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3. $A$ is a point on the parabola $x=-\frac{2}{7} y^{2}$, $F$ is the focus, $|A F|=14 \frac{7}{8}$, the equation of the line $l$ passing through point $F$ and perpendicular to $O A$ is $\qquad$ . | $3.8 x-4 y+7=0$ or $8 x+4 y+7=0$.
Let $A\left(x_{1}, y_{1}\right)$. By definition, $14 \frac{7}{8}=\left|x_{1}\right|+\frac{7}{8}$.
Then $A(-14,7)$ or $(-14,-7)$.
Thus $k_{O A}=\frac{1}{2}$ or $-\frac{1}{2}$. | 3.8 x-4 y+7=0 \text{ or } 8 x+4 y+7=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,411 |
4. From $\{1,2,3, \cdots, 20\}$, select 3 numbers such that no two numbers are adjacent, there are $\qquad$ different ways. | 4.816.
17 numbers produce 16 spaces plus 2 ends making a total of 18 spaces. According to the problem, there are $\mathrm{C}_{18}^{3}$ different ways to choose. | 816 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,412 |
5. In $\triangle A B C$, the lengths of the three sides $a, b, c$ satisfy $b+c \leqslant 2 a$ and $a+c \leqslant 2 b$. Then the range of $\frac{b}{a}$ is $\qquad$ | 5. $\frac{2}{3}a-b$, given $2 a-b<a+c<2 b$, i.e., $\frac{2}{3}<\frac{b}{a} \leqslant 1$.
When $a \leqslant b$, we have $1 \leqslant \frac{b}{a}<\frac{3}{2}$. Therefore, $\frac{2}{3}<\frac{b}{a}<\frac{3}{2}$. | \frac{2}{3}<\frac{b}{a}<\frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,413 |
6. The largest positive integer $n$ for which the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds for a unique integer $k$ is $\qquad$ . | 6.112 .
Obviously, $\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8} \Leftrightarrow \frac{6}{7}<\frac{k}{n}<\frac{7}{8}$
$$
\Leftrightarrow 48 n<56 k<49 n \text {. }
$$
There are at least $n-1$ integers between $48 n+1$ and $49 n-1$.
If $n-1 \geqslant 2 \times 56$, then there must be at least 2 multiples of 56 in this range, ... | 112 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,414 |
Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$
Find $P_{n}+\frac{1}{2} S_{n}$. | $$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,415 |
Four. (Total 20 points) Find the range of real number $a$ such that the inequality $\sin 2 \theta-(2 \sqrt{2}+\sqrt{2} a) \sin \left(\theta+\frac{\pi}{4}\right)-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)}>$ $-3-2 a$, holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$. | Let $x=\sin \theta+\cos \theta, \theta \in\left[0, \frac{\pi}{2}\right]$, then $x \in[1, \sqrt{2}]$.
$$
\begin{array}{l}
\because \sin 2 \theta=x^{2}-1, \\
\sin \left(\theta+\frac{\pi}{4}\right)=\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x,
\end{array}
$$
$\therefore$ the original inequality becomes
$$
x... | a>3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,416 |
Five. (Total 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the fixed point $A(1,0)$, and the foci are on the $x$-axis. The ellipse intersects the curve $|y|=x$ at points $B$ and $C$. There is a parabola with $A$ as the focus, passing through points $B$ and $C$, and opening to t... | $$
\begin{array}{l}
a=1, c=\sqrt{1-b^{2}}, e=\sqrt{1-b^{2}} . \\
\because 1>e^{2}>\frac{2}{3}, \therefore \frac{\sqrt{6}}{3}<b<1, \\
\because \frac{p}{2}=m-1, \\
\therefore y^{2}=4(1-m)(x-m) .
\end{array}
$$
Substituting $y=x, x \in\left(0, \frac{1}{2}\right)$ into equation (1) yields
$$
\begin{array}{l}
x^{2}+4(m-1) ... | 1<m<\frac{3+\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,417 |
Example 1 Express the positive integer $n$ as the sum of some positive integers $a_{1}, a_{2}, \cdots, a_{p}$, i.e., $n=a_{1}+a_{2}+\cdots+a_{p}$, where $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{p}$. Let $f(n)$ be the number of such representations (for example, $4=4, 4=1+3, 4=2+2, 4=1+1+2, 4=1+1+1+1$, so $f... | Analysis: Since this problem is to prove a non-strict inequality, it is necessary to construct an injection or a surjection to prove it.
Proof: This problem essentially requires proving
$f(n+1)-f(n) \leqslant f(n+2)-f(n+1)$.
By adding a "1" before each partition of $n$, we can obtain a partition of $n+1$, so $f(n+1)-f(... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,418 |
I. (Total 50 points) As shown in Figure 1, let $PQ$ be a fixed chord in circle $\odot O$. Take any points $A$ and $B$ on the arcs on either side of $PQ$. The extensions of $PA$ and $BQ$ intersect at point $M$, and the extensions of $QA$ and $BP$ intersect at point $N$. The angle bisector of $\angle M$ intersects $AQ$ a... | $$
\begin{array}{l}
\because \angle N P A \\
=\angle A Q B=\angle M Q C \\
=\angle M G A, \\
\angle N A P=\angle M A G, \\
\therefore \triangle M A G \sim \triangle N A P .
\end{array}
$$
$$
\text { Hence } \frac{M A}{M G}=\frac{N A}{N P} \text {, i.e., } \frac{M A}{M Q}=\frac{N A}{N P} \text {. }
$$
Since $M E$ and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,419 |
II. (Total 50 points) Let $x_{1}, x_{2}, \cdots, x_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ be two sets of real numbers satisfying the following conditions:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\sum_{i=1}^{n}\left|x_{i}\right|=1$;
(3) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}\left(a_{1}>a_{n}, n \geqslant 2\r... | Let $n=2$, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=0, \\
\left|x_{1}\right|+\left|x_{2}\right|=1 .
\end{array}\right.
\]
Then $x_{1}=\frac{1}{2}, x_{2}=-\frac{1}{2}$ or $x_{1}=-\frac{1}{2}, x_{2}=\frac{1}{2}$.
From $\left|\sum_{i=1}^{n} a_{i} x_{1}\right| \leqslant m\left(a_{1}-a_{n}\right)$,
we get $\frac{1}{2}... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,420 |
Three. (Total 50 points) In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies the above requirement. | Three, let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{i}<2$, then $\mathrm{C}_{x_{i}}^{2}=0$).
If an additional column is added to the side of a $13 \times 13$ square, a... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,421 |
Given three positive numbers $a, b, c$. Try to find all positive numbers $x, y, z, t$ that satisfy the following equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{a-z}=\frac{c-x-t}{c-x}, \\
\frac{x}{b-y}=\frac{a-x-z}{a-x}, \\
\frac{x}{c-t}=\frac{b-x-y}{b-x}, \\
x^{3}=y z t,
\end{array}\right. \\
\text { a... | Solution: From (1) we get $\frac{-x}{a-z}=\frac{-c+x+t}{c-x}, \frac{a-z-x}{a-z}=\frac{t}{c-x}$,
Substituting $a-z-x=\frac{(a-z) t}{c-x}$ into (2) yields
$$
\frac{t(a-z)}{c-x}=\frac{x(a-x)}{b-y} .
$$
Similarly, from (2) and (3) we get
$$
\frac{z(b-y)}{a-x}=\frac{x(b-x)}{c-t} \text {. }
$$
Again, from (3) and (1) we ge... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,422 |
Initial 122. Divide a rectangle with side lengths 3 and 4 into 5 convex shapes. Prove that for any method of division, there must be one of the shapes that contains two points, the distance between which is not less than $\sqrt{5}$. | Proof: As shown in Figure 3, $AE = BG = CM = DF = NH = EK = FT = 1$, and $EK, FT, NH$ are all parallel to side $AB$.
It is easy to see that in the above division, there are two points in each part, the distance between which is exactly $\sqrt{5}$, and the distance between any two points does not exceed $\sqrt{5}$.
Be... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,423 |
Example 2 Let $n, k$ be given positive integers, $k \leqslant n$, and call a permutation $\left\{x_{1}, x_{2}, \cdots,\right.$ $\left.x_{2 n}\right\}$ of the set $\{1,2, \cdots, 2 n\}$ "good" if there exists some $1 \leqslant i \leqslant 2 n-1$ such that $\left|x_{i}-x_{i+1}\right|=k$. Prove that the number of "good" p... | Analysis: This problem proves a strict inequality, that is, to prove that the number of "bad" sequences is less than the number of "good" sequences, so we need to construct an injective but not surjective mapping.
Proof: We call other permutations "bad".
