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Example 1. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{a+\lambda}+\frac{y}{b+\lambda}+\frac{z}{c+\lambda}=1, \\
\frac{x}{a+\mu}+\frac{y}{b+\mu}+\frac{z}{c+\mu}=1, \\
\frac{x}{a+\gamma}+\frac{y}{b+\gamma}+\frac{z}{c+\gamma}=1 .
\end{array}\right.
$$ | Given the system of equations, we know that $\lambda, \mu, \gamma$ are the roots of the equation in the variable $\theta$:
$$
\frac{x}{\theta+a}+\frac{y}{\theta+b}+\frac{z}{\theta+c}=1
$$
Let $\theta=t-a$, then (1) transforms into a cubic equation in the variable $t$:
$$
t^{3}+\cdots-(a-b)(a-c) x=0. \quad \text { (*) ... | \begin{array}{l}
x=\frac{(a+\lambda)(a+\mu)(a+\gamma)}{(a-b)(a-c)}, \\
y=\frac{(b+\lambda)(b+\mu)(b+\gamma)}{(b-a)(b-c)}, \\
z=\frac{(c+\lambda)(c+\mu)(c+\gamma)}{(c-a)(c-b)}
\end{array} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,320 |
Example 6. (1984 Nanjing City Mathematics Competition Question) On the coordinate plane, points whose both horizontal and vertical coordinates are integers are called integer points. Prove: The distance from the point $(\sqrt{2}, \sqrt{3})$ to any integer point is not equal. | Prove that if there exist two distinct integer points $(a, b)$ and $(c, b)$ equidistant from the point $(\sqrt{2}, \sqrt{3})$, i.e.,
$$
\begin{array}{l}
\sqrt{(a-\sqrt{2})^{2}+(b-\sqrt{3})^{2}} \\
=\sqrt{(c-\sqrt{2})^{2}+(d-\sqrt{3})^{2}} .
\end{array}
$$
Squaring both sides and simplifying, we get
$$
\begin{array}{l}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,321 |
Example 2. Try to solve the following system of linear equations:
(1)
$$
\begin{array}{l}
\overline{\lambda_{2}+x_{1}}+\frac{x_{2}}{\lambda_{2}+a_{2}}+\cdots+\frac{x_{\mathrm{a}}}{\lambda_{1}+a_{0}}=1, \\
\frac{x_{1}}{\lambda_{2}+a_{1}}+\frac{x_{0}}{\lambda_{2}+a_{2}}+\cdots+\frac{x_{\mathrm{a}}}{\lambda_{2}+a_{0}}=1, ... | (1) According to the structural characteristics of the given system of equations, construct an auxiliary equation about the variable $\theta$
$$
\frac{x_{1}}{\theta+a_{1}}+\frac{x_{2}}{\theta+a_{2}}+\cdots+\frac{x_{\mathrm{n}}}{\theta+a_{\mathrm{n}}}=1 \text {. (*) }
$$
Obviously, from the given system of equations, w... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,322 |
Example 3. Solve the system of equations
$$
\left\{\begin{array}{l}
x \sin a+y \sin 2 a+z \sin 3 a=\sin 4 a, \\
x \sin b+y \sin 2 b+z \sin 3 b=\sin 4 b, \\
x \sin c+y \sin 2 c+z \sin 3 c=\sin 4 c .
\end{array}\right.
$$ | By applying the double, triple, and quadruple angle formulas for $\sin \alpha$, equation (1) can be transformed into
$$
\begin{array}{l}
8 \cos ^{3} a-4 z \cos ^{2} a-(2 y+4) \cos a+z-x \\
=0 .
\end{array}
$$
Let $t=\cos a$, and divide both sides of the equation by 8, we get the equation in terms of $t$:
$$
t^{3}-\fra... | \left\{\begin{aligned}
x= & 2(\cos a+\cos b+\cos c) \\
& +8 \cos a \cos b \cos c, \\
y= & -2-4(\cos a \cos b+\cos a \cos c \\
& +\cos b \cos c), \\
z= & 2(\cos a+\cos b+\cos c) .
\end{aligned}\right.
} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,323 |
Example 4 - Let $\mu^{2}, \gamma^{2}, \rho^{2}$ be pairwise distinct, and the parameters $\mu, \gamma, \rho, b, c$ be chosen such that the following system of equations has a solution:
$$
\begin{array}{l}
x^{2}>0, y^{2}>0, z^{2}>0, \\
\frac{x^{2}}{\mu^{2}}+\frac{y^{2}}{\mu^{2}-b^{2}}+-\frac{z^{2}}{\mu^{2}-c^{2}}=1, \\
... | Considering the structure characteristics of the known system of equations, we can construct an auxiliary equation about the variable $t$:
$$
\frac{x^{2}}{t}+\frac{y^{2}}{t-b^{2}}+\frac{z^{2}}{t-c^{2}}=1.
$$
Obviously, from the known system of equations, we know that $t=\mu^{2}, \gamma^{2}, \rho^{2}$ are the three roo... | x^{2}+y^{2}+z^{2}=\mu^{2}+\gamma^{2}+\rho^{2}-b^{2}-c^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,324 |
Example 1. Let $a x+b y=c\left(a, b, c \in R^{+}, \boldsymbol{x}\right.$ , $\left.y \in \overline{R^{-}}\right)$, find the extremum of $f(x, y)=m \sqrt{x}+n \sqrt{y}(m$, $n>0)$. | Consider the points $A(\sqrt{a x},-\sqrt{b y})$, $B\left(\frac{n}{\sqrt{b}}, \frac{m}{\sqrt{a}}\right)$, and $\angle A O B=\theta$, then
$$
\begin{array}{l}
f(x, y)=O A \cdot O B \sin \theta \\
=\sqrt{(0-\sqrt{a x})^{2}+(0+\sqrt{b y})^{2}} \\
\cdot \sqrt{\left(0-\frac{n}{\sqrt{b}}\right)^{2}+\left(0-\frac{m}{\sqrt{a}}\... | f_{\text{max}}(x, y)=\sqrt{\frac{c}{a b}\left(a n^{2}+b m^{2}\right)}, \quad f_{\text{min}}(x, y)=\sqrt{c} k | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,326 |
Example 1. In space, there are three line segments of the same length. Try to prove that their projections are equal. (11th All-Russian Mathematics Competition Problem) | We use the direct proof method. If three line segments are parallel, then their projections on any plane are equal. Now, suppose at least two are not parallel, and translate the three line segments so that they have a common endpoint, denoted as $D A$, $D B$, $D C$. If points $A, B, C$ are collinear, then the plane pas... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,328 |
Example 2. For any tetrahedron \(ABCD\), there always exists a vertex such that the three edges meeting at this vertex can form a triangle. (10th IMO Problem) | Let's use proof by contradiction. Suppose $AB$ is the largest chord, then one of the points $A$, $B$ must meet the requirement. Otherwise, we have $AB \geqslant AC + AD, \quad AB \geqslant BC + BD$, but $2AB \geqslant (AC + BC) + (AD + BD)$ $> AB + AB = 2AB$. This is impossible. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,329 |
Example 3. Prove: There exists an integer-coefficient polynomial $P(x)$, such that for any $x$ in the interval $\left[\frac{1}{10}, \frac{9}{10}\right]$, it satisfies $\left|P(x)-\frac{1}{2}\right|<\frac{1}{100}$. | We use the construction method to prove. Consider the polynomial $P_{\mathrm{g}}(x) = \frac{1}{2}\left[(2 x-1)^{\mathrm{n}}+1\right], n \in N$. The coefficients of $P_{\mathrm{n}}(x)$ are indeed integers. When $x \in [0.1, 0.9]$,
$-0.8\frac{1 \mathrm{~g} 0.02}{\lg 0.8} = 27.8$. Taking $P(x) = P_{n}(x), n \geqslant 28$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,330 |
Example 4 - Whether a quadrilateral with one pair of equal diagonals and one pair of equal opposite sides is always a parallelogram.
(79 National Competition Question) | Not, just construct a counterexample: Construct an equilateral $\triangle A B C$, take a point $D$ on $B C$, such that $B D>D C$, in $\angle A D C$ draw ray $D E$, such that $\angle A D E=\angle D A C$, and take $D E=A C$, connect $A E$, then quadrilateral $A B D E$ is the one sought. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,331 |
Example 5 - Does there exist a natural number $n$, the sum of whose digits is $1000$, such that the sum of the digits of $n^{2}$ is $1000^{2}$. | We use mathematical induction to prove a conclusion: for any natural number $m$, there exists a natural number $n$ (consisting of 1 and 0), such that $S(n)=m$, and $S\left(n^{2}\right)=m^{2}$ (where $S(k)$ denotes the sum of the digits of $k$). When $n=1$, $n=1$ meets the requirement. Assume for $m$ there exists $n$ su... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,333 |
Example 8. (17th All-Union School Olympiad Problem) Prove that between the squares of two consecutive natural numbers there do not exist four natural numbers $a<b<c<d$ such that $a d=b c$.
