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 |
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Three. (16 points) As shown in Figure 3, the side length of square $ABCD$ is 1, and points $M$ and $N$ are on $BC$ and $CD$ respectively, such that the perimeter of $\triangle CMN$ is 2. Find:
(1) The size of $\angle MAN$;
(2) The minimum area of $\triangle MAN$. | (1) As shown in Figure 4, extend $C B$ to $L$ such that $B L = D N$, then
Rt $\triangle A B L \cong$ Rt $\triangle A D N$.
Thus, $A L = A N, \angle 1 = \angle 2$,
$$
\begin{array}{l}
\angle N A L = \angle D A B = 90^{\circ}. \\
\text{Also, } M N = 2 - C N - C M \\
= D N + B M \\
= B L + B M = M L,
\end{array}
$$
and $... | 45^{\circ}, \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,570 |
Four. (18 points) A student, in order to plot the graph of the function $y=a x^{2}+$ $b x+c(a \neq 0)$, took 7 values of the independent variable: $x_{1}<x_{2}<$ $\cdots<x_{7}$, and $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}$, and calculated the corresponding $y$ values, listing them in Table 1.
Table 1
\begin{tabular... | Let $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}=d$, and the function value corresponding to $x_{i}$ be $y_{i}$. Then
$$
\begin{array}{l}
\Delta_{k}=y_{k+1}-y_{k} \\
=\left(a x_{k+1}^{2}+b x_{k+1}+c\right)-\left(a x_{k}^{2}+b x_{k}+c\right) \\
=a\left[\left(x_{k}+d\right)^{2}-x_{k}^{2}\right]+b\left[\left(x_{k}+d\right)... | 551 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,571 |
Example 3 Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a maximum positive number $l(a)$, such that on the entire interval $[0, l(a)]$, the inequality $|f(x)| \leqslant 5$ holds. For what value of $a$ is $l(a)$ maximized? Find this maximum value of $l(a)$ and prove your conclusio... | Solution: $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$.
Therefore, $\max _{x \in \mathbf{R}} f(x)=3-\frac{16}{a}$.
We discuss in two cases.
(1) When $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a} \text {. }$
Thus, $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$. Therefore,
$$
\begin{array}{l}
l(a)=\fra... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,572 |
3. Three people, A, B, and C, are playing chess. After each game, if it is a draw, the two people continue to play until there is a winner. The loser then steps down, and the other person plays against the winner. After several games, A has won 4 games and lost 2 games; B has won 3 games and lost 3 games. If C has lost... | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,575 |
4. For the system of inequalities about $x$ $\left\{\begin{array}{l}\frac{2 x+5}{3}>x-5, \\ \frac{x+3}{2}<x+a\end{array}\right.$, there are only 5 integer solutions. Then the range of values for $a$ is ( ).
(A) $-6<a<-\frac{11}{2}$
(B) $-6 \leqslant a<-\frac{11}{2}$
(C) $-6<a \leqslant-\frac{11}{2}$
(D) $-6 \leqslant a... | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 714,576 |
6. A certain product is divided into 10 quality levels. Producing the lowest level product, a profit of 8 yuan is obtained per piece. For each increase in level, the profit per piece increases by 2 yuan. Using the same working hours, the lowest level product can produce 60 pieces per day, and increasing one level will ... | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,578 |
7. As shown in Figure 2, in Rt $\triangle A B C$, $\angle C=90^{\circ}$, $\angle A=30^{\circ}$, the angle bisector of $\angle C$ intersects the external angle bisector of $\angle B$ at point $E$, and $A E$ is connected, then $\angle A E B$ is ( ).
(A) $50^{\circ}$
(B) $45^{\circ}$
(C) $40^{\circ}$
(D) $35^{\circ}$ | 7. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,579 |
8. Given quadrilateral $A B C D$. From the following conditions:
(1) $A B / / C D$;
(2) $B C / / A D$
(3) $A B=C D$;
(4) $B C=A D$;
(5) $\angle A=\angle C$;
(6) $\angle B=\angle D$
If any two of these conditions are chosen, the number of ways to conclude that "quadrilateral $A B C D$ is a parallelogram" is ( ) kinds.
... | 8. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,580 |
9. Given $-1<a<0$. Simplify
$$
\sqrt{\left(a+\frac{1}{a}\right)^{2}-4}+\sqrt{\left(a-\frac{1}{a}\right)^{2}+4}
$$
to get $\qquad$ . | 二、 $9 .-\frac{2}{a}$
Two、 $9 .-\frac{2}{a}$ | -\frac{2}{a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,581 |
10. As shown in Figure 3, given $A D$ $=D B=B C$. If $\angle C=\alpha$, then $\angle A B C=$ $\qquad$ . | $10.180^{\circ}-\frac{3}{2} \alpha$ | 180^{\circ}-\frac{3}{2} \alpha | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,582 |
Example 8 Given real numbers $a \neq b$, and satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13
(2004, “TRULY ${ }^{\circledR}$ Xinli Cup” National Junior High School Mathematics Comp... | Given that $a$ and $b$ are the roots of the equation
$$
(x+1)^{2}+3(x+1)-3=0
$$
After rearranging, we get $x^{2}+5 x+1=0$. Since
$$
\Delta=25-4>0, a+b=-5, a b=1,
$$
it follows that $a$ and $b$ are both negative. Therefore,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=-\frac{b}{a} \sqrt{a b}-\frac{a}{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,583 |
Example 9 Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Then the minimum value of $7x+5y$ is $\qquad$
(15th "Five Sheep Forest" Junior High School Mathematics Competition (Initial $\exists)$) | Explanation: Let $x+2y=5A, x+y=3B, A, B$ be integers.
Since $x, y \geqslant 0$, hence $A, B \geqslant 0$, solving we get
$$
\begin{array}{l}
x=6B-5A, y=5A-3B, \\
2x+y=9B-5A \geqslant 99, \\
S=7x+5y=27B-10A .
\end{array}
$$
The problem is transformed to: integers $A, B \geqslant 0, 6B \geqslant 5A \geqslant$ $3B, 9B \g... | 366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,584 |
5. Connect the corresponding trisection points on two pairs of opposite sides of a square, as shown in Figure 5, then the number of rectangles formed by all the line segments in the figure is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the trans... | 5. 36 .
There are 9 rectangles with an area of $1 \times 1$, 4 rectangles with an area of $2 \times 2$, 1 rectangle with an area of $3 \times 3$, 12 rectangles with an area of $1 \times 2$, 6 rectangles with an area of $1 \times 3$, and 4 rectangles with an area of $2 \times 3$. In total, there are 36. | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,586 |
6. As shown in Figure 6, the maximum height of a parabolic arc is 15, and the span is 60. Then, the height of the arc at a point 12 units away from the midpoint $M$ is $\qquad$. | 6. $12 \frac{3}{5}$.
With $M$ as the origin, the line $AB$ as the $x$-axis, and the line $MC$ as the $y$-axis, establish a Cartesian coordinate system. The parabola is the graph of the quadratic function $y = a x^{2} + 15$.
Therefore, point $B(30,0)$ lies on the parabola, which means $900 a + 15 = 0$.
Thus, $a = -\fra... | 12 \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,587 |
Three, (16 points) Given that $n$ is a positive integer, the graph of the linear function $y=$ $\frac{n+1}{n} x+n+1$ forms a triangle with the coordinate axes. The area of the circumcircle of this triangle is $\frac{25}{4} \pi$. Find the equation of this linear function.
---
The graph of the linear function $y=$ $\fr... | Three, as shown in Figure 12, we easily get
$$
A(-n, 0), B(0, n+1) \text {. }
$$
From the given information,
$$
\left(\frac{|A B|}{2}\right)^{2} \pi=\frac{25}{4} \pi \text {, }
$$
we get $|A B|=5$.
By $|A O|^{2}+|B O|^{2}=|A B|^{2}$, we have
$$
n^{2}+(n+1)^{2}=25 \text {. }
$$
Thus, $n=3$.
Therefore, the equation of... | y=\frac{4}{3} x+4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,588 |
Four, (16 points) Let the four-digit number $\overline{a b c d}$ be a perfect square, and $\overline{a b}=2 \overline{c d}+1$. Find this four-digit number. | Let $\overline{a b c d}=m^{2}$, then $32 \leqslant m \leqslant 99$.
Suppose $\overline{c d}=x$, then $\overline{a b}=2 x+1$. Therefore,
$$
100(2 x+1)+x=m^{2} \text {, }
$$
which simplifies to $67 \times 3 x=(m+10)(m-10)$.
Since 67 is a prime number, at least one of $m+10$ and $m-10$ must be a multiple of 67.
(1) If $m... | 5929 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,589 |
Five. (17 points) As shown in Figure 7, given that $\triangle ABC$ is inscribed in $\odot O$, $AD$ and $BD$ are tangents to $\odot O$. Draw $DE \parallel BC$, intersecting $AC$ at $E$, and connect $EO$ and extend it to intersect $BC$ at $F$. Prove: $BF = FC$.
