problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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Example 6. In rhombus $A B C D$ with side length $a$, $\angle D A B$ $F$ is a moving point on $C D$, satisfying the condition $A E + C F = a$.
(1) Prove: Regardless of how $E, F$ move, $\triangle B E F$ is always an equilateral triangle.
(2) Find the minimum area of $\triangle B E F$.
(1990, Hefei Junior High School Co... | Proof (1) As shown in Figure 6, let
$AB = x$, then
$ED = FC = a - x$, $DF = x$.
In $\triangle ABE$,
$\triangle BCF$ and $\triangle DEF$,
by the Law of Cosines, we get
$$
\begin{aligned}
BE^2 & = a^2 + x^2 - 2ax \cos 60^\circ \\
& = a^2 + x^2 - ax, \\
BF^2 & = a^2 + (a-x)^2 - 2a(a-x) \cos 60^\circ \\
& = a^2 + x^2 - ax,... | \frac{3\sqrt{3}}{16} a^2 | Geometry | proof | Yes | Yes | cn_contest | false | 705,620 |
Example 7. If $X, Y$ are points on the upper and lower bases of trapezoid $ABCD$, respectively, and $XY$ bisects the area of this trapezoid. Prove that $XY$ must pass through a fixed point.
(1984, Guangzhou Junior Competition) | Proof As shown in Figure 7, let $M, N$ be the midpoints of the upper and lower bases of the trapezoid. It is easy to prove that $MN$ bisects the area of the trapezoid. Given that $XY$ also bisects the area of the trapezoid, thus $MN$ = $S \triangle OYN$. It is also evident that $\triangle OXM \sim \triangle OYN$, so
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,621 |
1. In $\triangle A B C$, the altitude is equal to the radius of $\odot O$, $\odot O$ rolls on $A B$, the point of tangency is $P, \odot O$ intersects $A C, B C$ at $M, N$. Then $\overparen{M T N}$ ( ).
(A) varies from $0^{\circ}$ to $30^{\circ}$.
(B) varies from $30^{\circ}$ to $60^{\circ}$.
(C) varies from $60^{\circ}... | (E)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 705,622 |
Example 1. Select 3 elements $i, j, k$ from the set $\{1,2, \cdots, n\}$, such that (i) $1 \leqslant i \leqslant j \leqslant k \leqslant n$, (ii) $1 \leqslant i<j<k \leqslant n$, and $j-i \geqslant m, k-j \geqslant m$. How many different ways are there to select them? | (i) Establish a correspondence: $(i, j, k) \xrightarrow{\mathrm{f}}\left(i^{\prime}, j^{\prime}, k^{\prime}\right)$, such that $i^{\prime}=i, j^{\prime}=j+1, k^{\prime}=k+2$, and $1 \leqslant i^{\prime} < j^{\prime} < k^{\prime} \leqslant n+2$. This is a one-to-one correspondence, so the original problem is transformed... | C_{\mathrm{n}-2(m-1)}^{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,625 |
Example 2: Let $1 \leqslant k<n$, consider all positive integer sequences of length $n$, and find the total number of terms $T(n, k)$ in these sequences that are equal to $k$. (29th IMO Shortlist) | The sequence has at least two terms, namely $n-k$ and $k$; at most $n-k+1$ terms, which include $k$ and $n-k$ ones. Let $T(p, n, k)$ denote the total number of terms equal to $k$ in all sequences of $p+1$ terms whose sum is $n$.
When $p=1$, there are only two sequences $\{n-k, k\}$, $\{k, n-k\}$, so $T(1, n, k)=2$. We ... | (n-k+3) \cdot 2^{n-k-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,626 |
For example. Let $a_{1}$ be the unit digit of $1^{2}+2^{2}+\cdots+n^{2}$, where $n=1,2,3, \cdots$. Try to prove that $\quad 0, a_{1} a_{2} \cdots a_{2} \cdots$ is a rational number. (1984, National High School League) | Solve: Prove $a_{2+20}=a_{a}$ using induction.
Specific verification shows that $a_{21}=a_{1}$. Assume $a_{x+20} = a_{x}$ holds, i.e., the unit digit of $1^{2}+2^{2}+\cdots+k^{2}+(k+1)^{2}+\cdots+(k+20)^{2}$ is the same as the unit digit of $1^{2}+2^{2}+\cdots+k^{2}$. Notice that in $1^{2}+2^{2}+\cdots+k^{2}+(k+1)^{2}+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,638 |
Example 5. Prove that the infinite decimal
$$
A=0.123456789101112 \cdots 9899100101 \cdots
$$
is an irrational number. | Let $A$ be an infinite decimal. Assume it is periodic, with the length of its repeating segment being $t$. Clearly, $t \neq 1$, and there must be a non-zero digit within the repeating segment, which means that at any position in $A$, there will not be a "segment" longer than $t$ that consists entirely of the digit 0.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,653 |
Example 7. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\left\{\begin{array}{l}
a_{1}=a_{2}=1, \\
a_{n+2}=a_{n+1}+a_{n}(n \geqslant 2) .
\end{array}\right.
$$
Let the last three digits of $a_{\mathrm{n}}$ be denoted as $b_{\mathrm{n}}$ (if there are fewer than three digits, pad with zeros on the left), then we g... | Solve the relationship by guessing this infinite decimal. Take the sequence $\left\{b_{a}\right\}$, form arrays $\left(b_{1}, b_{2}\right),\left(b_{2}, b_{8}\right), \cdots,\left(b_{\mathrm{B}-1}, b_{\mathrm{n}}\right), \cdots$, and only need to verify that there are two identical arrays, which means $B$ is a repeating... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,655 |
Example 8. Let $\left[(\sqrt{10})^{\mathrm{D}}\right]$ have the unit digit $\alpha_{\mathrm{n}}$, where $n$ is a natural number, and $[x]$ denotes the greatest integer not exceeding $x$. Prove that the infinite decimal
$$
A=0 . \alpha_{1} \alpha_{2} \cdots \alpha_{\mathrm{n}} \cdots
$$
is an irrational number. (Based ... | From $\alpha_{2n}=G\left(\left[(\sqrt{10})^{2n}\right]\right)=0$, we know that the digits at even positions after the decimal point of $A$ are all 0. And
$$
\begin{aligned}
\alpha_{2n+1} & =G\left(\left[(\sqrt{10})^{2n+1}\right]\right) \\
& =G\left(\left[10^{n} \cdot \sqrt{10}\right]\right)
\end{aligned}
$$
is the $n$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,656 |
Example 1. Two persons, A and B, color the 12 edges of a cube with two colors, red and green. First, A arbitrarily selects three edges and paints them red. Then, B selects three other edges and paints them green. Next, A paints another three edges red, and finally, B paints the remaining three edges green. Can A adopt ... | Solve by dividing the 12 edges into groups:
$$
\begin{array}{l}
\left\{A_{1} B_{1}, B_{2} B_{3},\right. \\
\left.A_{3} A_{4}\right\},\left\{A_{2} B_{2},\right. \\
\left.B_{3} B_{4}, A_{4} A_{1}\right\}, \\
\left\{A_{3} B_{3}, B_{4} B_{1},\right. \\
\left.A_{1} A_{2}\right\},\left\{A_{4} B_{4},\right.
\end{array}
$$
$\l... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,657 |
Example 2. On a plane, there are 5 points, among any 4 points, there are 3 points that are the 3 vertices of an equilateral triangle. Prove that among the 5 points, there must be 4 points that are the 4 vertices of a rhombus with an internal angle of $60^{\circ}$. (1985, Bulgarian Mathematical Competition) | Prove that 5 points can form 5 different sets of 4 points, each of which has 3 points that can form an equilateral triangle, hence there are at least 5 equilateral triangles (including repeated counts). Each equilateral triangle belongs to a team of different sets of 4 points, so it is known that 5 points form at least... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,658 |
Example 3. Given $2 n(n \geqslant 2)$ points in space, where no four points are coplanar. Try to prove that any $n^{2}+1$ segments connecting these points must form two triangles with a common edge. (1987, China Mathematical Olympiad Training Team Selection Competition) | Prove that when $n=2$, $n^{2}+1=5$, 4 points are connected by 5 line segments, which naturally form two triangles with one common side, i.e., the proposition holds when $n=2$.
