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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Three, (15 points) Let real numbers $a, b, c$ satisfy $a+b+c = 3$. Prove:
$$
\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4} \text {. }
$$
(Wang Jianwei provided) | Three, if $a, b, c$ are all less than $\frac{9}{5}$, then it can be proven that:
$$
\frac{1}{5 a^{2}-4 a+11} \leqslant \frac{1}{24}(3-a) \text {. }
$$
In fact,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow(3-a)\left(5 a^{2}-4 a+11\right) \geqslant 24 \\
\Leftrightarrow 5 a^{3}-19 a^{2}+23 a-9 \leqslant 0 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,645 |
Four. (15 points) Let $O$ be a point inside $\triangle ABC$. Prove: There exist positive integers $p, q, r$, such that
$$
|p O A+q O B+r O C|<\frac{1}{2007} .
$$ | Proof 1: First, we prove a lemma.
Lemma Let $\alpha, \beta$ be positive real numbers, and $N$ be any integer greater than $\max \left\{\frac{1}{\alpha}, \frac{1}{\beta}\right\}$. Then there exist positive integers $p_{1}, p_{2}, q$, such that $1 \leqslant q \leqslant N^{2}$, and
$$
\left|q \alpha-p_{1}\right||q \alpha|... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,646 |
Five. (15 points) Does there exist a triangle with integer side lengths such that the shortest side is 2007 and the largest angle is twice the smallest angle?
(Zhao Jiyuan provided) | Five, such a triangle does not exist.
Assume $\angle A \leqslant \angle B \leqslant \angle C$. Then $\angle C=2 \angle A$, and $a=2007$.
Draw the angle bisector $CD$ of $\angle ACB$ from point $C$, then $\angle BCD=\angle A$. Combining $\angle B=\angle B$, we know $\triangle CDB \sim \triangle ACB$. Therefore,
$$
\begi... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,647 |
Six. (15 points) Find all positive integers $n$, such that there exist non-zero integers $x_{1}, x_{2}, \cdots, x_{n}, y$, satisfying
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=0, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n y^{2} .
\end{array}\right.
$$
(Xiong Bin provided) | Obviously, $n \neq 1$.
When $n=2 k$ is even, let
$$
x_{2 i-1}=1, x_{2 i}=-1, i=1,2, \cdots, k, y=1 \text {, }
$$
then the conditions are satisfied.
When $n=3+2 k\left(k \in \mathbf{N}_{+}\right)$, let
$$
\begin{array}{l}
y=2, x_{1}=4, x_{2}=x_{3}=x_{4}=x_{5}=-1, \\
x_{2 i}=2, x_{2 i+1}=-2, i=3,4, \cdots, k+1,
\end{arr... | all positive integers except 1 and 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,648 |
Seven, (15 points) Let $P$ be a point inside an acute $\triangle ABC$, and $AP, BP, CP$ intersect sides $BC, CA, AB$ at points $D, E, F$ respectively. Given that $\triangle DEF \sim \triangle ABC$. Prove that $P$ is the centroid of $\triangle ABC$.
(Feng Zhigang provided) | Proof 1: Let $\angle E D C=\alpha, \angle A E F=\beta$, $\angle B F D=\gamma, \angle A, \angle B, \angle C$ be the three interior angles of $\triangle A B C$. Then
$$
\begin{array}{l}
\angle A F E=2 \angle B-(\angle D B E+\angle D E B) \\
=2 \angle B-\alpha .
\end{array}
$$
Similarly, $\angle B D F=2 \angle C-\beta$,
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,649 |
Eight. (15 points) Place $n$ white stones and $n$ black stones arbitrarily on a circle. Starting from a certain white stone, label the white stones $1, 2, \cdots, n$ in a clockwise direction. Then, starting from a certain black stone, label the black stones $1, 2, \cdots, n$ in a counterclockwise direction. Prove: Ther... | Eight, take one black and one white chess piece with the same label, so that the number of other points (excluding endpoints, regardless of color) in the minor arc determined by this pair of points is the least. Let's assume this label is 1.
In the aforementioned open minor arc, there is only one color of chess pieces... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 719,650 |
Example 5 Find the smallest real number $m$, such that for any positive numbers $a, b, c$ satisfying $a+b+c=1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(3rd China Southeast Mathematical Olympiad) | Explanation: First find the minimum value of $m$, then provide a proof.
When $a=b=c=\frac{1}{2}$, from the given inequality we have
$$
m \geqslant 27.
$$
Next, we prove that the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1,
$$
holds for any positive real numbers $a, b, c$... | 27 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,651 |
1. If point $P(x, y)$ moves on the line $x+3 y=3$, then the minimum value of the function $f(x, y)=3^{x}+9^{y}$ is equal to ( ).
(A) $5\left(\frac{27}{4}\right)^{\frac{1}{5}}$
(B) $7\left(\frac{27}{9}\right)^{\frac{1}{7}}$
(C) $7\left(\frac{16}{9}\right)^{\frac{1}{7}}$
(D) $3\left(\frac{5}{2}\right)^{\frac{1}{3}}$ | -1. A.
$$
\begin{array}{l}
f(x)=3^{x}+9^{y}=3^{x}+9^{\frac{3-x}{3}} \\
=3^{x}+3^{2\left(1-\frac{x}{3}\right)}=3^{x}+3^{2-\frac{2}{3} x} \\
=\frac{1}{2} \cdot 3^{x}+\frac{1}{2} \cdot 3^{x}+3^{1-\frac{2}{3} x}+3^{1-\frac{2}{3} x}+3^{1-\frac{2}{3} x} \\
\geqslant 5 \cdot \sqrt[5]{\frac{1}{2} \cdot 3^{x} \cdot \frac{1}{2} ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,652 |
2. The positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+3}+\sqrt{x+2007}$ ( ).
(A) There is only one pair
(B) There are exactly two pairs
(C) There are at least three pairs
(D) None exist | 2.B.
Let $a^{2}=x+3, b^{2}=x+2007$, where $a, b$ are natural numbers. Then
$$
\begin{array}{l}
y=a+b, \\
b^{2}-a^{2}=(b-a)(b+a) \\
=2004=2^{2} \times 3 \times 167 .
\end{array}
$$
Since $b+a$ and $b-a$ have the same parity, and $b+a > b-a$, we have
$$
\left\{\begin{array} { l }
{ b + a = 1 0 0 2 , } \\
{ b - a = 2 }... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,653 |
3. Let the sets be
$$
M=\{-2,0,1\}, N=\{1,2,3,4,5\},
$$
The mapping $f: M \rightarrow N$ ensures that for any $x \in M$, $x + f(x) + x f(x)$ is an odd number. The number of such mappings $f$ is ( ).
(A) 45
(B) 27
(C) 15
(D) 11 | 3. A.
When $x=-2$, $x+f(x)+x f(x)=-2-f(-2)$ is odd, then $f(-2)$ can take $1, 3, 5$, which gives three possibilities; when $x=0$, $x+f(x)+x f(x)=f(0)$ is odd, then $f(0)$ can take $1, 3, 5$, which gives three possibilities; when $x=1$, $x+f(x)+x f(x)=1+2 f(1)$ is odd, then $f(1)$ can take $1, 2, 3, 4, 5$, which gives ... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,654 |
4. The curve represented by the equation $\frac{x^{2}}{\sin \left(19^{2 \infty}\right)^{\circ}}+\frac{y^{2}}{\cos \left(19^{2 \infty}\right)^{\circ}}=1$ is ( ).
(A) Hyperbola
(B) Ellipse with foci on the $x$-axis
(C) Ellipse with foci on the $y$-axis
(D) None of the above answers is correct | 4.C.
$$
\begin{array}{l}
19^{2007}=19 \times\left(19^{2}\right)^{1003}=19 \times(360+1)^{1003} \\
=19 \times(360 n+1)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Therefore, $\sin \left(19^{2007}\right)^{\circ}=\sin (360 \times 19 n+19)^{\circ}=\sin 19^{\circ}$.
Similarly, $\cos \left(19^{2007}\right)^{\circ}=\co... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,655 |
5. Reverse the order of the three digits of a three-digit number, and add the resulting number to the original number. If none of the digits in the sum is even, then the number is called a "Odd Sum Number". Then, among all three-digit numbers, the number of Odd Sum Numbers is ( ).
(A) 100
(B) 120
(C) 160
(D) 200 | 5.A.
Let the three-digit number be $\overline{a_{1} a_{2} a_{3}}$. Then
$$
\begin{array}{l}
\overline{a_{1} a_{2} a_{3}}+\overline{a_{3} a_{2} a_{1}} \\
=100\left(a_{1}+a_{3}\right)+10\left(a_{2}+a_{2}\right)+\left(a_{1}+a_{3}\right) .
\end{array}
$$
If $a_{1}+a_{3}$ does not carry over, then the tens digit of the su... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 719,656 |
6. Let $a_{1}=6, a_{n+1}=\left[\frac{5}{4} a_{n}+\frac{3}{4} \sqrt{a_{n}^{2}-2}\right]$ $\left(n \in \mathbf{N}_{+}\right)$, where $[x]$ denotes the greatest integer not exceeding $x$. Then the units digit of $a_{1}+a_{2}+\cdots+a_{2000}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6.B.
Given $a_{1}=6=5 \times 2^{1-1}+1, a_{2}=11=5 \times 2^{2-1}+1, \cdots \cdots$, we conjecture: $a_{n}=5 \times 2^{n-1}+1$.
From the known recurrence relation, it is easy to prove by mathematical induction (omitted).
Thus, when $n>1$, $a_{n} \equiv 1(\bmod 10)$.
