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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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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3. Given the parabola $y=x^{2}+k x+4-k$ intersects the $x$-axis at integer points $A, B$, and intersects the $y$-axis at point $C$. Then $S_{\triangle A B C}=$ $\qquad$ . | 3.24.
Given the equation $x^{2}+k x+4-k=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$,
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$
Thus, $\left\{\begin{array}{l}... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,335 |
4. In $\triangle A B C$, $A B$ $=B C=12, \angle B=90^{\circ}$, as shown in the figure. Fold along $E F$ so that $A$ coincides with a point $D$ on $B C$. If $B D: D C=2: 1$, then the length of $A E$ is $\qquad$ | $$
\begin{array}{l}
\text { 4. } 8 \frac{2}{3} . \\
\because B D \cdot D C=2: 1, \\
\therefore B D=8 .
\end{array}
$$
Let $A E=x$, then $E D=A E=x, B E=12-x$. In Rt $\triangle B D E$, $x^{2}=(12-x)^{2}+8^{2}$. Solving for $x$ yields $x=8 \frac{2}{3}$. | 8 \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,336 |
One. (20 points) In quadrilateral $ABCD$, $O$ is the midpoint of $AB$. A semicircle with $O$ as its center (its diameter is less than $AB$) is tangent to sides $AD$, $DC$, and $CB$ at points $E$, $F$, and $G$ respectively. Prove: $AB^{2} = 4 AD \cdot BC$.
---
The translation maintains the original text's line breaks ... | Connect $O E, O D, C F, O C, O C$.
It is easy to know, Rt $\triangle A O E \cong$ Rt $\triangle B O G$.
We have $\angle A=\angle B$,
$\angle A O E=\angle B O G=\angle 1$.
Also, $\angle D O E=\angle D O F=\angle 2$,
$\angle C O F=\angle C O G=\angle 3$,
then $\angle 1+\angle 2+\angle 3=90^{\circ}$.
$$
\begin{array}{l}
\... | AB^{2} = 4 AD \cdot BC | Geometry | proof | Yes | Yes | cn_contest | false | 708,337 |
II. (25 points) Let $a, b$ be real numbers that make the equation $x^{4}+a x^{3}+b x^{2}+a x+1=0$ have at least one real root. For all such pairs $(a, b)$, find the minimum value of $a^{2}+b^{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translatio... | Obviously, $x \neq 0$, dividing both sides of the original equation by $x^{2}$, we get
$$
\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+(b-2)=0 \text {. }
$$
By the inequality $(a+b)^{2} \geqslant 4 a b$, we know
$$
\left|x+\frac{1}{x}\right| \geqslant 2 \text {. }
$$
Let $y=x+\frac{1}{x}$, then equation... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,338 |
Three, (25) 19 football teams hold a single round-robin tournament. It is known that each team has played at least 13 of the other teams. Prove: It is certain to find four teams, any two of which have played each other. | Three, from the hypothesis, each team has played against at most $19-(13+1)=5$ teams.
From 19 teams, choose one team and denote it as $A$, marking this team as $A Z A$.
In the first group, choose another team $B$ different from $A$, and classify the teams in the first group that have played against $B$ into the secon... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,339 |
Example 9. As shown in Figure 5, a circle intersects the three sides of an equilateral triangle at six points. Given that $A G=2, G F=13, F C=1, H J=7$. Find $DE$.
(33rd American High School Mathematics Examination) | Let $A H=x$. By the secant-tangent theorem, we have
$$
A H \cdot A J=A G.
$$
For $A F$,
$$
\begin{array}{l}
\because A G=2, G F \\
=13, H J=7, \\
\therefore A H(A H+7) \\
=2(2+13)=30 .
\end{array}
$$
Rearranging, we get $A H^{2}$
$$
+7 A H-30=0.
$$
Solving this, we get $A H_{1}=3, A H_{2}=-10$ (discard).
$$
\begin{ar... | 2 \sqrt{22} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,340 |
1. For tetrahedron $ABCD$, all dihedral angles are acute, and opposite edges are pairwise equal. The six dihedral angles of the tetrahedron are $\alpha_{i}(i=1,2, \cdots, 6)$. Then $\sum_{i=1}^{6} \cos \alpha_{i}=(\quad$.
(A) 1
(B) 2
(C) 4
(D) not a constant value | Suddenly, the four triangles of the tetrahedron are denoted as $S$.
Consider the projection of $A$ onto the plane $BCD$ as $O$, and the dihedral angle between the plane $ABC$ and the plane $BCD$ as $\alpha_{1}$, then
$$
S_{\triangle OBC}=S \cdot \cos \alpha_{1} .
$$
Thus, $S \cos \alpha_{1}+S \cos \alpha_{2}+S \cos \... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,341 |
3. For $\triangle A B C=$ with sides of length $a, b, c$, and incenter $I$, the value of $\frac{|I A|^{2}}{b c}+\frac{|I B|^{2}}{c a}+\frac{|I C|^{2}}{a b}$ is ( ).
(A) 1
(B) 2
(C) $\frac{1}{2}$
(D) not a constant value | 3. A.
Let $r$ be the radius of the incircle of $\triangle ABC$, then
$$
|IA|=\frac{r}{\sin \frac{A}{2}}.
$$
Also, $b=r\left(\operatorname{ctg} \frac{A}{2}+\operatorname{ctg} \frac{C}{2}\right)$,
$$
\begin{array}{l}
c=r\left(\operatorname{ctg} \frac{A}{2}+\operatorname{ctg} \frac{B}{2}\right). \\
\frac{|IA|^{2}}{bc}=
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,343 |
4. If the function $f(x)=\sin ^{t} x \cdot \sin t x+\cos ^{t} x \cdot \cos t x-$ $\cos ^{2} 2 x$ is a constant for all real numbers $x$, then the value of the positive integer $t$ is ().
(A) 3
(B) 1
(C) 3 or 1
(D) does not exist | 4. A.
From $x=0$ we know $f(0)=0$.
Also, by the problem statement, for all real numbers $x, f(x)$ is always a constant, so it must be that $f(x)=0$.
Taking $x=\frac{\pi}{t}$, we have
$$
\begin{array}{l}
f\left(\frac{\pi}{t}\right)=\sin ^{4} \frac{\pi}{k} \sin \pi+\cos ^{t} \frac{\pi}{k} \cos \pi-\cos ^{\prime} \frac{2... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,344 |
6. Let the set $T=\left\{x_{1}, x_{2}, \cdots, x_{10}\right\}$ have five-element subsets such that any two elements of $T$ appear in at most one subset. The maximum number of such subsets is $n$. Then $n$ equals ( ).
(A) 252
(B) 10
(C) 8
(D) 18 | 6. C.
An element has a chance to be in the same subset with 9 other elements, and the following is true for $S$ when $n \leqslant 4 \times 10, n \leqslant 8$.
Moreover, the sets $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\},\left\{x_{1}, x_{6}, x_{7}, x_{8}\right.$, $\left.x_{9}\right\},\left\{x_{1}, x_{3}, x_{5},... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,346 |
1. Given that the inverse function of $y=f(x)$ is $\varphi(x)$, and $\varphi(x)$ . $=\log _{m a_{0}{ }^{2}}\left(\frac{1996}{x}-\sin ^{2} 0\right), 0 \in\left(0, \frac{\pi}{2}\right)$. Then the solution to the equation $f(x)=$ 1996 is | $=, 1 .-1$.
Since $f(x)$ and $\varphi(x)$ are inverse functions of each other, solving $f(x) = 1996$ is equivalent to finding the value of $\varphi(1996)$.
$$
\varphi(1996)=\log _{\sec ^{2} \theta}\left(1-\sin ^{2} \theta\right)=-1 .
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,347 |
2. For any two points on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$, if the perpendicular bisector of the line segment joining these two points intersects the $x$-axis at point $P\left(x_{0}, 0\right)$, then the range of $x_{0}$ is $\qquad$ | 2. $(-3,3)$.
Let the coordinates of $A, B$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then $|P A|=$ $|P B|$, that is,
$$
\left(x_{1}-x_{0}\right)^{2}+y_{1}^{2}=\left(x_{2}-x_{0}\right)^{2}+y_{2}^{2} .
$$
Since $A, B$ are on the ellipse, we have
$$
y_{1}^{2}=4-\frac{1}{4} x_{1}^{2}, y_{2}^{2}=4-\f... | (-3,3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,348 |
$\begin{array}{l}\quad \text { 3. } \arcsin \frac{1}{\sqrt{10}}+\arccos \frac{7}{\sqrt{50}}+\operatorname{arctg} \frac{7}{31} \\ +\operatorname{arcctg} 10=\end{array}$ | 3. $\frac{\pi}{4}$.
