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One, (This question is worth 25 points) Prove: In the complex plane, the point set $S=\left\{z \in C: z^{3}+z+1=0\right\}$, except for one point, all other points lie in the annulus $\frac{\sqrt{13}}{3}<|z|<\frac{5}{4}$. | Let $f(x)=x^{3}+x+1$. Since $f\left(-\frac{9}{13}\right)=-\frac{53}{13^{3}} < 0$ and $f\left(-\frac{16}{25}\right) > 0$, it follows that $f(x)=0$ has a real root $x_{0} \in\left(-\frac{9}{13},-\frac{16}{25}\right)$. Furthermore, since $f(x)$ is an increasing function, $f(x)=0$ can only have this one real root $x_{0}$.
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,715 |
Example 12. Real numbers $x, y$ satisfy $x^{2}+y^{2}-4 x$ $-2 y+5=0$. Then the value of $\frac{\sqrt{x}+y}{\sqrt{3 y-2 \sqrt{x}}}$ is ( ).
(A) 1
(B) $\frac{3}{2}+\sqrt{2}$
(C) $3+2 \sqrt{2}$
(D) $3-2 \sqrt{2}$ | Solving, the given conditions can be rewritten as
$$
x^{2}-4 x+4+y^{2}-2 y+1=0 \text {. }
$$
which is $(x-2)^{2}+(y-1)^{2}=0$.
Thus, $x-2=0, y-1=0$,
which means $x=2 ; y=1$.
Therefore, the original expression $=\frac{\sqrt{2}+1}{\sqrt{3}-2 \sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$
$$
=3+2 \sqrt{2} \text {. }
$$
Hence,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,716 |
II. (This question is worth 25 points) Given the parabola $y^{2}=2 p x(p>0)$, with its focus at $F$. Question: Does there exist a chord $AB$ passing through $F$ (where $A$ and $B$ are both on the parabola, and $A$ is in the first quadrant, and a point $P$ on the positive $y$-axis), such that $P, A, B$ form a right tria... | II. Restate the problem, and the equation of the straight line $AB$ is $y=\sqrt[4]{\frac{4}{3}}\left(x-\frac{p}{2}\right)$, and the y-coordinate of point $P$ is $p \sqrt[4]{\frac{3}{4}}\left(\frac{3}{2}+\right.$ $\frac{\sqrt{3}}{2}$ ). The proof process is given below.
As shown in the figure, the focus $F\left(\frac{p... | y_{\mathrm{P}}=p \sqrt[4]{\frac{3}{4}}\left(\frac{3+\sqrt{3}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,717 |
46. Given that $\overline{a b}$ is a two-digit number, and $\overline{a b c d e}$ is a five-digit number. If $(\overline{a b})^{3}=\overline{a b c d e}$, find $\overline{a b c d e}$. | Solve $(\overline{a b})^{3}=\overline{a b c d e}=1000 \cdot \overline{a b}+\overline{c d e}$.
We have $\left[(\overline{a b})^{2}-1000\right] \cdot \overline{a b}=\overline{c d e}$.
Thus, $(\overline{a b})^{2}>1000$, which means $\overline{a b}>\sqrt{1000}>31$.
Therefore, $\bar{a} \bar{v} \geqslant 32$.
Also, $(\overli... | 32768 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,721 |
45. In $\triangle A B C$, $\angle A=90^{\circ}$, point $D$ is on $A C$, point $E$ is on $B D$, and the extension of $A E$ intersects $B C$ at
$F$. If $B E: E D=2 A C: D C$, then $\angle A D B=$ $\angle F D C$. | Prove that taking the midpoint $M$ of $BC$, and connecting $AM$ to intersect $BD$ at $G$. By Menelaus' theorem, we have
$$
\frac{B v}{F C} \cdot \frac{A C}{A D} \cdot \frac{E D}{R E}=1 \text{. }
$$
And
$$
\begin{array}{l}
\frac{B E}{E D}=\frac{2 A C}{D C} \cdot W, \\
\frac{B C}{F C}=\frac{2 A D}{D C}, \overrightarrow{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,722 |
Example 13. Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0, x y z \neq 0$. Then the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$ is equal to | Solve the system of equations by treating $\approx$ as a constant,
$$
\left\{\begin{array}{l}
4 x-3 y=6 z, \\
x-2 y=7 z,
\end{array}\right.
$$
we get
$$
\begin{array}{l}
x=3 z, y=2 z . \\
\text { Therefore, the original expression }=\frac{2 \cdot(3 z)^{2}+3 \cdot(2 z)^{2}+6 z^{2}}{(3 z)^{2}+5 \cdot(2 z)^{2}+7 z^{2}}=1... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,724 |
Example 14. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq$ $n$, then $m^{5}+n^{5}=$ $\qquad$ | $$
\begin{array}{l}
\text { Sol } \because(a-b)\left(a^{n}+b^{n-1}\right) \\
=a^{n}+b^{n}+a b\left(a^{n-2}+b^{n-2}\right), \\
\therefore a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right) \\
-a b\left(a^{n-2}+b^{n-2}\right) \text {. } \\
\end{array}
$$
Let $S_{n}=a^{n}-b^{n}$, we get the recursive formula
$$
S_{n}=(a+b) S_... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,725 |
Example 15. Given positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{1}{3} y^{2}=25, \\
z^{2}+\frac{1}{3} y^{2}=9, \\
x^{2}+x z+z^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 x z$. | Solve the original system of equations, i.e.,
$$
\left\{\begin{array}{l}
x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 x \\
\cdot \frac{y}{\sqrt{3}} \cos 150^{\circ}=5^{2}, \\
z^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}=3^{2}, \\
x^{2}+z^{2}-2 x z \cos 120^{\circ}=4^{2} .
\end{array}\right.
$$
From this, we can construct
... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,726 |
1. If $x=\sqrt{3}+1$, then $y=2 x^{5}-4 x^{4}-x^{3}-6 x^{2}$ -6
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | (Hint, from the known we get $x^{2}-2 x$
$$
-2=0 . y=6 \sqrt{3} \text {. }
$$ | 6 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,727 |
2. If $1=\frac{x y}{x+y}, 2=\frac{y z}{y+z}, 3=\frac{z x}{2+x}$, then the value of $x$ is | (Given: from the equations $\frac{1}{x}+\frac{1}{y}=1, \frac{1}{y}+$ $\frac{1}{z}$ $=\frac{1}{2} \cdot \frac{1}{z}+\frac{1}{x}=\frac{1}{3} \cdot x=\frac{12}{5}.$) | \frac{12}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,728 |
3. Let $x=\frac{\sqrt{5}+1}{2}$. Then $\frac{x^{3}+x+1}{x^{5}}=$ | （Hint: From the known, we can get $x^{2}-x=1$. Original expression $=\frac{\sqrt{5}-1}{2}$.）
| \frac{\sqrt{5}-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,729 |
5. Given the equation $a x^{2}+b x+c=0(a \neq 0)$, the sum of the roots is $s_{1}$, the sum of the squares of the roots is $s_{2}$, and the sum of the cubes of the roots is $s_{3}$. Then the value of $a s_{3}+$ $\left\langle s_{2}\right.$ $+c s_{1}$ is . $\qquad$ | (Tip: Let the two roots of the equation be $x_{1}, x_{2}$.
Then by definition, we have $a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0$. The original expression $=0$.) | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,731 |
Example 3. If $a$ is a root of $x^{2}-3 x+1=0$, try to find the value of $\frac{2 a^{5}-5 a^{4}+2 a^{3}-8 a^{2}}{a^{2}+1}$. | From the definition of the root, we know that $a^{2}-3 a+1=0$.
