problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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2. The two roots of the equation $x^{2}-3|x|-4=0$ are $x_{1}$, $x_{2}$, then $x_{1}+x_{2}=$ $\qquad$ | 2. 0
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,342 |
Five. (16 points) For any given $\triangle P Q R$, a convex quadrilateral can be drawn that simultaneously satisfies the following two conditions:
(1) The length of the shortest side is greater than the length of the longest side of $\triangle P Q R$;
(2) The area is less than the area of $\triangle P Q R$.
Please draw... | Five, there are many ways to solve this problem, one of which is as follows:
Let the longest side of $\triangle P Q R$ be $a$, and the height on this side be $h$, then $S=\frac{1}{2} a h$. Construct an isosceles trapezoid, with the upper base length being $2 a$, the lower base length being $6 a$, and the height being $... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,350 |
A-1. Prove that $f(n)=1-n$ is the unique integer-valued function defined on the integers and satisfying the following conditions:
(i) For all integers $n, f(f(n))=n$.
(ii) For all integers $n, f(f(n+2)+2)=n$.
(iii) $f(0)=1$. | Prove that if $f^{\prime}(n)=1-n$, then
$$
f(f(n))=f(1-n)=1-(1-n)=n .
$$
Therefore, (i) holds. Similarly,
$$
\begin{array}{l}
f(f(n+2)+2)=f((-n-1)+2) \\
=f(1-n)=n .
\end{array}
$$
Therefore, (ii) holds. Clearly, (iii) also holds.
Thus, $f(n)=1-n$ satisfies the given conditions.
Conversely, let $f$ satisfy the three g... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,351 |
A-3. For a given positive integer $m$, find all positive integer triples $(n, x, y)$, where $m, n$ are coprime, and satisfy
$$
\left(x^{2}+y^{2}\right)^{m}=(x y)^{n} .
$$ | When $m$ is odd, the equation has no solution. When $m$ is even, the only solution is $(n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right)$. We will prove this below.
If $(n, x, y)$ is a solution to the equation, then by the arithmetic-geometric mean inequality, we have $(x y)^{n}=\left(x^{2}+y^{2}\right)^{n} ... | (n, x, y)=\left(m+1,2^{\frac{m}{2}}, 2^{\frac{m}{2}}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,352 |
A-5. For every positive integer $n$, if the number of 1's in the binary representation of $n$ is even, then let $a_{n}=0$, otherwise let $a_{n}=1$. Prove that there do not exist positive integers $k$ and $m$ such that
$$
a_{k+j}=a_{k+m+j}=a_{k+2 m+j}, 0 \leqslant j \leqslant m-1 .
$$ | Proof: It is easy to observe that $a_{2 n}=a_{n}, a_{2 n+1}=1-a_{2 n}=1-a_{n}$. Suppose there exist $k, m$ satisfying the condition, we can assume $m$ is the smallest value that satisfies the condition.
Assume $m$ is odd, let's say $a_{k}=a_{k+m}=a_{k+2 m}=0$ (the case $a_{k}=1$ can be handled similarly). Because whet... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,353 |
A-6. Take any four points on a sphere to form a tetrahedron. What is the probability that the center of the sphere is inside the tetrahedron? (It can be assumed that each point is independently chosen and is uniformly distributed on the sphere) | On the sphere, consider three points $P, Q, R$, and we are interested in the position of another point $S$ on the sphere. Point $S$ must be such that the tetrahedron $P Q R S$ contains the center of the sphere $O$. Let $P^{\prime}, Q^{\prime}, R^{\prime}$ be the points symmetric to $P, Q, R$ with respect to the center ... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,354 |
$\mathrm{B}-1$. Let $S$ be a set of $n$ distinct real numbers, and $A_{s}$ be the set of all distinct averages of pairs of elements from $S$. For a given $n \geqslant 2$, what is the minimum number of elements that $A_{s}$ can have? | $A_{s}$ contains at least $2 n-3$ elements.
Let $x_{1}<x_{2}<\cdots<x_{n}$ denote the elements of $S$, then
$$
\begin{array}{l}
\frac{x_{1}+x_{2}}{2}<\frac{x_{1}+x_{3}}{2}<\cdots<\frac{x_{1}+x_{n}}{2}<\frac{x_{2}+x_{n}}{2}< \\
\frac{x_{3}+x_{n}}{2}<\cdots<\frac{x_{n-1}+x_{n}}{2}
\end{array}
$$
represents $(n-1)+(n-2)=... | 2n-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,355 |
B-4. $p(x)$ is a non-zero polynomial of degree less than 1992, and $p(x)$ has no common factor with $x^{3}-x$. Let
$$
\frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^{3}-x}\right)=\frac{f(x)}{g(x)},
$$
where $f(x), g(x)$ are polynomials. Find the smallest possible degree of $f(x)$. | The minimum possible degree of $f(x)$ is 3984.
By the division algorithm, we let $p(x) = (x^3 - x)q(x) + r(x)$, where $q(x)$ and $r(x)$ are polynomials, and the degree of $r(x)$ is less than 3, and the degree of $q(x)$ is less than 1989. Then,
$$
\cdot \frac{d^{1992}}{d x^{1992}}\left(\frac{p(x)}{x^3 - x}\right) = \fra... | 3984 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,357 |
1. (2nd Validation, 1st Question, Provided by Huang Yumin)
Let $0 \leqslant a \leqslant b \leqslant c \leqslant d \leqslant e$, and $a+b+c+d+e$ $=1$. Prove: $a d+d c+c b+b e+e a \leqslant \frac{1}{5}$. | Prove that because $a \leqslant b \leqslant c \leqslant d \leqslant e$, we have,
$$
d+e \geqslant c+e \geqslant b+d \geqslant a+c \geqslant a+b \text {. }
$$
Using Chebyshev's inequality, we have
$$
\begin{array}{l}
a(d+e)+b(c+e)+c(b+d)+d(a+c) \\
+e(a+b) \\
\leqslant \frac{1}{5}(a+b+c+d+e)[(d+e)+(c+e) \\
+(b+d)+(a+c)+... | a d+d c+c b+b e+e a \leqslant \frac{1}{5} | Inequalities | proof | Yes | Yes | cn_contest | false | 707,358 |
Example 7. Prove that there does not exist a point $P$ inside $\triangle A B C$ such that a line passing through $P$ divides the area of $\triangle A B C$ into two equal parts. | Prove the assumption that in $\triangle ABC$ there exists a point $P$, such that any line through $P$ divides the area of $\triangle ABC$ into two equal parts. Connect $AP, BP, CP$, and extend them to intersect the opposite sides at $D, E, F$. Since $AD$ divides $\triangle ABC$ into two equal areas, $D$ is the midpoint... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,359 |
2. (2nd Quiz, 2nd Question, provided by Huang Yumin) Given a sequence of positive numbers $a_{1}, a_{2}, a_{3}, \cdots$, satisfying $a_{n+1}$ $=\frac{1}{a_{1}+a_{2}+\cdots+a_{n}}(n \in N)$. Find $\lim _{n \rightarrow \infty} \sqrt{n} a_{n}$. | Solve: First, prove a limit equality:
$$
\lim _{n \rightarrow \infty} \frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=0 \text {. }
$$
Let \( T_{n}=\sum_{k=1}^{n} \frac{1}{k}, b_{n}=\frac{1}{n} T_{n} \). Then
$$
\begin{aligned}
T= & 1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,360 |
3. (Question 3 of the 3rd Decision, provided by Xuan Guo) $a$ is a non-fog real number, given a natural number $n \geqslant 2$. Find all $n$-degree real-coefficient polynomials $f(x)$ that satisfy the equation $f\left(x^{2}\right)=f(x) f(x+a)$. | From $f\left(x^{2}\right)=f(x) f(x+a)$, we know that if $x$ is a root of $f(x)$, then $x^{2}$ is also a root of $f(x)$. Assuming that the non-zero root $x_{0}$ of $f(x)$ has a modulus not equal to 1, then $x_{0}^{2}, x_{0}^{4}, x_{0}^{8}, \cdots$ have different moduli and are all roots of $f(x)$, which contradicts the ... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,361 |
4. (4th Quiz, Question 3, provided by Huang Xuan Guo)
Given a natural number $n \geqslant 3$ and a real constant $\lambda$, it is known that $x_{1}$, $x_{2}, \cdots, x_{n}$ are non-negative real numbers satisfying the condition $x_{1}+x_{2}+\cdots+x_{n}=1$. Find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\... | First, we prove two inequalities: under the given conditions,
\[ x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2} + n^{n-1}(n-1) x_{1} x_{2} \cdots x_{n} \leqslant 1, \]
(1)
\[ (n-1)\left(x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}\right) + n^{n-1} x_{1} x_{2} \cdots x_{n} \geqslant 1. \]
First, it is easy to see that
\[ x... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,362 |
5. (Question 1 of the 8th test, provided by Xia Xingguo)
Given a positive integer $n$, find all continuous functions $f(x)$, satisfying
$$
\begin{array}{l}
C_{n}^{0} f(x)+C_{n}^{1} f\left(x^{2}\right)+C_{n}^{2} f\left(x^{2^{2}}\right)+\cdots \\
+C_{n}^{n-1} f\left(x^{2^{2-1}}\right)+C_{n}^{n} f\left(x^{2^{n}}\right)=0... | Solution: It is obvious that $f(x)$ being constantly equal to zero is a solution to the problem.
