problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Proposition 2 If $n \in \mathbf{N}_{+}, x, y, z$ are all non-zero, then
$$
\begin{array}{l}
\frac{1}{x^{2 n+1}}+\frac{1}{y^{2 n+1}}+\frac{1}{z^{2 n+1}} \\
=\left(\frac{1}{x+y+z}\right)^{2 n+1}
\end{array}
$$
is true if and only if at least two of $x, y, z$ are opposite numbers. | Proof: The sufficiency is obvious. Next, we prove the necessity.
Since $x$, $y$, $z$ have different signs (otherwise, suppose $x>0$, $y>0$, $z>0$, then the left side of equation (1) is greater than the right side, leading to a contradiction), without loss of generality, assume $x>0$, $y>0$, $z<0$. When $x>0$, $\frac{1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,043 |
Proposition: For any two chords $AC$ and $BD$ intersecting at point $E$, connect $AB$ and $CD$. Any line $l$ intersects the circle at $M$ and $N$, and intersects $AB$, $AC$, $BD$, $DC$ (or their extensions) at points $R$, $Q$, $P$, $S$ respectively. Then
$$
\frac{PM \cdot QM}{PN \cdot QN} = \frac{RM \cdot SM}{RN \cdot ... | Proof: As shown in Figure 1, connect $A M$, $A N$, $B M$, $B N$, $C M$, $C N$, $D M$, and $D N$. According to the problem, we have
$$
\begin{array}{l}
\left(\frac{P M}{P N} \cdot \frac{Q M}{Q N}\right)^{2} \\
=\left(\frac{S_{\triangle M B D}}{S_{\triangle N B D}} \cdot \frac{S_{\triangle M A C}}{S_{\triangle N A C}}\ri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,044 |
1. Simplify the radical expression $\frac{\sqrt{2}(\sqrt{6}-\sqrt{3}-\sqrt{2}+1)}{\sqrt{3-2 \sqrt{2}} \cdot \sqrt{2-\sqrt{3}}}$, the result is ( ).
(A) 2
(B) -2
(C) $\sqrt{2}$
(D) $-\sqrt{2}$ | $\begin{array}{l}\text { I. 1. A. } \\ \text { Original expression }=\frac{\sqrt{2}(\sqrt{2}-1)(\sqrt{3}-1)}{\sqrt{(1-\sqrt{2})^{2}} \cdot \sqrt{\frac{4-2 \sqrt{3}}{2}}} \\ =\frac{2(\sqrt{2}-1)(\sqrt{3}-1)}{\sqrt{(1-\sqrt{2})^{2}} \cdot \sqrt{(1-\sqrt{3})^{2}}}=2 .\end{array}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,045 |
Example 12 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.
But due to carelessness, one... | Explanation: Let $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}$ $=d$, and the function value corresponding to $x_{i}$ is $y_{i}$. Then
$$
\begin{array}{l}
\Delta_{k}=y_{k+1}-y_{k} \\
=\left(a x_{k+1}^{2}+b x_{k+1}+c\right)-\left(a x_{k}^{2}+b x_{k}+c\right) \\
=a\left[\left(x_{k}+d\right)^{2}-x_{k}^{2}\right]+b\left[\lef... | 551 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,046 |
3. Given the quadratic equation in $x$
$$
a^{2} x^{2}+b^{2} x+c^{2}=0
$$
the sum of the roots is the sum of the squares of the roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$
Then the relationship between $a$, $b$, and $c$ is ( ).
(A) $a^{2}=b c$
(B) $b^{2}=a c$
(C) $c^{2}=a b$
(D) $a b c=1$ | 3.B.
First, we have $a \neq 0$. By Vieta's formulas, the sum of the roots of equation (1) is $-\frac{b^{2}}{a^{2}}$. The relationships between the roots of equation (2) are $x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$. Therefore, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=\frac{b^{2}}{a^{2}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,048 |
4. As shown in Figure 2, the diagonals $AC$ and $BD$ of square $ABCD$ intersect at point $M$, dividing the square into four triangles. $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$, and $\odot O_{4}$ are the incircles of $\triangle AMB$, $\triangle BMC$, $\triangle CMD$, and $\triangle DMA$, respectively. Given that $AB=... | 4.C.
Since $A C=B D=\sqrt{2}$, therefore, $A M=B M=\frac{\sqrt{2}}{2}$.
Let the inradius of the right triangle $\triangle A M B$ be $r$, then $\frac{\sqrt{2}}{2}-r+\frac{\sqrt{2}}{2}-r=1$, which means $r=\frac{\sqrt{2}-1}{2}$.
The area of the shaded part is equal to the area of the square $\mathrm{O}_{1} \mathrm{O}_{2... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,049 |
5. In a box with a height of $4 \mathrm{~cm}$ and a base side length of $18 \mathrm{~cm}$, standard size table tennis balls with a diameter of $40 \mathrm{~mm}$ are placed. Then the maximum number of table tennis balls that can be placed is ( ) .
(A) 16
(B) 18
(C) 20
(D) 23 | 5.C.
Let the base square be $A B C D$, the radius of the sphere be $r, r=2 \text{ cm}$. Since the height of the box is just enough to fit one ping-pong ball, we only need to consider the horizontal arrangement.
Now, consider a certain arrangement of ping-pong balls as shown in the top view in Figure 6, where the ball... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,050 |
6. Today is March 20, 2005, Sunday. Then, the day after today by $2005^{3}$ days is ( ).
(A) Wednesday
(B) Thursday
(C) Friday
(D) Saturday | 6.D.
Since $2005^{3}=(7 \times 286+3)^{3}=7 k+27=7 k_{1}+6$ $\left(k 、 k_{1} \in \mathbf{N}\right)$, therefore, it is Saturday. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,051 |
$1 . \underbrace{111111111^{2}}_{9 \uparrow 1}=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$1 . \underbrace{111111111^{2}}_{9 \uparrow 1}=$ | $$
\text { Ni, 1.12345 } 678987654321 \text {. }
$$
From $11^{2}=121,111^{2}=12321, \cdots$, then
$$
\underbrace{111111111^{2}}_{91}=12345678987654321 .
$$ | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,052 |
2. For the quadratic function $y=a x^{2}+b x+c$, the graph intersects the $x$-axis at two distinct points $A$ and $B$, and the vertex of the parabola is $C$. Then $S_{\triangle A B C}=$ $\qquad$ | 2. $\frac{\left(b^{2}-4 a c\right) \sqrt{b^{2}-4 a c}}{8 a^{2}}$.
Since $\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|}$, the coordinates of the vertex are $C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$, therefore,
$$
\begin{aligned}
& S_{\triangle B C}=\frac{1}{2} \cdot \frac{\sqrt{b^{2}-4 a c}}{|a|... | \frac{\left(b^{2}-4 a c\right) \sqrt{b^{2}-4 a c}}{8 a^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,053 |
3. Given $\sqrt{x^{2}+32}-\sqrt{65-x^{2}}=5$. Then $3 \sqrt{x^{2}+32}+2 \sqrt{65-x^{2}}=$ $\qquad$ | 3.35.
Observing the experiment, we know that $x^{2}=49$, so the original expression $=3 \times 9+2 \times 4=35$. | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,054 |
4. There are two positive integers $a$ and $b$, the sum of their squares is 585, and the sum of their greatest common divisor and least common multiple is 87. Then $a+b=$ $\qquad$ . | 4.33.
Let the required positive integers be $a<b$, and let $(a, b)=d$, then we have $a=d x, b=d y$, and $(x, y)=1$.
Thus, $[a, b]=d x y$. According to the problem, we have $\left\{\begin{array}{l}d^{2} x^{2}+d^{2} y^{2}=585, \\ d+d x y=87,\end{array}\right.$ which simplifies to $\left\{\begin{array}{l}x^{2}+y^{2}=\fra... | 33 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,055 |
5. As shown in Figure 3, on each side of the square $A B C D$ with side length 1, segments $A E=B F$ $=C G=D H=x$ are cut off, and lines $A F 、 B G 、 C H 、 D E$ are drawn to form quadrilateral $P Q R S$. Express the area $S$ of quadrilateral $P Q R S$ in terms of $x$, then $S=$ | 5. $\frac{(1-x)^{2}}{1+x^{2}}$.
Since $\angle B A F+\angle A E D=90^{\circ}$, therefore, $\angle A P E=90^{\circ}$. Hence $A F \perp D E$.
Similarly, $A F \perp B G, B G \perp C H, C H \perp D E$.
Since Rt $\triangle A P E \cong$ Rt $\triangle B Q F \cong \mathrm{Rt} \triangle C R G \cong$ Rt $\triangle D S H$, then $... | \frac{(1-x)^{2}}{1+x^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,056 |
Example 13 Given that $a$, $b$, $c$ are real numbers, $ac<0$, and $\sqrt{2}a + \sqrt{3}b + \sqrt{5}c = 0$. Prove: The quadratic equation $ax^2 + bx + c = 0$ has a root greater than $\frac{3}{4}$ and less than 1.