Therefore, $a_{1}+a_{2}+\cdots+a_{2007} \equiv 6+20... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,657 |
1. Given three positive integers $x, y, z$ whose least common multiple is 300, and $\left\{\begin{array}{l}x+3 y-2 z=0, \\ 2 x^{2}-3 y^{2}+z^{2}=0\end{array}\right.$. Then the solution to the system of equations $(x, y, z)=$ $\qquad$ . | 2. $(20,60,100)$.
Transform $x+3 y-2 z=0$ and substitute into $2 x^{2}-3 y^{2}+z^{2}=0$, eliminating $x$ we get
$$
5 y^{2}-8 y z+3 z^{2}=0 \text {, }
$$
which is $(5 y-3 z)(y-z)=0$.
Thus, $5 y-3 z=0$,
or $y-z=0$.
From equation (1) and $x+3 y-2 z=0$, we get $y=3 x, z=5 x$, i.e., $x: y: z=1: 3: 5$.
Therefore, by the gi... | (20,60,100) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,658 |
3. Let vectors $a, b, x, y$ in the plane satisfy the relations $a=x-y, b=2x+y$. Suppose the magnitudes of $a$ and $b$ are both 1, and they are perpendicular to each other. Then the angle between $\boldsymbol{x}$ and $\boldsymbol{y}$ is $\qquad$ | 3. $\pi-\operatorname{arcos} \frac{\sqrt{10}}{10}$.
Given $x=\frac{a+b}{3}, y=\frac{b-2 a}{3}$.
Let the angle between $\boldsymbol{x}$ and $\boldsymbol{y}$ be $\theta$, then $\cos \theta=\frac{\boldsymbol{x} \cdot \boldsymbol{y}}{|\boldsymbol{x}| \cdot|\boldsymbol{y}|}=-\frac{\sqrt{10}}{10}$.
Therefore, $\theta=\pi-\o... | \pi-\operatorname{arcos} \frac{\sqrt{10}}{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,660 |
Example 6 Let $a, b, c$ be positive numbers, and let $d$ be the minimum of
$$
(a-b)^{2}, (b-c)^{2}, (c-a)^{2}
$$
(1) Prove that there exists $\lambda(0<\lambda<1)$, such that
$$
d \leqslant \lambda\left(a^{2}+b^{2}+c^{2}\right) \text {; }
$$
(2) Find the smallest positive number $\lambda$ for which the inequality (1) ... | (1) From the definition of $d$, we know that
$$
d \leqslant(a-b)^{2}, d \leqslant(b-c)^{2}, d \leqslant(c-a)^{2}.
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
3 d \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
=2\left(a^{2}+b^{2}+c^{2}\right)-2 a b-2 b c-2 c a \\
0 \text { and } d=(b-c)^{2}.
\end{arra... | \frac{1}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,662 |
5. Given a unit cube $A B C D-E F C H$ with moving points $Q 、 P$ on edge $A D$ and line $B C$ respectively. If the line segment $M N$ obtained by the intersection of $\triangle P Q G$ and $\triangle B D E$ has a length of $y$, and $A Q$ $=x(0 \leqslant x \leqslant 1)$. Then the minimum value of $y$ expressed as a func... | 5. $y=\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3-x}$.
When $A Q=x$, let $G Q$ intersect the plane $B D E$ at point $N$, draw $N M \perp B D$ at point $M$, connect $Q M$ to intersect line $B C$ at point $P^{\prime}$, and take point $P^{\prime}$ as $P$. It is known that at this time $y=|M N|$ is minimized.
Establish a spatia... | y=\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3-x} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,663 |
6. Let $a_{1}, a_{2}, \cdots, a_{2007}$ be positive real numbers, and
$$
\frac{1}{2+a_{1}}+\frac{1}{2+a_{2}}+\cdots+\frac{1}{2+a_{200}}=\frac{1}{2} \text {. }
$$
Then the minimum value of $a_{1} a_{2} \cdots a_{200}$ is $\qquad$ | $6.4012^{2007}$.
Let $x_{i}=\frac{2}{2+a_{i}}$, then
$a_{i}=2 \cdot \frac{1-x_{i}}{x_{i}}$, and $\sum_{i=1}^{200} x_{i}=1$.
Thus, $a_{1} a_{2} \cdots a_{200}$
$$
\begin{array}{l}
=2^{2000} \cdot \frac{1}{x_{1} x_{2} \cdots x_{200}} \prod_{i=1}^{2007}\left(\sum_{j \neq i} x_{j}\right) \\
\geqslant 2^{2000} \cdot \frac{1... | 4012^{200} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,664 |
Three. (20 points) As shown in Figure 1, given that $\odot O_{1}$ and $\odot O_{2}$ intersect at two different points $A$ and $B$, points $P$ and $E$ are on $\odot O_{1}$, points $Q$ and $F$ are on $\odot O_{2}$, and $EF$ is a common tangent of the two circles. $PQ \parallel EF$, and $PE$ intersects $QF$ at point $R$. ... | $$
\begin{array}{l}
F Q=Y, P Q \cap E O_{1}=N_{1}, P Q \cap F O_{2}=N_{2}, B G \\
\cap E O_{1}=M_{1}, B H \cap F O_{2}=M_{2}, E M_{1}=F M_{2}= \\
a, E N_{1}=F N_{2}=b .
\end{array}
$$
It is easy to know that $E G=\sqrt{2 r_{1} a}, E P=\sqrt{2 r_{1} b}$,
$$
E X=\frac{a}{b} E P=\sqrt{\frac{a}{b}} \cdot \sqrt{2 r_{1} a} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,665 |
Four. (20 points) Given that the three sides of $\triangle A B C$ are $a, b, c$, and they satisfy
$$
a b c=2(a-1)(b-1)(c-1) .
$$
(1) Does there exist a $\triangle A B C$ with all sides being integers? If so, find the side lengths; if not, explain the reason.
(2) If $a>1, b>1, c>1$, find the minimum perimeter of $\trian... | (1) Suppose $a \geqslant b \geqslant c$. Clearly, $c \geqslant 2$.
If $c \geqslant 5$, then $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geqslant\left(... | \frac{3 \sqrt[3]{2}}{\sqrt[3]{2}-1} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,667 |
Five, it is known that the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the fixed point $A(1, 0)$, and its foci are on the $x$-axis. The ellipse intersects the curve $|y|=x$ at points $B$ and $C$. There is a parabola with $A$ as its focus, passing through points $B$ and $C$, and opening to the lef... | Five, the ellipse passes through the fixed point $A(1,0)$, then
$a=1, c=\sqrt{1-b^{2}}, e=\sqrt{1-b^{2}}$.
Since $\frac{2}{3}1,00, m>1) \text {. }
$$
Since $\frac{p}{2}=m-1$, then,
$$
\left\{\begin{array}{l}
f(0)=-4 m(m-1)0 .
\end{array}\right.
$$
Therefore, $\left\{\begin{array}{l}m>1 \text { or } m1) \text {. }
$$
... | m>1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,668 |
1. Given that $x, y, z$ are not all equal real numbers, and $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k$. Then all possible values of $k$ are ( ).
(A) $\pm 1$
(B) $\pm \frac{1}{2}$
(C) $\pm 2$
(D) cannot be determined | -、1.A.
Given $x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k$, we have
$$
\begin{array}{l}
\frac{1}{x}=\frac{y}{k y-1}, \\
z=\frac{1}{k-y} .
\end{array}
$$
Substituting equations (1) and (2) into $z+\frac{1}{x}=k$, we get
$$
\begin{array}{l}
\frac{1}{k-y}+\frac{y}{k y-1}=k \\
\Rightarrow k y-1+y(k-y)=k(k-y)(k y-1) \\
\Ri... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,669 |
2. As shown in Figure 1, in the square $A_{1} A_{2} A_{3} A_{4}$ with side length 1, points $A_{5}$ and $A_{6}$ are taken on the sides $A_{2} A_{3}$ and $A_{3} A_{4}$ respectively, such that $A_{3} A_{5}=A_{3} A_{6}=p$, and the lines $A_{1} A_{2}$ and $A_{4} A_{5}$, $A_{1} A_{4}$ and $A_{2} A_{6}$, $A_{1} A_{5}$ and $A... | 2. D
Let $A_{3} B_{3}=A_{3} B_{4}=q, A_{4} B_{2}=r$.
Since $\triangle B_{3} A_{3} A_{5} \sim \triangle B_{3} A_{4} A_{1}$, we have $\frac{q}{p}=\frac{q+1}{1}$.
Thus, $q=\frac{p}{1-p}$.
Also, since $\triangle B_{2} A_{4} A_{6} \sim \triangle B_{2} A_{1} A_{2}$, we have $\frac{r}{1-p}=\frac{r+1}{1}$.
Thus, $r=\frac{1-p}... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,670 |
3. Given that the difference between the two roots of the equation $x^{2}+p x+q=0$ is equal to the difference between the two roots of the equation $x^{2}+q x+p=0$. Then, apart from $p^{2}-4 q>0$ and $q^{2}-4 p>0$, the relationship that $p$ and $q$ should have is $(\quad)$.
(A) $p+q=4$
(B) $p=-q$
(C) $p+q=-4$ or $p=q$
... | 3.C.
Let the roots of the equation $x^{2}+p x+q=0$ be $\alpha_{1} 、 \alpha_{2}$, and the roots of the equation $x^{2}+q x+p=0$ be $\beta_{1} 、 \beta_{2}$. Then
$$
\begin{array}{l}
\alpha_{1}+\alpha_{2}=-p, \alpha_{1} \alpha_{2}=q ; \\
\beta_{1}+\beta_{2}=-q, \beta_{1} \beta_{2}=p .