保留了原文的换行和格式。 | Proof Assume there exists such a natural number, making $n^{2}q$, thus $p \geqslant q+1$. Also, $a, b$ are both multiples of $q$, and $b>a$, so $b \geqslant a+q$. Therefore, $\frac{p}{q}=\frac{d}{b}$ $\leqslant \frac{d}{a+q}<\frac{(n+1)^{2}}{n^{2}+q}$. Consequently, $1+\frac{1}{q}<\frac{(n+1)^{2}}{n^{2}+q}$. Multiplyin... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,337 |
Example 9・(US Olympiad Problem) If $2 a^{2}<5 b$, prove: the roots of the equation $x^{5}+a x^{4}+b x^{3}+c x^{2}+d x+e=0$ cannot all be real.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Example 9・(US Olympiad Probl... | Proof Assume all roots of the equation are real numbers, and let the left side of the equation be $(x+p)\left(x^{2}+q x+q\right)\left(x^{2}+s x+t\right)$, where $p, q, r, s, t$ are real numbers, and $q^{2}-4 r \geqslant 0$, $s^{2}-4 t \geq 0$. By comparing coefficients, we get
$$
a=p+q+s, b=p q+p s+q s+r+t .
$$
Substi... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,338 |
Example 10. (19th All-Soviet Union Middle School Olympiad Problem) A quadratic trinomial $a x^{2}+b x+c$, where $a>100$. How many different integer points can there be at which the absolute value of its value does not exceed 50? | Assume there are three different integer points satisfying the given condition, then there must be two points located on the same side of the axis of symmetry $x=-\frac{b}{2 a}$ of the quadratic function $y=a x^{2} +b x+c$, or one point is located on this axis of symmetry. Without loss of generality, assume
$-\frac{b}{... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,339 |
Column 1. A tour group is selecting visit locations from $A, B, C, D, E$ under the following constraints:
(1) If visiting $A$, must also visit $B$,
(2) At least one of $D, E$ must be visited,
(3) Only one of $B, C$ can be visited,
(4) Both $C, D$ must be visited or neither,
(5) If visiting $E$, both $A, D$ must be visi... | 堔 We adopt the貑 deduction method.
(1)
(4)
(2)
$1^{\circ}$ If going to $A \Rightarrow$ must go to $B \Rightarrow$ do not go to $C \Rightarrow$ do not go to $D \Rightarrow$ must not go to $A$. Going to $A, D$. This leads to the same contradiction as $1^{\circ}$, so do not go to $B$.
(5)
$3^{\circ}$ If going to $E \Righta... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,340 |
Example 2. A woman and her brother, son, and daughter are all chess players. The twin of the worst player (also one of the four players) and the best player are of opposite sexes. The worst player and the best player are of the same age. Who is the best player? | If this woman is the worst chess player, then her brother must be her twin brother, and her daughter must be the best chess player, but the woman and her daughter cannot be the same age, so the assumption is not true.
If the brother is the worst chess player, then the woman must be his twin sister, and the best chess ... | the\ son\ is\ the\ worst\ chess\ player | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,341 |
5. Five students $A, B, C, \ldots D, E$ participate in a competition. Someone guessed the result of the competition to be $A B C D E$, but did not guess any of the ranks correctly, nor did they guess any student's rank immediately following another's.
Another person guessed the competition result to be $D A E C B$, an... | (Answer: $E D A C B$. ) | E D A C B | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,343 |
Example 2. On a flat ground, there are $n \uparrow$ people, and the distance between each person is unique. Each person holds a water gun, and when a signal is given, each person shoots the person closest to them.
When $n$ is odd, prove that at least one person will remain dry.
When $n$ is even, is this conclusion sti... | Prove that when $n$ is odd, let $n=2m-1$. Apply mathematical induction on $m$.
(i) When $m=1$, $n=1$, there is only one person on the plain, and the person's body is dry, so the conclusion holds.
(ii) Assume the proposition holds for $m$, now consider the case for $m+1$, i.e., $n=2(m+1)-1=2m+1$.
Among these $2m+1$ peop... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,345 |
Example 4. Let $S$ be a finite set of points in the plane (with the number of points $\geqslant 5$), some of which are colored red and the rest are colored blue.
Assume that no three or more points of the same color are collinear.
Prove that there exists a triangle such that
(1) its three vertices are of the same colo... | Proof: For any five points colored red or blue, there must be three points of the same color, thus forming a triangle with three vertices of the same color, and conclusion (1) holds.
Now consider the triangle with vertices of the same color. If conclusion (2) does not hold, then each edge of every triangle with vertic... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,347 |
Example 5. Find the largest integer $A$, such that for any permutation of all natural numbers from 1 to 100, there are 10 consecutive numbers whose sum is greater than or equal to $A$.
[22nd Polish Mathematical Olympiad] | This problem, although it is to find the maximum integer $A$, is actually an existence problem, i.e., there exists a maximum integer $A$ such that the sum of 10 consecutive numbers is not less than $A$.
Solution: Let $T=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ be a permutation of the natural numbers from 1 to 100.
... | 505 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,348 |
Example 6. There are $n$ boys and $m$ girls $(n, m \geqslant 2)$, not all boys know all girls.
Prove: Among them, there must be two boys and two girls, where each boy knows exactly one of the girls, and each of these girls knows exactly one of the boys. | The set of boys and the set of girls are both non-empty sets, so there must be a girl who knows the most boys.
Let $a_{1}$ be the girl who knows the most boys.
By the problem statement, each girl does not know all $n$ boys, so $a_{1}$ must not know at least one boy, let's say $a_{1}$ does not know boy $b_{1}$.
Again, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,349 |
Example 8. On a plane, there are finitely many circles, assuming the total area they cover is 1 (they may intersect each other). The area covered is no less than $\frac{1}{9}$.
Put the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
---
Ex... | Proof: Since there are only a finite number of circles, there must exist a circle with the largest diameter. Let its center be $O_{1}$, radius be $r_{1}$, and area be $A_{1}$. Circles intersecting with circle $O_{1}$ must lie within a circle centered at $\mathrm{O}_{2}$ with a radius of $3 r_{1}$. Therefore, the total ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,351 |
Example 3. Four singers, A, B, C, and D, participated in a competition, and one of them won. Someone interviewed these four singers.
A said: "I won."
B said: "A did not win, and C did not win either."
C said: "It's either A or B who won."
D said: "It's B who won."
Among the four singers' statements, two are correct. Pl... | \begin{tabular}{|l|l|l|l|l|}
\hline True $(1)$ or False $(0)$ & Jia & Yi & Bing & Ding's statement \\
\hline Jia, Jia wins & 1 & 0 & 0 & 0 \\
\hline Yi: Jia doesn't win, Bing doesn't win & 0 & 1 & 0 & 1 \\
\hline Bing, Jia wins or Yi wins & 1 & 1 & 0 & 0 \\
\hline Ding: Yi wins & 0 & 1 & 0 & 0 \\
\hline
\end{tabular}
... | Jia | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,352 |
Inference
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof Assume there are only a finite number of prime numbers. Without loss of generality, assume they are given as
$$
F_{0}, F_{1}, \cdots, F_{n}
$$
the complete list of prime factors. Since Fermat numbers are infinite, there must exist $F_{n+1}$. According to the theorem mentioned above, $F_{n+1}$ has prime factors d... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,355 |
Lemma Let $\vec{a}, \vec{b}$ be arbitrary vectors, then
$$
\begin{array}{l}
(\vec{a} \times \vec{b}) \times \vec{a} \\
=(\vec{a})^{2} \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a}
\end{array}
$$ | Prove that if $\vec{a}$ and $\vec{b}$ are $\overrightarrow{0}$, then it is obvious. Suppose $\vec{a}$ and $\vec{b}$ are not $\overrightarrow{0}$, and $\vec{a} \times \vec{b} = \overrightarrow{0}$, then $\vec{a} = \lambda \vec{b}$ $(\lambda \in \mathbb{R})$, substituting into (1) shows that the scalar is also $\overrigh... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,357 |
Example 1. Given $x, y, z \in R^{+}$, and satisfying
$$
\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \frac{y^{2}}{3}+z^{2}=9, z^{2}+z x+x^{2} \\
=1 \text {, find } x y+2 y z+3 z x \text { . }
\end{array}
$$ | Solve: The original system of equations can be transformed into $x^{2}+\left(\sqrt{\frac{y}{3}}\right)^{2}$ $-2 x\left(-\frac{y}{3}\right) \cos 150^{\circ}=5^{2},\left(-\frac{y}{\sqrt{3}}\right)^{2}+z^{2}$ $=3^{2}, z^{2}+x^{2}-2 z x \cos 120^{\circ}=4^{2}$ (inspired by the cosine rule). From this, we can construct $\tr... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,358 |
In 1988, the Chinese Junior High School Mathematics League had the following problem: If natural numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, what is the maximum value of $x_{5}$? | This problem is novel and unique, and some students find it difficult to start. Below, I will first discuss the solution to this problem.
