-\infty,-\frac{b}{a}\right] \cup[0,+\infty) .
$$
However, the range of $f(x)$, $A \subseteq[0,+\infty)$, so $D \neq A$, which does not meet the requirement.
If $a>0$, so, $a=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,439 |
12. Given that the center of a hyperbola is at the origin, and its foci are on the coordinate axes, the distance from point $P(-2,0)$ to its asymptotes is $\frac{2 \sqrt{6}}{3}$. If a line with a slope of $\frac{\sqrt{2}}{2}$ passing through point $P$ intersects the hyperbola at points $A$ and $B$, and intersects the $... | 12. $\sqrt{3}$ or $\sqrt{21}$.
Let the equation of the asymptote be $y=k x$.
From the given condition, we have $\frac{|-2 k|}{\sqrt{1+k^{2}}}=\frac{2 \sqrt{6}}{3}$, solving for $k$ gives $k= \pm \sqrt{2}$. Therefore, the equations of the asymptotes of the hyperbola are $y= \pm \sqrt{2} x$. Suppose the equation of the ... | \sqrt{3} \text{ or } \sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,440 |
13. Draw a line parallel to the known line $l: y= \frac{1}{4} x+1$ through the point $Q(-1,-1)$, intersecting the hyperbola $\frac{x^{2}}{4}-y^{2}=1$ at points $M$ and $N$.
(1) Prove: $Q$ is the midpoint of segment $M N$;
(2) Draw the tangents $l_{1}$ and $l_{2}$ to the hyperbola at points $M$ and $N$ respectively, pro... | Three, 13. (1) The equation of the line $MN$ is $y=\frac{1}{4}(x-3)$.
Substituting into the hyperbola equation $\frac{x^{2}}{4}-y^{2}=1$, we get
$$
3 x^{2}+6 x-25=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$. Then $x_{1}$ and $x_{2}$ are the roots of the equation, so $x_{1}+x_{2}=-... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,441 |
14. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=\frac{1}{2} a_{n}^{2}-a_{n}+2(n \geqslant 1, n \in \mathbf{N}) \text {. }
$$
(1) If $a_{1}=4$, prove:
(i) When $n \geqslant 2$, $a_{n+1} \geqslant 2 a_{n}$;
(ii) When $n \geqslant 1$, $a_{n+1} \geqslant\left(\frac{3}{2}\right)^{n} a_{n}$.
(2) If $a_{1}=... | 14. Since $a_{n+1}-a_{n}=\frac{1}{2} a_{n}^{2}-2 a_{n}+2=$ $\frac{1}{2}\left(a_{n}-2\right)^{2} \geqslant 0$, therefore, $a_{n+1} \geqslant a_{n}$, which means the sequence $\left\{a_{n}\right\}$ is an increasing sequence.
(1) (i) From $a_{1}=4$ and $a_{n+1}=\frac{1}{2} a_{n}^{2}-a_{n}+2$, we get $a_{2}=6, a_{3}=14$.
T... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,442 |
15. Find all positive integers $n$, such that $n+36$ is a perfect square, and apart from 2 or 3, $n$ has no other prime factors. | 15. Let $n+36=(x+6)^{2}\left(x \in \mathbf{N}_{+}\right)$, then $n=x(x+12)$.
According to the problem, we can set $\left\{\begin{array}{l}x=2^{a_{1}} \times 3^{b_{1}}, \\ x+12=2^{a_{2}} \times 3^{b_{2}},\end{array}\right.$ where $a_{1}, a_{2}, b_{1}, b_{2}$ are all non-negative integers. Thus,
$$
2^{a_{2}} \times 3^{b_... | 64,108,288,864,1728,10368 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,443 |
Example 6 Convert $\frac{75-\sqrt{15}}{17}$ into a continued fraction. | Explanation: Since the largest integer not exceeding $\frac{75-\sqrt{15}}{17}$ is 4, we set $\frac{75-\sqrt{15}}{17}=4+\frac{1}{a_{1}}\left(0<\frac{1}{a_{1}}<1\right)$. Then
$$
\begin{array}{l}
a_{1}=\frac{17}{7-\sqrt{15}}=\frac{7+\sqrt{15}}{2} \\
=5+\frac{1}{a_{2}}\left(0<\frac{1}{a_{2}}<1\right), \\
a_{2}=\frac{2}{\s... | \frac{75-\sqrt{15}}{17}=[4,5, \overline{2,3}] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,444 |
1. Given the function $f(x)=|2 x-4|-3 x+8$ $(x \in \mathbf{R})$. Then the analytical expression of the inverse function $f^{-1}(x)$ is ( ).
(A) $f^{-1}(x)=-x+4(x \in \mathbf{R})$
(B) $f^{-1}(x)=-\frac{1}{5} x+\frac{12}{5}(x \in \mathbf{R})$
(C) $f^{-1}(x)=\left\{\begin{array}{l}-x+4, x \leqslant 2 ; \\ -\frac{1}{5} x+\... | $$
-1 . \mathrm{C} .
$$
Because $f(x)=\left\{\begin{array}{l}-5 x+12, x2 .
\end{array}\right.$
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,445 |
2. The arithmetic sequence $\left\{a_{k}\right\}$ has a total of $2 n+1$ terms $(n \in$ $\mathbf{N}_{+}$), where the sum of all odd-numbered terms is 310, and the sum of all even-numbered terms is 300. Then the value of $n$ is ( ).
(A) 30
(B) 31
(C) 60
(D) 61 | 2. A.
Since $\left\{a_{k}\right\}$ is an arithmetic sequence, and
$$
\left\{\begin{array}{l}
a_{1}+a_{3}+\cdots+a_{2 n+1}=310, \\
a_{2}+a_{4}+\cdots+a_{2 n}=300,
\end{array}\right.
$$
Therefore, $\left\{\begin{array}{l}(n+1) a_{n+1}=310, \\ n a_{n+1}=300 .\end{array}\right.$
Dividing the two equations yields $\frac{n... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,446 |
3. Let $a=\sin \left(\sin 2007^{\circ}\right)$,
$$
\begin{array}{l}
b=\sin \left(\cos 2007^{\circ}\right), \\
c=\cos \left(\sin 2007^{\circ}\right), \\
d=\cos \left(\cos 2007^{\circ}\right) .
\end{array}
$$
Then the size relationship of $a, b, c, d$ is ().
(A) $a<b<c<d$
(B) $b<a<d<c$
(C) $c<d<b<a$
(D) $d<c<a<b$ | 3. B.
Since $2007^{\circ}=5 \times 360^{\circ}+180^{\circ}+27^{\circ}$, therefore,
$$
\begin{array}{l}
a=\sin \left(-\sin 27^{\circ}\right)=-\sin \left(\sin 27^{\circ}\right)<0, \\
d=\cos \left(-\cos 27^{\circ}\right)=\cos \left(\cos 27^{\circ}\right)>0 .
\end{array}
$$
Also, $\sin 27^{\circ}<\cos 27^{\circ}$, hence ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,447 |
4. As shown in Figure 1, the diameter $A B=4$, $O$ is the center of the semicircle, and $C$ is any point on the semicircle different from $A$ and $B$. If $P$ is a moving point on the radius $O C$, then the minimum value of $(\boldsymbol{P A}+\boldsymbol{P B}) \cdot \boldsymbol{P C}$ is ( ).
(A) 2
(B) 0
(C) -1
(D) -2 | 4.D.
Since $O$ is the midpoint of $AB$, we have
$$
\begin{array}{l}
\boldsymbol{P A}+\boldsymbol{P B}=2 \boldsymbol{P O} . \\
\text { Then }(\boldsymbol{P A}+\boldsymbol{P B}) \cdot \boldsymbol{P C}=2 \boldsymbol{P O} \cdot \boldsymbol{P C} \\
=-2|\boldsymbol{P O}| \cdot|\boldsymbol{P C}| .
\end{array}
$$
Given that ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,448 |
5. The necessary and sufficient condition for line segments of lengths $1, a, a, a, a, a$ to be the 6 edges of the same tetrahedron is ( ).
(A) $0\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{3}}{3}<a<\sqrt{3}$ | 5.C.
In tetrahedron $ABCD$, $\triangle ABC$ and $\triangle BCD$ are both equilateral triangles with side length $a$, and $AD=1$. Taking the midpoint $E$ of $BC$, we have
$$
AE=DE=\frac{\sqrt{3}}{2} a.
$$
Clearly, the necessary and sufficient condition for the lengths $1, a, a, a, a, a$ to be the 6 edges of tetrahedro... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,449 |
6. Let $x$ and $y$ be integers, and satisfy
$$
x y + 2 = 2(x + y) \text{.}
$$
Then the maximum possible value of $x^2 + y^2$ is ( ).