Thus, $a^{2}+1=3 a$ or $a^{2}-3 a=-1$ or
$$
\begin{array}{l}
a^{3}=3 a^{2}-a \\
\therefore \text { the original expression }=\frac{a\left[2 a^{2}\left(a^{2}+1\right)-5 a^{3}-8 a\right]}{3 a} \\
=\frac{1}{3}\left(a^{3}-8 a\right)=\frac{1}{3}\left(3 a^{2}-9 a... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,732 |
6. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5=0$, $b^{2}+2 b-5=0$. Then $a b^{2}+a^{2} b=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Obviously, $a, b$ are the two distinct real roots of the equation $x^{2}+2 x-5=0$. The original expression $=$ 10. | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,733 |
$\begin{array}{l}\text { 8. Let } x>\frac{1}{4} \text {. Simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | Let $a$
$$
\begin{array}{l}
=x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}, b=x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1} . \text { Original } \\
\text { expression }=1 .)
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,735 |
9. Given
$$
\left\{\begin{array}{l}
1988(x-y)+1989(y-z)+1990(z-x)=0, \\
1988^{2}(x-y)+1989^{2}(y-z)+1990^{2}(z-x)=1989 .
\end{array}\right.
$$
Find the value of $y-z$. | $y-z=-1989$ | -1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,736 |
10. Let $x<0, x-\frac{1}{x}=5$. Find the value of $\frac{x^{10}+x^{6}+x^{4}+1}{x^{10}+x^{8}+x^{2}+1}$. | Prompt: Recurrence relation $S_{n}=-3 S_{n-1}-S_{n-2}$. Original expression $=\frac{42}{47}$. | \frac{42}{47} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,737 |
Example 1. Prove that there exist infinitely many natural numbers $a$ with the following property: for any natural number $n, z=n^{4}+a$ is not a prime number. | Analyzing $\approx$ is not a prime number, then it must be a composite number, so $\approx$ can certainly be decomposed into the product of two natural numbers greater than 1. Thus, the problem is reduced to: what kind of $a$ makes $n^{4}+a$ factorizable? Combining the experience of factorizing by completing the square... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,738 |
Example 2. Given that $n$ is a positive integer of group E, $r=f(k)$ is a function that maps integer $r$ satisfying $l \leqslant r \leqslant n$ to integer $k$ satisfying $1 \leqslant k \leqslant n$, and when $k_{1}<k_{2}$, it always holds that $f\left(k_{1}\right) \leqslant f\left(k_{2}\right)$. Prove: There exists an ... | Given that the size of $n$ is unknown and the mapping relationship of function $f$ is complex, it is difficult to determine the appropriate $m$. Let's try using proof by contradiction.
If for any $1 \leqslant m \leqslant n$, $f(m) \neq m$, then from $f(1) \geqslant 1$ and $f(1) \neq 1$ we know $f(1) \geqslant 2$. Thus... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,739 |
Example 3. Suppose $a, b, c, d$ and $m$ are integers such that $a m^{3}+b m^{2}+c m+d$ is divisible by 5, and the number $d$ is not divisible by 5. Prove: There always exists an integer $n$, such that $d n^{3}+c n^{2}+b n+a$ is also divisible by 5. | Let $f=d n^{3}+c n^{2}+b n+a, g=a m^{3}+b m^{2}+c m+d$. To find an integer $n$ such that $5 \mid f$, it is clear that $n$ is determined by $a, b, c, d, m$, and the relationship between $n$ and $a, b, c, d, m$ is linked by $f$ and $g$. With too many unknowns, we first eliminate $d$ from $f$ and $g$ (here we flexibly use... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,740 |
Example 4. Prove that between the squares of two consecutive natural numbers, there do not exist four natural numbers $a<b<c<d$ such that $a d=b c$.
---
The translation maintains the original text's line breaks and format. | To analyze and prove non-existence problems, we generally use proof by contradiction. If there exist $a, b, c, d$ such that $n^{2}a$. Then $b \geqslant a+q$.
Thus, $\frac{d}{b} \leqslant \frac{d}{a+q}q$, and we have $p \geqslant q+1$.
Therefore, $1+\frac{1}{q} \leqslant \frac{p}{q}$.
Hence, $1+\frac{1}{q}<\frac{(n+1)^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,741 |
Example $\mathbf{5}$. Consider every permutation of $1,2, \cdots, n$ as a natural number (all referring to the decimal system). Prove that when $n \geqslant 2$, among the natural numbers formed by the above method, there is at least one that is a multiple of 7. | Analysis: It is very difficult to find directly or use proof by contradiction. We start with special cases.
When $n=2$, $7 \mid 21$. When $n=3$, $7 \mid 231$. When $n=4$, a complete residue class modulo 7 can be selected: $P: 4123$, $2143$, $1234$, $2341$, $1243$, $1342$, $2134$.
When $n \geqslant 5$, it is very time-c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,742 |
Example 4. The value of $x$ that satisfies the following equations is
$$
\begin{array}{l}
(123456789) x+9=987654321, \\
(12345678) x+8=98765432 . \\
(1234567) x+7=9876543 . \\
\cdots \cdots .
\end{array}
$$ | Observe the numerical changes on both sides of each equation, it is easy to know that the last equation should be $x+1=9$, i.e., $x=8$. Upon verification, $x=8$ is the solution. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,743 |
Example 6. Let $n \geqslant 4$ be an integer, and $a_{1}, a_{2}, \cdots a_{n} \in(0$, $2 n)$ be $n$ distinct integers. Prove: The set $\left\{a_{1}, a_{2}\right.$, $\left.\cdots, a_{n}\right\}$ has a subset whose sum of elements is divisible by $2 n$. | First, we need to clarify the values of $a_{1}, a_{2}, \cdots, a_{n}$. Since the integers $a_{1}, a_{2}, \cdots, a_{n} \in (0, 2n)$, $a_{1}, a_{2}, \cdots, a_{n}$ are $n$ of the $2n-1$ numbers $1, 2, \cdots, 2n-1$. Notice that $1 + (2n-1) = 2n$. This reveals the special element $n$, with numbers equidistant from $n$ on... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,744 |
Example 8. Prove that there exists a rational number $\frac{n}{m}$, where $m<100$, and for $k=1,2, \cdots, 99$ we have $\left[k \cdot \frac{n}{m}\right]=\left[k \cdot \frac{73}{100}\right]$.
Here $[x]$ denotes the greatest integer not exceeding $x$. | Analysis: If $\frac{n}{m}=\frac{73}{100}$, it is obviously proven. But $\frac{n}{m} \neq$ $\frac{73}{100}$ (otherwise $m$ should be a multiple of 100). Now let's try:
For $k=1$, we have $\left[\frac{n}{m}\right\rfloor=\left[\frac{73}{100}\right]=0$,
thus $0 \leqslant \frac{n}{m}<1$.