Algorithm 1: Given the symmetry of the equation, we can assume without loss of generality that \( x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5} \), and then ... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,360 |
Example 1. As shown in the figure, $P, Q, R$ trisect the perimeter of $\triangle ABC$, and $P, Q \in AB$. Prove that $S_{\triangle PQR} > \frac{2}{9} S_{\triangle ABC}$. (88th National High School Mathematics Competition Question) | Solve as shown in the figure, divide each side of $\triangle ABC$ into three equal parts, dividing $\triangle ABC$ into 9 congruent smaller triangles. It is clear that (1) $PQ = \frac{1}{3} C_{\triangle ABC} = C_{\triangle B_1C_1} > 2AC_1 = \frac{2}{3} AB$, therefore, $P \in [AC_1), Q \in [BC_2)$.
(2) $R \notin [AB_1]$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,364 |
Example 2. As shown in the figure, $P \in BC$, $PE \| AB$ intersects $AC$ at $E$, $PF \| AC$ intersects $AB$ at $F$, and let $S_{\triangle ABC}=1$.
Prove: Among $S_{\triangle BPF}$, $S_{\triangle CPE}$, and $S_{\triangle EPF}$, at least one is not less than $\frac{4}{9}$.(84 National High School Mathematics Competitio... | Prove as shown in the figure, continuing from the previous example, when $P \in B P_{5}$, $S_{\triangle \mathrm{PCK}} \geqslant S_{\triangle \mathrm{P}_{4} \mathrm{C}_{P_{2}}}=\frac{4}{9}$. When $P \in P_{\mathrm{B}} C$, similarly, $S_{\triangle \mathrm{BPF}} \geqslant \frac{4}{9}$; when $P \in$ $P_{5} P_{8}$, it is ea... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,365 |
Example 3. $P$ is any point inside $\triangle ABC$, $AP$, $BP$, $CP$ intersect the opposite sides at $P_{1}$, $P_{2}$, $P_{3}$ respectively. Prove that $\frac{AP}{PP_{1}}$, $\frac{BP}{PP_{2}}$, $\frac{CP}{PP_{3}}$ are such that at least one is not less than 2, and at least one is not greater than 2. (3rd IMO problem) | Proof As shown in the figure, let $G$ be the centroid, $G A^{\prime} \| B C$, $G B^{\prime} \| A C$, $G C^{\prime} \| A B$, which divides $\triangle A B C$ into three trapezoids. When $P$ lies on $A A^{\prime} G C^{\prime}$ (including the interior), $\frac{A P}{P P_{1}} \leqslant \frac{A M}{M P_{I}}=\frac{A G}{G K}=2$,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,366 |
Example 1. The vertex of the parabola is at the origin, and the focus F is the center of the circle given by $x^{2}+y^{2}-4 x=0$. A line passing through point $F$ with a slope of 2 intersects the parabola at points $A$ and $D$, and intersects the circle at points $B$ and $C$.
Find $|A B|+|C D|$. | $$
\begin{array}{l}
\text { Given } F \text { as the pole, and the positive direction of the } x \text { axis as the polar axis, we establish the polar coordinate system. } \\
\left.\begin{array}{l}
\left.\therefore \text { The parabola equation is: } \rho=\frac{p}{1-\cos \theta}\right\} \Rightarrow \rho= \\
\because F... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,367 |
Example 3. Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with two moving points $P$ and $Q$ such that the angle subtended at the center (i.e., $\angle P O Q=\frac{\pi}{2}$). Find the maximum and minimum values of $\frac{1}{|O P|}+\frac{1}{|O Q|}$. | Given the pole as $O \%$ and the polar axis as $O X$, the coordinates of $P$ and $Q$ can be set as: $P( \rho_1, 0), Q\left(\rho_{2}, \frac{\pi}{2}+\theta\right)$.
In the polar coordinate system:
$$
\begin{array}{l}
P\left(\rho_{1} \cos \theta, \rho_{1} \sin \theta\right), \\
Q\left(\rho_{2} \cos \left(\frac{\pi}{2}+\th... | \left(\frac{1}{O P}+\frac{1}{O Q}\right)_{\max }=\frac{\sqrt{2\left(a^{2}+b^{2}\right)}}{a b}, \left(\frac{1}{O P}+\frac{1}{O Q}\right)_{\min}=\frac{a+b}{a b} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,369 |
Example 5. (Shanghai Junior High School Mathematics Competition, 1987) A school held a mathematics competition, and five students, Jia, Yi, Bing, Ding, and Cheng, placed in the top five. Before the awards were given out, the teacher asked them to guess the ranking of each person.
Jia said: Yi third, Bing fifth,
Yi sai... | 瞥：This example still uses the table method.
\begin{tabular}{|c|c|c|c|c|c|}
\hline If alive & & $Yi$ & Ling & $Ding$ & Cheng \\
\hline Jia: Yi(3): Bing(5) & & (3) & (5) & & \\
\hline Yi: Cheng(4), Guang(5) & & & & (5) & (4) \\
\hline Bing, Ping(1), Wu(4) & (1) & & & & (4) \\
\hline Ding: Liang(1), Yi(2) & & (2) & (1) & ... | Cheng | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,370 |
, Example 6. Find a four-digit number, which when multiplied by 4 is exactly equal to its reverse. | Assuming the four-digit number $x y z t(x \neq 0$, $1 \neq 0, y, z$ are integers) exists, at this point
$$
\text { 4. } x y z t=t z y x \text {. }
$$
Since $4 \cdot x y z t \geqslant 4 x \cdot 1000$,
$\overline{t z y x}7$. Thus, $t=8$ or $t=9$.
But $t=9$ is impossible, because if $t=9$, then the left-hand side of (*)... | 2178 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,371 |
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
保留源文本的换行和格式，翻译结果如下：
Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is ne... | Given that a junior high school student is not clear about the concept of "sequence", we can simply change the term "sequence" to "numbers arranged according to a certain rule".
Let the sequence be $a_{1}, a_{2}, \cdots a_{2}, \cdots, a_{0}$, i.e., $n \geq 17$. According to the problem:
$$
\begin{array}{l}
a_{k}+a_{k+... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,372 |
3. One day, three young men met at the library, and one of them said: "Starting today, I will come to the library once a day." The second person said: "I will come to the library once every two days." The third person stated that the library is closed on Sundays and does not open. One of the young men suggested that if... | （Hint: Use a "logic table" to solve this problem, and note that the conversation recorded above took place on a Saturday, 9 days before their second meeting,） | Saturday | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 704,375 |
Example 2. Let $a, b, c$ be the roots of the equation $x^{3}-x^{2}-x-1=0$. (1) Prove that $a, b, c$ are distinct, and $\frac{c^{1987}-a^{1987}}{c-a}$ is an integer. | Prove (1)(omitted).
(2) Let $f(n)=\frac{a^{\mathrm{n}}-b^{\mathrm{n}}}{a-b}+\frac{b^{\mathrm{n}}-c^{\mathrm{n}}}{b-c}+\frac{c^{\mathrm{n}}-a^{\mathrm{n}}}{c-a}$, we will prove that $f(n)$ is an integer for all $n$.
$f(0)=0 \in \mathbb{Z}$, and from $a^{3}=a^{2}+a+1$, we get $a^{k+3}=a^{k+2}+a^{k+1}+a^{k}$, similarly $b... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,377 |
Let $N$ be the set of natural numbers, $k \subseteq N$. If there is a function $f: N \rightarrow N$ that is strictly increasing, and for each $n \in N$, we have $f(f(n))=k n$. Prove that for every $n \in N$,
$$
\frac{2 k}{k+1} n \leqslant f(n) \leqslant \frac{k+1}{2} n
$$ | For $\forall n \in N$, let $f(n)=\alpha$. Then
$$
f(\alpha)=f(f(n))=k n .
$$
Therefore,
$$
\begin{array}{l}
f(k n)=f(f(\alpha))=k \alpha, \\
f(k \alpha)=f(f(k n))=k^{2} n
\end{array}
$$
Since $f: N \rightarrow N$ is strictly increasing,
$\therefore f(n) \in N$, and $f(n)$ is strictly increasing.