(A) 32
(B) 25
(C) 18
(D) 16 | 6.B.
From the given, we have $(x-2) y=2 x-2$, which is $y=\frac{2 x-2}{x-2}=2+\frac{2}{x-2}(x \neq 2)$.
Since $x, y \in \mathbf{Z}$, it follows that $x-2= \pm 1, \pm 2$. Therefore, $(x, y)=(0,1),(1,0),(3,4),(4,3)$. Hence, the maximum possible value of $x^{2}+y^{2}$ is $3^{2}+4^{2}=25$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,450 |
7. Given $0<k<4$, line $l_{1}: k x-2 y-2 k+$ $8=0$ and line $l_{2}: 2 x+k^{2} y-4 k^{2}-4=0$ form a quadrilateral with the two coordinate axes. The value of $k$ that minimizes the area of this quadrilateral is ( ).
(A)2
(B) $\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $\frac{1}{8}$ | 7.D.
It is known that the lines $l_{1}: y-4=\frac{k}{2}(x-2)$ and $l_{2}$: $y-4=-\frac{2}{k^{2}}(x-2)$ both pass through the fixed point $P(2,4)$.
In $l_{1}$, let $x=0$, we get $y_{A}=4-k$;
In $l_{2}$, let $y=0$, we get $x_{B}=2 k^{2}+2$. As shown in Figure 6, connect $O P$.
Then $S_{\text {quadrilateral PAOB }}$
$$
\... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,451 |
8. For a real number $t$, it is known that the first three terms of the geometric sequence $\left\{a_{n}\right\}$ are $2 t, 5 t-1, 6 t+2$, and the sum of the first $n$ terms of this sequence is $S_{n}$. Then the maximum integer value of $n$ that satisfies the inequality $\left|S_{n}-\frac{1}{65}\right|<1$ is ( ).
(A) 2... | 8. A.
From the given, we have $(5 t-1)^{2}=2 t(6 t+2)$.
Solving, we get $t=1$ or $t=\frac{1}{13}$. When $t=1$, $S_{n}=\frac{2\left(1-2^{n}\right)}{1-2}=2^{n+1}-2$. Then, for $\left|S_{n}-\frac{1}{65}\right|=\left|2^{n+1}-2-\frac{1}{65}\right|<1$, there is no positive integer $n$ that satisfies this condition.
When $t=... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,452 |
9. For non-empty sets $A$ and $B$, define the operation:
$A \oplus B=\{x \mid x \in A \cup B$, and $x \notin A \cap B\}$.
Given two open intervals $M=(a, b), N=(c,
d)$, where $a, b, c, d$ satisfy
$$
a+b<c+d, ab=cd<0 .
$$
Then $M \oplus N=(\quad)$.
(A) $(a, b) \cup(c, d)$
(B) $(a, c) \cup(b, d)$
$(\mathrm{C})(a, d) \c... | 9. B.
Let $a b=c d=t(t<0)$. Then
$$
a<0<b, c<0<d \text {. }
$$
Construct the functions
$$
\begin{array}{l}
f(x)=x^{2}-(a+b) x+t, \\
g(x)=x^{2}-(c+d) x+t .
\end{array}
$$
Then $a, b$ are the two roots of the equation $f(x)=0$, and $c, d$ are the two roots of the equation $g(x)=0$.
$$
\begin{array}{l}
\text { Also, } ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,453 |
10. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f(0)=$ $0, f(x)+f(1-x)=1, f\left(\frac{x}{5}\right)=\frac{1}{2} f(x)$, and when $0 \leqslant x_{1}<x_{2} \leqslant 1$, $f\left(x_{1}\right) \leqslant f\left(x_{2}\right)$. Then $f\left(\frac{1}{2007}\right)$ equals ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{16}$
(C) ... | 10.C.
Let $x=1$, then
$$
f(1)+f(0)=1, f\left(\frac{1}{5}\right)=\frac{1}{2} f(1) \text {. }
$$
Combining with $f(0)=0$, we get $f(1)=1, f\left(\frac{1}{5}\right)=\frac{1}{2}$.
Let $x=\frac{1}{5}$, then $f\left(\frac{1}{5}\right)+f\left(\frac{4}{5}\right)=1$.
Thus, $f\left(\frac{4}{5}\right)=\frac{1}{2}$.
When $0 \leq... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,454 |
Example 7 Convert $\sqrt{11}$ into a continued fraction, and find its first four convergents. | Explanation: Since the largest integer not exceeding $\sqrt{11}$ is 3, we set $\sqrt{11}=3+\frac{1}{a_{1}}\left(0<\frac{1}{a_{1}}<1\right)$. Then
$$
\begin{array}{l}
a_{1}=\frac{1}{\sqrt{11}-3}=\frac{\sqrt{11}+3}{2} \\
=3+\frac{1}{a_{2}}\left(0<\frac{1}{a_{2}}<1\right), \\
a_{2}=\frac{2}{\sqrt{11}-3}=\sqrt{11}+3 \\
=6+... | \frac{3}{1}, \frac{10}{3}, \frac{63}{19}, \frac{199}{60} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,455 |
11. Real numbers $x, y$ satisfy $\tan x=x, \tan y=y$, and $|x| \neq|y|$. Then the value of $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}$ is | II, 11.0.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y} \\
=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y .
\end{array}
$$
Similarly, $\frac{\sin (x-y)}{x-y}=\cos x \cdot \cos y$. Therefore, $\frac{\sin (x+y)}{x+y}-\frac{\sin (x... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,456 |
12. Definition: The length of the interval $\left[x_{1}, x_{2}\right]\left(x_{1}<x_{2}\right)$ is $x_{2}-x_{1}$. Given that the domain of the function $y=\left|\log _{\frac{1}{2}} x\right|$ is $[a, b]$, and the range is $[0,2]$. Then the difference between the maximum and minimum values of the length of the interval $[... | 12.3.
The graphs of the functions $y=$ $\left|\log _{\frac{1}{2}} x\right|$ and $y=2$ are shown in Figure 7.
From $y=0$, we get $x=1$;
From $y=2$, we get $x=\frac{1}{4}$ or $x=4$.
When $a=\frac{1}{4}, b=1$, $(b-a)_{\text {min }}=\frac{3}{4}$;
When $a=\frac{1}{4}, b=4$, $(b-a)_{\max }=\frac{15}{4}$.
Therefore, $(b-a)_... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,457 |
13. From the numbers $1,2, \cdots, 10$, if 3 numbers are randomly drawn, the probability that at least two of the numbers are consecutive integers is $\qquad$ | 13. $\frac{8}{15}$.
Suppose the 3 numbers drawn are such that no two are adjacent, denoted as $i, j, k(1 \leqslant i<j<k \leqslant 10)$. Then $i, j-1, k-2$ are distinct and can only take values from the set $\{1,2, \cdots, 8\}$, yielding a total of $C_{8}^{3}$ different combinations.
Additionally, the total number of... | \frac{8}{15} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,458 |
14. Two spheres $O_{1}$ and $O_{2}$, both with a radius of 1, are tangent to each other and are also tangent to the two half-planes of the right dihedral angle $\alpha-l-\beta$. There is another smaller sphere $O$ with radius $r (r<1)$ that is also tangent to the two half-planes of the dihedral angle, and is externally... | $14 \cdot 3-\sqrt{7}$.
It is known that the centers of the three spheres $O, O_{1}, O_{2}$ are all on the bisecting plane $\gamma$ of the right dihedral angle $\alpha-l-\beta$, as shown in Figure 8.
In the plane $\gamma$, draw $O_{1} A \perp l, O_{2} B \perp l, O C \perp l$, with the feet of the perpendiculars being ... | 3-\sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,459 |
15. As shown in Figure $2, A$ is a fixed point on the circle $\odot O$ with radius 1, and $l$ is the tangent line of $\odot O$ passing through point $A$. Let $P$ be a point on $\odot O$ different from $A$, and $P Q \perp$ $l$, with the foot of the perpendicular being $Q$. When point $P$ moves on $\odot O$, the maximum ... | 15. $\frac{3 \sqrt{3}}{8}$.
