problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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Example 6 There are two rivers, A and B, each with a flow rate of $300 \mathrm{~m}^{3} / \mathrm{s}$, which converge at a certain point and continuously mix. Their sediment contents are $2 \mathrm{~kg} / \mathrm{m}^{3}$ and $0.2 \mathrm{~kg} / \mathrm{m}^{3}$, respectively. Assume that from the convergence point, there... | Solution: Let the sand content of two water flows be $a \, \text{kg} / \text{m}^{3}$ and $b \, \text{kg} / \text{m}^{3}$, and the water volumes flowing through in a unit time be $p \, \text{m}^{3}$ and $q \, \text{m}^{3}$, respectively. Then the sand content after mixing is
$$
\frac{a p + b q}{p + q} \, \text{kg} / \te... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{arr... | Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.
When $1$ 0, if we reduce the number of balls in each drawer by 1, we ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-$ $a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics Competition) | Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.
Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$
Its ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $a_{n}$ be the coefficient of the $x$ term in the expansion of $(3-\sqrt{x})^{n}$ $(n=2,3,4, \cdots)$. Then
$$
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right)=
$$
$\qquad$ | 8.18
By the binomial theorem, $a_{n}=\mathrm{C}_{n}^{2} \cdot 3^{n-2}$. Therefore,
$$
\begin{array}{l}
\frac{3^{n}}{a_{n}}=\frac{3^{2} \times 2}{n(n-1)}=18\left(\frac{1}{n-1}-\frac{1}{n}\right) . \\
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right) \\
=\lim _{n... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. If:
(1) $a, b, c, d$ all belong to $\{1,2,3,4\}$;
(2) $a \neq b, b \neq c, c \neq d, d \neq a$;
(3) $a$ is the smallest value among $a, b, c, d$.
Then, the number of different four-digit numbers $\overline{a b c d}$ that can be formed is
$\qquad$ | 12.28.
When $\overline{a b c d}$ has exactly 2 different digits, it can form $C_{4}^{2}=6$ different numbers.
When $\overline{a b c d}$ has exactly 3 different digits, it can form $\mathrm{C}_{3}^{\mathrm{l}} \mathrm{C}_{2}^{\mathrm{d}} \mathrm{C}_{2}^{1}+$ $\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1}=12+4=16$ different numb... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given positive integers $k, m, n$, satisfying $1 \leqslant k \leqslant m \leqslant n$. Try to find
$$
\sum_{i=0}^{n}(-1)^{i} \frac{1}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}
$$
and write down the derivation process.
(Xu Yichao, provided) | II. The answer to this question is 0. Below, we will prove this combinatorial identity by constructing polynomials and using interpolation and difference methods.
Method 1: Construct the polynomial
$$
\begin{aligned}
f(x) & =\sum_{i=0}^{n} a_{i} x(x+1) \cdots(x+i-1)(x+i+1) \\
& \cdots(x+n)-(x-m-1) \cdots(x-m-n) .
\end{... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
'. Six, let $n$ be a positive integer, and let the set
$M=\{(x, y) \mid x, y$ are integers, $1 \leqslant x, y \leqslant n\}$.
Define a function $f$ on $M$ with the following properties:
(a) $f(x, y)$ takes values in the set of non-negative integers;
(b) When $1 \leqslant x \leqslant n$, we have $\sum_{y=1}^{n} f(x, y)... | Six, we first prove the following general lemma.
Lemma In each cell of an $m$-row $n$-column grid, fill in a non-negative integer, the number filled in the cell at the $i$-th row and $j$-th column is denoted by $a_{ij}$, $r_i (1 \leqslant i \leqslant m)$ and $s_j (1 \leqslant j \leqslant n)$ are non-negative integers, ... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. $A B$ is a chord of $\odot O$, $P$ is a point outside $\odot O$, $P B$ is tangent to $\odot O$ at $B$, $P A$ intersects $\odot O$ at $C$, and $A C = B C$, $P D \perp A B$ at $D$, $E$ is the midpoint of $A B$, $D E = 1000$. Then $P B=$ | 2.2000 .