That is,
$$
\begin{ali... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,378 |
For each positive integer $n$, let
$$
\begin{array}{l}
\quad S_{\mathrm{a}}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}, \quad T_{\mathrm{n}}=S_{1}+S_{2} \\
+S_{\mathrm{s}}+\cdots+S_{\mathrm{n}}, \quad U_{\mathrm{n}}=\frac{1}{2} T_{1}+\frac{1}{3} T_{2}+\frac{1}{4} T_{3} \\
+\cdots+\frac{1}{n+1} T_{\mathrm{n}}
\end{arr... | \[
\begin{array}{l}
\text { 1. Solve } T_{\mathrm{n}}=\sum_{k=1}^{n} \sum_{1=1}^{k} \frac{1}{i}=\sum_{1=1}^{n} \sum_{k=1}^{n} \frac{1}{i} \\
=\sum_{1=1}^{n} \frac{n+1-i}{i}=(n+1) S_{0}-n \\
\quad=(n+1) S_{n+1}-(n+1), \\
\begin{aligned}
\therefore \quad a=1989=b . \\
U_{0}=\sum_{1=1}^{n} \frac{1}{i+1} T_{1}=\sum_{i=2}... | a=1989, b=1989, c=1990, d=3978 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,383 |
2. In a tennis club in a certain area, 20 members hold 14 singles matches, with each person playing at least once. Prove that there must be six matches, the 12 participants of which are all different. | 2. Proof one: Consider a graph G with 20 vertices representing 20 players, and 14 edges representing 14 matches. Let the degree of vertex $i$ be $d_{i}(1 \leqslant i \leqslant 20)$. Then we have
$$
d_{i} \geqslant 1, \quad \sum_{i=1}^{20} d_{i}=2 \cdot 14=28 .
$$
Now, remove $d_{i}-1$ edges from each vertex (an edge c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,384 |
$$
\text { 3. } P(z)=z^{\mathrm{n}}+c_{1} z^{\mathrm{n}-1}+c_{2} z^{\mathrm{n}-2}+\cdots
$$
$+c_{0}$ is a polynomial with real coefficients in the complex variable $z$, and suppose $|P(i)|$ $<1$. Prove that there exist real numbers $a, b$ such that
$$
\begin{array}{l}
P(a+b i)=0 \text { and } \\
\left(a^{2}+b^{2}+1\rig... | 3. Prove that $p(z)$ has $n$ roots (including repeated roots), and non-real roots occur in conjugate pairs. Therefore, we can write
$$
\begin{aligned}
p(z) & =\prod_{i}(z-\alpha \jmath) \cdot \\
& \cdot \prod_{k}\left(z-\beta_{k}\right)\left(z-\bar{\beta}_{k}\right), \text { where } \alpha \text { are }
\end{aligned}
$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,385 |
4. In the acute triangle $\triangle ABC$, the three sides satisfy the inequality $AB < AC < BC$. If $I$ is the incenter of $\triangle ABC$ and $O$ is the circumcenter. Prove that the line $IO$ intersects the segments $AB$ and $BC$.
---
The translation is provided as requested, maintaining the original formatting and ... | 4. Proof of Acute Angle
The circumcenter $O$ of a triangle is inside the triangle. Extend $AO$, $CO$ to intersect the opposite sides at $D$, $E$. We now prove that $I$ is inside $\triangle AOE$, thus the line $IO$ intersects the segments $AE$ and $CD$.
Since $\angle AOC = 2 \cdot \angle B, \therefore \angle OAC$
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,386 |
1. A triangle has three sides of lengths $a, b, c$, satisfying $a>b>c$, and the heights to these sides are $h_{\mathrm{a}}, h_{\mathrm{b}}, h_{\mathrm{c}}$ respectively. Prove: $a+h_{\mathrm{a}}>b+h_{\mathrm{b}} \geqslant c+h_{\mathrm{c}}$ | Proof (Given the side lengths and the corresponding heights, it naturally leads to an association with the area). Let the area of this triangle be $S$, then
$$
\begin{array}{l}
a+h_{\mathrm{a}}=a+\frac{2 S}{a}, \\
b+h_{\mathrm{b}}=b+\frac{2 S}{b}, \\
c+h_{\mathrm{c}}=c+\frac{2 S}{c},
\end{array}
$$
Thus, $a+h_{\mathrm... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,388 |
3. For any real numbers $x_{1}, x_{2}, x_{3}$, if $x_{1}+x_{2}+x_{3}=0$, then $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$ $\leqslant 0$. Please prove it.
For what kind of $n(n \geq 4)$, if $x_{1}+x_{2}$ $+\cdots+x_{n}=0$, then $x_{1} x_{2}+x_{2} x_{3}+\cdots$ $+x_{n-1} x_{n}+x_{n} x_{1} \leqslant 0 ?$ | Prove that if $x_{1}+x_{2}+x_{3} \leqslant 0$, then
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1} \\
=\frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)}{2} \\
=-\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{2} \leqslant 0 .
\end{array}
$$
Then, when $n=4$, if $x_{1}+x_{2}+x_{3}+x_{4}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,391 |
5. Let $m, n$ be positive integers, prove that there exists a constant $\alpha>1$ independent of $m, n$, such that when $\frac{m}{n}<\sqrt{7}$, we have $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\alpha}{n^{2}}$. What is the largest value of $\alpha$? | Solve $\alpha=3$.
$$
\begin{array}{l}
\quad 7-\frac{m^{2}}{n^{2}} \geqslant \frac{a}{n^{2}} \Leftrightarrow 7 n^{2}-m^{2} \geqslant \alpha_{0} \\
\because \quad m+7 \equiv m(\bmod 7) \\
\therefore \quad(m+7)^{2} \equiv m^{2}(\bmod 7)
\end{array}
$$
$\therefore\left\{m^{2}(\bmod 7)\right\}$ is a purely periodic sequence... | 3 | Inequalities | proof | Yes | Yes | cn_contest | false | 704,392 |
6. The sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is defined by the recursive formula
$$
a_{\mathrm{n}+1}=\sqrt{a_{\mathrm{n}}^{2}+\frac{1}{a_{0}}} \quad(n \geqslant 1)
$$
with the first term $a_{1}=1$. Prove that for all $n \geqslant 1$, there exists such an $\alpha$ that
$$
\frac{1}{2} \leqslant \frac{a_{n}}{n^{... | Analysis: If $\frac{1}{2} \leqslant \frac{a_{n}}{n^{\alpha}} \leqslant 2$ holds for all natural numbers $n$, then
$$
\begin{array}{l}
\frac{n^{2} a}{4} \leqslant a_{\mathrm{n}}^{2} \leqslant 4 n^{2 a}, \quad(n \geqslant 1) \\
\frac{1}{2 n^{a}} \leqslant \frac{1}{a_{\mathrm{n}}} \leqslant \frac{2}{n^{\alpha}}, \quad(n \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,393 |
Three, in the "kite" $A B C D$, $A B=$ $A D, B C=C D$. Through the intersection point $O$ of $A C$ and $B D$, draw any two lines, one intersecting $A D$ at $E$, the other intersecting $B C$ at $F$, intersecting $A B$ at $G$, and intersecting $C D$ at $H$. $G F$ and $E H$ intersect $B O$ at $I$ and $J$ respectively. Pro... | Three, Proof: From the given, it is evident that $\triangle A B C \cong \triangle A D C$. Therefore, $\triangle A B O \cong \triangle A D O$. Hence, $\angle A O D=\angle A O B=90^{\circ}$, which means $A C \perp B D$. Taking $O$ as the origin, $B D$ as the $x$-axis, and $A C$ as the $y$-axis, we establish a Cartesian c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,394 |
Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ be $n$ distinct real numbers, $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$,
$$
M=\min _{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} .
$$
Prove: $\frac{S}{M} \geqslant \frac{n\left(n^{2}-1\right)}{12}$. | Let's assume $a_{1}>a_{2}>\cdots>a_{0}$, then
$$
\begin{array}{l}
a_{1}-a_{2} \geqslant \sqrt{M} . \\
a_{1}-a_{3}=a_{1}-a_{2}+a_{2}-a_{3} \geqslant 2 \sqrt{M}, \\
\cdots \cdots, a_{1}-a_{0} \geqslant(n-1) \sqrt{M} .
\end{array}
$$
That is, $\left(a_{1}-a_{2}\right)^{2} \geqslant M,\left(a_{1}-a_{s}\right)^{2} \geqslan... | \frac{S}{M} \geqslant \frac{n\left(n^{2}-1\right)}{12} | Inequalities | proof | Yes | Yes | cn_contest | false | 704,395 |
Five, suppose there are two identical gears $A$ and $B$, with $B$ placed flat on a horizontal plane, and $A$ placed on top of $B$ so that they are perfectly aligned (thus their projections on the horizontal plane are completely overlapping), and then any four pairs of overlapping teeth are removed.