For $k=90, \left[k \cdot \frac{73}{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,746 |
Example 9. Let $I_{k}=\underbrace{11 \cdots 1}_{k \uparrow}$. Prove: There exist infinitely many positive integers $n$, such that $I_{1}, I_{2}, \cdots, I_{n}$ give distinct remainders when divided by $n$. | Analyzing $I_{n}: 1,11,111,1111, \cdots, \underbrace{11 \cdots 1}_{k \uparrow}$, $\cdots$. We attempt to calculate: 2 does not fit; the remainders of 3 dividing $I_{1}, I_{2}, I_{3}$ are $1,2,0$; 4,5,6,7,8 do not fit; the remainders of 9 dividing $I_{1}$ $\sim I$ are $1 \sim 8$ and $0$; 10,11 are also not what we seek.... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,747 |
3. Does there exist an integer $k$, such that the image set of the mapping
$$
(x, y) \rightarrow x^{2}+k x y+y^{2}, x, y \in Z
$$
is the set of natural numbers $N$? | ( Hint: Consider modulo 4.) | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,750 |
4. Prove that there exist infinitely many positive integers $n$ such that: for every prime factor $p$ of $n^{2}+3$, there exists a positive integer $k$, such that $k^{2} < n$, and $p \mid k^{2}+3$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation... | (The problem is equivalent to selecting some terms $a_{i}$ from the sequence $\left\{n^{2}+3\right\}$ such that the prime factors of $a_{i}$ divide some previous term.) | null | Number Theory | other | Yes | Yes | cn_contest | false | 708,751 |
5. Besides $(x, y, z)=(n, n, n)$, does the indeterminate equation $(x+$ $y+z)^{3}=9\left(x^{2} y+y^{2} z+z^{2} x\right)$ have any other integer solutions? | (Make the transformation $y=x+u, z=x+v$. ) | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,752 |
Example 4. Let $x_{i}, y_{i}(i=1,2, \cdots, n, n \geqslant 2)$ be positive real numbers. Find:
$$
\begin{array}{l}
\left(x_{1} y_{2}+x_{2} y_{1}+x_{3} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right. \\
\left.+x_{1} y_{3}\right)^{2} \\
\geqslant 4\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)\left(y_{1} y_{2}+y_{2} y_{3}\right. ... | Prove that dividing both sides of the inequality by $\left(x_{1}+x_{2}+\right.$ $\left.x_{3}\right)^{2}\left(y_{1}+y_{2}+y_{3}\right)^{2}$, and using $\sqrt{x_{1}}, \sqrt{x_{2}}, \sqrt{x_{3}}$ and $\sqrt{y_{1}}, \sqrt{y_{2}}, \sqrt{y_{3}}$ as the three dimensions of two cuboids, and performing a trigonometric substitut... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,757 |
Example 5. If $a, b, c$ are positive numbers, $2 s=a+b+c, n \in$ $N$, then
$$
\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geqslant\left(\frac{2}{3}\right)^{n-2} s^{n-1} .
$$ | Prove the inequality by $(a+b+c)^{n-1}$, using the trigonometric substitution within the body of the inequality, we get
$$
\begin{array}{l}
\sum\left(\cos ^{2 n} \alpha / \sin ^{2} \alpha\right) \\
\geqslant \frac{1}{2} \cdot \frac{1}{3^{n-2}} .
\end{array}
$$
Assume without loss of generality that $\sin ^{2} \alpha \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,758 |
Example 1. Let the number of edges in a simple graph $G$ of order $n (n>1)$ that does not contain $k_{3}$ be maximized at $\left[\frac{n^{2}}{4}\right]$. | This is a well-known result, and the usual proofs are quite complex. In this paper, we provide a simpler proof using a counting method.
Let $A$ be the vertex of maximum degree in $G$, with degree $d$. The set of vertices adjacent to $A$ is $P=\left\{A_{1}, A_{2}, \cdots, A_{d}\right\}$, and the set of vertices not adj... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,759 |
Example 4. Let $G$ be a simple graph of order 10 and does not contain a cycle of length 4, $C_{4}$. Then the maximum number of edges in $G$ is 16. | Proof: Let $f(n)$ be the maximum number of edges in an $n$-order simple graph without $C_{4}$. Clearly, $f(n)=4$. Now we prove that $f(5)=6$. First, if $G$ is two triangles sharing exactly one common vertex, then $G$ is a 5-order graph with 6 edges and no $C_{4}$. Second, if a 5-order graph $G$ has 7 edges, then there ... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,762 |
Example 5. Let $G$ be an $n$-order simple graph without a cycle $C_{4}$ of length 4. Then the number of edges
$$
e \leqslant \frac{n(1+\sqrt{4 n-3})}{4} .
$$ | Proof: Let $V$ be the set of vertices of $G$, and take any $A \in V$, with its degree denoted as $d(A)$. Then, the number of vertex pairs $\{x, y\}$ that are both adjacent to $A$ is $C_{d(A)}^{2}$. When $A$ varies over $V$, all these counted pairs are distinct. Otherwise, there would be a pair $\{x, y\}$ counted in bot... | e \leqslant \frac{n(1+\sqrt{4 n-3})}{4} | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,763 |
Let $\triangle A B C$ have a circumradius $R=1$, and an inradius $r$. The inradius of its pedal triangle $A^{\prime} b^{\prime} C^{\prime}$ is $p$. Prove: $p \leqslant 1-\frac{1}{3}(1+r)^{2}$. | Prove that $B^{\prime}, C^{\prime}, B, C$ are concyclic.
$\angle A C^{\prime} B^{\prime}=\angle A C B$.
Thus, $\triangle A C^{\prime} B^{\prime} \backsim \triangle A C B$.
Therefore, $\frac{B^{\prime} C^{\prime}}{B C}=\frac{A B^{\prime}}{A B}=\cos A$.
So, $B^{\prime} C^{\prime}=B C \cos A$
$=2 R \sin A \cos A$.
B
From ... | p \leqslant \frac{1}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 708,764 |
Example 6. If $u, v$ satisfy
$$
v=\sqrt{\frac{2 u-v}{4 u+3 v}}+\sqrt{\frac{v-2 u}{4 u+3 v}}+\frac{3}{2} \text {. }
$$
then $u^{2}-\cdots v+v^{2}=$ | From the definition of the square root, we get
$$
2 u-v \geqslant 0 \text { and } v-2 u \geqslant 0 \text {, }
$$
which means $2 u-v=0$ or $v=2 u$.
Substituting $v=2 u$ into the given equation, we get
$$
v:=\frac{3}{2}, u=\frac{3}{4} \text {. }
$$
Therefore, $u^{2}-u v+v^{2}=\left(\frac{3}{4}\right)^{2}-\frac{3}{4} \... | \frac{27}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,765 |
1. Prove: $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S_{\triangle}$. | ```
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)^{2} \\
=\left[(y+z)^{2}+(z+x)^{2}+(x+y)^{2}\right]^{2} \\
\geqslant[4(y z+z x+x y)]^{2} \\
\geqslant 4^{2} \cdot 3(y z \cdot z x+z x \cdot x y+x y \cdot y z) \\
=16 \cdot 3(x+y+z) x y z=16 \cdot 3 S_{\Delta}^{2} .
\end{array}
```
Therefore, we get \(a^{2}+b^{2}+c^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,766 |
2. Prove: $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S_{\Delta}+(a-b)^{2}$ $+(b-c)^{2}+(c-a)^{2}$. (Finsler-Hadviger) | In the proof of $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S_{\Delta}$, it is not difficult to obtain
$$
[4(y z+z x+x y)]^{2} \geqslant 16 \cdot 3 S_{\Delta}^{2}
$$
or $4 y z+4 z x+4 x y \geqslant 4 \sqrt{3} S_{\triangle}$,
which is $4(s-b)(s-c)+4(s-c)(s-a)$
$$
+4(s-a)(s-b) \geqslant 4 \sqrt{3} S_{\Delta},
$$
or $(c+a-b... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,767 |
Theorem $l \geqslant \max \left\{h_{a}+\frac{\sqrt{3}}{2} a, h_{b}+\frac{\sqrt{3}}{2} b\right.$,
$$
\left.h_{c}+\frac{\sqrt{3}}{2} c\right\} \text {. }
$$
where, $h_{a}, h_{b}, h_{c}$ are the altitudes of $\triangle A B C$ on the sides $a, b, c$, respectively. | Prove that, as shown in the figure, construct an equilateral $\triangle A B^{\prime} C$ outside $\triangle A B C$ with $A C$ as one side, and draw the circumcircle of $\triangle A B^{\prime} C$. Connect $B B^{\prime}$ intersecting $\overparen{A C}$ at $P$. Connect $P A, P C$. It is easy to prove that $P$ is the Fermat ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,769 |
Inference $1 \quad l^{2} \geqslant 4 \sqrt{3} \triangle$.