(2005, National Junior High School Mathematics League) | Explanation: Let $y=f(x)=a x^{2}+b x+c$.
Since $a c<0, c<0$, so, $\sqrt{2} a>0$.
On the other hand, $f\left(\frac{3}{4}\right)=\frac{9}{16} a+\frac{3}{4} b+c$.
Because $a>0$, so, $\frac{9}{16} a<\frac{\sqrt{6}}{4} a$.
Also, because $c<0$, so, $c<\frac{\sqrt{15}}{4} c$.
Therefore, $f\left(\frac{3}{4}\right)<\frac{\sqrt{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,057 |
6. Given that $O$ is a point inside the equilateral $\triangle A B C$, the ratio of the angles $\angle A O B$, $\angle B O C$, and $\angle A O C$ is 6:5:4. Then, in the triangle formed by $O A$, $O B$, and $O C$, the ratio of the angles opposite these sides is | $6.5: 3: 7$.
As shown in Figure 7, from $\angle A O B$:
$\angle B O C: \angle A O C=6: 5: 4$.
We know
$$
\begin{array}{l}
\angle A O B=144^{\circ}, \\
\angle B O C=120^{\circ}, \\
\angle A O C=96^{\circ} .
\end{array}
$$
With point $A$ as the center, rotate $\triangle A O B$ counterclockwise by $60^{\circ}$ to get $\t... | 5: 3: 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,058 |
Three, (17 points) The quadratic trinomial $x^{2}-x-2 n$ can be factored into the product of two linear factors with integer coefficients.
(1) If $1 \leqslant n \leqslant 30$, and $n$ is an integer, how many such $n$ are there?
(2) When $n \leqslant 2005$, find the largest integer $n$.
| (1) Notice that
$$
x^{2}-x-2 n=\left(x-\frac{1+\sqrt{1+8 n}}{2}\right)\left(x-\frac{1-\sqrt{1+8 n}}{2}\right),
$$
then we should have
$$
1+8 n=9,25,49,81,121,169,225,289, \cdots \text {. }
$$
The corresponding solutions for $n$ are $1,3,6,10,15,21,28,36$ (discard the last one).
Therefore, when $1 \leqslant n \leqslan... | 1953 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,059 |
Four. (17 points) As shown in Figure 4, given the line
$$
l: y=k x+2-4 k \text{ ( } k \text{ is a real number). }
$$
- (1) Prove that regardless of the value of $k$, the line $l$ always passes through a fixed point $M$, and find the coordinates of point $M$;
(2) If the line $l$ intersects the positive $x$-axis and $y$-... | (1) Let $k=1$, we get $y=x-2$; let $k=2$, we get $y=2x-6$. Solving these simultaneously, we get $x=4, y=2$. Therefore, the fixed point is $M(4,2)$.
Substituting the coordinates of point $M(4,2)$ into the equation of line $l$, we get $2=2$, which is an identity independent of $k$. Hence, regardless of the value of $k$,... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,060 |
Five. (16 points) As shown in Figure 5, given that the orthocenter of $\triangle ABC$ is $H$, the circumcircle is $\odot O$, and $M$ is the midpoint of $AB$. Connect $MH$ and extend it to intersect $\odot O$ at $D$. Prove: $HD \perp CD$.
---
To prove that $HD \perp CD$, we will use properties of the orthocenter, the ... | As shown in Figure 8, draw the diameter $C E$ of $\odot O$, and connect $A E$, $B E$, $A H$, and $B H$.
Since $\angle C A E=90^{\circ}$,
it follows that $E A \perp A C$.
Also, since $H$ is the orthocenter,
$B H \perp A C$.
Therefore, $E A \parallel B H$.
Similarly, $E B \parallel A H$.
Thus, quadrilateral $B H A E$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,061 |
Six. (16 points) Given two quadratic functions $y_{A}=x^{2}+$ $3 m x-2$ and $y_{B}=2 x^{2}+6 m x-2$, where $m>0$. Construct the function $y$:
When $y_{A}>y_{B}$, set $y=y_{A}$;
When $y_{A} \leqslant y_{B}$, set $y=y_{B}$.
If the independent variable $x$ varies within the range $-2 \leqslant x \leqslant 1$, find the max... | Six, according to the problem, we have
$$
y=\left\{\begin{array}{ll}
x^{2}+3 m x-2, & \text { when } y_{A}>y_{B} \text {; } \\
2 x^{2}+6 m x-2, & \text { when } y_{A} \leqslant y_{B} \text {; }
\end{array}\right.
$$
It is easy to see that the graphs of the two given quadratic functions both open upwards, with a common... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,062 |
1. If $a=\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}, b=2+\sqrt{6}-\sqrt{10}$, then the value of $\frac{a}{b}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{1}{4}$
(C) $\frac{1}{\sqrt{2}+\sqrt{3}}$
(D) $\frac{1}{\sqrt{6}+\sqrt{10}}$ | - 1. B.
Since $a=\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$
$$
=\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5}) \cdot \sqrt{3}}{2 \sqrt{6}}=\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5}) \cdot \sqrt{2}}{4}=\frac{b}{4} \text {. }
$$
Therefore, $\frac{a}{b}=\frac{1}{4}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,063 |
2. As shown in Figure $1, I$ is the incenter of isosceles right $\triangle A B C$, and $IE$, $IF$ are the angle bisectors of $\angle A I C$ and $\angle A I B$ respectively. If the hypotenuse $B C$ $=2, A I$ intersects $E F$ at point $P$, then $I P=(\quad)$.
(A) $\sqrt{2}-1$
(B) $\sqrt{2-\sqrt{2}}$
(C) $1-\sqrt{2-\sqrt{... | 2.C.
As shown in Figure 1, let $A I$ intersect $B C$ at $D$. Then $\frac{I D}{A D}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. Given $A D=1$, we have
$$
\begin{array}{l}
I D=\sqrt{2}-1, I A=2-\sqrt{2}, \\
I C=\sqrt{I D^{2}+C D^{2}}=\sqrt{4-2 \sqrt{2}}, \\
\frac{A E}{C E}=\frac{I A}{I C}=\frac{2-\sqrt{2}}{\sqrt{4-2 \sqrt{2}}}=\f... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,064 |
3. Given $\alpha=3+2 \sqrt{2}, \beta=3-2 \sqrt{2}$. If $\alpha^{10}+\beta^{10}$ is a positive integer, then its last digit is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 3. B.
Since $\alpha+\beta=6, \alpha \beta=1$, then $\alpha, \beta$ are the two roots of the equation $x^{2}-6 x+1=0$. Therefore, we have
$$
\left\{\begin{array}{l}
\alpha^{2}=6 \alpha-1, \\
\beta^{2}=6 \beta-1 .
\end{array}\right.
$$
Thus, $\left\{\begin{array}{l}\alpha^{n}=6 \alpha^{n-1}-\alpha^{n-2}, \\ \beta^{n}=6... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,065 |
4. As shown in Figure 2, an equilateral triangle $\triangle A B C$ with side length 1, and circles with radius 1 are drawn with centers at vertices $A$, $B$, and $C$. Then the area of the region covered by these three circles is ( ).
(A) $\frac{3 \pi}{2}+\sqrt{3}$
(B) $\frac{5 \pi}{2}-\sqrt{3}$
(C) $\frac{7 \pi}{2}-2 \... | 4.A.
Given $S_{\triangle A B C}=S_{\triangle A B D}=\frac{\sqrt{3}}{4}$,
$S_{\text {sector }- \text { BC }}=\frac{\pi}{6}, S_{\text {sector } A}-\Omega C D=\frac{\pi}{3}$,
the area of the intersection of any two circles is
$$
2\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2} \text {, }
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,066 |
5. On the coordinate plane, a point $(x, y)$ where both the x-coordinate and y-coordinate are integers is called an integer point. If the area enclosed by the quadratic function $y=-x^{2}+8 x-\frac{39}{4}$ and the x-axis is colored red, then the number of integer points inside this red region and on its boundary is ( )... | 5.C.
$y=-\left(x-\frac{3}{2}\right)\left(x-\frac{13}{2}\right)$ intersects the $x$-axis at two points $M\left(\frac{3}{2}, 0\right)$ and $N\left(\frac{13}{2}, 0\right)$. Between $x=\frac{3}{2}$ and $x=\frac{13}{2}$, there are 5 integers: $2, 3, 4, 5, 6$.
Rewrite the function as
$$
y=-(x-4)^{2}+\frac{25}{4}.
$$
When $x... | 25 | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,067 |
1. When $k$ takes any real number, the vertex of the parabola $y=\frac{4}{5}(x-k)^{2}+$ $k^{2}$ lies on the curve ( ).
(A) $y=x^{2}$
(B) $y=-x^{2}$
(C) $y=x^{2}(x>0)$
(D) $y=-x^{2}(x>0)$
(2004, National Junior High School Mathematics Competition Hubei Province Preliminary Contest) | (Answer: A. ) | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,068 |
6. Add appropriate parentheses to $1 \div 2 \div 3 \div 4 \div 5$ to form a complete expression. Then, the number of different values that can be obtained is $(\quad)$.