We will use mathematical induction on $n$ to prove that the continuous function $f(x)$ satisfying the conditions of the problem must be constantly equal to zero.
When $n=1$, from the condition, we have $f(x)+f\left(x^{2}\r... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,363 |
5. The 3rd problem of the 1st test, *The American Mathematical Monthly*, 1952, p. 420
$$
\text{Let } f(x)=(x-x_{1})(x-x_{2}) \cdots(x-x_{n}),
$$
where $-1 \leqslant x_{i} \leqslant 1, i=1,2, \cdots, n$. Question: Does there exist a pair of real numbers $a, b$, simultaneously satisfying the following two conditions:
(1... | By symmetry, without loss of generality, let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$. Let
$g(x)=\left|x-x_{1}\right|+\left|x-x_{2}\right|+\cdots+\left|x-x_{n}\right|$, where $x \in[-1,1]$. When $x$ belongs to the intervals $\left[-1, x_{1}\right],\left[x_{1}, x_{2}\right]$, $\cdots,\left[x\left[\frac{4... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,364 |
8. (Question 2 of the 5th Test, *The American Mathematical Monthly*, pages 119 to 121, 1953)
Given a positive integer $n \geqslant 3$, for $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ with modulus 1, find
$$
\min _{x_{1}, z_{2} \cdots, s_{n}}\left[\max _{\omega \in C,|\omega|=1} \prod_{j=1}^{n}\left|\omega-z_{j}\... | $$
\begin{array}{l}
f(u)=\prod_{j=1}^{n}\left(u-x_{j}\right) \\
=u^{n}+C_{n-1} u^{n-1}+\cdots+C_{1} u+C_{0} .
\end{array}
$$
Here, $\left|C_{0}\right|=\left|z_{1} z_{2} \cdots z_{n}\right|=1$. Let $C_{0}=e^{i\theta}, 0 \leqslant \theta<2 \pi$. Take $n$ complex numbers of modulus 1, $\varepsilon_{k}=e^{i \frac{2 x+\the... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,366 |
9. (8th Decision Test 2nd Question, 22nd US Mathematical Olympiad 4th Question)
Let $a, b$ be positive odd numbers, and define the sequence $\left\{f_{n}\right\}$ as follows: $f_{1}$ $=a, f_{2}=b$, for $n \geqslant 3$, $f_{n}$ is the greatest odd divisor of $f_{n-1}+f_{n-2}$. Prove that when $n$ is sufficiently large,... | From the conditions of the problem, we know that if there are two adjacent terms $f_{A}$, $f_{k+1}$ that are equal, then when $n \geqslant k$, $f_{n}$ are all equal, forming a constant.
Using proof by contradiction, if $f_{n}$ is not a constant, then any two adjacent terms in the sequence $\left\{f_{n} \mid n \in N\ri... | (a, b) | Number Theory | proof | Yes | Yes | cn_contest | false | 707,367 |
Three, (35 points) A group of students took an exam with 3 multiple-choice questions, each with four options. It is known that any two students in this group have at most one answer in common, and if one more student is added, regardless of their answers, the above property no longer holds. How many students are there ... | Three, this group of students must have at least 8 people.
First, consider if this group of students has 8 people, then it can meet the conditions of the problem.
Let the answers of these 8 people be $(1,1,1),(1,2,2),(2,2,1)$, $(2,1,2),(3,4,3),(3,3,4),(4,3,3),(4,4,4)$, which will suffice.
On the other hand, assume thi... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,369 |
Example 8. Draw any line through the centroid of a triangle to divide it into two parts. Prove: The difference in the areas of these two parts is no greater than $\frac{1}{9}$ of the area of the entire triangle. (1979, Anhui Province Competition) | Prove that by trisecting each side of a triangle and drawing lines through each division point parallel to the other two sides, these lines divide $\triangle A B C$ into 9 smaller triangles of equal area. The internal intersection point $G$ is precisely the centroid of this triangle. Drawing any line through $G$ divide... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,370 |
1. $a, b$ are positive integer divisors of 1995, then the number of ordered pairs $(a, b)$ is ().
(A) 16
(B) 32
(C) 64
(D) 256 | 1. D $1995=3 \cdot 5 \cdot 7 \cdot 19$, the number of positive divisors is 24 $=16$. For the pair $(a, b)$, $a$ has 16 possible values, and for each $a$, $b$ also has 16 possible values. Therefore, the number of pairs $(a, b)$ is $16 \times 16=256$ (pairs). | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,371 |
2. Positive numbers $a, b, c, x, y, z$ satisfy $a+2 x=2 b+$ $3 y=3 c+4 z=k$. If $M=a x+b y+c z$, then the range of $M$ is ( ).
(A) $0<M \leqslant \frac{3}{16} k^{2}$
(B) $\frac{3 k^{2}}{16} \leqslant M \leqslant \frac{k^{2}}{2}$
(C) $0<M \leqslant \frac{k^{2}}{8}$
(D) $\frac{k^{2}}{3}=M \leqslant \frac{k^{2}}{2}$ | 2. Extreme case: When $a, b, c \rightarrow 0$, $M \rightarrow 0$, negating (B), (D).
By $\left\{\begin{array}{l}a \cdot 2 x \leqslant\left(\frac{k}{2}\right)^{2}, \\ 2 b \cdot 3 y \leqslant\left(\frac{k}{2}\right)^{2}, \\ 3 c \cdot 4 z \leqslant\left(\frac{k}{2}\right)^{2}\end{array}\right.$ we have $\left\{\begin{arr... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,372 |
3. If the number $x$ satisfies $x^{2}+144 \leqslant 25 x^{2}$, then the maximum and minimum values of the function $y=\frac{x-6}{x+5}$ are ( ).
(B) maximum value $-\frac{3}{2}$, minimum value -10
(C) maximum value $-\frac{2}{9}$, minimum value -10
(D) no maximum value, no minimum value | 3. C From $x^{4}-25 x^{2}+144<0$, we get $3 \leqslant x \leqslant 4$ or
$$
-4 \leqslant x \leqslant-3, y=\frac{x+5-11}{x+5}=1-\frac{11}{x+5} \text {. }
$$
When $x=4$, $y=-\frac{2}{9}$ is the maximum value,
When $x=-4$, $y=-10$ is the minimum value. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,373 |
4. As shown in the figure: In $\triangle A B C$, $D, E$ are points on $B C$, $B D: D E: E C=3: 2: 1, M$ is on $A C$, $C M: M A=1: 2, B M$ intersects $A D, A E$ at $H$, $G$. Then $B H: H G: G M$ equals ( ).
(A) $3: 2: 1$
(B) $5: 3: 1$
(C) $25: 12: 5$
(D) $51: 24: 10$ | 4. D Draw $M N / / B C$ intersecting $A D, A E$ at $Q, P$. Let $B H = x, H G = y, G M = z$. By the properties of similar triangles and proportional segments, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{y+z}=\frac{B D}{M Q}=\frac{D C}{M Q}=\frac{3}{2}, \\
\frac{x+y}{z}=\frac{B E}{M P}=\frac{5 E C}{M P}=... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,374 |
5. Three positive numbers $x, y, z$ satisfy the system of inequalities:
$$
\left\{\begin{array}{l}
\frac{11}{6} z < x + y < 2 \\
\frac{3}{2} x < y + z < \frac{5}{3} x \\
\frac{5}{2} y < x + z < \frac{11}{1} y .
\end{array}\right.
$$
Then the relationship in size of $x, y, z$ is ().
(A) $x < y < z$
(B) $y < x < z$
(C) ... | 5. B
From $\left\{\begin{array}{l}\frac{11}{6}\frac{5}{2}>2>\frac{x+y}{z}>\frac{11}{6}>\frac{5}{3}>\frac{y+z}{x}$.
Thus, $z(x+z)>y(x+y) \Rightarrow z>y$. (Otherwise, the inequality does not hold)
$x(x+y)>z(y+z) \Rightarrow x>z$. (For the same reason)
Therefore, $y<z<x$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,375 |
3. The equation $\frac{1}{x^{2}+1}+\frac{x+1}{x^{2}}=\frac{10}{3 x}$ is conceived as follows
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. Answer: $\frac{3 \pm \sqrt{5}}{2}$.
Let $y=x^{2}+1$, the equation is $\frac{1}{y}+\frac{y}{x^{2}}=\frac{10}{3 x}$. Clearing the denominators: $3 x^{2}-10 x y+3 y^{2}=0$.
We get $y_{1}=3 x, y_{2}=\frac{1}{3} x$.
From $y=3 x, x^{2}-1=3 x$, we get $x=\frac{3 \pm}{2}$.