---
The translation maintains the original text's formatti... | Five, as shown in Figure 13, connect $A O$ and $D O$, then
$$
\angle 1=\angle C \text{. }
$$
Since $D E / / B C$
$$
\begin{array}{l}
\Rightarrow \angle 2=\angle C \\
\Rightarrow \angle 1=\angle 2 \\
\Rightarrow D, A, E, O \text{ are concyclic } \\
\Rightarrow \angle D E O=\angle D A O=90^{\circ} \\
\Rightarrow D E \pe... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,590 |
1. Given real numbers $a, b$ satisfy $\frac{1}{a}+\frac{1}{b}=1, M=$ $\frac{b}{a}+\frac{a}{b}, N=\frac{b}{a^{2}}+\frac{a}{b^{2}}$. Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain | ,$- 1 . \mathrm{A}$.
Since $\frac{1}{a}+\frac{1}{b}=1$, therefore, $a+b=a b$. Hence
$$
\begin{array}{l}
M=\frac{a^{2}+b^{2}}{a b}, \\
N=\frac{a^{3}+b^{3}}{a^{2} b^{2}}=\frac{(a+b)\left(a^{2}-a b+b^{2}\right)}{a^{2} b^{2}} \\
=\frac{a^{2}-a b+b^{2}}{a b}=\frac{a^{2}+b^{2}}{a b}-1=M-1 .
\end{array}
$$
Therefore, $M>N$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,592 |
2. Let $A_{n}$ denote the product of multiples of $n$ not exceeding 100, where $n$ is a positive integer. For example, $A_{3}=3 \times 6 \times 9 \times \cdots \times 99$. Then the greatest common divisor of $A_{2}, A_{3}, A_{4}, \cdots, A_{17}$ is ( ).
(A)I
(B) 6
(C) 30
(D) 120 | 2. D.
$$
\begin{array}{l}
A_{2}=2 \times 4 \times 6 \times \cdots \times 100=2^{50}(1 \times 2 \times 3 \times \cdots \times 50), \\
A_{3}=3 \times 6 \times 9 \times \cdots \times 99=3^{33}(1 \times 2 \times 3 \times \cdots \times 33), \\
\cdots \cdots \\
A_{16}=16 \times 32 \times 48 \times 64 \times 80 \times 96 \\
... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,593 |
3. As shown in Figure 1, $A$ and $B$ are two points on the inverse proportion function $y=\frac{k}{x}(k>0)$, $A C \perp x$-axis at point $C$, $B D \perp y$-axis at point $D$, and $A C$ and $B D$ intersect at point $E$. The relationship between the areas of $\triangle A D E$ and $\triangle B C E$ is ().
(A) $S_{\triangl... | 3. B.
$$
\begin{array}{l}
S_{\text {AASE }}=\frac{1}{2}(O C \cdot A C-O C \cdot E C)=\frac{1}{2}(k-D E \cdot O D) \\
=\frac{1}{2}(D B \cdot O D-D E \cdot O D)=S_{\triangle B C E} .
\end{array}
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,594 |
4. As shown in Figure 2, $A B C D$ is a square with an area of 1, and $\triangle P B C$ is an equilateral triangle. Then the area of $\triangle B P D$ is ( ).
(A) $\frac{\sqrt{3}-1}{2}$
(B) $\frac{2 \sqrt{3}-1}{8}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{3}-1}{4}$ | 4.D.
As shown in Figure 6, construct $P E \perp C D$ at $E$, and $P F \perp B C$ at $F$. Then $P E=\frac{1}{2} C P$ $=\frac{1}{2}, P F=P C \sin 60^{\circ}=\frac{\sqrt{3}}{2}$. Therefore,
$$
\begin{array}{l}
S_{\triangle P P D}=S_{\text {quadrilateral } B C D P}-S_{\triangle B C D} \\
=S_{\triangle C P}+S_{\triangle A ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,595 |
6. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is
$\qquad$ .
(11th "Hope Cup" National Mathematics Invitational Competition (Grade 8)) | (Tip: Construct a symmetrical figure. Answer: 13) | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,596 |
5. As shown in Figure 3, the radii of the three semicircles are all $R$, and their centers $C_{1}, C_{2}, C_{3}$ are on the same straight line, with each center lying on the circumference of the other semicircle. $\odot C_{4}$ is tangent to these three semicircles. Let $r$ represent the radius of $\odot C_{4}$. Then, $... | 5.C.
As shown in Figure 7, connect $C_{1} C_{4}$ and $C_{2} C_{4}$. Then $C_{2} C_{4}=R-r$,
$$
\begin{array}{r}
C_{1} C_{4}=R+r, C_{1} C_{2}=R . \text { Therefore, } \\
(R-r)^{2}+R^{2}=(R+r)^{2} .
\end{array}
$$
Solving this, we get $R^{2}=4R$.
Thus, $R: r=4: 1$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,597 |
6. Let $x, y, z$ satisfy $x+y+2z=6, 2xy+yz+zx=4$. Then the maximum value of $z$ is ( ).
(A) 2
(B) $\frac{7}{3}$
(C) 3
(D) $\frac{8}{3}$ | 6.B.
$$
\begin{array}{l}
\text { From } x+y=6-2 z \text {, we get } \\
2 x y=4-z(x+y) \\
=4-z(6-2 z)=2 z^{2}-6 z+4,
\end{array}
$$
which means $x y=z^{2}-3 z+2$.
Therefore, $x, y$ are the two roots of the quadratic equation
$$
t^{2}+(2 z-6) t+z^{2}-3 z+2=0
$$
Thus,
$$
\Delta=(2 z-6)^{2}-4\left(z^{2}-3 z+2\right)=-12 ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,598 |
$\begin{array}{l}\text { 1. If } n \text { satisfies }(n-2003)^{2}+(2004-n)^{2} \\ =1 \text {, then }(2004-n)(n-2003)=\end{array}$ | $\begin{array}{l} \text { II.1.0. } \\ \text { From }(n-2003)^{2}+2(n-2003)(2004-n)+ \\ (2004-n)^{2}-2(n-2003)(2004-n) \\ =(n-2003+2004-n)^{2}-2(n-2003)(2004-n) \\ =1-2(n-2003)(2004-n)=1, \\ \text { we get }(n-2003)(2004-n)=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,599 |
2. Given that $n$ is a natural number, $11 n+10$ and $10 n+3$ are both multiples of some natural number $d$ greater than 1. Then $d=$ $\qquad$ | 2.67.
From $11 n+10$ and $10 n+3$ both being multiples of $d$, we know
$$
10(11 n+10)-11(10 n+3)=67
$$
is still a multiple of $d$. But 67 is a prime number, $d \neq 1$, so $d=67$. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,600 |
3. Several tourists are to board cars, with the requirement that the number of people in each car is equal. If each car carries 28 people, then 1 person is left without a car; if one car is reduced, then all tourists can be evenly distributed among the remaining cars. Given that each car can accommodate a maximum of 35... | $3.841,30$.
Suppose there are $k$ cars, and later each car carries $x$ people, then we have $28 k+1=n(k-1)$.
Thus, $n=\frac{28 k+1}{k-1}=28+\frac{29}{k-1}$.
Therefore, $k-1=1$ or $k-1=29$, which means $k=2$ or $k=30$. When $k=2$, $n=57$ is not reasonable, so we discard it.
Upon verification, we find that $k=30, n=29$.
... | 841, 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,601 |
4. As shown in Figure 4, in quadrilateral $ABCD$, $\angle ABC=135^{\circ}$, $\angle BCD=120^{\circ}$, $AB=\sqrt{6}$, $BC=5-\sqrt{3}$, $CD=6$. Then $AD=$ | $$
\begin{array}{l}
CF=\frac{1}{2} CD=3, \\
DF=CD \sin 60^{\circ}=3 \sqrt{3}, \\
BE=AB \sin 45^{\circ}=\sqrt{3}, \\
AE=BE=\sqrt{3}, \\
DG=DF-FG=DF-AE=2 \sqrt{3}, \\
AG=EF=EB+BC+CF=\sqrt{3}+5-\sqrt{3}+3=8. \\
\text{Therefore, } AD=\sqrt{AG^{2}+DG^{2}}=\sqrt{8^{2}+(2 \sqrt{3})^{2}}=2 \sqrt{19}. \\
\end{array}
$$ | 2 \sqrt{19} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,602 |
One, (20 points) Given the quadratic equation $x^{2}+a x+b=0$ has two consecutive integer roots, and the quadratic equation $x^{2}+b x+a=0$ has integer roots. Find the values of $a, b$.
| Let the two roots of $x^{2}+a x+b=0$ be $n$ and $n+1$ (where $n$ is an integer). Then
$$
2 n+1=-a, \quad n(n+1)=b.
$$
Thus, the equation $x^{2}+b x+a=0$ can be rewritten as
$$
x^{2}+n(n+1) x-(2 n+1)=0.
$$
(1) If $n \geqslant 0$, then equation (1) has one positive and one negative root. Let the positive integer root be... | (a, b) \text{ are } (-1,0), (-3,2), (5,6) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,603 |
II. (25 points) As shown in Figure 5, in $\triangle A B C$, $D$ and $E$ are two points on $B C$, $\angle B A D = \angle C A E$, and the incircles of $\triangle A B E$ and $\triangle C A D$ touch $B C$ at points $M$ and $N$ respectively. Prove:
$$
\frac{1}{M B}+\frac{1}{M D}=\frac{1}{N C}+\frac{1}{N D}
$$ | As shown in Figure 9, let the incircle of $\triangle ABE$ touch $BA$ and $AE$ at points $F$ and $G$, respectively, and let $BM = BF = x$, $AF = AG = y$, and $EG = EM = z$.