(A) 6
(B) 8
(C) 9
(D) 10
The denominator must be below the fraction line. By appropriately placing parentheses, the numbers $3, 4, 5$ can be placed ei... | 6. B.
No matter how parentheses are added, 1 must be in the numerator, 2 must | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,069 |
1. As shown in Figure 3, the side length of square $A B C D$ is 1, and an equilateral triangle $\triangle A B E$ is constructed inside the square with $A B$ as one side. Point $P$ is the center of this equilateral triangle, and $P D$ intersects $A E$ at point $F$. Then $P F=$ $\qquad$ | 2.1. $\sqrt{\frac{11}{6}-\sqrt{3}}$.
Let $PE$ intersect $CD$ at point $M$, and intersect $AB$ at point $N$. Then $PM=1-PN=1-\frac{\sqrt{3}}{6}, PE=\frac{\sqrt{3}}{3}$, $PD=\sqrt{DM^{2}+PM^{2}}=\sqrt{\frac{4-\sqrt{3}}{3}}$, $PE \parallel AD$,
$$
\frac{PF}{FD}=\frac{PE}{AD}=\frac{\sqrt{3}}{3}, \frac{PF}{PD}=\frac{\sqrt{3... | \sqrt{\frac{11}{6}-\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,070 |
2. Given $a, b$ are positive integers, $a=b-2005$. If the equation $x^{2}-a x+$ $b=0$ has positive integer solutions, then the minimum value of $a$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.95.
Let the two roots of the equation be $x_{1}$ and $x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{1} x_{2}=b .
\end{array}\right.
$$
Since $x_{1}$ and $x_{2}$ include one positive integer, the other must also be a positive integer. Without loss of generality, let $x_{1} \leqslant x_{2}$. From equat... | 95 | Number Theory | proof | Yes | Yes | cn_contest | false | 716,071 |
3. A person adds the page numbers of a book in the order $1,2,3, \cdots$, with one page number being added an extra time, resulting in an incorrect total sum of 2005. Then the page number that was added extra is $\qquad$ . | 3.52 pages.
Let the total number of pages in the book be $n$, and the page number that was added extra be $x(1 \leqslant x \leqslant n)$. We have $\frac{n(n+1)}{2}+x=2005$.
And $\frac{n(n+1)}{2}+1 \leqslant 2005 \leqslant \frac{n(n+1)}{2}+n$, which means $n^{2}+n+2 \leqslant 4010 \leqslant n(n+3)$.
Since $\sqrt{4010} ... | 52 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,072 |
4. Let $n$ be a natural number. If 2005 can be written as the sum of $n$ positive odd composite numbers, then $n$ is called a "good number". The number of such good numbers is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result di... | 4.111.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be odd composite numbers, and $a_{1}+a_{2}+\cdots+a_{n}=$ 2005, then $n$ is odd. Since 9 is the smallest odd composite number, and 2005 $<2007=9 \times 223$, hence $n<223$. Therefore, $n \leqslant 221$.
$$
2005=1980+25=\underbrace{9+9+\cdots}_{20 \uparrow}+9+25 \text {, }
$$
... | 111 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,073 |
1. (20 points) Given that $p$ and $q$ are integers, and are the roots of the equation in $x$
$$
x^{2}-\frac{p^{2}+11}{9} x+\frac{15}{4}(p+q)+16=0
$$
Find the values of $p$ and $q$. | $$
\text { Three, 1. According to }\left\{\begin{array}{l}
p+q=\frac{p^{2}+11}{9}, \\
p q=\frac{15}{4}(p+q)+16,
\end{array}\right.
$$
it is known that $p+q>0, p q>0$. Therefore, $p, q$ are positive integers.
From equation (2), we have $16 p q=60(p+q)+16^{2}$, which means
$$
\begin{array}{l}
(4 p-15)(4 q-15)=16^{2}+15^... | p=13, q=7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,074 |
2. (25 points) As
in Figure 4, in $\triangle ABC$, $AB < AC$, $I$ is the incenter, $M$ is the midpoint of $BC$, $P$ is a point on $BC$ such that $AP \parallel IM$, and $Q$ is a point on $AP$. If quadrilateral $IM PQ$ is a parallelogram, prove that $\triangle MPQ$ is a right triangle. | 2. As shown in Figure 5, let $BC = a$, $AC = b$, $AB = c$. Extend $IQ$ to intersect $AB$ and $AC$ at $N$ and $K$ respectively, and connect $IB$.
Notice that
$\angle NIB = \angle IBM = \angle IBN$, so $IN = BN$.
Similarly, $IK = CK$.
Let the incircle of $\triangle ABC$, $\odot I$, touch $BC$, $CA$, and $AB$ at points $D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,075 |
(1) Let $a \in \mathbf{R}$. Prove: the parabola $y=x^{2}+(a+2) x-2 a+1$ always passes through a fixed point, and the vertices all lie on a parabola; | (1) Let $f_{a}(x)=x^{2}+(a+2) x-2 a+1=x^{2}+2 x+1+a(x-2)$, therefore, the parabola passes through the fixed point $(2,9)$, and the coordinates of the vertex of the parabola are
$$
\begin{array}{l}
x=-\frac{a+2}{2}, \\
y=\frac{4(1-2 a)-(a+2)^{2}}{4}=\frac{-a^{2}-12 a}{4} .
\end{array}
$$
Eliminating $a$ yields $y=-x^{2... | y=-x^{2}+4 x+5 | Algebra | proof | Yes | Yes | cn_contest | false | 716,077 |
(2) If the equation $x^{2}+(a+2) x-2 a$ $+1=0$ has two distinct real roots with respect to $x$, find the range of values for its larger root.
(Wu Weizhao) | (2) The larger root of $f_{a}(x)=0$ is
$$
\begin{aligned}
x & =\frac{-(a+2)+\sqrt{(a+2)^{2}-4(1-2 a)}}{2} \\
& =\frac{-(a+2)+\sqrt{a^{2}+12 a}}{2} \\
& =\frac{-(a+2)+\sqrt{(a+6)^{2}-36}}{2} .
\end{aligned}
$$
Let $a+6=2 k$. Then
$$
x=\frac{-(2 k-4)+\sqrt{4 k^{2}-36}}{2}=\sqrt{k^{2}-9}-k+2 \text {. }
$$
By $\Delta>0$,... | (-1,2) \cup (5,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,078 |
II. $\odot O$ is separated from line $l$, and $OP \perp l$, with $P$ being the foot of the perpendicular. Let point $Q$ be any point on $l$ (not coinciding with point $P$). Draw two tangents $QA$ and $QB$ from point $Q$ to $\odot O$, with $A$ and $B$ being the points of tangency. $AB$ intersects $OP$ at point $K$. Draw... | II. As shown in Figure 1, construct $P I \perp A B$, with $I$ as the foot of the perpendicular, and let $J$ be the intersection of line $M N$ and segment $P K$.
It is easy to see that $\angle Q A O=\angle Q B O=\angle Q P O=90^{\circ}$.
Therefore, $O, B, Q, P, A$ all lie on the circumference of a circle with diameter $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,080 |
Three, let $n(n \geqslant 3)$ be a positive integer, and the set $M=$ $\{1,2, \cdots, 2 n\}$. Find the smallest positive integer $k$, such that for any $k$-element subset of $M$, there must be 4 distinct elements whose sum equals $4 n+1$. | Three: Consider the $(n+2)$-element subset of $M$
$$
P=\{n-1, n, n+1, \cdots, 2 n\} \text {. }
$$
The sum of any 4 different elements in $P$ is not less than
$$
n-1+n+n+1+n+2=4 n+2 \text {, }
$$
Therefore, $k \geqslant n+3$.
Pair the elements of $M$ into $n$ pairs,
$$
B_{i}=(i, 2 n+1-i), 1 \leqslant i \leqslant n .
$$... | n+3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,081 |
Find all triples of positive integers $(a, b, c)$ satisfying $a^{2}+b^{2}+c^{2}=2005$, and $a \leqslant b \leqslant c$. | Since any odd square number divided by 4 leaves a remainder of 1, and any even square number is a multiple of 4, and 2005 divided by 4 leaves a remainder of 1, it follows that among $a^{2}$, $b^{2}$, and $c^{2}$, there must be two even square numbers and one odd square number.
