problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Three. (50 points) We call non-empty sets $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if:
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j}=\varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 13-partition $A_{1}, A_{2}, \cdots, A_{13}$ o... | (1) First, prove that $m \geqslant 117$.
If not, $m1+\frac{13}{104} \geqslant \frac{9}{8}$.
This contradicts $b<a \leqslant \frac{9}{8} b$. Therefore, $m \geqslant 117$.
(2) Next, prove that $m=117$ satisfies the condition.
Because, in this case, the largest 14 numbers $104,105, \cdots, 117$ are distributed among 13 s... | 117 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,068 |
Initially, let $\odot O_{1}, \odot O_{2}$ intersect at $P, Q$, and both are internally tangent to $\odot O$, with the points of tangency being $M$ and $N$; the line $P Q$ intersects $\odot O$ at $A$ and $B$.
Prove: $A M \cdot B N = A N \cdot B M$. | Proof: $A M, B M$ intersect $\odot O_{1}$ at $E, G$, $A N, B N$ intersect $\odot O_{2}$ at $F, H$. Draw $O K \perp A M$ at $K$, other auxiliary lines see Figure 2.
By the property of the circumcenter, we know $\angle A O K = \angle A N M$. By
$$
\begin{array}{l}
A E \cdot A M = A P \cdot A Q = A F \cdot A N \\
\Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,069 |
For any four-digit number (digits can be the same), by changing the original order of its digits, one can always obtain the smallest four-digit number. If the difference between these two four-digit numbers is 999, find the number of such four-digit numbers.
Translate the above text into English, please keep the origi... | Solution: Let the original four-digit number be $\overline{a b c d}$ (where $a, b, c, d$ represent the digits in each place), then from the given information we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d-999 \\
=1000(a-1)+100 b+10 c+(d+1) .
\end{array}
$$
Obviously, when $d \leqslant 8, a>1$, $(a-1) b c(d+1)$ is the... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,070 |
Let $n \geqslant 2$. Real numbers $k \geqslant 1, a_{1} \geqslant 0, i=1,2, \cdots$. $n \cdot \sum_{i=1}^{n} a_{1}=n$. Prove:
$$
\sum_{i=1}^{n} \frac{\left(a_{i}+1\right)^{2 k}}{\left(a_{n+1}+1\right)^{k}} \geqslant n \cdot 2^{k} \text {. }
$$
where $a_{n+1}=a_{1}$, and determine the condition for equality. | Proof: Using the mean inequality, we get
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{\left(a_{i}+1\right)^{2 k}}{\left(a_{i+1}+1\right)^{k}}+\sum_{i=1}^{n}\left(a_{i+1}+1\right)^{k} \\
\geqslant 2 \sum_{i=1}^{n}\left(a_{i}+1\right)^{k} . \\
\because \sum_{i=1}^{n}\left(a_{i+1}+1\right)^{k}=\sum_{i=1}^{n}\left(a_{i}+1\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,072 |
3. $A$ and $B$ are two points in the plane $a$. The line passing through $A, B$ is $r$, and there are $n$ different points $P_{1}, P_{2}, \cdots, P_{n}$ on the same side of the line $r$. Prove: The set of distances from $A$ to $P_{1}(i=1,2, \cdots, n)$ and the set of distances from $B$ to $P_{1}(i=1,2, \cdots, n)$, at ... | 3. Consider the number of semicircles centered at $A$ or $B$ and passing through at least one point in $P_{1}$. Let the number of semicircles centered at $A$ and passing through at least one point in $P$ be $a$, and the number of semicircles centered at $B$ and passing through at least one point in $P$ be $b$. Since $P... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,073 |
4. Let $p_{1}, p_{2}, q_{1}, q_{2}$ be real numbers, and $p_{1} p_{2}=2\left(q_{1}+q_{2}\right)$. Prove: The equations $x^{2}+p_{1} x+q_{1}=0$ and $x^{2}+p_{2} x+q_{2}=0$ have at least one real root. | 4. Let the discriminants of two equations be $\Delta_{1}$ and $\Delta_{2}$, calculate the overall of $\Delta_{1}$ and $\Delta_{2}$, $\nu_{1}+\Delta_{2}=\left(p_{1}^{2}-4 q_{1}\right)+\left(p_{2}^{2}-4 q_{2}\right)=\left(p_{1}-\right.$ $\left.p_{2}\right)^{2} \geqslant 0$, then at least one of $\Delta_{1}$ and $\Delta_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,074 |
5. There are 1993 matches on the table. Two children, A and B, take turns to pick up 1, 2, or 3 matches each time. The one who picks up the last match wins. If A goes first, which child will win? How should he play this game? | 5. Player A can win. Note that as long as the number of matches left after Player A's turn is a multiple of 4. Since 1993 is a number of the form $4k+1$. Player A takes 1 match first, leaving 1992 matches, which is a multiple of 4. Thereafter, when Player B takes 1, 2, or 3 matches, Player A takes 3, 2, or 1 matches re... | Player A will win. | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,075 |
Example 1 In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=$ $\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$. T... | Solution: In $\triangle ABC$, from $\frac{\cos A}{\cos B}=\frac{\sin B}{\sin A}$ we can get
$$
\sin 2A=\sin 2B.
$$
Therefore, $\angle A+\angle B=\frac{\pi}{2}$, which means $\angle C=\frac{\pi}{2}$.
$$
\text{Given } \frac{b}{a}=\frac{4}{3}, c=10,
$$
it is easy to find $b=8, a=6$, and the radius of the inscribed circl... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,076 |
Example 2 If the ellipse $x^{2}+4(y-a)^{2}=4$ and the parabola $x^{2}=2 y$ have common points, then the range of the real number $a$ is $\qquad$
(1998, National High School Mathematics Competition) | Solution: From $x^{2}+4(y-a)^{2}=4$, we can set
$$
\begin{array}{l}
x=2 \cos \theta, y=a+\sin \theta . \\
\text { Substituting into } x^{2}=2 y, \text { we get } \\
4 \cos ^{2} \theta=2(a+\sin \theta) . \\
\text { Therefore, } a=2 \cos ^{2} \theta-\sin \theta \\
\quad=2-2 \sin ^{2} \theta-\sin \theta \\
\quad=-2\left(\... | -1 \leqslant a \leqslant \frac{17}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,077 |
Example 3 Given the family of curves $2(2 \sin \theta-\cos \theta+3) x^{2}$ $-(8 \sin \theta+\cos \theta+1) y=0$ ( $\theta$ is a parameter). Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
(1995, National High School Mathematics Competition) | Solution: The family of curves always passes through the origin, and the line $y=2 x$ also passes through the origin, so the length of the chord intercepted by the family of curves on $y=2 x$ depends on the coordinates of the other intersection point of the family of curves with $y=2 x$.
Substituting $y=2 x$ into the e... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,078 |
Example 4 For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the endpoints of the major axis are $A$ and $B$, and $C, D$ are two points on the ellipse on the same side of $A B$, with $C D / / A B$. Find the maximum value of the area of trapezoid $A B C D$. | Solution: From $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,
we can set $x=a \cos \theta, y=$ $b \sin \theta, \theta$ as the parameter. As shown in Figure 2, let $C(a \cos \theta, b \sin \theta)$. According to the problem, we have
$$
\begin{array}{l}
=a b(1+\cos \theta) \sqrt{1-\cos ^{2} \theta} \\
=a b \sqrt{(1+\cos \... | \frac{3 \sqrt{3}}{4} a b | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,079 |
Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition) | Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, an... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,080 |
Example 6 Given $A(-2,2)$, and $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, $F$ is the left focus. When $\mid A B \mid +\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of $B$.
(1999, National High School Mathematics Competition) | Let the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse be denoted as $a$, $b$, and $c$, respectively, and the eccentricity as $e$. Then, $a=5$, $b=4$, $c=\sqrt{a^{2}-b^{2}}=3$, $e=\frac{c}{a}=\frac{3}{5}$, and the left directrix is $x=-\frac{25}{3}$.