Assume the proposition holds for $n=k \geqslant 2$. When $n=k+1$, $2 k+2$ points are connected by $(k+1)^{2}+1=k^{2}+2 k+2$ line segments, and ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,659 |
Example 3. Let $A=A_{1} \cup A_{2} \cup \cdots \cup A_{k}$. (i) The intersection of any $k$ sets is not empty; (ii) The intersection of any $k+1$ sets is an empty set. Find: (1) What is the minimum value of $|A|$; (2) What is the minimum value of $\left|A_{1}\right|$ when $|A|$ is the smallest? | (1) Consider any $r$-element subset $\left\{i_{1}, i_{2}, \cdots, i_{r}\right\}$ of the index set $\{1,2, \cdots, k\}$. For any element $a$ in $A_{1} \cap A_{1} \cap \cdots \cap A_{1}$, the correspondence $\left\{i_{1}, i_{2}, \cdots, i_{r}\right\} \rightarrow a$ is an injection. This is because if $\left\{j_{1}, j_{2}... | C_{k-1}^{r-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,660 |
In $\triangle A B C$, extend $A B$, $A C$, and $B C$ successively, such that $B A^{\prime}=\lambda_{1} A B$, $C B^{\prime}=\lambda_{2} B C$, $A C^{\prime}=\lambda_{3} C A$. Find $S \triangle \triangle^{\prime} \mathrm{B}^{\prime} C^{\prime}: S \triangle \triangle B C$. | Solve for
$$
\begin{array}{l}
\frac{S \triangle A^{\prime} C^{\prime}}{S \triangle A B C}=\frac{A A^{\prime} \cdot A C^{\prime}}{A B \cdot A C} \\
\quad=\frac{\left(1+\lambda_{1}\right) A B \cdot \lambda_{3} C A}{A B \cdot A C} . \\
=\left(1+\lambda_{1}\right) \lambda_{3},
\end{array}
$$
Thus, $S \triangle A^{\prime} ... | 1+\left(\lambda_{1}+\lambda_{2}-\lambda_{3}\right)+\left(\lambda_{1} \lambda_{2}-\lambda_{2} \lambda_{3}-\lambda_{3} \lambda_{1}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,663 |
In $\triangle A B C$, extend $A B$, $C A$ so that $B A^{\prime}=\lambda_{1} A B, A C^{\prime}=\lambda_{3} C A$, and extend $B C$ so that $C B^{\prime}=\lambda_{2} B C$,
find $S \triangle A^{\prime} B^{\prime} C^{\prime}: S \triangle \triangle B C$. | \begin{array}{l}\text { Similarly, } S \triangle \triangle_{A^{\prime} B B^{\prime}}=\left(1-\lambda_{2}\right) \lambda_{1} S \triangle A B C, \\ S \triangle B^{\prime} B C^{\prime}=\left(1-\lambda_{2}\right)\left(1+\lambda_{3}\right) S \triangle \triangle B C C \\ \text { Also, } S \triangle \mathrm{A}^{\prime} \mathr... | 1+\left(\lambda_{1}-\lambda_{2}+\lambda_{3}\right)+\left(-\lambda_{1} \lambda_{2}-\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,664 |
Side 2. Given the sequence $a_{n}=2 a_{n-1}-a_{n-2}$, where $a_{1}=2, a_{2}=3, n=3,4, \cdots$. Try to find the sum of the first $n$ terms and the general term formula of this sequence. | Given the recursive sequence, where $p=2$, $q=-1$, by Theorem 1 we have
$$
\begin{aligned}
S_{\mathrm{n}} & =n a_{1}+\frac{n(n-1)}{2}\left(a_{2}-a_{1}\right) \\
& =\frac{1}{2} n^{2}+\frac{3}{2} n, \\
a_{\mathrm{n}} & =S_{\mathrm{n}}-S_{\mathrm{n}-1} \\
& =\frac{1}{2} n^{2}+\frac{3}{2} n-\frac{1}{2}(n-1)^{2}-\frac{3}{2}... | a_{\mathrm{n}} = n + 1, \quad S_{\mathrm{n}} = \frac{1}{2} n^{2} + \frac{3}{2} n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,666 |
3. Let $f\left(\frac{1-x^{2}}{1+x^{2}}\right)=x$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ is ( . ).
(A) $\frac{1-x}{1+x}$ or $\frac{x-1}{1+x}$.
(B) $\frac{1+x}{1-x}$ or $\frac{1+x}{x-1}$.
(C) $\frac{x-1}{x+1}$
(D) $\frac{x+1}{x-1}$ | 3. A (Hint: Let $x=\operatorname{tg} \theta)$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,669 |
2. Let $f(x)$ be an even function, $g(x)$ be an odd function, and $f(x)=-g(x+c)$ where $c>0$, then $f(x)$ is a periodic function with the smallest positive period of $\qquad$. | 2. $4 c_{.}$ | 4c | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,672 |
Four, $\odot O$ is the excircle of $\triangle ABC$ outside the side $BC$, and $D, E, F$ are the points of tangency of $\odot O$ with $BC, CA$, and $AB$ respectively (as shown in the figure).
(1) If $OD$ intersects $EF$ at $K$, prove that $AK$ bisects $BC$.
(2) Given $BC = a, AC = b, AB = c$, and $b > c$, find the ratio... | (1) Let $A K$ intersect $B C$ at point $M$, and connect $O E, O F$.
$$
\begin{array}{c}
\because \quad \angle A B C=\angle D O F=\angle B, \\
\angle A C B=\angle D O E=\angle C, \\
\therefore \frac{F K}{K E}=\sin \angle B \\
\sin \angle C .
\end{array}
$$
Also, $\frac{F K}{\sin \angle F A K}=\frac{A F}{\sin \angle A K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,674 |
Five, try to select 1999 integers from the set $\{1,2,3, \cdots, 3000\}$ to form a subset $A$, such that if $x \in A$, then $2 x \notin A$. | Five, the selection method is as follows:
1) Select 1500 integers from 1501 to 3000 to form $A_{1}$.
2) From 1 to 1500, remove numbers 751 to 1500 (these numbers multiplied by 2 are in $A_{1}$), and select 376 to 750, a total of 375 integers to form $A_{2}$.