By the tangent segment theorem, we have
$$
\begin{array}{c}
x + y = AB, \\
y + z = AE, \\
z + x = BE.
\end{array}
$$
Thus, $x = BM = \frac{1}{2}(AB... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,604 |
Three, (25 points) Prove that for any 2003 natural numbers $a_{1}, a_{2}, \cdots, a_{2003}$, it is always possible to find some of these numbers whose sum is divisible by 2003.
| Let $S_{1}=a_{1}, S_{2}=a_{1}+a_{2}, \cdots, S_{203}=a_{1}+a_{2}+\cdots+a_{2003}$, we know that the remainders of these 2003 numbers when divided by 2003 can be at most $0,1,2, \cdots, 2002$, a total of 2003 numbers.
If there exists some $S_{k}$ in $S_{i}(i=1,2, \cdots, 2003)$ that can be divided by 2003, then $S_{k}$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,605 |
1. If the set $A=\left\{\lg x, \lg y, \lg \left(x+\frac{y}{x}\right)\right\}$ has a subset $B=\{0,1\}$, then the maximum possible value of the largest element in $A$ is ( ). | $-、 1 . C$.
If $\lg x=0, \lg y=1$, then
$$
x=1, y=10, \lg \left(x+\frac{y}{x}\right)=\lg 11>1 \text {; }
$$
If $\lg x=1, \lg y=0$, then
$$
\lg \left(x+\frac{y}{x}\right)=\lg \left(10+\frac{1}{10}\right)<\lg 11 \text {; }
$$
If $\lg x=0, \lg \left(x+\frac{y}{x}\right)=1$, then $x=1, y=9, \lg y=\lg 9<\lg 11$;
If $\lg y... | \lg 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,606 |
Example 1 If the non-empty set $A=\{x \mid 2 a+1 \leqslant x \leqslant$ $3 a-5\}, B=\{x \mid 3 \leqslant x \leqslant 22\}$, then the set of all $a$ that makes $A \subseteq(A$ $\cap B)$ true is ( ).
(A) $\{a \mid 1 \leqslant a \leqslant 9\}$
(B) $\{a \mid 6 \leqslant a \leqslant 9\}$
(C) $\{a \mid a \leqslant 9\}$
(D) $... | Solution: According to
$$
A \subseteq(A \cap B)
$$
we know $A \subseteq B$,
as shown in Figure 1.
Thus,
$$
\left\{\begin{array}{l}
2 a+1 \geqslant 3, \\
3 a-5 \leqslant 22, \\
3 a-5 \geqslant 2 a+1
\end{array}\right.
$$
$\Rightarrow 6 \leqslant a \leqslant 9$, i.e., $a \in\{a \mid 6 \leqslant a \leqslant 9\}$.
Therefo... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,607 |
$\begin{array}{llll}\text { (A)1 } & \text { (B) } 2 \lg 3 & \text { (C) } \lg 11 & \text { (D) } \lg 15\end{array}$
2. $ABCD$ is a unit square, $P$ is the midpoint of side $AB$. By folding $ABCD$ along $PC$ and $PD$ to form a tetrahedron $PACD$ (where $A$ coincides with $B$), the volume of the tetrahedron $PACD$ is eq... | 2. B.
After folding, $A$ coincides with $B$.
Since $P A \perp A C, P A \perp A D$, therefore, $P A \perp$ plane $A C D$.
$$
=\frac{1}{3} \times \frac{1}{2} \times 1^{2} \times \frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}}{24} \text {. }
$$ | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,608 |
3. If the volume of a certain cylinder is numerically equal to its surface area, then the minimum possible value of the volume of the cylinder is ( ).
(A) $48 \pi$
(B) $50 \pi$
(C) $54 \pi$
(D) $66 \pi$ | 3.C.
Let the radius and height of the cylinder be $r$ and $h$, respectively. According to the problem, $\pi r^{2} h=2 \pi r^{2}+2 \pi r h$,
which simplifies to $h=\frac{2 r}{r-2}$.
Thus, $V=\pi r^{2} h=2 \pi \cdot \frac{r^{3}}{r-2}$.
Let $r-2=x^{2} (x>0)$, then
$$
\begin{array}{l}
V=2 \pi \cdot \frac{(2+x)^{3}}{x}=2 \... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,609 |
4. Let $f_{0}=0, f_{1}=1, f_{n+2}=f_{n+1}+f_{n}(n=$ $0,1,2, \cdots)$, and let $x_{1} \in \mathbf{R}, x_{n}=\frac{f_{n-1}+f_{n-2} x_{1}}{f_{n}+f_{n-1} x_{1}}$ $(n=2,3,4, \cdots)$. If $x_{2004}=\frac{1}{x_{1}}-1$, then $x_{1}$ equals
(A) $\frac{-1 \pm \sqrt{5}}{2}$
(B) $\frac{-1-\sqrt{5}}{2}$
(C) $\frac{-1+\sqrt{5}}{2}$
... | 4. A.
Let $x_{n}=\frac{1}{x_{1}}-1(n=2004)$, then
$$
\begin{array}{l}
\frac{f_{n-1}+f_{n-2} x_{1}}{f_{n}+f_{n-1} x_{1}}=\frac{1}{x_{1}}-1 \\
\Rightarrow \frac{1}{x_{1}}=\frac{f_{n+1}+f_{n} x_{1}}{f_{n}+f_{n-1} x_{1}} \\
\Rightarrow f_{n+1} x_{1}+f_{n} x_{1}^{2}=f_{n}+f_{n-1} x_{1} \\
\Rightarrow f_{n} x_{1}^{2}+f_{n} ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,610 |
5. Let $a, b, c, d$ be real numbers, and $a^{2}+b^{2}+c^{2}-$ $d^{2}+4=0$. Then the maximum value of $3 a+2 b+c-4|d|$ is ( ).
(A) $\sqrt{2}$
(B) 0
(C) $-\sqrt{2}$
(D) $-2 \sqrt{2}$ | 5.D.
From the problem, we have $a^{2}+b^{2}+c^{2}+2^{2}=d^{2}$, so, $4^{2} d^{2}=\left(a^{2}+b^{2}+c^{2}+2^{2}\right)\left[3^{2}+2^{2}+1^{2}+(\sqrt{2})^{2}\right]$ $\geqslant(3 a+2 b+c+2 \sqrt{2})^{2}$ (using the Cauchy-Schwarz inequality).
Thus, $4|d| \geqslant|3 a+2 b+c+2 \sqrt{2}|$
$$
\geqslant 3 a+2 b+c+2 \sqrt{2}... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,611 |
6. Among the 99 positive integers $1,2, \cdots, 99$, if $k$ numbers are arbitrarily taken, such that there must be two numbers $a$ and $b (a \neq b)$ satisfying $\frac{1}{2} \leqslant \frac{b}{a} \leqslant 2$. Then the smallest possible value of $k$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 6.B.
Divide the 99 positive integers from 1 to 99 into 6 groups so that the ratio of any two numbers in each group is within the closed interval $\left[\frac{1}{2}, 2\right]$, and the number of elements in each group is as large as possible. The groups are as follows:
$$
\begin{array}{l}
A_{1}=\{1,2\}, A_{2}=\{3,4,5,6... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,612 |
1. If $z$ is a complex number, and $(z+1)(\bar{z}+1)=2004$, then the range of $|z|$ is $\qquad$ . | 2.1. $[\sqrt{2004}-1, \sqrt{2004}+1]$.
Let $z=x+y \mathrm{i}, r=\sqrt{x^{2}+y^{2}}$, then $(x+1+y \mathrm{i})(x+1-y \mathrm{i})=2004$,
which means $(x+1)^{2}+y^{2}=2004$.
Thus, $r^{2}=2003-2 x$.
Also, $2004-(x+1)^{2}=y^{2} \geqslant 0$, so
$-\sqrt{2004}-1 \leqslant x \leqslant \sqrt{2004}-1$.
Therefore, $(\sqrt{2004}-1... | [\sqrt{2004}-1, \sqrt{2004}+1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,613 |
2. Let $m$ be a given positive integer, then the largest positive integer $n$ such that $m^{2}+n$ divides $n^{2}+m$ is $\qquad$ . | 2. $m^{4}-m^{2}+m$.
From the problem, we know that $\frac{n^{2}+m}{n+m^{2}} \in \mathbf{Z}$.