As shown in Figure 4, draw a perpendicular... | B\left(-\frac{5}{2} \sqrt{3}, 2\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,082 |
Example 7 In the Cartesian coordinate system, if the equation $m\left(x^{2}+y^{2}+2 y+1\right)=(x-2 y+3)^{2}$ represents an ellipse, then the range of values for $m$ is ( ).
(A) $(0,1)$
(B) $(1,+\infty)$
(C) $(0,5)$
(D) $(5,+\infty)$
(1997, National High School Mathematics Competition) | Solution: From the given, we have $\frac{\sqrt{x^{2}+(y+1)^{2}}}{\left|\frac{x-2 y+3}{\sqrt{1^{2}+(-2)^{2}}}\right|}=\sqrt{\frac{5}{m}}$, which indicates that the ratio of the distance from $(x, y)$ to the fixed point $(0,-1)$ to the distance from $(x, y)$ to the line $x-2 y+3=0$ is a constant $\sqrt{\frac{5}{m}}$. By ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,083 |
Example 8 The maximum value of the function $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$ $-\sqrt{x^{4}-x^{2}+1}$ is $\qquad$ .
(1992, National High School Mathematics Competition) | Given $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$
$$
\begin{aligned}
& -\sqrt{x^{4}-x^{2}+1} \\
= & \sqrt{(x-3)^{2}-\left(x^{2}-2\right)^{2}}-\sqrt{x^{2}+\left(x^{2}-1\right)^{2}},
\end{aligned}
$$
we can see that the geometric meaning of the function $y=f(x)$ is: as shown in Figure 5, for a point $P\left(x, x^{2}\right)$ on t... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,084 |
2. An ellipse is inscribed in a rectangle with length $m$ and width $n$. Find the maximum value of the perimeter of the rectangle that is tangent to this ellipse.
(1983, Shanghai Mathematics Competition) | (Tip: Take the major and minor axes of the ellipse as the $x$-axis and $y$-axis respectively, and set the equation of the ellipse as $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. From the given information, we have $2 \sqrt{a^{2}+b^{2}}=\sqrt{m^{2}+n^{2}}$. The coordinates of the vertex of the inscribed rectangle in the... | 2 \sqrt{m^{2}+n^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,086 |
3. Let $A$ and $B$ be two moving points on the ellipse $x^{2}+3 y^{2}=1$, and $O A \perp O B$ (where $O$ is the origin). Find the maximum and minimum values of $|A B|$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | (Tip: Establish a polar coordinate system with $O$ as the pole and the positive half-axis of the $x$-axis as the polar axis. Then the equation of the ellipse is $\rho^{2}\left(\cos ^{2} \theta+3 \sin ^{2} \theta\right)=1, \theta \in$ $[0,2 \pi)$. Let $A\left(\rho_{1}, \alpha\right), B\left(\rho_{2}, \alpha+\frac{\pi}{2... | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,087 |
4. Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=x^{2}+y^{2}$. Then the value of $\frac{1}{S_{\text {min }}}+\frac{1}{S_{\text {max }}}$ is $\qquad$
(1993, National High School Mathematics Competition) | (Let $x=\sqrt{S} \cos \theta, y=\sqrt{S} \sin \theta$, substituting into the known conditions yields $\sin 2 \theta=\frac{8 S-10}{5 S}$. Then $\left|\frac{8 S-10}{5 S}\right| \leqslant 1$. Solving this, we get $\frac{10}{13} \leqslant S$ $\leqslant \frac{10}{3}$. Therefore, $\frac{1}{S_{\max }}+\frac{1}{S_{\max }}=\fra... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,088 |
5. For a line segment $A B$ of fixed length 3, whose endpoints move on the parabola $y^{2}=x$, the midpoint of segment $A B$ is $M$. Try to find the shortest distance from point $M$ to the $y$-axis. | (Let the distances from points $A, B, M$ to the directrix be $d_{1}$, $d_{2}$, $d_{0}$ respectively. Then $d_{0}=\frac{d_{1}+d_{2}}{2}=\frac{|A F|+|B F|}{2} \geqslant \frac{|A B|}{2}=\frac{3}{2}$. The equality holds when $A B$ passes through point $F$, at which time the distance from $M$ to the $y$-axis is minimized, w... | \frac{5}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,089 |
6. When $s$ and $t$ take all real values, the minimum value that can be reached by $(s+5-3|\cos t|)^{2}$ $+(s-2|\sin t|)^{2}$ is $\qquad$
(1989, National High School Mathematics Competition) | (The original expression can be regarded as the square of the distance between any point on the line $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array}\right.$ and any point on the ellipse arc $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$. It is known that the square of the distance from the point... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,090 |
Example 1 Write two integers on the blackboard, then erase one of them and replace it with the sum or difference of the other two numbers. If this process is repeated several times, and the final result is (17.1999, 2015), then - what could the initial numbers on the blackboard be: (1)(2,2,2); (2)(3,3,3). | Solution: Due to the changes in the parity of the three numbers during the operation process as follows:
(even, even, even $) \rightarrow$ (even, even, odd $) \rightarrow$ (even, even, odd),
thus, (2,2,2) cannot reach $(17,1999,2015)$ after several operations.
Since (odd, odd, odd) $\rightarrow$ (odd, odd, odd) $\rig... | (3,3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,091 |
Example 4 Given a straight line and a circle with radius $n$ cm ($n$ is an integer) and $4n$ line segments of length 1 cm inside the circle. Prove: a chord can be drawn in the given circle that is parallel or perpendicular to the given straight line, and it intersects at least two of the given line segments. | Analysis: Due to the great randomness of these $4n$ segments, directly finding the chords that meet the requirements would make the problem overly complex. It might be better to first explore whether a simpler expression can be found.
Let $a$ be the given line, and line $b$ be perpendicular to line $a$. Chord $AB \par... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,092 |
Example 2 Write the number 1 at both ends of a circle's diameter, and let this diameter divide the circle into two semicircles. Write the sum of the adjacent two numbers (i.e., $1+1=2$) at each division point. Then, divide each of the resulting four $\frac{1}{4}$ circles, and write the sum of the adjacent two numbers (... | Solution: It is easy to find that $S_{1}=2, S_{2}=6, S_{3}=18, \cdots$. Let the sum after the $k$-th operation be $S_{k}$, then when performing the $(k+1)$-th operation, each number filled in is the sum of the two adjacent numbers that appeared after the $k$-th operation, so each of these numbers is added exactly twice... | 2 \times 3^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,093 |
Example 3 Given a chessboard, change the color of each square in a $2 \times 2$ square located inside it to another color. Can you end up with exactly one white square on the board? | Solution: In an $8 \times 8$ chessboard, there are 32 black squares and 32 white squares. We assign the value 1 to each black square and the value -1 to each white square. Initially, there are 32 ones and 32 negative ones, so their product $S_{0}=1^{32} \times(-1)^{32}=1$. Each operation changes the color of $2 \times ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,094 |
Example $4 \quad n(n \geqslant 3)$ students sit in a circle, numbered $1,2,3, \cdots, n$ in a specified direction. The teacher calls out numbers according to the following rule: if the number called at some point is $i$, then the next number called is the number of the student who is the $i$-th one behind the $i$-th st... | Solution: (1) When $i=n$, then $j=n$. Thus, when the $n$-th student is called for the first time, the rest of the students will never be called;
(2) When $i \neq n$, there are two cases:
(1) If $n$ is even, since $2i$ and $2i-n$ are both even, and $n \geqslant 3$. Therefore, at least one student with an odd number will... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,095 |
Example 5 Suppose there are $2^{n}$ balls divided into several piles. We can arbitrarily choose two piles, pile A and pile B, and move balls according to the following rule: if the number of balls in pile A, $p$, is not less than the number of balls in pile B, $q$, then take $q$ balls from pile A and put them into pile... | Proof: (1) When $n=1$, there are only two balls in total, which may form just one pile, so no movement is necessary; or they may be divided into two piles, which can be combined into one pile with just one move. That is, the proposition holds when $n=1$.
(2) Assume the proposition holds when $n=k$, i.e., $2^k$ balls ha... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,096 |
Let $a, b, c, d$ be integers, $a>b>c>$ $d>0$, and $a c+b d=(b+d+a-c)(b+d-a$ $+c)$. Prove: $a b+c d$ is not a prime. | First, let's discuss the following lemma.