3) From 1 to 375, remove 188 to 375 (these numbers multiplied... | 1999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,675 |
For $a \geqslant 1$, calculate the sum of the infinite series
$$
\begin{array}{l}
\frac{a}{a+1}+\frac{a^{2}}{(a+1)\left(a^{2}+1\right)} \\
+\frac{a^{4}}{\left.(a+1) a^{2}+1\right)\left(a^{4}+1\right)} \\
+\frac{a^{8}}{(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)}+\cdots
\end{array}
$$ | $\begin{array}{c}\text { I. When } a=1 \text {, } \Sigma=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \text {. When } a>1 \text {, we have } \Sigma_{1}=\frac{a^{2}-a}{a^{2}-1}, \\ \Sigma_{2}=\frac{a^{4}-a}{a^{4}-1}, \Sigma_{3}=\frac{a^{8}-a}{a^{8}-1}, \cdots, \Sigma_{n}=\frac{a^{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,676 |
II. Find the real solutions of the equation $4 x^{2}-40[x]+51=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | II. Observing the problem, we can see that $x$ cannot be zero or a negative number. Let $[x]=t$ ($t$ is a positive integer), substituting into the original equation we get $4 x^{2}-40 t+51=0$, which is $x=\sqrt{\frac{40 t-51}{4}}$. Also, since $0 \leqslant x-[x] < 1$. Therefore, the solutions to the original equation a... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,677 |
Three, as shown in the figure: ()
Square $ABCD$ has its diagonal intersection, $AE$ is the bisector of $\angle BAC$, intersecting $BC$ at $E$. $DH \perp AE$ at $H$, intersecting $AB$ at $F$, and intersecting $AO$ at $G$. Prove: $BF=2OG$. | Three, the three sides of $\triangle A O B$ are intercepted by the line $F \cdot G D$. By Menelaus' theorem, we have $\frac{A F}{F B} \cdot \frac{B D}{D O} \cdot \frac{O G}{G A}=1$.
$$
\begin{array}{l}
\because \angle F A H=\angle G A H \text{ and } \triangle A H \perp F G, \\
\therefore A F=A G . \\
\text { Substituti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,678 |
Four, let $A B C$ be an equilateral triangle, and $P$ a point on its incircle. Prove that $P A^{2}+P B^{2}+P C^{2}$ is a constant. | Four, as shown in the figure, establish
a rectangular coordinate system,
the coordinates of point $P$ are
$$
\begin{array}{l}
\left(\frac{\sqrt{3}}{3} \cos \theta,\right. \\
\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3} \\
\cdot \sin \theta) .
\end{array}
$$
Let $\Sigma=P A^{2}+P B^{2}+P C^{2}$, then
$$
\begin{array}{l}
\Sigm... | 5 | Geometry | proof | Yes | Yes | cn_contest | false | 705,679 |
Five, Triangular Number Table
Each number is equal to the sum of the number above it in the previous row and its adjacent numbers (if there is no number in its position, it is considered as zero).
Prove: From the third row onwards, each row must contain an even number. | Five, take the congruence modulo 2, the first eight rows of the triangular array are:
where the first 4 elements of the fourth row are 1101, which are exactly the same as the first 4 elements of the eighth row, thus forming
1101 as shown in the figure on the right, which repeats
1000
downward on the left side of the a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,680 |
Six, "Graceful Graph" is a type of graph that satisfies the following conditions: when all its vertices are labeled with different numbers (which can be integers starting from 0, but not necessarily consecutive), the labels of the edges (the difference between the two endpoints) form a continuous series of natural numb... | Six,
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,681 |
Example 5. A factory has $n \geqslant 3$ jobs, labeled from 1 to $n$ in increasing order of salary. $n$ job seekers are labeled from 1 to $n$ in increasing order of their abilities. A job seeker $i$ can only take on job $j$ if $i \geqslant j$.
Job seekers arrive in a random order, and each person is hired for the high... | Proof: Let $\left(x_{1}, x_{2}, \cdots, x_{\mathrm{n}}\right)$ be the sequence of job seekers arranged in the order they arrive at the factory, with $x_{\mathrm{p}}$ and $x_{\mathrm{q}}$ representing job seekers $n$ and $n-1$, respectively. Each sequence will have some number of people hired, with $x_{\mathrm{r}}$ deno... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,682 |
2. Given the set $P=\{(x, y) \mid x=\sin \theta + \cos \theta, y=\sin 2 \theta, \theta \in R\}, Q=\{(x, y) \mid x-y+1=0\}$. Then express $P \cap Q$ using the listing method $=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $\{(-1,0)\}$. | null | Algebra | proof | Yes | Yes | cn_contest | false | 705,686 |
3. Given $t \neq 0, \cos (x+b)=\frac{p^{5}+i}{2 p^{2}}$, $\cos (\alpha-\beta)=\frac{p-1}{2 p^{2}}$. Then express $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta$ in terms of $p$ $=$ $\qquad$ | 3. $-\frac{1}{p}$ | -\frac{1}{p} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,687 |
5. Given $\sin \theta+\cos \theta=\sqrt{2}$.
Then $\left(\log _{\frac{1}{2}} \sin \theta\right)\left(\log _{\frac{1}{2}} \cos \theta\right)$ is equal to | 5. $\frac{1}{4}$. | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,689 |
6. Given four points in the Cartesian plane $A(1,0)$, $B(3,0)$, $C(0,1)$, $D(6,1)$. Then the coordinates of the point in the Cartesian plane that minimizes the sum of the squares of the distances to these four points are $\qquad$ - | $6 .\left(\frac{5}{2}, \frac{1}{2}\right)$ | \left(\frac{5}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,690 |
7. When the point $(x, y)$ moves on the curve $\frac{(x-5)^{2}}{16}+\frac{y^{2}}{9}=1$, the sum of the maximum and minimum values that the algebraic expression $\frac{x^{2}}{16}+\frac{y^{2}}{9}$ can take is
| 7. $\frac{41}{8}$. | \frac{41}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,691 |
9. Given the function $y=\log _{4}(x-k a)$ $+\log _{x}\left(x^{2}-a^{2}\right)$ with the domain $x>a$, then the range of the real number $k$ is $\qquad$. | 9. $-1 \leqslant k \leqslant 1$ | -1 \leqslant k \leqslant 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,694 |
13. Given $n \in N, 2^{n}>n^{3}$. Then the range of values for $n$ is $\qquad$ - | 13. $n=1$ or $n \geqslant 10(n \in N)$. | n=1 \text{ or } n \geqslant 10(n \in N) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 705,698 |
18. For the old trigonometric equation $\sin \left(x+\frac{\pi}{4}\right)-\sin 2 x$ $=a$ to have real solutions, the range of real number $a$ is | 18. $-2 \leqslant a \leqslant \frac{9}{8}$. | -2 \leqslant a \leqslant \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,703 |
1. Divide each side of $\triangle A B C$ into $n$ equal parts, and draw lines parallel to the other two sides through each division point. How many parallelograms are there in total with these lines as sides? | ( Hint: For each parallelogram whose sides are not parallel to $BC$, extend the two pairs of opposite sides to intersect $BC$ at three or four points. Then use the result from Example 1. The answer is: $3 C_{n+2}^{4}$ ) | 3 C_{n+2}^{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,704 |
19. If $x_{1}, x_{2}$ are the roots of the equation $x^{2}-x \sin \frac{4 \pi}{7}$ $+\cos \frac{4 \pi}{7}=0$, then $\operatorname{arctg} x_{1}$ $+\operatorname{arctg} x_{2}$ is $\qquad$ - | 19. $\frac{3 \pi}{14}$ | \frac{3 \pi}{14} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,705 |
20. Given that the length of the major axis of an ellipse is 4, and the focal distance $\left|F_{1} F_{2}\right|=2$, the sum of the lengths of two perpendicular chords passing through the focus $F_{1}$ is $\frac{48}{7}$. Then the product of the lengths of these two chords is $\qquad$ | 20. $\frac{576}{49}$ | \frac{576}{49} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,706 |
7. Arrange the three numbers $10 a^{2}+81 a+207, a+2$, and $26-2a$ in such a way that their common logarithms form an arithmetic sequence with a common difference of 1. Then, at this point, $a=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the tra... | 7. $\frac{1}{2}$. | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 705,713 |
One, find all real numbers $x$ that satisfy $\left[x^{2}-2 x\right]=[x]^{2}-2[x]$, where [ $x$ ] denotes the greatest integer not exceeding $x$.
| Let $[x]=n, x-[x]=y$. Then $n \in Z$, $y \in[0,1)$. The original equation can be transformed into $\left[y^{2}+2(n-1) y\right]$ $=0$. Therefore,
$$
0 \leqslant y^{2}+2(n-1) y<1 \text {. }
$$
When $n \leqslant 0$, i.e., $n$ is a non-positive integer, it is evident that $y=0$ satisfies (1).