As shown in Figure 9, let the symmetric point of point $P$ with respect to line $AO$ be $P'$. Then,
$$
S_{\triangle PAQ}=\frac{1}{2} S_{\triangle PAP} .
$$
By plane geometry knowledge, when and only when $\triangle PAP'$ is an inscribed equilateral triangle in the circle, the area of $\tri... | \frac{3 \sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,460 |
One, (20 points) Let the function be
$$
f(x)=\cos 2 x+2 \sqrt{3} \sin x \cdot \cos x(x \in \mathbf{R})
$$
with the maximum value $M$ and the smallest positive period $T$.
(1) Find the values of $M$ and $T$, and write down the intervals where the function $f(x)$ is monotonically increasing;
(2) If 10 distinct positive ... | (1) Notice
$$
f(x)=2 \sin \left(2 x+\frac{\pi}{6}\right)(x \in \mathbf{R}) \text {. }
$$
Then $M=2, T=\frac{2 \pi}{2}=\pi$.
$$
\text { Also, } 2 k \pi-\frac{\pi}{2} \leqslant 2 x+\frac{\pi}{6} \leqslant 2 k \pi+\frac{\pi}{2}(k \in
$$
Z), then the monotonic increasing interval of the function $f(x)$ is
$$
\left[k \pi-\... | \frac{140 \pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,461 |
II. (20 points) As shown in Figure 3, in $\triangle A B C$, $A B>A C$. A tangent line to the circumcircle of $\triangle A B C$ is drawn through point $A$, intersecting the extension of $B C$ at point $D$. $E$ is the midpoint of $A D$, and $B E$ intersects the circumcircle of $\triangle A B C$ at point $F$. Prove:
$$
\a... | II. Since $E A$ is the tangent of the circumcircle of $\triangle A B C$, we have $E A^{2}=E F \cdot E B$.
Also, since $E$ is the midpoint of $A D$, then $D E^{2}=E F \cdot E B$, which means $\frac{D E}{B E}=\frac{E F}{E D}$.
Therefore, $\triangle D E F \backsim \triangle B E D \Rightarrow \angle F D E=\angle D B E$.
Si... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,462 |
Three. (20 points) Given $a, b, c \in \left[\frac{1}{2}, 1\right]$. Prove:
(1) $\frac{b}{a} + \frac{a}{b} \leqslant \frac{5}{2}$;
(2) $\frac{a b + b c}{a^2 + 2 b^2 + c^2} \geqslant \frac{2}{5}$. | Three, let $t=\frac{b}{a}$, then from $a, b \in\left[\frac{1}{2}, 1\right]$ we know, $t \in$ $\left[\frac{1}{2}, 2\right]$. Thus, $\frac{b}{a}+\frac{a}{b}=t+\frac{1}{t}$.
Since $f(t)=t+\frac{1}{t}$ is monotonically decreasing on $\left[\frac{1}{2}, 1\right]$ and monotonically increasing on $[1,2]$, therefore, when $t=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,463 |
Five. (30 points) Let the set
$$
X=\{(a, b, c) \mid a, b, c \in \mathbf{Z}\},
$$
and $f$ be a mapping from $X$ to $X$ such that
$$
f(a, b, c)=(a+b+c, ab+bc+ca, abc) .
$$
Find all triples $(a, b, c)$ such that
$$
f(f(a, b, c))=(a, b, c) .
$$ | Because $f(a, b, c)=(a+b+c, a b+b c+c a, a b c)$, so,
$$
\begin{array}{l}
f(f(a, b, c)) \\
=f(a+b+c, a b+b c+c a, a b c) .
\end{array}
$$
By $f(f(a, b, c))=(a, b, c)$, we get
$$
(a+b+c)+(a b+b c+c a)+a b c=a,
$$
which means $(1+a)(1+b)(1+c)=1+a$.
Thus, $1+a=0$ or $(1+b)(1+c)=1$.
(1) If $1+a=0$, i.e., $a=-1$, then
$$
... | (a, 0,0)(a \in \mathbf{Z}), (-1,-1,1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,465 |
Example 1 Prove: The circumcircle of an acute triangle is the smallest circle that can cover the triangle.
| Proof: As shown in Figure 1, the plane region of acute triangle \( \triangle ABC \) can be divided into three parts, namely \( \angle AOB \), \( \angle BOC \), and \( \angle COA \). Suppose the center \( D \) is within the region \( \angle BOC \), and connect \( AD \), \( BD \), and \( CD \). Clearly,
\[
\angle ABD \ge... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,466 |
1. Given $a=\frac{-\sqrt{6}+\sqrt{6-4 \sqrt{2}}}{2}$. Then $a^{3}+$ $\sqrt{6} a^{2}+\sqrt{2} a+\sqrt{6}$ is ( ).
(A) $-\sqrt{2}$
(B) $\sqrt{2}$
(C) $-\sqrt{6}$
(D) $\sqrt{6}$ | -、1.D.
Notice that $a=\frac{-\sqrt{6}+\sqrt{\sqrt{6}^{2}-4 \times 1 \times \sqrt{2}}}{2 \times 1}$, then $a$ is a root of the equation $x^{2}+\sqrt{6} x+\sqrt{2}=0$.
Therefore, $a^{2}+\sqrt{6} a+\sqrt{2}=0$.
Thus, $a^{3}+\sqrt{6} a^{2}+\sqrt{2} a+\sqrt{6}$
$$
=a\left(a^{2}+\sqrt{6} a+\sqrt{2}\right)+\sqrt{6}=\sqrt{6} .... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,467 |
4. Given $\frac{x}{x^{2}+x+1}=\frac{1}{8}(0<x<1)$. Then the value of $\sqrt{x}-\frac{1}{\sqrt{x}}$ is ( ).
(A) $-\sqrt{7}$
(B) $\sqrt{7}$
(C) $-\sqrt{5}$
(D) $\sqrt{5}$ | 4.C.
Obviously, $x \neq 0$. Thus, $\frac{x^{2}+x+1}{x}=8$, which means $x+\frac{1}{x}=7$.
Since $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}=x+\frac{1}{x}-2=5$, therefore, $\sqrt{x}-\frac{1}{\sqrt{x}}= \pm \sqrt{5}$.
Also, $0<x<1$, so $\sqrt{x}-\frac{1}{\sqrt{x}}=-\sqrt{5}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,470 |
5. A school conducted a math test for all students in Group 1 and Group 2 of Grade 8, with 10 students in each group. The median scores of Group 1 and Group 2 are 95 points and 97 points, respectively. If a score of 96 points or higher is considered excellent, then the number of excellent students in Group 1 and Group ... | 5.D.
Since the median score of 10 students in Group 1 is 95 points, the number of students in Group 1 with scores higher than 95 points does not exceed 5. Therefore, the number of students in Group 1 with scores greater than or equal to 96 points does not exceed 5, i.e., $m \leqslant 5$. Since the median score of 10 s... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 719,471 |
6. If $p$ is a prime number, and $p^{3}+3$ is also a prime number, then the last digit of $p^{33}+33$ is ( ).
(A) 5
(B) 7
(C) 9
(D) cannot be determined | 6.A.
From $p^{3}+3$ being a prime number, we know that $p$ is even. Since $p$ is also a prime number, then $p=2$.
Thus, $p^{33}+33=2^{33}+33=\left(2^{4}\right)^{8} \times 2+33$.
Since the last digit of $\left(2^{4}\right)^{8}$ is 6, the last digit of $\left(2^{4}\right)^{8} \times 2$ is 2.
Therefore, the last digit of... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,472 |
1. Let $[x]$ denote the greatest integer not exceeding $x$. Then
$$
\begin{array}{l}
\frac{2}{[\sqrt{1 \times 2}] \times[\sqrt{2 \times 3}] \times[\sqrt{3 \times 4}]}+ \\
\frac{2}{[\sqrt{2 \times 3}] \times[\sqrt{3 \times 4}] \times[\sqrt{4 \times 5}]}+\cdots+ \\
\frac{2}{[\sqrt{98 \times 99}] \times[\sqrt{99 \times 10... | $$
\text { II. } 1 . \frac{4949}{9900} \text {. }
$$
Let $n$ be a positive integer, then
$$
n<\sqrt{n(n+1)}<n+1 \text {. }
$$
Thus, $[\sqrt{n(n+1)}]=n$.
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{2}{1 \times 2 \times 3}+\frac{2}{2 \times 3 \times 4}+\cdots+\frac{2}{98 \times 99 \times 100} \\
=\... | \frac{4949}{9900} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,473 |
2. Given that $G$ is the centroid of $\triangle A B C$, $G A=5, G B$ $=12, G C=13$. Then the height from vertex $C$ to side $A B$ of $\triangle A B C$ is $\qquad$ | $2.13 \frac{11}{13}$.