Where, $\triangle$ is the area of $\triangle A B C$. | $$
\begin{array}{l}
l \geqslant h_{\Delta} + \frac{\sqrt{3}}{2} a \geqslant 2 \sqrt{h_{a} \cdot \frac{\sqrt{3}}{2} a} \\
\quad=2 \sqrt{\sqrt{3} \triangle}, \\
\text { i.e., } l^{2} \geqslant 4 \sqrt{3} \triangle .
\end{array}
$$ | l^{2} \geqslant 4 \sqrt{3} \triangle | Inequalities | proof | Yes | Yes | cn_contest | false | 708,770 |
1. Connect the common points of the circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}$ $=9$ with line segments, the resulting figure is
(A) line segment
(B) scalene triangle
(C) equilateral triangle
(D) quadrilateral | -、(C).
The circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}=9$ intersect at points $A(x, y)$ whose coordinates must satisfy the equation
$$
x^{2}+\frac{(y+1)^{2}}{9}=x^{2}+(y-1)^{2} \text {. }
$$
Solving this, we get $y=2$ or $\frac{1}{2}$.
Thus, the corresponding values of $x$ are $0$ or $x= \pm \frac{\s... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,771 |
2. For the geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=1536$, and the common ratio $q=$ $-\frac{1}{2}$. Let $\prod_{n}$ represent the product of its first $n$ terms. Then the largest $\prod_{n}(n \in N)$ is (. ).
(A) $\Pi$,
(B) $\prod_{11}$
(C) $\prod_{12}$
(D) $\prod_{13}$ | 2. (C).
The general term formula of the geometric sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1} \text {. }
$$
The product of the first $n$ terms is
$$
\prod_{n}=1536^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}} \text {. }
$$
It is easy to see that $\Pi_{9}, \Pi_{12}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,772 |
3. There exists an integer $n$, such that $\sqrt{p+n}+\sqrt{n}$ is an integer for the prime number $p(\quad)$.
(A) None
(B) Only one
(C) More than one, but a finite number
(D) Infinitely many
Translate the above text into English, please retain the original text's line breaks and format, and output the translation res... | 3. (D).
Let $p$ be any odd prime, and $p=2k+1$. Then, $n=k^{2}$. We have
$$
\begin{aligned}
\sqrt{p+n}+\sqrt{n} & =\sqrt{2 k+1+k^{2}}+\sqrt{k^{2}} \\
& =2 k+1 .
\end{aligned}
$$
Therefore, every odd prime has the property mentioned in the problem. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,773 |
4. Let $x \in\left(-\frac{1}{2}, 0\right)$, then $a_{1}=\cos (\sin x \pi)$, $a_{2}=\sin (\cos x \pi), a_{3}=\cos (x+1) \pi$ have the following relationship ( ).
(A) $a_{3}<a_{2}<a_{1}$
(B) $a_{1}<a_{3}<a_{2}$
(C) $a_{3}<a_{1}<a_{2}$
(D) $a_{2}<a_{3}<a_{1}$ | 4. (A).
Let $y=-x$, then $y \in\left(0, \frac{1}{2}\right)$, and $a_{1}=$ $\cos (\sin y \pi), a_{2}=\sin (\cos y \pi), a_{3}=\cos (1-y) \pi<0$.
Since $\sin y \pi+\cos y \pi$
$$
=\sqrt{2} \sin \left(y \pi+\frac{\pi}{4}\right) \leqslant \sqrt{2}<\frac{\pi}{2},
$$
therefore, $0<\cos y \pi<\frac{\pi}{2}-\sin y \pi<\frac{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,774 |
5. If on the interval $[1,2]$, the function $f(x)=x^{2}+p x+q$ and $g(x)=x+\frac{1}{x^{2}}$ take the same minimum value at the same point, then the maximum value of $f(x)$ on this interval is ( ).
(A) $4+\frac{11}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(B) $4-\frac{5}{2} \sqrt[3]{2}+\sqrt[3]{4}$
(C) $1-\frac{1}{2} \sqrt[3]{2}+\sq... | 5. (B).
On $(1,2)$,
$$
\begin{array}{l}
g(x)=x+\frac{1}{x^{2}}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^{2}} \\
\geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3}{2} \sqrt[3]{2} .
\end{array}
$$
Since $f(x)$ and $g(x)$ take the same minimum value at the same point,
$$
\therefore-\frac{p}{2}=\sqrt[3]{2}, \frac{4 q-p^{2}}{4}=\frac{3}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,775 |
Example 7. Let $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=0$ and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is $\qquad$. | From the given, we have $\left(\frac{1}{a}\right)^{2}-\frac{2}{a}-1=0$, $\left(b^{2}\right)^{2}-2 b^{2}-1=0$. By the definition of roots, $\frac{1}{a}, b^{2}$ are the roots of the equation $x^{2}-2 x-1=0$. Then
$$
\begin{aligned}
\frac{1}{a}+b^{2} & =2, \frac{1}{a} \cdot b^{2}=-1 . \\
\therefore \text { the original ex... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,776 |
6. A frustum of height 8 contains a sphere $O_{1}$ with a radius of 2, the center $O_{1}$ is on the axis of the frustum, and the sphere $O_{2}$ is tangent to the top base and the side of the frustum. The frustum can also contain another sphere $O_{2}$ with a radius of 3, such that the sphere $O_{2}$ is tangent to the s... | 6. (B).
Draw the axial section of the frustum through $\mathrm{O}_{2}$, as shown in Figure 1. Then, draw a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum; the intersection of this section with the axis of the frustum is a circle $O$. From Figure 1, it is easy to find that $O O_{2}=4$.
This ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,777 |
1. The number of proper subsets of the set $\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., x \in N\right\}$ is | $$
\begin{array}{l}
\text { II. 1. } 2^{90}-1 . \\
\text { The set }\left\{x \left\lvert\,-1 \leqslant \log _{\frac{1}{x}} 10<-\frac{1}{2}\right., x \in \mathbb{N}\right\} \\
=\left\{x \left\lvert\,-1 \leqslant \frac{1}{\lg \frac{1}{x}}<-\frac{1}{2}\right., x \in \mathbb{N}\right\} \\
=\{x \mid 1 \leqslant \lg x<2, x \... | 2^{90}-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,778 |
2. In the complex plane, non-zero complex numbers $z_{1}, z_{2}$ lie on a circle centered at $i$ with a radius of 1, the real part of $\overline{z_{1}} \cdot z_{2}$ is zero, and the principal value of the argument of $z_{1}$ is $\frac{\pi}{6}$. Then $z_{2}=$ $\qquad$ . | 2. $z_{2}=-\frac{\sqrt{3}}{2}+\frac{3}{2} i$.
Since $\arg z_{1}=\frac{\pi}{6}, z_{1}$ forms an angle of $\frac{\pi}{3}$ with the $y$-axis,
thus, $\left|z_{1}\right|=1$,
$$
z_{1}=\frac{\sqrt{3}}{2}+\frac{1}{2} i \text {. }
$$
Also, the real part of $\overline{z_{1}} \cdot z_{2}$ is zero,
$$
\therefore \arg z_{2}-\frac... | -\frac{\sqrt{3}}{2}+\frac{3}{2} i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,779 |
3. The polar equation of curve $C$ is $\rho=1+\cos \theta$, and the polar coordinates of point $A$ are $(2,0)$. When curve $C$ rotates once around $A$ in its plane, the area of the figure it sweeps through is $\qquad$. | 3. $\frac{16}{3} \pi$.
Let $\bar{P}(\rho, \theta)$ be any point on the curve $C$, then $|O P|=\rho$ $=1+\cos \theta$.
In $\triangle O A P$, by the cosine rule we have
$$
\begin{aligned}
|A P|^{2} & =|O P|^{2}+|O A|^{2}-2|O P| \cdot|O A| \cos \theta \\
& =(1+\cos \theta)^{2}+4-2 \times 2(1+\cos \theta) \cos \theta \\
&... | \frac{16}{3} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,780 |
4. Glue the bases of two given congruent regular tetrahedra together, precisely to form a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$ | 4. 3.