Let $z_{1}=3+i, z_{2}=7+i, z_{3}=31+7 i, z_{4}=10+i$.
Then $z_{1} z_{2} z_{3} z_{4}=5050(1+i)$.
Also, $\arcsin -\frac{1}{\sqrt{10}}, \arccos \frac{7}{\sqrt{50}}, \operatorname{arctg} \frac{7}{31}$,
$\operatorname{arcctg} 10$ all belong to $\left(0, \frac{\pi}{2}\right)$,
$$
\begin{array}{l}
\theref... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,349 |
4. Divide 1996 into the sum of 96 positive integers, the maximum value of the product of these 96 positive integers is | 4. $20^{20} \cdot 21^{76}$
Let $1996=a_{1}+a_{2}+\cdots+a_{96}$, and $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{96}$.
If $a_{06} \geqslant a_{1}+2$, then let $a^{\prime}{ }_{06}=a_{06}-1, a_{1}^{\prime}=a_{1}+1$, so $a_{96}^{\prime}+a_{1}^{\prime}=a_{96}+a_{1}$, and
$$
a_{1}^{\prime} \cdot a_{96}^{\prime}=\l... | 20^{20} \cdot 21^{76} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,350 |
Example 10. As shown in Figure 6-1, it is a part of a city's street map, with five roads running both longitudinally and latitudinally. If one walks from point $A$ to point $B$ (only from north to south, and from west to east), how many different ways are there to do so? | Solution: Because the number of different ways from the top-left vertex of each small square to the endpoint equals the sum of the number of different ways from the two adjacent vertices to the endpoint, we can deduce the data at the top-left corner of each small square in Figure 6-4, which represents the number of dif... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,351 |
6. For $\triangle A B C$ with sides $a, b, c$, if $|a-b| \leqslant|a-c|$, $|a-b| \leqslant|b-c|$, then the range of $\frac{b}{a}$ is $\qquad$ . | 6. $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.
Let's assume $a=1, b=q$.
If $\frac{a}{b}=\frac{1}{q} \geqslant 1$, by the problem's condition, then $c$ must be the longest side or the shortest side of $\triangle A B C$, which means
$q \leqslant 1 \leqslant c$ or $c \leqslant q \leqslant 1$.
In this case,... | \left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,353 |
t. 5 points, find all functions $f: Z \rightarrow Z$, such that for all $\therefore$ we have $f[f(n)]+f(n)=2 n+3$, and $f(0)=1$.
The text has been translated while preserving the original formatting and line breaks. | When $n=0$, $f[f(0)]+f(0)=f(1)+f(0)=$ $f(1)+1=2 \times 0+3$, so $f(1)=2$.
Let $n=1$, then
$$
f[f(1)]+f(1)=2 \times 1+3=5,
$$
which means $f(2)+2=5$, so $f(2)=3$.
From this, we conjecture: $f(n)=n+1$.
We need to prove this by induction for all $n$ in the set of integers $\mathbb{Z}$.
For $n \geqslant 0$, if $n \leqslan... | f(n)=n+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,356 |
Initially 37. Cities $A, B, C, D$ are exactly the four vertices of a square. If a highway system is to be established so that every two cities are connected by a highway, and the total length of the entire highway system is minimized, how should this highway system be constructed? | Solution: Since the length will not increase when the curve segments are replaced by straight segments, we can assume that the road system is as shown in Figure 1 (when necessary, $P$,
As shown in Figure 2, construct equilateral $\triangle A B E$ and $\triangle C D F$, then construct equilateral $\triangle B P M$. It ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,358 |
Tough 38. As shown in the figure,
$\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $A$. Through $A$, draw the diameters $A B$ and $A C$ of $\odot O_{1}$ and $\odot O_{2}$, respectively. $P$ is the midpoint of the external common tangent $M N$ of $\odot O_{1}$ and $\odot O_{2}$. The circle $\odot O$ passi... | Proof (1) Connect $P A, Q A, B M, A M, A B_{1}, A C_{1}$, AN.
$\because A B, A C$ are diameters,
$\therefore \angle A B_{1} B=\angle A C_{1} C=90^{\circ}$.
$\therefore A, C_{1}, Q, B_{1}$ are concyclic.
$$
\therefore \angle Q B_{1} C_{1}=\angle Q A C_{1} \text {. }
$$
Let the internal common tangent of $\odot O_{1}, \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,359 |
38. The function $f(x)$ defined on $(-1,1)$ satisfies:
(1) For any $x, y \in(-1,1)$, we have
$$
\begin{array}{l}
f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right), \\
\text { (2) } x \in(-1,0), f(x)>0 \text {. Prove: } \\
f\left(\frac{1}{5}\right)+f\left(\frac{1}{11}\right)+f\left(\frac{1}{19}\right)+\cdots+f\left(\frac{1}{n^{... | Prove that in $f(x)+f(y)=f\left(\frac{x+y}{1+x y}\right)$, by setting $x=y=0$, we have $f(0)=0$.
Then, let $y=-x$. Thus, $f(x)+f(-x)=f(0)=0$, which means $f(-x)=-f(x)$.
$\therefore$ The function $f(x)$ is an odd function.
Also, let $-1 < x_1 < x_2 < 1$. Then, $\frac{x_1 - x_2}{1 - x_1 x_2} < 0$.
$\therefore f(x_1) - f(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,361 |
When the expression changes, the minimum value of the fraction $\frac{3 x^{2}+6 x-5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$. (1993, National Junior High School Competition) | Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$
which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,362 |
Example 2. Let $x$ be a positive real number, then the minimum value of the function $y=x^{2}-x+\frac{1}{x}$ is $\qquad$ - (1995, National Junior High School Competition) | - $y=x^{2}-x+\frac{1}{x}$ $=(x-1)^{2}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+1$.
From the above equation, it is known that when $x-1=0$ and $\sqrt{x}-\frac{1}{\sqrt{x}}$ $=0$, i.e., $x=1$, $y$ takes the minimum value of 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,363 |
Example 11. Let $m, n, p$ be positive real numbers, and $m^{2}+n^{2}-p^{2}=0$. Find the minimum value of $\frac{p}{m+n}$. (1988, Sichuan Province Junior High School Competition) | Solve As shown in the figure, construct a square $ABCD$ with side length $m+n$. Points $E, F, G, H$ divide each side of the square into segments of lengths $m$ and $n$. Clearly, $EFGH$ is a square with side length $p$.
From the figure, it is evident that $EG = \sqrt{2} p$, and $EG \geqslant AD$ (equality holds when $D... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,364 |
Example 12. $a, b, c$ are all integers, and $a b c=1990$. Find the minimum value of $a b+b c+c a$. (1990, Jinyun Cup Junior Competition) | $$
\begin{array}{l}
\text { Sol } \because 1990=1 \times 1 \times 1990 \\
=(-1) \times(-1) \times 1990 \\
= 1 \times(-1) \times(-1990) \\
= 1 \times 2 \times 995 \\
=(-1) \times(-2) \times 995 \\
= \cdots=2 \times 5 \times 199=1 \times 10 \times 199,
\end{array}
$$
When $a=-1, b=-1, c=1990$ (letters can be swapped), ... | -3979 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,365 |
Example 13. Let $a, b, c$ be distinct integers from 1 to 9. What is the largest possible value of $\frac{a+b+c}{a b c}$? (1992, 1st Dannevirke-Shanghai Friendship Correspondence Competition) | Let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a, b$ remain unchanged temporarily, and only let $c$ vary, where $c$ can take any integer from 1 to 9. Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that when $c=1$, $P$ reaches its maximum value. Therefore, $c=1$.
Thus, $P=\frac{a+b+1}{a ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,366 |
Example 1. Let $x, y, z$ be positive numbers. Prove that:
$$
\frac{z^{2}-x^{2}}{x+y}+\frac{x^{2}-y^{2}}{y+z}+\frac{y^{2}-z^{2}}{z+x} \geqslant 0 .
$$
(W. Janous Conjecture) | $$
\begin{array}{l}
\frac{z^{2}}{x+y}+\frac{y^{2}}{x+z}+\frac{x^{2}}{y+z} \\
\geqslant \frac{x^{2}}{x+y}+\frac{y^{2}}{y+z}+\frac{z^{2}}{z+x} .
\end{array}
$$
Assume without loss of generality that $x \leqslant y \leqslant z$, then
$$
\begin{array}{l}
x^{2} \leqslant y^{2} \leqslant z^{2}, \quad x+y \leqslant x+z \leqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,367 |
Example 4. Let $a, b, c$ be positive numbers.
Prove: $\frac{a^{3}}{b c}+\frac{b^{3}}{a c}+\frac{c^{3}}{a b} \geqslant a+b+c$.