As shown in Figure 5, take the midpoint $F$ of $PA$, and connect $DF$, $EF$. Then $EF \parallel PB$. Also, $DF = \frac{1}{2} PA$, so $DF = AF$. Since $AC = BC$, $\triangle AFD$ is an isosceles triangle. Therefore, $\angle 2 = \angle 4 = \angle 1$. Since $EF \parallel PB$ and $DF \parallel BC$, $\angle 3 = \an... | 2000 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition) | Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a<b<c<d$. If variables $x, y, z, t$ are a permutation of $a, b, c, d$, then the expression
$$
\begin{array}{l}
n(x, y, z, t)=(x-y)^{2}+(y-z)^{2} \\
\quad+(z-t)^{2}+(t-x)^{2}
\end{array}
$$
can take different values. | $=、 1.3$.
If we add two more terms $(x-z)^{2}$ and $(y-t)^{2}$ to $n(x, y, z, t)$, then $n(x, y, z, t)+(x-z)^{2}+(y-t)^{2}$ becomes a fully symmetric expression in terms of $x, y, z, t$. Therefore, the different values of $n(x, y, z, t)$ depend only on the different values of $(x-z)^{2}+(y-t)^{2}=\left(x^{2}+y^{2}+z^{2... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the sides of $\triangle ABC$, and $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=999$, then $m$ $=$ $\qquad$ | 2.1999 .
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
\quad=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\cos A \cdot \sin B+\sin A \cdot \cos B)} \\
\quad=\cos C \cdot \frac{\sin A \cdot \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c... | 1999 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.
---
Please note that the translation pres... | Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 50 points) On a straight ruler of length $36 \mathrm{~cm}$, mark $n$ graduations so that the ruler can measure any integer $\mathrm{cm}$ length in the range $[1,36]$ in one go. Find the minimum value of $n$.
| Three, if the ruler is marked with 7 (or fewer than 7) graduations, we can prove that it is impossible to measure any integer length in the range [1, 6] cm in a single measurement.
In fact, 7 graduations, including the two end lines of the ruler, total 9 lines, which have 36 different combinations. Therefore, 7 gradua... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League) | Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.
Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Select $k$ edges and face diagonals from a cube such that any two line segments are skew lines. What is the maximum value of $k$? | (提示:考察线段 $A C 、 B C_{1} 、 D_{1} B_{1} 、 A_{1} D$, 它们所在的直线两两都是异面直线. 若存在 5 条或 5 条以上满足条件的线段, 则它们的端点相异, 且不少于 10 个, 这与正方体只有 8 个端点矛盾, 故 $k$ 的最大值是 4.)
(Translation: Consider the line segments $A C, B C_{1}, D_{1} B_{1}, A_{1} D$, the lines on which they lie are pairwise skew lines. If there are 5 or more line segments that s... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.5. Let $M$ be a finite set of numbers. It is known that from any 3 elements of it, two numbers can be found whose sum belongs to $M$. How many elements can $M$ have at most? | 10.5.7.
An example of a set of numbers consisting of 7 elements is: $\{-3,-2, -1,0,1,2,3\}$.
We will prove that for $m \geqslant 8$, any set of numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ does not have the required property. Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>\cdots>a_{m}$ and $a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. A biologist
observed a chameleon
for a while. During the rest periods, the chameleon can immediately catch a fly. Time:
(1) How many flies did the chameleon catch before its first 9-minute rest?
(2) How many minutes later did the chameleon catch the 98th fly?
(3) After 1999 minutes, how many flies did the chameleo... | Solution: Let the time the chameleon rests before catching the $m$-th fly be $r(m)$. Then $r(1)=1, r(2m)=r(m), r(2m+1)=r(m)+1$. This indicates that $r(m)$ is equal to the number of 1s in the binary representation of the number $m$.
Let $i(m)$ be the moment the chameleon catches the $m$-th fly, and $f(n)$ be the total ... | 462 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1.100 people share 1000 RMB, and the money of any 10 people does not exceed 190 RMB. Then, the most one person can have is ( ) RMB.
(A) 109
(B) 109
(C) 118
(D) 119 | -1 . (B).