When $n \geqslant 1$, solve t... | x = n \text{ (non-negative integers)}, \left[n, \sqrt{(n-1)^{2}+1}+1\right] (n=1,2,3,\cdots) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,714 |
2.11 people jointly manage an arsenal, and some locks are added to the door, and each person is given several keys. To ensure that any 6 people together can open all the locks, but any 5 people together cannot open all the locks, how many locks should be installed on the door at least? How should the keys be distribute... | （Hint 8
Use Example 3. The answer is: at least $C_{11}^{5}$ attempts). | C_{11}^{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,715 |
Let $S=\{1,2,3,4\}, a_{1}, a_{2}, \cdots$, be any permutation ending with 1, i.e., for any permutation $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$ of the four numbers in $S$ that does not end with 1, $\left(b_{4} \neq 1\right)$, there exist $i_{1}, i_{2}, i_{3}, i_{4}$, such that $1 \leqslant i_{1}<i_{2}<i_{3}<i_{4} \leq... | II. Hint: First consider the minimum value of the number of terms $k$ in any sequence containing a permutation of $S$.
1. For $S=\{1,2,3\}$, it can be proven that a sequence with only 6 terms cannot contain any permutation of $\{1,2,3\}$.
2. For $S=\{1,2,3,4\}$, prove that a sequence with only 11 terms cannot contain a... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,716 |
Given $\triangle A B C, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. In the plane of $\triangle A B C$, draw a line $l$ intersecting two sides of $\triangle A B C$. Cut $\triangle A B C$ along $l$ to form a spatial figure, and connect the smaller triangles formed by $l$ to create a two-si... | (1) Given the volume of the pyramid $=\frac{1}{3} \times$ height $\times$ base area, it is easy to know that to maximize the volume of the pyramid, the two parts of the broken line must be perpendicular.
$$
A H=\frac{2 S \Delta \Delta D B}{D E}=\frac{\sqrt{3} x y}{2 \sqrt{x^{2}+y^{2}-x y}} .
$$
Thus, when $\triangle A... | V = \frac{a^3}{12 \sqrt{3}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,718 |
5. If the roots of the equation $(x-a)(x-b) - c x=0$ are $x_{1}, x_{2}$, then the solutions of the equation $\left(x-x_{1}\right)\left(x-x_{2}\right)+c x=0$ are $\qquad$ | 5. $a, b$ | a, b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,723 |
3. A large cube is formed by $8 \times 8 \times 8$ unit cubes. How many lines can pass through the centers of 8 unit cubes? | (Prompt: Add another layer of unit cubes on the outside, prove that the line is in one-to-one correspondence with the added unit cubes. Answer: $\left.\frac{1}{2}\left(10^{3}-8^{3}\right)\right)$ | \frac{1}{2}\left(10^{3}-8^{3}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,726 |
In the Cartesian coordinate system XOY, we color the points with natural number horizontal coordinates and perfect square vertical coordinates (squares of natural numbers) as red points. Try to determine all the "red points" that the graph of the function $y=(x-36)(x-144)-1991$ passes through. | Let $y=m^{2}, m \in N$, then
$$
(x-90)^{2}-4907=m^{2} \text {. }
$$
Let $x-90=k$, then $k^{2}-m^{2}=4907$. Since $7 \times 701$ are both prime numbers, and $k-m<k+m$, we have
$$
\left\{\begin{array}{l}
k-m=1, \\
k+m=4907
\end{array}\right.
$$
which gives $k=2454, m=2453, x=2544$;
or $\left\{\begin{array}{l}k-m=7, \\ ... | (444,120409) \text{ and } (2544,6017209) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,731 |
Three, the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. On the face diagonal $A_{1} D$, take point $M$, and on $C D_{1}$, take point $N$, such that the line segment $M N$ is parallel to the diagonal plane $A_{1} A C C_{1}$. Calculate the minimum length of the line segment $M N$ that satisfies the abo... | Three, draw perpendiculars from $M$ and $N$ to the base, with the feet of the perpendiculars $M_{1}$ on $A D$ and $N_{1}$ on $C D$. Connect $M_{1} N_{1}$, then $M_{1} N_{1} \| A C$. As shown in the figure, let $M M_{1}=\boldsymbol{x}$, $N N_{1}=1-x$. Draw $M H \perp N N_{1}$ at $H$, then $N H=1-2 x, M H=\sqrt{2} x$. Th... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,732 |
Let $n$ be a non-negative integer; $M=\{0,1,2, \cdots\}$, then the following table defines $f_n$. Here, $a_{1}$ equals the number of times $k$ appears in $a_{0}, a_{1}, \cdots, a_{\mathrm{n}}$. For example: when $n=3$, $f_{3}$ can be defined as in the table on the right.
Determine for which non-negative integers $n, f_... | Because $a_{1} \geqslant 0$, each $a_{1}$ represents the number of times $i$ appears in the corresponding sequence, i.e., $\Omega_{1}$ boxes are filled with $i$, and there are a total of $n+1$ boxes, so
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{n}=n+1, \\
\left.a_{0} \times 1\right)+a_{1} \times 1+a_{2} \times 2+... | 0,1,2,5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,733 |
2. For any non-empty set of numbers $S$, let $\sigma(S)$ and $\pi(S)$ denote the sum and product of all elements in $S$, respectively. Prove:
$$
\begin{array}{l}
\sum_{n} \frac{\sigma(S)}{\pi(S)}=\left(n^{2}+2 n\right)-\left(1+\frac{1}{2}+\frac{1}{3}\right. \\
\left.\quad+\cdots+\frac{1}{n}\right)(n+1),
\end{array}
$$
... | 2. Since $\sigma(S)=\sum_{\mathrm{k} \in \mathrm{S}} k$, we have
$$
\sum_{\mathrm{S}} \frac{\sigma(S)}{\pi(S)}=\sum_{\mathrm{S}} \sum_{\mathrm{k} \in \mathrm{S}} \frac{k}{\pi(S)}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \sum_{\mathrm{S} \ni \mathrm{k}} \frac{k}{\pi(S)},
$$
where “ $\sum_{\mathrm{S} \mathrm{k}}$ ” indicates th... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 705,736 |
Example 1. Taking the last digit of the triangular numbers in sequence, an infinite decimal can be formed:
$$
N=0.1360518 \cdots \text {. }
$$
Discuss whether $N$ is a rational or irrational number. | So-called triangular numbers refer to
the $n$-th triangular number is $1+2+\cdots+n=\frac{n(n+1)}{2}$, denoted as $a_{n}$.
If we continue to concatenate $N$, the pattern becomes clearer:
$$
N=0.13605186556815063100136051 \cdots
$$
It seems that the unit digit of $a_{n+2}$ is the same as the unit digit of $a_{0}$. If... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,737 |
3. For any fixed integer $n \geqslant 1$, prove that the sequence
$$
2,2^{2}, 2^{2^{2}}, \cdots(\bmod n)
$$
is constant after some term. | 3. By induction. When $n=1$, the conclusion to be proved is obviously true. For $n>1$, assume that the conclusion to be proved holds for all natural numbers less than $n$. Let $n=2 \vee q$, where $q$ is an odd number. If $q=1$, then the conclusion to be proved is also obviously true. Without loss of generality, assume ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,738 |
4. Let $a=\frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}}$, where $m, n$ are positive integers. Prove:
$$
a^{m}+a^{n} \geqslant m^{m}+n^{n} .
$$ | 4. Let $N$ be a positive integer. Since
$$
\begin{array}{l}
a^{N}-N^{N}=(a-N)\left(a^{N-1}+a^{N-2} N+\cdots\right. \\
+a N^{N-2}+N^{N-1} \text {, } \\
\end{array}
$$
it follows that $\quad a^{N}-N^{N} \geqslant(a-N) N^{N}$.