Since $\frac{n^{2}+m}{n+m^{2}}$
$=\frac{\left(n+m^{2}\right)^{2}-2 m^{2}\left(n+m^{2}\right)+m^{4}+m}{n+m^{2}}$
$=n+m^{2}-2 m^{2}+\frac{m^{4}+m}{n+m^{2}}$,
therefore, $\frac{m^{4}+m}{n+m^{2}}$ is a positive integer.
Thus, $n+m... | m^{4}-m^{2}+m | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,614 |
3. The minimum distance from the origin to a point on the curve $x^{2}+x y-y^{2}=1$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $\sqrt{\frac{2 \sqrt{5}}{5}}$.
Let $x=r \cos \theta, y=r \sin \theta (r>0)$. Substituting into the curve equation, we get
$$
r^{2}\left(\cos ^{2} \theta+\cos \theta \cdot \sin \theta-\sin ^{2} \theta\right)=1 .
$$
Therefore, $r^{2}=\frac{1}{\cos 2 \theta+\frac{1}{2} \sin 2 \theta}$
$$
\geqslant \frac{1}{\sqrt{1^{2... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,615 |
4. If the solution set of the inequality $k x^{2}-2|x-1|+$ $6 k<0$ with respect to $x$ is an empty set, then the range of values for $k$ is $\qquad$. | 4. $k \geqslant \frac{1+\sqrt{7}}{6}$.
From the problem, for every $x \in \mathbf{R}$, we have
$$
k x^{2}-2|x-1|+6 k \geqslant 0,
$$
which means $k \geqslant f(x)=\frac{2|x-1|}{x^{2}+6}$,
Therefore, $k \geqslant \max f(x), x \in \mathbf{R}$.
When $x \geqslant 1$, let $t=x-1, t \geqslant 0$, then
$$
\begin{array}{l}
... | k \geqslant \frac{1+\sqrt{7}}{6} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,616 |
5. If three non-zero real numbers $x(y-z), y(z-$ $x), z(y-x)$ form a geometric sequence, then its common ratio is $\qquad$ | 5. $\frac{1 \pm \sqrt{5}}{2}$.
Notice that $x(y-z)+y(z-x)=z(y-x)$, so, $1+\frac{y(z-x)}{x(y-z)}=\frac{z(y-x)}{x(y-z)}=\frac{y(z-x)}{x(y-z)} \cdot \frac{z(y-x)}{y(z-x)}$,
i.e., $1+q=q^{2} (q$ is the common ratio $)$.
Solving for $q$ yields $q=\frac{1 \pm \sqrt{5}}{2}$. | \frac{1 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,617 |
Example 2 Let the set $M=\{u \mid u=12 m+8 n+$ $4 l, m, n, l \in \mathbf{Z}\}, N=\{u \mid u=20 p+16 q+$ $12 r, p, q, r \in \mathbf{Z}\}$. Prove: $M=N$. | Solution: For any element $u$ in $N$, we have
$$
\begin{array}{l}
u=20 p+16 q+12 r \\
=12 r+8(2 q)+4(5 p) \in M,
\end{array}
$$
thus, $N \subseteq M$.
On the other hand, for any element $u$ in $M$, we have
$$
\begin{array}{l}
u=12 m+8 n+4 l \\
=20 n+16 l+12(m-n-l) \in N,
\end{array}
$$
thus, $M \subseteq N$.
In concl... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,618 |
6. If real numbers $x, y$ satisfy $x \geqslant 0$, and
$$
\max \{1-x, x-1\} \leqslant y \leqslant x+2 \text {, }
$$
then the minimum value of the bivariate function $u(x, y)=2 x+y$ is
$\qquad$ . | 6.1.
From the given information, we have
$$
\begin{array}{l}
u(x, y)=2 x+y \geqslant 2 x+\max \{1-x, x-1\} \\
=\max \{2 x+(1-x), 2 x+(x-1)\} \\
=\max \{x+1,3 x-1\} \\
\geqslant \max \{1,-1\}=1,
\end{array}
$$
i.e., $u(x, y) \geqslant 1$, with equality holding if and only if $x=0, y=1$. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,619 |
Three, (20 points) Given that the two foci of an ellipse are $F_{1}(-3,0), F_{2}(3,0)$, and the line $y=x-1$ is one of its normals (a line passing through the point of tangency and perpendicular to the tangent). Try to find the length of the chord (normal chord) intercepted by this normal on the ellipse.
| Three, it is easy to get that the symmetric point of $F_{2}(3, 0)$ about the line $y=x-1$ is $F_{2}^{\prime}(1,2)$, as shown in Figure 1. Let $A$ and $B$ be the points of intersection of the ellipse with the normal line $y=x-1$. Then $A F_{1}$ and $A F_{2}$ are symmetric about the line $A B$. Since $F_{2}^{\prime}$ is ... | \frac{80}{9} \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,620 |
Four, (20 points) Let $a \in \mathbf{R}, A=\left\{x \mid 2^{1+x}+2^{1-x}\right.$ $=a\}, B=\{\sin \theta \mid \theta \in \mathbf{R}\}$. If $A \cap B$ contains exactly one element, find the range of values for $a$. | Four, first, $B=[-1,1], 0 \in B$.
Obviously, $a=2^{1+x}+2^{1-x} \geqslant 2 \sqrt{2^{1+x} \times 2^{1-x}}=4$, equality holds if and only if $2^{1+x}=2^{1-x}$, i.e., $x=0$.
Therefore, $a=4$ satisfies the problem's requirements.
Next, we prove that $a>4$ does not satisfy the problem's requirements.
Assume $a>4$, and ther... | a=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,621 |
Five. (20 points) Given $a_{1}=1, a_{n+1}=2 a_{n}+(-1)^{n} n+1(n=1,2,3, \cdots)$. Find the general term formula $a_{n}$ (require it to be written as a finite sum, and the number of terms should be independent of $n$).
| Let $a_{n}=2^{n}-1+b_{n}$, then $b_{1}=0$, and
$$
2^{n+1}-1+b_{n+1}=2^{n+1}-2+2 b_{n}+(-1)^{n} n+1 \text {, }
$$
i.e., $b_{1}=0, b_{n+1}=2 b_{n}+(-1)^{n} n$.
Let $b_{n}=(-1)^{n} c_{n}$, then
$$
c_{1}=0, c_{n+1}=-2 c_{n}-n \text {. }
$$
Adding $s(n+1)+t$ to both sides of the above equation, we get
$$
\begin{array}{l}
... | \frac{8}{9} \times 2^{n}+\left(\frac{1}{9}-\frac{n}{3}\right)(-1)^{n}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,622 |
One, (50 points) In $\triangle ABC$, $AB=AC$. Try to find a necessary and sufficient condition that $\triangle ABC$ must satisfy so that there exists a point $P$ inside $\triangle ABC$ that meets the following conditions:
(1) $PB=AB$;
(2) $\angle ABP=2 \angle ACP$.
---
The translation preserves the original text's fo... | As shown in Figure 2, let the circles with radius $AB$ and centers at $A$ and $B$ intersect at another point $Q$. Clearly, $\triangle QAB$ is an equilateral triangle, and $AQ = AC, \angle AQC = \angle ACQ$.
Since point $P$ is inside $\triangle ABC$ and $PB = AB$, point $P$ lies on the arc $\widehat{QQ}$. Also, $\angle... | 60^\circ < \angle A < 120^\circ | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,623 |
(1) Try to find the largest positive integer $k$, such that there exists a triangle with side lengths all being positive integers not greater than $n$, and the difference between any two sides (larger minus smaller) is not less than $k$;
(2) Try to find all positive integers $n(n \geqslant 4)$, such that the triangle c... | (1) Let the lengths of the three sides of a triangle be positive integers $a, b, c$, and
$1 \leqslant a < b < c \leqslant n, a + b \geqslant c + 1$,
$c - b \geqslant k, b - a \geqslant k$.
Then, $b \leqslant c - k, a \leqslant b - k \leqslant c - 2k$.
Thus, $(c - 2k) + (c - k) \geqslant a + b \geqslant c + 1$.