As shown in the figure, construct $CE \perp AD$, connect $EF$, it is easy to prove that $EF \perp AD$. Therefore, $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,781 |
5. Choose several colors from the given six different colors to color the six faces of a cube, with each face being colored with exactly one color, and any two faces sharing a common edge must be colored differently. Then, the number of different coloring schemes is $\qquad$. (Note: If we color two identical cubes and ... | 5. 230 ways.
(1) Using 6 colors, the division is $\frac{6}{5 \times} \times \frac{4 \times 3 \times 2 \times 1}{6 \times 4}=$ 30 ways;
(3) Using 4 colors, the division is $C_{6}^{4} \cdot C_{4}^{2} \cdot \frac{2 \times 1}{2}=90$ ways;
(4) Using 3 colors, the painting method is $C_{6}^{3}=20$ ways; therefore, the total ... | 230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,782 |
6. In the Cartesian coordinate plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ . | 6. 4 .
Let $A(x, y)$ be an integer point on circle $O$. As shown in the figure, the equation of circle $O$ is $y^{2}+(x-199)^{2}=$ $199^{2}$.
$$
\begin{array}{l}
\text { Clearly, } x=0, y=0 ; \\
x=199, y=199 ; \\
x=199, y=-199 ; x=389, y=0
\end{array}
$$
These are 4 solutions to the equation. However, when $y \neq 0,... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,783 |
One. (This question is worth 25 points) Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}=2 a_{n}-1(n=1,2, \cdots)$, and the sequence $\left\{b_{n}\right\}$ satisfies $b_{1}=3, b_{k+1}=$ $a_{k}+b_{k}(k=1,2, \cdots)$. Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\righ... | $$
\begin{array}{l}
\text { -、 } \because S_{n}=2 a_{n}-1, a_{1}=S_{1}=2 a_{1}-1, \\
\begin{array}{l}
\therefore a_{1}=1 . \\
\text { Also, } a_{k}=S_{k}-S_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right) \\
=2 a_{k}-2 a_{k-1},
\end{array}
\end{array}
$$
$\therefore a_{k}=2 a_{k-1}$. Therefore, $\left\{a_{n}\righ... | 2^{n}+2 n-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,784 |
II. (This question is worth 25 points) Find the range of real numbers $a$ such that for any real number $x$ and any $\theta \in \left[0, \frac{\pi}{2}\right]$, the following inequality always holds:
$$
(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8} .
$$ | $\therefore$ The known inequality $\Leftrightarrow(3+2 \sin \theta \cos \theta-a \sin \theta$
$-a \cos \theta)^{2} \geqslant \frac{1}{4}$, for any $\theta \in\left[0, \frac{\pi}{2}\right]$.
From (1), we get $a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}$, for any $0 \in\left[0, \fr... | a \geqslant \frac{7}{2} \text{ or } a \leqslant \sqrt{6} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,785 |
Three. (This question is worth 35 points) As shown in the figure, circles $O_{1}$ and $O_{2}$ are tangent to the three sides of $\triangle A B C$. Points $E, F, G, H$ are the points of tangency, and the extensions of $E G$ and $F H$ intersect at point $P$. Prove that line $P A$ is perpendicular to $B C$.
---
The tran... | Extend $P A$ to intersect $B C$ at $D$, connect $O_{1} A, O_{1} E, O_{1} G$, $\mathrm{O}_{2} A, \mathrm{O}_{2} \mathrm{~F}, \mathrm{O}_{2} H$. Then
$$
\begin{aligned}
\frac{E D}{D F}=\frac{S_{\triangle P E D}}{S_{\triangle P D F}}= & \frac{\frac{1}{2} P E \cdot P D \cdot \sin \angle 1}{\frac{1}{2} P F \cdot P D \cdot \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,786 |
Example 8. Calculate
$$
\begin{array}{l}
\sqrt{3633 \times 3635 \times 3639 \times 3641+36} \\
-3636 \times 3638=
\end{array}
$$ | Let $3637=a$. Then
the original expression $=$
$$
\begin{array}{l}
\sqrt{(a-4)(a-2)(a+2)(a+4)+36} \\
-(a+1)(a-1) \\
=\sqrt{\left(a^{2}-10\right)^{2}}-\left(a^{2}-1\right) \\
=a^{2}-10-a^{2}+1=-9 . \\
\end{array}
$$ | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,787 |
Four. (This question is worth 35 points) There are $n(n \geqslant 6)$ people at a meeting, and it is known:
(1) Each person knows at least $\left[\frac{n}{2}\right]$ other people, or among the remaining people, there are 2 people who know each other.
Prove: Among these $n$ people, there must be 3 people who all know ea... | Assume that among these $n$ people, no 3 people know each other.
Let $a, b$ be 2 people among these $n$ people who know each other. By the proof by contradiction, it can be deduced that among the remaining $n-2$ people, no one knows both $a$ and $b$. Therefore, there are at least $2\left[\frac{n}{2}\right]$ different p... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,788 |
1. (Mamania) Let $k$ be a positive integer, prove that there are infinitely many perfect squares of the form $n \cdot 2^{k}-7$, where $n$ is a positive integer. | Proof: First, we prove that for any given $k$, there exists a positive integer $a_{4}$, satisfying $a_{k}^{2} \equiv-7\left(\bmod 2^{k}\right)$. We use mathematical induction on $k$ to prove this.
Direct observation shows: when $k$: $\left\{3\right.$, it holds. Taking $a_{4}=1$ satisfies the condition. Suppose for som... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,789 |
2. (Russia) Let $Z$ denote the set of all integers. Prove that for any integers $A, B$, there exists an integer $C$ such that the sets $M_{1}=$ $\left\{x^{2}+A x+B: x \in Z\right\}$ and $M_{2}=\left\{2 x^{2}+2 x+C: x \in Z\right\}$ are disjoint. | If $A$ is odd, $M_{1}$ can consist of numbers of the form $x(x+A)+B \equiv B(\bmod 2)$, and $M_{2}$ can consist of numbers of the form $2 x(x+1)+C \equiv C(\bmod 2)$. To ensure that these two sets do not intersect, we can choose $C=B+1$.
If $A$ is even, $M_{1}$ can consist of numbers of the form $\left(x+\frac{A}{2}\r... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,790 |
4. (Bulgaria) Determine all positive integers $x, y$, where $z$ is the greatest common divisor of $x, y$, that satisfy the equation $x+y^{2}+z^{3}$ $=x y z$.
保留了原文的换行和格式。 | Let $x = zc, y = zb$, where $c, b$ are coprime integers. Then the given Diophantine equation can be transformed into $c + zb^2 + z^2 = z^2cb$. Therefore, there exists some integer $a$ such that $c = za$. Thus, we get: $a + b^2 + z = z^2ab$, i.e., $a = \frac{b^2 + z}{z^2b \cdots 1}$. If $z = 1$, then $a = \frac{b^2 + 1}... | (4, 2), (4, 6), (5, 2), (5, 3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,791 |
5. (Ireland) A conference is attended by $12 k$ people, each of whom has greeted exactly $3 k+6$ others. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference? | For any two people, let the number of other people who have greeted these two people be $n$. By the problem, $n$ is fixed. For a specific person $a$, let $\mathrm{B}$ be the set of all people who have greeted $a$, and $\mathrm{C}$ be the set of people who have not greeted $a$. Then, $\mathbf{B}$ contains $3 k+6$ people... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,792 |
8. (Maki Garden) Let $p$ be an odd prime. Determine the positive integers $x$, $y$, $x \leqslant y$, such that $\sqrt{2 p}-\sqrt{x}-\sqrt{y}$ is as small a non-negative number as possible. | Let $p=2n+1$, where $n$ is a positive integer.