Let the person with the most money have $x$ yuan, and the rest are divided into 11 groups, each with 9 people, and this person together with each group of 9 people does not exceed 190 yuan. Therefore,
$$
190 \times 11 \geqslant 1000-x+11 x .
$$
This gives $x \leqslant 109$.
So, one person can have at most 10... | 109 | Inequalities | MCQ | Yes | Yes | cn_contest | false |
5. Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$, and the quadratic equation with $\alpha^{2}, \beta^{2}$ as roots is still $x^{2}-p x+q=0$. Then the number of pairs $(p, q)$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 0 | 5. (B).
From the problem,
$\left\{\begin{array}{l}\alpha+\beta=p, \\ \alpha \beta=q\end{array}\right.$ and $\left\{\begin{array}{l}\alpha^{2}+\beta^{2}=p, \\ \alpha^{2} \beta^{2}=q .\end{array}\right.$
Thus, $q^{2}=q$. Therefore, $q=0$ or $q=1$.
And we have $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$, whi... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Real numbers $a, b, c$ are all non-zero, and $a+b+c=$
0. Then
$$
=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)
$$ | \begin{aligned} \text { II.1. } & -3 . \\ \text { Original expression }= & a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \\ & +c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ = & \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c)-3=-3 .\end{aligned} | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2,
Square $A B C D$ has a side length of $1, E$ is a point on the extension of $C B$, connect $E D$ intersecting $A B$ at $P$, and $P E$ $=\sqrt{3}$. Then the value of $B E-P B$ is $\qquad$ | 4.1.
Let $B E=x, P B=y$, then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=(\sqrt{3})^{2}, \\
\frac{y}{1}=\frac{x}{x+1} .
\end{array}\right.
$$
From (2), we have $x-y=x y$.
From (1) and (3), we have
$$
\begin{array}{l}
(x-y)^{2}+2(x-y)-3=0, \\
(x-y+3)(x-y-1)=0 .
\end{array}
$$
Clearly, $x>y, x-y+3>0$.
Therefore, $x-y-1=0,... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a_{n}=\log _{n}(n+1)$, let $\sum_{n=2}^{1023} \frac{1}{\log _{100} a_{n}}$ $=\frac{q}{p}$, where $p, q$ are integers, and $(p, q)=1$. Then $p+q=(\quad)$.
(A) 3
(B) 1023
(C) 2000
(D) 2001 | -1 (A).
$$
\begin{array}{l}
\sum_{n=2}^{1023} \frac{1}{\log _{a_{n}} 100}=\sum_{n=2}^{1023} \log _{100} a_{n} \\
=\log _{100}\left(a_{2} a_{3} \cdots a_{1023}\right) . \\
\text{Since } a_{n}=\log _{n}(n+1)=\frac{\lg (n+1)}{\lg n}, \\
\text{therefore } a_{2} a_{3} \cdots a_{1023}=\frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 1 Given that the polynomial $3 x^{3}+a x^{2}+b x+1$ can be divided by $x^{2}+1$, and the quotient is $3 x+1$. Then, the value of $(-a)^{i}$ is $\qquad$
（Dinghe Hesheng Junior High School Math Competition） | Solution: According to the polynomial identity, we have
$$
3 x^{3}+a x^{2}+b x+1=\left(x^{2}+1\right)(3 x+1) \text {. }
$$
Taking $x=1$ gives $a+b+4=8$.
Taking $x=-1$ gives $a-b-2=-4$.
Solving these, we get $a=1, b=3$.
$$
\therefore(-a)^{b}=(-1)^{3}=-1 \text {. }
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Height 97 Find the last three digits of $2^{2^{2} 1100}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $1000=2^{3} \times 5^{3}$ divided by the remainder.
Obviously, $x==1\left(\bmod ^{1} 2^{1}\right)$.
Next, consider the remainder of $x$ divided by $5^{3}$.
By Euler's theorem, $2^{2 / 5^{3}} \equiv \mathrm{I}\left(\bmod 5^{3}\right)$.
That is, $2^{+5^{2}}=1\left(\operatorname{mxl} 55^{3}\right)$.