(*) | Assume $a \leqslant b \leqslant c$. Then $\frac{1}{b c} \leqslant \frac{1}{a c} \leqslant \frac{1}{a b}$. The left side of (*) is the sum in order.
$$
\begin{array}{l}
\text { Left side }=\frac{a}{b c} \cdot a^{2}+\frac{b}{a c} \cdot b^{2}+\frac{c}{a b} \cdot c^{2} \\
\geqslant \frac{a}{b c} \cdot b^{2}+\frac{b}{a c} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,368 |
Example 5. Let $a, b, c$ be positive numbers. Prove:
$$
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{1}{2}(a+b+c) \text {. (*) }
$$
(2nd Friendship Cup International Competition Problem) | Proof: Let $a \leqslant b \leqslant c$. Then $a^{2} \leqslant b^{2} \leqslant c^{2}, \frac{1}{b+c} \leqslant$ $\frac{1}{c+a} \leqslant \frac{1}{a+b}$. It follows that the left side of (*) is an ordered sum, denoted as $S$. By the rearrangement inequality, we have
$$
\begin{array}{l}
S \geqslant \frac{b^{2}}{b+c}+\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,369 |
Example 6. (1) If $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant$ $b_{n}$. Then we have
$$
\begin{array}{l}
\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right) \\
\geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(b_{1}+b_{2}+\cdots\right. \\
\... | Prove (1) The left side is the ordered sum, denoted as $S$, then
$$
\begin{array}{l}
S=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}, \\
S \geqslant a_{1} b_{2}+a_{2} b_{3}+\cdots+a_{n} b_{1}, \\
S \geqslant a_{1} b_{3}+a_{2} b_{4}+\cdots+a_{n} b_{2}, \\
\cdots \cdots \\
S \geqslant a_{1} b_{n}+a_{2} b_{1}+\cdots+a_{n} b_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,370 |
Example 7. Let $a, b, c$ be positive real numbers, and $abc=1$. Prove that:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }
$$
(Problem from the 36th IMO) | Proof Let the left side be $S$, then
$$
\begin{aligned}
S & =\frac{(a b c)^{2}}{a^{3}(b-1 c)}+\frac{(a b c c)^{2}}{b^{3}(c+c)}+\frac{(a b c)^{2}}{c^{3}(a+b)} \\
& =\frac{b c}{2 b+c)} \cdot b c+\frac{a c}{b(c+a)} \cdot a c \\
& +\frac{a b}{c(a+b)} \cdot a b .
\end{aligned}
$$
Let $a \leqslant b \leqslant c$, then $a b ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,371 |
Example 8. $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers. Prove:
$$
\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \text {. }
$$
(Arithmetic Mean Theorem) | To prove that if the proposition holds for $n-1$ positive numbers, it also holds for $n$ positive numbers, we need to show that:
$$
\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}}
$$
or equivalently,
$$
\frac{a_{1}+a_{2}+\cdots+a_{n}}{\sqrt[n]{a_{1} a_{2} \cdots a_{n}}} \geqslant n.
$$
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,372 |
Example 9. $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$
\begin{array}{l}
\frac{x_{1}}{\sqrt{1-x_{1}}}+\frac{x_{2}}{\sqrt{1-x_{2}}}+\cdots+\frac{x_{n}}{\sqrt{1-x_{n}}} \\
\geqslant \frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{n-1}} .
\end{ar... | Proof: Let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$. It is easy to see that the left side of (1) is an ordered sum, denoted as $S$. Then
$$
\begin{array}{l}
S \geqslant \frac{x_{2}}{\sqrt{1-x_{1}}}+\frac{x_{3}}{\sqrt{1-x_{2}}}+\cdots+\frac{x_{1}}{\sqrt{1-x_{n}}}, \\
S \geqslant \frac{x_{3}}{\sqrt{1-x_{1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,373 |
Example 3. The maximum value of the function $y=-x-\frac{9}{x}+18(x>0)$ is ( ).
(A) 24
(B) 18
(C) 12
(D) 2
(1991～1992, Guangzhou, Fuzhou, Luoyang, Wuhan, Chongqing Junior High School League) | According to the basic inequality $a+b \geqslant 2 \sqrt{a b}(a, b >0)$, we know
$$
x+\frac{9}{x} \geqslant 2 \sqrt{x \cdot \frac{9}{x}}=2 \sqrt{9}=6,
$$
the equality holds when $x=\frac{9}{x}$, i.e., $x=3$.
$$
\begin{aligned}
\therefore y & =-x-\frac{9}{x}+18 \\
& =-\left(x+\frac{9}{x}\right)+18 \leqslant-6+18=12 .
\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,374 |
Example 10. In $\triangle A B C$, prove that:
$$
\frac{\pi}{3} p \leqslant a A+b B+c C<\frac{\pi}{2} p
$$ | Prove that if $M = aA + bB + cC$ (where $M$ is a cyclic sum), then
$$
\begin{array}{l}
M \geqslant aB + bC + cA, \\
M \geqslant aC + bA + cB.
\end{array}
$$
Adding the three inequalities, we get
$$
\begin{aligned}
3M \geqslant & a(A + B + C) + b(A + B + C) \\
& + c(A + B + C) \\
= & \pi(a + b + c), \\
\therefore M \ge... | \frac{\pi}{3} p \leqslant M < \frac{\pi}{2} p | Inequalities | proof | Yes | Yes | cn_contest | false | 708,375 |
Example 11. In $\triangle A B C$, prove that:
$$
\frac{\sin A}{h}+\frac{\sin B}{h}+\frac{\sin C}{h} \geqslant \frac{\sqrt{3}}{R} \text {. }
$$ | Prove that the left side of the above equation is the ordered sum, denoted as $M$. By Chebyshev's inequality, we have
$$
M \geqslant \frac{1}{3}\left(\frac{1}{h_{c}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)(\sin A+\sin B+\sin C) \text {. }
$$
By the inequality $\frac{1}{h_{d}}+\frac{1}{h_{b}}+\frac{1}{h_{c}} \geqslant \... | \frac{\sqrt{3}}{R} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,376 |
Example 1. If $x, y, z$ satisfy $z+4 y+z=1$ and are non-negative real numbers, prove: $0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27}$.
(25th IMO Problem) | Prove that for $f\left(\frac{1}{2}\right)=\left(\frac{1}{2}-x\right)\left(\frac{1}{2}-y\right)\left(\frac{1}{2}-z\right)$, $u=x y+y z+z x-2 x y z$, then
$$
u=2 f\left(\frac{1}{2}\right)+\frac{1}{4}=2\left[f\left(\frac{1}{2}\right)+\frac{1}{8}\right] .
$$
When $x, y, z$ are all no more than $\frac{1}{2}$,
$$
\begin{arr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,378 |
3. Given $\triangle A B C$, let $I$ be its incenter, and the internal angle bisectors of $\angle A, \angle B$, and $\angle C$ intersect the opposite sides at $A^{\prime}, B^{\prime}, C^{\prime}$, respectively. Prove that:
$$
\frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} ... | Let $B C=a, C A=b, A B=c$. Let $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$, then $\frac{A I}{A A^{\prime}}=1-x, \frac{B I}{B B^{\prime}}=1-y, \frac{C I}{C C^{\prime}}=1-z$. $\therefore \frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}}=x y+y z+z x-x y z .$.
From $a, b, c$ b... | \frac{1}{4}<\frac{A I \cdot B I \cdot C I}{A A^{\prime} \cdot B B^{\prime} \cdot C C^{\prime}} \leqslant \frac{8}{27} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,380 |
Example 4. Positive numbers $a, b, c, A, B, C$ satisfy the conditions $a+A=b+B$ $=c+C=k$. Prove: $a B+b C+c A<k^{2}$.
(21st All-Soviet Union Mathematical Competition) | Let $f(k)=(k-a)(k-b)(k-c)$.