Extend $B G$ to intersect $A C$ at point $D$, then $B D$ is a median,
$$
\begin{array}{l}
G D=\frac{1}{2} B G \\
=6 .
\end{array}
$$
Extend $G D$ to point $E$ such that $D E=G D=6$, and connect $A E$.
It is easy to see that $\triangle A D E \cong \triangle C D G$. Therefore,
$$
A E=G C=13 \text {... | 13 \frac{11}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,474 |
3. If the equation about $x$
$$
\frac{2 k}{x+2}-\frac{x-1}{x^{2}+2 x}=\frac{k x+1}{x}
$$
has only one solution (equal solutions are counted as one), then the value of $k$ is $\qquad$ | 3.0 or 1 or $\frac{3}{4}$.
The original fractional equation can be transformed into
$$
k x^{2}+2 x+1=0 \text {. }
$$
(1) When $k=0$, $x=-\frac{1}{2}$ is a solution to the fractional equation.
(2) When $k \neq 0, \Delta=2^{2}-4 k=0$, $k=1$, $x=-1$ is a solution to the fractional equation.
(3) When $k \neq 0, \Delta>0$, ... | 0 \text{ or } 1 \text{ or } \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,475 |
4. As shown in Figure $2, P A$ is tangent to
$\odot O$ at point $A, P C$ intersects
$\odot O$ at points $B$ and $C, P O$ intersects
$\odot O$ at point $D, A E \perp P O$
at $E$, connect $B E$ and extend it
to intersect $\odot O$ at point $F$, connect $O C, O F, A D, A F$. If $\angle B C O=30^{\circ}, \angle B F O=$ $20... | $4.115^{\circ}$.
Connect $A O$. Since $P A \perp A O, A E \perp P O$, we get $P A^{2}=P E \cdot P O$.
Also, $P A^{2}=P B \cdot P C$, so $P E \cdot P O=P B \cdot P C$.
Thus, $\frac{P E}{P C}=\frac{P B}{P O}$.
Since $\angle E P B=\angle C P O$, therefore,
$\triangle P E B \backsim \triangle P C O$.
Hence $\angle B E P=\a... | 115^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,476 |
Example 2: Prove that two circles with diameters less than 1 cannot cover a circle with a diameter of 1. | Proof: Let the circle with a diameter of 1 be $\odot O$, and the two circles with diameters less than 1 be $\odot O_{1}$ and $\odot O_{2}$. To prove that $\odot O_{1}$ and $\odot O_{2}$ cannot cover $\odot O$, it is only necessary to prove that a diameter of $\odot O$ cannot be covered. First, use $\odot O_{1}$ to cove... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,477 |
One. (20 points) Given the parabola $y=x^{2}+3 x+c$ passes through two points $(m, 0)$ and $(n, 0)$, and
$$
m^{3}+3 m^{2}+(c-2) m-2 n-c=8,
$$
the parabola intersects the hyperbola $y=\frac{k}{x}$ at the point $(1, d)$.
(1) Find the equations of the parabola and the hyperbola;
(2) Given points $P_{1}, P_{2}, \cdots, P_... | (1) According to the problem, we have
$$
\left\{\begin{array}{l}
m\left(m^{2}+3 m+c\right)-2(m+n)-c=8 \\
m^{2}+3 m+c=0, \\
m+n=-3 .
\end{array}\right.
$$
Substituting equations (2) and (3) into equation (1) gives $c=-2$.
Therefore, the equation of the parabola is $y=x^{2}+3 x-2$.
Since the intersection point of the pa... | 2017035 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,478 |
II. (25 points) As shown in Figure 3, in trapezoid $ABCD$, $AD \parallel BC$, $CE$ bisects $\angle ACB$, $AE \perp CE$ at $E$, $(BC + AB)(BC - AB) + (AD + CD)(AD - CD) = 2BC \cdot AD$, $\angle BCD = 60^\circ$, $\angle BAE = 40^\circ$. Prove:
$$
\angle DAC = 2 \angle DCA
$$ | As shown in Figure 7, draw $A F / / D C$ intersecting $B C$ at point $F$, and extend $A E$ to intersect $B C$ at point $G$. From the given information, we have
$$
\begin{array}{l}
A B^{2}+C D^{2} \\
=(B C-A D)^{2} .
\end{array}
$$
It is easy to see that quadrilateral $A F C D$ is a parallelogram, thus
$$
A F=C D, C F=... | \angle DAC = 2 \angle DCA | Geometry | proof | Yes | Yes | cn_contest | false | 719,479 |
Three. (25 points) Given that $\sqrt{1-|x|}+\sqrt{2-2|y|}$ is meaningful, $M=\sqrt{x^{2}+2 x y+y^{2}}+\sqrt{x^{2}+2 x+1}+|2 x-y-4|$ has maximum and minimum values of $a$ and $b$ respectively. If the five-digit integer $1 a b c d$ is divisible by 99, find the values of $c$ and $d$.
---
Given that $\sqrt{1-|x|}+\sqrt{2... | Three, it is known that $1-|x| \geqslant 0, 2-2|y| \geqslant 0$.
Therefore, $-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1$.
Then $-2 \leqslant x+y \leqslant 2, 0 \leqslant x+1 \leqslant 2$,
$$
-7 \leqslant 2 x-y-4 \leqslant -1 \text{. }
$$
When $x+y \geqslant 0$,
$$
\begin{array}{l}
M=|x+y|+|x+1|+|2 x-y-4| \... | c=2, d=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,480 |
1. Let $p$ be a prime number greater than 3, and $p+8$ is also a prime number. Let $S_{1}=p(p+8)$, and $S_{i+1} (i=1,2, \cdots, k-1)$ denote the sum of the digits of $S_{i}$. If $S_{k}$ is a single-digit number, for all possible prime numbers $p$, $S_{k}$ has ( ) different values.
(A) 1
(B) 2
(C) 3
(D) 4 | $-1 . A$.
Since $p$ is a prime number greater than 3, and $p+8$ is also a prime number, therefore, $p$ is a prime of the form $3M+2$. So,
$$
p(p+8)=(p+4)^{2}-16 \equiv 2(\bmod 9) \text {. }
$$
Also, $S_{i+1} \equiv S_{i}(\bmod 9)$, then $S_{k}=2$, which means $S_{k}$ has only one value. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,481 |
2. If the 2007th degree equation
$$
x^{2007}-4014 x^{2006}+a_{2} 005 x^{2005}+\cdots+a_{1} x+a_{0}=0
$$
has 2007 positive real roots, then, for all possible equations, the maximum value of $\sum_{k=0}^{2005}\left|a_{k}\right|$ is ( ).
(A) $2^{2007}-4014$
(B) $2^{2007}-4015$
(C) $3^{2007}-4014$
(D) $3^{2007}-4015$ | 2. D.
Let the 2007 positive real roots of the original equation be $x_{1}$, $x_{2}, \cdots, x_{2007}$. We can obtain the identity
$$
\begin{array}{l}
x^{200}-4014 x^{206}+a_{20 \sigma} x^{2 \omega 6}+\cdots+a_{1} x+a_{0} \\
=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{2007}\right) .
\end{array}
$$
By Vi... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,482 |
3. The number of positive integers $n$ that satisfy $\mathrm{C}_{n}^{i}$ is coprime with 2007 for all $i=0,1, \cdots, n$ is ( ) .
(A) 3
(B) 5
(C) 7
(D) 9 | 3. D.
If the prime $p$ is coprime with all $\mathrm{C}_{n}^{i}(i=0,1, \cdots, n)$, since the maximum power of the prime factor $p$ in $\mathrm{C}_{n}^{i}$ is
$$
\sum_{k=1}^{+\infty}\left(\left[\frac{n}{p^{k}}\right]-\left[\frac{i}{p^{k}}\right]-\left[\frac{n-i}{p^{k}}\right]\right),
$$
then for every $i$ we have
$$
\... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,483 |
4. The largest solution and the smallest solution of the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ sum to $(\quad)$ (where, $[x]$ denotes the greatest integer not exceeding $x$, the same applies below).
(A) 85
(B) -85
(C) 42
(D) -42 | 4.B.
Let $x=42 p+q(p, q$ be integers, $0 \leqslant q \leqslant 41)$.