\end{array}
$$
From the given condi... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,671 |
4. Given that $H$ is the orthocenter of acute $\triangle A B C$, the rays $A H, B H, C H$ intersect $B C, C A, A B$ at points $D, E, F$ respectively. Let $B C=a, C A=b, A B=c$. If line segment $E F$ bisects the area of $\triangle A B C$, then $D E^{2}+E F^{2}+F D^{2}=$ ( ).
(A) $a^{2}+b^{2}+c^{2}$
(B) $a^{2}$
(C) $b^{2... | 4.B.
Since line segment $E F$ bisects the area of $\triangle A B C$, and it is easy to prove that $\triangle A E F \backsim \triangle A B C$, we have
$$
\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\left(\frac{E F}{B C}\right)^{2}=\frac{1}{2},
$$
which means $E F=\frac{\sqrt{2}}{2} B C$.
Also, $\cos \angle B A C=\... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,672 |
Example 7 Let $n(n \geqslant 2)$ be a fixed integer.
(1) Determine the smallest constant $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$
holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n... | Explain: First, find the range of values for $c$, then provide a proof, and finally point out the necessary and sufficient conditions for the equality to hold when $c$ takes its minimum value.
$$
\begin{array}{l}
\text { When } x_{1}=x_{2}=1, x_{3}=x_{4}=\cdots=x_{n}=0 \text {, } \\
c \geqslant \frac{1}{16} \times 1 \t... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,673 |
5. Given the quadratic function $y = a x^{2} + b x + c (a \neq 0)$ whose graph is shown in Figure 2, the following 5 conclusions are given:
(1) $a b c > 0$;
(2) $b < 0$;
(3) $c > 0$;
(4) $2 c > m(a m + b) ($ real number $m \neq 1)$. Among these, the correct conclusions are ( ) in number.
(A) 2
(B) 3
(C) 4
(D) 5 | 5.B.
From the graph, we can see that $a0, c>0$.
When $x=1$, $a+b+c>0$;
When $x=-1$, $a-b+c>0$;
When $x=2$, $4 a+2 b+c>0$;
When $x=3$, $y=m(a m+b)$ can be verified to be correct. Therefore, (3)(4)(5) are correct. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,674 |
6. Let $a$, $b$, $c$ and $m_{a}$, $m_{b}$, $m_{c}$ represent the lengths of the three sides and the lengths of the medians to these sides of $\triangle ABC$, respectively. Given the following conclusions:
(1) $m_{a}+m_{b}+m_{c} \leq \frac{3}{4}(a+b+c)$;
(4) $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^... | 6.C.
In Figure 6, let the three medians of $\triangle A B C$ be $A D=m_{a}$,
$$
B E=m_{b}, C F=m_{c},
$$
and they intersect at the centroid $G$.
Then
$$
\begin{array}{l}
G A+G B>A B, \\
G B+G C>B C, \\
G C+G A>C A .
\end{array}
$$
Adding the three inequalities, we get
$$
\begin{array}{l}
2(G A+G B+G C) \\
>A B+B C+C... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,675 |
1. Given real numbers $x, y$ satisfy the equation
$$
x^{2}-3 x y+3 y^{2}+4 x-18 y+52=0 \text {. }
$$
then the units digit of $y^{x}$ is $\qquad$ . | $=1.4$.
From the given equation, we have
$$
x^{2}-(3 y-4) x+\left(3 y^{2}-18 y+52\right)=0 \text {. }
$$
Since $x$ is a real number, then
$$
\Delta=(3 y-4)^{2}-4\left(3 y^{2}-18 y+52\right) \geqslant 0 \text {, }
$$
which simplifies to $-3(y-8)^{2} \geqslant 0$.
Thus, $y=8$.
Substituting $y=8$ into the original equat... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,676 |
2. As shown in Figure 3, select a point $P$ inside $\triangle A B C$, and draw line segments $D E, F G$, and $H M$ through $P$ parallel to the three sides of $\triangle A B C$. The areas of the resulting $\square A F P M, \square B D P G$, and $\square C E P H$ are $28, 12$, and $42$ respectively. Then the area of $\tr... | 2.144.
Let the areas of $\triangle D P M$, $\triangle P G H$, and $\triangle P E F$ be $S_{1}$, $S_{2}$, and $S_{3}$, respectively. According to the problem,
$\triangle D P M \backsim \triangle P E F \backsim \triangle D E A$.
Therefore, $\frac{\sqrt{S_{1}}}{\sqrt{S_{\triangle A D E}}}=\frac{D P}{D E}$, $\frac{\sqrt{S... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,677 |
3. Let $M=\frac{8}{\sqrt{2008}-44}$, $a$ is the integer part of $M$, and $b$ is the fractional part of $M$. Then
$$
a^{2}+3(\sqrt{2008}+37) a b+10=
$$ | 3.2008 .
Since $M=\frac{8}{\sqrt{2008}-44}=\frac{\sqrt{2008}+44}{9}$
$$
\begin{array}{l}
=9+\frac{\sqrt{2008}-37}{9}, \\
0<\frac{\sqrt{2008}-37}{9}<1,
\end{array}
$$
Therefore, $a=9, b=\frac{\sqrt{2008}-37}{9}$.
Then $a b=\sqrt{2008}-37$.
Hence $a^{2}+3(\sqrt{2008}+37) a b+10$
$$
\begin{array}{l}
=9^{2}+3(\sqrt{2008}... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,678 |
4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=... | 4.12.
From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,679 |
One, (20 points) Given prime numbers $p$ and $q$ such that the algebraic expressions $\frac{2 p+1}{q}$ and $\frac{2 q-3}{p}$ are both natural numbers. Try to find the value of $p^{2} q$.
---
The translation is provided as requested, maintaining the original format and line breaks. | Let $p \geqslant q$, then we have $1 \leqslant \frac{2 q-3}{p}<2$.
Thus, it can only be $\frac{2 q-3}{p}=1$, which means $p=2 q-3$.
At this point, $\frac{2 p+1}{q}=\frac{4 q-5}{q}=4-\frac{5}{q}$.
To make $\frac{2 p+1}{q}$ a natural number, it can only be $q=5$, hence $p=7$.
Now, let $p<q$, in this case, $1 \leqslant \... | 245 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,680 |
II. (25 points) As shown in Figure 5, in $\odot O$, $AB$ and $CD$ are two perpendicular diameters. Point $E$ is on radius $OA$, and point $F$ is on the extension of radius $OB$, such that $OE = BF$. Lines $CE$ and $CF$ intersect $\odot O$ at points $G$ and $H$, respectively. Lines $AG$ and $AH$ intersect line $CD$ at p... | As shown in Figure 7, connect $D G$ and $D H$, and extend $C G$ to point $P$. Since $A B$ and $C D$ are perpendicular diameters,
we have
$$
\overparen{A C}=\overparen{A D}=\overparen{B C}=\overparen{B D}.
$$
Thus,
$$
\begin{array}{l}
\angle A H C=\angle A H D \\
=\angle A G C \\
=\angle P G N=45^{\circ}.
\end{array}
$$... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 719,681 |
Three. (25 points) Given five numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ $\left(x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}\right)$, find the sum of every two numbers, resulting in 10 numbers, denoted as $a_{1}, a_{2}, \cdots, a_{10}$. Can the original five numbers $x_{1}, x_{2}, x_{3}, x_{4}... | Three, let $S=\frac{1}{4}\left(a_{1}+a_{2}+\cdots+a_{10}\right)$.
From the problem, $S$ is equal to the sum of the original 5 numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$.
Given $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$, we have
$$
x_{1}+x_{2} \leqslant x_{1}+x_{3} \leqslant \cdots \leqslan... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,682 |
1. Given the function $f(x)=\frac{4}{|x|+2}-1$ with the domain $[a, b]$, and the range $[0,1]$. Then the number of integer pairs $(a, b)$ that satisfy the condition is $(\quad)$.
(A)2
(B) 5
(C) 6
(D) infinitely many | -、1.B.
It is known that the graph of $f(x)$ is symmetric about the $y$-axis, and $f(x)$ is decreasing in the interval $[0,+\infty)$.
When $x \in \mathbf{R}$, $f(x)_{\max }=f(0)=1$.
Also, $f(-2)=f(2)=0$, then the integer pairs $(a, b)$ are
$$
(-2,0),(-2,1),(-2,2),(-1,2), (0,2)
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,683 |
Example 8 Find the largest real number $\lambda$ such that for a polynomial with real coefficients $f(x)=x^{3}+a x^{2}+b x+c$ whose all roots are non-negative real numbers, if $x \geqslant 0$, then
$$
f(x) \geqslant \lambda(x-a)^{3},
$$
and ask when the equality holds in this inequality?
(1999, China Mathematical Olym... | Explanation: First, sort the three roots of $f(x)$, then classify and solve for $x$ by intervals.
Let the three roots of $f(x)$ be $\alpha, \beta, \gamma (0 \leqslant \alpha \leqslant \beta \leqslant \gamma)$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
\alpha+\beta+\gamma=-a, \\
\a... | -\frac{1}{27} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,684 |
2. Point $P$ is on the plane of the base $A B C D$ of the cube $A_{1} B_{1} C_{1} D_{1}-A B C D$, $E$ is the midpoint of $A A_{1}$, and $\angle E P A=\angle D_{1} P D$. Then the locus of point $P$ is ( ).
(A) line
(B) circle
(C) parabola
(D) hyperbola | 2.B.
$$
\begin{array}{l}
\angle E P A=\angle D_{1} P D \\
\Rightarrow \tan \angle E P A=\tan \angle D_{1} P D \\
\Rightarrow \frac{E A}{A P}=\frac{D D_{1}}{P D} \Rightarrow P D=2 P A .