For each "bad" permutation $\left\{x_{1}, x_{2}, \cdots, x_{2 n}\... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,426 |
Example 3 Connect all diagonals of a convex $(4k+3)$-gon, such that no three lines intersect at a point inside the shape. Let $P$ be a point inside the shape, and not on any diagonal. Prove: Among the quadrilaterals formed by the $(4k+3)$ vertices of the $(4k+3)$-gon, the number of quadrilaterals that contain point $P$... | Analysis: The conclusion of this problem suggests that we need to construct a $m$-fold mapping to achieve the purpose of the proof.
Proof: Let's assume that the number of triangles and quadrilaterals containing point $P$, with vertices from the vertices of a $4k+3$-gon, are $y$ and $x$ respectively.
When $k=0$, it is ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,427 |
Example 5 In a local area network consisting of several computers, for any $m(m \geqslant 3)$ computers there is a unique one that is connected to all of them. How many computers at most can be connected to the same computer in this local area network? | Analysis: We should first intuitively guess that the answer to this problem is $k = m$. To clarify the problem, it is best to use proof by contradiction.
Solution: Let the maximum number of computers that can be connected to the same computer be $k$. Without loss of generality, assume that computers $B_{1}, B_{2}, \cd... | m | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,429 |
Example 6 Write a positive integer $n$ as the sum of several 1s and several 2s, considering different orders of summands as different ways, and denote the number of all such ways as $\alpha(n)$; write $n$ as the sum of several positive integers greater than 1, considering different orders of summands as different ways,... | Proof: Let the set of all sequences where each term is 1 or 2, and the sum of the terms is $n$, be denoted as $A_{n}$; the set of all sequences where each term is a positive integer greater than 1, and the sum of the terms is $n$, be denoted as $B_{n}$.
For any $\left(a_{1}, a_{2}, \cdots, a_{m-1}, a_{m}\right)=a \in ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,430 |
Example 7 gives the definition:
(1) A subset $A \subseteq M=\{1,2,3, \cdots, 2 m\},(m$ $\left.\in \mathbf{Z}_{+}\right)$ is called good if it has the following property:
(1) If $2 k \in A(1 \leqslant k<m)$, then $2 k-1 \in A$ and $2 k+1 \in A$;
(2) If $2 m \in A, 2 m-1 \in A(\varnothing$ and $M$ are both good).
(2) A s... | \begin{array}{l}a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) \\ \Rightarrow a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] .\end{array} | a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+2}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,431 |
1. Let $S=\{1,2,3, \cdots, n\}, A$ be an arithmetic sequence with at least two terms,
with a positive common difference, all of whose terms are in $S$, and such that when the other elements of $S$ are added to $\boldsymbol{A}$, they cannot form an arithmetic sequence with the same common difference as $\boldsymbol{A}$... | (When $n=2 k$, in each sequence $A$ that satisfies the problem's requirements, there are two consecutive terms, such that the former term is in the set $\{1,2, \cdots, k\}$, and the latter term is in the set $\{k+1, k+2, \cdots, 2 k\}$; conversely, by selecting one number from $\{1,2, \cdots, k\}$ and one from $\{k+1, ... | \left[\frac{1}{4} n^{2}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,433 |
2. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1,2,3, \cdots, 14$, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. Find the number of all different ways to choose such numbers. | (Generally, we solve the following problem:
Find the number of different ways to choose $r$ numbers $a_{1}, a_{2}, \cdots, a_{r}$ from the set $\{1,2, \cdots, n\}$ that satisfy the following conditions:
(1) $1 \leqslant a_{1}<a_{2}<\cdots<a_{r} \leqslant n$;
(2) $a_{k+1}-a_{k} \geqslant m, k=1,2, \cdots, r-1$, where $m... | \mathrm{C}_{14-2 \cdot 2}^{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,434 |
Example 1 Solve the system of equations
$$
\left\{\begin{array}{l}
a x^{2}+b x+c=y, \\
a y^{2}+b y+c=z, \\
a z^{2}+b z+c=x .
\end{array}\right.
$$
where $a \neq 0$, and $(b-1)^{2}=4 a c$.
(1979, Hubei Province High School Mathematics Competition) | We find that the parabola $y=a x^{2}+b x+c$ is tangent to the line $y=x$ at the point $\left(\frac{1-b}{2 a}, \frac{1-b}{2 a}\right)$. This leads us to think that $(x, y)=(p, p)$ is a solution to the first equation of the system (where $p=\frac{1-b}{2 a}$). Similarly, $(y, z)=(p, p)$ and $(z, x)=(p, p)$ are solutions t... | (p, p, p) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,436 |
Example 2 Try to find a solution to the system of equations
$$
\begin{array}{l}
\frac{x\left(y^{2}+z^{2}\right)}{a \sqrt{p}}=\frac{y\left(z^{2}+x^{2}\right)}{b \sqrt{q}}=\frac{z\left(x^{2}+y^{2}\right)}{c \sqrt{r}} \\
=\sqrt{p q r}
\end{array}
$$
where $a, b, c, p, q, r$ are positive numbers satisfying
$$
\frac{s-a}{p... | Let $x=\sqrt{q r} u, y=\sqrt{p p} v, z=\sqrt{p q} w$, then the original system of equations transforms to
$$
\begin{array}{l}
\frac{u\left(r v^{2}+q w^{2}\right)}{a}=\frac{v\left(p w^{2}+r u^{2}\right)}{b} \\
=\frac{w\left(q u^{2}+p v^{2}\right)}{c}=1 .
\end{array}
$$
If $s-a \leqslant 0$, then $s-b \leqslant 0, s-c \... | (x, y, z)=(k \sqrt{q r}, k \sqrt{p}, k \sqrt{p q}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,437 |
Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-a}+\sqrt{z-a}=1, \\
\sqrt{z-b}+\sqrt{x-b}=1, \\
\sqrt{x-c}+\sqrt{y-c}=1 .
\end{array}\right.
$$
where $a, b, c$ are positive numbers, and
$$
\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{3}}{2} \text{. }
$$
(2nd National Mathematical Olympiad Training... | Solution 1: Construct an equilateral triangle $PQR$ with side length 1, then the sum of the distances from any point $O$ inside the triangle to the three sides is always equal to the height $\frac{\sqrt{3}}{2}$ (proof omitted). Due to equation (4), we can choose point $O$ such that the distances from $O$ to the three s... | (x, y, z)=\left(\frac{4}{3}(b+\sqrt{bc}+c), \frac{4}{3}(c+\sqrt{ca}+a), \frac{4}{3}(a+\sqrt{ab}+b)\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,438 |
Example 4 Solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-q r}+\sqrt{z-q r}=\sqrt{1-p}, \\
\sqrt{z-p}+\sqrt{x-p p}=\sqrt{1-q}, \\
\sqrt{x-p q}+\sqrt{y-p q}=\sqrt{1-r} .
\end{array}\right.
$$
where $p, q, r$ are positive numbers, and satisfy
$$
p+q+r+2 \sqrt{p q r}=1 .
$$ | Solution 1: Treating equation (4) as a quadratic equation in $\sqrt{p}$, we get $\sqrt{p}=\sqrt{(1-q)(1-r)}-\sqrt{q r}$.
Then $1-p$
$$
\begin{array}{c}
=q+r-2 q r+2 \sqrt{q r(1-q)(1-r)}, \\
\sqrt{1-p}=\sqrt{q(1-r)}+\sqrt{r(1-q)} .
\end{array}
$$
Therefore, $(x, y, z)=(p, q, r)$
satisfies equation (1).
Similarly, equat... | (x, y, z) = (p, q, r) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,439 |
Example 7 Given $0<a, b, c<1$, and
$$
\begin{array}{l}
\sqrt{a(1-b)(1-c)}+\sqrt{b(1-c)(1-a)} \\
+\sqrt{c(1-a)(1-b)}=\sqrt{a b c} .
\end{array}
$$
Try to solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-b c}+\sqrt{z-b c}=\sqrt{a}, \\
\sqrt{z-c a}+\sqrt{x-c a}=\sqrt{b}, \\
\sqrt{x-a b}+\sqrt{y-a b}=\sqrt... | Solution 1: (Algebraic Solution) Equation (1) can be transformed as follows:
$$
\begin{array}{l}
\sqrt{1-a}[\sqrt{b(1-c)}+\sqrt{c(1-b)}] \\
=\sqrt{a}[\sqrt{b c}-\sqrt{(1-b)(1-c)}] . \\
\because a>0, \\
\therefore \sqrt{\frac{1-a}{a}}=\frac{\sqrt{b c}-\sqrt{(1-b)(1-c)}}{\sqrt{b(1-c)+\sqrt{c(1-b)}}} . \\
\text { Hence } ... | (x, y, z) = (a, b, c) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,440 |
Example 4 As shown in Figure 3, in the rectangular coordinate system, the coordinates of vertex $B$ of rectangle $O A B C$ are $(15,6)$, and the line $y=\frac{1}{3} x+b$ exactly divides rectangle $O A B C$ into two parts of equal area. Therefore, $b=$ $\qquad$ (2000, National Junior High School Mathematics Competition) | Solution: By the symmetry of the rectangle, the line that bisects its area must pass through its center (i.e., the intersection of the diagonals).