Hence, $... | n \equiv 1 \pmod{3} \text{ and } n \geqslant 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,624 |
Three. (50 points) Given $x_{1}=\frac{1}{2}, x_{n+1}=x_{n}+\left(\frac{x_{n}}{n}\right)^{2}(n=1,2,3, \cdots)$. Prove:
(1) $x_{n}>\frac{6 n}{5 n+6}(n \geqslant 3)$;
(2) $x_{n}<\frac{3}{2}(n \geqslant 1)$. | Three, (1) From the recursive formula, we get
$$
\frac{1}{x_{i}}-\frac{1}{x_{i+1}}=\frac{1}{x_{i}}-\frac{i^{2}}{i^{2} \cdot x_{i}+x_{i}^{2}}=\frac{1}{i^{2}+x_{i}} \text {. }
$$
By mathematical induction, it is easy to prove
$$
0<\frac{1}{x_{i}}-\frac{1}{x_{i+1}}=\frac{1}{i^{2}+x_{i}}<\frac{1}{i^{2}+i}=\frac{1}{i}-\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,625 |
Let $x, a_{1}(i=0,1, \cdots, 2004)$ be integers, and $51\left(a_{0} x^{2004}+a_{1} x^{2003}+\cdots+a_{2003} x+a_{2004}\right)$, but $5 \nmid a_{2004}$. Prove: there exists an integer $y$, such that
$$
51\left(a_{2004} y^{2004}+a_{2003} y^{2003}+\cdots+a_{1} y+a_{0}\right) .
$$ | Proof: From $51\left(a_{0} x^{2004}+a_{1} x^{2003}+\cdots+a_{2004}\right)$, but $5 \times a_{2004}$ has
$$
\begin{array}{l}
5 \nmid\left(a_{0} x^{2004}+a_{1} x^{2003}+\cdots+a_{20 \beta} x\right) \text {, hence } 5 \nmid x \text {. } \\
y^{2004}\left(a_{0} x^{2004}+a_{1} x^{2003}+\cdots+a_{2004}\right)- \\
\left(a_{200... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,626 |
Find all the side lengths of triangles that satisfy the following conditions:
(1) The side lengths of the triangle are integers;
(2) The inradius of the triangle is 2. | Let the three sides of the triangle be $a, b, c$, the semi-perimeter be $p$, the inradius be $r$, and $S$ represent the area of the triangle. By Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$ and $S=pr=2p$, we have
$$
2_{p}=\sqrt{p(p-a)(p-b)(p-c)}.
$$
Therefore, $4 p=(p-a)(p-b)(p-c)$, which means
$$
4 \cdot \frac{a+b+c}{... | (6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,627 |
139 Let $a, b, \lambda > 0$. Prove:
(1) When $\lambda \geqslant 3$,
$$
\sqrt{\frac{a}{a+\lambda b}}+\sqrt{\frac{b}{\lambda a+b}} \geqslant \frac{2}{\sqrt{1+\lambda}} \text{; }
$$
(2) When $0<\lambda \leqslant 2$,
$$
\sqrt{\frac{a}{a+\lambda b}}+\sqrt{\frac{b}{\lambda a+b}} \leqslant \frac{2}{\sqrt{1+\lambda}} .
$$ | Proof: (1) Let $x_{1}=\frac{b}{a}, x_{2}=\frac{a}{b}$, then $x_{1}, x_{2}>0, x_{1} x_{2}=1$. Therefore,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow \frac{1}{\sqrt{1+\lambda x_{1}}}+\frac{1}{\sqrt{1+\lambda x_{2}}} \geqslant \frac{2}{\sqrt{1+\lambda}} \\
\Leftrightarrow \sqrt{1+\lambda}\left(\sqrt{1+\lamb... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,628 |
Example 3 The number of proper subsets of the set $\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., x\right.$ $\left.\in \mathbf{N}_{+}\right\}$ is $\qquad$ .
(1996, National High School Mathematics Competition) | $$
\begin{array}{l}
\text { Solution: } A=\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., x \in \mathbf{N}_{+}\right\} \\
=\left\{x \left\lvert\,-1 \leqslant-\log _{x} 10<-\frac{1}{2}\right., x \in \mathbf{N}_{+}\right\} \\
=\left\{x \mid 1 \leqslant \lg x<2, x \in \mathbf{N}_{+}\right\} ... | 2^{90}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,629 |
Given $x 、 y$ are positive numbers satisfying $x+y=1$. Prove: $\left(x^{3}-\frac{1}{x^{3}}\right)\left(y^{3}-\frac{1}{y^{3}}\right) \geqslant 62 \frac{1}{64}$. | Proof: Since $0 < xy \leqslant \left(\frac{x+y}{2}\right)^{2} = \frac{1}{4}$, hence
$$
\begin{array}{l}
\left(x^{3}-\frac{1}{x^{3}}\right)\left(y^{3}-\frac{1}{y^{3}}\right) \\
=x^{3} y^{3}-\frac{x^{6}+y^{6}}{x^{3} y^{3}}+\frac{1}{x^{3} y^{3}} \\
=x^{3} y^{3}-\frac{\left(x^{3}+y^{3}\right)^{2}}{x^{3} y^{3}}+\frac{1}{x^{... | 62 \frac{1}{64} | Inequalities | proof | Yes | Yes | cn_contest | false | 714,630 |
Example 4 Given the set $\{1,2,3,4,5,6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
(1996, Shanghai High School Mathematics Competition) | Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that have the given property. Then, the subsets of $\{1,2, \cdots, k, k+1, k+2\}$ that have the given property can be divided into two categories: the first category does not contain $k+2$, and there are $a_{k+1}$ such subsets; the second category con... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,631 |
Example 5 Let $M=\{1,2, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. The maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Solution: From the given, we know that at least one of the numbers $k$ and $15k$ does not belong to $A$.
Since $\left[\frac{1995}{15}\right]=133$, when $k=134,135$, $\cdots, 1995$, $15k$ definitely does not belong to $A$.
Similarly, $\left[\frac{133}{15}\right]=8$, when $k=9,10, \cdots, 133$, $k$ and $15k$ cannot bot... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,632 |
Example 6 Given that for any real number $x$, the function $f(x)$ is defined, and $f^{2}(x) \leqslant 2 x^{2} f\left(\frac{x}{2}\right)$. If $A=\{a \mid$ $\left.f(a)>a^{2}\right\} \neq \varnothing$, prove: $A$ is an infinite set.
(1994, Jiangsu Province High School Mathematics Competition) | Solution: In $f^{2}(x) \leqslant 2 x^{2} f\left(\frac{x}{2}\right)$, let $x=0$, we get $f^{2}(0) \leqslant 0$. Therefore, $f(0)=0,0 \notin A$.
Thus, there must be a real number $a(a \neq 0)$ such that $a \in A$, i.e., $f(a)>a^{2}$. From the given,
$$
f\left(\frac{a}{2}\right) \geqslant \frac{f^{2}(a)}{2 a^{2}}>\frac{a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,633 |
Example 7 Let the set $M=\{1,2, \cdots, 1000\}$, and for any non-empty subset $X$ of $M$, let $a_{x}$ denote the sum of the largest and smallest numbers in $X$. Then, the arithmetic mean of all such $a_{x}$ is $\qquad$
(1991, National High School Mathematics Competition) | Analysis: For any subset $X=\left\{b_{1}, b_{2}, \cdots, b_{k}\right\}$ of set $M$, assume $b_{1}<b_{2}<\cdots<b_{k}$, then there must exist another subset $X^{\prime}=\left\{1001-b_{1}, 1001-\right.$ $\left.b_{2}, \cdots, 1001-b_{k}\right\}$, at this time
$$
\begin{array}{l}
a_{x}=b_{1}+b_{k}, \\
a_{x^{\prime}}=\left(... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,634 |
Example 8 Let the set $S_{n}=\{1,2, \cdots, n\}$. If $X$ is a subset of $S_{n}$, the sum of all numbers in $X$ is called the "capacity" of $X$ (the capacity of the empty set is defined as 0). If the capacity of $X$ is odd (even), then $X$ is called an odd (even) subset of $S_{n}$. Prove: The number of odd subsets of $S... | Analysis: If a one-to-one correspondence can be established between the odd and even subsets of $S_{n}$, then it indicates that the number of odd and even subsets are equal.
Proof: For any even subset $B$ of $S_{n}$, let
$A=\left\{\begin{array}{ll}B \cup\{1\}, & 1 \notin B \text { when, } \\ B \backslash\{1\}, & 1 \in ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,635 |
Example 10 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text {. }
$$
Find the minimum value of $b^{2}-4 a c$.
(2004, "TRULY ${ }^{\circledR}$ Xinli Cup" National Junior High School Mathematics Competition) | Let $y=a x^{2}+b x+c$.
From $a<0$, we know $\Delta=b^{2}-4 a c>0$.
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
Since $x_{1} x_{2}=\frac{c}{a}<0$, let's assume $x_{1}<x_{2}$.
The... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,636 |
Example 9 Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. Then, what is the maximum number of elements that $S$ can have?
(43rd American High School Mathematics Examination) | Solution: For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of the remainders when they are divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7. In $A_{i}$, each element has a remainder of $i$ when divided... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,637 |
Example 10 Let the set $A=\{1,2, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17 | $(a+$ $b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) Find the number of a set of binary subsets of $A$, which are pairwise disjoint and have prope... | Solution: (1) Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: $[0],[1], \cdots,[16]$.
Since $366=17 \times 21+9$, the classes $[1],[2], \cdots$, [9] each contain 22 numbers; [10], [11], $\cdots$, [16] and [0] each contain 21 numbers.
(i) When $a, b \in[0]$, the number of subsets wi... | 3928 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,638 |
1. Let set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$ be a collection of non-empty subsets of $A$, and when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Then the maximum value of $k$ is $\qquad$
(1999, National High School Mathematics League Guangxi Preliminary Contest (High) Three)... | (It is easy to see that the family of all subsets of $A$ containing at most three elements meets the requirements of the problem, where the number of subsets is $\mathrm{C}_{10}^{\mathrm{l}}+$ $\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{3}=175$. It remains to prove that this is the maximum value.) | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,639 |
2. Among all the subsets of the set $M=\{1,2, \cdots, 10\}$, there is a family of distinct subsets such that the intersection of any two of them is not empty. Then, the maximum number of subsets in this family is ( ) .