Let $a=2m$, $b=2n$, $c=2k-1$, where $m$, ... | (23,24,30),(12,30,31),(9,18,40),(9,30,32),(4,15,42),(15,22,36),(4,30,33) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,082 |
Five, Given that line $l$ is tangent to the unit circle $\odot O$ at point $P$, point $A$ is on the same side of line $l$ as $\odot O$, and the distance from $A$ to line $l$ is $h(h>2)$, two tangents are drawn from point $A$ to $\odot O$, intersecting line $l$ at points $B$ and $C$. Find the product of the lengths of s... | Five, As shown in Figure 2, let the lengths of $PB$ and $PC$ be $p$ and $q$, respectively, $\angle ABP = \beta$, $\angle ACP = \gamma$, and the point where $AC$ touches $\odot O$ be $E$, with the length of $AE$ being $t$. Connecting $AO$ and $OE$, in the right triangle $\triangle AOE$, we have $\angle AOE = \frac{1}{2}... | pq = \frac{h}{h - 2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,083 |
Six, let the arithmetic mean of all elements in the set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be denoted as $P(A)$ $\left(P(A)=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)$. If $B$ is a non-empty subset of $A$ and $P(B)=P(A)$, then $B$ is called a "balanced subset" of $A$. Try to find the number of all "balanced ... | $$
\begin{array}{l}
\text { Six, since } P(M)=5 \text {, let } \\
M^{\prime}=\{x-5 \mid x \in M\} \\
=\{-4,-3,-2,-1,0,1,2,3,4\},
\end{array}
$$
then $P\left(M^{\prime}\right)=0$. According to this translation relationship, the balanced subsets of $M$ and $M^{\prime}$ can be one-to-one correspondence. Let $f(k)$ denote... | 51 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,084 |
(1) Discuss the number of roots of the equation with respect to $x$
$$
|x+1|+|x+2|+|x+3|=a
$$
(2) Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence, and
$$
\begin{array}{l}
\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right| \\
=\left|a_{1}+1\right|+\left|a_{2}+1\right|+\cdots+\left|a_{n}+1\right... | (1) According to the graph of the function $y=|x+1|+|x+2|+|x+3|$ (as shown in Figure 3), we can see that:
When $a2$, the equation has two solutions.
(2) Since the equation $|x|=|x+1|=|x-2|$ has no solution, it follows that $n \geqslant 2$ and the common difference is not 0. Let's assume the terms of the sequence are $... | 26 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,085 |
Eight, let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+\sin ^{3} \gamma=1$.
Prove: $\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2}$. | $$
\text { Eight, let } a=\sin \alpha, b=\sin \beta, c=\sin \gamma \text {, then } a, b, c \in
$$
$(0,1)$ and
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}=1, \\
a-a^{3}=\frac{\sqrt{2}}{2} \times \sqrt{2 a^{2}\left(1-a^{2}\right)^{2}} \\
\leqslant \frac{\sqrt{2}}{2} \times \sqrt{\left(\frac{2 a^{2}+1-a^{2}+1-a^{2}}{3}\right)^{... | \tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,086 |
1. A university has 10001 students, some of whom join and form several clubs (a student can belong to different clubs), and some clubs join together to form several associations (a club can belong to different associations). It is known that there are $k$ associations in total. Assume the following conditions are met:
... | 1. Replace 10001 with $n$, and use two methods to calculate the number of ordered triples $(a, R, S)$, where $a, R, S$ represent a student, a club, and a society, respectively, and satisfy $a \in R, R \in S$. We call such triples "acceptable".
Fix a student $a$ and a society $S$. By condition (2), there is a unique cl... | 5000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,087 |
2. Let $n, k$ be positive integers. It is known that there are $n$ circles on a plane, each pair of which has two distinct intersection points, and all the intersection points determined by them are pairwise distinct. Each intersection point must be colored with one of $n$ different colors, such that each color is used... | $2 . k, n$ should satisfy $2 \leqslant k \leqslant n \leqslant 3$ and $3 \leqslant k \leqslant n$.
Label the $n$ colors and $n$ circles as $1,2, \cdots, n$. For any coloring of the intersection points, let $F(i, j)$ be the set of colors of the intersection points of the $i$-th circle and the $j$-th circle, then $F(i, j... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,088 |
3. For a finite graph, the following operation can be performed: choose any cycle of length 4, select any edge in this cycle, and remove it from the graph. For a fixed integer $n(n \geqslant 4)$, if the complete graph with $n$ vertices is repeatedly subjected to the above operation, find the minimum number of edges in ... | 3. The minimum value is $n$.
If a graph can be obtained from a complete graph of $n$ vertices through operations, then this graph is called "permissible"; if any two points in a graph are connected by a path, then this graph is called "connected"; if the vertices of a graph can be divided into two sets $V_{1}$ and $V_... | n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,089 |
3. If $a$ and $b$ are the x-coordinates of the points where the parabola $y=(x-c)(x-c-d)-2$ intersects the x-axis, $a<b$, then the value of $|a-c|+|c-b|$ is $\qquad$ . | (Answer: $b-a$.) | b-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,090 |
4. Consider an $n \times n$ matrix, where each element is a real number with absolute value not exceeding 1, and the sum of all elements is 0. Let $n$ be a positive even number. Find the minimum value of $c$ such that for every such matrix, there exists a row or a column whose sum of elements has an absolute value not ... | 4. The minimum value of $c$ is $\frac{n}{2}$.
Define the matrix $\left(a_{i j}\right)_{1 \leqslant i, j \leqslant n}$ as follows:
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant \frac{n}{2} ; \\
-1, & i, j>\frac{n}{2} ; \\
0, & \text { otherwise. }
\end{array}\right.
$$
Then the sum of all elements is 0, and ... | \frac{n}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,091 |
5. Let $N$ be a positive integer. Two players, A and B, take turns writing numbers from the set $\{1,2, \cdots, N\}$ on a blackboard. A starts and writes 1 on the blackboard. Then, if a player writes $n$ on the blackboard during their turn, their opponent can write $n+1$ or $2n$ (not exceeding $N$). The player who writ... | 5. $N$ is of type B if and only if all the digits on the odd positions in the binary representation of $N$ are 0, with the odd positions determined from right to left.
Assume that a number $n \in\{1,2, \cdots, N \mid$ is written on the blackboard at some moment. We call $n$ "winning" or "losing" based on the next play... | 2048 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,092 |
8. Given a finite graph $G$, let $f(G)$ be the number of 3-cliques in $G$, and $g(G)$ be the number of 4-cliques in $G$. Find the smallest constant $c$ such that for every graph $G$,
$$
(g(G))^{3} \leqslant c(f(G))^{4} .
$$ | 8. Let the vertices of a finite graph $G$ be $V_{1}, V_{2}, \cdots, V_{n}$, and let $E$ be the set of its edges, with $|E|$ denoting the number of elements in the set $E$. Our first task is to establish the relationship between $f(G)$ and $|E|$.
Assume the degree of vertex $V_{i}$ is $x_{i}$, meaning $x_{i}$ edges ema... | \frac{3}{32} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,094 |
1. Given real numbers $a b c \neq 0$, and the three quadratic equations $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ have a common root. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=(\quad)$.
(A) 3
(B) 2
(C) 1
(D) 0 | $-1 . A$.
Let the common root of the three quadratic equations be $x_{0}$. Substituting $x_{0}$ into the three equations and adding them yields
$$
(a+b+c)\left(x_{0}^{2}+x_{0}+1\right)=0 .
$$
Since $x_{0}^{2}+x_{0}+1=\left(x_{0}+\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4}>0$.
Therefore, $a+b+c=0$.
$$
\beg... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,095 |
2. Given positive integers $a, b (a>b)$ such that the sum of $a+b$, $a-b$, $a b$, and $\frac{a}{b}$ is 243. Then the number of pairs $(a, b)$ with this property ( ).
(A) has exactly 1 pair
(B) has exactly 2 pairs
(C) has a finite number of pairs, but more than 1 pair
(D) does not exist | 2. B.
From the given, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=243,
$$
which simplifies to $2 a+a b+\frac{a}{b}=243$.
Since $a, b, 243$ are all positive integers, $a$ must be a multiple of $b$.
Let's assume $a=b k$, substituting into equation (1) we get
$$
2 b k+b^{2} k+k=243 \text {. }
$$
Completing the square, we ha... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,096 |
3. Figure 1 is a part of the graph of the quadratic function $y=a x^{2}+b x+c$, and let $M$ $=a+b$. Then the range of values for $M$ is ( ).
(A) $-1<M<0$
(B) $0<M<1$
(C) $-1<M<1$
(D) cannot be determined
Translate the above text into English, please retain the original text's line breaks and format, and output the tra... | 3.C.
According to the graph of the quadratic function,
$$
\begin{array}{l}
a<0, -\frac{b}{2 a}>0, \\
a-b+c=0, c=1 .
\end{array}
$$
From this, we know that $b=a+1, -1<a<0, 0<b<1$.
Therefore, $-1<M<1$. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,097 |
4. Let $a, b, c$ be the lengths of the sides of any triangle, and let $s=a+b+c, t=ab+bc+ca$. Then $(\quad)$.
(A) $t<s^{2} \leqslant 2 t$
(B) $2 t \leqslant s^{2}<3 t$
(C) $3 t<s^{2} \leqslant 4 t$
(D) $3 t \leqslant s^{2}<4 t$ | 4.D.
$$
\begin{array}{l}
\text { Since } s^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\
=a^{2}+b^{2}+c^{2}+2 t,
\end{array}
$$
It is easy to know that $a^{2}+b^{2}+c^{2}-a b-b c-c a \geqslant 0$, hence $a^{2}+b^{2}+c^{2} \geqslant t$.