From $y=\frac{1}{3} x, x^{2}+1=\frac{x}{3}, \Delta... | \frac{3 \pm \sqrt{5}}{2} | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,379 |
4. As shown in the figure, in rectangle $A B C D$, $A B=\frac{2}{5} B C, D E$ $=B G, \angle B E C=$ $\angle D G A=90^{\circ}$, the area of rectangle $E F G H$ is $S$. Then the area of rectangle $A B C D$ is $\qquad$ . | 4. Answer: $\frac{25 S}{4}$.
Let $A B=a, B C=b, B G=D E=x, G F=m, G H=$ $n$. According to the geometric mean theorem in right triangles, we have
$$
x^{2}=n \cdot A G \text{ (in } \operatorname{Rt} \triangle A B G \text{)}
$$
and $(b-x)^{2}=m \cdot G D$ (in Rt $\triangle D C G$).
Multiplying the two equations, we get ... | \frac{25 S}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,380 |
One, (20 points) For positive numbers $a, b$, define the operation * as follows: when $a \leqslant b$, $a * b=a b$; when $b<a<2 b$, $a * b=b^{2}+a-b$; when $a \geqslant 2 b$, $a * b=a^{2}+b-a$. If $3 * x=8$, find the value of $\left[x^{3}-4 x\right]$ (where $[x]$ represents the greatest integer not exceeding $x$). | If $3 \leqslant x$, then $3 * x=3 x \geqslant 3 \cdot 3>8$. This does not satisfy $3 * x=8$.
If $x<3<2 x$, then $3 * x=x^{2}+3-x=8$, which means $x^{2}-x-5=0$.
Thus, $x=\frac{1+\sqrt{21}}{2}$ (taking the positive root) (since $\sqrt{21}<5$ satisfies $x<3<2 x$).
If $3 \geqslant 2 x$, then $3 * x=3^{2}+x-3=6+x \leqslan... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,382 |
II. (20 points) In the acute $\triangle ABC$, three semicircles are constructed outside the triangle with each side as the diameter, as shown in the figure. $O$ is a point in $\triangle ABC$, and the extensions of $AO$, $BO$, and $CO$ intersect the opposite semicircles at $A_{1}$, $B_{1}$, and $C_{1}$, respectively. If... | Let $\angle A C O=\angle 1, \angle A B O=\angle 2, \angle C A O=\angle 3, \angle C B O=\angle 4, \angle B C O=\angle 5, \angle B A O=\angle 6$. From $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6$ we have
$$
\begin{array}{l}
\angle 1+\angle 3+\angle 5=\angle 2+\angle 4+\angle 6=99^{\circ} \\
\Rightarrow A A_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,383 |
The number of lines, and the number of all triangles formed with the vertices of a nonagon (nine-sided polygon).
(2) Assign an arbitrary natural number to each vertex of a convex nonagon. Among the triangles formed with the vertices of this nonagon, if the sum of the three numbers at the vertices is odd, the triangle i... | (1) From each vertex, there are 6 diagonals, and with nine vertices, a total of 54 diagonals are emitted. Since each diagonal is counted twice, the total number of unique diagonals is $\frac{54}{2}=27$.
Given that the total number of sides and diagonals is 36 $\left(C_{\theta}^{2}=\right.$ number of sides + number of ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,384 |
1. If $x, y \in R$, then the relationship between the sets of real numbers $P=\left\{s \mid s=x^{2}+3 x\right.$ $+1\}$ and $Q=\left\{t \mid t=y^{2}-3 y+1\right\}$ is ( ).
(A) $P \cap Q=\varnothing$
(B) $P \subset Q$
(C) $P \supset Q$
(D) $P=Q$ | 1. (D)
$$
\because P=\left\{s \left\lvert\, s=\left(x+\frac{3}{2}\right)^{2}-\frac{5}{4}\right.\right\} \text {, }
$$
$\therefore$ The range of values for the element $s$ in $P$ is $\left[-\frac{5}{4},+\infty\right)$. $X \because Q=\left\{t \left\lvert\, t=\left(y-\frac{3}{2}\right)^{2}-\frac{5}{4}\right.\right\}$,
$\t... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,385 |
3. In $\triangle A B C$, $\cos 2 B>\cos 2 A$ is a ( ) for $A>B$.
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient nor necessary condition | 3. (C).
$$
\begin{array}{l}
\therefore \text { In } \triangle A B C, A>B \Leftrightarrow a>b \Leftrightarrow 2 R \sin A>2 R \sin B \\
\Leftrightarrow \sin A>\sin B>0 \Leftrightarrow \sin ^{2} A>\sin ^{2} B \Leftrightarrow-2 \sin ^{2} A1-2 \sin ^{2} A \Leftrightarrow \cos 2 B>\cos 2 A .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,387 |
4. The radii of the two bases of a frustum are $R$ and $r(R > r)$, respectively. A plane parallel to the bases divides the lateral surface of the frustum into two equal areas. Then, the radius of the cross-sectional circle is
(A) $\sqrt{R r}$
(B) $\sqrt{\frac{R^{2}+r^{2}}{2}}$ | (C) $\frac{R+r}{2}$
(D) cannot be determined
4. (B)
Since the lateral areas of the two frustums are equal, and the angle formed by the slant height with the base and the section is equal, according to $S^{\prime}=S \cos \theta$, it can be concluded that the projection area of the upper frustum on the section is equal ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,388 |
6 . For the parabola $y^{2}=4 x$, a line is drawn through its focus intersecting the parabola at points $P, Q$. Then, the equation of the locus of the midpoint of segment $P Q$ is (
(A) $y=2 x-1$
(D) $y^{2}=2 x-2$
(C) $y'=-2 x+1$
(D) $y^{2}=-2 x+2$ | G. (B)
Since the focus $F(1,0)$ should lie on the trajectory, (A) and (C) are incorrect. Also, because the parabola opens to the right, the trajectory cannot open to the left, so (D) is incorrect. Therefore, the answer is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,390 |
Given the function
$$
f(x)=\left\{\begin{array}{lc}
-x, & 0 \leqslant x \leqslant 1, \\
1+x, & -1 \leqslant x<0 .
\end{array}\right.
$$
Then its inverse function is $\qquad$ | $$
\text { 2. } f^{-1}(x)=\left\{\begin{array}{l}
x-1, x \in[0,1), \\
-x, x \in[-1,0]
\end{array}\right.
$$
Draw the graph of the function $f(x)$. Since its graph is symmetric to the graph of its inverse function about $y=x$, we can obtain the graph of $f^{-1}(x)$, and thus get the analytical expression of the functio... | f^{-1}(x)=\left\{\begin{array}{l} x-1, x \in[0,1), \\ -x, x \in[-1,0] \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,391 |
3. In $\triangle A B C$, it is known that $B C=10, A C>A B$, and the medians $B E, C F$ are perpendicular to each other, and the distance from the centroid $G$ to $B C$ is 3. Try to find the lengths of $A B, A C$. (1987, Shanghai Junior High School Competition) | (Solution: $A B=\sqrt{130}, A C=\sqrt{370}$ ) | A B=\sqrt{130}, A C=\sqrt{370} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,392 |
3. A city's phone numbers are being increased from six to seven digits, which can add $\qquad$ more dialing combinations.
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。 | 3. $10^{7}-10^{6}$.
Six-digit numbers can form $10^{6}$ dialings, seven-digit numbers can form $10^{7}$ dialings. After the upgrade, $10^{7}-10^{6}$ more dialings can be added. | 10^{7}-10^{6} | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 707,394 |
4. $\operatorname{arctg} x+\operatorname{arctg} \frac{1-x}{1+x}$ 的值是
4. The value of $\operatorname{arctg} x+\operatorname{arctg} \frac{1-x}{1+x}$ is | 4. $-\frac{3}{4} \pi$ or $\frac{\pi}{4}$.
Let $\operatorname{arctg} x=\alpha, \operatorname{arctg} \frac{1-x}{1+x}=\beta$. Then
$$
\begin{array}{l}
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{x+\frac{1-x}{1+x}}{1-x \c... | \frac{\pi}{4} \text{ or } -\frac{3}{4} \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,395 |
6. $\triangle A B C$ vertices lie on a parabola with the $x$-axis as its axis of symmetry and the origin as its focus. It is known that $A(-6,8)$, and the centroid of $\triangle A B C$ is at the origin. Then the equation of the line passing through points $B$ and $C$ is $\qquad$
Translate the above text into English, ... | $6.4 x+y-8=0$.
From the given information, the parabola equation is $y^{2}=32(x+8)$. Since the origin is the centroid, and point $A$ is given, we can deduce that the midpoint $M$ of $E C$ is $(3, -4)$.