Thus, $\quad\left(a^{m}+a^{n}\right)-\left(m^{m}+n^{n}\right)$
$$
\begin{array}{l}
=\left(a^{m}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 705,739 |
5. Let $D$ be a moving point on side $AB$ of the given $\triangle ABC$, and let point $E$ be inside the triangle and the intersection of the external common tangent of the incircles of $\triangle ACD$ and $\triangle BCD$ with $CD$. Prove that the locus of point $E$ is a segment of a circle. | 5. As shown in the figure, let $M, N, R, T, S$, $Q, O$ and $P$ be the points of tangency of the corresponding lines and the circle.
Since $C E=C O-E O=C Q-S E$,
$$
C E=C P-E P=C R-E T \text {, }
$$
thus $2 C E=C Q+C R-S T$.
Also,
$$
\begin{aligned}
C Q+C R & =C A+C B-(Q A+R B) \\
& =C A+C B-(A M+N B),
\end{aligned}
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,740 |
Example 1. Write the numbers $1, 2, 3, \cdots, 1974$ on the blackboard. It is allowed to erase any two numbers and immediately write down their sum or difference. Repeat this procedure until only one number remains on the blackboard. Prove that this number cannot be even. (1974, Soviet Kyiv Mathematical Competition) | Consider the number of odd numbers. After one operation, the number of odd numbers either remains unchanged or decreases by 2 (if the erased numbers are both odd), meaning the number of odd numbers remains odd, unchanged. Thus, after repeating the operation 1973 times, the resulting number is still odd, which means it ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,748 |
Example 2. A square grid paper with 8 rows and 8 columns, arbitrarily color 32 of the cells in black, and the remaining 32 cells in white. Then, perform "operations" on the colored grid paper, where each operation involves simultaneously changing the color of all cells in any row or column. Can you end up with a grid p... | Cannot. Suppose a row (or column) contains $k$ black squares and $8-k$ white squares, where $0 \leqslant k \leqslant 8$. After changing the colors of the squares, we get $8-k$ black squares and $k$ white squares. Therefore, each operation "increases" the number of black squares by $(8-k) - k = 8 - 2k$, which is an even... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,749 |
Example 3. Let the function $f(x)=-x^{2}+(2 m-4) x$ $-(m+1)$ have a maximum value of $g(m)$. Try to find the minimum value of $g(m)$.
保留了原文的换行和格式。 | Given $f(x)=-\left[x-(m-2]^{2}+m^{2}\right.$ $-5 m+3$, we have
$$
\begin{aligned}
g(m) & =m^{2}-5 m+3 \\
& =\left(m-\frac{5}{2}\right)^{2}-\frac{13}{4} \geqslant-\frac{13}{4}
\end{aligned}
$$
Therefore, the minimum value of $g(m)$ is $-\frac{13}{4}$. | -\frac{13}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,750 |
To 4. Let $f(x)=x^{2}-2 a x+a$ have the minimum value $g(a)$ on the interval $[-1$, 1]. Find the maximum value of $g(a)$.
| Solve $f(x)=(x-a)^{2}+a-a^{2}$, and
$-1 \leqslant x \leqslant 1$. Therefore, when $a \geqslant 1$,
$$
g(a)=f(1)=1-a_{3}
$$
When $-11, \\ a-a^{2}, & -1<a<1, \\ 1+3 a, & a \leqslant-1 .\end{array}\right.$
- Since when $a \geq 1$, the maximum value of $g(a)$ is $g(1)=0$
When $-1<a<1$,
$$
g(a)=\frac{1}{4}-\left(\frac{1}{2... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,751 |
List 5. Try to find the maximum value of $f(x)=\min \{4 x+1, x+2, -2 x+4\}(x \in R)$.
| Solve: In the same Cartesian coordinate system, draw the graphs of the functions
$$
\begin{array}{l}
y_{1}=4 x+1, \\
y_{1}=x+2, \\
y_{3}=-2 x+4 \text { . }
\end{array}
$$
Since for each $x$ value, $f(x)$ is the minimum of $y_{1}, y_{2}, y_{3}$, we have
$$
f(x)=\left\{\begin{array}{ll}
4 x+1, & x<\frac{1}{3}, \\
x+2, &... | \frac{8}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,752 |
In quadrilateral $ABCD$ that is circumscribed around $\odot O$, $AC \perp BD$ at point $E$. Points $P, Q, R, S$ are the points of tangency on sides $AB, BC, CD, DA$ respectively. Connect $PR, QS$. Prove: Quadrilateral $ABCD$ is a symmetric figure with one diagonal as the axis of symmetry. (1979, Hunan Mathematics Compe... | $$
\begin{array}{l}
AB=a, BC=b, \\
CD=c, DA=d . \\
\because \odot O \text { is inscribed in }
\end{array}
$$
quadrilateral $ABCD$,
$$
\begin{array}{l}
\therefore a+c \\
=b+d \text {. } \\
\therefore e^{2}+c^{2}=AE^{2}+BE^{2}+CE^{2}+DE^{2} \\
=b^{2}+d^{2} \text {. } \\
\end{array}
$$
Also, $\because AC \perp BD$ at $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,754 |
8. Given $a, b \in R$. Try to find $s=\max \{|a+b|,|a-b|, |1-a|, |1-b|\}$, the minimum value. | If $a b>0$, then
$$
|a-b| \leqslant|a|+|b|=|a+b| \text {. }
$$
Thus, $s=\max \{|a+b|,|1-a|,|1-b|\}$
$$
\begin{array}{l}
>\frac{1}{3}[|a+b|+|1-a|+|1-b|] \\
>\frac{1}{3}|(a+b)+(1-a)+(1-b)| \\
=\frac{2}{3},
\end{array}
$$
Equality is achieved when $a=b=\frac{1}{3}$.
If $a b<0$, then $|a-b|>|a|+|b|$. Thus,
$$
S \geqslant... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,756 |
Example 2. Let $I$ be the incenter of $\triangle ABC$, and $ID \perp BC$ at $D$. If $AB \cdot AC = 2 BD \cdot DC$. Prove: $BA \perp AC$. (1984, Fujian Junior High School Competition) | Prove: As shown in the figure, since $I$ is the incenter of $\triangle ABC$, we have $BD = BE$. Let $BD = BE = x$, $CD = CF = y$.
$AE = AF = z$. Also, let the sides opposite to angles $A, B, C$ in $\triangle ABC$ be $a, b, c$, respectively. Then we have
$$
\left\{\begin{array}{l}
x + y = a, \\
y + z = b, \\
z + z = c
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,757 |
Example 3. The area of $\triangle ABC$ is 3 times the area of its inscribed rectangle $PQRS$, and the values of side $BC$ and altitude $AD$ are rational numbers. Under what conditions is the perimeter of rectangle $PQR$ a rational number? Under what conditions is it an irrational number?
(1985, Joint Mathematics Compet... | As shown in the figure, let $BC = a$, $AD = h$, $PQ = m$, $PS = n$. We have $\frac{h-n}{h} = \frac{m}{a}$, which means
$$
\begin{array}{l}
mh + an = ah. \\
\text{Also, } mn = \frac{1}{3} \cdot \frac{1}{2} ah = \frac{1}{6} ah, \\
\therefore \quad mh \cdot an = \frac{1}{6} a^2 h^2.