Let the center be point $O^{\prime}$.
Given $B(15,6)$, then $O^{\prime}\left(\frac{15}{2}, 3\right)$.
Substituting into $y=\frac{1}{3} x+b$, we get $3=\frac{1}{3} \times \fr... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,443 |
Given $c \in\left(\frac{1}{2}, 1\right)$. Find the smallest constant $M$, such that for any integer $n \geqslant 2$ and real numbers $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, if $\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k}$, then $\sum_{k=1}^{n} a_{k} \leqslant M \sum_{k=1}^{m} a_{k}$, w... | Solution: The required minimum constant $M=\frac{1}{1-c}$.
$$
\begin{array}{l}
\because m=[c n], \text { and } c \in\left(\frac{1}{2}, 1\right), \\
\therefore c n-1c-\frac{2}{n}>\cdots>c-\frac{m}{n} \geqslant 0,
\end{array}
$$
and when $00, a_{m+1}$ $=a_{m+2}=\cdots=a_{n}>0$, and it can satisfy the given condition $\f... | \frac{1}{1-c} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,445 |
Let $a, b, c \in \mathbf{R}_{+}$. Prove that:
$$
\frac{a}{b^{2}}+\frac{b}{c^{2}}+\frac{c}{a^{2}} \geqslant \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text {. }
$$
This article generalizes inequality (1), obtaining the following
proposition: Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}, n>1$, $\alpha \beta>0$. Then
$$... | Proof: (Using Mathematical Induction)
(1) When $n=2$, equation (2) is
$$
\begin{array}{l}
\text { Left }- \text { Right }=\frac{x_{1}^{\alpha}}{x_{2}^{\beta}}+\frac{x_{2}^{\alpha}}{x_{1}^{\beta}}-x_{1}^{\alpha-\beta}-x_{2}^{\alpha-\beta} \\
=\frac{\left(x_{1}^{\alpha}-x_{2}^{\alpha}\right)\left(x_{1}^{\beta}-x_{2}^{\be... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,446 |
Proposition Let $D$, $E$, and $F$ be the midpoints of the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, and let
$$
\begin{array}{l}
a, CA=b, AB=c, \\
s=\frac{1}{2}(a+b+c),
\end{array}
$$
The areas of $\triangle AEF$, $\triangle BDF$, $\triangle CDE$, and $\triangle ABC$ are denoted by $\triangle \triangl... | Proof: By the definition of the midpoints of the sides of a triangle, we have
$$
\begin{array}{l}
s=AB + AE = c + AE, \\
s=AC + AF = b + AF,
\end{array}
$$
Thus, \( AE = s - c \) and \( AF = s - b \).
Since \(\sin A = \frac{a}{2R}\), \(\sin B = \frac{b}{2R}\), and \(\sin C = \frac{c}{2R}\),
$$
\begin{aligned}
\therefo... | \frac{\triangle^2}{2R} | Geometry | proof | Yes | Yes | cn_contest | false | 713,447 |
Proposition: Let the area of $\triangle ABC$ be $\triangle$, and the lengths of its three sides be $a$, $b$, and $c$. Then the minimum area of the inscribed equilateral triangle in $\triangle ABC$ is $\frac{\Delta^{2}}{\frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}\right)+2 \triangle}$. | Prove: As shown in Figure 1, the equilateral $\triangle PQR$ is inscribed in
$$
\begin{array}{l}
\triangle ABC, BC=a, CA \\
=b, AB=c.
\end{array}
$$
Let $\angle BRP=\theta$, then it is easy to find that $\angle PQC=\angle A + 60^\circ - \theta$. Let the side length of $\triangle PQR$ be $x$, then by the Law of Sines i... | \frac{\Delta^2}{\frac{\sqrt{3}}{6}(a^2 + b^2 + c^2) + 2 \Delta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,448 |
Let points $I$ and $H$ be the incenter and orthocenter of acute $\triangle ABC$, respectively. Points $B_{1}$ and $C_{1}$ are the midpoints of sides $AC$ and $AB$, respectively. It is known that ray $B_{1} I$ intersects side $AB$ at point $B_{2} \left(B_{2} \neq B\right)$, and ray $C_{1} I$ intersects the extension of ... | First, prove that
$A 、 I 、 A_{1}$ are collinear $\Leftrightarrow$ $\angle B A C=60^{\circ}$.
As shown in Figure 1, let $O$ be the circumcenter of $\triangle A B C$, and connect $B O, C O$. Then
$$
\begin{aligned}
& \angle B H C \\
= & 180^{\circ}-\angle B A C, \\
& \angle B A_{1} C \\
=2 & \left(180^{\circ}-\angle B H... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,449 |
II. Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor (gcd) of $a$ and $c$ is 1, and the gcd... | II. 72.
Express each positive integer $n$ not exceeding 100 as
$$
n=2^{\alpha_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdot q \text {. }
$$
where $q$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-negative ... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,450 |
Three, given a positive integer $n$, find the smallest positive number $\lambda$, such that for any $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$, as long as $\tan \theta_{1} \cdot \tan \theta_{2} \cdots \tan \theta_{n}=2^{\frac{n}{2}}$, then $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}$ is... | When $n=1$, $\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right)^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}$, we have $\lambda=\frac{\sqrt{3}}{3}$. When $n=2$, it can be proven that
$$
\cos \theta_{1}+\cos \theta_{2} \leqslant \frac{2 \sqrt{3}}{3},
$$
and the equality holds when $\theta_{1}=\theta_{2}=\arctan \sqrt{2}$. In fac... | n-1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,451 |
Find all triples of positive integers $(a, m, n)$ satisfying $a \geqslant 2, m \geqslant 2$ such that $a^{n}+203$ is divisible by $a^{m}+1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
---
Find all triples of positive int... | For $n, m$, we discuss in three cases.
(i) When $nm$, from $a^{m}+1 \mid 203\left(a^{m}+1\right)$, we have $a^{m}+11 a^{n}+203-\left(203 a^{m}+203\right)$ $=a^{m}\left(a^{n-m}-203\right)$.
Also, $\left(a^{m}+1, a^{m}\right)=1$, so $a^{m}+1 \mid a^{n-m}-203$.
(1) If $a^{n-m}203$, let $n-m=s \geqslant 1$, then
$$
a^{m}+1... | \begin{array}{l}
(2,2,4 k+1),(2,3,6 k+2),(2,4,8 k+8), \\
(2,6,12 k+9),(3,2,4 k+3),(4,2,4 k+4), \\
(5,2,4 k+1),(8,2,4 k+3),(10,2,4 k+2), \\
( | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,452 |
Five, a company needs to hire a secretary, with a total of 10 people applying. The company manager decides to interview them in the order of application, and the first 3 people will definitely not be hired. Starting from the 4th person, he will be compared with those who have been interviewed before. If his ability sur... | Let the rank of the strongest candidate among the first 3 interviewees be denoted as $a$. Clearly, $a \leqslant 8$. Let the set of permutations where the candidate ranked $k$ is selected be denoted as $A_{k}(a)$, and the number of such permutations be denoted as $\left|A_{k}(a)\right|$.
(1) It is easy to see that when ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,453 |
Example 5 Let the line $n x+(n+1) y=\sqrt{2}(n$ be a natural number) form a triangle with the two coordinate axes, with the area of the triangle being $S_{n}(n=1,2, \cdots, 2000)$. Then the value of $S_{1}+S_{2}+\cdots+S_{2000}$ is ( ).
(A) 1
(B) $\frac{1999}{2000}$
(C) $\frac{2000}{2001}$
(D) $\frac{2001}{2002}$
(2001... | Solution: The intersection points of the line $n x+(n+1) y=\sqrt{2}$ with the two coordinate axes are $\left(0, \frac{\sqrt{2}}{n+1}\right)$ and $\left(\frac{\sqrt{2}}{n}, 0\right)$, respectively.