The parametric equation of line $2 C$ is $x=3+t \cos \theta, y=-4+t \sin \theta$ (where $t$ is a parameter). Substitu... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,397 |
Three, (20 points) As shown in the figure,
the projection of the apex of the pyramid $S-ABCD$ onto the base is exactly the intersection point $O$ of the diagonals of the base. It is known that the height of the pyramid $S-ABCD$ is always $3$, $V_{s-ADO}=m^{2}$, $V_{s-BCO}=n^{2}\left(m, n \in R^{+}, m \neq n\right)$. Wh... | $$
\text { Three, } V_{S-A D O}=\frac{1}{3} S O \cdot S_{\triangle A D O}=S_{\triangle A D O}=m^{2} \text {. }
$$
Similarly, $S_{\triangle B C O}=n^{2}$.
Let $S_{\triangle D O C}=x, S_{\triangle A O B}=y$, then
$$
V_{S-A B C D}=\frac{1}{3} S O \cdot S_{\triangle B C D}=m^{2}+n^{2}+x+y \text {. }
$$
And $\frac{S_{\tri... | A B C D \text{ is a trapezoid} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,398 |
Four, (20 points) The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$. Question: Does there exist an inscribed isosceles right triangle, one of whose legs passes through the point $F$? If it exists, | Let one of the legs be $P_{1} P_{2}$, then the equation of the line $P_{1} P_{2}$ is $y=k\left(x-\frac{p}{2}\right)$.
$$
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
y=k\left(x-\frac{p}{2}\right)
\end{array}\right.
$$
we get $k^{2} x^{2}-p\left(2+k^{2}\right) x$
$$
+\left(\frac{k p}{2}\right)^{2}=0 \text {. }... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,399 |
Five. (20 lines) The first term of the sequence $\left\{a_{n}\right\}$ is $a \neq 0$, and the sequence is a geometric sequence with a common ratio of $-a$. Let
$$
b_{n}=a_{n} \cdot \lg \left|a_{n}\right|, S_{n}=\sum_{i=1}^{n} b_{i} .
$$
(1) Prove: When $a \neq-1$, for all $n \in N$, we have
$$
\begin{aligned}
S_{n}= & ... | (1) Using mathematical induction:
When $n=1$, $S_{1}=b_{4}=u i g|a|$; On the other hand, in
$$
\frac{a \operatorname{g}|a|}{(1+a)^{2}}\left[1+(-1)^{n+1} \cdot(1+n+n a) a^{n}\right] \text {, }
$$
let $n=1$, then we have $\frac{a \lg |a|}{(1+a)^{2}} \cdot(1+a)^{2}$. It is clear that the proposition holds for $n=1$.
Assu... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,400 |
One, (35 points) $H$ is the orthocenter of $\triangle ABC$, $D, E, F$ are the midpoints of $BC, CA, AB$ respectively. A circle $\odot H$ centered at $H$ intersects the lines $EF, FD, DE$ at $A_{1}, A_{2}, B_{1}$, $B_{2}, C_{1}, C_{2}$. Prove:
$$
A A_{1}=A A_{2}=B B_{1}=B B_{2}=C C_{1}=C C_{2} .
$$ | Given the figure, $H$ is the center, $A H \perp B C$ at $H_{1}, E F$ is the midline of $\triangle A B C$, so $E F \parallel B C, A H \perp E F$ at $M$. Therefore, $A A_{1} = A A_{2}$. Similarly,
$$
\begin{array}{l}
B B_{1}=B B_{2}, \\
C C_{1}=C C_{2} .
\end{array}
$$
Thus, we only need to prove $A A_{1}=B B_{1}=C C_{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,401 |
21. As shown, point $D$ is a point inside right triangle $\triangle ABC$ such that $\angle ADB - \angle ACB = \alpha$, and $AC \cdot BD = AD \cdot BC$.
Prove: $\frac{AB \cdot CD}{AC \cdot BD} = 2 \sin \frac{\alpha}{2}$. | Proof As shown in the figure, rotate $BD$ clockwise around point $D$ by $a$ to $ED$, and connect $EA$, $EB$.
From the given, we have
$$
\begin{array}{l}
\angle ADE = \angle ADB - \alpha \\
= \angle ACB, \\
\frac{AD}{DE} = \frac{AD}{DB} = \frac{AC}{CB}. \\
\therefore \triangle ADE \sim \triangle ACB.
\end{array}
$$
Thu... | 2 \sin \frac{\alpha}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 707,402 |
Example 1. Solve the system of equations
$$
\left\{\begin{array}{l}
x(y+z-x)=39-2 x^{2}, \\
y(z+x-y)=52-2 y^{2}, \\
z(x+y-z)=78-2 z^{2} .
\end{array}\right.
$$ | Solving by expanding the brackets and factoring, the original system of equations can be transformed into
\[
\left\{\begin{array}{l}
x(x+y+z)=39, \\
y(x+y+z)=52, \\
z(x+y+z)=78 .
\end{array}\right.
\]
By adding (1) + (2) + (3) and then taking the square root, we get
\[
x+y+z= \pm 13.
\]
Substituting (4) into (1), (2),... | x_1=3, y_1=4, z_1=6; \quad x_2=-3, y_2=-4, z_2=-6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,403 |
Example 2. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \\ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} .\end{array}\right.$ | First, find a common denominator for the left side of each equation, then take the reciprocal of both sides of each equation, and let $x+y+z=k$. The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
x y + x z = 2 k, \\
y z + y x = 3 k, \\
z x + z y = 4.6.
\end{array}\right.
$$
(1) + (2) + ... | x = \frac{23}{10}, y = \frac{23}{6}, z = \frac{23}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,404 |
Example 11. Find the real solutions of the system of equations
$$
\left\{\begin{array}{l}
1-x^{2}=y, \\
1-y^{2}=z, \\
1-z^{2}=x
\end{array}\right.
$$ | (i) If $x=y=z$, it is easy to get
$$
x=y=z=\frac{-1 \pm \sqrt{5}}{2} .
$$
(ii) If two of the unknowns are equal, let's assume $x=y$, we can similarly get $x=y=z=\frac{-1 \pm \sqrt{5}}{2}$.
(iii) If the three unknowns are not equal, let's assume $x>y>z$, then
$1-x^{2}=y>z=1-y^{2}$, which implies $x^{2}y$, hence $yz$, so... | \begin{array}{l}
x_{1}=y_{1}=z_{1}=\frac{-1+\sqrt{5}}{2} ; \\
x_{2}=y_{2}=z_{2}=\frac{-1-\sqrt{5}}{2}
\end{array} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,405 |
Example 12. Solve the system of equations
$$
\left\{\begin{array}{l}
x z=3 u, \\
u^{2}+y^{2}+y u=x^{2}, \\
y^{2}+z^{2}-y z=9, \\
x^{2}=u^{2}+z^{2}+2 u z+9
\end{array}\right.
$$
for positive solutions. | The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
\frac{x}{y}=\frac{3}{u}, \\
u^{2}+y^{2}-2 u y \cos 120^{\circ}=x^{2}, \\
y^{2}+z^{2}-2 y z \cos 60^{\circ}=3^{2}, \\
x^{2}=(z+u)^{2}+3^{2} .
\end{array}\right.
$$
According to this, we can construct $\triangle A B C$, such that $\angle... | x=6, y=2\sqrt{3}, z=\sqrt{3}, u=2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,406 |
Example 1. Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$. Prove: For every natural number $n$, $(a+b)^{n}-a^{n}-b^{n}$ $>2^{2 n}-2^{n+1}$. (1988, National High School Competition) | $$
\begin{array}{l}
\text { Prove that } F=(a+b)^{n}-a^{n}-b^{n} \\
=C_{n}^{1} a^{n-1} b+C_{n}^{2} a^{n-2} b^{2}+\cdots+C_{n}^{n-1} a b^{n-1}, \\
F^{\prime}=C_{n}^{n-1} a b^{n-1}+C_{n}^{n-2} a^{2} b^{n-2}+\cdots+C_{n}^{1} a^{n-1} b .
\end{array}
$$
Clearly, $F=F^{\prime}$.
Adding the two equations, we get
$$
\begin{a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,407 |
Example 2. Compare $\left(\frac{n+1}{2}\right)^{n}$ and $n!$ $(n \in$ $N)$. | When $n=1$, it is easy to see that $\left(\frac{n+1}{2}\right)^{n}=n!$.
When $n>1, n \in N$, let
$$
\begin{array}{l}
F=1 \cdot 2 \cdot 3 \cdot \cdots \cdot(n-1) \cdot n, \\
F^{\prime}=n \cdot(n-1) \cdot \cdots \cdot 3 \cdot 2 \cdot 1 .