Thus, $16^{35}=1\left... | 136 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Third question: There are $n$ people, and it is known that any two of them make at most one phone call. Any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
The translation maintains the original text's format and line... | Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:
If... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$. Then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$
(1990, National High School Mathematics Competition) | Solution: $\because a, b>0$,
$$
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1 \text {. }
$$
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}}$ $\geqslant 1$.
When $a=b=1$, the above expression $=1$, hence the minimum value is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 When $s$ and $t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is $\qquad$
(1989, National High School Mathematics Competition) | Solution: As shown in Figure 1, the distance squared between any point on the line
$$
\left\{\begin{array}{l}
x=s+5, \\
y=s
\end{array}\right.
$$
and any point on the elliptical arc
$$
\left\{\begin{array}{l}
x=3|\cos t|, \\
y=2|\sin t|
\end{array}\right.
$$
is what we are looking for.
Indeed, the shortest distance ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$, such that for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
（Supplied by Zhang Zhusheng） | To divide $X$ into the following 5 subsets for examination:
$$
\begin{array}{l}
2001=1024+977 \geqslant x \geqslant 1024-977=47, \\
46=32+14 \geqslant x \geqslant 32-14=18, \\
17=16+1 \geqslant x \geqslant 16-1=15, \\
14=8+6 \geqslant x \geqslant 8-6=2, \\
x=1 .
\end{array}
$$
To construct an example that does not sat... | 999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let $a$, $b$, $c$, $a+b-c$, $a+c-b$, $b+c-a$, and $a+b+c$ be 7 distinct prime numbers, and suppose that the sum of two of $a$, $b$, and $c$ is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$.
(Liang Darong, problem contributor) | $$
\begin{array}{l}
\text { Let's assume } a<b<c<d \text {, and } a, b, c, d \text { are all prime numbers. } \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, and $b+c$ is 800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \times 47$ and 798 are not prime numbers, but 797 is a prime num... | 1594 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five, divide a circle with a circumference of 24 into 24 equal segments, and select 8 points from the 24 points; such that the arc length between any two points is not equal to 3 and 8. How many different ways are there to select such 8-point groups? Explain your reasoning.
（Supplied by Li Chengzhang） | Solution 1: Number the 24 points as $1,2, \cdots, 24$, and arrange them in a $3 \times 8$ number table according to their "bad relationship":
$1,4,7,10,13,16,19,22$,
$9,12,15,18,21,24,3, \quad 6$,
$17,20,23, \quad 2, \quad 5, \quad 8, \quad 11, \quad 14$.
It is easy to see that the arc length between two points represe... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x_{1}, x_{2}, \cdots, x_{7}$ all be integers, and
$$
\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6} \\
+49 x_{7}=1, \\
4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6} \\
+64 x_{7}=12, \\
9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6} \\
+81 x_{7}=123 .
\end{array}
$$
Find $1... | Solution: Since the coefficients of the same letters in the four equations are the squares of four consecutive natural numbers, i.e., $n^{2}, (n+1)^{2}, (n+2)^{2}, (n+3)^{2}$, to find the value of (4), it is necessary to express (4) using (1), (2), and (3), i.e., to express $(n+3)^{2}$ using $n^{2}, (n+1)^{2}, (n+2)^{2... | 334 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (Problem 16) The permutation $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ of $1,2,3,4,5$ has the following property: for $1 \leqslant i \leqslant 4$, $a_{1}$, $a_{2} \cdots, a_{i}$ does not form a permutation of $1,2, \cdots, i$. Find the number of such permutations. | Obviously, $a_{1} \neq 1$.
When $a_{1}=5$, all $4!$ permutations meet the requirements.
When $a_{1}=4$, the $3!$ permutations where $a_{5}=5$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!)$.
When $a_{1}=3$, permutations in the form of $3 \times \times \times 5$ and $... | 71 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (This question is worth 16 points) How many ordered pairs of positive integers $(x, y)$ have the following properties:
$y<x \leqslant 100$, and $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are both integers? | Three, let $\frac{x+1}{y+1}=m(m \in \mathbf{N}, m>1)$, then $x=m y+(m-1)$.