If each gear origin... | To prove that the original proposition is equivalent to: for four different numbers, does there exist a $k$ such that $a_{1}+k \neq a_{i}(\bmod 14) \quad(1 \leqslant i, \quad i \leqslant 4)$
If such a $k$ exists, then in the original proposition, we only need to label the teeth of wheels $A$ and $B$ with 1, 2, ..., 14... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,396 |
In the convex quadrilateral $A B C D$, $A B$ is not parallel to $C D$. Circle $O_{1}$ passes through $A, B$ and is tangent to side $C D$ at $P$, circle $O_{2}$ passes through $C, D$ and is tangent to side $A B$ at $Q$, circles $O_{1}$ and $O_{2}$ intersect at $E, F$. Prove that $E F$ bisects segment $P Q$ if and only i... | First, prove that $B C / A D \Leftrightarrow P D \cdot P C = Q A \cdot Q B$.
As shown in the figure, extend $C D$ and $B A$, and let their intersection point be $S$. Let $a = S A, l n = S B, q = S Q$,
$$
b = S D, t b = S C, p = S P \text{. }
$$
Obviously, $l, t > 1$, and $B C \| A D \Leftrightarrow \Rightarrow l = t$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,397 |
Let $x$ be a natural number. If a sequence of natural numbers $x_{0}=1, x_{1}, x_{2}, \cdots, x_{\mathrm{t}-1}, x_{\mathrm{t}}=x$, satisfies $x_{1-1}<x_{1}, x_{1-1} \mid x_{1}, i=1,2, \cdots, t$. Then $\left\{x_{0}, x_{1}, x_{2}, \cdots, x_{\mathrm{t}}\right\}$ is called a factor chain of $x$, and $t$ is the length of ... | For any natural number
$$
x=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \cdots \cdot p_{n}^{\alpha_{n}},
$$
where $p_{1}, p_{2}, \cdots, p_{n}$ are distinct prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are positive integers. Clearly, the number of factor chains of $x$ is finite, and thus there... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,398 |
Three, let the function $f(x)$ be defined for $x \geqslant 0$, and satisfy the following conditions:
(1) For any $x, y \geqslant 0$,
$$
f(x) f(y) \leqslant y^{2} f\left(\frac{x}{2}\right)+x^{2} f\left(\frac{y}{2}\right)_{3}
$$
(2) There exists a constant $M>0$ such that when $0 \leqslant x \leqslant 1$,
$$
|f(x)| \leqs... | Three, Proof by contradiction, suppose there exists $x_{0} \geqslant 0$ such that
$$
f\left(x_{0}\right)>x_{0}^{2} \text {. }
$$
From (1), we get $f^{2}(0) \leqslant 0 \Rightarrow f(0)=0$. Therefore, $x_{0}>0$. Thus, $f\left(\frac{x_{0}}{2}\right) \geq \frac{f^{2}\left(x_{0}\right)}{2 x_{0}^{2}}>\frac{1}{2} x_{0}^{2}$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,399 |
Let $a$ be a given positive integer, and $A$ and $B$ be two real numbers. Determine the necessary and sufficient condition for the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=(13 a)^{2}, \\
x^{2}\left(A x^{2}+B y^{2}\right)+y^{2}\left(A y^{2}+B z^{2}\right) \\
+z^{2}\left(A z^{2}+B x^{2}\right)=\fr... | $$
\begin{array}{l}
\left(A-\frac{1}{2} B\right)\left(x^{4}+y^{4}+z^{4}\right) \\
=\frac{1}{2}\left(A-\frac{1}{2} B\right)(13 a)^{4} .
\end{array}
$$
If $A \neq \frac{1}{2} B$, then
$$
\text { (1) } \Leftrightarrow \Rightarrow 2\left(x^{4}+y^{4}+z^{4}\right)=(13 a)^{4} \text {. (2) }
$$
Assume the original system of ... | A=\frac{1}{2} B | Algebra | proof | Yes | Yes | cn_contest | false | 704,400 |
Five, let $X$ be a finite set. The rule $f$ assigns to each even subset $E$ (a subset consisting of an even number of elements) of $X$ a real number $f(E)$, and satisfies the following conditions:
(1) There exists an even subset $D$ such that $f(D)>1990$;
(2) For any two disjoint even subsets $A, B$ of $X$, $f(A \cup B... | Five, Proof: Since $X$ is a finite set, the number of all subsets of $X$ is finite. Let $P$ be the even subset of $X$ with the maximum value of $f$ and the least number of elements, and let $Q$ be the complement of $P$ relative to $X$. It can be proven that the above $P, Q$ satisfy the given properties.
(1) Clearly, $P... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,401 |
Example 3. Let $S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$, prove: $\frac{1}{p}<S_{\mathrm{P}}+S_{\mathrm{q}}-S_{\mathrm{pq}} \leqslant 1(p, q$ are positive integers, and $p \geqslant 2, q \geqslant 1$ ). | Prove by induction on $p$: When $p=2$, $S_{2}+S_{q}-S_{2 q}=1+\frac{1}{2}-\left(\frac{1}{2 q}+\cdots+\frac{1}{q+1}\right)$.
Since $\frac{1}{2} \leqslant \frac{1}{2 q}+\cdots+\frac{1}{q+1} \leqslant \frac{q}{q+1}<1$,
we have $\frac{1}{2}<S_{2}+S_{0}-S_{29} \leqslant 1$. Assume that when $p=k$, $\frac{1}{k}<S_{k}+S_{\mat... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 704,402 |
Given that $f(x)$ is a function defined on the set of real numbers, and $f(x+2)(1-f(x))=1+f(x)$.
(1) Prove that $f(x)$ is a periodic function.
(2) If $f(1)=2+\sqrt{3}$, find the value of $f(1989)$. | (1) Proof
$$
f(x+2)(1-f(x))=1+f(x)
$$
defined on $R$, we know $f(1) \neq 1$.
$$
\begin{array}{l}
\therefore \quad f(x+2)=\frac{1+f(x)}{1-f(x)}, \\
f(x+4)=-\frac{1}{f(x)}, \\
f(x+8)=f[(x+4)+4]=-\frac{1}{f(x+4)} \\
=f(x) \text {. } \\
\end{array}
$$
Therefore, $f(x)$ is a periodic function with a period of 8.
(2) Solut... | \sqrt{3}-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,409 |
Three, as shown, in the triangular prism $A B C A_{1} B_{1} C_{1}$, all nine edges are equal to 1, and $\angle A_{1} A B$ $=\angle A_{1} A C$ $=\angle B A C$. Point $P$ is on the diagonal $A_{1} B$ of the side face $A_{1} A B B_{1}$, with $A_{1} P=\frac{\sqrt{3}}{3}$. Connect $P C_{1}$. Find the degree measure of the a... | $$
\begin{array}{l}
\text{ Given } A B=A C \\
= B C=1, \\
\angle B A C=60^{\circ} .
\end{array}
$$
So,
$$
\begin{array}{l}
\angle A_{1} A B \\
=\angle A_{1} A C \\
=\angle B A C=60^{\circ} .
\end{array}
$$
Thus, from $A_{1} A=A B=A C$ we know that $\triangle A_{1} A B$ and $\triangle \dot{A}_{1} A C$ are both equila... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,410 |
Four, place coins of the same size with radius $r$ on a circular tabletop with radius $R$. Ensure that the coins do not protrude beyond the edge of the tabletop, and that the coins do not overlap with each other. After placing $n$ coins, no more coins of this size can be placed on the tabletop.
Prove that: $\sqrt{n} r<... | Four, Proof: Since $n$ coins of radius $r$ are placed without overlapping inside a circular table of radius $R$, the total area covered by the $n$ coins of radius $r$ is less than the area of the circle of radius $R$, i.e., $n \pi r^{2} < \pi(R-r)^{2}$.
Simplifying, we get $R < (2 \sqrt{n}+1) r$.
Combining, we have $\s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,411 |
Five, a positive integer, if each of its prime factors is at least double (i.e., the exponent of each prime factor is no less than 2), then the positive integer is called a "beautiful number." If two consecutive positive integers are both "beautiful numbers," they are called a pair of "twin beautiful numbers." For exam... | Five, Solution Given that $(8,9)$ is a pair of “twin beautiful numbers”.
Let $(n, n+1)$ be a pair of “bent beautiful numbers”. From this, we know that $4 n(n+1)$ is also a beautiful number.
And $4 n(n+1)+1=4 n^{2}+4 n+1=(2 n+1)^{2}$ is a perfect square, which is easily known to be a beautiful number.
Therefore, $(4 n... | (288, 289), (332928, 332929) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,412 |
Example 4. Let the function $f$ be defined for all natural numbers $n$, and (1) $f(n)$ is an integer,
(2) $f(2)=2$,
$f(m n)=f(m) \cdot f(n)$, for all natural numbers $m$ and $n$;
(4) $f(m)>f(n)$, when $m>n$. Prove:
$f(n)=n_{0}$ | Prove: From $(2),(3), 2=f(2)=$ $f(1 \cdot 2)=f(1) \cdot f(2)=2 f(1)$, we get $f(1)=1$. Assume for all $n \leqslant k, f(n)=n$. We need to prove $f(k+1)=k+1$, then we must discuss $k+1$. (1) If $k+1$ is even, $k+1=2 j, j \leqslant k$. Then $f(j)=$, $f(k+1)=f(2 j)=f(2) \cdot f(j)=2 j=k+1$. (ii) If $k+1$ is odd, $k+1=2 j+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,413 |
5. $p^{2} \geqslant 4 q$ is ( ) for the real-coefficient equation $x^{4}+p x^{2}+q=0$ to have real roots.