From $a^{2}<a b+a c, b^{2}<b c+b a, c^{2}<c a+c b$, we get $a^{2}+b^{2}+c^{2}<2(a b+b c+c... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,098 |
5. Given real numbers $a, b, c, d$ satisfy
$$
\frac{a c-b^{2}}{a-2 b+c}=\frac{b d-c^{2}}{b-2 c+d} \text {. }
$$
Then $\frac{(a-b)(c-d)}{(b-c)^{2}}=(\quad)$.
(A) 1
(B) $\pm 1$
(C) 0
(D) Cannot be determined | 5.A.
Let $\frac{a c-b^{2}}{a-2 b+c}=\frac{b d-c^{2}}{b-2 c+d}=k$. Then
$$
\begin{array}{l}
a c-k(a+c)=b^{2}-2 b k, \\
b d-k(b+d)=c^{2}-2 c k . \\
\text { Hence }(a-k)(c-k)=(b-k)^{2}, \\
(b-k)(d-k)=(c-k)^{2} .
\end{array}
$$
Thus, $(a-k)(d-k)=(b-k)(c-k)$.
Simplifying, we get $k=\frac{a d-b c}{a-b-c+d}$.
Also, $k=\frac... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,099 |
6. Let $I$ be the incenter of $\triangle A B C$, $r$ be the radius of its incircle, and $R$ be the radius of its circumcircle. If the extension of $A I$ intersects the circumcircle of $\triangle A B C$ at point $P$, then $I A \cdot I P=(\quad)$.
(A) $2 R_{r}$
(B) $2 R^{2} r$
(C) $2 R r^{2}$
(D) $2 R^{2} r^{2}$ | 6. A.
As shown in Figure 3, construct the diameter $PQ$ of the circumcircle $\odot O$ of $\triangle ABC$, and connect $BQ$, $BP$. Let $\odot I$ be tangent to $AB$ at point $D$, and connect $IB$, $ID$. It is easy to see that
$$
\begin{array}{l}
\angle PQB = \angle PAB \\
= \angle PAC = \angle PBC.
\end{array}
$$
There... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,100 |
4. Given that $m$ and $n$ are positive integers. If the two real roots of the equation $4 x^{2}-2 m x+n=0$ are both greater than 1 and less than 2, find the values of $m$ and $n$.
$(2003$, Shanghai (Yuzhen Cup) Junior High School Mathematics Competition) | (Solution: $m=6, n=9$.) | m=6, n=9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,101 |
1. If $x=\frac{13}{4+\sqrt{3}}$, then $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+3}{x^{3}-7 x^{2}+5 x+15}=$ $\qquad$ | $$
\text { II.1. }-5 \text {. }
$$
Since $x=\frac{13}{4+\sqrt{3}}=4-\sqrt{3}$, then $x^{2}-8 x+13=0$.
$$
\begin{array}{l}
\text { Hence } x^{4}-6 x^{3}-2 x^{2}+18 x+3 \\
=\left(x^{2}+2 x+1\right)\left(x^{2}-8 x+13\right)-10=-10, \\
x^{3}-7 x^{2}+5 x+15 \\
=(x+1)\left(x^{2}-8 x+13\right)+2=2 .
\end{array}
$$
Therefore... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,102 |
2. A natural number minus 69 is a perfect square, and this natural number plus 20 is still a perfect square. Then this natural number is $\qquad$ . | 2.2005 .
Let this natural number be $x$. According to the problem, we have
$$
\left\{\begin{array}{l}
x-69=m^{2}, \\
x+20=n^{2},
\end{array}\right.
$$
where $m, n$ are both natural numbers.
Subtracting the two equations gives $n^{2}-m^{2}=89$, which is $(n-m)(n+m)=89$.
Since $n>m$, and 89 is a prime number, we have
$... | 2005 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,103 |
3. Given real numbers $x_{1}, x_{2}, y_{1}, y_{2}$ satisfy
$$
\begin{array}{l}
x_{1}^{2}+25 x_{2}^{2}=10, \\
x_{2} y_{1}-x_{1} y_{2}=25, \\
x_{1} y_{1}+25 x_{2} y_{2}=9 \sqrt{55} .
\end{array}
$$
Then $y_{1}^{2}+25 y_{2}^{2}=$ $\qquad$ | 3.2008 .
Notice that
$$
\begin{array}{l}
\left(x_{1}^{2}+a x_{2}^{2}\right)\left(y_{1}^{2}+a y_{2}^{2}\right) \\
=\left(x_{1} y_{1}+a x_{2} y_{2}\right)^{2}+a\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2},
\end{array}
$$
where $a$ is any real number.
For this problem, take $a=25$, and it is easy to find that $y_{1}^{2}+25 ... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,104 |
4. In $\triangle A B C$, $C D$ is the altitude, $C E$ is the angle bisector, and $C D=12 \text{ cm}, A C=15 \text{ cm}, B C=20 \text{ cm}$. Then $C E=$ $\qquad$ . | 4. $\frac{60 \sqrt{2}}{7} \mathrm{~cm}$ or $12 \sqrt{2} \mathrm{~cm}$.
Since the foot of the altitude $C D$ from $\triangle A B C$ can be on side $A B$ or on the extension of side $B A$, we need to consider two cases.
(1) As shown in Figure 4, if
$\triangle A B C$ is an acute triangle,
then the foot $D$ is on side $A ... | \frac{60 \sqrt{2}}{7} \mathrm{~cm} \text{ or } 12 \sqrt{2} \mathrm{~cm} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,105 |
One, (20 points) Prove: regardless of the real value of $x$, we always have
$$
\frac{1}{7} \leqslant \frac{x^{2}-3 x+4}{x^{2}+3 x+4} \leqslant 7 \text {. }
$$ | Let $y=\frac{x^{2}-3 x+4}{x^{2}+3 x+4}$. From the discriminant of the denominator being less than 0, we know that the denominator is not zero in the real number range.
Multiplying both sides by $x^{2}+3 x+4$, and simplifying, we get $(y-1) x^{2}+3(y+1) x+4(y-1)=0$.
Considering this equation as a quadratic equation in $... | \frac{1}{7} \leqslant \frac{x^{2}-3 x+4}{x^{2}+3 x+4} \leqslant 7 | Inequalities | proof | Yes | Yes | cn_contest | false | 716,106 |
II. (25 points) As shown in Figure 2, given that $D$ is any point on side $BC$ of $\triangle ABC$, $I_{1}$ and $I_{2}$ are the incenters of $\triangle ABD$ and $\triangle ACD$ respectively, $I_{1}^{\prime}$ and $I_{2}^{\prime}$ are the excenters of these two triangles (tangent to side $BC$), and $P$ and $P^{\prime}$ ar... | As shown in Figure 6, let $\odot I_{1}$ and $\odot I_{2}$ be tangent to side $BC$ at points $E$ and $F$, respectively. Connect $D I_{1}$, $D I_{2}$, $E I_{1}$, and $F I_{2}$.
It is easy to see that $\angle I_{1} D I_{2}=\frac{1}{2}(\angle A D B+\angle A D C)=90^{\circ}$,
$$
\begin{array}{l}
D E=\frac{1}{2}(A D+B D-A B)... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,107 |
Three, (25) Let $x_{1}, x_{2}, \cdots, x_{p}$ all be natural numbers, and $x_{1}<x_{2}<\cdots<x_{p}$, satisfying $2^{x_{1}}+2^{x_{2}}+\cdots+2^{x_{p}}$ $=2008$. Try to find the values of $p$ and $x_{1}, x_{2}, \cdots, x_{p}$.
| Three, since $x_{1}, x_{2}, \cdots, x_{p}$ are all natural numbers, therefore, $2^{x_{1}}$, $2^{x_{2}}, \cdots, 2^{x_{p}}$ are all positive integers.
Also, since $x_{1}<x_{2}<\cdots<x_{p}$, it follows that
$$
2^{x_{1}}<2^{x_{2}}<\cdots<2^{x_{p}} \text {. }
$$
From $2^{x_{1}}+2^{x_{2}}+\cdots+2^{x_{p}}=2008$, we get
$$... | p=7, x_{1}=3, x_{2}=4, x_{3}=6, x_{4}=7, x_{5}=8, x_{6}=9, x_{7}=10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,108 |
1. A parabolic pond is shown in Figure 1, where $A$ and $B$ are two corresponding points on the edge of the pond, and $O$ is the center of the pond bottom (i.e., the vertex of the parabola). Then for point $C$ in the pond,
$$
\tan \angle C A B + \tan \angle C B A
$$
should satisfy ( ).
(A) When point $C$ is closer to ... | - 1.C.