Then $S_{n}=\frac{1}{n(n+1)}$. Therefore,
$$
\begin{array}{l}
S_{1}+S_{2}+\cdots+S_{2000} \\
=\frac{1}{1 \times 2}+\frac{1}... | \frac{2000}{2001} | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,454 |
Six, let $a, b, c, d$ be positive real numbers, satisfying $a b + a d = 1$. Points $P_{i}(x_{i}, y_{i}) (i=1,2,3,4)$ are four points on the unit circle centered at the origin. Prove:
$$
\begin{array}{l}
\left(a y_{1} + b y_{2} + c y_{3} + d y_{4}\right)^{2} + \left(a x_{4} + b x_{3} + c x_{2} + d x_{1}\right)^{2} \\
\l... | Let $u=a y_{1}+b y_{2}, v=c y_{3}+d y_{4}, u_{1}=a x_{4}+ b x_{3}, v_{1}=c x_{2}+d x_{1}$. Then
$$
\begin{aligned}
u^{2} & \leqslant\left(a y_{1}+b y_{2}\right)^{2}+\left(a x_{1}-b x_{2}\right)^{2} \\
& =a^{2}+b^{2}+2 a b\left(y_{1} y_{2}-x_{1} x_{2}\right),
\end{aligned}
$$
i.e., $x_{1} x_{2}-y_{1} y_{2} \leqslant \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,455 |
2. Given that $x, y$ are positive numbers, and $\frac{\sin \theta}{x}=\frac{\cos \theta}{y}, \frac{\cos ^{2} \theta}{x^{2}}$ $+\frac{\sin ^{2} \theta}{y^{2}}=\frac{10}{3\left(x^{2}+y^{2}\right)}$. Then the value of $\frac{x}{y}$ is $\qquad$. | $2 . \sqrt{3}$ or $\frac{1}{\sqrt{3}}$ | \frac{x}{y} = 2\sqrt{3} \text{ or } \frac{1}{\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,457 |
4. Given that the ratio of the areas of the upper base, lower base, and the four isosceles trapezoidal sides of a regular frustum is $2: 5: 8$. Then the angle between the side and the base is $\qquad$ . | 4. $\arccos \frac{3}{8}$ | \arccos \frac{3}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,459 |
5. If for all $x$ such that $|x| \leqslant 1$, $t+1>(t^2-4)x$ always holds, then the range of values for $t$ is $\qquad$ | 5. $\left(\frac{\sqrt{13}-1}{2}, \frac{\sqrt{21}+1}{2}\right)$ | \left(\frac{\sqrt{13}-1}{2}, \frac{\sqrt{21}+1}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,460 |
6. Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2}=5$. Then the maximum value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}$ $+(c-d)^{2}$ is $\qquad$ . | 6. 20 | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,461 |
Example 6 As shown in Figure 4, the line $y=-\frac{\sqrt{3}}{3} x +1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. An isosceles right triangle $\triangle ABC$ is constructed in the first quadrant with $AB$ as one of the legs, and $\angle BAC=90^{\circ}$. If there is a point $P\left(a, \frac... | Solution: From the given, we have $A(\sqrt{3}, 0) 、 B(0,1), O A=$ $\sqrt{3} 、 O B=1$.
Thus, $A B=\sqrt{O A^{2}+O B^{2}}=2$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 2 \times 2=2$.
Connect $P O$. Then
$$
\begin{array}{l}
S_{\triangle A B P}=S_{\triangle P B O}+S_{\triangle A B O}-S_{\triangle A P O} \\
=\frac{... | \frac{\sqrt{3}-8}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,465 |
II. (16 points)
Given that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, $S_{n}=a_{1}+a_{2}+\cdots+a_{n}, T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, and for all positive integers $n, \frac{S_{n}}{T_{n}}=\frac{3 n+31}{31 n+3}$.
(1) Find the value of $\frac{b_{28}}{a_{28}}$;
(2) Fin... | $$
\begin{array}{l}
\text { II. (1) } \frac{a_{n}}{b_{n}}=\frac{\frac{a_{1}+a_{2 n-1}}{2}}{\frac{b_{1}+b_{2 n-1}}{2}}=\frac{S_{2 n-1}}{T_{2 n-1}}=\frac{3(2 n-1)+31}{31(2 n-1)+3} \\
=\frac{3 n+14}{31 n-14}, \\
\end{array}
$$
Therefore, $\frac{b_{28}}{a_{28}}=\frac{31 \times 28-14}{3 \times 28+14}=\frac{31 \times 2-1}{3... | 1, 18, 35, 154 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,467 |
Three. (This question is worth 16 points)
Let $F$ be the set of all ordered $n$-tuples $\left(A_{1}, A_{2}, \cdots, A_{n}\right)$, where $A_{i}(1 \leqslant i \leqslant n)$ are subsets of the set $\{1,2,3, \cdots, 2002\}$. Let $|A|$ denote the number of elements in the set $A$. For all elements $\left(A_{1}, A_{2}, \cdo... | Three, $\because\{1,2, \cdots, 2002\}$ has $2^{2002}$ subsets,
$$
\therefore|F|=\left(2^{2002}\right)^{n}=2^{2020 n} \text {. }
$$
Let $a$ be a fixed element in $\{1,2, \cdots, 2002\}$. To ensure that $a$ is not in $A_{1} \cup A_{2} \cup \cdots \cup A_{n}$, each $A_{i}(1 \leqslant i \leqslant n)$ must not contain $a$.... | 2002\left(2^{2002 n}-2^{2001 n}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,468 |
Four. (This question is worth 18 points)
On a piece of paper, there are $1,2, \cdots, n$ these $n$ positive integers. In the first step, the first 4 numbers $1,2,3,4$ are crossed out, and the sum of the 4 crossed-out numbers, 10, is written at the end of $n$; in the second step, the first 4 numbers $5,6,7,8$ are crosse... | (1) With each step, the number of numbers on the paper decreases by 3. If $n$ numbers are reduced to 1 number after $p$ steps, then $n-3p=1$, i.e., $n=3p+1$.
$\therefore$ The necessary and sufficient condition for $n$ is that $n$ leaves a remainder of 1 when divided by 3.
(2) Suppose there are $4^{k}$ numbers initially... | 12880878 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,469 |
1. The inverse function of the function $y=f(-x)$ defined on the set of real numbers $\mathbf{R}$ is $y=f^{-1}(-x)$, then ( ).
(A) $y=f(x)$ is an odd function
(B) $y=f(x)$ is an even function
(C) $y=f(x)$ is both an odd and an even function
(D) $y=f(x)$ is neither an odd nor an even function | -1 (A).
From $y=f^{-1}(-x)$, we get $f(y)=-x$. Therefore, $y=-f(x)$ is the inverse function of $y=f^{-1}(-x)$, i.e., $-f(x)=f(-x)$.
This shows that $y=f(x)$ is an odd function. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,470 |
2. The graph of the quadratic function $f(x) = a x^{2} + b x + c$ is shown in Figure 1. Let
$$
\begin{array}{c}
N=|a+b+c| \\
+|2 a-b|, \\
M=|a-b+c| \\
+|2 a+b| \text {. Then ( ). }
\end{array}
$$
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) The relationship between $M$ and $N$ cannot be determined | $$
\begin{array}{l}
\text { 2. (C). } \\
\text { As shown in Figure 1, } f(1)=a+b+c0, a>0,-\frac{b}{2 a}>1 \text {. }
\end{array}
$$
2. (C).
As shown in Figure 1, $f(1) = a + b + c0, a > 0, -\frac{b}{2a} > 1$.
Thus, $b < 0$.
Also, $f(0) = c0$, hence
$M - N = |a - b + c| + |2a + b| - |a + b + c| - |2a - b| = (a - b + c... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,471 |
$$
\begin{array}{l}
\quad \text { 4. In } \triangle A B C \text {, if }(\sin A+\sin B)(\cos A+\cos B) \\
=2 \sin C \text {, then } \triangle A B C(\quad .
\end{array}
$$
4. In $\triangle A B C$, if $(\sin A+\sin B)(\cos A+\cos B)$ $=2 \sin C$, then $\triangle A B C(\quad)$.
(A) is an isosceles triangle but not necessar... | 4. (A).
$$
\begin{array}{l}
\text { Left side }=\sin A \cdot \cos A+\sin A \cdot \cos B+\cos A \cdot \sin B+ \\
\sin B \cdot \cos B=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B) \\
=\sin (A+B) \cos (A-B)+\sin (A+B). \\
\text { Right side }=2 \sin \left[180^{\circ}-(A+B)\right]=2 \sin (A+B). \\
\quad \text { Therefore, } \s... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,473 |
5. In $\triangle A B C$, $\angle C=90^{\circ}$. If $\sin A$ and $\sin B$ are the two roots of the quadratic equation $x^{2}+p x+q=0$, then which of the following relationships is correct? ( ).
(A) $p= \pm \sqrt{1+2 q}$ and $q>-\frac{1}{2}$
(B) $p=\sqrt{1+2 q}$ and $q>-\frac{1}{2}$
(C) $p=-\sqrt{1+2 q}$ and $q>-\frac{1}... | 5. (D).
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
\sin A+\sin B=-p>0, \\
\sin A \sin B=q>0 .
\end{array}\right.
$$
That is,
$$
\left\{\begin{array}{l}
\sin A+\cos A=-p>0, \\
\sin A \cdot \cos A=q>0 .