(A) $2^{10}$
(B) $2^{9}$
(C) $10^{2}$
(D) $9^{2}$ | (提示: 对 $M$ 的任一子集 $A$, 易知 $A$ 与 $M-A$ 至多有一个在题设的子集族, 力中, 故 $\mid$. $b \mid \leqslant 2^{9}$.)
( Hint: For any subset $A$ of $M$, it is easy to see that at most one of $A$ and $M-A$ is in the given family of subsets, so $\mid$. $b \mid \leqslant 2^{9}$. ) | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,640 |
3. Given two sets of real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\}$. If the mapping $f$ from $A$ to $B$ makes every element in $B$ have a preimage, and $f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant$ $f\left(a_{100}\right)... | (Tip: Let $b_{1}<b_{2}<\cdots<b_{50}$, and divide the elements $a_{1}, a_{2}, \cdots, a_{100}$ of $A$ into 50 non-empty groups in order. Define the mapping $f$: $A \rightarrow B$, such that the image of the elements in the $i$-th group under $f$ is $b_{i}(i=1,2, \cdots, 50)$. It is easy to see that such an $f$ satisfie... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,641 |
4. Determine the truth of the statement: "Let $A$ and $B$ be two sets of points in the coordinate plane, and $C_{r}=\left\{(x, y) \mid x^{2}+y^{2} \leqslant r^{2}\right\}$. If for any $r \geqslant 0$ we have $\left(C_{r} \cup A\right) \subseteq\left(C_{r} \cup B\right)$, then it must be that $A \subseteq B$."
(1984, Na... | (Let \( A = \{(x, y) | x^{2} + y^{2} \leqslant 1\} \), \( B \) be the set \( A \) with \((0,0)\) removed. It is clear that \( \left(C_{r} \cup A\right) \subseteq \left(C_{r} \cup B\right) \), but \( A \) is not contained in \( B \).) | not found | Geometry | proof | Yes | Yes | cn_contest | false | 714,642 |
5. Let $S=\{1,2,3,4\}$, and the sequence $a_{1}, a_{2}, \cdots, a_{n}$ has the following property: for any non-empty subset $B$ of $S$, there are consecutive $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition) | (Since a binary subset containing a fixed element of $S$ has 3 elements, any element of $S$ appears at least twice in the sequence. Therefore, the minimum value of $n$ is estimated to be 8. On the other hand, an 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the minimum value of $n$ is 8.) | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,643 |
6. Given the set $M=\{1,2, \cdots, k\}$, for $A \subseteq M$, denote the sum of all elements in $A$ as $S(A)$. If $M$ can be divided into two disjoint subsets $A$ and $B$, and $A \cup B=M, S(A)=2 S(B)$. Find all values of $k$.
(1994, Sichuan Province High School Mathematics Competition) | (Tip: Since $A \cup B=M, A \cap B=\varnothing, S(A)=$ $2 S(B)$, thus $S(M)=3 S(B)=\frac{k(k+1)}{2}$ is a multiple of 3, i.e., $3 \mid k$ or $3 \mid(k+1)$. (1) When $k=3 m$, $A=\{1,3,4,6$, $\cdots, 3 m-2,3 m\}, B=\{2,5,8, \cdots, 3 m-1\}$ meets the requirements;
(2) When $k=3 m-1$, $A=\{2,3,5,6,8, \cdots, 3 m-3,3 m$ $-1... | k=3m \text{ or } k=3m-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,644 |
Example 1 As shown in Figure $1, M$ and $N$ are the internal division points of the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, and $\frac{A M}{A C}=$
$\frac{C N}{C E}=\lambda$. If points $B$, $M$, and $N$ are collinear, find the value of $\lambda$.
(23rd IMO) | Solution: Extend $E A$ and $C B$ to intersect at point $P$. Let the side length of the regular hexagon be 1, then $P B=2$, and $A$ is the midpoint of $E P$, so $E A=A P$ $=\sqrt{3}$.
Since $\boldsymbol{A M}=\lambda A C$, then $C M=(1-\lambda) C A$.
Since $C P=3 C B$, and $C A$ is the median of $\triangle P C E$ on side... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,645 |
Example 2 In a cyclic pentagon $A B C D E$, let the centroid of any triangle formed by three of its vertices be $G$. Draw perpendiculars from $G$ to the lines connecting the other two vertices. Prove: these ten perpendiculars intersect at one point. | Proof: As shown in Figure 2, let $G$ be the centroid of $\triangle ABC$, and $P$ be the midpoint of $DE$. Draw $GQ \perp DE$ at point $Q$. Take a point $H$ on $GQ$ such that $\boldsymbol{GH} = \frac{2}{3} \boldsymbol{OP}$. Then,
$$
\begin{array}{l}
OH = OG + GH \\
= \frac{1}{3}(OA + OB + OC) + \frac{2}{3} OP \\
= \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,646 |
Example 11 Let $x_{1}, x_{2}, \cdots, x_{19}$ all be positive integers, and satisfy $x_{1}+x_{2}+\cdots+x_{19}=95$. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{19}^{2}$.
(1995, Hebei Province Junior High School Mathematics Competition) | Explanation: $\bar{x}=\frac{1}{19}\left(x_{1}+x_{2}+\cdots+x_{19}\right)=5$.
From the variance formula $S^{2}=\frac{1}{19}\left(\sum_{i=1}^{19} x_{i}^{2}-19 \bar{x}^{2}\right)$, we know that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}$ and $S^{2}$ reach their maximum values simultaneously.
For the positive integers $x_{1},... | 5947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,647 |
Example 3 Let $P$ be a moving point inside $\triangle ABC$, and the distances from $P$ to the three sides $a, b, c$ are denoted as $r_{1}, r_{2}, r_{3}$, respectively. Try to find the point $P$ that minimizes the value of $\frac{a}{r_{1}}+\frac{b}{r_{2}}+\frac{c}{r_{3}}$.
(22nd IMO) | Let the area of $\triangle ABC$ be $S$, then $a r_{1}+b r_{2}+c r_{3}=2 S$ is a constant, and the perimeter $a+b+c$ is a constant. Let
$\alpha=\left(\sqrt{a r_{1}}, \sqrt{b r_{2}}, \sqrt{c r_{3}}\right)$,
$\boldsymbol{\beta}=\left(\sqrt{\frac{a}{r_{1}}}, \sqrt{\frac{b}{r_{2}}}, \sqrt{\frac{c}{r_{3}}}\right)$.
Since $(\... | r_{1}=r_{2}=r_{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,648 |
Example 4 Let $A_{1} A_{2} A_{3} A_{4}$ be a cyclic quadrilateral inscribed in $\odot O$, and let $H_{1}, H_{2}, H_{3}, I_{4}$ be the orthocenters of $\triangle A_{2} A_{3} A_{4}$, $\triangle A_{3} A_{4} A_{1}$, $\triangle A_{4} A_{1} A_{2}$, and $\triangle A_{1} A_{2} A_{3}$, respectively. Prove that $H_{1}, H_{2}, H_... | Proof: As shown in Figure 3, draw the diameter \(A_{2} B\) of \(\odot O\), and connect
\[
A_{3} B, A_{4} B, A_{3} H_{1}.
\]
Since \(A_{3} H_{1} \perp A_{2} A_{4}\),
\(B A_{4} \perp A_{2} A_{4}\), we have
\[
A_{3} H_{1} \parallel B A_{4}.
\]
Similarly, \(A_{4} H_{1} \parallel B A_{3}\).
Therefore, \(A_{3} B A_{4} H_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,649 |
For example, $5 B C$ is the diameter of circle $\Gamma$, the center of $\Gamma$ is $O$, and $A$ is a point on $\Gamma$ with $0^{\circ}<\angle A O B<120^{\circ}$. $D$ is the midpoint of the arc $\overparen{A B}$ (excluding $C$), and the line through $O$ parallel to $D A$ intersects $A C$ at $I$. The perpendicular bisect... | Proof: As shown in Figure 4, connect $OD$, $OF$, $AE$, $AF$, and $BE$.
Since $OA$ and $EF$ bisect each other perpendicularly, quadrilateral $AE OF$ is a rhombus, $\overparen{AE}=\overparen{AF}$, and $\angle ACE = \angle ACF$. Therefore, $AC$ is the angle bisector of $\angle ECF$.