Lemma Let $a, b, c$ be positive integers, $c0$,
and from $a>b>c>d>0$, it is easy to see that $b+d+a-$ $c>0$. Therefore, $b+d-a+c>0$, i.e., $a0$,
$$
\therefore a b+c d>a c+b d \text {. }
$$
Using equation (2), we easily get
$$
\begin{array}{l}
a^{2}-a c+c^{2}>a d+b c . \\
\text... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,097 |
Proposition: Let the area of $\triangle ABC$ be $\triangle$, and the lengths of its three sides be $a$, $b$, and $c$. Then the maximum area of the circumscribed equilateral triangle of $\triangle ABC$ is
$$
\frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}\right)+2 \triangle .
$$ | Proof: As shown in Figure 1, the equilateral triangle \( \triangle DEF \) is circumscribed around the given \( \triangle ABC \), with the area of \( \triangle ABC \) being \( \triangle \), \( BC = a \), \( CA = b \), and \( AB = c \). Perpendiculars are drawn from \( B \) and \( C \) to \( DE \), with the feet of the p... | \frac{\sqrt{3}}{6} (a^2 + b^2 + c^2) + 2 \triangle | Geometry | proof | Yes | Yes | cn_contest | false | 713,098 |
Text [1] provides the following two elegant inequalities:
$$
\begin{array}{l}
\searrow \frac{h_{a} h_{b}}{r_{o} r_{b}} \geqslant 3 ; \\
\triangle \frac{r_{a} r_{b}}{h_{a} h_{b}} \geqslant 3 .
\end{array}
$$
This article will prove that the optimal form of the above two inequalities is
$$
\begin{array}{l}
\simeq \frac{... | Proof: $\because \triangle=\frac{1}{2} a h_{a}=r_{a}(s-a)$,
$$
\begin{array}{l}
\therefore \sum \frac{h_{a} h_{b}}{r_{a} r_{b}}=\sum \frac{4(s-a)(s-b)}{a b} \\
=\frac{4}{a b c} \sum c(s-a)(s-b) \\
=\frac{4}{a b c}\left(s^{2} \sum a-2 S \sum b c+3 a b c\right) .
\end{array}
$$
Substitute the known equations
$$
a b c=4 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,099 |
1. Let $n$ be any given positive integer. $T$ is the set of all points $(x, y)$ on the plane that satisfy $x+y<n$, where $x, y$ are non-negative integers. Each point $(x, y)$ in $T$ is colored either red or blue, satisfying: if $(x, y)$ is red, then all points $\left(x^{\prime}, y^{\prime}\right)$ in $T$ that satisfy $... | 1. Let the number of red points be $m$, the number of $X$-sets be $t$, and the number of $Y$-sets be $s$. We will prove by induction on $m, 0 \leqslant m \leqslant \frac{n(n+1)}{2}$.
(1) When $m=0$, by the multiplication principle, $s=t=n$!, and the proposition holds.
(2) Assume the proposition holds for $m=k\left(0 \l... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,100 |
2. $BC$ is the diameter of circle $\Gamma$, the center of $\Gamma$ is $O$, $A$ is a point on $\Gamma$, $0^{\circ}<\angle AOB<120^{\circ}$, $D$ is the midpoint of arc $AB$ (the arc not containing $C$), the line through $O$ parallel to $DA$ intersects $AC$ at $I$, the perpendicular bisector of $OA$ intersects $\Gamma$ at... | 2. As shown in Figure 1, given that $A$ is the midpoint of $\overparen{E A F}$, thus $C A$ is the angle bisector of $\angle E C F$.
Since $O A=O C$,
$$
\begin{array}{l}
\angle A O D=\frac{1}{2} \angle A O B \\
=\angle O A C,
\end{array}
$$
then $O D \parallel I A$.
And $A D \parallel I O$,
therefore, $A D O I$ is a pa... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,101 |
3. Find all pairs of positive integers $m, n \geqslant 3$, such that there exist infinitely many positive integers $a$ for which $\frac{a^{m}+a-1}{a^{n}+a^{2}-1}$ is an integer. | 3. Suppose $(m, n)$ is the solution, it is clear that $n\beta, \beta=0.618 \cdots$ is the positive root of $h(x)=x^{2}+x-1$. Since $f(x)$ is monotonically increasing in $(0,1)$, and
$$
f(\beta)<h(\beta)=0=f(\alpha) .
$$
On the other hand, if $m \geqslant 2 n$, then
$$
1-\alpha=\alpha^{m} \leqslant\left(\alpha^{n}\righ... | (5,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,102 |
Example 5 Let real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=$ 1. Prove: $|a-b|, |b-c|, |c-a|$ must have one that does not exceed $\frac{\sqrt{2}}{2}$. | Analysis: Considering the whole, the aim is to prove
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leqslant \frac{3}{2} \text {. }
$$
If equation (1) holds, then the conclusion must be true.
Let $m$ be the smallest of $|a-b|, |b-c|, |c-a|$.
Now consider $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$.
$$
\begin{array}{l}
(a-b)^{2}+(b-c)^{2}+(c-a... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,103 |
5. Find all functions $f$ from the set of real numbers $\mathbf{R}$ to $\mathbf{R}$ such that for all $x, y, z, t \in \mathbf{R}$, we have
$$
\begin{array}{l}
(f(x)+f(z))(f(y)+f(t)) \\
=f(x y-z t)+f(x t+y z) .
\end{array}
$$ | 5. Suppose for all $x, y, z, t \in \mathbf{R}$,
$$
\begin{array}{l}
f(x y-z t)+f(x t+y z) \\
=(f(x)+f(z))(f(y)+f(t)) .
\end{array}
$$
Let $x=y=z=0$. Then
$$
2 f(0)=2 f(0)(f(0)+f(t)) \text {. }
$$
In particular, $2 f(0)=4 f^{2}(0)$, so,
$$
f(0)=0 \text { or } f(0)=\frac{1}{2} \text {. }
$$
If $f(0)=\frac{1}{2}$, then... | f(x)=0, f(x)=\frac{1}{2}, f(x)=x^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,105 |
6. Let $n \geqslant 3$ be an integer, and $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}, \cdots, \Gamma_{n}$ be circles of radius 1 in the plane, with centers $O_{1}, O_{2}, O_{3}, \cdots, O_{n}$. Assume that any line intersects or is tangent to at most two of these circles. Prove:
$$
\sum_{1 \leqslant i<j \leqslant n} \frac{1}{... | 6. As shown in Figure 2, for circles $\Gamma_{1}$, $\Gamma, 1 \leqslant i < j \leqslant n$, draw two tangents from the center $O$ to $\Gamma$. This divides the plane into 4 parts. Color the arc on $\Gamma$ that is in the same part as $\Gamma$ and the opposite arc. Similarly, for $\Gamma$, two arcs of the same length ar... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,106 |
3. Let $x_{1}, x_{2}, \cdots, x_{n}$ be any real numbers. Prove:
$$
\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+\cdots+x_{n}^{2}}<\sqrt{n} .
$$ | Prove: By Cauchy's inequality, for any real numbers \( u_{1}, a_{2}, \cdots, a_{n} \) we have
\[
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n} \leqslant \sqrt{n} \cdot \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} . \\
\text { Let } a_{k}=\frac{x_{k}}{1+x_{1}^{2}+\cdots+x_{k}^{2}}, k=1,2, \cdots, n .
\end{array}
\]
It is su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,109 |
4. Find all functions $f: \mathscr{E} \rightarrow \mathscr{B}$, for any $x, y \in \mathscr{F}, f$ satisfies
$$
f(x y)(f(x)-f(y))=(x-y) f(x) f(y) \text {. }
$$
where $\mathscr{B}$ is the set of real numbers. | Let $y=1$, then
$f(x)(f(x)-f(1))=(x-1) f(x) f(1)$,
which means $f^{2}(x)=x f(x) f(1)$.
If $f(1)=0$, then $f(x)=0$ for all $x$, which satisfies the given condition.