\end{array}\right.
$$
Thus,
$$
\left\{\begin{array}{l}
p^{2}-2 q=1, \\
p^{2}-4 q \geqslant 0, \\
p<0 .... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,474 |
6. Given points $A(-7,0), B(7,0), C(2,-12)$. If one focus of an ellipse is $C$, and the ellipse passes through points $A$ and $B$, the locus of the other focus of the ellipse is ( ).
(A) Hyperbola
(B) Ellipse
(C) Part of an ellipse
(D) Part of a hyperbola | 6. (D).
Let the other focus of the ellipse be $F(x, y)$.
Since $A$ and $B$ are points on the ellipse, by the definition of an ellipse, we have $|A C|+|A F|=|B C|+|B F|$,
thus $|B F|-|A F|=|A C|-|B C|$.
Given $|A C|=15,|B C|=13$, we get $|B F|-|A F|=2$, hence the locus of point $F$ is part of a hyperbola. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,475 |
Example 7 A home appliance manufacturing company, based on market research analysis, has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be ... | Solution: Let the number of air conditioners produced each week be $x$, the number of color TVs be $y$, and the number of refrigerators be $z$, with the total output value being $w$ thousand yuan. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=360, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=120, \... | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,476 |
7. The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ . | ii. 7.8 items.
$X$ must include the 3 elements $1,2,3$, while the numbers 4,5,6 may or may not belong to $X$. Each number has 2 possibilities, so the total number of different $X$ is $2^{3}=8$. | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,477 |
10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
（1）For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} 、 x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$ | 10.0 .
From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefo... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,479 |
11. Given the complex number $z$ satisfies $z \cdot \bar{z}-z-\bar{z}=3$, and $\arg (z-1)$ $=\frac{\pi}{3}$. Then $z=$ $\qquad$ . | $$
11.2+\sqrt{3} i \text {. }
$$
Let $w=z-1$, and $|w|=r$, then $z=w+1$.
$$
\text { Hence }(w+1)(\bar{w}+1)-(w+1)-(\bar{w}+1)=3 \text {. }
$$
From this, we get $r^{2}=4, r=2$. Therefore,
$$
\begin{array}{l}
w=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=1+\sqrt{3} i \\
\therefore z=2+\sqrt{3} i
\end{array}
$... | 2+\sqrt{3} i | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,480 |
12. Given a moving point $P(x, y)$ satisfies the quadratic equation $10 x-2 x y-2 y+1=0$. Then the eccentricity of this quadratic curve is $\qquad$. | $12 \cdot \sqrt{2}$.
From $10 x-2 x y-2 y+1=0$ we have
$$
\begin{array}{l}
x^{2}+6 x+y^{2}-6 y-2 x y+9 \\
=x^{2}-4 x+4+y^{2}-4 y+4 .
\end{array}
$$
That is, $\sqrt{(x-2)^{2}+(y-2)^{2}}=|x-y+3|$,
which means $\frac{\sqrt{(x-2)^{2}+(y-2)^{2}}}{\frac{|x-y+3|}{\sqrt{2}}}=\sqrt{2}$, hence $e=\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,481 |
13. (12 points) As shown in Figure 3, in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ and $F$ are the midpoints of edges $A B$ and $B C$ respectively.
(1) Find the size of the dihedral angle $B F B_{1}-E$;
(2) Find the distance from point $D$ to the plane $B_{1} E F$;
(3) Can a point $M$ be found ... | (1) As shown in Figure 6, draw $B H \perp B_{1} F$, with the foot of the perpendicular at $H$, and connect $E H$.
By the properties of a cube, $E B \perp$ plane $B B_{1} F$, so $B H$ is the projection of $E H$ in plane $B B_{1} F$.
By the theorem of three perpendiculars, $E H \perp B_{1} F$.
Thus, $\angle E H B$ is t... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,482 |
14. (13 points) Let the quadratic equation in $x$, $2 x^{2}-t x-2=0$, have two roots $\alpha, \beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove that $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minim... | $$
\begin{aligned}
14. (1) & \text{ From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text{ Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2} \\
& -t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$
Th... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,483 |
15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
| 15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\r... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,484 |
16. (13 points) A
schematic diagram of a computer device is shown in Figure 4, where $J_{1}$ and $J_{2}$ represent data inputs, and $C$ is the output for the computation result. The computation process involves inputting natural numbers $m$ and $n$ through $J_{1}$ and $J_{2}$ respectively, and after computation, a nat... | 16. When $J_{1}$ inputs $m, J_{2}$ inputs $n$, the output result is denoted as $f(m, n)$. Let $f(m, n)=k$, then $f(1,1)=1, f(m, n+1)$ $=f(m, n)+2, f(m+1,1)=2 f(m, 1)$.
(1) Since $f(1, n+1)=f(1, n)+2$, the sequence $f(1,1)$, $f(1,2), \cdots, f(1, n), \cdots$ forms an arithmetic sequence with the first term $f(1,1)$ and ... | 2^{2001}+16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,485 |
Example 8 Given the line $y=2 x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and points $P$ and $Q$ have coordinates $P(-2,0)$ and $Q(0, k)$, where $k<6$. A circle is drawn with point $Q$ as the center and $PQ$ as the radius. Then
(1) For what value of $k$ does the circle $\odot Q$ touch ... | Solution: (1) From the given, we have $A(-3,0)$, $B(0, 6)$. Therefore,
$$
\begin{array}{l}
B Q=6-k, \\
P Q=\sqrt{k^{2}+4}, \\
A B=3 \sqrt{5} .
\end{array}
$$
As shown in Figure 5, draw $Q R \perp A B$ at $R$, then
Rt $\triangle B Q R \backsim$ Rt $\triangle B A O$, so
$$
\frac{B Q}{A B}=\frac{Q R}{O A} \text {. }
$$
... | k=1 \text{ or } k=-4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,486 |
17. (13 points) With $A$ as the center and $2 \cos \theta\left(\frac{\pi}{4}<\theta<\frac{\pi}{2}\right)$ as the radius, there is a point $B$ outside the circle. It is known that $|A B|=2 \sin \theta$. Let the center of the circle passing through point $B$ and externally tangent to $\odot A$ at point $T$ be $M$.
(1) Wh... | 17. (1) As shown in Figure 7, connect $M T$, $M A$, and $M B$. Clearly, points $M$, $T$, and $A$ are collinear, and $|M A| - |M T| = |A T| = 2 \cos \theta$.
Also, $|M T| = |M B|$, so $|M A| - |M B| = 2 \cos \theta$, where $A B = 2 \sin \theta$.
Therefore, the locus of point $M$ is the branch of a hyperbola with foci ... | 0 < f(\theta) < 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,487 |
18. (14 points) Let the length, width, and height of a rectangular prism be $a$, $b$, and $c$, respectively, and its body diagonal be $l$. Prove:
$$
\left(l^{4}-a^{4}\right)\left(l^{4}-b^{4}\right)\left(l^{4}-c^{4}\right) \geqslant 512 a^{4} b^{4} c^{4} .
$$ | 18. The original inequality is equivalent to
$$
\left(\frac{l^{4}}{a^{4}}-1\right)\left(\frac{l^{4}}{b^{4}}-1\right)\left(\frac{l^{4}}{c^{4}}-1\right) \geqslant 512 \text {. }
$$
Let \( x=\frac{a^{2}}{l^{2}}, y=\frac{b^{2}}{l^{2}}, z=\frac{c^{2}}{l^{2}} \), then \( x+y+z=1 \), and the original inequality can be rewrit... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,488 |
1. Given the sets $P=\left\{x \mid x^{2}=1\right\}$ and $Q=\{x \mid m x=1\}$. If $Q \subset P$, then the number of possible values for the real number $m$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | $-1 .(\mathrm{D})$.
$$
\because P=\{-1,1\}, Q \subset P \text {, }
$$
$\therefore Q$ could be $\varnothing,\{-1\},\{1\}$.
When $Q=\varnothing$, $m=0$;
When $Q=\{-1\}$, $m=-1$;
When $Q=\{1\}$, $m=1$.
Thus, $m$ has 3 values: $0,-1,1$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,489 |
2. If $a, b$ are any real numbers, and $a>b$, then the inequality that must be true is ( ).
(A) $a^{2}>b^{2}$
(B) $\frac{b}{a}<1$
(C) $a-b>0$
(D) $\left(\frac{1}{2}\right)^{a}<\left(\frac{1}{2}\right)^{b}$ | 2. (D).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Inequalities | MCQ | Yes | Yes | cn_contest | false | 713,490 |
3. If the circle $x^{2}+y^{2}=k^{2}$ covers at least one maximum point and one minimum point of the function $f(x)$ $=\sqrt{3} \sin \frac{\pi x}{k}$, then the range of $k$ is ( ).