Since $k-a, k-b, k-c>0$, we have $f(k)>0$.
$$
\begin{array}{l}
\text { Also, } k^{3}-(a+b+c) k^{2}+(a b+b c+c a) k-a b c>0, \\
(a+b+c) k-(a b+b c+c a)<k^{2}-\frac{a b c}{k}<k^{2},
\end{array}
$$
Therefore, $(a k-a b)+(b k-b c)+(c k-a c)<k^{2}$,
which means $a B+b C+c A<k^{2}$. | a B+b C+c A<k^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,381 |
Example 1. Place four equal small balls on a plane so that the line connecting their centers forms a square, with each side having two balls touching, and place one more equal ball on top of these four balls so that it touches all of them. Given the highest point of the top ball is a. Find the radius $x$ of the small b... | Analyzing with the five ball centers as vertices, forming a regular tetrahedron $A-B C D E$ with edge lengths of $2 x$ (this figure), the height $A O$ $=\sqrt{2} x$. Extending upwards and downwards by the radius $x$ of the small balls, the distance from the highest point $P$ of the upper small ball to the ground is the... | x=\left(1-\frac{\sqrt{2}}{2}\right) a | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,382 |
Example 2. In a cylindrical barrel with a radius of $2 R$, there are 6 small balls with diameters all equal to $2 R$, and they are in a fixed state. Water is poured into the barrel. What is the minimum height of the water level to completely submerge these 6 balls? What is the volume of water in the barrel at this time... | (1) The diameter of the cylinder's base is $4 R$, and the diameter of the sphere is $2 R$. To ensure the small balls are in a stable state, the 6 small balls are arranged in three layers. The line connecting the centers of the balls in the first layer, $A B$, is parallel to the base, and the line connecting the centers... | 8 \sqrt{2} \pi R^{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,383 |
Example 4. Let $x$ be a real number, $y=|x-1|+|x-1|$. Among the following four conclusions:
(1) $y$ has no minimum value;
(2) there is only one $x$ that makes $y$ reach its minimum value;
(3) there are a finite number of $x$ (more than one) that make $y$ reach its minimum value;
(4) there are infinitely many $x$ that m... | Solution 1
$$
y=|x-1|+|x+1|=\left\{\begin{array}{l}
-2 x, x \leqslant-1, \\
2,-1 \leqslant x \leqslant 1, \\
2 x, x \geqslant 1 .
\end{array}\right.
$$
Its graph is shown in the figure.
From the graph, it can be seen that when $x$ takes any value in the range $-1 \leqslant x \leqslant 1$, $y$ achieves its minimum valu... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,385 |
Example 4. Let $A_{1} A_{2} A_{3} A_{4}$ be a tetrahedron, and $S_{1}, S_{2}, S_{3}, S_{4}$ be spheres with centers at $A_{1}, A_{2}, A_{3}, A_{4}$, respectively, such that they are pairwise tangent. If there exists a point $O$ such that a sphere with radius $r$ centered at this point is tangent to $S_{1}, S_{2}, S_{3}... | Prove that $S_{1}, S_{2}, S_{3}$, $S_{4}$ can only be pairwise externally tangent. Let the sphere with center $O$ and radius $R$ be tangent to the edges $A_{2} A_{3}, A_{3} A_{6}, A_{4} A_{2}$ at points $P_{1}, P_{2}, P_{3}$ (as shown in the figure).
Then, in the right triangles $\triangle O P_{1} A_{3} \cong \mathrm{R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,386 |
Example 5. In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, construct an inscribed sphere $O$, and then construct a small sphere at each of the eight corners of the cube, such that they are all externally tangent to sphere $O$, and each is tangent to three faces of the cube. Find the radius of the small spheres.
Tr... | (1) By symmetry, it is known that the eight small balls are equal. The diagonal plane $A C C_{1} A_{1}$ of the cube passes through 5 ball centers and 10 tangent points, as well as the edges and diagonals of the cube, containing its main elements. By dissecting this diagonal plane (as shown in the figure), focusing on t... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,387 |
Example 6. The base edge length of a regular triangular pyramid $P-ABC$ is 1, and the height $PH=2$. On the inscribed sphere of this pyramid, stack a sphere that is externally tangent to it and also tangent to each side face of the pyramid. Continue stacking spheres in this manner. Find the sum of the volumes of these ... | (1) The section passing through the side vertex $P$ and the height $P H$ goes through the center of the sphere and the tangent point, including the main elements of the regular triangular prism. Dissect it (as shown in the figure), and analyze it point by point, reducing it to a plane geometry problem to solve.
(2) Le... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,388 |
7. There are 120 equally distributed spheres inside a regular tetrahedron $A-B C D$. How many spheres are placed at the bottom of this regular tetrahedron? | Analysis (1) The base of the regular tetrahedron $A-BCD$ is the equilateral $\triangle BCD$. Assuming the number of spheres closest to the edge $BC$ is $n$, then the number of rows of spheres tangent to the base $\triangle BCD$ is also $n$, with the number of spheres in each row being $n, n-1, \cdots, 3,2,1$. Thus, the... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,389 |
The necessary and sufficient condition for two positive integers $6s \pm 1$ to both be prime is that $s \neq p k \pm \frac{p+1}{6}(p=-1(\bmod 6))$, and $s \neq p k \pm \frac{p-1}{6}$ $(p \equiv 1(\bmod 6))$, where $p>3$ is a prime number, and $s, k \in N$. | Proof: Let $p$ be a prime of the form $6n+1$. Then, when $s=pk+\frac{p-1}{6}$, $6s+1=(6k+1)p$ is a composite number.
Conversely, if $6s+1$ contains a prime factor $p=6n+1$, then the other factor is $6k+1$. Thus,
$$
\begin{array}{l}
6s+1=(6n+1)(6k+1) \\
=p(6k-1)=6\left(pk+\frac{p-1}{6}\right)+1 .
\end{array}
$$
Then $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,392 |
Theorem In the extension of side $BC$ and on sides $CA$, $AB$ of $\triangle ABC$, take points $D, E, F$, such that $\frac{AF}{FB}=\lambda_{1}, \frac{BD}{DC}=\lambda_{2}, \frac{CE}{EA}=\lambda_{3}$, then $S_{\triangle DEF}=\left|\frac{\lambda_{1} \lambda_{2} \lambda_{3}-1}{\left(\lambda_{1}+1\right)\left(\lambda_{2}-1\r... | $$
\begin{array}{c}
\text { Prove } \because \frac{C D}{B C} \\
=\frac{1}{\lambda_{2}-1}, \frac{C E}{C A}=\frac{\lambda_{3}}{\lambda_{3}+1}, \\
\therefore \frac{S_{\triangle C D E}}{S_{\triangle A B C}}=\frac{1}{\lambda_{2}-1} \\
\cdot \frac{\lambda_{3}}{\lambda_{3}+1} .
\end{array}
$$
Similarly, we have
$$
\begin{ali... | S_{\triangle DEF}=\left|\frac{\lambda_{1} \lambda_{2} \lambda_{3}-1}{\left(\lambda_{1}+1\right)\left(\lambda_{2}-1\right)\left(\lambda_{3}+1\right)}\right| S_{\triangle ABC} | Geometry | proof | Yes | Yes | cn_contest | false | 708,393 |
i. Real numbers $a, b$ satisfy $a b=1$, let $M=\frac{1}{1+a}+\frac{1}{1+b}$. $N=\frac{a}{1+a}+\frac{b}{1+b}$. Then the relationship between $M, N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain | \[
\begin{array}{l}
-1 . \text { (B). } \\
M=\frac{1}{1+a}+\frac{1}{1+b}=\frac{b}{b+a b}+\frac{a}{a+a b}=\frac{b}{b+1}+\frac{a}{a+1} \\
=N \text {. }
\end{array}
\] | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,394 |
Example 5. If $x \neq 0$, then the maximum value of $\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}$ is $\qquad$ . (1992, National Junior High School Competition) | Solution 1 $\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}$
$$
\begin{array}{l}
=\frac{x}{\sqrt{1+x^{2}+x^{4}}+\sqrt{1+x^{4}}} \\
=\frac{x}{|x|\left(\sqrt{x^{2}+\frac{1}{x^{2}}+1}+\sqrt{x^{2}+\frac{1}{x^{2}}}\right)} \\
=\frac{x}{|x|\left(\sqrt{\left.x-\frac{1}{x}\right)^{2}+3}+\sqrt{\left.\left(x-\frac{1}{x}\right)^{2}... | \sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,396 |
3. If $A$ is a point outside circle $\odot O$ with radius 1, $O A=2$, $A B$ is a tangent to $\odot O$ at point $B$, chord $B C \parallel O A$, and $A C$ is connected, then the area of the shaded part is
(A) $\frac{2 \pi}{9}$
(B) $\frac{\pi}{6}$
(C) $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
(D) $\frac{\pi}{4}-\frac{\sqrt{3}}{8... | 3. (B).
Connect $O B$,
$$
\begin{array}{l}
\because B C / / O A, \\
\therefore S_{\triangle O B C}=S_{\triangle A B C} .
\end{array}
$$
$\because A B$ is the tangent of the circle,
$$
\therefore O B \perp A B \text {. }
$$
In $\mathrm{Rt} \triangle A O B$, $A O=2, O B=1$,
$$
\begin{array}{l}
\therefore \angle A O B=6... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,397 |
4. Let $x_{1}, x_{2}$ be the two roots of the quadratic equation $x^{2}+x-3=0$. Then, the value of $x_{1}^{3}-4 x_{2}^{2}+19$ is ().
(A) -4
(B) 8
(C) 6
(D) 0 | 4. (D).