First, we prove: for any positive integers $x, y, D \equiv \sqrt{2p} - \sqrt{x} - \sqrt{y}$ is never zero. Otherwise, we would have $2p = x + y + 2\sqrt{xy}$. Let $b^2, c^2$ be the largest square numbers that divide $x, y$ respectively, then $b + c \geq 2$. If we take $x =... | (x, y) = \left(\frac{p-1}{2}, \frac{p+1}{2}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,794 |
1. (Ukraine) Does there exist a sequence of non-negative integers $F(1)$, $F(2)$, $F(3)$, $\cdots$, that satisfies the following three conditions:
(a) Any integer $0, 1, 2, \cdots$ can appear in this sequence;
(b) Each positive integer appears infinitely many times in this sequence;
(c) For any $n \geqslant 2$,
$$
F\le... | Let $F(1)=0, F(361)=1$. Then, condition $(c)$ can be transformed into: when $n \geqslant 2$, $F\left(F\left(n^{103}\right)\right)=F(F(n))$. For $2 \leqslant n \leqslant 360$, take $F(n)=n$. For $n \geqslant 362$, define $F(n)$ recursively as follows:
$1^{\circ}$ If for some $m, n=m^{163}$, then take $F(n)=F(m)$.
$2^{\c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,795 |
3. (Flowing Orchid) For $x \geqslant 1$, let $p(x)$ be the smallest prime number that does not divide $x$, and $q(x)$ be the product of all primes less than $p(x)$. Specifically, $p(1)=2$. If some $x$ makes $p(x)=2$, then define $q(x)=1$. The sequence $x_{0}, x_{1}, x_{2}, \cdots$ is defined by the following formula, w... | Obviously, from the definitions of $p(x)$ and $q(x)$, it can be derived that for any $x$, $q(x)$ divides $x$. Therefore,
$$
x_{n=1}=\frac{x_{i}}{q\left(x_{x}\right)} \cdot p(x ;)
$$
Moreover, it is easy to prove by induction that for all $n$, $x_{n}$ has no square factors. Thus, a unique encoding can be assigned to $x... | 142 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,796 |
Example 9. Let $x=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$, $y=\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}(n$ is a natural number).
Then when $n=$ $\qquad$, the value of the algebraic expression $19 x^{2}+123 x y+$ $19 y^{2}$ is 1985. | From the given information, we have $x+y=4n+2, xy=1$. Therefore,
$$
\begin{array}{l}
19x^2 + 123xy + 19y^2 \\
= 19(x+y)^2 - 38xy + 23xy \\
= 19(4n+2)^2 + 85.
\end{array}
$$
According to the problem, $19(4n+2)^2 + 85 = 1985$. | 19(4n+2)^2 + 85 = 1985 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,798 |
5. (Finland) For a positive integer $n$, the number $f(n)$ is recursively defined as follows: $f(1)=1$, and for any positive integer $n$, $f(n+1)$ is the largest integer $m$ such that there exists an increasing arithmetic sequence of positive integers, $a_{1} < a_{2} < \cdots < a_{m} = n$, and $f\left(a_{1}\right) = f\... | Prove that by calculating the first few values of $f(n)$, the following relations can be inductively derived, with exceptions for the initial few terms.
$$
\begin{array}{l}
f(4 k)=k, \\
f(4 k+1)=1 \text {, but } f(8)=3 ; \\
f(4 k+2)=k-3 \text {, but } f(2)=f(13)=2 ; \\
2, f(14)=f(18)=3, f(26)=4 ; \\
f(4 k+3)=2 .
\end{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,799 |
6. (India) Let $N$ be the set of all positive integers. Prove that there exists a unique function $f: \mathrm{N} \rightarrow \mathrm{N}$, satisfying
$$
f(m+f(n))=n+f(m+95),
$$
where $m, n$ are any elements of $N$. And calculate the value of $\sum_{k=1}^{19} f(k)$. | Prove that for all $n \geqslant 1$, let $F(n)=f(n)-95$. By substituting $k$ for $m+95$, the given condition becomes
$$
F(k+F(n))=n+F(k),
$$
where $n \geqslant 1, k \geqslant 96$. In (1), substitute $m$ for $k$, then add $k$ to both sides, and apply the function $F$, we get $F(k+n+F(m))=F(k+F(m+F(n)))$. Using (1) again... | 1995 | Algebra | proof | Yes | Yes | cn_contest | false | 708,800 |
1. In $\triangle A B C$, $\angle C=90^{\circ}, \angle A$'s bisector intersects $B C$ at $D$. Then $\frac{A B-A C}{C D}$ equals ( ).
(A) $\sin A$
(B) $\cos A$
(C) $\operatorname{tg} A$
(D) $\operatorname{ctg} A$ | -1 (C).
As shown in the figure, draw $D E \perp A P$ at $F$. Since $D$ is on the angle bisector of $\angle A$, we have $D E$
$$
\begin{array}{l}
=C D . A E=A C \\
\quad \therefore \frac{A B-A C}{C D}=\frac{A B-A E}{D E} \\
\quad=\frac{B E}{D E}=\operatorname{tg} \angle B D E=\operatorname{tg} A .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,801 |
2. Let the two roots of the equation $x^{2}+x-1=0$ be $\alpha, \beta$. Then the linear equation with roots $\alpha^{2}+\beta^{2}$ and $(\alpha+\beta)^{2}$ is ( ).
(A) $x^{2}+4 x-3=0$
(B) $x^{2}-4 x+3=0$
(C.) $x^{2}+4 x-3=0$
(D) $x^{2}-4 x-3=0$ | 2. (B).
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 \Rightarrow(\alpha+\beta)^{2}=1 . \\
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=1-2(-1)=3 .
\end{array}
$$
Therefore, the equation with roots 1 and 3 is $x^{2}-4 x+3=0$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,802 |
3. If $x-y=2, x^{2}+y^{2}=4$, then the value of $x^{1998}+y^{1996}$ is ( ).
(A) 4
(B) $1996^{2}$
(C) $2^{1996}$
(D) $4^{1996}$ | 3. (C).
From $x-y=2$
squaring gives $x^{2}-2 x y+y^{2}=4$.
Also, $x^{2}+y^{2}=4$.
(3) - (2) gives $2 x y=0 \Rightarrow x y=0$.
Therefore, at least one of $x, y$ is 0, but since $x^{2}+y^{2}=4$, only one of $x, y$ can be 0, the other being 2 or -2. In either case,
$$
x^{1996}+y^{1996}=2^{1996} .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,803 |
4. If the equations $x^{2}+2 a x+b^{2}=0$ and $x^{2}+2 c x-$ $b^{2}=0$ have a common root, and $a, b, c$ are exactly the three sides of a triangle, then this triangle is ( ).
(A) Acute triangle
(B) Obtuse triangle
(C) Right triangle with $a$ as the hypotenuse
(D) Right triangle with $c$ as the hypotenuse | 4. (C).
$x^{2}+2 a x+b^{2}=0, x^{2}+2 c x-b^{2}=0$ subtract to get $2 a x-2 c x+2 b^{2}=0$.
Obviously, $c \neq a$. (Otherwise, if $a=c$, then we have $b=0$, contradicting the assumption $b$
$>($ transition)
Thus $x=\frac{b^{2}}{c-a}$. This is the common root of the two equations.
Substitute $x=\frac{b^{2}}{c-a}$ into o... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,804 |
5. Write down the two-digit numbers from 19 to 96 in sequence to form a natural number \( N, N=19202122 \cdots 949596 \).
If the highest power of 3 in the prime factorization of \( N \) is \( 3^{k} \). Then \( k=(\quad) \).
(A) 0
(B) 1
(C) 2
(D) 3 | 5. (B).