Assume $f(1)=c \neq 0$, then $f(0)=0$. Let $G=\{x \mid x \in \mathscr{B}, f(x) \neq 0\}$. Clearly, $0 \notin G$, and for all $x \in G$, we have $f(x) = x f... | f(x)=\left\{\begin{array}{cc} C x, & x \in G, \\ 0, & x \notin G . \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,110 |
5. Find all positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$
\begin{array}{l}
\quad \frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}, \\
\text { where } a_{0}=1,\left(a_{k+1}-1\right) a_{k-1} \geqslant a_{k}^{2}\left(a_{k}-1\right), k=1,2, \cdots, \\
n-1 .
\end{array}
$$ | Solution: Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers satisfying the given conditions.
Since $a_{0}=1$, we have $a_{1} \neq 1$. Thus, $\frac{a_{0}}{a_{1}}=1>\frac{99}{100}$. Therefore, $a_{1} \geqslant 2$, and $a_{1}>a_{0}$. Assume $a_{k}>a_{k-1}, a_{k} \geqslant 2$. Then
$$
a_{k-1} \geqslant \frac{a_{i}^{2}... | a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=78400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,111 |
3. Define a “$k$-clique” as a set of $k$ people such that every pair of them know each other. At a certain gathering, every two “3-cliques” have at least one person in common, and there is no “5-clique”. Prove: at this gathering, there exist two (or fewer) people, such that when they leave, no “3-clique” remains. | Proof: For convenience, we use the language of graph theory, representing each person at the gathering as a point. If two people know each other, we connect the two vertices with an edge. Thus, an “$m$-clique” corresponds to a set of $m$ points, with each pair of vertices connected by an edge. In other words, such an “... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,113 |
Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$. | Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,114 |
4. A set of three non-negative integers $\{x, y, z\}(x<y<z)$ is called "historical" if $z-y, y-x=1776, 2001$. Prove: The set of all non-negative integers can be partitioned into pairwise disjoint "historical" sets. | Proof: For convenience, let $a=1776, b=2001$. In fact, as long as $0<a<b$.
Define $A=\{0, a, a+b\}, B=\{0, b, a+b\}$, then $A$ and $B$ are both "historical", and a set $X$ is "historical" if and only if $X = x + A$ or $x + B$, where $x$ is some non-negative integer, and $x + S = \{x + s \mid s \in S\}$.
In fact, if $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,115 |
5. Find all finite sequences $x_{0}, x_{1}, \cdots, x_{n}$, such that for each $j, 0 \leqslant j \leqslant n, x_{j}$ is equal to the number of times $j$ appears in the sequence. | Solution: Let $\left.\mid x_{0}, x_{1}, \cdots, x_{n}\right\}$ be any sequence that satisfies the conditions. Since each $x_{j}$ is the number of times $j$ appears, each term in the sequence is a non-negative integer, and $x_{0}>0$.
Let $m$ denote the number of positive terms in $x_{1}, x_{2}, \cdots, x_{n}$, and let ... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,116 |
6. For a positive integer $n$, if a $0-1$ sequence has $n$ zeros and $n$ ones, then this sequence is called "balanced". For two balanced sequences $a, b$, if you can move one number in $a$ to another position to get $b$, then $a$ and $b$ are called "adjacent". For example, when $n=4$, the balanced sequences $a=01101001... | Prove: For each balanced sequence $a=\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}$, let $f(a)$ be the sum of the indices of 1s in $a$. For example, $f(01101001)=2+3+5+8=18$. The $\mathrm{C}_{2 n}^{n}$ balanced sequences are divided into $n+1$ classes based on the remainder of $f$ modulo $n+1$. Let $S$ be the class with... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,117 |
7. Choose one of the following two questions.
(1) Place $n$ pebbles in a vertical column, and perform operations according to the following rules. If a pebble is at the top of a column and the number of pebbles in this column is at least two more than the number of pebbles in the adjacent column to its right (if there ... | Proof: (1) For each state, let $p_i$ be the number of pebbles in the $i$-th column, where $i=1,2, \cdots$. The first column represents the leftmost column. We will prove that in the "final state," for all $i$, each $p_i > 0$, except for at most one $i$ such that $p_i = p_{i+1}$, all $p_i = p_{i+1} + 1$.
This state is ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,118 |
9.1 Can the natural numbers from 1 to $2002^{2}$ be filled into a 2002 $\times$ 2002 square grid so that for any cell, it is possible to find three numbers either in its row or in its column, where the product of two of these numbers equals the third? | 9.1 Impossible.
Numbers 1 to 2001 are distributed at most in 2001 rows and 2001 columns, so it is certain to find one row and one column where all the numbers filled are not less than 2002. Thus, the product of any two numbers in that row (or that column) will be greater than \(2002^{2}\). Therefore, for the cell at t... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,119 |
9.2 On one of the sides of the angle with vertex $O$, take a point $A$; on the other side, take two points $B$ and $C$, where point $B$ is between points $O$ and $C$. Construct the incircle of $\triangle OAB$, with center $O_{1}$; then construct an excircle of $\triangle OAC$, with center $O_{2}$, that is tangent to si... | $9.2 \angle A O_{1} O_{2}$ is the exterior angle of $\triangle O A O_{1}$, so,
$$
\begin{array}{l}
\angle A O_{1} O_{2}=\angle A O O_{1}+\angle O A O_{1} \\
=\frac{1}{2} \angle A O B+\frac{1}{2} \angle O A B \\
=90^{\circ}-\frac{1}{2} \angle A B O=\frac{1}{2} \angle A B C .
\end{array}
$$
As shown in Figure 1, let $M$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,120 |
9.3 In the plane, 6 red points, 6 blue points, and 6 green points are given, with no three points being collinear. Prove: the sum of the areas of all triangles formed by points of the same color does not exceed one quarter of the sum of the areas of all triangles formed by these given points. | 9.3 Consider 3 blue points $A, B, C$ and 1 non-blue point $D$. It is easy to see that
$$
S_{\triangle B C} \leqslant S_{-\triangle B D}+S_{\text{C } A C D}+S_{B C D}.
$$
Summing up all such inequalities for these sets of four points, we find that in the sum, each "blue" triangle is counted 12 times, while each "blue-b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,121 |
9.4 In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$. However, head $A$ immediately grows new necks connecting to all the heads it was not previously connec... | 9.410 .
We will reformulate the problem using graph theory terminology, with heads as vertices, necks as edges, and a strike that cuts the necks connected to head $A$ as a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is sufficient to perform a "reversal" on ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,122 |
9.5 On a chessboard, 8 rooks are placed such that they cannot attack each other. Prove: There must be two pairs of rooks whose distances are equal (the distance between two rooks is the distance between the centers of the squares they occupy). | 9.5 Consider 7 pairs of “rooks” distributed in adjacent columns. The differences in the y-coordinates of each pair of “rooks” are between 1 and 7. Therefore, either there are two pairs of “rooks” with the same difference in their y-coordinates, in which case the distances between these two pairs of “rooks” are already ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,123 |
9.6 There is a red card box and $k$ blue card boxes $(k>1)$, and a deck of cards, totaling $2n$ cards, which are numbered from 1 to $2n$. Initially, this deck of cards is stacked in the red card box in any order. From any card box, the top card can be taken out, either placed in an empty box, or placed on top of a card... | $9.6 \quad n=k-1$.
First, we prove that for larger $n$, it is not necessarily possible. Suppose at the beginning, the cards in the red box are placed from top to bottom first with odd-numbered cards (in any order), then the card numbered $2n$, and finally the remaining even-numbered cards (in any order). In this case, ... | n=k-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,124 |
Example 7 Given
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}... | Analysis: If we consider the four known equations as a system of four equations in $x, y, z, w$, solving this system to find the values of $x, y, z, w$ would be quite difficult. However, if we view these four equations as a whole and temporarily treat $x, y, z, w$ as known numbers, then these four equations can be tran... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,125 |
9.7 Let the circumcenter of $\triangle A B C$ be $O$. Take points $M$ and $N$ on its sides $A B$ and $B C$, respectively, such that $2 \angle M O N = \angle A O C$. Prove: The perimeter of $\triangle M B N$ is not less than the length of side $A C$.