Substituting $x$ into the original equation, we get
$$
p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \text {. }
$$
For each different $q$, a unique ordered pair $(p, q)$ is determined, and thus, $x$ is also unique. To... | -85 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,484 |
5. If the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies
$$
a_{0}=0, a_{n}=a_{\left[\frac{n}{2}\right]}+\left[\frac{n}{2}\right](n \geqslant 1),
$$
then the number of positive integers $n$ such that $a_{n}=2007$ is ( ) .
(A) 0
(B) 1
(C) 2
(D) greater than 2 | 5.A.
By induction, we can obtain $a_{n}=n-t_{n}$, where $t_{n}$ represents the number of 1s in the binary representation of $n$.
In fact, when $n=0,1,2$, $a_{0}=0, a_{1}=0$, $a_{2}=1$, the conclusion holds true.
Assume that when $n \leqslant k-1$, $a_{n}=n-t_{n}$.
If $k=2 l$, then
$$
\begin{array}{l}
a_{k}=a_{l}+l=l-... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,485 |
6. A car starts from point $A$ and drives to point $B$ (no reverse driving allowed), the distance between $A$ and $B$ is 2007. Some stations are set up along the way, with stations established at all points where the distance to $A$ is a multiple of 100 (the set of these stations is denoted as $X$), and stations establ... | 6. C.
Let $k$ represent the station that is $k$ units away from $A$, to 223 there are 3 ways, to $300, 400$ there are 4 ways each, to 446 there are 11 ways, to 500, 600 there are 15 ways each, to 669 there are 41 ways, to 700, 800 there are 56 ways each, to 892 there are 153 ways, to $900, 1000, 1100$ there are 209 wa... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,486 |
1. Let the circumcenter, incenter, and orthocenter of non-isosceles $\triangle ABC$ be $O$, $I$, and $H$, respectively, with the circumradius being $1$ and $\angle A=60^{\circ}$. Then the circumradius of $\triangle OIH$ is $\qquad$. | If $\triangle A B C$ is an acute triangle, since
$$
\angle B O C=\angle B I C=\angle B H C=120^{\circ} \text {, }
$$
then, $O$, $I$, $H$, $B$, and $C$ are concyclic;
if $\triangle A B C$ is an obtuse triangle, since
$$
\angle B O C=\angle B I C=120^{\circ}, \angle B H C=60^{\circ} \text {, }
$$
and $H$ is on the oppo... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,487 |
Example 3 On a plane, there are 25 points, and among any three points, the distance between two of them is less than 1. Prove: a circular paper with a radius of 1 can cover at least 13 of these points. | Proof: In 25 points, take any point $O$, and then make the center of a circular paper with a radius of 1 coincide with $O$. If there are at least 13 points within this circle, the proposition holds.
Assume that there are at most 12 points within this circle, then there are at least 13 points outside this circle. Clear... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,488 |
2. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}(n \geqslant 1)$ satisfy
$$
a_{n+1}=2 b_{n}-a_{n}, b_{n+1}=2 a_{n}-b_{n}
$$
$(n=1,2, \cdots)$. If $a_{1}=2007, a_{n}>0(n=2,3$, $\cdots$ ), then $b_{1}$ equals $\qquad$ . | 2.2007.
Since $b_{n}=\frac{1}{2}\left(a_{n}+a_{n+1}\right)$, we have
$$
a_{n+2}+2 a_{n+1}-3 a_{n}=0 \text{. }
$$
By the method of characteristic roots, we get
$$
a_{n}=\frac{2007+b_{1}}{2}+\frac{b_{1}-2007}{6}(-3)^{n} \text{. }
$$
If for any positive integer $n$, we have $a_{n}>0$, then it must be that $b_{1}=2007$. | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,489 |
3. $\sum_{k=1}^{207}\left[\sqrt[4]{\frac{2007}{k}}\right]$ The value is $\qquad$ | 3.2167 .
The value sought is the number of lattice points within the region bounded by the curve $y=\sqrt[4]{\frac{2007}{x}}$ and $x>0$, $y>0$.
Changing the column-wise counting to row-wise counting, the value sought is
$$
\begin{array}{l}
\sum_{i=1}^{6}\left[\frac{2007}{i^{4}}\right]=2007+125+24+7+3+1 \\
=2167
\end{a... | 2167 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,490 |
4. Let the function be
$$
\begin{array}{l}
f(x)=\sqrt{10-6 \cos x}+\sqrt{\frac{17}{8}-\frac{3 \sqrt{2}}{2} \sin x}+ \\
\sqrt{19-2 \sqrt{2} \cos x-8 \sin x} \text {. } \\
\end{array}
$$
For all real numbers $x$, the minimum value of $f(x)$ is
$\qquad$ . | 4. $\frac{21 \sqrt{2}}{4}-1$.
Rewrite $f(x)$ as
$$
\begin{array}{l}
\sqrt{(\cos x-3)^{2}+\sin ^{2} x}+ \\
\sqrt{\cos ^{2} x+\left(\sin x-\frac{3 \sqrt{2}}{4}\right)^{2}}+ \\
\sqrt{(\cos x-\sqrt{2})^{2}+(\sin x-4)^{2}} .
\end{array}
$$
Let $A(3,0)$, $B\left(0, \frac{3 \sqrt{2}}{4}\right)$, $C(\sqrt{2}, 4)$, and $P(\co... | \frac{21 \sqrt{2}}{4}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,491 |
5. Given an even function $f: \mathbf{Z} \rightarrow \mathbf{Z}$ that satisfies $f(1)=1$, $f(2007) \neq 1$, and for any integers $a, b$,
$$
f(a+b) \leqslant \max \{f(a), f(b)\} \text {. }
$$
Then the possible value of $f(2008)$ is $\qquad$ | 5.1.
Since $f(2) \leqslant \max \{f(1), f(1)\}=1$, assume for a positive integer $k$ greater than or equal to 2, we have $f(k) \leqslant 1$, then
$$
f(k+1) \leqslant \max \{f(k), f(1)\}=1 \text {. }
$$
Therefore, for all positive integers $n$, we have $f(n) \leqslant 1$.
Since $f$ is an even function, we can conclude... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,492 |
6. For a positive integer $n \geqslant 2007$, a complex number $z$ satisfies
$$
(a+1) z^{n+1}+a \text { i } z^{n}+a \text { i } z-(a+1)=0 \text {, }
$$
where the real number $a>-\frac{1}{2}$. Then the value of $|z|$ is $\qquad$ . | 6.1.
Given $z^{n}[(a+1) z+a \mathrm{i}]=a+1-a \mathrm{i} z$, then $|z|^{n}|(a+1) z+a \mathrm{i}|=|a+1-a \mathrm{i} z|$. Let $z=x+y \mathrm{i}$, where $x, y$ are real numbers. Then
$$
\begin{array}{l}
|(a+1) z+a \mathrm{i}|^{2}-|a+1-a \mathrm{i} z|^{2} \\
=(2 a+1)\left(|z|^{2}-1\right) .
\end{array}
$$
If $|z|>1$, sin... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,493 |
Three. (20 points) Given positive real numbers $a, b, c$ satisfying $ab + bc + ca = 1$. Prove:
$$
\frac{1}{\sqrt{a^{2}+1}}+\frac{2}{\sqrt{b^{2}+1}}+\frac{3}{\sqrt{c^{2}+1}}<\frac{3 \sqrt{14}}{2} .
$$ | $$
\begin{array}{l}
\text { Three, let } a=\tan \frac{A}{2}, b=\tan \frac{B}{2}, c=\tan \frac{C}{2} \\
(0<\angle A, \angle B, \angle C<\pi) .
\end{array}
$$
From $a b+b c+c a=1$, we get
$$
\angle A+\angle B+\angle C=\pi \text {. }
$$
Thus, $\cos A+\cos B+\cos C \leqslant \frac{3}{2}$.
Using the Cauchy-Schwarz inequal... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,494 |
Four. (20 points) Given a positive integer $n(n \geqslant 5)$, let the side lengths of rectangle $ABCD$ be $AB=2a, BC=2b (a>b)$. Points $P_{1}, P_{2}, \cdots, P_{n}=C$, $Q_{1}, Q_{2}, \cdots, Q_{n}=D$, $R_{0}=D, R_{1}, R_{2}, \cdots, R_{n}=C$ satisfy
$$
\begin{array}{c}
BP_{1}=P_{1}P_{2}=\cdots=P_{n-1}P_{n}=\frac{2b}{n... | (1) Establish a Cartesian coordinate system with $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $A P_{i}$ and $B R_{n-i}$ intersect at point $E$, and let the projections of $E$ and $R_{n-i}$ on $AB$ be $F$ and $G$, respectively, with $E(x, y)$.