\end{array}
$$
Then
$$
\begin{array}{l}
F \cdot F^{\prime}=(n!)^{2}=[1 \cdot n] \c... | not found | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 707,408 |
Example 3. Let $a, b \in \mathbb{R}^{+}$, and $a > b, \sin \theta = \frac{2 a b}{a^{2} + b^{2}}$ $\left(0 < \theta < \frac{\pi}{2}\right), A_{n} = \left(a^{2} + b^{2}\right)^{n} \sin \theta$, then for all natural numbers $n, A_{n}$ are integers. (1990, National High School Competition) | Prove that for $F=A_{n}=\left(a^{2}+b^{2}\right)^{n} \sin n \theta$, $F^{\prime}=\left(a^{2}+b^{2}\right)^{x} \cos n \theta$. Given $\sin \theta=\frac{2 a b}{a^{2}+b^{2}}$, then
$$
\begin{aligned}
\cos \theta & =\frac{a^{2}-b^{2}}{a^{2}+b^{2}} . \\
& F^{\prime}+F i=\left(a^{2}+b^{2}\right)^{n}(\cos n \theta+i \sin n \t... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,409 |
Example 4. Prove that the smallest integer greater than $(\sqrt{3}+1)^{2 n}$ is divisible by $2^{n+1}$, where $n \in N$. (6th Putnam Mathematical Competition)
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Prove that if $F=(\sqrt{3}+1)^{2 n}, F^{\prime}=(\sqrt{3}-1)^{2 n}$, then $F+F^{\prime}=(\sqrt{3}+1)^{2 n}+(\sqrt{3}-1)^{2 n}$. Since $0<\sqrt{3}-1<1$, then $0<(\sqrt{3}-1)^{2 x}<1$. If we can prove that $F+F^{\prime}$ is an integer, then $F+F^{\prime}$ is the smallest integer greater than $(\sqrt{3}+1)^{2 n}$.
Since ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 707,410 |
Example 5. Let the sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ satisfy $x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}$. Find the sum of the first 1994 terms of the sequence $\left\{x_{n}\right\}$, $S_{1994}$. | $\begin{array}{l}\text { Let } \begin{array}{l} \text { } F:=x_{n}+y_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n}, F^{\prime} \\ =x_{h}-j_{n} i=\left(\frac{\sqrt{3} i-1}{2}\right)^{n} . \text { Then } F+F^{\prime}=2 x_{n} \\ =\omega^{n}+(\bar{\omega})^{n}\left(\text { where } \omega=\frac{-1+\sqrt{3} i}{2}\right) . \t... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,411 |
Example 6. Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Try to find the value of $f\left(\frac{1}{1001}\right)$ $+f\left(\frac{2}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$. (1986, National High School Competition) | Let $F=f(x)=\frac{4^{x}}{4^{x}+2}, F^{\prime}=f(y)=$ $\frac{4^{y}}{4^{y}+2}$. If $x+y=1$, then $f(x)+f(y)=1$. In fact, $f(x)+f(y)=f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}$ $+\frac{4^{1-x}}{4^{1-x}+2}=1$. Therefore, $f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)$ $+\cdots+f\left(\frac{999}{1001}\right)+f\left(\frac... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,412 |
Example 7. Let $n=1990$. Find the value of $\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}\right)$. (1990, National High School Competition) | \[
\begin{array}{l}
\quad \text { Solve } F=\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots\right. \\
\left.+3^{1994} C_{n}^{1998}-3^{995} C_{n}^{1990}\right), F^{\prime}=\frac{1}{2^{n}}\left(-\sqrt{3} C_{n}^{1}+\right. \\
\sqrt{3^{3}} C_{n}^{3}-\sqrt{3^{5}} C_{n}^{5}+\cdots+\sqrt{3^{1987}} C_... | -\frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,413 |
Example 8. In a ping-pong tournament, $p_{1}, p_{2}, p_{3}, \cdots$, $p_{n}$ are $n$ players, each pair of whom plays exactly one match, with no ties. If the number of wins for $p_{i}$ is denoted by $a_{i}$, and the number of losses by $b_{i}(1 \leqslant i \leqslant n)$, prove that $\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^... | Analysis: It is evident that calculating $a_{i}, b_{i}(1 \leqslant i \leqslant n)$ to prove the equation would be futile. If we consider the winning set $F$ and losing set $F^{\prime}$ of the number of wins and losses $n(F)$ and $n\left(F^{\prime}\right)$, the problem is easily resolved.
Proof First, for any $p_{i}(1 ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,414 |
Example 3. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x y}{x+y}=\frac{1}{9}, \\
\frac{y z}{y+z}=\frac{1}{11}, \\
\frac{x z}{x+z}=\frac{1}{12} .
\end{array}\right.
$$ | Obviously, $x, y, z$ are all non-zero. Therefore, we can take the reciprocal of both sides of each equation and convert them into partial fractions, obtaining
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=9 \\
\frac{1}{y}+\frac{1}{z}=11 \\
\frac{1}{z}+\frac{1}{x}=12
\end{array}\right.
$$
$$
\text { (1) }+ \text { ... | x=\frac{1}{5}, \quad y=\frac{1}{4}, \quad z=\frac{1}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,415 |
Example 9. For the set $\{1,2,3, \cdots, n\}$ and any of its non-empty subsets, define the "alternating sum" as follows: rearrange the elements in decreasing order, then alternately add and subtract the subsequent numbers starting from the largest. For example, for $\{1,2,4,6,9\}$, rearranged as $(9,6,4,2,1)$, the alte... | Solution: Let's define the alternating sum of the elements of the empty set to be 0. Divide the $2^{n}$ subsets of $\{1,2, \cdots, n\}$ into two categories: one category contains $n$ and has $2^{n-1}$ subsets, and the other category does not contain $n$, also having $2^{n-1}$ subsets. Pair the subset $F=\left\{n, a_{1}... | 448 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,416 |
1. Find the value of $\cos ^{2} 73^{\circ}+\cos ^{2} 47^{\circ}+\cos 47^{\circ} \cos 43^{\circ}$. (1987, Jiangsu Youth Mathematics Summer Camp) | $\begin{array}{l}\text { 1. Let } F=\cos ^{2} 73^{\circ}+\cos ^{2} 47^{\circ}+\cos 47^{\circ} \cdot \cos 73^{\circ}, F^{\prime} \\ \text {- } \sin ^{2} 73^{\circ}+\sin ^{2} 47^{\circ}-\sin 47^{\circ} \sin 73^{\circ} \text {. Then } F+F^{\prime}=2 \text { , } \\ \cos 120^{\circ}=\frac{3}{2}, F-F^{\prime}=2 \cos 120^{\ci... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,417 |
2. Find the sum of the coefficients of the integer power terms of $x$ in the expansion of $(\sqrt{x}+2)^{2 n+1}$. (1988, National High School Competition)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. For $(-\sqrt{x}+2)^{2 n+1}$. By the binomial theorem, the sum of the integer powers of $x$ in the expansions of $(\sqrt{x}+2)^{2 n+1}$ and $(2-\sqrt{x})^{2 n+1}$, denoted as $P(x)$, are the same. The sums of the non-integer powers are opposite in sign, so $P(x)=(\sqrt{x}+2)^{2 n+1}+(2-\sqrt{x})^{2 n+2}$. Therefore, ... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 707,418 |
4. Let $a_{k}>0(k=1,2,3, \cdots, n)$, and $a_{1}+a_{2}+\cdots$ $+a_{n}=1$. Prove: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$. (24th All-Soviet Union Mathematical Olympiad (10th grade)). | $$
\begin{array}{l}
\text { 4. Let } F=\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}}, \\
Y'=\frac{a_{2}^{2}}{a_{1}+a_{2}}+\frac{a_{3}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{1}^{2}}{a_{n}+a_{1}} .
\end{array}
$$
Then $F-H'-\frac{a_{1}^{2}-a_{2}^{2}}{a_{1}+a_{2}}+\frac{a_{2}... | F \geqslant \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 707,420 |
6. Let $E=\{1,2, \cdots, 200\}, G=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ $\subset E$, and $G$ has the following two properties:
(i) For any $1 \leqslant i \leqslant j \leqslant 100$, it always holds that $a_{i}+a_{j} \neq 201$;
(ii) $\sum_{i=1}^{130} a_{i}=10080$.
Prove: The number of odd numbers in $G$ is a mul... | 6. Divide the elements of $E$ into 100 subsets such that the sum of the two numbers in each subset is 201: $E_{1}=\{1,200\}, E_{2}=\{2,199\}, \cdots, E_{100}=\{100, 101\}$. According to property (i) of $G$, for any $1 \leqslant i \leqslant 100$, the two elements in $E_{i}$ cannot both be in $G$. Since the number of ele... | 1349380 | Number Theory | proof | Yes | Yes | cn_contest | false | 707,422 |
Example. Given $f(x)=[3 x]+[4 x]+[12 x]+[16 x]$.
(1) Find the number of positive integer values $M$ that $f(x)$ takes when $0<x \leqslant 1$.
(2) Is 97 a value of $f(x)$? | $$
\begin{array}{l}
\text { (1) } \sum n_{i}=3+4+12+16=35 . \\
\sum_{1<i<j<1}\left(n_{i}, n_{j}\right)=(3,4)+(3,12)+(3,16)+(4,12)+(4,16)+(12,16)=17. \\
\sum_{1<i<j<k<t}\left(n_{i}, n_{j}, n_{k}\right)=(3,4,12)+(3,4,16)+(3,12,16)+(4,12,16)=7. \\
\left(n_{i}, n_{2}, n_{3}, n_{!}\right)=(3,4,12,16)=1, \\
\end{array}
$$
B... | 97 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,425 |
Example 4. Solve the system of equations
$$
\left\{\begin{array}{l}
x\left(1+\frac{x}{y}\right)=4, \\
y\left(1+\frac{y}{x}\right)=9 .