\end{array}
$$
$D$ and $A$ are fixed points. From equation (1) and the definition of the Apollonian circle, the locus of point $P$ is a... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,685 |
4. Given that $F$ is the right focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the line $l$ passes through point $F$ and intersects the two asymptotes $l_{1} 、 l_{2}$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at points $M 、 N$, and intersects the ellipse at points $A 、 B$. If... | 4.C.
As shown in Figure 3, from
$$
O M \cdot M N=0,
$$
we get $O M \perp M N$.
Given $k_{O M}=-\frac{b}{a}$, the slope of line $l$ is $\frac{a}{b}$, and the focus $F(c, 0)$. Therefore, the equation of line $l$ is $y=\frac{a}{b}(x-c)$.
From $\left\{\begin{array}{l}y=\frac{a}{b}(x-c), \\ y=\frac{b}{a} x,\end{array}\rig... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,687 |
5. Let the function $y=f(x)$ satisfy: for all $x \in \mathbf{R}$,
$$
\begin{array}{l}
f(x) \geqslant 0 \text {, and } f(x+1)=\sqrt{9-f^{2}(x)} \text {. } \\
\text { When } x \in[0,1) \text {, }
\end{array}
$$ | 5.D.
From the given, we have
$$
\begin{array}{l}
f^{2}(x+1)=9-f^{2}(x), \\
f^{2}(x+2)=9-f^{2}(x+1) .
\end{array}
$$
Therefore, $f^{2}(x+2)=f^{2}(x)$.
Since $f(x) \geqslant 0$, then $f(x+2)=f(x)$.
Thus, $f(x)$ is a periodic function with a period of 2.
Given $31\frac{1}{2}$, then
$$
f(10 \sqrt{10}-31)=\lg 10 \sqrt{10}... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,688 |
6. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $\angle B=60^{\circ}, O$ is the circumcenter of $\triangle A B C$, point $P$ is on the plane of $\triangle A B C$, $O P=O A + O B + O C$, and $B P \cdot B C=8$. Then the maximum value of the altitude $h$ on side... | 6. A.
From the given, $O P-O A=O B+O C$, which means
$$
\begin{array}{l}
A P=O B+O C . \\
\text { Also, }(O B+O C) \cdot B C \\
=(O C+O B) \cdot(O C-O B) \\
=O C^{2}-O B^{2}=|O C|^{2}-|O B|^{2}=0,
\end{array}
$$
Thus, $(O B+O C) \perp B C$.
From equation (1), we know $\boldsymbol{A P} \perp B C$.
Similarly, $B P \per... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,689 |
1. Let the geometric sequence $z_{1}, z_{2}, \cdots, z_{n}, \cdots$ be such that $z_{1}=$ $1, z_{2}=a+b \mathrm{i}, z_{3}=b \mathrm{i}(a, b \in \mathbf{R}, ab>0)$. Then the smallest natural number $n$ for which $z_{1} z_{2} \cdots z_{n}<0$ is $\qquad$ . | $Ni, 1.8$.
$$
\begin{array}{l}
\text { Given } z_{2}^{2}=z_{1} z_{3} \Rightarrow(a+b \mathrm{i})^{2}=b \mathrm{i} \\
\Rightarrow a^{2}-b^{2}+2 a b \mathrm{i}=b \mathrm{i} \\
\Rightarrow\left\{\begin{array} { l }
{ a ^ { 2 } = b ^ { 2 } , } \\
{ 2 a b = b }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=\frac{1}{2},... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,690 |
2. The sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with all terms being positive integers, and the common difference $d \in \mathbf{N}_{+}$. If the sum of any two terms in $\left\{a_{n}\right\}$ is still a term in the sequence, and $a_{1}=6$, then $d=$ $\qquad$ | $$
2.1,2,3,6 \text {. }
$$
For any two terms $a_{m} 、 a_{k}$, their sum $a_{m}+a_{k}=a_{n}$, i.e.,
$$
2 a_{1}+(m+k-2) d=a_{1}+(n-1) d \text {. }
$$
Rearranging gives $a_{1}=(n-m-k+1) d=6$.
Since $d \in \mathbf{N}_{+}, n-m-k+1$ is an integer, thus, $d=1,2,3,6$. | 1,2,3,6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,691 |
3. Person A has a box, inside there are 4 balls in total, red and white; Person B has a box, inside there are 2 red balls, 1 white ball, and 1 yellow ball. Now, A randomly takes 2 balls from his box, B randomly takes 1 ball from his box. If the 3 balls drawn are all of different colors, then A wins. To ensure A has the... | 3.2.
Suppose box A contains $n(n \geqslant 1)$ red balls, then it has $4-n$ white balls.
Therefore, the probability of A winning is
$$
P=\frac{\mathrm{C}_{n}^{1} \mathrm{C}_{4-n}^{1}}{\mathrm{C}_{4}^{2} \mathrm{C}_{4}^{1}}=\frac{1}{24} n(4-n) \text {. }
$$
Since $\sqrt{n(4-n)} \leqslant \frac{n+4-n}{2}=2$, that is,
$... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,692 |
4. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the ellipse $\frac{y^{2}}{a^{2}}+$ $\frac{x^{2}}{b^{2}}=1(a>b>0)$, $m=\left(\frac{x_{1}}{b}, \frac{y_{1}}{a}\right)$, $n$ $=\left(\frac{x_{2}}{b}, \frac{y_{2}}{a}\right)$, and $\boldsymbol{m} \cdot \boldsymbol{n}=0$. The eccentricity ... | 4.1.
Given $e^{2}=\frac{c^{2}}{a^{2}}=\frac{3}{4}, b=1$, then $a^{2}=b^{2}+c^{2}=4$.
Also, $\boldsymbol{m} \cdot \boldsymbol{n}=0 \Rightarrow y_{1} y_{2}=-4 x_{1} x_{2}$.
Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{4}+x_{1}^{2}=1 \\
\... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,693 |
5. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies for any real numbers $x, y$
$$
f(x+y)+f(x-y)=2 f(x) \cos y,
$$
and $f(0)=0, f\left(\frac{\pi}{2}\right)=1$.
Consider the following conclusions:
(1) $f\left(\frac{\pi}{4}\right)=\frac{1}{2}$; (2) $f(x)$ is an odd function;
(3) $f(x)$ is a periodic functi... | 5.(2),(3).
Let $x=0, y \in \mathbf{R}$, we get $f(y)+f(-y)=0$, so $f(x)$ is an odd function, hence (2) is correct.
Let $y=\frac{\pi}{2}$, we get $f\left(x+\frac{\pi}{2}\right)+f\left(x-\frac{\pi}{2}\right)=0$, i.e.,
$$
\begin{array}{l}
f(x+\pi)=-f(x), \\
f(x+2 \pi)=-f(x+\pi)=f(x) .
\end{array}
$$
Hence (3) is correct... | (2),(3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,694 |
Example 9 Find the smallest real number $M$ such that for all real numbers $a, b, c$, we have
$$
\begin{array}{l}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
\leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{array}
$$
(47th IMO) | Consider:
$$
P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+a\left(c^{2}-t^{2}\right) \text {. }
$$
It is easy to see that $P(b)=P(c)=P(-c-b)=0$.
Then $\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+a a\left(c^{2}-a^{2}\right)\right|$
$$
\begin{array}{l}
=|P(a)| \\
=|(b-c)(a-b)(a-c)(a+b+c)... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,695 |
6. Let the function $f(x)=(2 a-b) x+2 b-8$ $(b \leqslant 6)$, when $x \in[-4,2]$, $f(x) \geqslant 0$ always holds. Then the range of $\frac{a^{4}-2 b^{4}}{a^{3} b}$ is $\qquad$ . | 6. $\left[-\frac{161}{3},-\frac{39071}{4116}\right]$.
From the given, we have
$$
\left\{\begin{array}{l}
f(-4) \geqslant 0, \\
f(2) \geqslant 0, \\
b \leqslant 6,
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
-4 a+3 b-4 \geqslant 0, \\
a \geqslant 2, \\
b \leqslant 6 .
\end{array}\right.
$$
The region ... | \left[-\frac{161}{3},-\frac{39071}{4116}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,696 |
Three. (20 points) Let the function $f(x)=x|x-a|+b$, where the constant $b<2 \sqrt{2}-3$, and for any $x \in[0,1], f(x) < 0$ always holds. Find the range of the real number $a$.
---
The function is given by $f(x)=x|x-a|+b$, with the constant $b<2 \sqrt{2}-3$. It is required that for any $x \in [0,1]$, $f(x) < 0$ alwa... | Three, $b\left(x+\frac{b}{x}\right)_{\max },
$
and $a\left(x+\frac{b}{x}\right)_{\text {max }}=f(1)=1+b .
$
For equation (2), when $-1 \leqslant b<0$, $x \in(0,1]$,
$
x-\frac{b}{x}=x+\frac{-b}{x} \geqslant 2 \sqrt{-b} \text {, }
$
with equality if and only if $x=\sqrt{-b}$, $\left(x-\frac{b}{x}\right)_{\min }=2 \sqr... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,697 |
Four. (20 points) Let points $A, B, C, D$ all lie on the right branch of the hyperbola $x^{2}-y^{2}=1$.
(1) If $\boldsymbol{A B}=\lambda C D$ (where $\lambda \neq 0$ is a real number), prove:
$O A \cdot O B=O C \cdot O D$ (where $O$ is the origin);
(2) If $|A B|=2$, and $P$ is the midpoint of segment $A B$, draw perpen... | (1) From $\boldsymbol{A B}=\lambda C D$, we get $\boldsymbol{A B} / / \boldsymbol{C D}$.
(i) When the inclination angle of line $A B$ is $90^{\circ}$, let its equation be $x=m(m>1)$.