From $\triangle A E F \backsim \triangle ... | a \sqrt{\frac{a^{2}+2 b^{2}}{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,495 |
Five. (20 points) Given $n(n \geqslant 3)$ is a positive integer. Prove: there exist $n$ points in the coordinate plane such that the distance between any two points is an irrational number, no three points are collinear, and the area of the triangle formed by any three points is an integer; the centroid of any $k(2 \l... | Five, $n$ points $P_{i}\left(a_{i}, b_{i}\right)(i=1,2, \cdots, n)$ satisfy the condition, where,
$$
a_{i}=(n+i)!, b_{i}=[(n+i)!]^{2} .
$$
In fact, for any two points $P_{i} 、 P_{j}$, we have
$$
\begin{aligned}
P_{i} P_{j}= & |(n+i)!-(n+j)!| \cdot \\
& \sqrt{1+[(n+i)!+(n+j)!]^{2}}
\end{aligned}
$$
is an irrational nu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,496 |
One, (50 points) Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $E$, and the extensions of sides $BA$ and $CD$ intersecting at point $F$. Let the circumcircles of $\triangle EAD$, $\triangle EBC$, $\triangle FAD$, and $\triangle FBC$ be $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$... | As shown in Figure 1.
Let the center of the circumcircle of quadrilateral $ABCD$ be $O$, connect $BQ$, intersecting $\odot O$ at point $K$, and connect $KA$. Then
$$
\angle AKB = \angle ACB = \angle ECB = \angle EQB.
$$
Thus, $KA \parallel PQ$.
Connect $PD$, intersecting $\odot O$ at point $L$, and connect $CL$. Then
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,497 |
II. (50 points) If there exist real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying
$$
\begin{array}{l}
a_{n} \leqslant a_{n-1} \leqslant \cdots \leqslant a_{1}<2, \\
a_{1}+a_{2}+\cdots+a_{n} \geqslant n, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant n^{2},
\end{array}
$$
Find all possible values of the positive ... | Let $a_{1}, a_{2}, \cdots, a_{i}>0, a_{i+1}, a_{i+2}, \cdots, a_{n} \leqslant 0, j=n-i$. Then
$$
\begin{array}{l}
2 i>a_{1}+a_{2}+\cdots+a_{i} \\
\geqslant n-a_{i+1}-a_{i+2}-\cdots-a_{n},
\end{array}
$$
i.e., $\square$
$$
\begin{array}{l}
i-j>-a_{i+1}-a_{i+2}-\cdots-a_{n} . \\
\text { Also, }\left(-a_{i+1}-a_{i+2}-\cd... | 1, 2, 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,498 |
Example 1 Use the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once each to form two numbers, such that one number is twice the other. Find all solutions that satisfy the condition. | Show a four-digit number. Thus, the original number $M=2N$, or
as an odd number, there must be $D=5$, which then leads to $H=$ being 7. But $C=$, so $H=2, 2 \times H$ should not have a carry,
this contradiction indicates $C \neq 7$. (3) If $D=7$, knowing $H \neq 8$, we have $H=\frac{7-1}{2}=3$.
as an odd number, there... | 6729 \times 2=13458, 6792 \times 2=13584, 6927 \times 2=13854, 7269 \times 2=14538, 7293 \times 2=14586, 7329 \times 2=14658, 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,499 |
Example 1 Given $\triangle A B C$, the necessary and sufficient condition for the perpendiculars drawn from points $A_{1}, B_{1}, C_{1}$ to the sides $B C, C A, A B$ respectively to be concurrent is
$$
A_{1} B^{2}-B C_{1}^{2}+C_{1} A^{2}-A B_{1}^{2}+B_{1} C^{2}-C A_{1}^{2}=0 .
$$ | Necessity.
As shown in Figure 2, let the perpendiculars from points $A_{1}$, $B_{1}$, and $C_{1}$ to sides $BC$, $CA$, and $AB$ respectively intersect at a point $M$, with the feet of the perpendiculars being $H_{1}$, $H_{2}$, and $H_{3}$.
By the theorem of equal difference of squares,
we have
$$
\begin{array}{l}
A_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,500 |
2. Construct squares outwardly on the four sides of quadrilateral $A B C D$, and let the centers of these four squares be $L$, $M$, $N$, and $K$. Prove that $L N \perp M K$. | It is known that $A K^{2}=\frac{1}{2} A D^{2}, A L^{2}=\frac{1}{2} A B^{2}$, $\angle K A L=90^{\circ}+\angle D A B$. By the cosine rule, we have
$$
\begin{array}{l}
K L^{2}=\frac{1}{2} A D^{2}+\frac{1}{2} A B^{2}+A D \cdot A B \sin A, \\
L M^{2}=\frac{1}{2} A B^{2}+\frac{1}{2} B C^{2}+A B \cdot B C \sin B, \\
M N^{2}=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,501 |
Five. (20 points) Given a moving point $P(1,1)$ inside the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$, a chord $A B$ of the ellipse is drawn through $P$. Prove: $|P A|, |P B|$ must have one that does not exceed
$$
\frac{1}{b} \sqrt{a^{2} b^{2}-a^{2}-b^{2}} .
$$ | Five, the parametric equation of the line on which the chord $AB$ lies is
$$
\left\{\begin{array}{l}
x=1+t \cos \alpha, \\
y=1+t \sin \alpha .
\end{array}\right.
$$
Substituting into the ellipse equation and simplifying, we get
$$
\begin{array}{l}
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2(b \c... | \frac{1}{b} \sqrt{a^{2} b^{2}-a^{2}-b^{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 719,502 |
One, (50 points) As shown in Figure $1, \triangle A B C$ has a circumcircle $\odot O$. The tangents through $B$ and $C$ intersect at point $D$. Connect $A D$. A line through $A$ perpendicular to $A D$ intersects the extension of $B C$ at point $E$. Connect $E D$. Take two points $B_{1}$ and $C_{1}$ on line $D E$ such t... | As shown in Figure 3, take the midpoint $M$ of $BC$, and connect $AM$ and $MD$. Then
$MD \perp BC$.
Therefore, $M, D, E, A$ are concyclic.
First, we prove:
$$
\angle BAM = \angle DAC.
$$
Notice that
$$
\begin{array}{c}
\frac{\sin \angle BAD}{\sin \angle DAC} = \frac{\frac{BD}{AD} \sin \angle ABD}{\frac{CD}{AD} \sin \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,503 |
II. (50 points) $a>1, a \in \mathbf{Z}$, given $a_{1}=1$, $a_{2}=a, a_{n+2}=a a_{n+1}-a_{n}(n \geqslant 1)$. Prove: For any $n, m \in \mathbf{Z}_{+}, a_{(m, n)}=\left(a_{m}, a_{n}\right)$. Where, $(m, n)$ represents the greatest common divisor of $m$ and $n$. | Second, first prove:
$$
a_{n+m+1}=a_{n+1} a_{m+1}-a_{n} a_{m}(n 、 m \geqslant 0) \text {. }
$$
Let $a_{0}=0$.
Proceed by induction on $m$.
When $m=0,1$, equation (1) clearly holds.
Assume that equation (1) holds for all numbers less than $m$, consider $m$. By the induction hypothesis,
$$
\begin{array}{l}
a_{n+m-1}=a_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,504 |
$\begin{array}{l} \text { Three. (50 points) } A_{1}, A_{2}, \cdots, A_{n} \text { are finite sets, satisfying } \\ \left|A_{i} \cap A_{i+1}\right|>\frac{n-2}{n-1}\left|A_{i+1}\right|(i=1,2, \cdots, n), \\ A_{n+1}=A_{1} . \text { Prove: } \bigcap_{i=1}^{n} A_{i} \neq \varnothing .\end{array}$ | Three, Proof by Contradiction.
Let $I=\bigcup_{i=1}^{n} A_{i}$, for any $\alpha \in I$, by the contradictory assumption, there must exist an $i$ such that $\alpha \notin A_{i-1}, \alpha \in A_{i}$, we say $\alpha$ has the “$i$ property”.
Suppose for each $i$ there are $a_{i}$ elements $\alpha_{i}$ that have the $i$ pro... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,505 |
Let $k$ be a given positive integer. Try to find all real numbers $a$, such that the quadratic equation $a x^{2}+a x+a+2006=0$ has two integer roots, and the absolute value of the difference between the two roots is not less than $2 k-1$.