(A) A necessary condition but not a sufficient condition.
(B) A sufficient condition but not a necessary condition.
(C) A necessary and sufficient condition.
(D) Neither a sufficient condition nor a necessary condi... | $\begin{array}{r}5 \\ \mathrm{~A}\end{array}$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 704,418 |
Three planes $\alpha, \beta$ intersect the trihedral angle with vertex $S$, resulting in two tetrahedrons $S-ABC$ and $S-A_{1}B_{1}C_{1}$, whose volumes are $V$ and $V_{1}$, respectively. Prove:
$$
\frac{V}{V_{1}}=\frac{A S \cdot B S \cdot C S}{A_{1} S \cdot B_{1} S \cdot C_{1} S} .
$$ | $$
\begin{array}{l}
\text{III. Proof: Draw perpendiculars } C H \text{ and } C_{1} H_{1} \text{ from } C \text{ and } C_{1} \text{ to the plane } A S B, \text{ with feet of the perpendiculars at } H \text{ and } H_{1}, \text{ respectively. Then, } S, H, H_{1} \text{ are collinear. Let } \angle A S B \\
=\angle A_{1} S ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,427 |
Four, prove that for all positive numbers $x$, the inequality $x^{2}+\pi x+\frac{15}{2} \pi \sin x>0$ holds.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 1) When $x \in(0, \pi)$, it obviously holds.
2) When $x \in\left[\frac{3 \pi}{2},+\infty\right)$, it obviously holds.
3) When $x \in\left(\pi, \frac{3}{2} \pi\right)$, using the inequality
we get
$$
y>\sin y, y=\pi-x \text {, }
$$
$$
\begin{array}{c}
x^{2}+\pi x+\frac{15}{2} \pi \sin x=x^{2}+\pi x-\frac{15}{2} \pi \\
... | null | Inequalities | proof | Yes | Yes | cn_contest | false | 704,428 |
Let $a_{\mathrm{a}}=(1+\sqrt{2})^{\mathrm{n}},[A]$ denote the greatest integer not exceeding $A$. Prove that the sequence $\left[a_{1}\right],\left[a_{2}\right]$, $\cdots,\left[a_{n}\right], \cdots$ is a sequence where even and odd numbers alternate. | $$
\begin{array}{l}
(1+\sqrt{2})^{\mathrm{a}}+(1-\sqrt{2})^{\mathrm{a}} \\
=2 \sum_{k=0}^{\left[\frac{a}{2}\right]} C_{4}^{2} b^{b}(\sqrt{2})^{2 k},
\end{array}
$$
$$
(1+\sqrt{2})^{n}=\text { number }-(1-\sqrt{2})^{n} \text {. }
$$
Since $-1 < -\sqrt{2} < 0$, we know that $\left|(1-\sqrt{2})^{n}\right| < 1$ and $(1-\s... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,429 |
Given sphere $O$ is tangent to each face of the pyramid $S-ABCD$ at the centroid of that face.
(1) Prove that this pyramid is a regular pyramid,
(2) If the radius of sphere $O$ is $r$, find the volume of this pyramid. | II. Solve $(1 ;$ construct the medians $S M, S N$ of $\triangle A S E$ and $\triangle B S C$, and denote their corresponding centroids as $C_{1}$ and $S_{2}$. Prove that $\triangle S^{\prime} G_{1} B=\triangle S G_{2} B$. From this, it follows that $\triangle S M B \cong \triangle S N B, \triangle A S B \cong \triangle... | \frac{32}{3} r^{3} | Geometry | proof | Yes | Yes | cn_contest | false | 704,430 |
Three, cut out a 2 $\times$ 2 square from one corner of an 8 $\times$ 8 grid paper. Can the remaining 60 squares be divided into 15 pieces shaped like “ $\qquad$ ”? | Three, Solution Fill in the numbers as shown in the figure, then the sum of the numbers in any qualified four-cell block is either 2 or -2. If it can be divided into 15 four-cell blocks, where $x$ blocks have a sum of 2, and $y$ blocks have a sum of -2, then we have
$$
\left\{\begin{array}{l}
x+y=15 \\
2 x-2 y=0
\end{a... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,431 |
Example 6. Let $n$ be a positive integer, and $k$ be an integer no less than 2. Prove: $n^{\mathrm{k}}$ can be expressed as the sum of $n$ consecutive odd numbers. | Prove (by induction on $k$). When $k=2$, $n^{2}=1+3+\cdots+(2 n-1)$. Assume $n^{\mathrm{s}}=(a+1)$ $+\cdots+(a+2 n-1)=n a+n^{2}(a$ is a non-negative even number $)$, then $n^{\mathrm{k}+1}=n^{\mathrm{k}} \cdot n=\left(n a+n^{2}\right) \cdot n=n\left(n a+n^{2}\right.$ $-n)+n^{2}$ (let $\left.b=n a+n^{2}-n\right)=n b+n^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,435 |
2. Given the equation $x^{2}+\left(a^{2}-9\right) x+a^{2}-5 a$ $+6=0$ has one root as negative, and the other root is greater than 2, then the range of $a$ is | 2. $2<a<\frac{8}{3}$ | 2<a<\frac{8}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,439 |
Three, let $S_{\mathrm{n}}$ denote the sum of the elements of all subsets of the set of natural numbers $\{1,2,3, \cdots, n\}$ (the sum of elements of the empty set is defined as 0). Find $S_{1980}$. | Three, Solution: Divide all subsets of the set $\{1,2,3, \cdots, n\}$ (a total of $2^{\text {D}}$ subsets) into two categories:
1. All subsets that do not contain the element $n$ have $2^{\mathrm{n}-1}$ subsets.
2. All subsets that contain the element $n$ also have $2^{\mathrm{n}-1}$ subsets.
Establish a one-to-one cor... | 2^{1978} \cdot 1980 \cdot 1981 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,443 |
Four, points $A_{1}, B_{1}$, $C_{1}$ are on the sides $B C, A C, A B$ of $\triangle A B C$, respectively, and $A C_{1}: C_{1} B=B A_{1}$ : $A_{1} C=C B_{1}: B_{1} A$ $=1: k$. If the perimeter of $\triangle A B C$ is $p$, and the perimeter of $\triangle A_{1} B_{1} C_{1}$ is $s$.
Prove: $\frac{k-1}{k+1} p<s<\frac{k}{k+1... | (1) We have $\left\{\begin{array}{l}\frac{k}{k+1} a + \frac{1}{k+1} bc_{1} \text {. }
\end{array}\right.$
Similarly, we have $\frac{k-1}{k+1} b + \frac{1}{k+1} a > a_{1}$,
$$
\frac{k-1}{k+1} c + \frac{1}{k+1} b > b_{1} \text {. }
$$
Adding the three inequalities, we get $\frac{k}{k+1} p > s$.
Combining (1) and (2), w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,444 |
Five, glue the bases of two congruent regular tetrahedra together. In the resulting hexahedron, all dihedral angles are equal, and the vertices can be divided into two categories: in the first category, each vertex emits three edges; and in the second category, each vertex emits four edges. Try to find the ratio of the... | Let $D-A B C$ and $E-A B C$ be congruent regular tetrahedra, with their base being the equilateral $\triangle A B C$.
If their bases are glued together, then the vertices $D$ and $E$ of the resulting hexahedron should pass through the centroid of $\triangle A B C$, and lie on a line perpendicular to the plane $A B C$.... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,445 |
Example 7. Let $f(x)$ be an $n$-degree polynomial in $x$. If for integers $k$ satisfying $0 \leqslant k \leqslant n$, $f(k)$ takes integer values, then for any integer $m$, $f(m)$ also takes integer values. | Prove (by induction on $n$). When $n=1$, we can set $f(x)=a x+b$, so $f(0)=b, f(1)=a+b$ are both integers $\Rightarrow a=f(1)-f(0)$ and $b$ are both integers, thus for $m \in \mathbb{Z}, f(m)$ is an integer, and the proposition holds in this case. Assume the proposition is true for $(n-1)$-degree polynomials, and let f... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,446 |
Six The terms of the sequence $x_{1}, x_{2}, x_{3}, \cdots$ are non-zero real numbers, and satisfy $x_{\mathrm{a}+2}=\frac{x_{\square} \cdot x_{\mathrm{n}+1}}{2 x_{\mathrm{a}}-x_{\mathrm{a}+1}} \quad(n=1,2, \cdots)$,
(1) Find the necessary and sufficient conditions for $x_{1}$ and $x_{2}$ such that all $x_{0}$ are inte... | Six, Solution (1) Hint: By mathematical induction, it is easy to prove
$$
x_{\mathrm{n}}=\frac{x_{1} x_{2}}{(n-1) x_{1}-(n-2) x_{2}},
$$
i.e., $x_{\mathrm{n}}=\frac{x_{1} x_{2}}{n\left(x_{1}-x_{2}\right)+\left(2 x_{2}-x_{1}\right)}$.