Let $A\left(m, y_{1}\right)$, then $B\left(m,-y_{1}\right)$, and point $A\left(m, y_{1}\right)$ satisfies $m^{2}-y_{1}^{2}=1$.
$$
O A \c... | 1 | Algebra | proof | Yes | Yes | cn_contest | false | 719,698 |
Five. (20 points) If the sum of the greatest odd divisors of the terms in the natural number sequence $1,2,3, \cdots, 2^{n}$ is $a_{n}$. Prove:
$$
\sum_{i=1}^{n} \frac{1}{a_{i} a_{i+1}}<\frac{3}{20} .
$$ | Obviously, $a_{1}=2$.
Below, we consider the relationship between $a_{n}$ and $a_{n+1}$.
The largest odd factor of an odd number $2 k+1$ is $2 k+1$; the largest odd factor of a number of the form $2^{k}(k \in \mathbf{N})$ is 1.
Since the sum of the largest odd factors of the terms $1,2,3, \cdots, 2^{n}, \cdots, 2^{n+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,699 |
One, (50 points) As shown in the figure
$2, A D$ is an altitude of acute $\triangle A B C$, $P$ is a point on $A D$, extending $B P$ intersects $A C$ at point $M$, extending $C P$ intersects $A B$ at point $N$, and $M N$ intersects $A P$ at point $Q$. A line through $Q$ intersects $P N$ at point $E$ and $A M$ at point ... | As shown in Figure 5, connect $D M$ and $D N$. Draw $R K / / B C$ through point $A$.
First, prove: $\angle N D A = \angle M D A$.
From $\triangle A R N \backsim \triangle B D N$ and $\triangle A K M \backsim \triangle C D M$, we have
$$
A R = \frac{A N}{B N} \cdot B D,
$$
$$
A K = \frac{A M}{M C} \cdot D C.
$$
By Ceva'... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,700 |
Sure, here is the translated text:
```
II. (50 points) Let
$$
\boldsymbol{F}(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+1 \text {, }
$$
where, $n \in \mathbf{N}_{+}, a_{1}, a_{2}, \cdots, a_{n-1} \geqslant 0$, and all roots of $F(x)=0$ are real. Prove: For any $x \geqslant 0$, we have
$$
F(x)(F(x)+1) \geqslant 4^{n} x^{... | When $x \geqslant 0$, $F(x) \geqslant 1>0$ holds, then all real roots of $F(x)=0$ are negative. Since all roots of $F(x)=0$ are real, $F(x)=0$ has exactly $n$ negative roots, denoted as $-x_{1},-x_{2}, \cdots,-x_{n}$, where $x_{1}, x_{2}$, $\cdots, x_{n}>0$. Thus,
$$
F(x)=\left(x+x_{1}\right)\left(x+x_{2}\right) \cdots... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,701 |
For any $n \in \mathbf{N}_{+}$, define
$f(n)=\left\{\begin{array}{l}1, \text { if the number of } 1 \text {s in the binary representation of } n \text { is odd; } \\ 0, \text { if the number of } 1 \text {s in the binary representation of } n \text { is even. }\end{array}\right.$
Prove: There do not exist positive inte... | Three, from the given conditions, we easily get $f(2 n)=f(n)$,
$$
f(2 n+1)=1-f(2 n)=1-f(n) \text {. }
$$
Assume there exist positive integers $k, m$ that satisfy the requirements, and assume $m$ is the smallest value that makes the conclusion true.
If $m$ is odd, without loss of generality, let
$$
f(k)=f(k+m)=f(k+2 m)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,702 |
Find all integers $n$ such that $20n+2$ divides $2003n+2002$.
The original text has been translated into English while preserving the original line breaks and format. | Solution: Since $20 n+2$ is even and can divide $2003 n +2002$, therefore, $2003 n+2002$ is even. Thus, $n$ is even.
Let $n=2 m$ (where $m$ is an integer), then
$$
\begin{array}{l}
\frac{2003 n+2002}{20 n+2} \in \mathbf{Z} \\
\Leftrightarrow \frac{2003 m+1001}{20 m+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{100(20 m+1)... | 0 \text{ or } -42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,703 |
As shown in Figure $3, \odot O$ is inside the rectangle $ABCD$. Tangents are drawn from vertices $A, B, C, D$ to $\odot O$, with points of tangency being $A_{1}, B_{1}, C_{1}, D_{1}$, respectively. If $AA_{1}=3, BB_{1}=4$, $CC_{1}=5$, find the length of $DD_{1}$. | Solution: Let the radius of $\odot O$ be $r$. As shown in Figure 3, connect $O A_{1}, O B_{1}, O C_{1}, O D_{1}$. Then
$$
\begin{array}{l}
O A_{1} \perp A A_{1}, \\
O B_{1} \perp B B_{1}, \\
O C_{1} \perp C C_{1}, \\
O D_{1} \perp D D_{1} .
\end{array}
$$
Connect $A O, B O, C O, D O$. Then $\triangle A O A_{1}$, $\tri... | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,704 |
221 Given that $a, b, c$ are positive numbers satisfying $abc=1$. Prove:
$$
\frac{1}{a^{2}+2 b^{2}+3}+\frac{1}{b^{2}+2 c^{2}+3}+\frac{1}{c^{2}+2 a^{2}+3} \leqslant \frac{1}{2} \text {. }
$$ | Prove: Notice
$$
\begin{array}{l}
a^{2}+2 b^{2}+3 \geqslant a^{2}+b^{2}+2 b+2 \\
\geqslant 2(a b+b+1)
\end{array}
$$
Then $\frac{1}{a^{2}+2 b^{2}+3} \leqslant \frac{1}{2(a b+b+1)}$.
Similarly, $\frac{1}{b^{2}+2 c^{2}+3} \leqslant \frac{1}{2(b c+c+1)}$,
$$
\frac{1}{c^{2}+2 a^{2}+3} \leqslant \frac{1}{2(c a+a+1)} \text ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,705 |
Example 1 As shown in Figure 1, in quadrilateral $ABCD$, $AC$ intersects $BD$ at point $E$. Prove:
$$
\frac{S_{\triangle ABD}}{S_{\triangle BCD}}=\frac{AE}{CE} .
$$ | Prove: Draw $A F \perp$
$B D, C G \perp B D$, the feet of the perpendiculars are $F, G$.
Then $A F / / C G$, so $\frac{A F}{C G}=\frac{A E}{C E}$.
Therefore, $\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\frac{1}{2} B D \cdot A F}{\frac{1}{2} B D \cdot C G}=\frac{A F}{C G}=\frac{A E}{C E}$.
Note: This conclusi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,706 |
Example 2 As shown in Figure 2, in $\triangle A B C$, $D$ is a point on $A B$, and $E$ is a point on $A C$. Prove:
$$
\frac{S_{\triangle A D E}}{S_{\triangle A B C}}=\frac{A D \cdot A E}{A B \cdot A C} .
$$ | $$
\begin{array}{l}
\text { Prove: } \frac{S_{\triangle A D E}}{S_{\triangle A B C}} \\
=\frac{\frac{1}{2} A D \cdot A E \sin A}{\frac{1}{2} A B \cdot A C \sin A}=\frac{A D \cdot A E}{A B \cdot A C} .
\end{array}
$$
Figure 2
Note: The conclusion of this problem is known as the "Angle-Sharing Proportionality Theorem".
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,707 |
Example 2 The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{1}=2, a_{n+1}=a_{n}^{2}-a_{n}+1 \text {. }
$$
Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2000}}<1$. | Explanation: From the given condition, we have
$$
a_{n+1}-1=a_{n}\left(a_{n}-1\right) \text {. }
$$
By decomposing and transforming the above equation, we get
$$
\begin{array}{l}
\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2008}} \\
=\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,708 |
3. Given $\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0$. Then $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=$ $\qquad$ | 3. -3 or 1.
From the given we have
$$
\begin{array}{l}
\left(\frac{x^{2}}{y+z}+x\right)+\left(\frac{y^{2}}{z+x}+y\right)+\left(\frac{z^{2}}{x+y}+z\right) \\
=x+y+z
\end{array}
$$
which means $\square$
$$
\begin{array}{l}
\frac{x(x+y+z)}{y+z}+\frac{y(x+y+z)}{z+x}+\frac{z(x+y+z)}{x+y} \\
=x+y+z . \\
\text { Then }\left... | -3 \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,709 |
4. In the Cartesian coordinate system, there is a parabola $y=$ $x^{2}-(5 c-3) x-c$ and three points $A\left(-\frac{1}{2} c, \frac{5}{2} c\right)$, $B\left(\frac{1}{2} c, \frac{9}{2} c\right)$, $C(2 c, 0)$, where $c>0$. There exists a point $P$ on the parabola such that the quadrilateral with vertices $A$, $B$, $C$, an... | 4.3.
(1) If $A B$ is the diagonal, then $P_{1}(-2 c, 7 c)$. For $P_{1}$ to be on the parabola, it must satisfy
$$
7 c=(-2 c)^{2}-(5 c-3)(-2 c)-c \text {. }
$$
Solving this, we get $c_{1}=0$ (discard), $c_{2}=1$.
Thus, $c=1$, and at this point, $P_{1}(-2,7)$.
(2) If $B C$ is the diagonal, then $P_{2}(3 c, 2 c)$. Simila... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,710 |
$$
\begin{array}{l}
\text { I. (20 points) Let } 0<x<1, \\
y=\sqrt{x^{2}+1}+\sqrt{1+(1-x)^{2}} .