$\because x_{1}+x_{2}$ are the roots of the quadratic equation $x^{2}+x-3=0$,
$$
\therefore x_{1}^{2}+x_{1}-3=0, x_{2}^{2}+x_{2}-3=0 \text {, }
$$
i.e., $x_{1}^{2}=3-x_{1}, x_{2}^{2}=3-x_{2}$.
By the relationship between roots and coefficients, we know $x_{1}+x_{2}=-1$, thus we have
$$
\begin{array}{l}
x_{1}^{... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,398 |
5. If a line bisects both the area and the perimeter of a triangle, then the line must pass through the ( ) of the triangle.
(A) incenter
(B) circumcenter
(C) centroid
(D) orthocenter | 5. (A)
Let the line that bisects the area and perimeter of $\triangle A B C$ intersect $A B$ and $A C$ at $D, E$, and let $I$ be the incenter of $\triangle A B C$, and $r$ be the radius of its incircle. Then we have
$$
S_{\text {IEAD }}=\frac{1}{2} r(A E+
$$
$A D)$,
$$
S_{B C E I D}=\frac{1}{2} r(B D+B C+C E) \text {.... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,399 |
6. If 20 points divide a circle into 20 equal parts, then the number of regular polygons that can be formed with vertices only among these 20 points is ( ) .
(A) 4
(B) 8
(C) 12
(D) 24 | 6. (C).
Let the regular $k$-sided polygon satisfy the condition, then the 20 $-k$ points other than the $k$ vertices are evenly distributed on the minor arcs opposite to the sides of the regular $k$-sided polygon.
Thus, $\frac{20-k}{k}=\frac{20}{k}-1$ is an integer, so $k \mid 20$.
But $k \geqslant 3$,
$\therefore k=4... | 12 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,400 |
1. Given real numbers $x_{0}, y_{0}$ are solutions to the system of equations $\left\{\begin{array}{l}y=\frac{1}{x}, \\ y=|x|+1\end{array}\right.$. Then $x_{0}+y_{0}=$ $ـ$. $\qquad$ | $$
=\ldots 1 . \sqrt{5} .
$$
From the given, $\frac{1}{x}=|x|+1$, since $x \neq 0$, we get (1) $\left\{\begin{array}{l}x<0, \\ x^{2}+x+1=0 .\end{array}\right.$ Equation (I) has no solution.
Solving (I) yields $x_{0}=\frac{\sqrt{5}-1}{2}$, thus we get $y_{0}=\frac{\sqrt{5}+1}{2}$.
$$
\therefore x_{0}+y_{0}=\sqrt{5} \te... | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,401 |
. Let $1995 z^{3}=1996 y^{3}=$
$1997 z^{3} \cdot x y z>0$, and
$\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=\sqrt[3]{1995}+\sqrt[3]{1996}+$
$\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$ $\qquad$ . | 3.1.
Let $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, obviously, $k \neq 0$, then $1995=\frac{k}{x^{3}}, 1996=\frac{k}{y^{3}}, 1997=\frac{k}{z^{3}}$.
From the given information,
$$
\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}}>0,
$$
which means
$$... | 3.1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,403 |
4. As shown in the figure, the square $A B C D$ with side length 1 is rotated counterclockwise around point $A$ by $60^{\circ}$ to the position $A B^{\prime} C^{\prime} D^{\prime}$. Then the area of the overlapping part of the two squares is $\qquad$ | $4.2-\sqrt{3}$.
Draw $M N / / A D$ through $B^{\prime}$, intersecting $A B, C D$ at points $M, N$ respectively, and let $B^{\prime} C^{\prime}$ intersect $C D$ at $K$. Then
$$
\begin{aligned}
B^{\prime} M & =A B^{\prime} \cdot \sin 60^{\circ} \\
& =\frac{\sqrt{3}}{2}, \\
\therefore B^{\prime} N & =1-\frac{\sqrt{3}}{2}... | 2-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,404 |
One, (20 points) In a school's donation activity for the "Hope Project", the total donation amount from $m$ boys and 11 girls in Class A is equal to the total donation amount from 9 boys and $n$ girls in Class B, which is $(m \cdot n + 9 m + 11 n + 145)$ yuan. It is known that each person's donation amount is the same ... | $$
-\because m n+9 m+11 n+145=(m+11)(n+9)+
$$
46, by the given
$$
\begin{array}{l}
m+11 \mid(m n+9 m+11 n+145), \\
n+9 \mid(m n+9 m+11 n+145)
\end{array}
$$
and
$$
\begin{array}{l}
m+11=n+9, \\
\therefore m+11|46, n+9| 46 .
\end{array}
$$
$\because m, n$ are non-negative integers,
$$
\begin{array}{l}
\therefore m+11 \... | 47 \text{ yuan or } 25 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,405 |
II. (25 points) Let the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $M$. Draw a line through point $M$ parallel to $AD$, intersecting $AB$ and $CD$ at points $E$ and $F$ respectively, and intersecting the extension of $BC$ at point $O$. $P$ is a point on the circle with center $O$ and ... | $$
\begin{array}{l}
\because O M / / A K, \\
\therefore \frac{O F}{D K}=\frac{C O}{C K}=\frac{O M}{A K} . \\
\text { So } \frac{O M}{O F}=\frac{A K}{D K} . \\
\because O E / / A K, \\
\therefore \frac{O E}{A K}=\frac{B O}{B K}=\frac{O M}{D K} . \\
\text { So } \frac{O E}{O M}=\frac{A K}{D K} .
\end{array}
$$
From (1) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,406 |
Example 6. When $|x+1| \leqslant 6$, the maximum value of the function $y=x|x|$ $-2 x+1$ is $\qquad$. (1994, National Junior High School Competition) | Solve: From $|x+1| \leqslant 6$ we get $-7 \leqslant x \leqslant 5$.
When $0 \leqslant x \leqslant 5$,
$$
y=x^{2}-2 x+1=(x-1)^{2} \text {. }
$$
Since $x=1 \in[0,5]$, and $1>\frac{-7+0}{2}, \\
\therefore y_{\text {max }}=f(5)=(5-1)^{2}=16; \\
y_{\text {min }}=f(1)=(1-1)^{2}=0 .
$$
When $-7 \leqslant x < 0$,
$$
y=x^{2}... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,407 |
Three, (25 points) Given that $a, b, c$ are positive integers, and the parabola $y=$ $a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
Translate the above text into English, please retain the origin... | Three, let the coordinates of $A, B$ be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$ and $x_{1}<0, x_{2}>0$,
\[
\begin{array}{l}
\therefore x_{1} x_{2}=\frac{c}{a}<0, \\
\therefore a c<0, \text { i.e., } a>0, c<0 .
\end{array}
\]
\[
\begin{array}{l}
\because x_{1}+x_{2}=-\frac{b}{a}>0, \\
\therefore b<0 .
\end{array}
... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,408 |
1. Given $a-\frac{1}{a}=\frac{1}{b}-b=3$, and $a+b \neq 0$. Then, $\frac{a}{b^{3}}-\frac{b}{a^{3}}$ is ( ).
(A) $21 \sqrt{5}$
(B) $21 \sqrt{13}$
(C) $33 \sqrt{5}$
(D) $13 \sqrt{13}$ | ,- 1. (D).
Given $a^{2}-3 a-1=0,(-b)^{2}-3(-b)-1=0$, and $a \neq -b$. Therefore, $a$ and $(-b)$ are the roots of the equation $x^{2}-3 x-1=0$. Hence, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a - b = 3, \\
a(-b) = -1
\end{array} \quad \text { i.e., } \left\{\begin{array}{l}
a - b = 3, \\
a b = 1
\end{array}\r... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,409 |
2. If the quadratic function $y=x^{2}+(k+2) x+k+5$ intersects the $x$-axis at two different points with both coordinates being positive, then the value of $k$ should be ( ).
(A) $k>4$ or $k<-5$
(B) $-5<k<-4$
(C) $k \geqslant-4$ or $k \leqslant-5$
(D) $-5 \leqslant k \leqslant-4$ | 2. (B).
According to the problem, we have
$$
\left\{\begin{array}{l}
k+5>0, \\
(k+2)^{2}-4(k+5)>0 .
\end{array}\right.
$$
Solving this, we get $-5<k<-4$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,410 |
3. As shown in the figure, $\triangle A B C$ is an acute triangle, $B E \perp A C, C F \perp A B$. Then the value of $S_{\triangle A E F}=S_{\triangle A B C}$ is ( ).
(A) $\sin A$
(B) $\cos A$
(C) $\sin ^{2} A$
(D) $\cos ^{2} A$ | 3. (D).
$$
\begin{array}{l}
\text { In Rt } \triangle A B E, \frac{A E}{A B}=\cos A, \\
\text { In Rt } \triangle A C F, \frac{A F}{A C}=\cos A, \\
\therefore \frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\frac{\frac{1}{2} A E \cdot A F \sin A}{\frac{1}{2} A B \cdot A C \sin A}=\cos ^{2} A .