When $x_{1} \neq x_{2}$ and $n$ is sufficiently large, there must be $x_{0}<1$. When ... | 27 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,447 |
3. If the circumference of the top base of a frustum is $4 \pi$, and the circumference of the bottom base is $9 \pi$, and the circumference of a section parallel to the bases is the geometric mean of the circumferences of the top and bottom bases, then the ratio of the two segments of the height of the frustum divided ... | 3. $2: 3$, | 2: 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,455 |
The first question: Given in $\triangle A B C$, $A B>A C$, the bisector of an exterior angle of $\angle A$ intersects the circumcircle of $\triangle A B C$ at point $E$, and a perpendicular line $E F \perp A B$ is drawn through $E$, with the foot of the perpendicular being $F$. Prove: $2 A F=A B-A C$ | Prompt one As shown in Figure 1, draw $E D \perp C A$, intersecting the extension of $C A$ at $D$, and connect $B E, E C$. Prove that R $\mathrm{t} \triangle E B F \cong \mathrm{Rt} \triangle E C D$, and the conclusion follows.
Prompt two As shown in Figure 2, draw $C M \| E A$, intersecting $A B$ at $H$ and the circl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,457 |
5. Given a set $A$ composed of functions, there are 10 odd functions, 8 increasing functions defined on the interval $(-\infty,+\infty)$, and 12 functions whose graphs pass through the origin. Then the set $A$ can have at most $\qquad$ elements. | 5. 14.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,458 |
Let $G$ be any convex region tightly sandwiched between parallel lines $l_{1}$ and $l_{2}$ (i.e., the line segment connecting any two points on its boundary is contained within its region), and its boundary $C$ intersects with $l_{1}$ and $l_{2}$.
Parallel to $l_{1}$, the line $l$ divides $G$ into two parts $A$ and $B$... | (1) When $G$ is a triangle with one side on $l_{1}$ and another vertex on $l_{2}$, $S_{\mathrm{A}} / S_{\mathrm{a}}$ reaches its maximum.
(2) The maximum value of $S_{A} / S_{B}$ is $\frac{a^{2}+2 a b}{b^{2}}$. | \frac{a^{2}+2 a b}{b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,460 |
1. Insert “+” or “- -” between $1,2,3, \cdots, 1989$, what is the smallest non-negative number that the sum can achieve? | Except for 995, the numbers $1,2,3, \cdots, 1989$ can all be divided into 994 pairs: $(1,1989),(2,1988)$, $\cdots$, (994,996). Since the parity of the two numbers in each pair is the same, the result of the operation, regardless of how “+” or “-” signs are placed before each pair, can only be an even number. And 995 is... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,461 |
2. Does a ten-digit number exist that is divisible by 11 and has all different digits on each of its positions?
翻译完成，保留了原文的换行和格式。 | Solve Write a three-digit number that can be divided by 11 and has different digits on each place, such as 275, 396, 418. Using these three numbers, it is not difficult to construct a ten-digit number that can be divided by 11. For example:
$$
\begin{array}{l}
2753964180 \\
=275 \cdot 10^{7}+396 \cdot 10^{4}+418 \cdot ... | 1427385960 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,462 |
3. Find the relationship between the coefficients $a, b, c$ such that the system of equations
$$
\left\{\begin{array}{l}
a x^{2}+b x+c=0, \\
b x^{2}+c x+a=0, \\
c x^{2}+a x+b=0
\end{array}\right.
$$
has a solution. | Solving: Add the equations in the system, we get
$$
\begin{array}{l}
(a+b+c) x^{2}+(a+b+c) x+(a+b \\
+c)=0 .
\end{array}
$$
Thus, $(a+b+c)\left(x^{2}+x+1\right)=0$.
If $a+b+c \neq 0$, then for any $x, x^{2} + x + 1 \neq 0$. Therefore, the equation has no solution. Hence, for the system of equations to have a solution,... | a+b+c=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,463 |
5. Using a one-sided ruler and a set square, divide the given line segment: a) into two equal parts, b) into three equal parts. | a) Using a ruler, extend line segment $AB$ in both directions. Then, using a set square, construct two equilateral triangles on either side of line $AB$ (as shown in Figure 1). Connect segment $MN$ intersecting segment $AB$ at its midpoint $C$.
b) Complete the construction indicated in Figure 2 to obtain points $C$ and... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,465 |
6. In a rectangle of $2.15 \times 4 \mathrm{~cm}^{2}$, what is the maximum number of $\frac{1}{4}$ sectors of circles with a radius of $1 \mathrm{~cm}$ that can be placed without overlapping?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
Note: The note after the translation is not part of the translated text and is provid... | The area of each placed figure is equal to $0.25 \pi$ $\mathrm{cm}^{2}$. Given the rectangular area $=2.15 \times 4=8.6 \mathrm{c}_{\mathrm{m}}{ }^{2}$. If $n$ is the number of figures required, then according to the non-overlapping agreement, the inequality $0.25 \pi \times n \leqslant 8.6$ should hold. Thus, $n \leqs... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 704,466 |
${ }^{*} 7$. Find the smallest term in the sequence $\left\{a_{n}\right\}$. Where, $a_{\mathrm{n}}=n+\frac{1989}{n^{2}}$. | First, use a calculator to find the first few terms of the sequence $\left\{a_{\mathrm{n}}\right\}$:
$$
\begin{array}{l}
a_{1}=1990, a_{2}=499.25, a_{\mathrm{s}}=224, \\
\cdots, a_{15}=23.84, a_{18}=23.769531, \cdots, \\
a_{17}=23.882352, \cdots
\end{array}
$$
We conjecture: The smallest term of the sequence $\left\{a... | 23.769531 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,467 |
*8. In decimal, find the smallest natural number: its square number starts with 19 and ends with 89. | The last digit of the required number can only be 3 or 7. We write out all two-digit numbers ending in 3 and 7, and we get $17, 33, 67, 83$. To ensure that the first two digits of the natural number $x$ are 19, the inequality $19 \leqslant x^{2} \cdot 10^{-N}<20$ must hold, where $N$ is a natural number. When $N = 2k$ ... | 1383 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,469 |
1. There are three piles of stones, with 1989, 989, and 89 stones respectively. The following operations can be performed: either take the same number of stones from each pile (the number can vary each time), or take half of the stones from any one pile (if the pile has an even number of stones) and place them into ano... | a) The desired result can be achieved. Just follow the operations as shown below (arrows indicate the transition from one operation to the next according to the problem conditions).
$$
\begin{aligned}
& (1989,989,89) \rightarrow(1900,900,0) \\
\rightarrow & (950,900,950) \rightarrow(50,0,50) \\
\rightarrow & (25,25,50)... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,470 |
2. Prove: for each natural number $n, 13 \cdot(-50)^{\mathrm{n}}$ $+17 \cdot 40^{n}-30$ is always divisible by 1989. | Let $f(n)=13 \cdot(-50)^{n}+17 \cdot 40^{\mathrm{n}} - 30$.
$$
\begin{array}{l}
g(n)=f(n+1)-f(n) \\
g(n)=(-1)^{n}+1 \cdot 13 \cdot 50^{n} \cdot 51+17 \cdot 40^{n} \cdot 39 \\
= 3 \times 13 \times 17 \times 10^{n}\left[(-1)^{n+1} \cdot 5^{n}+4^{n}\right]
\end{array}
$$
Since $f(1)=0$, and for $n \geqslant 2$, $f(n)=f(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 704,471 |
3. Prove: When $0 < x < \frac{\pi}{2}$, $4 - \cos x$ holds.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that in the interval $\left(0, \frac{\pi}{2}\right)$, the original inequality is equivalent to $3 x>4 \sin x-0.5 \sin 2 x$.
Consider $f(x)=3 x-4 \sin x+0.5 \sin 2 x$, $f^{\prime}(x)=3-4 \cos x+\cos 2 x=2(1-\cos x)^{2}$.