Let the radius of $\odot O$ be $R$, an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,650 |
Example 6 In the quadrilateral pyramid $S-ABCD$, the side length of the square base $ABCD$ is $a$, and the side edges $SA=SB=SC=SD=2a$, $M$ and $N$ are the midpoints of edges $SA$ and $SC$ respectively. Find the distance $d$ between the skew lines $DM$ and $BN$ and the cosine value of the angle $\alpha$ they form. | Solution: Taking the center $O$ of the square base as the origin, establish a rectangular coordinate system as shown in Figure 5, then
$$
\begin{array}{l}
B\left(\frac{a}{2},-\frac{a}{2}, 0\right), \\
D\left(-\frac{a}{2}, \frac{a}{2}, 0\right), \\
M\left(-\frac{a}{4},-\frac{a}{4}, \frac{\sqrt{14} a}{4}\right), \\
N\lef... | d=\frac{\sqrt{10} a}{5}, \cos \alpha=\frac{1}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,651 |
Example 7 As shown in Figure 6, given $\lambda>1$, let point $P$ be a moving point on the arc $\overparen{B A C}$ of the circumcircle of $\triangle A B C$. Take points $U$ and $V$ on the rays $B P$ and $C P$ respectively, such that $B U=\lambda B A$, $C V=\lambda C A$. On the ray $U V$, take point $Q$ such that $U Q=\l... | Solution: Let $B A$ and $C A$ correspond to the complex numbers $a$ and $b$ respectively,
$$
\begin{array}{l}
\angle A B U=\angle A C V=\alpha, \\
\boldsymbol{B} \boldsymbol{U}=\lambda a(\cos \alpha+\mathrm{i} \sin \alpha), \\
\boldsymbol{C V}=\lambda b(\cos \alpha+\mathrm{i} \sin \alpha) .
\end{array}
$$
Since $B Q=U... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,652 |
Example 8 As shown in Figure 7, in $\triangle A B C$, $\angle A=60^{\circ}$, $A B>A C$, point $O$ is the circumcenter, altitudes $B E$ and $C F$ intersect at
point $H$, points $M$ and $N$ are
on segments $B H$ and $H F$ respectively,
and satisfy $B M=C N$. Find
the value of $\frac{M H+N H}{O H}$.
(2002, National High S... | Given $\angle A=60^{\circ}, B M=C N$, we know
$$
M H+N H=B H-H C=2(F H-H E) .
$$
Since $O H=O A+O B+O C=O A+A H$,
thus, $|\boldsymbol{A H}|=|\boldsymbol{O B}+\boldsymbol{O C}|=R$, where $R$ is the circumradius of $\triangle A B C$.
Also, $\angle A H F=\angle A B C, \angle A H E=\angle A C B$, so $F H-H E=R(\cos B-\cos... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,653 |
Example 9 In an acute triangle $\triangle ABC$, $AD$ and $BE$ are two altitudes, $AP$ and $BQ$ are two angle bisectors, $I$ and $O$ are the incenter and circumcenter of the triangle, respectively. Prove that points $D$, $E$, and $I$ are collinear if and only if points $P$, $Q$, and $O$ are collinear.
(38th IMO Shortlis... | Proof: As shown in Figure 8, take $C$ as the origin, and use $C A=a, C B=b$ as the basis vectors to establish an affine coordinate system.
Let $|a|=a$, $|\boldsymbol{b}|=b$, $|\boldsymbol{A B}|=c$, $s=a+b+c$, $\boldsymbol{C D}=\lambda \boldsymbol{b}$, then $\boldsymbol{A D}=\lambda \boldsymbol{b}-\boldsymbol{a}$.
Sinc... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,654 |
Example 10 As shown in Figure 9, in tetrahedron $A B C D$, $E$ and $F$ are the midpoints of $A C$ and $B D$ respectively. Prove that $E F$ is the common perpendicular of $A C$ and $B D$ if and only if $A B=C D$ and $A D=$ $B C$. | Proof: Since
$$
\begin{array}{l}
E F=A F-A E \\
=\frac{1}{2}(A B+A D-A C) \\
=\frac{1}{2}(B A+B C-B D),
\end{array}
$$
Therefore, $\boldsymbol{E F} \cdot \boldsymbol{A C}=\boldsymbol{E F} \cdot \boldsymbol{B D}=0$
$$
\begin{array}{l}
\Leftrightarrow A C \cdot 2(A B+A D-A C) \\
=B D \cdot 2(B A+B C-B D)=0 \\
\Leftright... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,655 |
Title: Some numbers are placed along a circle. If there are four consecutive numbers $a, b, c, d$ that satisfy the inequality $(a-d)(b-c)>0$, then the positions of $b$ and $c$ can be swapped, which is called one operation.
(1) If the numbers $1, 2, 3, 4, 5, 6$ are placed consecutively along the circle, can it be achiev... | Original Solution (1): Perform the four operations as shown in Figure 1.
Figure 1
Below are other solutions discovered through research.
1. Two operations
Observation reveals that, with the line through 2 and 5 as the axis, the 4 adjacent numbers on both sides, $(2,3,4,5)$ and $(5,6,1,2)$, both satisfy $(a-d)(b-c)>0$.... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,656 |
Example 1 A water tank is equipped with 9 inlet and outlet pipes numbered $1, 2, \cdots, 9$. Some are only for inlet, and some are only for outlet. It is known that the pipe numbers opened and the time required to fill the tank with water are as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline Pi... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,657 |
Example 12 Let $a$, $b$, $c$ be real numbers, consider the following propositions:
(1) If $a^{2}+a b+c>0$, and $c>1$, then $01$, and $00$;
(3) If $00$, then $c>1$.
Determine which propositions are correct and which are incorrect. Provide proofs for the propositions you believe to be correct; for the propositions you be... | Explanation: Proposition (1) is incorrect. A counterexample is constructed as follows:
Let $b=4, c=5$, at this time $a^{2}+a b+c=a^{2}+$ $4 a+5=(a+2)^{2}+1>0$ and $c>1$, satisfying the conditions, but the conclusion $00$, satisfying the conditions, but the conclusion $c>1$ does not hold. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,658 |
Proposition $1^{[1]}$ Given $n(n>3)$ points on a plane, where no three points are collinear. Arbitrarily connect some points with line segments (these line segments are called edges), resulting in $x$ edges. If it is ensured that the figure contains a triangle with the given points as vertices, prove:
$$
x \geqslant \f... | Proof: Let the given $n$ points on the plane be $A_{1}, A_{2}, \cdots, A_{n}$, and let $A=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$. Then we have
$$
\left[\frac{n}{2}\right] \leqslant \frac{n}{2} \leqslant n-\left[\frac{n}{2}\right] \leqslant\left[\frac{n}{2}\right]+1 .
$$
(1) Necessity.
If the graph contains a tria... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,659 |
The 44th IMO Problem 4:
Let $A B C D$ be a cyclic quadrilateral. From point $D$ draw perpendiculars to the lines $B C, A C$, and $A B$, with feet $P, Q$, and $R$ respectively. Prove that $P Q=Q R$ if and only if the angle bisectors of $\angle A B C$ and $\angle A D C$ and the line $A C$ intersect at one point.
Now prov... | Proof: As shown in Figure 1,
connect $Q R$, $Q P$, $A D$,
$D C$.
Since $D R \perp$
$A R$, $A Q \perp Q D$,
therefore, $A$, $R$, $D$, $Q$ are
concyclic, and $A D$ is
the diameter of this circle.
Thus, $Q R = A D \sin \angle Q D R = A D \sin \angle B A C$.
Similarly, $Q P = D C \sin \angle A C B$.
By $\triangle A B C$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,660 |
Each time, A puts out 1, 2, or 3 stones, and B guesses. If B guesses correctly, the stones A puts out go to B; if B does not guess correctly, B pays A 1 stone. This process is repeated, and the one with more stones at the end wins.
Obviously, this is a zero-sum game problem.
Let the probabilities of A putting out 1, 2,... | Prove: When A adopts the above probabilities of showing the stones, no matter what guessing probabilities B adopts $\left(\beta_{1} 、 \beta_{2} 、 \beta_{3}, \beta_{j} \geqslant 0, \beta_{1}+\right.$ $\left.\beta_{2}+\beta_{3}=1\right)$, there is always $\mathrm{E}(X)=-\frac{1}{13}$. However, if A adopts any other proba... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,662 |
1. Given that $\triangle A B C$ is an acute triangle, $A B \neq A C$, the circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $M$ and $N$, respectively. Let $O$ be the midpoint of $B C$, and the angle bisector of $\angle B A C$ and the angle bisector of $\angle M O N$ intersect at point $R$. Prove tha... | Proof: (Revised based on Peng Mindou's solution) As shown in Figure 1, first, we need to prove that points $A, M, R, N$ are concyclic. Since $\triangle ABC$ is an acute triangle, points $M, N$ lie on segments $AB, AC$ respectively. On the ray $AR$, take a point $R_{1}$ such that points $A, M, R_{1}, N$ are concyclic. S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,663 |
2. Find all real-coefficient polynomials $P(x)$ such that for all real numbers $a, b, c$ satisfying $a b+b c+c a=0$, we have
$$
P(a-b)+P(b-c)+P(c-a)=2 P(a+b+c) .
$$ | Solution: (Based on the solution by Huang Zhiyi) For any given $a, b, c \in \mathbf{R}$, satisfying $ab + bc + ca = 0$, we have
$$
P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c).
$$
In equation (1), let $a = b = c = 0$, then $P(0) = 0$.