As shown in Figure 4, based on the given conditions, establish a rectangular coordinate system $x O^{\prime} y$ with the midpoint of $A B$ as the origin and $A B$ as the $x$-axis. Suppose the equation of the parabola is
$$
y=a x^{2}+b(a>0, b<0).
$$
We have $A\left(-\sqrt{\frac{-b}{a}}, 0\right)$,
$B\left(\sqrt{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,109 |
2. Define the sequence $\left\{a_{n}\right\}$ :
$$
a_{i} \in\left(0, \frac{1}{2}\right](i=1,2, \cdots) \text {, }
$$
and. $a_{1}=\frac{1}{2}, \frac{a_{n-1}}{a_{n}}+4 a_{n}^{2}=3$.
Then $\lim _{n \rightarrow+\infty} a_{n}=(：$.
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{1}{12}$
(C) $\frac{\sqrt{2}}{4}$
(D) 0 | 2.1).
Let $a_{n}=\sin \theta$. Since $a_{n}$ satisfies $a_{n-1}=3 a_{n}-4 a_{n}^{3}>0$, and $a_{n}<1$, it follows from the double angle formula that $a_{n-1}=\sin 3 \theta$.
Considering that $f(x)=3 x-4 x^{3}$ is monotonically increasing on $\left(0, \frac{1}{2}\right]$, for a given $\dot{a}_{n-1}, a_{n}$ is uniquely... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,110 |
3. As shown in Figure 2, in a regular tetrahedron $A B C D$ with edge length 1, $M$ and $N$ are the midpoints of $A D$ and $B C$, respectively. The minimum area of the section cut by the plane through $M N$ in this tetrahedron is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{12}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{\sqrt{3}}{8}... | 3. A.
First, the plane through $M N$ must intersect a pair of edges at one point each, and we should set it to intersect $A B$ at $E$ and $C D$ at $F$, as shown in Figure 5.
Notice that
$$
S_{\triangle E M V}=\frac{1}{2} M N \cdot h_{E-M V}
$$
and $h_{\kappa-u v} \geqslant h_{A H-u v}$. Since $A B$ and $M N$ are comm... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,111 |
Example 1 If \(a^{x}=b^{y}=1994^{x}\) (where \(a, b\) are natural numbers), and \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\). Then all possible values of \(2a+b\) are ( ).
(A) 1001
(B) 1001,3989
(C) 1001,1996
(D) \(1001,1996,3989\)
(1994, National Junior High School Mathematics League) | Given: From the known information, we have $a=1994^{\frac{x}{x}}, b=1994^{\frac{x}{y}}$. Therefore, $a b=1994^{\frac{x}{x}+\frac{z}{y}}=1994$ $=2 \times 997$ (prime factorization).
From the conditions, we know $a \neq 1, b \neq 1$, so, $a=2, b=997$ or $a=997, b=2$.
Thus, $2 a+b$ is 1001 or 1996. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,112 |
Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last part of the text is a note about the translation instruction and is not part of the content to be translated. Here is ... | Solution: Let the greatest common divisor be $d$, and the 10 numbers be $b_{1} d, b_{2} d, \cdots, b_{10} d$. According to the problem, we have
$$
\left(b_{1}+b_{2}+\cdots+b_{10}\right) d=1001 .
$$
Since $b_{1}+b_{2}+\cdots+b_{10} \geqslant 10$, it follows that $d \leqslant$ 100 and is a divisor of 1001.
Given $1001=... | 91 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,113 |
Example 11 Express $\frac{x}{(x-2)\left(x^{2}+2 x+2\right)}$ as partial fractions. | Solution: Since the discriminant of $x^{2}+2 x+2$ in the denominator is less than zero, it cannot be factored into a product of linear factors. We call it a quadratic prime factor. Let
$$
\begin{array}{l}
N=\frac{x}{(x-2)\left(x^{2}+2 x+2\right)} \\
=\frac{A}{x-2}+\frac{B x+C}{x^{2}+2 x+2} .
\end{array}
$$
Then $x=A\l... | N=\frac{1}{5(x-2)}+\frac{-x+1}{5\left(x^{2}+2 x+2\right)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,114 |
Example 12 Express $\frac{2 x^{2}+2 x+13}{(x-2)\left(x^{2}+1\right)^{2}}$ as partial fractions. | Solution: Let $N=\frac{2 x^{2}+2 x+13}{(x-2)\left(x^{2}+1\right)^{2}}$
$$
\equiv \frac{A}{x-2}+\frac{B x+C}{x^{2}+1}+\frac{D x+E}{\left(x^{2}+1\right)^{2}} \text {. }
$$
By comparing coefficients, we get $A=1, B=-1$,
$$
C=-2, D=-3, E=-4 \text {. }
$$
Therefore, $N=\frac{1}{x-2}+\frac{-x-2}{x^{2}+1}+\frac{-3 x-4}{\lef... | N=\frac{1}{x-2}+\frac{-x-2}{x^{2}+1}+\frac{-3 x-4}{\left(x^{2}+1\right)^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,115 |
2. If $1 \times 2 \times \cdots \times 100=12^{n} M$, where $M$ is a natural number, and $n$ is the largest natural number that makes the equation true, then $M$ ( ).
(A) is divisible by 2 but not by 3
(B) is divisible by 3 but not by 2
(C) is divisible by 4 but not by 3
(D) is not divisible by 3, nor by 2
(1991, Natio... | (Tip: In the prime factorization of 100!, the exponent of 2 is 97, and the exponent of 3 is 48. Let $100!=2^{97} \times 3^{48} \times M_{1}=12^{18} \times 2 M_{1}$. Choose (A).) | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,117 |
3. For what values of $m$ and $n$, can the polynomial $x^{4}-5 x^{3}+11 x^{2}+$ $m x+n$ be divisible by $x^{2}-2 x+1$? | Hint: Method of undetermined coefficients. $m=-11, n=4$.) | m=-11, n=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,118 |
4. Let the polynomial $f(x)=a_{0} x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$. If $f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72$, find $f(x)$. | (Tip: By the factor theorem, let $f(x)=(x-1)(x-2)$ $(x-3)(a x+b)$. Then $f(4)=3 \times 2 \times 1 \times(4 a+b)=6$, $f(5)=4 \times 3 \times 2 \times(5 a+b)=72$. Solving yields $a=2, b=$ -7.) | f(x) = (x-1)(x-2)(x-3)(2x-7) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,119 |
5. Express $M=\frac{8 x}{(x+1)^{2}(x-1)}$ as partial fractions. | (Given: $M=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-1}$, we get $A=$ $-2, B=4, C=2$.) | A=-2, B=4, C=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,120 |
6. Express $M=\frac{4 x^{2}+2 x+6}{x^{4}+x^{2}+1}$ as partial fractions. | (Hint: $x^{4}+x^{2}+1=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$. Let $M=\frac{A x+B}{x^{2}+x+1}+\frac{C x+D}{x^{2}-x+1}$, we get $A=1, B=2, C=-1$, $D=4$. ) \quad(-1$ | M=\frac{x+2}{x^{2}+x+1}+\frac{-x+4}{x^{2}-x+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,121 |
Example 1 Proof: The sum of the squares of the distances from any point on the circumcircle of an equilateral triangle to its three sides is a constant.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 1 Proof: The su... | As shown in Figure 1, let the side length of the equilateral $\triangle ABC$ be $a$ (fixed length), and point $P$ lies on the circumcircle of $\triangle ABC$. $P P_{1} \perp AC$ at $P_{1}$, $P P_{2} \perp AB$ at $P_{2}$, and $P P_{3} \perp BC$ at $P_{3}$. Connect $PA$, $PB$, and $PC$.
Let the radius of $\odot O$ be $R$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,122 |
Example 2 As shown in Figure $2, \triangle A B C$ is an equilateral triangle, and point $D$ is a moving point on line $B C$. Through point $D$, draw $D E \perp A C$ at $E$ and draw a perpendicular to $B C$ through point $D$ intersecting the perpendicular to $A B$ through $E$ at $F, C F$ intersects $A B$ at $P$. Prove: ... | As shown in Figure 2, construct $P S / / F E, P Q / / F D$, and connect $Q S$.
Obviously, $\triangle D E F$ is an equilateral triangle. At this point,
$$
\frac{P Q}{F D}=\frac{C P}{C F}=\frac{P S}{F E}.
$$
Thus, $P Q=P S$.
It is easy to see that $\angle Q P S=\angle D F E=60^{\circ}$. Therefore, $\triangle P Q S$ is ... | \frac{A P}{P B}=\frac{1}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 716,123 |
Example 3 Find the smallest natural number $n$ such that $\frac{n-13}{5 n+6}$ is a non-zero reducible fraction.
(6th IMO) | Solution: Let the common divisor of the numerator and denominator be $d(d>1)$, and let
$$
\begin{array}{l}
n-13=k d, \\
5 n+6=l d .
\end{array}
$$
(2) $-5 \times$ (1) gives
$$
(l-5 k) d=71 \text{. }
$$
Since 71 is a prime number and $d>1$, we have
$$
d=71 \text{ and } l-5 k=1 \text{. }
$$
Thus, $n=13+k \times 71$.