The remainder when $N$ is divided by 3 is the same as the remainder when the sum $19+20+21+22+\cdots+95+96$ is divided by 3. It is easy to see that the remainder is $0$. The remainder when $N$ is divided by 9 is the same as the remainder when the sum $19+20+21+22+\cdots+95+96$ is divided by 9. It is easy to see... | B | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,805 |
1. $\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16}=$ | Let $x=1994$, then
$$
\begin{array}{l}
\sqrt{1991 \cdot 1993 \cdot 1995 \cdot 1997+16} \\
=\sqrt{(x-3)(x-1)(x+1)(x+3)+16} \\
=\sqrt{\left(x^{2}-1\right)\left(x^{2}-9\right)+16}=\sqrt{x^{4}-10 x^{2}+25} \\
=\sqrt{\left(x^{2}-5\right)^{2}}=1994^{2}-5 \\
=3976031 .
\end{array}
$$ | 3976031 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,807 |
Example 10. If $1+x+x^{2}+x^{3}+x^{4}+x^{5}=0$, then $x^{6}=$ $\qquad$ . | From the given conditions, we know that $x \neq 1$. Therefore,
$$
(x-1)\left(x^{5}+x^{4}+x^{3}+x^{2}+x+i\right)=0 \text {. }
$$
$x^{6}-1=0$, i.e., $x^{6}=1$ | x^{6}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,809 |
3. In $\triangle A B C$, $A B=B C, \angle A B C=$ $20^{\circ}$. Take a point $M$ on side $A B$ such that $B M=A C$. Then the degree measure of $\angle A M C$ is $\qquad$ | 3. $\left(30^{\circ}\right)$.
Construct an equilateral triangle $K B C$ outside $\triangle A B C$ with $B C$ as a side, and connect $M K$. It is easy to prove that $\triangle A B C \cong \triangle K B M$. Therefore, $K B=K M=K C . \angle B K M=20^{\circ}$.
With $K$ as the center and $K B$ as the radius, the circle pas... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,810 |
4. Let $a, b$ be unequal real numbers, and $a^{2}+2 a=$ $5 \cdot b^{2}+2 b=5$. Then $a^{2} b+a b^{2}=$ | 4. (10).
It is known that $a, b$ are two unequal real numbers which are exactly the two real roots of the equation $x^{2}+2 x-5=0$. Therefore, $a+b=-2, ab=-5$.
Thus, $a^{2} b+ab^{2}=ab(a+b)=(-2) \cdot(-5)=$
10. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,811 |
5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is | 5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,812 |
6. In the figure, $P$ is a point inside square $A B C D$, $P D \doteq 1$, $P A=\sqrt{2}, P B=\sqrt{3}$. Then the area of the shaded region is $\qquad$ | 6. $\left(\frac{1}{2}(2+\sqrt{3})\right)$.
As shown in the figure, construct $\triangle A P^{\prime} B \cong \triangle A P D$,
connect $P P^{\prime}, \angle P A P^{\prime}=90^{\circ}$,
$$
A P=A P^{\prime}=\sqrt{2} \text {, }
$$
then $P P^{\prime}=2$.
In $\triangle P P^{\prime} B$, $P^{\prime} P^{2}=4=3+1=P B^{2}+P^{\... | \frac{1}{2}(2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,813 |
One, (Full marks 20 points) Given that positive integers $p, q$ are both prime numbers, and $7 p+q$ and $p q+11$ are also prime numbers. Calculate the value of $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)$. | Given that $p q+11$ is a prime number, we know that $p q+11$ must be odd. Therefore, $p q$ is even. So, at least one of $p, q$ must be even. But since $p, q$ are both primes, one of $p, q$ must be 2. If $p=q=2$, then $p q+11=15$ is not a prime number. Therefore, $p, q$ cannot both be 2. So, one and only one of $p, q$ i... | 221 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,814 |
II. (Full marks 25 points) As shown in the figure, in quadrilateral $ABCD$, $AB \parallel CD$, $AD=DC=DB=p$, $BC=q$. Find the length of diagonal $AC$.
---
The translation maintains the original text's format and line breaks. | $$
\text { Given, } \because A D=D C=D B=p \text {, }
$$
$\therefore$ With $D$ as the center and $P$ as the radius, draw a circle, C, $R, A$ all lie on the circle ( p). Extend $C J$ to intersect at $E$, and connect $A E$. It is easy to know that $C D E$ is the diameter of $\odot(D, p)$, $C E=$ $2 p$.
Also, $B A / / C D... | AC=\sqrt{4p^2-q^2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,815 |
Three. (Full marks 25 points) Color every point in the plane either red or blue. Prove: There exists a right-angled triangle with a hypotenuse of length 1996 and one acute angle of $30^{\circ}$, such that all three vertices are the same color.
| In the two-colored plane, construct an equilateral $\triangle A P D$ with a side length of 1996. Among the three vertices $A$, $P$, $D$, there must be two points of the same color. Without loss of generality, assume $A$ and $D$ are both red.
Using $A D=1996$ as the diameter, construct a circle and complete the inscrib... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,816 |
2. If a circle with radius $r$ can be drawn above the parabola $y=a x^{2}(a>0)$, tangent to the parabola at the origin $O$, and the circle has no other common points with the parabola, then the maximum value of $r$ is ( ).
(A) $\frac{1}{2 a}$
(B) $\frac{1}{a}$
(C) $a$
(D) $2 a$ | 2. (A).
From the problem, we know the system of equations
$$
\left\{\begin{array}{l}
x^{2}+(y-r)^{2}=r^{2}(r>0) . \\
y=a x^{2}(a>0) .
\end{array}\right.
$$
has only one solution $x=y=0$.
Substituting (2) into (1) yields $y \cdot\left(y,-\frac{2 a r-1}{a}\right)=0$. Therefore, it must be that $2 a r-1 \leqslant 0$, i.... | \frac{1}{2 a} | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,818 |
3. Consider the three diagonals on the three pairwise adjacent faces and the space diagonal of a rectangular cuboid (a total of four line segments). Then the correct proposition is ( ).
(A) There must be some three segments that cannot form the three sides of a triangle
(E) Any three segments can form a triangle, each ... | 3. (B).
Let the three dimensions (i.e., length, width, height) of the rectangular prism be $a, b, c$, then the four line segments are $\sqrt{a^{2}+b^{2}}, \sqrt{b^{2}+c^{2}}, \sqrt{c^{2}+a^{2}}, \sqrt{a^{2}+b^{2}+c^{2}}$. By the triangle existence theorem (the sum of any two sides is greater than the third side) and r... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,819 |
Example 1 Find all integer values of $a$ such that the equation $(a+1) x^{2}-\left(a^{2}+1\right) x+2 a^{3}-6=0$ has integer roots. (1996, Huanggang Region, Hubei Junior High School Mathematics Competition) | When $a+1=0$, i.e., $a=-1$, the original fraction becomes
$-2 x-2-6=0$.
Solving this, we get $x=-4$;
When $a+1 \neq 0$, i.e., $a \neq -1$, let the roots of the equation be $x_{1}, x_{2}$. Then
$$
x_{1}+x_{2}=\frac{a^{2}+1}{a+1}=a-1+\frac{2}{a+1} .
$$
When $a=0,1,-2,-3$, $\frac{2}{a+1}$ is an integer, and we discuss th... | a=-1,0,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,820 |
Example 2 Let the quadratic function $y=-x^{2}+(m-2) x +3(m+1)$ have its graph intersecting the $x$-axis at points $A$ and $B$ ($A$ to the left of $B$), and the $y$-axis at point $C$. The product of the lengths of segments $AO$ and $OB$ equals 6 ($O$ is the origin). Connect $AC$ and $BC$. Find the value of $\sin C$.