Translate the above text into English, please keep the original text'... | 9.7 As shown in Figure 2, take points $K$ and $L$ inside $\angle A O C$ such that $\triangle A K O \cong \triangle B M O$,
$$
\begin{array}{c}
\triangle C L O \cong \triangle B N O . \\
\text { Therefore, } K O=O M, \\
L O=O N, \\
\text { and } \angle K O L=\angle A O C- \\
\angle M O B-\angle B O N=\angle M O N .
\end... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,126 |
9.8 Take $2^{2 n-1}+1$ odd numbers in the interval $\left(2^{2 n}, 2^{3 n}\right)$. Prove: Among the taken numbers, there must be two numbers such that the square of each number cannot be divided by the other. | 9.8 It is easy to see that among the chosen numbers, there exist $a$ and $b$ which leave the same remainder when divided by $2^{2n}$. We will prove that these are the desired numbers.
Assume $b \mid a^{2}$. Then
$b \mid\left(a^{2}-2 a b+b^{2}\right)=(a-b)^{2}$.
Let $a=p \cdot 2^{2 n}+r, b=q \cdot 2^{2 n}+r$.
Then $b \m... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,127 |
10.1 Let $P, Q, R$ be polynomials with real coefficients, among which there are both quadratic and cubic polynomials, and they satisfy the relation $P^{2}+Q^{2}=R^{2}$. Prove: There must be a quadratic polynomial whose roots are all real. | 10.1 From the problem, we know that $R$ and one of $P$ or $Q$ are cubic polynomials. Suppose $R$ and $Q$ are cubic polynomials, and $P$ is a quadratic polynomial. If necessary, we can change the signs to ensure that the coefficients of the $x^3$ terms in $R$ and $Q$ are positive. Thus, $R+Q$ is a cubic polynomial. From... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,128 |
10.2 Let quadrilateral $ABCD$ be inscribed in circle $\omega$. The tangent to circle $\omega$ at $A$ intersects the extension of side $BC$ at point $K$, with point $B$ lying between points $K$ and $C$: the tangent to circle $\omega$ at $B$ intersects the extension of side $AD$ at point $M$, with point $A$ lying between... | 10.2 We have
$$
\begin{array}{l}
M B^{2}=M A \cdot M D=\frac{1}{2} M D^{2}, \\
K A^{2}=K B \cdot K C=\frac{1}{2} K C^{2} .
\end{array}
$$
Using the Law of Sines, we know that
$$
\frac{1}{\sqrt{2}}=\frac{A K}{K C}=\frac{\sin \angle A C K}{\sin \angle C A K}, \frac{1}{\sqrt{2}}=\frac{B M}{M D}=\frac{\sin \angle B D M}{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,129 |
10.3 Proof: For any natural number $n>10000$. There exists a natural number $m$, where $m$ can be expressed as the sum of two perfect squares, and satisfies the condition $0<m-n<3 \sqrt[4]{n}$. | 10.3 Let $x$ be the largest integer whose square does not exceed $n$, i.e., $x^{2} \leqslant n$.
That is, $m$ can be expressed as the sum of two perfect squares, and
$$
m-n>0 \text {. }
$$
On the other hand, we have
$$
\begin{array}{l}
m-n=x^{2}+y^{2}-n=y^{2}-\left(n-x^{2}\right) \\
\leqslant y^{2}-(y-1)^{2}=2 y-1 \le... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,130 |
10.4 A country originally had 2002 cities, some of which were connected by roads. It is known that if travel through any one of these cities is prohibited, it is still possible to travel from any other city to any other city. Each year, the authorities select a non-self-intersecting road network, order the construction... | 10.4 Construct a graph where the vertices and edges correspond to the initial cities and roads of the country, respectively. According to the problem, a series of transformations are performed on the graph: each transformation involves removing all edges of a simple cycle and connecting all vertices of the cycle to a n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,131 |
10.5 Let \(a, b, c\) be positive numbers, with \(a+b+c=3\). Prove:
\[
\sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant a b+b c+c a .
\] | 10.5 Observe the equation $(a+b+c)^{2}=9$, we have
$$
a b+b c+c a=\frac{9-a^{2}-b^{2}-c^{2}}{2} .
$$
Thus, it suffices to prove
$$
2 \sqrt{a}+2 \sqrt{b}+2 \sqrt{c}+a^{2}+b^{2}+c^{2} \geqslant 9 \text {. }
$$
To do this, we first prove $2 \sqrt{a}+a^{2} \geqslant 3 a$.
In fact, we have
$$
2 \sqrt{a}+a^{2}=\sqrt{a}+\sq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,132 |
10.7 Let the excircle of triangle $ABC$ opposite to vertex $A$ touch side $BC$ at point $A'$. Draw a line $a$ through point $A'$ parallel to the angle bisector of $\angle A$, and similarly draw lines $b$ and $c$. Prove: The lines $a$, $b$, and $c$ intersect at the same point. | 10.7 As shown in Figure 4, let the incircle of $\triangle ABC$ touch its sides at points $A_{1}, B_{1}, C_{1}$. Draw a line $a_{1}$ through $A_{1}$ parallel to the angle bisector of $\angle A$. Since $\triangle AB_{1}C_{1}$ is an isosceles triangle, the angle bisector of $\angle A$ is perpendicular to $B_{1}C_{1}$, thu... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,133 |
10.8 In the plane, a finite number of red lines and blue lines are given, where no two are parallel, and at the intersection of any two lines of the same color, there is a line of the opposite color passing through. Prove: All the given lines intersect at the same point. | 10.8 Using proof by contradiction.
Assume not all lines intersect at the same point. As shown in Figure 6, take any blue line \( l \), and observe the intersection points of \( l \) with the red lines. Let points \( A \) and \( B \) be the two farthest intersection points, and let lines \( m \) and \( n \) be the red ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,134 |
11.2 On the plane, a finite number of points are given. For any three of these points, there exists a Cartesian coordinate system (i.e., two coordinate axes perpendicular to each other, and the length units on both axes are the same), such that their coordinates in this system are integers. Prove: There exists a Cartes... | 11.2 Solution 1: Consider any three points $A, B, C$ that are not collinear (if all points are collinear, the proposition is obviously true). Let $T_{1}$ be a Cartesian coordinate system where their coordinates are all integers, with unit length $t_{1}$. Observe any other point, denoted as $D$. Let $T_{2}$ be a Cartesi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,135 |
Example 8 The curve $L$ divides an equilateral $\triangle A B C$ with side length $a$ into two equal areas. Prove: the length $l$ of the curve $L$ satisfies $l \geqslant \frac{\sqrt{\pi} a}{2 \sqrt[4]{3}}$. | Analysis: To facilitate thinking, we can complete the curve $L$ into a closed curve. For this purpose, we can continuously flip the equilateral $\triangle ABC$ five times to form a regular hexagon, as shown in Figure 2. This way, the curve divides the area of the regular hexagon into two equal parts, and the area enclo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,136 |
11.4 A city has several squares, some of which are connected by one-way driving routes, and from each square, there are exactly two outgoing routes. Prove: The city can be divided into 1014 districts such that each route connects two squares belonging to different districts, and for any two districts, all routes connec... | 11.4 First, prove that if no more than two roads lead out of each square, then all squares can be colored with 13 different colors such that from any square, it takes at least three roads to reach any square of the same color. To do this, observe the following auxiliary directed graph: the vertices of the graph are the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,138 |
11.5 Find the smallest positive integer that can be expressed as the sum of 2002 positive integers with equal sums of their digits, and also as the sum of 2003 positive integers with equal sums of their digits. | 11.510010 .
Assume for a positive integer $n$ the following expression holds:
$$
n=a_{1}+a_{2}+\cdots+a_{2002}=b_{1}+b_{2}+\cdots+b_{2000} \text {. }
$$
Notice that the sum of the digits of $a_{1}, a_{2}, \cdots, a_{2002}$ is the same, so they have the same remainder when divided by 9. Let this remainder be $r(0 \leq... | 10010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,139 |
11.6 Let \(ABCD\) be a cyclic quadrilateral, its two diagonals \(AC\) and \(BD\) intersect at point \(O\). The circumcircle of \(\triangle ABO\) intersects the circumcircle of \(\triangle COD\) at point \(K\). Given that point \(L\) makes \(\triangle BLC\) similar to \(\triangle AKD\). Prove: If \(BLCK\) is a convex qu... | 11.6 As shown in Figure 8, assume
point $K$ is inside $\triangle AOD$ (other cases can be proven similarly). Let $L'$ be the reflection of $L$ across $BC$. Then we have
$$
\begin{array}{l}
\angle L'BO \\
=\angle OBC - \angle L'BC \\
=\angle OBC - \angle LBC.