\end{array}
$$
Find the range of $y$. | $$
\begin{array}{l}
-\sqrt{1+x^{2}}+\sqrt{1+(1-x)^{2}} \geqslant \sqrt{5} \\
\Leftrightarrow \sqrt{1+(1-x)^{2}} \geqslant \sqrt{5}-\sqrt{1+x^{2}} \\
\Leftrightarrow 1+(1-x)^{2} \geqslant 5+1+x^{2}-2 \sqrt{5\left(1+x^{2}\right)} \\
\Leftrightarrow \sqrt{5\left(1+x^{2}\right)} \geqslant x+2 \\
\Leftrightarrow 5+5 x^{2} \... | \sqrt{5} \leqslant y < 1 + \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,711 |
II. (25 points) As shown in Figure 2, $EF$ intersects the diagonal $AC$ of $\square ABCD$ at point $G$,
intersects $AB$, $AD$ at points
$E$, $F$, and connects $CE$,
$CF$, $BG$. If $\frac{1}{S_{\triangle ACE}}+$
$\frac{1}{S_{\triangle ACF}}=\frac{\lambda}{S_{\triangle ABG}}$, find
the value of $\lambda$. | II. As shown in Figure 8, draw $EM \parallel BC$ intersecting $AC$ at $M$, then $\frac{AB}{AE}=\frac{AC}{AM}$.
From $\frac{AD}{AF}=\frac{BC}{AF}$
$=\frac{BC}{EM} \cdot \frac{EM}{AF}$
$=\frac{AC}{AM} \cdot \frac{GM}{AG}$
$\Rightarrow \frac{AB}{AE}+\frac{AD}{AF}=\frac{AC}{AM}+\frac{AC}{AM} \cdot \frac{GM}{AG}$
$=\frac{AC... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,712 |
Three, (25 points) In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
---
In $\tr... | Three, as shown in Figure 9, let the sides of $\triangle ABC$ be $BC = a$, $CA = b$, and $AB = c$. Let $CD = h$, $AD = x$, and $BD = y$. The radii of the two incircles are $r_1$ and $r_2$. Then,
$$
\begin{array}{l}
RS = |RD - SD| \\
= |r_1 - r_2|. \\
\text{Since } b = AC = AP + CP = AE + CR \\
= (x - r_1) + (h - r_1),
... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 719,713 |
1. The table tennis training team has 10 players, and every two players form a pair to practice. The number of different pairing methods is ( ).
(A) 972
(B) 945
(C) 864
(D) 891 | 1.B.
According to the problem, there are a total of $9 \times 7 \times 5 \times 3 \times 1=945$ ways. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,714 |
2. Let $\angle A$, $\angle B$, $\angle C$ all be acute angles. Given the following three propositions:
(1) $\tan \frac{A}{2}=\tan ^{3} \frac{C}{2}$, (2) $\tan C=2 \tan B$,
(3) $\angle A$, $\angle B$, $\angle C$ form an arithmetic sequence.
Then the number of true propositions derived from any two of (1), (2), (3) is ( ... | 2.A.
Notice
$$
\begin{array}{l}
\tan B=\frac{1}{2} \tan C=\frac{\tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}} \\
=\frac{\tan \frac{C}{2}\left(1+\tan ^{2} \frac{C}{2}\right)}{1-\tan ^{4} \frac{C}{2}}=\frac{\tan \frac{C}{2}+\tan ^{3} \frac{C}{2}}{1-\tan ^{4} \frac{C}{2}} \\
=\frac{\tan \frac{C}{2}+\tan \frac{A}{2}}{1-\tan... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,715 |
3. Given constants $a>0, b>0$. Then the range of the function
$$
f(x)=a|\sin x|+b|\cos x|
$$
is ( ).
(A) $\left[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}\right]$
(B) $\left[a, \sqrt{a^{2}+b^{2}}\right]$
(C) $\left[b, \sqrt{a^{2}+b^{2}}\right]$
(D) $\left[a, \sqrt{a^{2}+b^{2}}\right]$ or $\left[b, \sqrt{a^{2}+b^{2}}\righ... | 3. D.
Since $f(x)$ is an even function, and when $a \neq b$, $T=$ $\pi$, when $a=b$, $T=\frac{\pi}{2}$, so, without loss of generality, let $x \in[0$, $\left.\frac{\pi}{2}\right]$.
Then $f(x)=a \sin x+b \cos x$
$$
=\sqrt{a^{2}+b^{2}} \sin (x+\varphi),
$$
where $\sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}, \cos \varphi=... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,716 |
4. On the basketball court, there are 5 players practicing. Their tactic is: starting with player A, after 6 passes (with each player having an equal chance of passing and receiving the ball), the ball returns to A, who then attempts a 3-point shot. The number of different passing methods is ( ).
(A) 4100
(B) 1024
(C) ... | 4.C.
Convert to coloring the vertices of a hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ using 5 colors, each point gets one color, the color of $A_{1}$ is fixed, and adjacent points must have different colors. How many coloring methods are there?
$$
\begin{array}{l}
\text { Given } a_{n}=\frac{m-1}{m}\left[(m-1)^{n-1... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 719,717 |
5. For the central conic section $\frac{x^{2}}{m}+\frac{y^{2}}{n}=1(m>|n|>0)$, there is a point $P$ on it such that the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ are perpendicular to each other. Then the area of $\mathrm{Rt} \triangle P F_{1} F_{2}$ is $(\quad)$.
(A) $|n|$
(B) $m$
(C) $n^{2}$
(D) Uncerta... | 5.A.
$n>0$, the curve is an ellipse. Then
$$
\begin{array}{l}
\left|P F_{1}\right|+\left|P F_{2}\right|=2 \sqrt{m}, \\
P F_{1}^{2}+P F_{2}^{2}=4(m-n) .
\end{array}
$$
(1)$^{2}$ - (2) gives
$$
\left|P F_{1}\right|\left|P F_{2}\right|=2 n, S_{\triangle P F_{1} F_{2}}=|n| \text {. }
$$
Similarly, when $n<0$, $S_{\triangl... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 719,718 |
Example $\mathbf{3}$ For $n \in \mathbf{N}_{+}$, let $T_{n}=\frac{1}{1^{4}}+\frac{1}{2^{4}}+\cdots+\frac{1}{n^{4}}$. Prove: $T_{n} \leqslant \frac{11}{10}$. | Explanation 1: When $n \geqslant 4$,
$$
\begin{array}{l}
\frac{1}{n^{4}}<\frac{1}{n^{4}-1}=\frac{1}{2}\left(\frac{1}{n^{2}-1}-\frac{1}{n^{2}+1}\right) \\
<\frac{1}{2}\left(\frac{1}{n^{2}-1}-\frac{1}{n^{2}+n}\right) \\
=\frac{1}{4}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,719 |
6. $[x]$ represents the greatest integer not exceeding the real number $x$. Let $N$ be a positive integer. Then the number of all solutions to the equation $x^{2}-\left[x^{2}\right]=(x-[x])^{2}$ in the interval $1 \leqslant x \leqslant N$ is ( ).
(A) $N^{2}+N+1$
(B) $N^{2}-N$
(C) $N^{2}-N+1$
(D) $N^{2}-N+2$ | 6.C.
Obviously, $x=N$ is a solution to the equation.
Suppose $1 \leqslant x < N, m = [x]$,
$$
p = x - m = \{x\}.
$$
Then $x = m + p$.
The original equation becomes $(m + p)^2 - [(m + p)^2] = p^2$, which simplifies to $2mp = [2mp + p^2]$.
Since $0 \leqslant p < 1$ and $2mp$ is an integer, then $p = 0, \frac{1}{2m}, \f... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 719,720 |
1. It is known that the inequality $a^{x} \geqslant x \geqslant \log _{a} x$ holds for all $x$ in the interval $(0,+\infty)$. Then the range of positive real number $a$ is $\qquad$ | II. $1 . a \geqslant \mathrm{e}^{\frac{1}{6}}$.
Obviously, a necessary condition for the inequality to hold is $a>1$.
Let $f(x)=a^{x}-x$, then
$$
\begin{array}{l}
f^{\prime}(x)=a^{x} \ln a-1 \geqslant 0 \Leftrightarrow x \geqslant-\frac{\ln (\ln a)}{\ln a}, \\
f^{\prime}(x) \leqslant 0 \Leftrightarrow x \leqslant-\frac... | a \geqslant \mathrm{e}^{\frac{1}{e}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,721 |
2. Let $1 \leqslant r \leqslant n$. Then the arithmetic mean of the largest numbers in the $r$-element subsets of the set $M=\{1,2, \cdots, n\}$ is $\qquad$ . | 2. $\frac{r(n+1)}{r+1}$.
The number of subsets of $M$ containing $r$ elements is $\mathrm{C}_{n}^{r}$. The number of subsets of $M$ containing $r$ elements with the largest number being the positive integer $k$ is $\mathrm{C}_{k-1}^{r-1}$. Therefore, the arithmetic mean of the largest numbers in the subsets of $M$ con... | \frac{r(n+1)}{r+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,722 |
3. Let $z \in \mathbf{C}, f(z)=a z^{2}+b z+c$ be a polynomial with complex coefficients, and suppose that when $|z| \leqslant 1$, $|f(z)| \leqslant 1$. Then, when $|z| \leqslant 1$, the range of $|a z+b|$ is | 3. $[2,+\infty)$.
It is easy to see that when $|z| \leqslant 1$,
$$
\left|\frac{z+1}{2}\right| \leqslant 1, \left|\frac{z-1}{2}\right| \leqslant 1 \text{. }
$$
Therefore, $\left|f\left(\frac{z+1}{2}\right)\right| \leqslant 1, \left|f\left(\frac{z-1}{2}\right)\right| \leqslant 1$.
Then $|a z+b|$
$$
\begin{array}{l}
=\... | [2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,723 |
4. The number of ways to choose 5 different and non-adjacent numbers from the set $\{1,2, \cdots, 25\}$ is $\qquad$ kinds. | 4.20349 .