(H... | Solution: According to the problem, we have $A O \cdot O B=6$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, then $A O=\left|x_{1}\right|, O B=\left|x_{2}\right|$. Therefore, $\left|x_{1}\right| \cdot\left|x_{2}\right|=6$, which means $\left|x_{1} x_{2}\right|=6$.
We consider two cases:
(1) $x_{1} x_{2}=6$,
i.... | \sin C=\frac{\sqrt{2}}{10} \text{ or } \sin C=\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,821 |
Example 1 Let $a, b, c$ be positive numbers. Prove:
$$
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2} \text {. }
$$
(1988, Friendship Cup Competition) | Proof: From (*) we get
$$
\text { Left } \geqslant \frac{(a+b+c)^{2}}{(b+c)+(c+a)+(a+b)}=\text { Right. }
$$ | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,822 |
Example 2 Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$. To make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National Competition) | Solution: According to the problem, all logarithms are positive. Therefore, from (*) we get
$$
\begin{aligned}
\text { LHS } & =\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
& \geqslant \frac{(1+1+1)^{2}}{\log _{1993}\left(\frac{x_{0}... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 708,823 |
2. The sum of all solutions of the equation $\cos 2x=0$ in the interval $[0,100]$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $1024 \pi$ | 1024 \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,825 |
4. Given a right triangle $\triangle A B C$ with the altitude from the right angle to the hypotenuse being $C D$, and $A D=\frac{1}{3} A B$. If $\triangle A C D$ is rotated around $C D$ to $\triangle A_{1} C D$, such that the dihedral angle $A_{1}-C D-B$ is $60^{\circ}$. Then the angle between the skew lines $A_{1} C$ ... | 4. $\arccos \frac{\sqrt{3}}{6}$ | \arccos \frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,827 |
7. Connect the right focus $F_{2}$ of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ with a moving point $A$ on the ellipse, and construct a square $F_{2} A B C\left(F_{2} 、 A 、 B 、 C\right.$ are arranged in a clockwise direction). Then, as point $A$ moves around the ellipse once, the trajectory equation of the moving... | $\begin{array}{l}\text { 7. } \frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9} \\ =1\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 7. } \frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9} \\ =1\end{array}$ | \frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,830 |
8. Four spheres, each with a radius of 1, are pairwise externally tangent and all are inside a larger sphere, each also tangent to the larger sphere. Then the radius of the larger sphere is $\qquad$ . | $8.1+\frac{\sqrt{6}}{2}$ | 1+\frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,831 |
Solve the equation: $|x-| 3 x+1||=4$.
(1994, Tianjin City Junior High School Mathematics Competition)
Analysis: This is an equation involving absolute values. Absolute values are defined by classification, so we should discuss three cases: $3 x+1=0$, $3 x+1>0$, and $3 x+1<0$. | (1) When $x=-\frac{1}{3}$, the left side of the equation $=\frac{1}{3}$, the right side $=4$. Contradiction, no solution.
(2) When $x-\frac{1}{3}$, the original equation becomes
$$
|x-(3 x+1)|=4 \text{, }
$$
i.e. $\left\{\begin{array}{l}-2 x-1=4, \\ -2 x-1=-4 .\end{array}\right.$
Solving, we get $x=-\frac{5}{2}$ (disc... | x=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,832 |
II. (15 points) Let $k_{1}<k_{2}<k_{3}<\cdots$ be positive integers, none of which are consecutive, and for $m=1,2,3, \cdots, s_{m}=k_{1}+k_{2}+\cdots+k_{m}$. Prove that for every positive integer $n$, the interval $\left[s_{n}, s_{n+1}\right)$ contains at least one perfect square. | For the interval $\left[s_{n}, s_{n+1}\right)$ to contain at least one perfect square, the necessary and sufficient condition is that $\left[\sqrt{s_{n}}, \sqrt{s_{n+1}}\right)$ contains at least one integer.
Therefore, to prove this, we only need to show that for each $n \in N$, we have $\sqrt{s_{n+1}}-\sqrt{s_{n}} \... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,834 |
Three. (This question is worth 16 points) Given the set $\{1,2,3,4,5, 6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
| Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that have the given property.
The set $\{1,2, \cdots, n, n+1, n+2\}$ has subsets with the given property divided into two categories: the first category of subsets contains the element $n+2$, and there are $a_{n}+n$ such subsets (i.e., the union of e... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,835 |
Four, (This question is worth 18 points) Pingxi Given $n$ points $A_{1}$, $A_{2}$, ..., $A_{n}(n \geqslant 3)$, no three points are collinear. By selecting $k$ pairs of points to determine $k$ lines (i.e., drawing a line through each pair of points among the $k$ pairs), ensure that these $k$ lines do not intersect to f... | If a straight line $l$ connects two points (let's denote them as $A_{1}$ and $A_{2}$), the restriction that no such two lines intersect to form a triangle means that these two points cannot simultaneously connect to any of the remaining $n-2$ points. That is, the number of lines passing through $A_{1}$ and $A_{2}$ is a... | \left\{\begin{array}{ll}\frac{n^{2}}{4}, & \text { if } n \text { is even; } \\ \frac{n^{2}-1}{4}, & \text { if } n \text { is odd. }\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,836 |
9. 1. Divide all natural numbers from 1 to 1000000 into two mutually exclusive categories: one category consists of numbers that are the sum of a perfect square and a perfect cube, and the other category consists of the remaining numbers. Which category has more numbers? Prove your conclusion. | 9. 1. There are many numbers that cannot be expressed in this form.
Proof: Let $n=k^{2}+m^{3}$, where $k, m, n \in N$, and $n \leqslant 1000000$. Clearly, in this case, $k \leqslant 1000$, $m \leqslant 100$. Therefore, we need to consider no more than 100000 pairs $(k, m)$. The numbers $n$ that satisfy the condition a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,837 |
9. 2. As shown in the figure, several circles with equal radii do not intersect, and their centers $O_{1}$, $O_{2}$, $O_{3}$ are not collinear. Tangents are drawn from each circle's center to the other two circles, forming a convex hexagon. The adjacent sides of this hexagon are alternately colored red and blue (in the... | 9. 2. As shown in the figure, since the diameters of the three circles are equal, we have
$$
\begin{array}{l}
X_{1} O_{2}=O_{1} Y_{2}, \\
Y_{1} O_{3}=O_{2} Z_{2}, \\
Z_{1} O_{1}=O_{3} X_{2} .
\end{array}
$$
That is,
$$
\begin{array}{l}
X_{1} A+A B+B O_{2} \\
=O_{1} B+B C+C Y_{2}, \\
Y_{1} C+C D+D O_{3} \\
=O_{2} D+D E... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,838 |
9. 3. $x$, $y$, $p$, $n$, $k$ are all natural numbers, and satisfy: $x^{n}+y^{n}=p^{k}$. Prove: If $n$ is an odd number greater than 1, and $p$ is an odd prime, then $n$ can be expressed as a power of $p$ with a natural number as the exponent. | 9. 3. Let $m$ be a prime. Let $x, y$ have the greatest common divisor. Let $x=m x_{1}, y=m y_{1}$, from the given condition we have $m^{\prime \prime}\left(x_{1}{ }^{\prime \prime}+y_{1}{ }^{\prime \prime}\right)=p^{2}$. Then, for any non-zero integer $\alpha$, we have
$$
x_{1}{ }^{n}+y_{1}{ }^{n}=p^{k-n o} .
$$
For $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,839 |
9. 4. A parliament consisting of 1600 members is divided into 16000 committees, each committee comprising 80 members. Prove: there must exist two committees that have at least 4 members in common. | There are more than a certain number of common members in each committee. Now, two members of the parliament are to compile three lists of chairpersons for the parliamentary meetings. The first member believes that any member of the parliament can be the chairperson of any of these meetings, so he writes $1600^{3}$ lis... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,840 |
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