\end{array}
$$
Since $ABCD$ is a cyclic quadrilateral, $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,140 |
1. In a company, nine-tenths of the total salary is earned by one-tenth of the employees. The company has several departments. Is it possible that in each department, any one-tenth of the employees earn no more than eleven percent of the department's total salary? | 1. A proposition may be true or false. For example:
Suppose this company has 100 employees, among whom more than 10 employees each have a salary of 81 yuan, and the rest each have 1 yuan. Then the total salary is $10 \times 81 + 90 \times 1 = 900$ yuan, and its nine-tenths is exactly what one-tenth of the employees ea... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,142 |
2. Three piles of stones, one with 51 stones, another with 49 stones, and the last with 5 stones. Any two piles can be combined into one pile. Also, a pile with an even number of stones can be divided into two piles of equal number of stones. Can these two steps be used to transform the three piles into 105 piles, each... | 2. Impossible.
Due to the first step having three scenarios:
(i) $(51,49.5) \rightarrow(100.5)$ are all multiples of 5;
(ii) $(51,49,5) \rightarrow(51,54)$ are all multiples of 3;
(iii) $(51.49 .5) \rightarrow(56,49)$ are all multiples of 7.
In summary, according to the two steps given in the problem, a parent pile w... | Impossible | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,143 |
3. Point $A$ is inside $\angle K M N$, point $B$ is on $K M$, and point $C$ is on $M N$. If $\angle C B M = \angle A B K$ and $\angle B C M = \angle A C N$, prove that the circumcenter of $\triangle B C M$ lies on $A M$.
---
Translation:
3. Point $A$ is inside $\angle K M N$, point $B$ is on $K M$, and point $C$ is ... | 3. As shown in Figure 1, extend $A B$ and $A C$, and connect $A M$.
Figure 1
By the property of vertical angles being equal and $\angle C B M = \angle A B K$, we know that $B M$ is the external angle bisector of $\triangle A B C$.
Similarly, we know that $C M$ is another external angle bisector of $\triangle A B C$.
T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,144 |
4. A convex polygon is divided into triangles by diagonals that intersect only at vertices. Each vertex is marked with the number of triangles it belongs to. If all the diagonals are erased, can the diagonals be redrawn using only these marks? | 4. OK.
Let a convex $n$-sided polygon be divided into $y$ triangles by $x$ diagonals. Since the sum of the interior angles of all triangles equals the sum of the interior angles of the convex $n$-sided polygon, i.e., $y \cdot 180^{\circ} = (n-2) \cdot 180^{\circ}$. Therefore, $y = n-2$.
Since each diagonal is the com... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,145 |
Example 9 Solve the equation $\sqrt{x-2}+\sqrt{x-7}=5$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text provided is already in English, here is the equivalent Chinese text translated back to English as ... | Analysis: Directly solving this equation requires squaring twice. However, if we consider the equation paired with it,
$$
\sqrt{x-2}-\sqrt{x-7}=m \text{, }
$$
treating the two as a system of equations with $x, m$ as unknowns.
Multiplying equation (1) and equation (2) eliminates the radicals, yielding $(x-2)-(x-7)=5m$.... | x=11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,147 |
6. $A D 、 B E 、 C F$ are the altitudes of $\triangle A B C$. $K 、 M 、 N$ are the orthocenters of $\triangle A E F 、 \triangle B F D 、 \triangle C D E$ respectively. Prove that $\triangle D E F$ and $\triangle K M N$ are congruent triangles. | 6. As shown in Figure 3, let the foot of the altitude from $G$ to $B F$ in $\triangle B F D$ be $G$, and the foot of the altitude from $H$ to $C E$ in $\triangle C D E$ be $H$.
$$
\begin{aligned}
\because & B E \perp A C, \\
& A D \perp B C, \\
\therefore & A, B, D, E \text{ are concyclic.}
\end{aligned}
$$
$$
\theref... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,148 |
7. Player B has to guess a two-digit number from Player A. If both digits are correct, or one is correct and the other is off by 1, Player A says “near,” otherwise, Player A says “far.” For example, if Player A’s number is 65, and Player B guesses 65, 64, 66, 55, or 75, Player A says “near,” otherwise, Player A says “f... | 7. (a) The first 17 guesses can exclude at most $17 \times 5=85$ numbers. There are 5 numbers left; the 18th guess cannot determine the number from the remaining 5 numbers.
(c) Construct a $10 \times 9$ table, where the cell in the $i$-th column and $j$-th row contains $(10 i+j-1)$, where $1 \leqslant i \leqslant 9,1 \... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,149 |
2. If the roots of the equation $(x-a)(x-b)=M$ are $\alpha, 3$, then, the sum of the squares of the roots of the equation $(x-\alpha)(x-\beta)=-M$ is ).
(A) $a^{2}+b$
(B) $a+b^{2}$
(C) $a^{2}+b+b^{2}$
(D) $a^{2}+b^{2}$ | 2. (D):
$x_{1} x_{2}=M+2 \beta, x_{1}+x_{2}=\alpha+\beta$, thus.
$$
\begin{aligned}
x_{1}^{2}+x_{2}^{2} & =\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
& =(\alpha+\beta)^{2}-2(M+2 \beta) .
\end{aligned}
$$
And $a+\beta=a+b, \alpha \beta=a b-M$.
Therefore, $x_{1}^{2}+x_{2}^{2}=a^{2}+b^{2}$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,151 |
3. As shown in Figure 1, EF//BC, $S_{\text {EEF }}$ $=S_{\triangle E B C}$. If $S_{\triangle A B C}=1$, then $S_{\triangle C E F}=$ ( ).
(A) $\sqrt{5}-2$
(B) $\sqrt{3}-\frac{3}{2}$
(C) $\frac{1}{4}$
(D) $\frac{1}{5}$ | 3. (A).
As shown in Figure 6, draw $A Q \perp E F, E P$
$$
\begin{array}{c}
\perp B C \text{. By } A Q \cdot E F=E P \cdot B C \Rightarrow \\
\frac{E F}{B C}=\frac{E P}{A Q}=\frac{E B}{A E} . \\
\text{Also } \frac{E F}{B C}=\frac{A E}{A B} \\
\Rightarrow A E^{2}=A B \cdot E B,
\end{array}
$$
$E$ is the golden section ... | \sqrt{5}-2 | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,152 |
5. The positive integer solutions to the equation $x y-3 x^{2}=12$ are $(x, y)$. Then the number of all such solutions $(x, y)$ is ( ).
(A) 5
(B) 3
(C) 4
(D) 6 | 5.(D).
$y=3 x+\frac{12}{x}$. When $x=1,2,3,4,6,12$, $y=15,12,13,15,20,37$. Therefore, there are 6 solutions. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,154 |
6. As shown in Figure 2. $AB$ is the diameter of $\odot O$, $\odot O_{1}$ and $\odot O_{2}$ are both internally tangent to $\odot O$, with points of tangency $C_{1}$ and $C_{2}$, and both are tangent to $AB$. Points of tangency are $D_{1}$ and $D_{2}$. The intersection point $M$ of lines $C_{1}D_{1}$ and $C_{2}D_{2}$ h... | 6.(B).
As shown in Figure 7, it is clear that $O, O_{1}, C_{1}$ are collinear. Let $C_{1} D_{1}$ intersect $\odot O$ at $M_{1}$, and connect $O_{1} D_{1}, O M_{1}$. It is easy to prove that $O M_{1} / / O_{1} D_{1} \Rightarrow O M_{1} \perp A B$. $M_{1}$ is the midpoint of $\overparen{A B}$. Similarly, $C_{2} D_{2}$ a... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,155 |
1. As shown in Figure $3 . \triangle P A B 、 \triangle P C D$ are both equilateral triangles. $M 、 N$ are the midpoints of $P B 、 C D$ respectively, and $P Q \perp A C$ at $Q$. Then $\angle M Q N=$ | $=.1 .30^{\circ}$.