Wh... | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,124 |
Example 3 As shown in Figure $3, A B$ is the diameter of the semicircle $\odot O$, and the moving point $C$ is on the semicircle $\odot O$, $C D \perp A B$ at point $D . \odot O_{1}$ is tangent to $\overparen{A C} , C D , A D$, and $\odot O_{2}$ is tangent to $\overparen{B C} , C D , D B$, with the points of tangency $... | Explanation: Let $\odot \mathrm{O}_{2}$
be tangent to $\overparen{B C}$ at point $M$ and to $C D$ at point $N$. Auxiliary lines are shown in Figure 3.
It is easy to see that $O, O_{2}, M$ are collinear.
From $O A=O M$, we get $\angle O A M=\angle O M A$;
From $O_{2} N=O_{2} M$, we get $\angle O_{2} N M=\angle O_{2} ... | 45^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 716,125 |
Example 4 As shown in Figure 4, two circles are concentric, with radii $R$ and $r (R>r)$. An equilateral $\triangle A B C$ is inscribed in the smaller circle, and a moving point $P$ is on the larger circle. Prove: The area of the triangle with sides $P A, P B, P C$ is a constant.
Translate the above text into English,... | As shown in Figure 4, construct
the equilateral $\triangle A P P^{\prime}$, then the sides of $\triangle P C P^{\prime}$ are equal to
$P A, P B, P C$. Let
$\angle P O B=\alpha, \angle P O C$
$=\beta$, it is easy to see that $\alpha+\beta=$
$120^{\circ}, \frac{\alpha-\beta}{2}=60^{\circ}-\beta$. Therefore, we have
$$
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,126 |
Example 6 As shown in Figure 6, point $O$ is the circumcenter of acute $\triangle A B C$, and ray $A O$ intersects $B C$ at point $D$. A moving line $l$ intersects $A B$ and $A C$ at points $E$ and $F$. If points $A, E, D, F$ are concyclic, then the orthogonal projection of line segment $E F$ on $B C$ is always a const... | As shown in Figure 6, draw $D M \perp A B$ at point $M$, $D N \perp A C$ at point $N$, $M M^{\prime} \perp B C$ at point $M^{\prime}$, and $N N^{\prime} \perp B C$ at point $N^{\prime}$. Clearly, $M^{\prime} N^{\prime}$ is a constant. Draw $E E^{\prime} \perp B C$ and $F F^{\prime} \perp B C$. It is easy to prove that ... | E^{\prime} F^{\prime} = M^{\prime} N^{\prime} | Geometry | proof | Yes | Yes | cn_contest | false | 716,128 |
Example 7 As shown in Figure 7, in $\odot O$, chord $M N=a$ (fixed length), a moving line $X Y$ passes through point $O$ and intersects $M N$. $X Y$ intersects $\odot O$ at points $A$ and $B$, and through the midpoint $Q$ of $M N$, $Q C \perp M B$ at point $C$, and $Q D \perp M A$ at point $D$. Prove: $A M \cdot Q C + ... | Explanation: As shown in Figure 7, connect $A N$ and $B N$. It is easy to prove that $\triangle A B N \sim \triangle M Q C$. At this point, $\frac{A B}{M Q}=\frac{A N}{M C}=\frac{B N}{Q C}$.
Since quadrilateral $D M C Q$ is a rectangle, $M C=Q D$, so we have $\frac{A B}{M Q}=\frac{A N}{Q D}=\frac{B N}{Q C}$.
Let the r... | \frac{1}{2} a^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 716,129 |
Example 8 As shown in Figure 8, in $\triangle ABC$, $AB=AC$, a moving line $l$ passes through point $A$ (line $l$ does not pass through the interior of $\angle BAC$). It is known that $\odot O_{1}$ is tangent to line $l$, $AB$, and $BC$, and $\odot O_{2}$ is tangent to line $l$, $AC$, and $BC$. Prove: regardless of the... | Let $\odot O_{1} 、 \odot O_{2}$ have radii $R_{1} 、 R_{2}$. As shown in Figure 8, construct the altitude $A D$ of $\triangle A B C$.
We consider two cases:
(1) When $l \parallel B C$, it is easy to see that $\odot O_{1} 、 \odot O_{2}$ are two equal circles, and $R_{1}=R_{2}=\frac{1}{2} A D$, so,
$$
R_{1}+R_{2}=A D \tex... | R_{1}+R_{2}=A D | Geometry | proof | Yes | Yes | cn_contest | false | 716,130 |
Example 9 As shown in Figure $9, \odot Q$ has a diameter $A B=d$ (a constant). $\odot O$ and $\odot O^{\prime}$ are two moving circles, both internally tangent to $\odot Q$ and tangent to $A B$. A tangent line to $\odot Q$ through point $B$ intersects the rays $A O$ and $A O^{\prime}$ at points $E$ and $F$; a tangent l... | Explanation: As shown in Figure 9, let
$\odot O$ be tangent to $AB$ at point $D$ and tangent to
$\odot Q$ at point $M$. Clearly, points $Q$,
$O$, and $M$ are collinear, and $OD \perp$
$AB$.
Let $AD = a$, $BD = b$ (where $a > b$). Then
$AB = a + b$,
$QM = QA = QB = \frac{1}{2}(a + b)$,
$QD = \frac{1}{2}(a - b)$.
Let $O... | d^2 | Geometry | proof | Yes | Yes | cn_contest | false | 716,131 |
Example 10 As shown in Figure 10, square $ABCD$ is a piece of cardboard with side length $a$. Two lines $l_{1} \parallel l_{2}$ in the plane, and the distance between them is also $a$. Now, place this square piece of paper flat on the two parallel lines so that $l_{1}$ intersects $AB$ and $AD$, with the intersection po... | As shown in Figure 10, connect $A C$, draw the altitudes $A A^{\prime}$ and $C C^{\prime}$, and draw $C S \perp A A^{\prime}$ at point $S$. Let $E F=t, \angle A E F=\alpha$.
It is easy to see that
$$
\begin{array}{l}
m_{1}=E F+A F+A E \\
=t(1+\sin \alpha+\cos \alpha) .
\end{array}
$$
Notice that $A A^{\prime}=A E \cdo... | m_{1}+m_{2}=2a | Geometry | proof | Yes | Yes | cn_contest | false | 716,132 |
1. As shown in Figure 11, in trapezoid $ABCD$, the two bases $AD$ and $BC$, and the two legs $AB$ and $CD$ are all of fixed length. Point $E$ moves on $AB$, and $O_{1}, O_{2}$ are the circumcenters of $\triangle ADE$ and $\triangle BCE$, respectively. Prove: the line segment $O_{1}O_{2}$ is of fixed length. | (First prove $\triangle O_{2} O_{1} E \sim \triangle C D E$, then we have $\frac{O_{2} O_{1}}{C D}=\frac{O_{2} E}{C E}=\frac{1}{2 \sin B}=\frac{1}{2 \times \frac{h}{A B}}(h$ is the height of the trapezoid). Therefore, $O_{1} O_{2}=\frac{A B \cdot C D}{2 h}$.) | O_{1} O_{2}=\frac{A B \cdot C D}{2 h} | Geometry | proof | Yes | Yes | cn_contest | false | 716,133 |
Example 4 Given that $x, y$ are integers, and
$$
15 x^{2} y^{2}=35 x^{2}-3 y^{2}+412 \text {. }
$$
then $15 x^{2} y^{2}=$ $\qquad$ . | Solution: Transform the given equation into
$$
\left(5 x^{2}+1\right)\left(3 y^{2}-7\right)=405=3^{4} \times 5 \text {. }
$$
Since $5 x^{2}+1$ is not a multiple of 5, and $3 y^{2}-7$ is not a multiple of 3, it can only be
$$
5 x^{2}+1=3^{4}, 3 y^{2}-7=5 \text {. }
$$
Therefore, $5 x^{2}=80,3 y^{2}=12$.