(A) $|k| \geqslant 3$
(B) $|k| \geqslant 2$
(C) $|k| \geqslant 1$
(D) $1 \leqslant|k| \leqslant 2$ | 3. (B).
$\because f(x)=\sqrt{3} \sin \frac{\pi x}{k}$ is an odd function, the graph is symmetric about the origin,
$\therefore$ the circle $x^{2}+y^{2}=k^{2}$ only needs to cover one of the extreme points of $f(x)$.
Let $\frac{\pi x}{k}=\frac{\pi}{2}$, solving gives the nearest maximum point $P\left(\frac{k}{2}, \sqrt... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,491 |
4. Given $\overrightarrow{O P}=(2,1), \overrightarrow{O A}=(1,7), \overrightarrow{O B}=(5,1)$. Let $X$ be a point on the line $O P$ (where $O$ is the origin). Then, the value of $\angle A X B$ when $\overrightarrow{X A} \cdot \overrightarrow{X B}$ is minimized is ( ).
(A) $90^{\circ}$
(B) $\arccos \frac{4 \sqrt{17}}{17... | 4. (C).
Let $\overrightarrow{O X}=\left(x_{0}, y_{0}\right), \overrightarrow{O P}=(2,1)$, then $\frac{x_{0}}{2}=\frac{y_{0}}{1}$.
$\therefore x_{0}=2 y_{0}$, then
$$
\begin{array}{c}
\overrightarrow{X A}=\left(1-2 y_{0}, 7-y_{0}\right), \overrightarrow{X B}=\left(5-2 y_{0}, 1-y_{0}\right) . \\
\therefore \overrightarr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,492 |
5. A triangular pyramid has three lateral faces, two of which are isosceles right triangles, and the other is an equilateral triangle with a side length of 1. Then, the volume of this triangular pyramid ( ).
(A) has a unique value
(B) has 2 different values
(C) has 3 different values
(D) has more than 3 different value... | 5. (C).
(1) As shown in Figure 2 (1), if $S A=S B=S C=B C=1$, $S A \perp S B, S A \perp S C$, then $V_{S A B C}=V_{A S B C}=\frac{\sqrt{3}}{12}$;
(2) As shown in Figure 2 (2), if $S B=S C=B C=1, S A=\frac{\sqrt{2}}{2}$, $S A \perp A B, S A \perp A C$, then $V_{S A B C}=\frac{\sqrt{2}}{24}$;
(3) As shown in Figure 2 (3)... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,493 |
6. On a plane, there are 2 fixed points $A$ and $B$, and 4 other points $C_{1}, C_{2}, C_{3}, C_{4}$ that do not coincide with $A$ or $B$. If $\sin \angle A C_{i} B - \sin \angle A C_{j} B \left\lvert\, \leqslant \frac{1}{3}(i \neq j, i, j=1,2,3,4)\right.$, then $(C_{i}, C_{j})$ is called a good point pair. Then, such ... | 6. (B).
$$
\begin{array}{l}
\because \angle A C_{i} B \in[0, \pi], \\
\therefore \sin \angle A C_{i} B \in[0,1](i=1,2,3,4) .
\end{array}
$$
Divide the interval $[0,1]$ into three segments: $\left[0, \frac{1}{3}\right],\left(\frac{1}{3}, \frac{2}{3}\right],\left(\frac{2}{3}, 1\right]$. Then, among $\sin \angle A C_{1} ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,494 |
7. Given the functions $f(x)=x^{2}+2 b x+1$ and $g(x)=2 a(x+b)$, where $x$, $a$, and $b$ are all real numbers. The set of real number pairs $(a, b)$ that make the graphs of $y=f(x)$ and $y=g(x)$ not intersect in the $x O y$ plane is denoted as $A$. Then, the area of the figure $S$ represented by $A$ in the $a O b$ plan... | Let the pair of numbers $(a, b)$ satisfy the requirement, such that the equation $x^{2}+2 b x+1$ $=2 a(x+b)$ has no real roots.
$$
\begin{array}{l}
\therefore \Delta=[2(b-a)]^{2}-4(1-2 a b)<0, \text { which gives } \\
a^{2}+b^{2}<1 .
\end{array}
$$
In the $a O b$ plane, the set of points $(a, b)$ that satisfy this con... | \pi | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,495 |
8. Given the sequence $\left\{a_{n}\right\}$, where $a_{n}$ is a real number, and for $n \geqslant 3, n \in$ $\mathbf{N}$, we have $a_{n}=a_{n-1}-a_{n-2}$. If the sum of the first 1985 terms is 1000, and the sum of the first 1995 terms is 4000, then the sum of the first 2002 terms is $\qquad$. | 8.3000
Let $a_{1}=a, a_{2}=b$, then the first 6 terms of the sequence are $a, b, b-a$, $-a, -b, a-b$. Also, $a_{7}=a, a_{8}=b$, it is easy to verify that the sum of any consecutive 6 terms of this sequence is 0.
$$
\begin{aligned}
\therefore S_{1995} & =S_{332 \times 6+3}=0+a+b+(b-a) \\
& =2 b=4000 . \\
S_{1 \text { g... | 3000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,496 |
1. The graph of a linear function is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. It also passes through the point $(-1, -25)$. Then, on the line segment $AB$ (including endpoints $A$ and $B$), the number of points with both integer c... | (
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,497 |
9. Two people, A and B, agree to meet at a certain place within 10 days. The person who arrives first will wait for the other, but can leave after 3 days. If they are equally likely to arrive at the destination within the limit, then the probability that the two will meet is $\qquad$ . | 9. $\frac{51}{100}$.
Let A and B arrive at a place on the $x$-th and $y$-th day respectively, $0 \leqslant x \leqslant 10, 0 \leqslant y \leqslant 10$. The necessary and sufficient condition for them to meet is $|x-y| \leqslant 3$. The point $(x, y)$ is distributed within the square $O A B C$ as shown in Figure 3, and... | \frac{51}{100} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,498 |
10. For a line segment $AB$ of fixed length $m$ whose endpoints move on the right branch of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $\left(m>\frac{2 b^{2}}{a}\right)$. Then, the minimum value of the x-coordinate of the midpoint $M$ of $AB$ is (expressed in terms of $a$, $b$, and $m$). | 10. $\frac{a(m+2 a)}{2 \sqrt{a^{2}+b^{2}}}$.
As shown in Figure 4, let the projections of $A$, $B$, and $M$ on the right directrix of the hyperbola be $A^{\prime}$, $B^{\prime}$, and $M^{\prime}$, respectively, and the right focus be $F$, with the eccentricity being $e$.
By the definition of the hyperbola, we have
$$
\... | \frac{a(m+2 a)}{2 \sqrt{a^{2}+b^{2}}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,499 |
11. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ are points on the diagonals $A C$ and $B F$ respectively, and $A M = F N$. If $A B=1$, then the range of values for $M N$ is $\qquad$ | 11. $\left[\frac{\sqrt{3}}{2}, 1\right]$.
As shown in Figure 5, draw $MP \perp AB$ at $P$, and connect $PN$. By the given conditions, we have $PN \perp AB$.
Therefore, $\angle MPN$ is the plane angle of the dihedral angle $D \cdot AB-E$, i.e., $\angle MPN = 120^{\circ}$.
Let $AM = FN = x$, then $MP = \frac{\sqrt{2}}{... | \left[\frac{\sqrt{3}}{2}, 1\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,500 |
13. Let the quadratic function $f(x)$ satisfy $f(x+2)=f(-x+2)$, and its graph intersects the $y$-axis at the point $(0,1)$, and the segment it intercepts on the $x$-axis has a length of $2 \sqrt{2}$. Find the analytical expression of $f(x)$. | Three, 13.
$f(x)$ is symmetric about $x=2$, let $f(x)=a(x-2)^{2}+b$. According to the problem, $4 a+b=1$. Suppose the graph of $f(x)$ intersects the $x$-axis at $x_{1}$ and $x_{2}$, then $\left|x_{1}-x_{2}\right|=2 \sqrt{2}$, i.e., $2 \sqrt{-\frac{b}{a}}=2 \sqrt{2}$. Solving this, we get $\left\{\begin{array}{l}a=\frac... | f(x)=\frac{1}{2}(x-2)^{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,501 |
14. Let $a \in \mathbf{N}_{+}, a \geqslant 2$, and the sets $A=\left\{y \mid y=a^{x}\right.$, $\left.x \in \mathbf{N}_{+}\right\}, B=\left\{y \mid y=(a+1) x+b, x \in \mathbf{N}_{+}\right\}$. Does there exist $b$ in the closed interval $[1, a]$ such that $A \cap B \neq \varnothing$? If it exists, find all possible value... | 14. Let $b \in [1, a]$ such that $A \cap B \neq \varnothing$, i.e., there exists $y_{0} \in A$ and $y_{0} \in B$, such that $y_{0} = a^{m}$ ($m \in \mathbf{N}_{+}$) and $y_{0} = (a+1)n + b$ ($n \in \mathbf{N}_{+}$). Then there should exist $m, n \in \mathbf{N}_{+}$ such that $a^{m} = (a+1)n + b$ ($1 \leqslant b \leqsla... | b = 1 \text{ or } b = a | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,502 |
15. As shown in Figure $1, \odot O$ is the circumcircle of $\triangle A B C$, and point $I$ is its incenter. The rays $A I, B I, C I$ intersect the opposite sides at points $D, E, F$, respectively, and the rays $A D, B E, C F$ intersect $\odot O$ at points $A^{\prime}, B^{\prime}, C^{\prime}$, respectively.