Let $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ be a subset of 5 non-adjacent numbers from $\{1,2, \cdots, 25\}$, and
$a_{i+1}-a_{i} \geqslant 2(i=1,2,3,4)$.
Let $a_{i+1}^{\prime}-a_{i}^{\prime}=a_{i+1}-a_{i}-1$, i.e.,
$\left(a_{i+1}-a_{i+1}^{\prime}\right)-\left(a_{i}-a_{i}^{\prime}\right)=1$.
Take $... | 20349 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,724 |
5. Given the increasing sequence $1,3,4,9,10,12,13, \cdots$ where each term is either a power of 3 or the sum of several different powers of 3. Then the 100th term of this sequence is $\qquad$ . | 5.981.
The terms of this sequence are
$$
a_{0}+a_{1} 3+a_{2} 3^{2}+\cdots+a_{n} 3^{n},
$$
where $a_{i} \in\{0,1\}, i=1,2, \cdots, n$.
When $n=5$, there can be $2^{6}-1=63$ numbers, the 64th term is $3^{6}=729$. Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}=$ 243, the 96th term i... | 981 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,725 |
6. Let $f(x)$ be defined on $\mathbf{N}_{+}$, with its range $B \subseteq$ $\mathbf{N}_{+}$, and for any $n \in \mathbf{N}_{+}$, we have
$$
f(n+1)>f(n) \text {, and } f(f(n))=3 n \text {. }
$$
Then $f(10)+f(11)=$ $\qquad$ | 6.39.
Given $f(f(1))=3$, we know $f(f(f(1)))=f(3)$.
If $f(1)=1$, then $3=f(f(1))=f(1)=1$, which is a contradiction.
Therefore, $2 \leqslant f(1)<f(2) \leqslant f(f(1))=3$.
Thus, $f(2)=3, f(1)=2$,
$f(3)=f(f(2))=6$,
$f(6)=f(f(3))=9$.
Also, $6=f(3)<f(4)<f(5)<f(6)=9$, so
$f(4)=7, f(5)=8$,
$f(7)=f(f(4))=12$,
$f(12)=f(f(7))... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 719,726 |
Three. (20 points) Given that $a$ and $b$ are real constants not equal to 0, and the variable $\theta$ satisfies the inequality system
$$
\left\{\begin{array}{l}
a \sin \theta + b \cos \theta \geqslant 0, \\
a \cos \theta - b \sin \theta \geqslant 0 .
\end{array}\right.
$$
Try to find the maximum value of $\sin \theta... | From
\[
\left\{\begin{array}{l}
a \sin \theta + b \cos \theta \geqslant 0, \\
a \cos \theta - b \sin \theta \geqslant 0,
\end{array}\right.
\]
we know
\[
\left\{\begin{array}{l}
\sin (\theta + \varphi) \geqslant 0 \\
\cos (\theta + \varphi) \geqslant 0
\end{array}\right.
\]
where, \(\sin \varphi = \frac{b}{\sqrt{a^2 +... | \frac{a}{\sqrt{a^2 + b^2}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,727 |
Four, (20 points) If a hexagon inscribed in a conic section $\Gamma$ (including degenerate conic sections) has three pairs of opposite sides that are not parallel, then the three points of intersection of the lines containing these pairs of opposite sides are collinear. | Let the equation of the curve $\Gamma$ be $F(x, y)=0$, simply denoted as $F=0$ (the same below), and the sides $A_{i} A_{i+1}$ of the inscribed hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ on the curve $\Gamma$ have the equations $f_{i}=0(i=1,2, \cdots, 6)$, and the diagonal $A_{1} A_{4}$ has the equation $g=0$. Then ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,728 |
Five. (20 points) Try to find all functions $f(x)$ that satisfy the following conditions:
(1) $f(x)$ is a monotonic function defined on $\mathbf{R}$;
(2) For any real numbers $x, y$,
$$
f(x+y)=f(x) f(y) \text{; }
$$
(3) $f(1)=a(0<a \neq 1)$. | Lemma: Let $A$ and $B$ be constants. If for any $n \in \mathbf{N}_{+}$, we have $|A-B| \leqslant \frac{1}{n}$, then $A=B$.
Proof of the Lemma: Suppose $A \neq B$, then according to the problem,
$|A-B| \leqslant 1$.
Let $|A-B|=\frac{1}{m} (m \geqslant 1)$. Take $n=[m]+1$. Then $|A-B|=\frac{1}{m}>\frac{1}{n}$, which con... | f(x)=a^{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,729 |
Example 4 Given the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ $3^{n}-(-2)^{n}$. Prove: $\sum_{k=1}^{n} \frac{1}{a_{k}}<\frac{1}{2}$. | Explanation: If we directly apply the bounding technique to the general term, it would be very difficult to achieve the goal. However, if we adopt the technique of combining two terms and then apply the bounding, it can lead to a surprising victory. Using the analytical method, it is easy to prove that when $k$ is a po... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,730 |
One, (50 points) As shown in Figure $1, \odot O_{1}$ is internally tangent to $\odot O_{2}$ at point $P, \odot O_{1}$'s chord $A B$ is tangent to $\odot O_{2}$ at point $C$.
(1) If the radii of $\odot O_{1}$ and $\odot O_{2}$ are constants $R$ and $r$, respectively, find $\frac{A C^{2}}{A P^{2}}$;
(2) If $P C$ intersec... | (1) As shown in Figure 2, connect $O_{1} P$. Note that $A C^{2}=A E \cdot A P$, thus,
$$
\begin{array}{c}
\frac{A C^{2}}{A P^{2}}=\frac{A E}{A P} \\
=\frac{O_{1} O_{2}}{O_{1} P} \\
=\frac{R-r}{R} .
\end{array}
$$
(2) Extend $P A$ to $M$ such that $A M=F B$. The other auxiliary lines are as shown in Figure 2.
Since $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,731 |
II. (50 points) Let $x_{i} \in[1,3], i=1,2, \cdots$, 2007, and $\sum_{i=1}^{2007} x_{i}=5988$. Try to find the maximum value of $\sum_{i=1}^{2007} x_{i}^{2008}$ and the values of $x_{i}(i=1,2, \cdots, 2007)$ at this time. | $$
\begin{array}{l}
\text { For } 1 \leqslant x_{k} \leqslant x_{j} \leqslant 3, \text { let } \\
h=\min \left\{x_{k}-1,3-x_{j}\right\}, \\
x_{k}^{\prime}=x_{k}-h, x_{j}^{\prime}=x_{j}+h,
\end{array}
$$
the rest of $x_{i}^{\prime}=x_{i}$. Then $x_{i}^{\prime}, x_{k}^{\prime}, x_{j}^{\prime} \in[1,3]$, and we have
$$
\... | 16+1990 \times 3^{2008}+2^{2008} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 719,732 |
Three, (50 points) Let $A=\{1,2, \cdots, 30\}$. Find the smallest positive integer $n$, such that for any 11 subsets of $A$, if the union of any 5 of them has at least $n$ elements, then there must exist 3 of these 11 subsets whose intersection is non-empty. | Three, the minimum value of $n$ is 22.
First, prove: $n \geqslant 22$.
$$
\begin{array}{l}
\quad \text { Let } A_{i}=\{i, i+5, i+10, i+15, i+20, i+25\}, \\
i=1,2, \cdots, 5, \\
\quad B_{j}=\{j, j+6, j+12, j+18, j+24\}, j= \\
1,2, \cdots, 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Clearly, }\left|A_{i}\right|=6(i=1,... | 22 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 719,733 |
As shown in Figure 3, in $\triangle A B C$, points $E$ and $F$ are on sides $A B$ and $A C$ respectively, $B F$ intersects $C E$ at point $P$, points $M$ and $N$ are the midpoints of $B F$ and $C E$ respectively, line $M N$ intersects $A B$ and $A C$ at points $Q$ and $S$. If $\angle C B F=\angle B C E=\frac{1}{2} \ang... | Proof: From Figure 3, we have
$$
\begin{array}{l}
\angle B E C + \angle B F C \\
= \angle A + \angle A C E + \angle A + \angle A B F \\
= \angle A + \angle A C E + \angle B C E + \angle A B F + \angle C B F \\
= \angle A + \angle A C B + \angle A B C \\
= 180^{\circ}.
\end{array}
$$
From the area relationship, we get
... | B Q = F S | Geometry | proof | Yes | Yes | cn_contest | false | 719,734 |
Let $a$ be a given real number, solve the system of inequalities about $x$
$$
a \cdot \frac{1-x}{2} \leqslant \frac{x+2}{3} \leqslant a \cdot \frac{3-x}{4} \text {. }
$$ | Solution: The original system of inequalities is equivalent to
$$
\left\{\begin{array}{l}
(3 a+2) x \geqslant 3 a-4, \\
(3 a+4) x \leqslant 9 a-8 .
\end{array}\right.
$$
We classify and discuss based on the coefficient of $x$ in each inequality.