When $0<x<\frac{\pi}{2}$, $f^{\prime}(x)>0$, knowing that $f(x)$ is strictly increasing in t... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 704,472 |
4. In a convex quadrilateral $ABCD$, the line segments connecting the midpoints of opposite sides intersect at point $O$. Prove:
$$
S_{\triangle O B}+S_{\mathrm{DOC}}=0.5 \cdot S_{\mathrm{ABCD}} .
$$ | Proof: Let $K, L, M, N$ be the midpoints of the sides of a quadrilateral. $h_{1}, h, h_{2}$ are the altitudes of $\triangle K B O, \triangle K L M$, and $\triangle O C M$. Thus, we have $h=0.5\left(h_{1}+h_{2}\right)$. Since $K L\|A C\| N M, L M\|B D\| K N$, quadrilateral $K L M N$ is a parallelogram, and point $O$ is ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,473 |
*7. Find all natural numbers $n \in [1000, 2000]$
such that
$$
a_{0}=\sqrt{57121+35 n}
$$
is a natural number. | Solving based on the condition $1000 \leqslant n \leqslant 2000$, we get
$$
a_{0} \geqslant \sqrt{57121+35 \cdot 1000}=303.51441 \cdots
$$
and $a_{n} \leqslant \sqrt{57121}+35 \cdot 2000=356.54032 \cdots$, i.e., $304 \leqslant a_{a} \leqslant 356$.
Since $57121=1632 \cdot 35+1$, we have $a_{\mathrm{n}}=\sqrt{35 i n+1... | 1096, 1221, 1749, 1888, 1185, 1312, 1848, 1999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,475 |
*8. Find all natural numbers $x$, such that $x^{2}$ \% is a 12-digit number of the form $x^{2}=2525 * * * * * * 9$ (the six unknown digits represented by $*$ do not necessarily have to be the same).
The translation is provided as requested, preserving the original text's line breaks and format. | The last two digits of the required number $x$ can be 17, 33, 67, 83 (see Question 8 of Grade 8). To ensure that the first four digits of the square of the number are 2525, the inequality must be formed as follows:
$$
2525 \cdot 10^{8}<x^{2}<2526 \cdot 10^{8} \text {. }
$$
This simplifies to
$$
\sqrt{2525} \cdot 10^{4... | 502517, 502533, 592567, 502583 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,476 |
Third question: There is an $n \times n (n \geqslant 4)$ blank grid table, and in each of its cells, one of the two numbers 1 and -1 is filled in arbitrarily. The product of the numbers in $n$ cells that are pairwise in different rows (horizontal) and different columns (vertical) is called a basic term. Prove that for ... | To distinguish the basic terms equal to $1$ and $-1$ from the grid, we denote the basic term equal to $1$ as $1^{*}$ and the basic term equal to $-1$ as $-1^{*}$. There are $n!$ terms of $1^{*}$ and $-1^{*}$.
When all $n^{2}$ cells are filled with $1$, the number of $1^{*}$ is $n!$, which is even, and the number of $-... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,479 |
4. The vertex $C$ of $\triangle A B C$ has coordinates $(x, y)$ that satisfy the inequalities $x^{2}+y^{2} \leqslant 8+2 y, y \geqslant 3$. The side $A B$ lies on the x-axis. If it is known that the point $Q(0,1)$ is at a distance of 1 from the lines $A C$ and $B C$, find the maximum area of $\triangle A B C$. | From the given information, point $C$ lies on the circle with center $Q$ and a radius of 3, within the shaded area, i.e., the horseshoe shape $K L M N$ (as shown in the figure). The circle $S$ with center $Q$ and a radius of 1 is the incircle of $\triangle A B C$. The equation of the tangent line to $\odot S$ at point ... | 6 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 704,481 |
5. On the edges $S A, S B$, $S C$ of the tetrahedron $S-A B C$, there are points $A_{1}, B_{1}, C_{1}$. And $8 A \cdot S A_{1}$ $=S B \cdot S B_{1}=S C \cdot S C_{1}$, prove that points $A, B$, C, $A_{1}, B_{1}, C_{1}$ lie on the same sphere. | Solve: Obviously, if points
$A, B, C$ and $A, \mathrm{r}$ are not coplanar,
construct a sphere, which
intersects with lines $SB$ and $SC$
at points $B_{1}^{\prime}$ and $C_{1}^{\prime}$ (as
shown in the figure). These intersections
are the points where
the sphere intersects with planes
$SAB$ and $SAC$. The intersection... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,482 |
Find the smallest natural number $x$, such that in decimal notation, the number $x^{3}$ has the first three digits and the last four digits as 1. | Solving the problem, the last four digits are only related to the last four digits of $x$.
Number $x^{3}$ ends with 1, so the last digit of number $x$ can only be 1. Cubing the numbers 01, 11, 21, ..., 91, we confirm that only $71^{3}=357911$. Therefore, the last two digits of number $x$ are 71. Cubing the numbers 071... | 1038471 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 704,484 |
Disk. In this square, there are $2^{3}$ pieces, each in a cell, and these pieces are also symmetrical about a diagonal of the square, call this diagonal $d$. Prove: there is a method to move the pieces so that there are no pieces on $d$.
Keep the original text's line breaks and format, output the translation result di... | 1. Proof a) This is correct: the $j$ shape has a total of $n^{2}$ rows, each row has the same number of chess pieces, and with $n^{2}$ rows, there are a total of $n^{3}$ chess pieces, so each row clearly has $n$ chess pieces.
By symmetry, both sides of $d$ obviously have the same number of chess pieces. Let the number... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 704,485 |
3. Prove: Every octagon with all side lengths being rational numbers and all interior angles being equal is centrally symmetric.
| 3. Prove that the sum of the interior angles of an $n$-sided polygon is $(n-2) \cdot 130^{\circ}$, then the sum of the interior angles of an octagon is $1080^{\circ}$. From this, we know that for an octagon with equal interior angles, each interior angle is $135^{\circ}$, and each exterior angle is $45^{\circ}$. This o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 704,487 |
4. Starting from four given integers $a_{1}, b_{1}, c_{1}, d_{1}$, for all $n \in \mathbb{N}$, recursively define:
$$
\begin{array}{l}
a_{n+1}:=a_{n}-b_{n}, \\
b_{n+1}:=b_{n}-c_{n} \mid, \\
\left.c_{n+1}:=\mid c_{n}-d_{n}\right\}, \\
d_{n+1}:=\left|d_{n}-a_{n}\right| .
\end{array}
$$
Prove: There exists a natural numb... | 4. Prove that for every $n \in N$, let $M_{a}$ denote the maximum of $a_{\mathrm{a}+1}, b_{\mathrm{a}+1}, c_{\mathrm{B}+1}, d_{\mathrm{B}+1}$. The sequence $\left(M_{1}, M_{2}, M_{\mathrm{s}}, \cdots\right)$ is (in the non-strict sense) monotonically decreasing, and since $M \in N$, this sequence must eventually become... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 704,488 |
1. The integers $1,2, \cdots$, n can be arranged such that each number is either greater than all the numbers before it or less than all the numbers before it. How many such arrangements are there? | 1. Let the number of permutations be denoted as $a_{n}$, then $a_{1}=1$. For $n \geqslant 2$, if the number $n$ is placed at the $i$-th position, then the $n-i$ numbers after it are completely determined (they can only be $n-i, n-i-1, \cdots, 1$). The $i-1$ numbers before it have $a_{i-1}$ ways of being arranged, hence... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 704,489 |
Example 1. If $p, q$ are both natural numbers, and the two roots of the equation $p x^{2}-$ $q x+1985=0$ are both prime numbers, then what is the value of $12 p^{2}+q$? (85 Beijing Mathematics Competition Question) | Since $1985=5 \times 397$, and the two roots $x_{1}, x_{2}$ are both prime numbers, and $x_{1} x_{2}=\frac{1985}{p}=\frac{5 \times 397}{p}$, thus $p=1$, then $x_{1}, x_{2}$ are 5 and $397, x_{1}+x_{2}=\frac{q}{p}$ $=q=402$, therefore $12 p^{2}+q=12+402=414$. | 414 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,490 |
Example 2. For a natural number $n$, let the two roots of the quadratic equation $x^{2}+(2 n+1) x+n^{2}=0$ be $\alpha, \beta$. Find the value of the following expression:
$$
\frac{1}{\left.\left(\alpha_{3}+1\right)\left(\beta_{3}+1\right)\right)}+\frac{1}{\left(\alpha_{4}+1\right)\left(\beta_{4}+1\right)}+\cdots
$$
$+\... | $$
\begin{array}{l}
\alpha_{\mathrm{n}}+\beta_{\mathrm{n}}=-(2 n+1), \quad \alpha_{\mathrm{a}} \beta_{\mathrm{a}}=n^{2} . \\
\text { Hence } \frac{1}{\left(\alpha_{\mathrm{a}}+1\right)\left(\beta_{\mathrm{a}}+1\right)}=\frac{1}{n^{2}-(2 n+1)+1} \\
=\frac{1}{n(n-2)}=\frac{1}{2}\left(\begin{array}{c}
1 \\
n-2
\end{array}... | \frac{531}{760} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 704,491 |
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