---
The translation maintains the original text's format and line breaks as req... | Solution: Let the two integer roots of $a x^{2}+a x+a+2006=0$ be $x_{1}$ and $x_{2}$. Then
$$
a x_{1}^{2}+a x_{1}+a+2006=0.
$$
Thus, $a=-\frac{2006}{x_{1}^{2}+x_{1}+1}=-\frac{2006}{t^{2}+t+1}(t \in \mathbf{N})$.
Conversely, if $a=-\frac{2006}{t^{2}+t+1}(t$ is any integer), then the original equation can be rewritten a... | a=-\frac{2006}{t^{2}+t+1}(t=k-1, k, \cdots) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,506 |
220 As shown in the figure
$3, D$ is a point on side $BC$ of $\triangle ABC$, $\frac{BD}{DC} = \frac{m}{n}, P$ is any point on $AD$, and circles $\odot O_{1}$ and $\odot O_{2}$ are drawn through points $A, P$.
$\odot O_{1}$ intersects $AB, AC$ at points $M, N$ respectively, and $\odot O_{2}$ intersects $AB, AC$ at poin... | Prove: As shown in Figure 3, construct a circle passing through points $B$, $M$, and $P$, intersecting the extension of $AP$ at point $Q$. Draw $BK \parallel AC$, intersecting the extension of $AP$ at point $K$. Connect $BQ$, $MP$, and $PN$. Then,
$$
AM \cdot AB = AP \cdot AQ.
$$
Since $\angle BQK = \angle BMP = \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,507 |
Let $\left\{a_{n}\right\}$ be a positive arithmetic sequence with common difference $d(d>0)$, $n \in \mathbf{N}_{+}$. Prove that:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant \frac{n}{d}\left[\left(\frac{a_{n+1}}{a_{1}}\right)^{\frac{1}{n}}-1\right] .
$$ | Prove: When $n=1$,
Left side $=\frac{1}{a_{1}}$,
Right side $=\frac{1}{d}\left(\frac{a_{2}}{a_{1}}-1\right)=\frac{1}{d}\left(\frac{a_{1}+d}{a_{1}}-1\right)=\frac{1}{a_{1}}$.
Therefore, when $n=1$, the original inequality holds.
When $n>1$, note that $a_{n+1}>a_{n-1}>0$, then
$$
a_{n-1} a_{n+1}n \sqrt[n]{\frac{a_{2}}{a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,508 |
As shown in Figure 4, from a point $P$ outside $\odot O$, two tangents $PA$ and $PB$ are drawn to $\odot O$, touching the circle at points $A$ and $B$. From point $P$, two secants $PCD$ and $PEF$ are also drawn, intersecting $\odot O$ at points $C$, $D$, $E$, and $F$. The chords $CF$ and $DE$ intersect at point $G$. Pr... | Prove: Draw a secant line $PG$ through point $G$ intersecting $\odot O$ at points $M$ and $N$, and connect $AG$, $PO$. It is easy to see that
$\angle PGE > \angle PDE = \angle PFC$.
Take a point $R$ on $GN$, connect $FR$, such that $\angle RFP = \angle PGE$, and connect $AR$.
Thus, $E, F, R, G$ are concyclic.
Then $PG ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,509 |
3. Given a cyclic quadrilateral $ABCD$, $BA$ and $CD$ intersect at point $P$, $AD$ and $BC$ intersect at point $Q$, $AC$ and $BD$ intersect at point $M$. Prove: The center $O$ of the circle is the orthocenter of $\triangle PQM$. | (提示: As shown in Figure 11, $\angle Q M D > \angle C B D = \angle D A M$. Extend $Q M$ to point $F$ such that $\angle F A D = \angle Q M D$. Let the radius of $\odot O$ be $r$, then $A$, $D$, $M$, and $F$ are concyclic. It is easy to know that $B$, $F$, $D$, and $Q$ are also concyclic. By the power of a point theorem, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,510 |
Proposition Let points $B$ and $C$ lie on two rays emanating from point $A$. Then the circumcircle of $\triangle ABC$ always passes through a fixed point $D$ inside $\angle A$ $\left(0^{\circ}<\angle A<180^{\circ}\right)$, independent of points $B$ and $C$, if and only if there exist positive real numbers $\lambda_{1}$... | Proof: Necessity.
As shown in Figure 1, if the circumcircle of $\triangle ABC$ always passes through a fixed point $D$ inside $\angle A$ that is independent of points $B$ and $C$, then $\angle BAD$, $\angle CAD$, $\angle BAC$, and line segment $AD$ are all constants.
By Ptolemy's theorem, we have
$$
\begin{array}{l}
AB... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,511 |
Example 2 As shown in Figure 3, given $\triangle A B C$ and lines $l_{1}, l_{2}$, points $D$, $E$, and $F$ are on sides $B C$, $C A$, and $A B$ respectively, and satisfy $D E \parallel l_{1}$, $D F \parallel l_{2}$. Prove: The circumcircle of $\triangle A E F$ passes through a fixed point $P(P \neq A)$. | Analysis: As shown in Figure 3, it is easy to see that the movement of points $E$ and $F$ causes the circumcircle of $\triangle AEF$ to move. Therefore, the essence of the problem is to find the constraint condition between segments $AE$ and $AF$.
Proof: It is easy to know that $\angle BDF$, $\angle BFD$, $\angle CDE$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,512 |
Example 3 Given a convex quadrilateral $A B C D, B C=\lambda A D$ ( $\lambda$ is a positive constant), and $B C$ is not parallel to $A D$. Let points $E$ and $F$ be on the interiors of sides $B C$ and $A D$, respectively, and satisfy $B E=$ $\lambda D F$. Line $A C$ and $B D$ intersect at point $P$, line $E F$ and $B D... | Proof: Let the lengths of segments $PA, PB, PC, PD$ be $a, b, c, d$ (constants), and the lengths of segments $PQ, PR$ be $x, y$.
As shown in Figure 4, first consider the case where points $R, Q$ lie on the interiors of segments $PA, PB$ respectively.
From the given conditions, it is easy to see that
$$
\frac{BC}{AD}=\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,513 |
Example 4 Given a convex quadrilateral $A B C D$, the ray $B A$ and $C D$ intersect at point $P, E, F$ are moving points on sides $A D, B C$ respectively, and satisfy $\frac{A E}{E D}=\frac{B F}{F C}$. The ray $F E$ intersects $B A, C D$ at points $Q, R$ respectively. Prove: the circumcircle of $\triangle P Q R$ passes... | Proof: As shown in Figure 5, connect $R A$, $R B$, $Q C$, and $Q D$. Let $P Q=x$, $P R=y$. The lengths of line segments $P A$, $P B$, $P C$, and $P D$ are denoted as $a$, $b$, $c$, and $d$, respectively. It is easy to see that $a$, $b$, $c$, and $d$ are constants, and $b>a$, $c>d$. Using the area ratio theorem, we get
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,514 |
Example 1 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 .
\end{array}
$$
(2003, USA Mathematical Olympiad) | Analysis: Let's assume $a+b+c=1$. Then the original inequality transforms into
$$
\begin{array}{l}
\frac{(1+a)^{2}}{2 a^{2}+(1-a)^{2}}+\frac{(1+b)^{2}}{2 b^{2}+(1-b)^{2}}+ \\
\frac{(1+c)^{2}}{2 c^{2}+(1-c)^{2}} \leqslant 8 .
\end{array}
$$
Let $f(x)=\frac{(1+x)^{2}}{2 x^{2}+(1-x)^{2}}$
$$
=\frac{1}{3}\left[1+\frac{8 x... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false | 719,515 |
Example 2 Let $n$ be a positive integer, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$. Prove:
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \sum_{j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2} \\
\leqslant \frac{2\left(n^{2}-1\right)}{3} \sum_{i=1}^{n} \sum_{j=... | Analysis: Notice that
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \sum_{j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2}=\left[2 \sum_{1<i<j<n}\left(x_{j}-x_{i}\right)\right]^{2}, \\
\frac{2\left(n^{2}-1\right)}{3} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(x_{i}-x_{j}\right)^{2} . \\
=\frac{4\left(n^{2}-1\right)}{3} \sum_{1<i<j<n}\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,516 |
Example 3 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$
\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 \text{. }
$$
(41st IMO) | Analysis: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}(x, y, z \in$ $\mathbf{R}_{+}$). Then, it suffices to prove
$$
(x-y+z)(y-z+x)(z-x+y) \leqslant x y z \text {. }
$$
Let $u=x-y+z, v=y-z+x, w=z-$ $x+y$. Then
$$
x=\frac{u+v}{2}>0, y=\frac{v+w}{2}>0, z=\frac{w+u}{2}>0 .
$$
Thus, equation (1)
$$
\Leftrightarrow 8 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,517 |
Example 4 Let $a, b, c \in \mathbf{R}_{+}$, and $a+b+c=1$. Prove:
$$
\begin{array}{l}
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \\
\geqslant \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} .