Prove: $A ... | 15. As shown in Figure 6, draw the diameter \( A^{\prime} G \) of \( \odot O \), and \( I K \perp B C \) at point \( K \).
Let the radius of \( \odot O \) be \( R \), and the radius of the incircle of \( \triangle A B C \) be \( r \), then \( A^{\prime} G = 2 R \), \( I K = r \).
\[
\because I \text{ is the incenter o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,503 |
12. The expansion of $(a+b+c)^{10}$, after combining like terms, has
$\qquad$ terms. | 12.66.
The form of each term after merging is $a^{k_{1}} b^{k_{2}} c^{k_{3}}$, where $k_{1}+k_{2}+k_{3}=10\left(k_{1} 、 k_{2} 、 k_{3} \in \mathrm{N}\right)$, and when $k_{1} 、 k_{2}$ are determined, $k_{3}$ is uniquely determined.
If $k_{1}=10, k_{2}=0$, there is 1 way;
If $k_{1}=9, k_{2}=0$ or 1, there are 2 ways;
If... | 66 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,504 |
1. Let $k$ and $n$ be positive integers, and $n>2$. Prove: The equation $x^{n}-y^{n}=2^{k}$ has no positive integer solutions.
（53rd Romanian Mathematical Olympiad Final） | Proof: Proof by contradiction.
Let $n_{0}>2$ be the smallest one satisfying $x^{n_{0}}-y^{n_{0}}=2^{m}(m>0)$.
If $n_{0}$ is even, let $n_{0}=2 l, l \in \mathbf{N}$, then $x^{2 l}-y^{2 l}=\left(x^{l}-y^{l}\right)\left(x^{l}+y^{l}\right)$, so $x^{l}-y^{l}$ is a power of 2, which contradicts the minimality of $n_{0}$.
I... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,505 |
2. Let $a, b$ be positive real numbers. For any positive integer $n$, let $x_{n}$ be the sum of the digits in the decimal representation of $[a n+b]$. Prove: The sequence $\left\{x_{n}\right\}$ contains a subsequence that is constant.
(2002, Romania for IMO and Balkan Mathematical Olympiad selection test (second round)... | Proof: For any integer $k$, define
$$
n_{k}=\left[\frac{10^{k}+a-b}{a}\right] \text {. }
$$
Then $10^{k}=a\left(\frac{10^{k}+a-b}{a}-1\right)+b$
$$
a$. Therefore, $x_{n_{k}}$ is the sum of the digits of some number $t$ in the set $\{0,1, \cdots,[b]\}$ plus 1. Since $k$ can take infinitely many values, and $t$ is finit... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,506 |
4. Three sticks each have a length no less than $n$, where $n$ is a positive integer. If the sum of the lengths of these three sticks is $\frac{n(n+1)}{2}$, prove: the three sticks can be cut into $n$ segments, with lengths 1, $2, 3, \cdots, n$.
(2002, Bulgarian Winter Mathematical Competition) | Proof: Let the lengths of the three sticks be $a_{1}, a_{2}$, and $a_{3}$, and assume without loss of generality that $a_{1} \leqslant a_{2} \leqslant a_{3}$. Let $a_{1}=n+p, a_{2}=n+q$, and $a_{3}=n+r$, then $p \leqslant q \leqslant r$. If $r \geqslant n-1$, we can cut $a_{3}$ into two segments of lengths $n$ and $r$.... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,507 |
2. As shown in Figure 6, the line $y=$ $-2 x+10$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. If $\triangle A O B$ is folded along $A B$, point $O$ lands at point $C$. The coordinates of point $C$ are $\qquad$
(1999, Huanggang City, Hubei Province, Junior High School Mathematics Competition... | (Answer: $C(8,4)$ ) | C(8,4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,508 |
3. Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy that for all $n \geqslant 2, a_{n+1}$ is the smallest prime factor of $a_{n-1}+a_{n}$, and the fractional part of a real number $x$ is written in the order of $a_{1}, a_{2}, \cdots, a_{n}, \cdots$. Prove: $x$ is a rational number.
(2002, Romania fo... | Proof: By parity, one of the first five numbers must be 2.
If $a_{i}=a_{i+1}=2$, then $a_{n}=2, n \geqslant i$.
If there are two adjacent numbers 2,3 or 3,2, then a periodic sequence can be obtained: $2,3,5,2,7,3,2,5,7,2,3, \cdots$
If $a_{i}=2$ and $a_{i+1}$ is an odd prime greater than 3, then either $a_{i+1} \equiv ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,509 |
5. Given the sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ defined as follows: $x_{1}=3, y_{1}=4$, $x_{n+1}=3 x_{n}+2 y_{n}, y_{n+1}=4 x_{n}+3 y_{n}, n \geqslant 1$. Prove: $x_{n} 、 y_{n}$ cannot be expressed as the cube of an integer.
(2002, Bulgarian Winter Mathematical Competition) | Prove: Since $2 x_{n+1}^{2}-y_{n+1}^{2}=2\left(3 x_{n}+2 y_{n}\right)^{2}-\left(4 x_{n}+3 y_{n}\right)^{2}=2 x_{n}^{2}-y_{n}^{2}=\cdots=2 x_{1}^{2}-y_{1}^{2}=2$, it means to prove that the equations $2 x^{6}-y^{2}=2$ and $2 x^{2}-y^{6}=2$ have no positive integer solutions.
Assume $2 x^{6}-y^{2}=2$ has a solution, let... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,510 |
6. Find all positive integers that have 16 divisors, and the sum of all their divisors is equal to 4032.
(2002, Bulgarian Spring Mathematical Competition) | Solution: If a positive integer has 16 divisors, it can only be one of the following cases: $p^{15}, p^{7} q, p^{3} q^{3}, p^{3} q r, p q r s$, where $p, q, r$, and $s$ are all prime numbers. The sum of all divisors are respectively
$$
\begin{array}{l}
1+p+p^{2}+\cdots+p^{15} \\
=(1+p)\left(1+p^{2}\right)\left(1+p^{4}\... | 1722, 1794, 2002, 2145 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,511 |
7. Find all positive integers $n$ such that $n-[n\{\sqrt{n}\}]=2$ holds.
$(2002$, Bulgarian Spring Mathematical Competition) | Solution: If $n=p^{2}$, then $\{\sqrt{n}\}=0$, which implies $n=2$, contradicting $n=p^{2}$. Therefore, there exists an integer $t$ such that $(t-1)^{2}<n<t^{2}$, hence $[\sqrt{n}]$
$$
\begin{aligned}
=t & -1 \text {. Also, } \\
& n-[n\{\sqrt{n}\}]=n-[n(\sqrt{n}-[\sqrt{n}])] \\
& =n-[n(\sqrt{n}-t+1)]=n-[n \sqrt{n}]+n t... | 2,8,15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,512 |
8. Let $p$ and $n$ be positive integers, and $p$ be a prime number, such that $1 + np$ is a perfect square. Prove that $n + 1$ can be expressed as the sum of $p$ perfect squares.
(19th Iranian Mathematical Olympiad (First Round)) | Proof: Let $n p+1=k^{2}$, then $n p=(k-1)(k+1)$. If $p \mid(k-1)$, set $k=p l+1$, then
$$
\begin{array}{l}
n p+1=k^{2}=(p l+1)^{2}=p^{2} l^{2}+2 p l+1, \\
n+1=p l^{2}+2 l+1=(p-1) l^{2}+(l+1)^{2} .
\end{array}
$$
If $p \mid (k+1)$, set $k=p l-1$, then
$$
\begin{array}{l}
n p+1=k^{2}=(p l-1)^{2}=p^{2} l^{2}-2 p l+1 \\
n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,513 |
9. (1) Positive integers $p, q, r, a$ satisfy $pq = ra^2$, and $r$ is a prime, $p, q$ are coprime. Prove: one of $p, q$ is a perfect square?
(2) Does there exist a prime $p$ such that $p(2^{p+1}-1)$ is a perfect square.