\end{array}\right.
$$ | Let $x=k y$, substituting into the original system of equations, we get
$$
\left\{\begin{array}{l}
k y(1+k)=4, \\
y\left(1+\frac{1}{k}\right)=9 .
\end{array}\right.
$$
(1) $\div$ (2) yields $k^{2}=\frac{4}{9}$ or $k= \pm \frac{2}{3}$.
Substituting $k= \pm \frac{2}{3}$ into (1) and (2), the solutions to the original sy... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,426 |
Example 1. Point $P$ inside rectangle $A B C D$ is at distances of $3, 4, 5$ from $A, B, C$ respectively. Find the length of $P D$.
(1982 Shanghai Junior High School Mathematics Competition, 1987 Zhejiang Normal University Junior High School Mathematics Basic Knowledge Competition) | Connect $A C, B D$ intersecting at point $O$, and connect $P O$.
$$
\begin{array}{l}
\because A O=O C, B O=D O, \\
\therefore P O^{2}=\frac{1}{2} P A^{2} \\
+\frac{1}{2} P C^{2}-\frac{1}{4} A C^{2}, \\
P O^{2}=\frac{1}{2} P B^{2} \\
+\frac{1}{2} P D^{2}-\frac{1}{4} B D^{2} . \\
\text { Also } A C=B D, \\
\therefore P A... | 3 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,427 |
Given $D$ is a point on side $AC$ of $\triangle ABC$, $AD, DC=2+1, \angle C=45^{\circ}, \angle ADB=$ $60^{\circ}$. Prove: $AB$ is the tangent to the circumcircle of $\triangle BCD$. | Proof As shown in the figure, let $E$ be the midpoint of $AD$, and connect $BE$. We can set $AE = ED = DC = 1$, $BC = x$, $BD = y$, $BE = z$, and $AB = u$. By the median theorem, we have
$$
\begin{array}{l}
y^{2}=\frac{1}{2} x^{2}+\frac{1}{2} z^{2}-1, \\
z^{2}=\frac{1}{2} y^{2}+\frac{1}{2} u^{2}-1 .
\end{array}
$$
In ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,428 |
Example 3. On the side $BC$ of the equilateral $\triangle ABC$, take a point $D$ such that $\frac{BD}{DC}=\frac{1}{2}$. Draw $CH \perp AD$, with $H$ as the foot of the perpendicular, and connect $BH$. Prove that $\angle DBH = \angle DAB$.
---
The translation maintains the original text's formatting and line breaks. | Proof As shown in the figure, $E$ is the midpoint of $DC$, and $AE$ is connected. We can set $BD=DE=EC=a, AB=AC=b$. By the symmetry of the isosceles triangle, we can set $AD=AE=x$. By the median theorem, we get
$$
\begin{array}{l}
x^{2}=\frac{1}{2} b^{2}+\frac{1}{2} x^{2}-\frac{1}{4}(2 a)^{2}, \\
\Rightarrow x^{2}=b^{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,429 |
Example 4. In a right triangle $\triangle ABC$, the hypotenuse $BC$ is divided into $n$ equal parts, where $n$ is an odd number. The angle $\alpha$ is the angle of view from $A$ to the segment containing the midpoint of $BC$. Let the height of the right triangle be $h$, and the hypotenuse be $a$. Prove that: $\operator... | Proof As shown in the figure, let $D$ be the midpoint of $BC$, and $P_{k}$ and $P_{k+1}$ represent the two division points that include $D$. It is easy to see that $D$ is also the midpoint of $P_{\star} P_{k+1}$, $P_{k} P_{k+1}=\frac{a}{n}$, $AD=\frac{1}{2} BC=\frac{a}{2}$. Let $AP_{\star}=x, AP_{k+1}=y$.
In $\triangle... | \operatorname{tg} \alpha=\frac{4 n h}{a\left(n^{2}-1\right)} | Geometry | proof | Yes | Yes | cn_contest | false | 707,430 |
Example 5. Given the side lengths $a, b, c$ and the area $S$ of $\triangle ABC$. Prove: $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S$. (3rd IMO) | Proof: Let $m_{a}, m_{b}, m_{c}$ be the medians and $h_{a}, h_{b}, h_{c}$ be the altitudes to the sides of a triangle.
Since $m_{a} \geqslant h_{a}, m_{b} \geqslant h_{b}, m_{c} \geqslant h_{c}$,
then $a^{2} m_{a}^{2} \geqslant a^{2} h_{a}^{2}, b^{2} m_{b}^{2} \geqslant b^{2} h_{b}^{2}, c^{2} m_{c}^{2} \geqslant c^{2} ... | a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S | Inequalities | proof | Yes | Yes | cn_contest | false | 707,431 |
Example 6. In $\triangle A B C$, $M, N$ are on $A C$, $A B$ respectively, and $A M=A N, D, E$ are the midpoints of $C M, B N$ respectively, and $B D=C E$. Prove: $A B=A C$.
---
The translation maintains the original text's line breaks and format. | Prove: As shown in the figure, draw $M Q \perp A N$ at $Q$, $N P \perp A M$ at $P$, and connect $B M, C N$.
Since $A M = A N$, then $A Q = A P$, $M Q = N P$.
Assume $A B > A C$, then $B N > C M$, $B Q > C P$. Thus, $B M > C N$. By the median theorem, we have
$$
\begin{array}{l}
B D^{2} = \frac{1}{2} B C^{2} + \frac{1... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,432 |
2. Given the following two propositions:
(1) Let $a, b, c$ be complex numbers. If $a^{2}+b^{2}>c^{2}$, then $a^{2}+b^{2}-c^{2}>0$;
(2) Let $a, b, c$ be complex numbers. If $a^{2}+b^{2}-c^{2}$ $>0$, then $a^{2}+b^{2}>c^{2}$.
Then, the correct statement is ().
(A) Proposition (1) is correct, and proposition (2) is also c... | 2. (B)
Statement (1) is correct. $a^{2}+b^{2}>c^{2}$ indicates that $a^{2}+b^{2}$ and $c^{2}$ are real numbers, therefore, according to the rule of transposition, we have $a^{2}+b^{2}-c^{2}>0$.
Statement (2) is incorrect. $a^{2}+b^{2}-c^{2}>0$ only indicates that $a^{2}+b^{2}-c^{2}$ is a real number, but it does not ... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,434 |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n$ $\geqslant 1$ ), and $a_{1}=9$, the sum of its first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
(A) 5
(B) 6
(C) 7
(D) 8 | 3. (C)
From $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$ we know that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$.
Therefore, $S_{n}-n=\frac{8 \times\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,435 |
Example 5. Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=\sqrt{x+y+z+1}+5, \\
\frac{x}{2}=\frac{y}{3}=\frac{z}{4} .
\end{array}\right.
$$ | Let $\sqrt{x+y+z+1}=t$, then (1) can be transformed into
$$
i^{3}-t-6=0 \text {. }
$$
We get $t=3$ or $t=-2$ (discard $t=-2$)
F: So, $\sqrt{x}-y+z+1=3, x+y+z=8$.
From (2), we get $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{8}{9}$.
Therefore, the solution to the original system of equations is
$$
x=\... | x=\frac{16}{9}, \quad y=\frac{8}{3}, \quad z=\frac{32}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,437 |
6. In the Cartesian coordinate system, the equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$
( $a, b$ are two different positive numbers) represents the curve ( ).
(A) Triangle
(B) Square
(C) Non-square rectangle
(D) Non-square rhombus | 6. (D)
Method one: The lines $l_{1}: x-y=0$ and $l_{2}: x+y=0$ divide the plane $X O Y$ into 4 regions:
( I) $\left\{\begin{array}{l}x+y \geqslant 0, \\ x-y \geqslant 0,\end{array}\right.$
( I ) $\left\{\begin{array}{l}x+y \geqslant 0, \\ x-y \leqslant 0,\end{array}\right.$
(1) $\left\{\begin{array}{l}x+y \leqslant 0,... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,439 |
2. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=0$. Then $\cos (x+2 y)=$ $\qquad$ . | 2.1.
Given that $x^{3}+\sin x=2 a=(-2 y)^{3}+\sin (-2 y)$. Let $f(t)=t^{3}+\sin t$, then $f(x)=f(-2 y)$. The function $f(t)=t^{3}+\sin t$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, $x=-2 y, x + 2 y=0$. Hence, $\cos (x+2 y)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,441 |
3. Given the point sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-3)^{2}+(y-4)^{2} \leqslant\left(\frac{5}{2}\right)^{2}\right.\right\}, \\
B=\left\{(x, y) \left\lvert\,(x-4)^{2}+(y-5)^{2}>\left(\frac{5}{2}\right)^{2}\right.\right\} .
\end{array}
$$
Then the number of integer points (i.e., points with both c... | 3. 7.