Thus, $15 x^{2... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,135 |
The radii of the two circles are $R$ and $R^{\prime}\left(R>R^{\prime}\right)$, a moving line $l \perp O O^{\prime}$, and intersects the two circles at $B$ and $B^{\prime}$. Prove: The circumradius of $\triangle A B B^{\prime}$ is a constant. | (提示: Let the circumradius of $\triangle A B B^{\prime}$ be $r, O O^{\prime}$ intersect the larger and smaller circles at points $C$ and $C^{\prime}$, respectively, and $l$ intersects $O O^{\prime}$ at point $K$. First prove that $A B^{2} \cdot A B^{\prime 2}=4 R R^{\prime} \cdot A K^{2}$. Then, since $A B \cdot A B^{\p... | r = \sqrt{R R^{\prime}} | Geometry | proof | Yes | Yes | cn_contest | false | 716,136 |
5. As shown in Figure 15, given a circle $\odot O$ with radius $R$ that is tangent to lines $l_{1}$ and $l_{2}$ at points $A$ and $B$, respectively, and $l_{1} / / l_{2}$. Two externally tangent moving circles $\odot O_{1}$ and $\odot O_{2}$ are both externally tangent to $\odot O$, and are tangent to $l_{1}$ and $l_{2... | (提示: From $O_{1} K^{2}+O_{2} K^{2}=O_{1} O_{2}^{2}$, we know $\left[2 R-\left(r_{1}+\right.\right.$ $\left.\left.r_{2}\right)\right]^{2}+\left(2 \sqrt{R r_{2}}-2 \sqrt{R r_{1}}\right)^{2}=\left(r_{1}+r_{2}\right)^{2}$, thus leading to $2 \sqrt{r_{1} r_{2}}=R$. At this point, $C D=2 \sqrt{r_{1} r_{2}}=R$ (a constant).) | R | Geometry | proof | Yes | Yes | cn_contest | false | 716,138 |
Example 1 Let $f(x), g(x)$ be functions on $[0,1]$. Prove: There exist $x_{0}, y_{0} \in [0,1]$, such that
$$
\left|x_{0} y_{0}-f\left(x_{0}\right)-g\left(y_{0}\right)\right| \geqslant \frac{1}{4} \text {. }
$$
Analysis: To find specific $x_{0}, y_{0}$, it is difficult to proceed, so we might consider using proof by c... | Proof: Suppose such $x_{0}$ and $y_{0}$ do not exist. Take special values $x_{0}=0, y_{0}=0$, we get
$$
|f(0)+g(0)|<\frac{1}{4}.
$$
Similarly, $|f(0)+g(1)|<\frac{1}{4}$,
$$
\begin{array}{l}
|f(1)+g(0)|<\frac{1}{4}, \\
|1-f(1)-g(1)|<\frac{1}{4}.
\end{array}
$$
Thus, $1=|1-f(1)-g(1)|+|f(1)+g(0)|+|f(0)+g(1)|-|f(0)+g(0)|... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,139 |
Example 2 Given a finite number of weights, their total weight is $1 \mathrm{~kg}$, and they are numbered as $1,2, \cdots$. Prove: From these finite weights, there must be a weight with a number $n$ whose weight is greater than $\frac{1}{2^{n}} \mathrm{~kg}$. | Proof: Assume there does not exist an index $n$ such that the corresponding weight of the weight $f(n) > \frac{1}{2^{n}}$, and assume there are $m$ weights, $m > 0$. Thus, we have
$$
\begin{array}{l}
f(1) \leqslant \frac{1}{2}, \\
f(2) \leqslant \frac{1}{2^{2}}, \\
\cdots \cdots \\
f(m) \leqslant \frac{1}{2^{m}} .
\end... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,140 |
Example 3 Prove: It is impossible for any three real numbers to simultaneously satisfy the following three inequalities:
$$
|x|<|y-z|,|y|<|z-x|,|z|<|x-y| \text {. }
$$
Analysis: This problem requires proving that all objects have the same property, which cannot be considered from a direct perspective, so it is advisab... | Proof: Assume there exist three real numbers $x$, $y$, and $z$ that simultaneously satisfy the given three inequalities. Squaring both sides of each inequality and then rearranging and factoring, we get
$$
\begin{array}{l}
(x-y+z)(x+y-z)<0, \\
(y-z+x)(y+z-x)<0, \\
(z+x-y)(z-x+y)<0 .
\end{array}
$$
$$
\begin{array}{l}
\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,141 |
Example 4 In space, 8 known points are given, where no four points are coplanar. It is known that 17 line segments are connected with them as endpoints. Prove: these line segments must form at least one triangle.
Analysis: It is not feasible to specifically find a triangle, so it is advisable to consider the opposite.... | Proof: Assume that none of the 17 line segments form a triangle. Let point $A$ be the point among the 8 points that has the most line segments connected to it. Suppose point $A$ is connected to $n$ line segments: $A B_{1}$, $A B_{2}, \cdots, A B_{n}$. Thus, no line segment connects any two points among $B_{1}, B_{2}, \... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,142 |
Example 5: It is desired to cut out a triangle of arbitrary shape with an area of 1 from an infinitely long strip of paper. What is the minimum width of the strip?
Analysis: Since it is necessary to estimate the width of the strip, let's start with special triangles first.
Translate the above text into English, pleas... | Prove: For an equilateral triangle with an area of 1, the side length is $\frac{2}{\sqrt{3}}$, and the height is $\sqrt[4]{3}$. Therefore, it is conjectured that the width of the paper strip should be at least $\sqrt[4]{3}$. Assume that an equilateral $\triangle ABC$ with an area of 1 can be cut from a paper strip narr... | \sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,143 |
Example 1 Given that $a, b, c$ are the lengths of the three sides of $\triangle ABC$, and satisfy $a^{2}=b^{2}+bc$. Prove: $\angle A=2 \angle B$.
Analysis: Considering the characteristics of the given relationship in the known condition
$$
\begin{array}{l}
a^{2}=b^{2}+bc=b(b+c), \\
a \cdot a=b \cdot b+b \cdot c .
\end... | Proof: As shown in Figure 1, extend $CA$ to $D$ such that $AD = AB = c$. Then, $CB^2 = CA \cdot CD$. Therefore, $CB$ is the tangent to the circumcircle of $\triangle ABD$ at point $B$. Notice that $\angle ABC = \angle ADB$ $= \angle ABD$, hence
\[
\angle CAB = 2 \angle ABC,
\]
\[
\angle CAB = 2 \angle ABC.
\] | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,144 |
Example 5 Factorization:
$$
(x-1)(x-2)(x-3)(x-4)-120 \text {. }
$$
(1991, Jilin Province Participates in the National Junior High School Mathematics League Preliminary Selection Competition) | Solution: Let $f(x)$
$$
=(x-1)(x-2)(x-3)(x-4)-120 \text {. }
$$
Notice that
$$
\begin{array}{l}
f(-1)=(-2) \times(-3) \times(-4) \times(-5)-120=0, \\
f(6)=5 \times 4 \times 3 \times 2-120=0,
\end{array}
$$
Thus, we know that $f(x)$ has factors $x+1$ and $x-6$.
Expanding and performing division, we get
$$
\begin{array... | (x+1)(x-6)\left(x^{2}-5 x+16\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,146 |
Example 7 Fill the positive integers from 1 to 100 into a $10 \times 10$ grid arbitrarily, with each cell containing one number. Prove: there must be two adjacent cells (i.e., cells sharing a common edge) where the difference between the numbers is at least 6. | Proof: Suppose we can find a way to fill the grid such that the difference between the numbers in any two adjacent cells does not exceed 5 (i.e., is less than 6). Observe the number \( a \) in the cell that is in the same row as 1 and the same column as 100. Since \( a \) is at most 8 cells away from 1, we have
$$
a \l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,147 |
Example 2 As shown in Figure 2, let the circumradius of $\triangle A B C$ be $R, A D \perp B C, D E \perp A B, D F \perp A C$. Prove: $S_{\triangle A B C}=R \cdot E F$.
Analysis: Considering the characteristics of the relationship in the conclusion and the relationship with the area of the triangle, we have
$$
\begin{... | Proof: Draw diameter $A H$ intersecting $E F$ at $G$, as shown in the auxiliary line in Figure 2. Then,
$$
\angle A H B=\angle A C B \text {. }
$$
Since $A, E, D, F$ are concyclic, we have,
$$
\angle A E F=\angle A D F=\angle A C B=\angle A H B \text {. }
$$
Thus, $E, B, H, G$ are concyclic.
Since $\angle A B H=90^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,148 |
Example 3: There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.
(2000, National High School Mathematics Competition)
Analysi... | Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\r... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,149 |
Example 4: Circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $M$ and $N$. Let $l$ be the common tangent of circles $\Gamma_{1}$ and $\Gamma_{2}$ that is closer to point $M$, with $l$ tangent to circle $\Gamma_{1}$ at point $A$ and to circle $\Gamma_{2}$ at point $B$. Let the line through point $M$ parallel to $... | Prove: Connect $A M, B M$, let $K$ be the intersection of $M N$ and $A B$. Then
$$
A K^{2}=K N \cdot K M=B K^{2} .
$$
Thus, $K$ is the midpoint of $A B$.
Since $A B \parallel C D$, $M$ is the midpoint of $P Q$.
Since $C D \parallel A B$, $A, B$ are the midpoints of $\overparen{C M}$ and $\overparen{M D}$, respectively... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,150 |
Example 5 Hexagon $A B C D E F$ is inscribed in $\odot O$, and $A B=B C=C D=\sqrt{3}+1, D E=E F=F A=1$. Find the area of this hexagon.
Analysis: To find the area of any hexagon, a direct approach can be quite troublesome. Observing Figure 4, we notice that it has two sets of congruent triangles, and the three triangle... | Solution: Connect $O A, O B, O C, O D, O E, O F$, it is easy to see that
$$
\begin{array}{l}
S_{\triangle M O B}=S_{\triangle B O C}=S_{\triangle C O D}, \\
S_{\triangle D O E}=S_{\triangle E O F}=S_{\triangle F O A} .
\end{array}
$$
Reorganize the six triangles to form a cyclic hexagon $A^{\prime} B^{\prime} C^{\prim... | \frac{9}{4}(2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,151 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.