(19th Greek Mathematical Olympiad) | Solution: (1) Let $p=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{m}^{k_{m}}, q=q_{1}^{3} q_{z}^{3} \cdots q_{n}^{\prime}, a=$ $a_{1}^{t_{1}} a_{2}^{i} \cdots a_{t}^{t}$, where $p_{i}, q_{j}, a_{h}$ are all prime numbers, and $\left(p_{i}, q_{j}\right)=1$. Then we have $p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{m}^{k_{m}} q_{1}^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,514 |
10. Solve the equation $\left(x_{5}\right)^{2}+\left(y^{4}\right)_{5}=2 x y^{2}+51$, where $n_{5}$ denotes the closest multiple of 5 to the integer $n$, and $x, y$ are integers.
(51st Czech and Slovak Mathematical Olympiad (Open Question)) | Solution: Since $\left(x_{5}\right)^{2}+\left(y^{4}\right)_{5}$ can be divisible by 5, and $2 x y^{2}$ has a remainder of 4 when divided by 5, i.e., $51\left(2 x y^{2}-4\right)$, then $y=5 k \pm 1$ or $5 k \pm 2$, at this time $y_{5}=5 k$.
If $y=5 k \pm 1$, by $51\left(y^{2}-1\right)$, we get $51(2 x-4)$, i.e., $51(x-... | (52,6), (52,-6), (8,2), (8,-2), (3,3), (3,-3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,515 |
11. Prove: A positive integer $A$ is a perfect square if and only if for any positive integer $n$, $(A+1)^{2}-A,(A+2)^{2}-A$, $\cdots,(A+n)^{2}-A$ contains at least one term that is divisible by $n$.
(51st Czech and Slovak Mathematical Olympiad (Final)) | Solution: If $A=d^{2}$, then $(A+j)^{2}-A=\left(d^{2}+j\right)^{2}-d^{2}=$ $\left(d^{2}-d+j\right)\left(d^{2}+d+j\right)$. Since $d^{2}-d+j$ for $j=1$, $2, \cdots, n$ are $n$ consecutive positive integers, there must be some $j$ such that $(A+j)^{2}-A$ can be divisible by $n$.
If $A$ is not a perfect square, then $A$ ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,516 |
12. Given $p$ is a prime number, $r$ is the remainder when $p$ is divided by 210. If $r$ is a composite number and can be expressed as the sum of two perfect squares, find
(52nd Belarusian Mathematical Olympiad (Final C Category)) | Let $p=210n+r$, then $07$.
Let $q$ be the smallest prime dividing $r$, i.e., $r=qm, q \leqslant m$, we get $210>r=qm \geqslant q^{2}$, so, $q \leqslant 13$.
On the other hand, $r$ cannot be divisible by 2, 3, 5, 7, otherwise $p$ would also be divisible by the same number, but $p$ is a prime, so $q>7$, hence $q=11$ or $... | 169 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,517 |
13. Prove: a positive integer can be written as the sum of consecutive positive integers if and only if this number is not an integer power of 2.
(2002, Croatian National Mathematical Competition) | Proof: Suppose $n$ can be written as the sum of several consecutive positive integers, i.e., $\exists m, k \in \mathbf{N}$, such that $n=m+(m+1)+\cdots+(m+k)$. Then
$$
\begin{aligned}
n & =(k+1) m+1+2+\cdots+k \\
& =(k+1) m+\frac{k(k+1)}{2}=(k+1)\left(m+\frac{k}{2}\right) .
\end{aligned}
$$
If $k$ is even, then $n$ ca... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,518 |
Example 1 If $x, y, z$ are positive real numbers, and satisfy $xyz = 1$, then the minimum value of the algebraic expression $(x+1)(y+1)(z+1)$ is ( . ).
(A) 64
(B) 8
(C) $8 \sqrt{2}$
(D) $\sqrt{2}$
(2002, Huanggang City, Hubei Province, Junior High School Mathematics Competition) | Solution: Let $M=(x+1)(y+1)(z+1)$,
then
$$
\begin{aligned}
M & =\frac{M}{1}=\frac{(x+1)(y+1)(z+1)}{x y z} \\
& =\frac{x+1}{x} \cdot \frac{y+1}{y} \cdot \frac{z+1}{z} \\
& =\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) .
\end{aligned}
$$
From (1) and (2) we get
$$
\begin{aligned}
M^{2}=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,519 |
Example 2 Let the side length of the equilateral $\triangle ABC$ be $2, M$ is the midpoint of $AB$, $P$ is any point on $BC$, and $PA+PM$ are denoted as $s$ and $t$ for their maximum and minimum values, respectively. Then $s^{2}-t^{2}=$ $\qquad$
(2000, National Junior High School Mathematics League) | Solution: Since $P A \leqslant A C$, and when $P$ is at the vertex $C$ of $\triangle A B C$, we get the maximum value of $P M$, which is $P M=C M=\sqrt{3}$.
Thus, $s=P A+P M=2+\sqrt{3}$.
As shown in Figure 1, construct an equilateral $\triangle A^{\prime} B C$, and let $M^{\prime}$ be the midpoint of $A^{\prime} B$. Th... | 4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,520 |
Example 11 A certain project, if contracted by Team A and Team B, can be completed in $2 \frac{2}{5}$ days, costing 180000 yuan; if contracted by Team B and Team C, it can be completed in $3 \frac{3}{4}$ days, costing 150000 yuan; if contracted by Team A and Team C, it can be completed in $2 \frac{6}{7}$ days, costing ... | Solution: Let the time required for A, B, and C to complete the work alone be $x$, $y$, and $z$ days, respectively, then
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{5}{12} \\
\frac{1}{y}+\frac{1}{z}=\frac{4}{15} \\
\frac{1}{z}+\frac{1}{x}=\frac{7}{20}
\end{array} .\right.
$$
Solving, we get $x=4, y=6, z=1... | 177000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,521 |
13. In a mathematics competition,
(i) the number of problems is $n(n \geqslant 4)$;
(ii) each problem is solved by exactly 4 people;
(iii) for any two problems, exactly 1 person solves both problems.
If the number of participants is greater than or equal to $4 n$, find the minimum value of $n$ such that there always e... | Solution: Let $n \geqslant 14$, and represent each problem as a rectangle, with its 4 vertices representing the 4 contestants who solved the problem. Let $P$ be any problem.
Each problem different from $P$ shares a common vertex with $P$, since $n \geqslant 14$. By the pigeonhole principle, there is a vertex of $P$, d... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,522 |
14. There are $n$ natives living on an island, and any two of them are either friends or enemies. One day, the chief asked each resident (including himself) to make a stone necklace according to the following principle: for every two friends, their necklaces must have at least one stone in common; for every two enemies... | Proof: By mathematical induction.
When $n=1$ and $n=2$, it is obvious that the residents can make necklaces according to the rules. Assume that the conclusion holds for $n=k$, we need to prove that the conclusion also holds for $n=k+2$. If every two residents are enemies, no stones will be needed. Now assume that at le... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,523 |
15. For a lattice point \(X\) on the coordinate plane, if the line segment \(OX\) does not contain any other lattice points, then the point \(X\) is said to be visible from the origin \(O\). Prove: For any positive integer \(n\), there exists a square \(ABCD\) with area \(n^2\) such that there are no lattice points vis... | Proof: Note that a lattice point $(x, y)$ is visible from the origin if and only if $x$ and $y$ are coprime. Therefore, the problem is reduced to finding a square $ABCD$ such that every lattice point $(x, y)$ inside it satisfies that $x$ and $y$ are not coprime. Let $p_{1}, p_{2}, \cdots$ be a sequence of distinct prim... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,524 |
16. Given positive integers $m, n, m<2001, n<2002$, there are $2001 \times 2002$ different real numbers. Place these numbers into a $2001 \times$ 2002 rectangular chessboard (the chessboard has 2001 rows and 2002 columns), so that each cell contains one number, and each number is placed in one cell. When the number in ... | Solution: Let $p, q$ replace $2001, 2002 (m \leqslant p, n \leqslant q)$, and strengthen the problem to an induction on $p+q$, proving that
$S \geqslant (p-m)(q-n)$.
Clearly, when $p+q=2,3,4$, inequality (1) holds.
Assume that when $p+q=k$, inequality (1) holds.
When $p+q=k+1$, consider the $(p, q)-$ chessboard. If $m=... | (2001-m)(2002-n) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,525 |
17. 14 people participate in a Japanese chess round-robin tournament, where each person plays against the other 13 people, and there are no draws in the matches. Find the maximum number of "triangles" (here, a "triangle" refers to a set of three people where each person has one win and one loss).
(2002, Japan Mathemat... | Solution: Let the 14 people be represented as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{4}$ win $a_{4}$ games. In any group of three people (i.e., selecting 3 out of 14 people), if there is no "double corner," then one person must have won the other two. Therefore, the number of groups of three that do not form a "tr... | 112 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,526 |
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