\end{array}
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,411 |
4. The number of positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{1997}$ is
(A) 1
(B) 2
(C) 3
(D) greater than or equal to 4 | 4. (C).
It is easy to know that $x>1997, y>1997$. Let $x=1997+t_{1}$, $y=1997+t_{2}\left(t_{1}, t_{2}\right.$ be natural numbers $)$, then we have
$$
\frac{1}{1997+t_{1}}+\frac{1}{1997+t_{2}}=\frac{1}{1997} .
$$
Simplifying, we get $t_{1} t_{2}=1997^{2}$.
$\because t_{1}, t_{2}$ are natural numbers,
$$
\therefore t_{... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,412 |
5. $P$ is a point inside $\triangle A B C$, connecting $P A, P B, P C$, dividing the triangle's area into three equal parts. Then $P$ is the ( ) of $\triangle A B C$.
(A) Incenter
(B) Circumcenter
(C) Orthocenter
(D) Centroid | 5. (D).
As shown in the figure, extend $A P$ to intersect $B C$ at $D$. By the ratio of the areas of triangles with the same height, which is equal to the ratio of their bases, we have
$$
\begin{array}{l}
\frac{S_{\triangle A P B}}{S_{\triangle A B C}}=\frac{A P}{A D}=\frac{S_{\triangle A P C}}{S_{\triangle A C D}} . ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,413 |
-6. The parabola $y=x^{2}+23 x+1$ intersects the line $y=2 a x+2 a b$ at most one point. Then the maximum value of $a^{2}+b^{2}$ is ).
(A) 1
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{2}}{2}$
(D) 0 | 6. (A).
From $\left\{\begin{array}{l}y=x^{2}+2 b x+1, \\ y=2 a x-2 a b\end{array}\right.$ we get
$$
x^{2}+2(b-a) x+1-2 a b=0 .
$$
Since the parabola and the line have at most one intersection point,
$$
\therefore \Delta=4(b-a)^{2}-4(1-2 a b) \leqslant 0 \text {, }
$$
which means $a^{2}+b^{2} \leqslant 1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,414 |
1. Given the product of four real numbers is 1, and the sum of any one of the numbers with the product of the other three is 1000. Then the sum of these four numbers is | $=、 1.2000$.
These four real numbers are all non-zero and not all equal, denoted as $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$. Taking any one of these numbers $x_{i} (i=1,2,3,4)$, the other three numbers are $x_{j}$, $x_{k}$, $x_{r}$, then we have
$$
\left\{\begin{array}{l}
x_{i}+x_{j} x_{k} x_{r}=1000, \\
x_{i} x_{j} x_{k} ... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,415 |
2. If $x y=a, x z=b, y z=c$, and none of them are equal to 0, then, $x^{2}+y^{2}+z^{2}=$ | 2. $\frac{(a b)^{2}+(a c)^{2}+(b c)^{2}}{a b c}$.
From the known three equations, multiplying both sides, we get $x y z= \pm \sqrt{a b c}$.
Dividing (*) by the known three equations respectively, we get
$$
\begin{array}{l}
x= \pm \sqrt{\frac{b c}{a}}, y= \pm \sqrt{\frac{a c}{b}}, z= \pm \sqrt{\frac{a b}{c}} . \\
\ther... | \frac{(a b)^{2}+(a c)^{2}+(b c)^{2}}{a b c} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,416 |
Example 7. For the quadratic function $y=-x^{2}+6 x+7$, when $x$ takes values in the range $t \leqslant x \leqslant t+2$, the maximum value of this function is $y^{\prime}=-(t-3)^{2}+2$. Then the range of values for $t$ is ).
(A) $t=0$
(B) $0 \leqslant t \leqslant 3$
(C) $t \geqslant 3$
(D) None of the above | $$
\text{Solve} \begin{aligned}
\because y & =-x^{2}+6 x+7 \\
& =-(x-3)^{2}+2, \\
\therefore x=-\frac{b}{2 a} & =3, \text{ and } a=-1<0 .
\end{aligned}
$$
When $3 \in[t, t+2]$, i.e., $1 \leqslant t \leqslant 3$,
$$
y=f(3)=2 \text{ and } y \text{ here }=-(t-3)^{2}+
$$
2 is contradictory.
When $3 \geqslant t+2$, i.e., $... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,418 |
One, (20 points) As shown in the figure, given that $AB, CD$ are perpendicular chords in a circle $\odot O$ with radius 5, intersecting at point $P$. $E$ is the midpoint of $AB$, $PD=AB$, and $OE=3$. Try to find the value of $CP + CE$.
---
The translation is provided as requested, maintaining the original text's form... | By the intersecting chords theorem, we have
$$
C P \cdot P D=A P \cdot P B \text {. }
$$
Also, $A E=E B=\frac{1}{2} A B$,
$$
\begin{aligned}
\because C P \cdot P D & =\left(\frac{1}{2} A B+E P\right)\left(\frac{1}{2} A B-E P\right) \\
& =\frac{1}{4} A B^{2}-E P^{2} .
\end{aligned}
$$
And $P D=A B$,
$$
\therefore \qua... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,420 |
II. (25 points) Given that $a, b$ are integers. If the quadratic equation
$$
x^{2}+a x^{a-b}+(2 a-b-1) x+a^{2}+a-b-4=0
$$
has roots that are all integers. Find the values of $a, b$. | Given is a quadratic equation in $x$, so $a-b$ can have three cases:
$a-b=0, a-b=1, a-b=2$.
(1) When $c-b=0$, i.e., $a=b$, the original equation becomes
$$
x^{2}+(a-1) x+a^{2}+a-4=0 \text {. }
$$
When the discriminant is a perfect square, the equation has integer roots.
$$
\begin{aligned}
\Delta & =(a-1)^{2}-4\left(a^... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,421 |
Three. (25 points) Prove: In any 11 integers, there must be 6 integers whose sum is divisible by 6, but this property does not necessarily hold for any 10 integers. | Three, let $x_{1}, x_{2}, \cdots, x_{11}$ be 11 arbitrary integers. Any integer divided by 2 leaves a remainder of 0 or 1. By the pigeonhole principle, among the given 11 integers, any three chosen will have at least two with the same remainder when divided by 2. Therefore, the sum of these two integers can definitely ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,422 |
1. The function defined on $(-\infty,-2) \cup(2,+\infty)$
$$
f(x)=\frac{x+2+\sqrt{x^{2}-4}}{x+2-\sqrt{x^{2}-4}}+\frac{x+2-\sqrt{x^{2}-4}}{x+2+\sqrt{x^{2}-4}}
$$
( $|x|>2)$ has the following property regarding its odd and even nature ( ).
(A) It is an odd function but not an even function
(B) It is an even function but ... | $-、 1 .(A)$.
For $|x|>2$, we have
$$
\begin{array}{r}
4=x^{2}-\left(x^{2}-4\right)=\left(x+\sqrt{x^{2}-4}\right)\left(x-\sqrt{x^{2}-4}\right) \\
\frac{1}{1} \frac{x+\sqrt{x^{2}-4}}{2}=\frac{2}{x-\sqrt{x^{2}}-4}=\frac{x+2+\sqrt{x^{2}-4}}{x+2-\sqrt{x^{2}-4}} .
\end{array}
$$
Thus, $f(x)=\frac{x+\sqrt{x^{2}-4}}{2}+\frac{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,423 |
3. In the tetrahedron $A-B C D$, $A B \perp$ plane $B C D$. Then the relationship between $\angle C A D$ and $\angle C B D$ is ( ).
(A) $\angle C A D>\angle C B D$
(B) $\angle C A D=\angle C B D$
(C) $\angle C A D<\angle C B D$
(D) Uncertain | 3. (D).
A popular misconception is to choose (C), in fact, the relationship between these two angles is uncertain. Consider Example 1, where the right triangle $\triangle ACD$ with side lengths $1, \sqrt{3}, 2$ is folded along the altitude $AB$ from the hypotenuse, forming the relevant edges of the three-story structu... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,425 |
4. Let the sequence of positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ be the irreducible proper fractions with a denominator of 60. Then $\sum_{i=1}^{\pi} \cos \frac{a_{i} \pi}{2}=(\quad)$.
(A) 0
(B) 8
(C) 16
(D) 30 | 4. (A).