(1) If $3 a+2>0$, i.e., $a>-\frac{2}{3}$, then $3 a+4>0$. Thus, the origi... | \frac{3 a-4}{3 a+2} \leqslant x \leqslant \frac{9 a-8}{3 a+4} \text{ for } -\frac{2}{3}<a \leqslant 0 \text{ or } a \geqslant \frac{1}{3}; \, x \leqslant \frac{9 a-8}{3 a+4} \ | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 719,735 |
223 Let $p$ be a prime. Prove:
(1) For all primes $p (p \geqslant 2), p^{2} \mid \sum_{k=1}^{p-1}\left(\mathrm{C}_{p}^{k}\right)^{2}$;
(2) For all primes $p (p \geqslant 5), p^{3} \mid \sum_{k=1}^{p-1}\left(\mathrm{C}_{p}^{k}\right)^{2}$. | Proof: (1) Note that $k!(p-k)!\mathrm{C}_{p}^{k}=p$ ! and when $1 \leqslant k \leqslant p-1$, $(p, k!)=(p,(p-k)!)$ $=1$, we know
$$
p \mid \mathrm{C}_{p}^{k}(1 \leqslant k \leqslant p-1) .
$$
Thus, $p^{2} \mid\left(\mathrm{C}_{p}^{k}\right)^{2}(1 \leqslant k \leqslant p-1)$.
Therefore, $p^{2} \mid \sum_{k=1}^{p-1}\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 719,736 |
Given 224 As shown in Figure 4, from a point $P$ outside $\odot O$, two tangents to $\odot O$ are drawn, touching the circle at points $A$ and $B$. Point $C$ is on the minor arc $\overparen{A B}$. The tangent to $\odot O$ at $C$ intersects $PA$ and $PB$ at points $D$ and $E$, respectively. Let $O^{\prime}$ be the circu... | Proof: Auxiliary lines as shown in Figure 4. Let $OP$ intersect $AB$ at $H$ and the circumcircle of $\triangle PDE$ at $R$. Let $HC$ intersect $OO'$ at $G'$. Denote $PD = m$, $PE = n$, $DE = p$, $\angle DPE = 2\theta$, and the circumradius of $\triangle PDE$ as $R$.
Applying the Three Chords Theorem to the chords $PD$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,737 |
Example 5 The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta, \\
a_{n+2}-a_{n+1}+a_{n} \sin ^{2} \theta \cdot \cos ^{2} \theta=0, \theta \in\left(0, \frac{\pi}{2}\right) .
\end{array}
$$
Prove: $\frac{1}{2^{n-1}} \leqslant a_{n} \leqslant 1-\sin... | Given the characteristic equation
$$
x^{2}-x+\sin ^{2} \theta \cdot \cos ^{2} \theta=0
$$
the roots are $x_{1}=\sin ^{2} \theta, x_{2}=\cos ^{2} \theta$.
Let $a_{n}=A\left(\sin ^{2} \theta\right)^{n}+B\left(\cos ^{2} \theta\right)^{n}$.
Given $a_{1}=1, a_{2}=1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta$, we can solve f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,738 |
Example 6 In the Cartesian coordinate system $x O y$, the sequence of points $\left\{A_{n}\right\}$ on the positive $y$-axis and the sequence of points $\left\{B_{n}\right\}$ on the curve $y=\sqrt{2 x}(x \geqslant 0)$ satisfy $\left|O A_{n}\right|=\left|O B_{n}\right|=\frac{1}{n}$, the intercept of the line $A_{n} B_{n... | (1) From $| O B_{n} \left\lvert\,=\frac{1}{n}\right.$, we get
$$
b_{n}^{2}+2 b_{n}=\frac{1}{n^{2}} \text {. }
$$
Solving for $b_{n}$, we get $b_{n}=\sqrt{1+\frac{1}{n^{2}}}-1$.
Thus, for $n \in \mathbf{N}_{+}$, we have $b_{n}>b_{n+1}>0$.
Notice that the equation of the line $A_{n} B_{n}$ is
$$
\left(\sqrt{2 b_{n}}-\fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,739 |
Example 7 Given the sequence $\left\{a_{n}\right\}$ where the adjacent terms $a_{2 k-1}$ and $a_{2 k}$ are the roots of the equation
$$
x^{2}-\left(2^{k}+3 k\right) x+3 k \cdot 2^{k}=0
$$
and $a_{2 k-1} \leqslant a_{2 k}(k=1,2, \cdots)$.
$$
\begin{array}{l}
\text { Let } f(n)=\frac{1}{2}\left(\frac{|\sin n|}{\sin n}+3... | Notice the equation
$$
x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0
$$
has two roots $x_{1}=3 k, x_{2}=2^{k}$, from which we can find
$$
a_{1}=2, a_{2}=3, a_{3}=4, \dot{a}_{4}=6 \text {. }
$$
When $k \geqslant 4$, it is easy to prove that $2^{k}>3 k$.
Since $a_{2 k-1} \leqslant a_{2 k}(k=1,2, \cdots)$, we have
$$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,740 |
Example 8 Let $a_{n}=\sum_{k=1}^{n} \frac{1}{k(n+1-k)}$. Prove that for any positive integer $n \geqslant 2$, $a_{n+1}<a_{n}$.
(2007, National High School Mathematics Competition) | Notice that
$$
\frac{1}{k(n+1-k)}=\frac{1}{n+1}\left(\frac{1}{k}+\frac{1}{n+1-k}\right) \text {. }
$$
Therefore, $a_{n}=\frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k}$.
Thus, for any positive integer $n \geqslant 2$, we have
$$
\begin{array}{l}
\frac{1}{2}\left(a_{n}-a_{n+1}\right) \\
=\frac{1}{n+1} \sum_{k=1}^{n} \frac{1}{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,741 |
Example 9 Prove:
\[
\begin{array}{l}
\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\cdots+ \\
\frac{n}{(n-2)!+(n-1)!+n!}<\frac{1}{2} .
\end{array}
\] | Explanation: First, apply the "combination" strategy to the denominator of the fraction on the left side of the inequality, obtaining
$$
\begin{array}{l}
(n-2)!+(n-1)!+n! \\
=[1+n-1+n(n-1)](n-2)! \\
=n^{2}(n-2)!.
\end{array}
$$
Substitute this back into the original expression and then "decompose" each term, yielding
... | \frac{1}{2!}-\frac{1}{n!}<\frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 719,742 |
$\begin{array}{l}\text { 1. Prove: } \frac{1}{n+1}\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}\right) \\ >\frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2 n}\right)(n \geqslant 2, n \in \mathbf{N}) .\end{array}$ | (提示: From $\frac{1}{2}>\frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)$, $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}>\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}$, we get $1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}$ $>\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}+\frac{1}{n}\le... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,743 |
2. Let $a>0$ and $a \neq 1, f(x)=\frac{a\left(a^{2 x}-1\right)}{a^{x}\left(a^{2}-1\right)}$. Prove: $f(n)>n\left(n \in \mathbf{N}_{+}\right)$. | $\begin{array}{l}\text { (Hint: } f(n)=\frac{1}{a^{n}} \cdot \frac{a\left[1-\left(a^{2}\right)^{n}\right]}{1-a^{2}} \\ =\frac{1}{a^{n}}\left(a+a^{3}+a^{5}+\cdots+a^{2 n-1}\right) \\ =\frac{1}{2 a^{n}}\left[\left(a+a^{2 n-1}\right)+\left(a^{3}+a^{2 n-3}\right)+\cdots+\right. \\ \left.\quad\left(a^{2 n-1}+a\right)\right]... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 719,744 |
Example 3 As shown in Figure 3, in $\triangle A B C$, $D$ is a point on side $B C$, $\angle B A D=\alpha, \angle C A D=\beta$. Prove:
$$
\begin{array}{l}
\frac{\sin \beta}{A B}+\frac{\sin \alpha}{A C} \\
=\frac{\sin (\alpha+\beta)}{A D} .
\end{array}
$$ | Prove: From $S_{\triangle A B D}+S_{\triangle A C D}=S_{\triangle A B C}$, we get
$$
\begin{array}{l}
\frac{1}{2} A B \cdot A D \sin \alpha+\frac{1}{2} A C \cdot A D \sin \beta \\
=\frac{1}{2} A B \cdot A C \sin (\alpha+\beta) .
\end{array}
$$
Dividing both sides by $\frac{1}{2} A B \cdot A C \cdot A D$ completes the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 719,745 |
3. Let any real numbers $x, y$ satisfy $|x|<1,|y|<1$. Prove: $\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geqslant \frac{2}{1-x y}$. | $\begin{array}{l}\text { (Hint: } \frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \\ =\left(1+x^{2}+x^{4}+\cdots\right)+\left(1+y^{2}+y^{4}+\cdots\right) \\ =2+\left(x^{2}+y^{2}\right)+\left(x^{4}+y^{4}\right)+\cdots \\ \left.\geqslant 2+2 x y+2 x^{2} y^{2}+\cdots=\frac{2}{1-x y} .\right)\end{array}$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,746 |
$\begin{array}{l}\text { 4. Prove: } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^{n}-1}+\frac{1}{2^{n}} \\ >1+\frac{n}{2}(n \geqslant 2, n \in \mathbf{N}) \text {. }\end{array}$ | $$
\begin{array}{l}
= 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+ \\
\left(\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}\right)+\cdots+ \\
\left(\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\cdots+\frac{1}{2^{n}}\right) \\
>1+\frac{1}{2}+\left(\frac{1}{4}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 719,747 |
5. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
1=a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n} \leqslant \cdots,
$$
The sequence $\left\{b_{n}\right\}$ is defined as: $b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \cdot \frac{1}{\sqrt{a_{k}}}$. Prove: $0 \leqslant b_{n}<2$. | $\begin{array}{l}\text { (Hint: }\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}=\frac{a_{k-1}}{\sqrt{a_{k}}}\left(\frac{1}{a_{k-1}}-\frac{1}{a_{k}}\right) \\ =\frac{a_{k-1}}{\sqrt{a_{k}}}\left(\frac{1}{\sqrt{a_{k-1}}}+\frac{1}{\sqrt{a_{k}}}\right)\left(\frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_{k}}}\right)... | 0 \leqslant b_{n} < 2 | Algebra | proof | Yes | Yes | cn_contest | false | 719,748 |
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