\end{array}
$$
From (1) and (2), we kn... | 2(m+n) = a + h \pm \frac{\sqrt{3}}{3}(a - h) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,758 |
Example 4. Let $A M$ be the median of $\triangle A B C$ on $B C$, and draw a line intersecting $A B, A C, A M$ at $P$, $Q$, $N$ respectively. Prove that $\frac{A B}{A P}, \frac{A M}{A N}, \frac{A C}{A Q}$ form an arithmetic sequence. (1978, Liaoning Mathematics Competition) | $$
\begin{array}{l}
\text { Let } \angle B A M=\alpha, \\
\angle M A C=\beta, A B \\
=c, A C=b, \\
A P=p, A Q=q, \\
A M=m, \quad A N
\end{array}
$$
$=n$. From the area relationship of triangles and $M B=M C$, we get
$$
\left\{\begin{array}{r}
p q \sin (\alpha+\beta)-p n \sin \alpha-q n \sin \theta=0, \\
b c \sin (\alph... | \frac{A B}{A P}+\frac{A C}{A Q}=2 \frac{A M}{A N} | Geometry | proof | Yes | Yes | cn_contest | false | 705,759 |
Example 3. Place a match at the origin of a square. Start by placing 1 match at one of the vertices, leaving the other three vertices empty. It is allowed to remove any number of matches from a vertex, and then place matches at all adjacent vertices, with the number of matches being twice the number removed. Is it poss... | Consider the general case of the number of matchsticks at the vertices. Let the number of matchsticks be $a_{1}, a_{2}, a_{3}, a_{4}$, respectively. Then $S_{1}=a_{1}-a_{2}+a_{8}-a_{4}$, after one operation it becomes $S_{2}=S_{1} \pm 3 k$ (where $k$ is a non-negative integer), thus $S_{2}-S_{1}$ is a multiple of 3. Th... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,760 |
Example 5. Given that $\odot O$ passes through the vertices $A, C$ of $\triangle ABC$, and intersects $AB, BC$ at $K, N$ respectively. The circumcircles of $\triangle ABC$ and $\triangle KBN$ intersect at $B, M$. Prove: $\angle OMB = 90^{\circ}$. (26th IMO) | To prove that with $B$ as the origin and $BM$ as the $x$-axis, we establish a coordinate system (as shown in the figure), let the equations of the circumcircle $\odot O$ and the triangles $\triangle BKN$, $\triangle ABC$ be:
$$
\begin{array}{l}
x^{2}+y^{2}+m_{1} x+n_{1} y+l=0, \\
x^{2}+y^{2}+m_{2} x+n_{2} y=0, \\
x^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,761 |
Example 1. Given $\triangle A B C$ is inscribed in the unit circle $\odot O$, and the three altitudes $A D, B E, C F$ intersect at point $H$, and the product of the distances from the orthocenter $H$ to the three sides is $\frac{1}{4}$. Prove: the distances from the orthocenter $H$ to the three vertices of $\triangle A... | Prove in $\triangle ABC$,
$$
BC=2R\sin A \stackrel{B_{m}-1}{=} 2\sin A.
$$
In $\triangle BHC$,
$$
\begin{aligned}
BC & =2R' \sin \angle BHC \\
& =2R' \sin (\pi-A)=2R' \sin A,
\end{aligned}
$$
where $R'$ is the circumradius of $\triangle BHC$.
Combining (1) and (2), we get $R'=1$. Thus, the circumcircle passing throug... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,762 |
Example 4. As shown in the figure, through any point $O$ inside $\triangle ABC$, draw lines parallel to the three sides $BC=a$, $CA=b$, $AB=c$, and let the lengths of the segments intercepted inside $\triangle ABC$ be
$$
\begin{array}{l}
B_{1} C_{1}=a^{\prime}, \\
C_{2} A_{2}=b^{\prime}, \\
A_{8} B_{3}=c^{\prime}.
\end... | To prove this problem, we first prove the preparatory proposition: Given that the three sides of $\triangle ABC$ are $a, b, c$. Also, through any point $M$ on side $BC$, draw parallel lines to the other two sides, with the lengths of these parallel segments being $b'$ and $c'$, then
$$
\frac{b'}{b} + \frac{c'}{c} = 1.
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,765 |
Example 5. Through the altitude of the regular tetrahedron $D-ABC$, any plane is drawn intersecting the three lateral faces of the tetrahedron in three lines. These three lines form angles $\alpha, \beta, \gamma$ with the base $\triangle ABC$ of the tetrahedron. Prove that among $\alpha, \beta, \gamma$, there is at lea... | Proof: Let the height of the regular tetrahedron \(D-ABC\) be \(DH\). Clearly, \(H\) is the centroid of the base \(\triangle ABC\).
Let the intersection points of any plane passing through the height \(DH\) with the sides \(BC\), \(CA\), and \(AB\) of the base \(\triangle ABC\) be \(M\), \(N\), and \(P\) respectively ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,766 |
Example 1. Factorize the polynomial $x^{8}+x^{7}+1$. | Let $x=10$, then
$$
\begin{aligned}
x^{8}+x^{7}+1 & =10^{8}+10^{7}+1 \\
& =110000001 \\
& =3 \times 37 \times 990991 .
\end{aligned}
$$
These three numbers are all factors. Substituting $x=10$ again, 3 must be $x-7$, 37 must be $3x+7$ or $4x-3$. However, because the constant term of the original expression is 1, the c... | \left(x^{2}+x+1\right)\left(x^{6}-x^{4}+x^{3}-x+1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,767 |
Example 2. Factor the polynomial $x^{4}-3 x^{2}+9$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $x=10$, then
$$
\begin{aligned}
x^{4}-3 x^{2}+9 & =10^{4}-3 \times 10^{2}+9 \\
& =9709 \\
& =7 \times 19 \times 73 .
\end{aligned}
$$
These three numbers are all prime. Substituting $x=10$ back, 7 is $x-3$, 19 is $x+9$ or $2x-1$. Dividing the original expression by $x-3$ does not result in an integer, so $x-3$ is ... | x^{4}-3x^{2}+9 = \left(x^{2}+3x+3\right)\left(x^{2}-3x+3\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,768 |
Example 4. Given the following number table, if all numbers in any row or any column $\left(\begin{array}{ccrr}-3 & 4 & -2.7 & 67 \\ 0.2 & -3.14 & 3.8 & 98 \\ -7 & 20 & -101 & -1\end{array}\right)$ are operated on. Can all the numbers in the table be made positive after several operations?
(1987, Anhui Province: Traini... | It cannot be done. Because each operation changes the signs of 4 numbers in the table. Let there be $k$ positive numbers in any row (or any column), then there are $4-k$ negative numbers. Before the operation, the product of the 4 numbers has a sign of $(-1)^{4-k}$, and after one operation, the product of the 4 numbers... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,771 |
1. Let real numbers $A, B, C$ be such that the following inequality (＊) holds for any real numbers $x, y, z$. What conditions should $A$, $B, C$ satisfy? (The conditions should be necessary and sufficient, and are not limited to being expressed by equations or inequalities involving only $A, B, C$)
$$
\begin{array}{l}
... | Let $x=y$ in ( * ), we get
$$
C(z-x)^{2} \geqslant 0 \text {. }
$$
By the arbitrariness of $x, z$, we have $C \geqslant 0$. Similarly, by symmetry, we get
$$
A \geqslant 0, B \geqslant 0, C \geqslant 0 .
$$
Let $x-y=s, y-z=t$, then $x-z=s+t$, ( * ) is equivalent to
$$
\begin{array}{l}
A_{s}(s+t)+B_{t}(-s)+C(-s-t)(-t)... | A^{2}+B^{2}+C^{2} \leqslant 2(A B+B C+C A) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 705,773 |
Define a function $f_{2}, Q \rightarrow C$ that satisfies the following conditions:
(1) For any 1988 rational numbers $x_{1}, x_{2}, \cdots, x_{1988}$, we have
$$
\begin{array}{l}
f\left(x_{1}+x_{2}+\cdots+x_{1988}\right) \\
=f\left(x_{1}\right) \cdot f\left(x_{2}\right) \cdot \cdots \cdot f\left(x_{1988}\right) ;
\end... | Obviously, $f \equiv 0$ is a solution. Suppose $f \neq 0$, then there exists a rational number $x_{0}$, such that $f\left(x_{0}\right) \neq 0$. In (1), let $x_{1}=x_{0}, x_{2}=x_{3}=\cdots=x_{1883}=0$, we get
$$
f\left(x_{0}\right)=f\left(x_{0}\right) \cdot(f(0))^{1987} \text {. }
$$
Since $f\left(x_{0}\right) \neq 0$... | f(x)=\omega \cdot a^{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,774 |
3. In $\triangle A B C$, $\angle C=30^{\circ}, O$ is the circumcenter, $I$ is the incenter, and points $D$ on $A C$ and $E$ on $B C$ are such that $A D=B E=A B$.