Given $60=2^{2} \times 3 \times 5$, the numerators of $a$ in order are
$$
1,7,11,13,17,19,23,29,31,37,41,43,47,49 \text {, }
$$
53,59, a total of 16 (which can be calculated using the principle of inclusion and exclusion), and $a_{i}+a_{17-i}=1$ (the sum of terms equidistant from the start and end is 1), we ha... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,426 |
5. The number of ways to choose 3 vertices from the 8 vertices of a cube such that at least two vertices are on the same edge is ( ).
(A) 44
(B) 48
(C) 50
(D) 52 | 5. (B).
From 8 points, taking 3 has $C_{8}^{3}=56$ ways. The number of ways to take any two points not on the same edge is what we are looking for.
Since any two points of the three points are not on the same edge, the lines connecting each pair are face diagonals, and thus each way of taking corresponds to a regular... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,427 |
Example 8. If the polynomial $P=2a^{2}-8ab+17b^{2}$ $-16a-4b+1990$, what is the minimum value of $P$? (1990, Junior High School Correspondence Contest in Some Provinces and Cities) | $$
\begin{array}{r}
\text { Sol } P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b \\
+1990=2(a-2 b-4)^{2}+9(b-2)^{2}+1922 .
\end{array}
$$
From this, for all real numbers $a, b, P \geqslant 1922$. The equality holds only when $a-2 b-4=0, b-2=0$, i.e., $a=$ $8, b=2$. Therefore, the minimum value of $P$ is $1922$. | 1922 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,429 |
1. Given $f(x)=\operatorname{tg}\left(\frac{\pi}{4}-\operatorname{arctg} x\right)$. Then $f\left(\frac{1-\sqrt[3]{3}-\sqrt{2}}{1+\sqrt[3]{3}+\sqrt{2}}\right)=$ $\qquad$ 2. | $$
=\therefore \sqrt[3]{3}+\sqrt{2} \text {. }
$$
Given the function simplified as $f(x)=\frac{1-x}{1+x}$. By the properties of ratios, we get $\quad \frac{1-f(x)}{r f(x)} x$, that is $\quad f[f(x)]=\ldots$ Thus, $f\left(\frac{1-\sqrt[3]{3} \cdot \sqrt{2}}{1+\sqrt[3]{3}+\sqrt{2}}\right.$ $\Rightarrow f[f(\sqrt[3]{3}+\... | \sqrt[3]{3}+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,430 |
2. The Dao sequence $\left\{\begin{array}{l}x_{1}=x_{2}=1, \\ x_{n+2}=a x_{n+1}+b x_{n}(n \in N) .\end{array}\right.$
If $T=1996$ is the smallest natural number such that $x_{T+1}=x_{T+2}=1$, then $\sum_{i=1}^{1006} x_{i}=$ $\qquad$ . | 2. 0 .
From $T=1996$ being the smallest natural number that makes $x_{1+1}=x_{T+2}=1$, we know that $a+b \neq 1$. Otherwise, if $a+b=1$, then $x_{3}=a x_{2}+b x_{1}=a+b=1$. This would mean that when $T=1$, $x_{T+1}=x_{T+2}=1$, which contradicts the fact that $T=1996$ is the smallest value. Furthermore, we have
$$
\beg... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,431 |
3. In the plane $\alpha$ there is a $\triangle A B C, \angle A B C=105^{\circ}$, $A C=2(\sqrt{6}+\sqrt{2})$. On both sides of the plane $\alpha$, there are points $S, T$, satisfying $S A=S B=S C=\sqrt{41}, T A=T B=T C=$ 5. Then $S T=$ $\qquad$. | 3. 8 .
From Figure 4, since $S A=S B=S C$, we know that the projection $D$ of $S$ on plane $\alpha$ is the circumcenter of $\triangle A B C$. Similarly, the projection of $T$ on plane $\alpha$ is also the circumcenter of $\triangle A B C$. By the uniqueness of the circumcenter, we have $S T \perp a$.
Connect $A D$. By... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,432 |
4. Given functions $f(x)$ and $g(x)$ are defined on $R$, and $f(x-y)=f(x) g(y)-g(x) f(y)$. If $f(1)=f(2) \neq$ 0, then $g(1)+g(-1)=$ $\qquad$ . | 4. 1 .
For $x \in R$, let $x=u-v$, then we have
$$
\begin{array}{l}
f(-x)=f(v-u) \\
=f(v) g(u)-g(v) f(u) \\
=-[f(u) g(v)-g(u) f(v)] \\
=-f(u-v)=-f(x) .
\end{array}
$$
Therefore, $f(x)$ is an odd function, and we have
$$
\begin{array}{l}
f(2)=f[1-(-1)] \\
=f(1) g(-1)-g(1) f(-1) \\
=f(1) g(-1)+g(1) f(1) .
\end{array}
$... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,433 |
5. Take any three points $A, B, C$ on the hyperbola $x^{2}-y^{2}=2$, then the locus equation of the orthocenter $H$ of $\triangle A B C$ is $\qquad$. | 5. $x^{2}-y^{2}=2$.
Let the coordinates of the three vertices of $\triangle ABC$ be $A\left(\frac{t_{1}^{2}+1}{\sqrt{2} t_{1}}, \frac{t_{1}^{2}-1}{\sqrt{2} t_{1}}\right)$,
$$
B\left(\frac{t_{2}^{2}+1}{\sqrt{2} t_{2}}, \frac{t_{2}^{2}-1}{\sqrt{2} t_{2}}\right), C\left(\frac{t_{3}^{2}+1}{\sqrt{2} t_{3}}, \frac{t_{3}^{2... | x^{2}-y^{2}=2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,434 |
6. For the complex number $x$, the minimum value of the expression $u=|x|+|x-i|+|x-\sqrt{3}|$ is $\qquad$ . | 6. $\sqrt{7}$.
Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$, then
$$
\omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i, 1+\omega+\omega^{2}=0 .
$$
We get
$$
\begin{aligned}
u & =|x|+|x-i||\omega|+|x-\sqrt{3}|\left|\omega^{2}\right| \\
& =|x|+|(x-i) \omega|+\left|(x-\sqrt{3}) \omega^{2}\right| \\
& \geqslant\left|x+(x-i)... | \sqrt{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,435 |
One. (25 points) On side $AB$ of $\triangle ABC$, take any point $D$. Draw $DE \parallel AC$ intersecting $BC$ at $E$, and connect $CD$. Prove that the area of $\triangle CDE$ does not exceed $\frac{1}{4}$ of the area of the original triangle. | $$
\begin{array}{l}
\frac{D F}{A H}=\frac{D D}{B A}=\frac{B E}{B C} . \\
\text { Also } \frac{S_{\triangle C D E}}{S_{\triangle A B C}}=\frac{C E}{B C} \cdot D F \\
=\frac{C E \cdot B E}{B C^{2}} \\
=\frac{C E(B C-C E)}{B C^{3}} \\
\leqslant \frac{1}{B C^{2}}\left[\frac{C E+(B C-C E)}{2}\right]^{2}=\frac{1}{4} \text {,... | S_{\triangle C D E} \leqslant \frac{1}{4} S_{\triangle A B C} | Geometry | proof | Yes | Yes | cn_contest | false | 708,436 |
II. (25 points) Prove that for any given positive number $a$, there must exist a natural number $N$, such that for every natural number $n$ greater than $N$, there is a unique natural number $f(n)$, satisfying $\frac{10^{n}}{f(n)+1}<a \leqslant \frac{10^{n}}{f(n)}$. | For any positive number $a$, there must exist a natural number $N$ such that $10^{x} > a$.
For $n > N$, we have $10^{n} > 10^{N} > a$. Or $\frac{10^{N}}{a} > 1$.
For every $n$ greater than $N$, from the above inequality, there exists a unique natural number $f(n)$ equal to the integer part of $\frac{10^{n}}{a}$
$$
f(n)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,437 |
Three. (35 points) For the set of points on the coordinate plane
$$
S=\{(x, y) \mid 1 \leqslant x<y \leqslant 6, x \in N, y \in N\},
$$
Prove: When any 11 points are taken from it, there must exist 3 points such that the slopes of the lines connecting each pair of these points are distinct and non-zero. | Three, the slope of the line connecting two points exists and is not zero, which is equivalent to the coordinates of the two points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ being unequal: $x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$.