Connect $A M, P N$. It is easy to prove that $A, P, M, Q$ are concyclic. Therefore, $\angle P Q M=\angle P A M=30^{\circ}$. Also, $P, N, C, Q$ are concyclic, so we get $\angle C Q N=\angle C P N=30^{\circ}$. | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,156 |
Example 10 As shown in Figure 3, in $\triangle A B C$, $D$ and $E$ are points on sides $B C$ and $A B$ respectively, and $\angle 1=\angle 2=\angle 3$. If the perimeters of $\triangle A B C$, $\triangle E B D$, and $\triangle A D C$ are $m$, $m_{1}$, and $m_{2}$ respectively, prove: $\frac{m_{1}+m_{2}}{m} \leqslant \fra... | Analysis: For the sake of convenience in narration, let $BC = a, AC = b, AB = c$. Note that the condition given in the problem is $\angle 1 = \angle 2 = \angle 3$, which means the three triangles involved are similar, i.e.,
$\triangle ABC \backsim \triangle EBD \backsim \triangle DAC$.
Considering the ratio of the peri... | \frac{m_{1} + m_{2}}{m} \leq \frac{5}{4} | Geometry | proof | Yes | Yes | cn_contest | false | 713,158 |
3. The number of intersections of the function $y=x \cdot|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is $\qquad$ | 3.3 .
$$
y=\left\{\begin{array}{ll}
x^{2}-2 \sqrt{3} x+2, & x>0, \\
2, & x=0 . \\
-x^{2}-2 \sqrt{3} x+2, & x<0
\end{array}\right.
$$
When $x>0$, $y=x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 2 points;
When $x=0$, $y=2$ does not intersect the x-axis;
When $x<0$, $y=-x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 1 p... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,159 |
4. As shown in Figure $4, \odot O$ is a unit circle.
$A B 、 C D$ are both its diameters, $\overparen{A D}=$ $60^{\circ}, P$ is a point on $\overparen{D B}$. Then $(P A+$ $P C)_{\text {max }}=$ $\qquad$ .$(P A+P C)_{\text {min }}$ $=$ | $4.2 \sqrt{3}, 3$.
As shown in Figure 8, connect $B C$. It is easy to prove that
$\triangle A M O \subseteq \triangle C N B$.
Thus, $A M = C N, M O = N B$.
It is easy to prove that $\triangle A O M \backsim \triangle A P N$,
$\triangle C O N \backsim \triangle C P M$.
Therefore, $A M \cdot A P = A O \cdot A N$,
$C N \c... | 2 \sqrt{3}, 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,160 |
One. (20 points) The graph of the quadratic function $y=x^{2}+p x+q$ passes through the point $(2,-1)$ and intersects the $x$-axis at two distinct points $A(a, 0)$ and $B(b, 0)$. Let the vertex of the graph be $M$. Find the analytical expression of the quadratic function that minimizes the area of $\triangle A M B$.
-... | From the problem, we know that $4+2 p+q=-1$, which means $2 p+q=-5$.
Also, $|a-b|=\sqrt{(a+b)^{2}-4 a b}=\sqrt{p^{2}-4 q}$, and the point $M\left(-\frac{p}{2}, \frac{4 q-p^{2}}{4}\right)$, then
$$
\begin{array}{l}
S_{-A Y B}=\frac{1}{2}|A B| \cdot\left|\frac{4 q-p^{2}}{4}\right| \\
=\frac{1}{8}|a-b| \cdot\left(p^{2}-4 ... | y=x^{2}-4 x+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,161 |
II. (25 points) As shown in Figure 5, in $\triangle ABC$, $AD$ is the angle bisector, and $M, N$ are the midpoints of $AB, AC$ respectively. It is known that $\angle MDN=\frac{1}{2}(\angle B + \angle C)$. Prove:
$$
BC=\frac{1}{2}(AB+AC).
$$ | When $A B=A C$, the conclusion is easy to prove.
When $A B \neq A C$, two different proof methods are provided:
Proof 1: As shown in Figure 9, connect
$M N$ and take the incenter $I$ of $\triangle A M N$, $I$ must
be on $A D$, then connect $I M, I N$.
$$
\begin{array}{l}
\because \angle M D V \\
=\frac{1}{2}(\angle B+\... | BC=\frac{1}{2}(AB+AC) | Geometry | proof | Yes | Yes | cn_contest | false | 713,162 |
Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.
| Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.
Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red lin... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,163 |
1. In space, there are $n(n \geqslant 3)$ planes, among which any three planes do not have a common perpendicular plane. Then, among the following four conclusions:
（1）No two planes are parallel to each other;
（2）No three planes intersect in a single line;
(3) Any two lines of intersection between planes are not parall... | - 1. (D).
If two planes $\alpha_{1}, \alpha_{2}$ are parallel, then the common perpendicular plane of $\alpha_{1}, \alpha_{3}$ is also a perpendicular plane of $\alpha_{2}$, which contradicts the condition that there is no common perpendicular plane with any plane. Conclusion (1) is true.
If three planes intersect at... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,164 |
2. If the graph of the function $y=f(x)$ over the interval $[a, b]$ can be approximately regarded as a line segment, then when $c \in(a, b)$, the approximate value of $f(c)$ can be expressed as ( ).
(A) $\frac{f(a)+f(b)}{2}$
(B) $f\left(\frac{a+b}{2}\right)$
(C) $\frac{(b-c) f(a)+(c-a) f(b)}{(b-a)}$
(D) $f(a)-\frac{c-a... | 2. (C).
Idea 1: Find the equation of the line passing through the points $(a, f(a)),(b, f(b))$
$$
f(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a) .
$$
Substitute $x=c$ to get (C).
Idea 2: By the section formula, we have
$$
\left\{\begin{array}{l}
c=\frac{a+\lambda b}{1+\lambda}, \\
f(c)=\frac{f(a)+\lambda f(b)}{1+\lambda} .
\en... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,165 |
3. Let $a>b>c, a+b+c=1$, and $a^{2}+b^{2}+c^{2}=1$. Then ( )
(A) $a+b>1$
(B) $a+b=1$
(C) $a+b<1$
(D) Cannot be determined, depends on the specific values of $a, b$ | 3. (A).
Take $a=\frac{1+\sqrt{5}}{4}, b=\frac{1}{2}, c=\frac{1-\sqrt{5}}{4}$, which can negate (B) and (C), but cannot negate (D). To determine (A), we can use proof by contradiction. Assume there exist $a_{0} 、 b_{0}$ such that $a_{0}+b_{0} \leqslant 1$, then there exists $c_{0} \geqslant 0$, and further $a_{0}>b_{v}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,166 |
4. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ have an eccentricity of $e=\frac{\sqrt{3}}{2}$. Given that the point $P\left(0, \frac{3}{2}\right)$ is at a maximum distance of $\frac{7}{4}$ from a point on the ellipse. Then the length of the semi-minor axis $b=(\quad)$.
(A) $\frac{1}{16}$
(B) $\frac{1}{8... | 4. (C).
From $e=\frac{c}{a}=\frac{\sqrt{3}}{2}$, we get $\frac{a^{2}-b^{2}}{a^{2}}=\frac{3}{4}$, which implies $a^{2}=4 b^{2}$.
The equation of the ellipse is $\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1$.
I
Let the distance from a point $(x, y)$ on the ellipse to point $P$ be $d$, then
$$
\begin{aligned}
d^{2} & =x^{... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,167 |
5. $S=\{1,2, \cdots, 2003\} . A$ is a three-element subset of $S$. Satisfies: The elements in $A$ can form an arithmetic sequence. Then, the number of such three-element subsets $A$ is ( ).
(A) $C_{2 \propto \infty}^{3}$
(B) $\mathrm{C}_{1001}^{2}+\mathrm{C}_{1000}^{2}$
(C) $P_{1 \infty 1}^{2}+P_{1 \infty}^{2}$
(D) $\m... | 5. (B).