As shown in the figure, circles $E$ and $F$ intersect at points $M$ and $N$. The entire figure is symmetric about the line connecting the centers $E F$. Among them, $A \cap B$ is a crescent shape $S$ near the origin $O$ on the lower left. The x-coordinate of the points in $S$ can be 1, 2, 3, 4, and the y-coordin... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,442 |
4. Let $0<\theta<\pi$, then the maximum value of $\sin \frac{\theta}{2}(1+\cos \theta)$ is $\qquad$ . | $$
\begin{array}{l}
\text { 4. } \frac{4}{9} \sqrt{3} \text {. } \\
\sin \frac{\theta}{2}(1+\cos \theta)=2 \sin \frac{\theta}{2} \cos ^{2} \frac{\theta}{2} \\
=\sqrt{2} \cdot \sqrt{2 \sin ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}} \\
\leqslant \sqrt{2} \cdot \sqrt{\left[\frac{2 \sin ^{... | \frac{4 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,443 |
6. Given 95 numbers $a_{1}, a_{2}, \cdots, a_{95}$, each of which can only take one of the two values +1 or -1. Then, the minimum positive value of the sum of their pairwise products $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{81} a_{23}$ is $\qquad$ . | 6.13.
Let $N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{98} a_{95}$.
Suppose $a_{1}, a_{2}, \cdots, a_{95}$ contain $m \uparrow+1, n \uparrow-1$, then $m+n=95$.
Multiplying (1) by 2 and adding $a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}=95$, we get
$$
\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95 \text{. }
$$
Also, $a_{1}+a_{2... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,444 |
One, (This section is 25 points) In the quadratic equation of $x$, $x^{2} + z_{1} x + z_{2} + m = 0$, $z_{1}, z_{2}, n$ are complex numbers, and $x_{1}^{2} - 4 z_{2} = 1 f + 20 i$.
Assume the two roots of this equation, $\alpha, \beta$, satisfy $|\alpha - \beta| = 2 \sqrt{7}$.
Find the maximum and minimum values of $|... | $$
\begin{array}{l}
\left\{\begin{array}{l}
\alpha+\beta=-z_{1}, \\
\alpha \beta=z_{2}+m .
\end{array}\right. \\
\because(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m, \\
\therefore|\alpha-\beta|^{2}=\mid 4 m- \\
\left(z_{1}^{2}-4 z_{2}\right) \mid=28 . \\
\quad \therefore\left|m-\frac{1}{4... | 7-\sqrt{41} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,445 |
II. (This question is worth 25 points) Arrange all positive integers that are coprime with 105 in ascending order. Find the 1000th term of this sequence. | $$
\begin{array}{r}
\text { 2. Let } S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S, \text { and } 3 \mid \\
x\}, A_{2}=\{x \mid x \in S, 45 \mid x\}, A_{3}=\{x \mid x \in S, \text { and } 7 \mid
\end{array}
$$
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=| S \mid-\left(... | 2186 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,446 |
Example 6. Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+\sqrt{(x+2)(y+3)}=34, \\
(x+2)^{2}+(y+3)^{2}=741-(x+2)(y+3) .
\end{array}\right.
$$ | Let $x+2=s, y+3=t$, the original system of equations can be transformed into a symmetric system of equations
$$
\left\{\begin{array}{l}
s+t+\sqrt{s t}=39, \\
s^{2}+t^{2}=741-s t .
\end{array}\right.
$$
Set $s+t=u, s t=v$, we get
$$
\left\{\begin{array}{l}
u+\sqrt{v}=39, \\
u^{2}-v=741 .
\end{array}\right.
$$
(2) $\div... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,447 |
Three. (This question is worth 35 points) As shown in the figure, let the circumcircle $O$ of $\triangle ABC$ have a radius $R$, the incenter be $I$, $\angle B=60^{\circ}$, $\angle A<\angle C$, and the external angle bisector of $\angle A$ intersects the circle $O$ at $E$. Prove:
$$
\begin{array}{l}
\text { (1) } I O=A... | Three, Proof Method One (1) As shown in the figure, let the ray $B I$ intersect the circle $O$ at $M$, and connect $M A, M C$. It is easy to know that $M$ is the midpoint of $\overparen{A C}$, and $M A = M C$. Connect $O A, O C, O M$. It is easy to know that $\angle A O M = \angle B = 60^{\circ}$. Therefore, $\triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,448 |
Four. (This question is worth 35 points) Given a set of points on a plane $P=\left\{P_{1}\right.$, $\left.P_{2}, \cdots, P_{1004}\right\}$, where no three points in $P$ are collinear. Divide all the points in $P$ arbitrarily into 83 groups, such that each group has at least 3 points, and each point belongs to exactly o... | (1) Let $m(G)=m_{0}, G$ be obtained by the partition $X_{1}, X_{2}, \cdots, X_{\text {в3 }}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots$, 83.
$$
\begin{array}{l}
\text { Let }\left|X_{i}\right|=x_{i}(i=1,2, \cdots, 83) \text {, then } x_{1}+x_{2}+\cdots \\
+x_{83}=1994, \text { and } \\
m_... | 168544 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,449 |
On a smooth table, there are three wooden balls with radii of $1, 2, 4$ respectively, all externally tangent to each other. Additionally, there is another small wooden ball on the table with a radius less than 1, which is tangent to the table and externally tangent to all three balls. Find the radius of this small wood... | It is known that the distance between the two points of tangency of two spheres with radii $r_{1}, r_{2}$, which are tangent to the table and externally tangent to each other, is $2 \sqrt{r_{1} r_{2}}$. Let the radius of the small wooden ball be $x$, and the points of tangency of the wooden balls with radii $1, 2, 4, x... | x=2-\frac{2}{7} \sqrt{21} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,450 |
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
保留源文本的换行和格式，这里的翻译结果应该是：
Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?
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。
Throug... | Let the regular dodecagon be $A_{1} A_{2} \cdots A_{12}$.
First, prove that the 4 diagonals $A_{1} A_{5}, A_{2} A_{6}, A_{3} A_{8}, A_{4} A_{11}$ intersect at a non-central point inside the regular dodecagon.
Since $\overparen{A_{5} A_{8}}=\overparen{A_{8} A_{11}}$, $\overparen{A_{11} A_{1}}=\overparen{A_{1} A_{3}}$,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,451 |
Given 9 points in space, where any 4 points are not coplanar. Connect several line segments among these 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?
Translate the above text into English, please keep the original text's line breaks and format, and output ... | First, prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the maximum, say $k$ edges $A_{1} A_{n}, A_{1} A_... | 27 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,452 |
Let $P(z)$ denote the point corresponding to the complex number $z$ on the complex plane, and let the complex number $\alpha = p + i q$ (where $p, q$ are real numbers). Given that the points $P(z_1), P(z_2), P(z_3), P(z_4), P(z_5)$ are the vertices of a convex pentagon $Q$, and that the origin and the points $P(\alpha ... | Proof of conclusion (*): Suppose
angles $a_{1}, \alpha_{2}, a_{3}, a_{4}, a_{5}$ are all greater than $\frac{3 \pi}{10}$, then $A=\sum_{j=1}^{5} a_{j}>\frac{3}{2} \pi$.
By the sine rule of triangles, we have
$\frac{\sin \alpha_{1}}{\sin \beta_{2}} \cdot \frac{\sin \alpha_{2}}{\sin \beta_{3}} \cdot \frac{\sin \alpha_{3... | p + q \tan \frac{\pi}{5} \leqslant 1 | Geometry | proof | Yes | Yes | cn_contest | false | 707,453 |
14. (Question 1 of the 9th Minghadi, provided by Guo Boxiang)
Given that $\triangle D E F$ is an inscribed similar triangle of $\triangle A B C$, where $D, E, F$ are on $B C, C A, A B$ respectively, and $\angle D=\angle B, \angle E=\angle C$.
(1) Construct an inscribed similar $\triangle D E F$ of $\triangle A B C$.
(... | (1) On $AB, AS$, take points $P', E'$, and point $D'$, such that $\angle D' F' E' = \angle A, \angle D' E' I'' = \angle C$.
Connect $AD'$, and extend it to intersect $BC$ at $D$. Through point $D$, draw lines parallel to $D' F'$ and $D' E'$, intersecting $AB$ and $AC$ at $F$ and $E$ respectively. Thus,
$$
\frac{A F'}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,454 |
15. The 9th problem of human measurement, provided by Guo Boxiang)
Given $F, R$ are two points on line $A X$, $Q, S$ are points on ray $B Y$, and satisfy $\frac{A P}{B Q}=\frac{A R}{B S}=\lambda . M, N, T$ are points on segments $A B, P Q, R S$ respectively, and satisfy $\frac{A M}{M B}=\frac{P N}{N Q}=\frac{R T}{T S}... | Solve: As shown in the figure, set up a Cartesian coordinate system, and use $x_{D}, y_{D}$ to represent the abscissa and ordinate of point $D$ respectively.