Prove: $O I \perp D E$
$$
\text { and } O I=D E \text {. }
$$ | Proof one: As shown in the figure, extend $A I$ to intersect the circumcircle of $\triangle A B C$ at $M$, and connect $B D$, $O M$, $O B$, $B A$.
$\because I$ is the incenter,
$\therefore M$ bisects $\overparen{B M C}$. Thus, $O M \perp B C$, and $\angle M O B = \frac{1}{2} \overparen{B M C} = \angle B A C$.
By the La... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,775 |
4. Let $k$ be a natural number, and let $S_{\mathrm{k}}=\{(a, b) \mid a, b = 1, 2, \cdots, k\}$. For two elements $(a, b), (c, d)$ in $S_{k}$, if
$a - c \equiv 0$ or $\pm 1 \pmod{k}$ and $b - d \equiv 0$ or $\pm 1 \pmod{k}$, then $(a, b)$ and $(c, d)$ are said to be indistinguishable in $S \times 4^{7}$ (for example, (... | Solve: Let's make the problem more straightforward. Use a $k \times k \uparrow$ grid to represent the $k^2$ elements in $S_{\mathrm{k}}$, naturally with the square in the $i$-th row and $j$-th column representing $(i, j)$. Then, two elements $(a, b)$ and $(c, d)$ in $S_{\mathbf{x}}$ are indistinguishable if and only if... | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 705,776 |
5. Let $f(x)=3x+2$. Prove: there exists a positive integer $m$, such that $f^{(100)}(m)$ is divisible by 1988.
$$
(f^{(k)}(x) \text { denotes } f(\underbrace{f(\cdots f(x) \cdots)}_{\mathbf{k} \uparrow}))
$$ | Prove that from $f(x)=3 x+2$, we get
$$
(f(x)+1)=3(x+1).
$$
Thus, for any given $x$, we have
$$
(f(n+1)(x)+1)=3(f(a)(x)+1),
$$
$n=0,1,2, \cdots$, which means $\left\{f^{(n)}(x)+1\right\}$ is a geometric sequence with a common ratio of 3, so
$$
f^{(n)}(x)+1=3^{n} \cdot(x+1), n=0,1,
$$
$3, \cdots$ To prove that there ex... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,777 |
8. There is a damaged calculator that only retains the original data c, 1, and -1. Each operation can only compute $uv + v$ for $u$ and $v$, and display the result. In the first operation, $u$ and $v$ can only take the original data $c, 1$, or -1. For subsequent operations, $u$ and $v$ can only take $c, 1, -1$, or the ... | Proof: First, we prove that if $a$ can be computed, then $a \pm 1$ can also be computed. In fact, let $\boldsymbol{x}=a, \boldsymbol{y}=1$, then we can compute $a+1$. Let $u=a, y=-1$, then we can compute $-a-1$, thus we can compute $-a$. Then, let $u=-a$ and $y=-1$, we can compute $a-1$.
We use induction to prove the ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,780 |
Example 5. Given three numbers 89, 12, 3. Perform the following operations: take any two of these numbers, find their sum and divide by $\sqrt{2}$, and simultaneously find their difference and divide by $\sqrt{2}$. Can we obtain the three numbers 90, 10, 14 after several such operations? Prove your conclusion. (1990, Z... | It cannot be done. If starting from specific numbers, it is difficult to identify the invariants. For this, let $a, b, c$ be any three numbers. After one operation, the three numbers become $\frac{a+b}{\sqrt{2}}, \frac{a-b}{\sqrt{2}}, c$. Observing, we find $\left(\frac{a+b}{\sqrt{2}}\right)^{2} +\left(\frac{a+b}{\sqrt... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 705,781 |
Example 6. The circular dial of a clock (with the numbers $1,2,3, \cdots, 12$ written on the circumference) is fixed on a blackboard at its center axis. The dial can rotate by any integer multiple of $30^{\circ}$. Initially, the number at each position on the blackboard corresponding to the clock numbers is zero. Then ... | It is not possible. After each rotation of the clock face, the sum of all numbers added to the blackboard is $1+2+3+\cdots+12=78$, meaning that after one rotation, the sum of the numbers on the blackboard has a remainder of zero when divided by 78. After several rotations, the sum of the numbers on the blackboard still... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,782 |
Example 7. There are three piles of stones with the numbers being 19, 8, and 9. Now, the following operation is performed: each time, 1 stone is taken from any two of the three piles, and then these 2 stones are added to the other pile. Can the following be achieved after several such operations?
(1) The number of ston... | (1) It can be achieved. Since the original minimum pile has 8 stones, to make it 2, it requires at least six operations. In fact, it can succeed in six operations:
\[
\begin{aligned}
& (19,8,9) \rightarrow(21,7,8) \\
\rightarrow & (23,6,7) \rightarrow(22,5,9) \rightarrow(24,4,8) \\
\rightarrow & (23,3,10) \rightarrow(2... | (1) 6 \text{ operations}, (2) \text{cannot be achieved} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,783 |
Example 1. Given that $a, b$ are positive integers, and $a>b$,
$$
\begin{array}{l}
\sin \theta=\frac{2 a b}{a^{2}+b^{2}} \text { (where } 0<\theta<\frac{\pi}{2}\text{)}, A_{\mathrm{a}}=\left(a^{2} \\
\left.+b^{2}\right)^{n} \sin \theta \text {. } \\
\end{array}
$$
Prove: For all natural numbers $n, \frac{A_{n}}{2 a b}... | $\because$ Proof $\because \sin \theta=\frac{2 a b}{a^{2}+b^{2}}, a>b>0$, $02) \text {, }
$$
Therefore, substituting $\sin n \theta=\frac{A_{n}}{\left(a^{2}+b^{2}\right)^{n}}$, and rearranging, we get
$$
A_{n}=2\left(a^{2}-b^{2}\right) A_{n-1}-\left(a^{2}+b^{2}\right)^{2} A_{n-2}
$$
$(n>2)$.
Given $a, b \in N$, we us... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 705,784 |
Example 1. Find the minimum value of the function $f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \quad(x \in R)$. | $$
\begin{array}{l}
\text { Sol } \because\left(x^{2}+1\right)-\cos x \\
=x^{2}+(1-\cos x) \geqslant 0, \\
\left(x^{2}+1\right)-2 x=(x-1)^{2} \geqslant 0, \\
\therefore \quad x^{2}+1 \geqslant \cos x, x^{2}+1 \geqslant 2 x \\
\text { Therefore, } f(x)=\max \left\{x^{2}+1, \cos x, 2 x\right\} \\
=x^{2}+1 \geqslant 1,
\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,785 |
Example 1. In $\triangle A B C$, the angle bisector of $\angle C$ intersects side $A B$ and the circumcircle of the triangle at points $D$ and $K$ respectively. Find:
$$
\begin{aligned}
\quad \text { (1) } & \frac{1}{I D}-\frac{1}{I K} \\
= & \frac{1}{C I}
\end{aligned}
$$
(2) $\frac{C I}{I D}-\frac{I D}{D K}=1$.