Let $b = c = 0$, for any real number $a_1$ we have
$$
P(-a) = P(a).
$$
Therefore, all the c... | P(x) = \alpha x^4 + \beta x^2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,664 |
4. Let $n(n \geqslant 3)$ be an integer, and $t_{1}, t_{2}, \cdots, t_{n}$ be positive real numbers, satisfying
$$
n^{2}+1>\left(t_{1}+t_{2}+\cdots+t_{n}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\cdots+\frac{1}{t_{n}}\right) .
$$ | Proof: For all integers $i, j, k$ satisfying $1 \leqslant i < j < k \leqslant n$, positive real numbers $t_{i}, t_{j}, t_{k}$ can always form the three sides of a triangle.
Proof: (Based on the solution by Zhu Qingsan) Assume that among $t_{1}, t_{2}, \cdots, t_{n}$, there are three numbers that cannot form the sides ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,666 |
5. In a convex quadrilateral $ABCD$, the diagonal $BD$ is neither the angle bisector of $\angle ABC$ nor the angle bisector of $\angle CDA$. Point $P$ is inside the quadrilateral $ABCD$ and satisfies $\angle PBC = \angle DBA$ and $\angle PDC = \angle BDA$. Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP ... | Proof: As shown in Figure 11, let's assume $P$ is inside $\triangle ABC$ and $\triangle BCD$.
Let $ABCD$ be a cyclic quadrilateral, and lines $BP$ and $DP$ intersect $AC$ at points $K$ and $L$, respectively. Since
$$
\begin{array}{c}
\angle PBC = \angle DBA, \\
\angle PDC = \angle BDA, \\
\angle ACB = \angle ADB, \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,667 |
Example 13 Unload several boxes from a cargo ship, with a total weight of $10 \mathrm{t}$, and the weight of each box does not exceed $1 \mathrm{t}$. To ensure that these boxes can be transported away in one go, how many trucks with a carrying capacity of $3 \mathrm{t}$ are needed at least?
(1990, Jiangsu Province Juni... | First, note that the weight of each box does not exceed $1 \mathrm{t}$, so the weight of the boxes that each vehicle can carry at one time will not be less than $2 \mathrm{t}$; otherwise, another box can be added.
Let $n$ be the number of trucks needed, and the weights of the boxes they carry be $a_{1}, a_{2}, \cdots,... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 714,669 |
1. Let real numbers $a_{i}$ satisfy: when $i=j$, $a_{y}$ is positive; when $i \neq j$, $a_{y}$ is negative, where $i=1,2,3; j=1,2,3$. Prove: There exist positive real numbers $c_{1}, c_{2}, c_{3}$, such that the following three numbers
$$
\begin{array}{l}
a_{11} c_{1}+a_{12} c_{2}+a_{13} c_{3}, a_{21} c_{1}+a_{22} c_{2... | Proof: Let in the spatial rectangular coordinate system, $O(0,0,0)$, $P\left(a_{11}, a_{21}, a_{31}\right)$, $Q\left(a_{12}, a_{22}, a_{32}\right)$, $R\left(a_{13}, a_{23}, a_{33}\right)$. We need to prove that in $\triangle P Q R$, there exists a point whose coordinates are either all negative, all positive, or all ze... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,670 |
3. Consider two sequences of positive real numbers $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant \cdots, b_{1} \geqslant b_{2}$ $\geqslant b_{3} \geqslant \cdots$. Let $A_{n}=a_{1}+a_{2}+\cdots+a_{n}, B_{n}=b_{1}+b_{2}+$ $\cdots+b_{n}, n=1,2,3, \cdots$. Set $c_{i}=\min \left\{a_{i}, b_{i}\right\}, C_{n}=c_{1}+$ $c_{... | Solution: (1) Exists.
Let $\{c_i\}$ be any sequence of positive numbers, and satisfies $c_i \geqslant c_{i+1}$ and $\sum_{i=1}^{+\infty} c_i = \sum_{i>1} a_i = +\infty$. Contradiction.
If there are infinitely many $i$ such that $b_i = c_i$, let the integer sequence $\{k_m \mid$ satisfy $k_{m+1} \geqslant 2k_m$, and $b... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 714,672 |
5. Let $\mathbf{R}_{+}$ be the set of positive real numbers, find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, satisfying the conditions:
(1) For all $x, y, z \in \mathbf{R}_{+}$, we have
$$
\begin{array}{l}
f(x y z)+f(x)+f(y)+f(z) \\
=f(\sqrt{x y}) f(\sqrt{y z}) f(\sqrt{z x}) ;
\end{array}
$$
(2) For ... | Solution: We prove that $f(x)=x^{2}+x^{-\lambda}$ satisfies the conditions, where $\lambda$ is any positive real number. For this, we first prove a lemma.
Lemma There exists a unique function $g:[1,+\infty) \rightarrow[1,+\infty)$ such that
$$
f(x)=g(x)+\frac{1}{g(x)}.
$$
Proof of the lemma: Let $x=y=z=1$, then condi... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,673 |
6. Given that $n$ is a positive integer, $\left|x_{1}, x_{2}, \cdots, x_{n}\right|$ and $\left\{y_{1}, y_{2}, \cdots, y_{n}\right\}$ are two sequences of positive real numbers. If $\left\{z_{1}, z_{2}, \cdots, z_{2 n}\right\}$ is a sequence of positive real numbers satisfying $z_{i+j}^{2} \geqslant x_{i} y_{j}$ for $1 ... | Proof: Let $X=\max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}, Y=\max \left\{y_{1}, y_{2}, \cdots, y_{n}\right\}$. By substituting $x_{i}^{\prime}=\frac{x_{i}}{X}$ for $x_{i}$, $y_{i}^{\prime}=\frac{y_{i}}{Y}$ for $y_{i}$, and $z_{i}^{\prime}=\frac{z_{i}}{\sqrt{X Y}}$ for $z_{i}$, the original inequality remains unchan... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,674 |
2. As shown in Figure 3, given three fixed points on a line in sequence as $A, B$, and $C$, $\Gamma$ is a circle passing through $A$ and $C$ with its center not on $AC$. The tangents to the circle $\Gamma$ at points $A$ and $C$ intersect at point $P$, and $PB$ intersects the circle $\Gamma$ at point $Q$. Prove: The ang... | Proof: Suppose the angle bisector of $\angle A Q C$ intersects $A C$ at point $R$, and intersects circle $\Gamma$ at point $S$, where $S$ and $Q$ are distinct points.
Since $\triangle P A C$ is an isosceles triangle, we have
$$
\frac{A B}{B C}=\frac{\sin \angle A P B}{\sin \angle C P B} .
$$
Similarly, in $\triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,675 |
3. As shown in Figure 4, given a point $P$ inside $\triangle ABC$, let $D$, $E$, and $F$ be the projections of point $P$ onto sides $BC$, $CA$, and $AB$ respectively. Suppose $AP^{2} + PD^{2} = BP^{2} + PE^{2} = CP^{2} + PF^{2}$, and the excenters of $\triangle ABC$ are $I_{A}$, $I_{B}$, and $I_{C}$. Prove: $P$ is the ... | Proof: From the given conditions, we have
$B F^{2}-C E^{2}=\left(B P^{2}-P F^{2}\right)-\left(C P^{2}-P E^{2}\right)$
$=\left(B P^{2}+P E^{2}\right)-\left(C P^{2}+P F^{2}\right)=0$.
Thus, $B F=C E$.
Let $x=B F=C E$. Similarly, we can set
$y=C D=A F, z=A E=B D$.
If one of the points $D$, $E$, $F$ is on the extension of ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,676 |
4. As shown in Figure $5, \Gamma_{1}, \Gamma_{2}, \Gamma_{3}, \Gamma_{4}$ are four different circles, and $\Gamma_{1}$ is externally tangent to $\Gamma_{3}$ at point $P$, $\Gamma_{2}$ is externally tangent to $\Gamma_{4}$ also at point $P$. Suppose $\Gamma_{1}$ intersects $\Gamma_{2}$, $\Gamma_{2}$ intersects $\Gamma_{... | Proof: Let the intersection of $AB$ with the common internal tangents of circles $\Gamma_{1}$ and $\Gamma_{3}$ be $Q$, then
$\angle A P B=\angle A P Q+\angle B P Q=\angle P D A+\angle P C B$.
Thus, $\theta_{2}+\theta_{3}+\angle A P B=\theta_{2}+\theta_{3}+\theta_{5}+\theta_{8}=180^{\circ}$.
Similarly, from $\angle B P ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,677 |
5. As shown in Figure 7, in isosceles $\triangle ABC$, $AC = BC$, and $I$ is its incenter. Let $P$ be a point on the arc of the circumcircle of $\triangle AIB$ inside $\triangle ABC$. The lines through $P$ parallel to $CA$ and $CB$ intersect $AB$ at points $D$ and $E$, respectively. The line through $P$ parallel to $AB... | Proof: Since the corresponding sides of $\triangle P D E$ and $\triangle C F G$ are parallel, and $D F$ and $E G$ are not parallel, the two triangles are similar. Therefore, $D F$, $E G$, and $C P$ intersect at one point, which is the center of similarity.
Below is the proof: If $C P$ intersects the circumcircle of $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,678 |
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