According to the problem, we can set
$$
\frac{A R}{A P}=\frac{B S}{B Q}=u \text {, }
$$
Let $x_{p}=x_{A}+a_{1}$,
$$
y_{p}=y_{A}+b_{1} \text {, }
$$
$x_{Q}=x_{B}+... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,455 |
16. (Question 3 of Decision Test 9, provided by Guo Boxiang)
Given two non-intersecting circles $S_{1}, S_{2}$ with unequal radii. $R$ is the intersection point of the internal common tangents of $S_{1}, S_{2}$, and $S$ is the intersection point of the external common tangents of $S_{1}, S_{2}$. Take a point $C$ on th... | Proof: Let the centers of circles $S_{1}, S_{2}$ be $O_{1}, O_{2}$, and their radii be $r_{1}, r_{2}$, respectively.
As shown in Figure A, it is easy to see that $R, S$ are both on the line segment $O_{1} O_{2}$, and $\frac{R O_{1}}{R O_{2}}=\frac{S O_{1}}{S O_{2}}=\frac{r_{1}}{r_{2}}$. Therefore, the locus of points ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,456 |
17. (1st Decision Test, Question 1, American Mathematical Monthly, page 117, 1947)
Given $A_{1} A_{2} \cdots A_{2 n+1}$ is a regular $(2 n+1)$-sided polygon in the plane ($n \in \cdots$), and $O$ is any point inside this regular $(2 n+1)$-sided polygon. Prove that there must exist an $\angle A_{i} O A_{j}$ such that $... | Prove that in fact, this proposition holds for any regular $k$-sided polygon for any positive integer $k \geqslant 3$. Below is the proof for a regular $k$-sided polygon.
Its vertices are sequentially $A_{1}, A_{2}, \cdots A_{4}$.
$A_{i}$ is the vertex closest to point $O$:
a) Draw the line $A_{i} O$ and extend it to ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,457 |
Example 7. Given that $a \cdot b, c$ are distinct real numbers, and $a b c \neq 0$, solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{b^{3}}-\frac{y}{b^{2}}+\frac{z}{b}=1, \\
\frac{x}{c^{3}}-\frac{y}{c^{2}}+\frac{z}{c}=1, \\
\frac{x}{a^{3}}-\frac{y}{a^{2}}+\frac{z}{a}=1 .
\end{array}\right.
$$ | From the form of the original system of equations and the definition of the roots, we know that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are the three roots of the equation $x t^{3}-y t^{2}+z t=1$. Therefore,
$$
x t^{3}-y t^{2}+z t-1
=x\left(t-\frac{1}{a}\right)\left(t-\frac{1}{b}\right)\left(t-\frac{1}{c}\right),
$$
w... | \left\{\begin{array}{l}
x=a b c, \\
y=a b+b c+c a, \\
z=a+b+c
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,458 |
18. (Test 5, 3rd time, *American Mathematical Monthly* 1941, pp. 275-277)
Given $\triangle ABC$, construct three similar rectangles $BCA_1A_2$, $CAB_1B_2$, and $ABC_1C_2$ outside $\triangle ABC$, such that $\frac{CA_1}{BC} = \frac{AB_1}{C} = \frac{BC_1}{AB} = k$. Points $A^*$, $B^*$, and $C^*$ lie on segments $A_1A_2$... | Prove that, as shown in the figure, according to the conditions of the problem, in the right triangles $A B_{1} B^{*}, C A_{1} A^{*}$, $B C_{1} C^{*}$, we have $\frac{B_{1} B^{*}}{A B_{1}} = \frac{A_{1} A^{*}}{C A_{1}} = \frac{C_{1} C^{*}}{B C_{1}} = \frac{\lambda}{k}$. Therefore, $\angle B_{1} A B^{\circ} = \angle A_{... | \frac{[\lambda + k (\operatorname{ctg} A + \operatorname{ctg} B + \operatorname{ctg} C)]^2}{\lambda^2 + k^2} | Geometry | proof | Yes | Yes | cn_contest | false | 707,459 |
1. If $a, b, c$ are distinct real numbers, and the coordinates of three points are $A(a+b, c), B(b+c, a), C(c+$ $a, b)$, then the positional relationship of these three points is ( ).
(A) on the same straight line
(B) form a right triangle
(C) form an obtuse triangle
(D) form an equilateral triangle | ,- 1. (A).
$$
\begin{aligned}
\because A B & =\sqrt{(a-c)^{2}+(c-a)^{2}}=\sqrt{2}|a-c|, \\
A C & =\sqrt{(b-c)^{2}+(c-b)^{2}}=\sqrt{2}|b-c|, \\
B C & =\sqrt{(b-a)^{2}+(a-b)^{2}}=\sqrt{2}|a-b|,
\end{aligned}
$$
Without loss of generality, assume $a>b>c$, then
$$
A B=\sqrt{2}(a-c), A C=\sqrt{2}(b-c),
$$
$B C=\sqrt{2}(a-b... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,460 |
2. The number of pairs of positive integer solutions $(x, y)$ for the equation $x^{3}-y^{3}+x^{2} y-x y^{2}=32$ is ( ).
(A) 0
(B) 1
(C) 2
(D) greater than or equal to 3 | 2. (B).
$$
\begin{array}{l}
\because x^{3}-y^{3}+x^{2} y-x y^{2} \\
\quad=(x-y)(x+y)^{2}=2^{5},
\end{array}
$$
Since $x, y$ are positive integers, we have
$$
\left\{\begin{array}{l}
x+y=2^{n}, \\
x-y=2^{m}, \\
2 n+m=5 .
\end{array} \quad(n, m \text { are non-negative integers) }\right.
$$
Thus,
$$
\left\{\begin{array... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,461 |
3. The maximum value of the shortest side of a cyclic quadrilateral inscribed in a circle with radius 1 is ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) 1
(C) $\sqrt{2}$
(D) 2 | 3. (C).
As shown in the figure, let $AB$ be the shortest side, then $\overparen{AB} \leqslant 90^{\circ}, \angle ADB = \alpha \leqslant 45^{\circ}$.
$$
\begin{array}{c}
\therefore AB = 2R \sin ADB \leqslant \\
2 \sin 45^{\circ} = \sqrt{2},
\end{array}
$$
Thus, the maximum value of $AB$ is $\sqrt{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,462 |
5. If $\triangle A B C$ is a right triangle with integer sides, $B C$ is the hypotenuse and $2 A C=B C+A B$, then which of the following numbers could be a side length of this triangle? ( ).
(A) 1993
(B) 1994
(C) 1995
(D) 1997 | 5. (C).
From the given conditions and the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
B C^{2}=A C^{2}+A B^{2}, \\
2 A C=B C+A B .
\end{array}\right.
$$
Solving these equations, we get
$$
B C: A C \text { : } A B=5: 4: 3 \text {. }
$$
$\therefore$ The sides of $\triangle A B C$ are multiples of 5, 4, and 3... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,464 |
6. $P$ is a point inside $\triangle A B C$, and three equal-length line segments $D E, F G, H I$ are parallel to sides $A B, B C$, and $C A$ respectively, and all pass through point $P$. Given that $A B=12, B C=8$, $C A=6$. Then $A I: I F: F B$ equals $(\quad)$.
(A) $1: 2: 3$
(B) $1: 4: 3$
(C) $1: 5: 3$
(D) $1: 3: 4$ | 6. (C).
Let $A I=x, I F=y, F B=z$. From $D E / / A B, F G / /$ $B C, I H / / A C$ and $A I P D$ and $F B E P$ are parallelograms, we get $D P=I A=x, P E=F B=z$. Therefore, $D E=F G=H I$ $=x+z$.
Also, from $F G / / B C, H I / / C A$, we have
$$
\frac{F G}{B C}=\frac{A F}{A B}, \frac{H I}{C A}=\frac{B I}{A B},
$$
which... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 707,465 |
The bright image of $\frac{h}{x}$ is a time-scale graph, one of its axes of symmetry is the graph of which of the following direct proportion functions? ().
(A) $y=k x$
(B) $y=-k x$
(C) $y=|k| x$
(D) $y=-\frac{k}{|k|} x$ | 7. (D).
When $k>0$, the axis of symmetry is $y=-x$; when $k<0$, the axis of symmetry is $y=x$. Therefore, it is $y=-\frac{k}{|k|} x$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,466 |
1. The equation $x^{2}+m x+1=0$ and the equation $x^{2}-x$ $-m=0$ have one common root, then $m=$ $\qquad$ | Two equations have a common root $x=-1$. Substituting this root into either equation, we get $m=2$.
| 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,468 |
Example 8. Solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{x(1-y)}+\sqrt{y(1-x)}=\frac{1}{2}, \\
\sqrt{x y}+\sqrt{(1-x)(1-y)}=\frac{\sqrt{3}}{2} .
\end{array}\right.
$$ | Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta(0 \leqslant \alpha, \beta \leqslant \frac{\pi}{2})$, then the original system of equations can be transformed into
$$
\left\{\begin{array}{l}
\sin \alpha \cos \beta+\cos \alpha \sin \beta=\frac{1}{2}, \\
\sin \alpha \sin \beta+\cos \alpha \cos \beta=\frac{\sqrt{3}}{2} .
\end{a... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,469 |
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