Wh... | $$
\begin{array}{l}
\text { Analysis (1) } \frac{1}{I D}-\frac{1}{I K}=\frac{1}{C I} \\
\Leftrightarrow \frac{D K}{I D \cdot I K}=\frac{1}{C I} \Leftrightarrow \Rightarrow \frac{C I}{I D}=\frac{I K}{D_{K}} \cdot I_{D K} . \\
\end{array}
$$
$$
I K=K B=A K,
$$
Therefore, $\frac{C I}{I D}=\frac{B C}{B D}, \frac{I K}{D K}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 705,787 |
For example. Let $m$ be an integer, for all $4<m<40$, the equation $x^{2}-2(2 m-3) x+4 m^{2}-14 m+8=0$ has two integer roots. Find the value of $m$ and the roots of the equation. (1984, Wuhan Preparatory Autumn Competition) | Solving, we get $\Delta=4(2 m+1)$. Given that $\Lambda=4(2 m +1)$ is a perfect square and $4<m<40$, it can be deduced that $m=12$ or 24, and the corresponding roots are $26,16$ and $52, 38$ (all integer roots). Therefore, the required values of $m$ are 12,24.
Note: If the discriminant is not a perfect square and its st... | m=12,24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,789 |
1. The number of real solutions to the equation $|x-2|+| x-3| = 1$ is ( ).
(A) 2.
(B) 3.
(C) 4.
(D) Infinitely many. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,791 |
2. The sides of a triangle are all integers, and the perimeter is 8. Then the area of this triangle is ( ).
(A) $2 \sqrt{2}$.
(B) $2 \sqrt{5}$.
(C) $2 \sqrt{10}$
(D) cannot be determined. | A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 705,792 |
3. A person is sailing a sailboat downstream along a river, and encounters a motorboat coming upstream. The sailboat asks the motorboat driver, "Is there a ship coming from behind you?" The motorboat driver replies, "I passed a ship half an hour ago." The sailboat continues to sail for another half hour and meets a shi... | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 705,793 |
6. Trapezoid $A B C D, A B \| C D, \angle A+\angle B$ $=90^{\circ}, A B=a, C D=b, E, F$ are the midpoints of $A B$, $C D$ respectively. Then $E F$ equals ( ).
(A) $\frac{1}{2}(a+b)$.
(B) $\frac{1}{2}(a-b)$.
(C) $\sqrt{a^{2}+b^{2}}$.
(D) $\sqrt{a^{2}-b^{2}}$. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 705,796 |
1. The average score of the participants in the Zu Chongzhi Cup Mathematics Invitational at a certain school is 75. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore, the av... | 1. 84 | 84 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,798 |
Example 4. Find all integer values of $a$ such that the equation $(a+1) x^{2}-\left(a^{2}+1\right) x$ $+2 a^{3}-6=0$ has integer roots. (1990,
untranslated text retained as requested. | When $a=-1$, the original equation has an integer root $x=-4$. If $a \neq-1$, then by Vieta's formulas,
$$
x_{1}+x_{2}=a-1+\frac{2}{a+1} \text {. }
$$
Thus, $a+1= \pm 1, \pm 2$. That is, $a=0,-2,1,-3$.
$\because$ When $a=0$, the solutions of the original equation are $x_{1}=-2, x_{2}=3$; when $a=-2$, the original equat... | -1,0,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,800 |
4. $A, B$ are two fixed points on a plane, find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles triangle. There are $\qquad$ such points $C$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The blank space represented by $\qquad$ in the original text is kept as is in the translation. | 4.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 6 | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 705,802 |
5. As shown in the figure, points $D, E$ are on $BC$, $AC$ of the equilateral $\triangle ABC$, and $CD = AE, AD$, $BE$ intersect at point $P$, $BQ \perp AD$. Given $PE = 1, PQ = 3$. Then $AD=$ | 5.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,803 |
7. Definition: If $n$ different positive integers, for any two numbers among them, the product of these two numbers can be divided by the sum of these two numbers, then, this set of numbers is called an $n$-number Zu Chongzhi array.
For example: $60, 120, 180$ these three numbers form a three-number Zu Chongzhi array ... | $7.420,840,1260,1680$ | 420,840,1260,1680 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 705,805 |
Three, (1) Prove: $\sqrt{a^{2}+\frac{1}{b^{2}}+\frac{a^{2}}{(a b+1)^{2}}}$ $=\left|a+\frac{1}{b}-\frac{a}{a b+1}\right|$
(2) Using or not using (1), calculate
$$
\sqrt{1+1990^{2}+\frac{1990^{2}}{1991^{2}}}-\frac{1}{1991}
$$ | Three, Hint: Square both sides of the equation, noting that
$$
\frac{a}{b}-\frac{a^{2}}{a b+1}-\frac{a}{b(a b+1)}=0,
$$
and then take the square root of both sides.
Note: In fact, the following equation also holds:
$$
\sqrt{a^{2}+b^{2}+\frac{a^{2} b^{2}}{(a+b)^{2}}}=\left|a+b-\frac{a b}{a+b}\right|
$$
Using equation ... | 1990 | Algebra | proof | Yes | Yes | cn_contest | false | 705,806 |
Four, As shown in the figure, points $M, N$ are on the sides $B C$, $C D$ of square $A B C D$, respectively. It is known that the perimeter of $\triangle M C N$ is equal to half the perimeter of square $A B C D$. Find $\angle M A N$ | Four, Solution: Rotate $\triangle A B M$ counterclockwise by $90^{\circ}$ around point $A$, so that $A B$ falls on $A D$, and point $M$ moves to point $M^{\prime}$.
Since $\angle A D M^{\prime}=\angle A B M=90^{\circ}$,
thus $\angle N D M^{\prime}=\angle N D A+\angle A D M^{\prime}$ $=180^{\circ}$,
which means $N, D, ... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 705,807 |
Six, Table A is an English letter electronic display panel.
Each operation can change 4 letters in a row at the same time, or change 4 letters in a column at the same time. The rule of change is, according to the order of the English alphabet, each English letter becomes the next letter (i.e., $A$ becomes $B$, $B$ be... | Six, Solution: It is not possible.
For convenience, replace the English letters in the table with their order numbers in the alphabet (i.e., $A$ is $1$, $B$ is $2$, ..., $Z$ is 26). Thus, Table A and Table B can be considered as two $4 \times 4$ number tables. Each operation is equivalent to adding 1 to the remainder w... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 705,809 |
1. The number of solutions to the equation $|2 x-1|-1|=2$ is ( ).
(A.) 1, ( B) 2.
( C ) 3 .
(D) 4 . | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,810 |
Example 5. Determine the value of the natural number $n$, such that the two roots of the quadratic equation $2 x^{2}-8 n x+10 x-n^{2}+35 n-76$ $=0$ are prime numbers, and find these two roots. (1990, Shanxi Province Junior High School Mathematics Competition) | Solve: From the given equation, we get $x_{1}+x_{2}=4 n-5$. Since $x_{1}$ and $x_{2}$ are both prime numbers, and $4 n-5$ is an odd number, it follows that one of $x_{1}$, $x_{2}$ must be even, and the only even prime number is 2. Without loss of generality, let $x_{1}=2$, substituting into the original equation yields... | n=3, x_1=2, x_2=5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 705,811 |
3. The graph of the equation $|x y|+|x-y+1|=0$ is ( ).
(A) Three lines: $x=0, y=0$, $x-y+1=0$.
(B) Two lines: $x=0, x-y+1=0$.
(C) One point and one line: $(0,0)$, $x-y+1=0$.
(D) Two points: $(0,1),(-1,0)$. | D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,813 |
4. Given $a+b+c=10, a^{2}+b^{2}+c^{2}=38$, $a^{a}+b^{3}+c^{8}=160$. Then, the value of $a b c$ is ( ).
(A) 24.
(B) 30.
(C) 36.
(D) 42. | B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 705,814 |
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