\end{array}
$$
(29th Russian Mathematical Olympiad) | Analysis: Substituting the conditions, the original inequality can be transformed into: Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\begin{array}{l}
\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \\
\geqslant \frac{2}{2 a+b+c}+\frac{2}{a+2 b+c}+\frac{2}{a+b+2 c} .
\end{array}
$$
From $\frac{1}{x}+\frac{1}{y} \geqslant \fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,518 |
Example 2 As shown in Figure 3, in $\triangle A B C$, $A B=A C$, $D$ is the midpoint of $B C$, $D E$ $\perp A C, F$ is the midpoint of $D E$. Prove: $A F \perp B E$. | $$
\begin{array}{c}
\text { Explanation: Let } AB= \\
a, \angle ABC=\alpha. \text { Then } \\
\angle ADF \\
=\angle ADE=\alpha.
\end{array}
$$
Thus, $AD=a \sin \alpha, BD=a \cos \alpha$,
$$
\begin{array}{l}
AE=AD \sin \alpha=a \sin ^{2} \alpha, \\
DF=FE=\frac{1}{2} DE=\frac{1}{2} DC \sin \alpha \\
=\frac{1}{2} a \sin ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,519 |
Example 5 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (36th IMO) | Analysis: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}(x, y, z \in$ $\mathbf{R}_{+}$). Then, the original problem is transformed into:
Given $x, y, z \in \mathbf{R}_{+}$, and $x y z=1$. Prove:
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geqslant \frac{3}{2} \text {. }
$$
Proof 1: The above inequality... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,520 |
Example 6 Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real roots $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.
(1) Prove: $\frac{2 a^{3}+27 c-9 a b}{\cdot \lambda^{3}}=0$ if and only if $x_{1}, x_{2}, ... | Analysis: The structure of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$ is relatively complex and difficult to solve problems (1) and (2) directly. Recalling the method for solving a cubic equation, we first make a substitution, let $x=y-\frac{a}{3}$, and get $g(y)=y^{3}+p y+q$.
Thus, the original problem can be transforme... | \frac{3 \sqrt{3}}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 719,521 |
Example 7 Given $a<b$. Prove:
$$
a^{3}-3 a \leqslant b^{3}-3 b+4 \text {. }
$$ | Analysis: Let $b=a+t(t>0)$. It is sufficient to prove
$$
t^{3}+3 a t^{2}+3\left(a^{2}-1\right) t+4 \geqslant 0 \text {. }
$$
Taking $a$ as the main variable, we need to prove
$$
3 t a^{2}+3 t^{2} a+\left(t^{3}-3 t+4\right) \geqslant 0 \text {. }
$$
And $\Delta=-3 t\left(t^{3}-12 t+16\right)$, so we only need to prove... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,522 |
Example 8 The sequences $a_{0}, a_{1}, \cdots$ and $b_{0}, b_{1}, \cdots$ are defined as follows:
$$
\begin{array}{l}
a_{0}=\frac{\sqrt{2}}{2}, a_{n+1}=\frac{\sqrt{2}}{2} \sqrt{1-\sqrt{1-a_{n}^{2}}}(n=0,1, \cdots) ; \\
b_{0}=1, b_{n+1}=\frac{\sqrt{1+b_{n}^{2}}-1}{b_{n}}(n=0,1, \cdots) .
\end{array}
$$
Prove that for e... | Analysis: Clearly, $0<a_{n} \leqslant \frac{\sqrt{2}}{2}, 0<b_{n} \leqslant 1$. Let
$$
\begin{array}{l}
a_{n}=\sin \beta_{n}\left(0<\beta_{n} \leqslant \frac{\pi}{4}\right), \\
b_{n}=\tan \gamma_{n}\left(0<\gamma_{n} \leqslant \frac{\pi}{4}\right) .
\end{array}
$$
Then $a_{n+1}=\sin \frac{\beta_{n}}{2}$,
$$
b_{n+1}=\f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,523 |
1. Let $\triangle ABC$ be an acute triangle with unequal side lengths, and let $O$ be its circumcenter. Point $A'$ lies on the extension of line segment $AO$ such that $\angle B A' A = \angle C A' A$. Draw $A' A_1 \perp AC$ and $A' A_2 \perp AB$, with the feet of the perpendiculars being $A_1$ and $A_2$, respectively. ... | 1. First, it is evident that points $A^{\prime}, B, O, C$ are concyclic.
In fact, as shown in Figure 1,
construct the circumcircle of $\triangle B O C$; let it intersect $A O$ at point $P$ (different from point $A^{\prime}$). Then
$$
\begin{array}{l}
\angle B P A \\
=\angle B C O \\
=\angle C B O \\
=\angle C P A .
\e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,526 |
2. Given an integer $n(n \geqslant 3)$. Prove: The set $X=$ $\left\{1,2, \cdots, n^{2}-n\right\}$ can be written as the union of two non-empty, disjoint subsets, such that each subset does not contain $n$ elements $a_{1}, \ddot{a}_{2}, \cdots, a_{n}, a_{1}<a_{2}<\cdots<a_{n}$, satisfying
$$
a_{k} \leqslant \frac{a_{k-1... | 2. Definition: $S_{k}=\left\{k^{2}-k+1, \cdots, k^{2}\right\}$,
$T_{k}=\left\{k^{2}+1, \cdots, k^{2}+k\right\}$,
$k=1,2, \cdots, n-1$.
Let $S=\bigcup_{k=1}^{n-1} S_{k}, T=\bigcup_{k=1}^{n-1} T_{k}$.
We will prove that $S$ and $T$ are the two subsets that satisfy the requirements of the problem.
First, $S \cap T=\varnot... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,527 |
3. Given a positive integer $n$, and real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant$ $x_{n}, y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{n}$, satisfying
$$
\sum_{i=1}^{n} i x_{i} \geqslant \sum_{i=1}^{n} i y_{i} .
$$
Prove: For any real number $\alpha$, we have
$$
\sum_{i=1}^{n} x_{i}[i \alpha]... | 3. First, prove a lemma.
Lemma For any real number $\alpha$ and positive integer $n$, we have
$$
\sum_{i=1}^{n-1}[i \alpha] \leqslant \frac{n-1}{2}[n \alpha].
$$
Proof of the lemma: It suffices to sum $[i \alpha]+[(n-i) \alpha] \leqslant[n \alpha]$ for $i=1,2, \cdots, n-1$.
Back to the original problem.
We use induct... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,528 |
4. Let $A$ be an infinite subset of the set of positive integers, and let $n (n >$ 1) be a given integer. It is known that for any prime $p$ that does not divide $n$, the set $A$ contains infinitely many elements not divisible by $p$. Prove: For any integer $m (m > 1), (m, n) = 1$, the set $A$ contains finitely many di... | 4. Let $p^{\alpha} . \| m$, then the set $A$ has an infinite subset $A_{1}$, all of whose elements are not divisible by $p$.
By the pigeonhole principle, the set $A_{1}$ has an infinite subset $A_{2}$, all of whose elements are congruent to $a \mod m n$, where $a$ is a number not divisible by $p$.
Since $(m, n)=1$, we... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,529 |
Example 3 Given that $Q$ is a point on the circle with diameter $AB$, $Q \neq A, B$, and the projection of $Q$ on $AB$ is $H$. The circle with center $Q$ and radius $QH$ intersects the circle with diameter $AB$ at points $C, D$. Prove: $CD$ bisects the segment $QH$.
(2006, Turkish National Team Selection Exam) | Explanation: As shown in Figure 4, let
$$
O Q=1, Q H=d \leqslant 1 \text {. }
$$
Since $O Q$ is the line connecting the centers of $\odot O$ and $\odot Q$, we have
$$
\begin{array}{l}
O Q \perp C D . \\
\text { Therefore, } O E+E Q \\
=1, \\
O E^{2}-E Q^{2}=O D^{2}-D Q^{2}=1-d^{2} .
\end{array}
$$
From equations (1) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,530 |
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