If $a_{1}, a_{2}, a_{3}$ form an arithmetic sequence, then $a_{1}+a_{3}=2a_{2}$, which means the parity of the first and last terms $a_{1}, a_{3}$ is the same, and once the first and last terms are determined, the arithmetic sequence is also determined. It is worth noting that although $a_{1}, a_{2}, a_{3}$ fo... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 713,168 |
Example 1 Given that the sum of several positive integers is 1976. Find the maximum value of their product. | Explanation: This is a problem from the 18th International Mathematical Olympiad held in 1976, but the year 1976 is not crucial. Replacing 1976 with 2002, 2003, the method remains the same.
Since there are only a finite number of different sets of positive integers whose sum is 1976, and their products are also finite... | 2 \times 3^{658} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,169 |
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{67}$ are positive integers, and their sum equals 110. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$. | Explanation: Start with any set of 67 positive integers $x_{1}, x_{2}, \cdots, x_{67}$ whose sum is 110. Without loss of generality, assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{66} \leqslant x_{67}.
$$
First, freeze $x_{2}, x_{3}, \cdots, x_{66}$ and only study $x_{1}$ and $x_{67}$. Since
$$
\begin{a... | 2002 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,170 |
Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$
Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only i... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,171 |
2. As shown in Figure $1, A$ and $C$ are any two points on the graph of the function $y=\frac{1}{x}$ that are symmetric with respect to the origin. $AB$ and $CD$ are both perpendicular to the $x$-axis, with feet of the perpendiculars at $B$ and $D$, respectively. Let the area of quadrilateral $ABCD$ be $S$, then ).
(A)... | 2. (B).
Let the coordinates of point $A$ be $\left(x_{0}, \frac{1}{x_{0}}\right)\left(x_{0}>0\right)$, then $O B=x_{0}$, $B A=\frac{1}{x_{0}}$. Therefore, $S_{\text {בAOB }}=\frac{1}{2} \cdot O B \cdot B A=\frac{1}{2} \cdot x_{0} \cdot \frac{1}{x_{0}}=\frac{1}{2}$. Hence, $S_{G \text { ver } D}=4 S_{\dot{S}^{4} / B B}... | null | Number Theory | proof | Yes | Yes | cn_contest | false | 713,172 |
3. If the lengths of three line segments $a, b, c$ satisfy $\frac{b}{a}=\frac{c}{b}=\frac{\sqrt{5}-1}{2}$, then $(a, b, c)$ is called a "golden segment group". The three line segments in a golden segment group ( ).
(A) can form an acute triangle
(B) can form a right triangle
(C) can form an obtuse triangle
(D) cannot f... | 3. (D).
Let $k=\frac{\sqrt{5}-1}{2}$, then $k^{2}+k=1$.
Given $\frac{b}{a}=\frac{c}{b}=k$, we have $b=k a, c=k^{2} a$. Since $b+c=\left(k+k^{2}\right) a=a$,
therefore $a, b, c$ cannot form a triangle. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,173 |
4. Given the equation about $x$
$$
x^{2}-6 x+(a-2)|x-3|+9-2 a=0
$$
has exactly two distinct real roots. Then the range of the real number $a$ is ( ).
(A) $a=-2$
(B) $a>0$
(C) $a=-2$ or $a>0$
(D) $a \leqslant-2$ or $a>0$ | 4. (C).
Solution 1: The original equation can be transformed into
$$
\begin{array}{l}
|x-3|^{2}+(a-2)|x-3|-2 a=0 . \\
\text { Let } t=|x-3|(t \geqslant 0), \text { then } \\
t^{2}+(a-2) t-2 a=0 .
\end{array}
$$
By the problem statement, equation (1) about $t$ has only one positive root. We discuss in two cases:
(1) W... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,174 |
5. Given that the three altitudes of $\triangle A B C$ are $A D=3, B E=4, C F=5$, and the lengths of the three sides of this triangle are all integers. Then the minimum value of the length of the shortest side is ( ).
(A) 10
(B) 12
(C) 14
(D) 16 | 5. (B).
From $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{1}{2} C A \cdot B E=\frac{1}{2} A B \cdot C F$,
we get $3 B C=4 C A=5 A B$.
It is clear that $A B$ is the shortest side.
From $B C=\frac{5}{3} A B, C A=\frac{5}{4} A B$ and the lengths of $B C, C A, A B$ are all
integers, we know that $3 \mid A B$ and ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false | 713,175 |
6. Write 10 pairwise distinct positive integers $a_{1}, a_{2}, \cdots, a_{10}$ in the form of the following chart, where the number pointed to by two arrows equals the sum of the two numbers at the starting points of these arrows. For example $\stackrel{a_{1} \rightarrow a_{2}}{a_{5}}$ indicates $a_{2}=$ $a_{1}+a_{5}$.... | 6. (B).
\[
\begin{aligned}
a_{4} & =a_{3}+a_{7}=\left(a_{2}+a_{6}\right)+\left(a_{6}+a_{9}\right) \\
& =a_{2}+2 a_{6}+a_{9} \\
& =\left(a_{1}+a_{5}\right)+2\left(a_{5}+a_{8}\right)+\left(a_{8}+a_{10}\right) \\
& =a_{1}+3 a_{5}+3 a_{8}+a_{10} .
\end{aligned}
\]
To minimize \(a_{4}\), we can first consider the values of... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,176 |
1. Given the inequality about $x$: $(2 a-3 b) x>7 a-5 b$, the solution set is $x>2 a-$ $5 b$ is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
=.1 . x>-\frac{67}{12} \text {. }
$$
From the given, we know $2 a-3 b0$.
Therefore, $7 a-3 b=\frac{63}{17} b-3 b=\frac{12}{17} b>0$.
Thus, the solution to the inequality is
$$
x>\frac{2 a-5 b}{7 a-3 b}=-\frac{67}{12} .
$$ | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,177 |
3. It is known that the ages of A, B, and C
are all positive integers. A's age is twice B's age, B is 7 years younger than C. If the sum of the ages of these three people is a prime number less than 70, and the sum of the digits of this prime number is 13, then the ages of A, B, and C are | 3. A is 30 years old, B is 15 years old, C is 22 years old.
Let the ages of A, B, and C be $x$ years, $y$ years, and $z$ years respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
x=2 y, \\
y=z-7, \\
x+y+z \text { is a prime number, and } x+y+z<70 . \\
\because 13=4+9=5+8=6+7,
\end{array}\right.
... | x=30, y=15, z=22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,179 |
4. Let $x^{2}-y^{2}-z^{2}=0$. Then the factorization of $x^{3}-y^{3}-z^{3}$ into a product of linear factors is $\qquad$ . | 4. $(x-y)(x-z)(2 x+y+z)$.
Let $y=a x, z=b x$, then by the given condition we get $a^{2}+b^{2}=1$.
$$
\begin{array}{l}
\therefore x^{3}-y^{3}-z^{3}=x^{3}\left(1-a^{3}-b^{3}\right) \\
=x^{3}\left(a^{2}+b^{2}-a^{3}-b^{3}\right) \\
=x^{3}(1-a)(1-b)(2+a+b) \\
=(x-a x)(x-b x)(2 x+a x+b x) \\
=(x-y)(x-z)(2 x+y+z) .
\end{arra... | (x-y)(x-z)(2 x+y+z) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,180 |
One. (20 points) As shown in Figure 3, let $O$ be the circumcenter of acute $\triangle ABC$, $BE$ and $CF$ be two altitudes, and $M$, $N$ be the midpoints of $BC$ and $EF$ respectively. Prove that $OA \parallel MN$. | Proof 1: As shown in Figure 5, connect $O B, M E, M F$.
$$
\begin{array}{c}
\because \angle A O B=2 \angle A C B, \\
\therefore \angle 1=\frac{1}{2}\left(180^{\circ}-\right. \\
\angle A O B)=90^{\circ}-\angle A C B . \\
\because B E \perp A C, C F \perp A B,
\end{array}
$$
$\therefore B, C, E, F$ are concyclic.
$$
\the... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,181 |
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