Now, divide 15 points in the set $S$ of integer points into 5 groups such t... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,438 |
Initial 39. In $\triangle A B C$, place any 4 points $D, E, F, G$ different from the vertices (including the boundary). Prove that among these 7 points, the distances between all pairs of points take at least 4 different values. | Proof divided into two cases:
(1) If one of the points $D, E, F, G$, for example $D$, does not lie on any perpendicular bisector of a side, then the three segments $A D, B D, C D$ are all of different lengths. Suppose $A B$ is the longest side, then it is easy to prove that the four segments $A B, A D, B D, C D$ are al... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,441 |
Given $x, y, z \in R^{+}$. Prove:
$$
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} .
$$ | Prove $\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{x}{x+y+2 z}$
$$
\begin{array}{l}
=\frac{x}{(x+y)+(x+z)}+\frac{y}{(x+y)+(y+z)} \\
+\frac{z}{(z+x)+(z+y)} \\
\leqslant \frac{x}{2 \sqrt{(x+y)(x+z)}}+\frac{y}{2 \sqrt{(y+x)(y+z)}} \\
+\frac{z}{2 \sqrt{(z+x)(z+y)}} \\
=\frac{1}{4}\left(2 \sqrt{\frac{x}{x+y} \cdot \frac{x}{x+... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 708,442 |
39. Given $a, b, c$ are non-negative real numbers, and $a b + b c + a c = 1$. Prove: $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geqslant \frac{5}{2}$. | Proof: By symmetry, we can assume $a \geqslant b \geqslant c > 0$. From the given conditions, it is easy to see that $a \geqslant b > 0$.
$$
\begin{array}{l}
\frac{1}{b+c}+\frac{1}{a+c} \geqslant \frac{2}{\sqrt{(b+c)(a+c)}} \\
=\frac{2}{\sqrt{a b+a c+b c+c^{2}}}=\frac{2}{\sqrt{1+c^{2}}},
\end{array}
$$
The equality ho... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,443 |
40. Let $N$ be the set of all natural numbers, and let $k \in N$, where $k$ is odd. Prove: there exists a function $f: N \rightarrow N$, such that for every $n \in N$ we have $f(f(n))=k \cdot n$, and $f$ is strictly increasing. | Prove that when $k=1$, take $f(n)=n(n \in N)$. Now let $k$ be an odd number greater than 1.
When $k=3$, we find and can prove by mathematical induction that the function $f: N \rightarrow N$ that meets the requirements must be as follows:
$$
\begin{array}{l}
(i=0,1,2, \cdots) \\
\end{array}
$$
That is, when $k=3$, th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,444 |
Example 10. Let $x, y$ be real numbers, and $x^{2}+x y+y^{2}$ $=3$. Find the maximum and minimum values of $x^{2}-x y+y^{2}$. (1994, Huanggang Region Junior High School Competition) | Let $x^{2}-x y+y^{2}=P$,
and $x^{2}+x y+y^{2}=3$.
From (1) and (2), we get
$$
x+y= \pm \sqrt{\frac{9-P}{2}}, x y=\frac{3-P}{2} .
$$
$\therefore x, y$ are the two real roots of the equation $t^{2} \pm \sqrt{\frac{9-P}{2}} t+\frac{3-P}{2}=0$.
From $\Delta=\left( \pm \sqrt{\frac{9-P}{2}}\right)^{2}-4 \cdot \frac{3-P}{2} ... | 1 \leqslant P \leqslant 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,445 |
Example 1. If the equation $m x^{2}-2(m+2) x+m$ $+5=0$ has no real roots with respect to $x$, then the number of real roots of the equation $(m-5) x^{2}$ $-2(m+2) x+m=0$ with respect to $x$ is ( ).
(A) 2
(B) 1
(C) 0
(D) Uncertain
(1989, National Junior High School Mathematics Competition) | Solve: For $m x^{2}-2(m+2) x+m+5=0$ to have no real roots, then
$$
\Delta_{1}=4(m+2)^{2}-4 m(m+5)4$.
Also, the discriminant of $(m-5) x^{2}-2(m+2) x+m=0$ is
$$
\Delta_{2}=4(m+2)^{2}-4 m(m-5)=36 m+16 \text {. }
$$
When $m>4$ and $m \neq 5$, $\Delta_{2}>0$, the equation has two real roots; when $m=5$, the equation is a ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,446 |
Example 2. Given real numbers $a, b, c, r, p$ satisfy the conditions $p r>1, p c - 2 b + r a=0$. Prove: The quadratic equation $a x^{2}+2 b x+c=0$ must have real roots.
(1987, Guangzhou and Four Other Cities Junior High School Mathematics Competition) | Prove that $\Delta=(2 b)^{2}-4 a c$,
and from the given condition $2 b=p c+r a$,
we can get $\Delta=(p c+r a)^{2}-4 a c$
$$
\begin{array}{l}
=(p c-r a)^{2}+4 a c(p r-1) . \\
\because(p c-r a)^{2} \geqslant 0 . \hat{r} r>\text {, } \\
\end{array}
$$
$\therefore$ if $a c>0$,
when $a x \geqslant 0$, from the known conditi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,447 |
Example 11. Solve the equation
$$
5 x^{2}+10 y^{2}-72 x y-6 x-4 y+13=0 \text {. }
$$
(1994, Tianjin Junior High School Mathematics Competition) | Solve: The original equation can be transformed into a quadratic equation in terms of $x$
$$
5 x^{2}+(-12 y-6) x+10 y^{2}-4 y+13=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\begin{aligned}
\Delta & =(-12 y-6)^{2}-4 \times 5\left(10 y^{2}-4 y+13\right) \\
& =-(y-2)^{2} \geqslant 0 .
\end{aligned}
$$
Solving... | x=3, y=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,448 |
1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\fra... | II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,449 |
2. Let $a_{1}, a_{2}, \cdots, a_{k}$ be $k$ distinct positive integers, and $a_{1}+a_{\varepsilon}+\cdots+a_{k}=1995$. Then, the maximum value of $k$ is $\qquad$ | 2. $\because a_{1}, a_{2}, \cdots, a_{4}$ are all distinct, without loss of generality, let $a_{1}<a_{2}<\cdots<a_{k}$. Since $a_{1}, a_{2}, \cdots, a_{k}$ are all positive integers, we have $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{k} \geqslant k$, thus $1+2+\cdots+k \leqslant a_{1}+a_{2}+\cdots+a_{k}=$ 1995, ... | 62 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,450 |
$\begin{array}{l}\text { 3. Let } \frac{x}{x^{2}+x+1}=a \text {, where } a \neq 0 \text {. Then } \frac{x^{2}}{x^{4}+x^{2}+1} \\ =\end{array}$ | $$
\begin{array}{l}
\text { 3. } \because a \neq 0 \text {, hence } x \neq 0 \text {, then we have } \frac{1}{a}=\frac{x^{2}+x+1}{x}=x+\frac{1}{x} \\
+1, \therefore x+\frac{1}{x}=\frac{1}{a}-1 \text {. Thus, } \frac{x^{4}+x^{2}+1}{x^{2}}=x^{2}+1+\frac{1}{x^{2}} \\
=\left(x+\frac{1}{x}\right)^{2}-1=\left(\frac{1}{a}-1\r... | \frac{a^{2}}{1-2 a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,451 |
4. In $\triangle A B C$, $B C=2, C A=3, A B=4, P$ is a point inside $\triangle A B C$, $D, E, F$ are on $A B, B C, C A$ respectively, and $P D \parallel B C, P E \parallel A C, P F \parallel A B$. If $P D=P E=P F=l$, then $l=$ | 4. Connect $A P, B P, C P$ and extend them to intersect $B C, C A, A B$ at $G, H, K$ respectively. Since $P D / / B C$, we have $\frac{P D}{B C}=\frac{P K}{C K}=\frac{S_{\triangle B P K}}{S_{\triangle B C K}}=\frac{S_{\triangle A P K}}{S_{\triangle A C K}}$. By the property of equal ratios, we have $\frac{l}{2}=\frac{S... | \frac{12}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,452 |
5. If $\sqrt{a^{2}-1996}$ is an integer, then the minimum value of the integer $a$ is | 5. Consider the case of natural number $a$: Let $\sqrt{a^{2}-1996} = m$, then $a^{2}-m^{2}=1996$, $(a+m)(a-m)=1996$. Since $a+m$ and $a-m$ are both even numbers, and $1996=2 \times 998$ is the only way to split it into the product of two even numbers, we have $a+m=998, a-m=2$, solving for $a=500, m=498$. Since the larg... | -500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,453 |
Three, (Full marks 12 points) Find all integer values of $a$ such that the equation $(a+1) x^{2}$ $-\left(a^{2}+1\right) x+2 a^{3}-6=0$ has integer roots.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | When $a=-1$, the equation becomes $-2 x-2-6=0$, at this time $x=-4$;
When $a \neq-1$, let the two roots be $x_{1}, x_{2}$, then
$$
\begin{aligned}
x_{1}+x_{2} & =\frac{a^{2}+1}{a+1} \\
& =\frac{a^{2}-1+2}{a+1}=a-1+\frac{2}{a+1} ;
\end{aligned}
$$
When $a=0,1,-2,-3$, $\frac{2}{a+1}$ is an integer, $x_{1}+x_{2}$ is an i... | a=-1,0,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,454 |
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