Since $y \mid x$, it follows that $y \mid(m-1)$.
Let $m-1=k y(k \in \mathbf{N})$, then $m=k y+1$,
Substituting into (1), we get
$$
\begin{array}{l}
x=(k y+1) y+k y=k y(y+1)+y \leqslant 100, \\
k \leqslant \frac{100-y}{y(y+1)} .
\end{array}
$$
T... | 85 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The ground floor of a hotel has 5 fewer rooms than the second floor. A tour group has 48 people. If all are arranged to stay on the ground floor, each room can accommodate 4 people, but there are not enough rooms; if each room accommodates 5 people, some rooms are not fully occupied. If all are arranged to stay on t... | 5. (B).
Let the number of guest rooms on the ground floor be $x$, then the number of rooms on the second floor is $x+5$. According to the problem, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{48}{5}<x<\frac{48}{4}, \\
\frac{48}{4}<x+5<\frac{48}{3} .
\end{array}\right.
$$
which simplifies to $\l... | 10 | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, in the acute triangle $\triangle ABC$, the three altitudes $AD$, $BE$, and $CF$ intersect at $H$. Connect $DE$, $EF$, and $FD$. Then the number of triangles in the figure is ( ).
(A) 63
(B) 47
(C) 45
(D) 40 | 6. (B).
The number of triangles with $A$ as a vertex is 13 (excluding $\triangle A B C$); similarly, the number of triangles with $B, C$ as vertices is also 13 each. In total, there are 39.
In $\triangle D E F$, the number of triangles with $D$ as a vertex is 8 (excluding $\triangle D E F$); similarly, the number of ... | 47 | Geometry | MCQ | Yes | Yes | cn_contest | false |
6. In space, there are 4 non-coplanar fixed points. The number of parallelepipeds that can be formed with these 4 points as vertices is ( ).
(A) 20
(B) 32
(C) 25
(D) 29 | 6. (D).
Each parallelepiped is uniquely determined by the designated 1 vertex and 3 median planes (each of these median planes is equidistant from all vertices of the parallelepiped). For any given 4 non-coplanar points, there exist 7 planes equidistant to these 4 points. Choosing any 3 out of these 7, there are $\math... | 29 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Three, (50 points) A conference was attended by $12 k$ people $(k \in$ $\mathbf{N}$ ), where each person has greeted exactly $3 k+6$ other people. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference? | Three, connecting the points of people and their acquaintances, each point forms $\mathrm{C}_{3 k+6}^{2}$ angles, each simple connection corresponds to a vertex, so the graph has a total of $12 k \cdot C_{3 k+6}^{2}$ angles. On the other hand, suppose for any two people, there are $n$ people who know both of them, thus... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Cut a wire of length $143 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, with each segment no less than $1 \mathrm{~cm}$. If no three segments can form a triangle, the maximum value of $n$ is | 2.1.10.
Each segment should be as small as possible, and the sum of any two segments
should not exceed the third segment. A $143 \mathrm{~cm}$ wire should be divided into 10
such segments:
$1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}$,
$13 \mathrm{~cm}, 21 \mathrm{~cm}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$, $b$, $c$, $d$ are the thousands, hundreds, tens, and units digits of a four-digit number, respectively, and the digits in lower positions are not less than those in higher positions. When $|a-b|+|b-c|+|c-d|+|d-a|$ takes the maximum value, the maximum value of this four-digit number is $\qquad$ . | $$
\begin{array}{l}
\text { 3. } 1999 . \\
\because a \leqslant b \leqslant c \leqslant d, \\
\therefore a-b \leqslant 0, b-c \leqslant 0, c-d \leqslant 0, d-a \geqslant 0 . \\
\therefore|a-b|+|b-c|+|c-d|+|d-a| \\
=b-a+c-b+d-c+d-a \\
=2(d-a) .
\end{array}
$$
When $d=9, a=1$,
$$
|a-b|+|b-c|+|c-d|+|d-a|
$$
has a maximu... | 1999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A monkey is climbing an 8-rung ladder, each time it can climb one rung or jump two rungs, and at most jump three rungs. From the ground to the top rung, there are $\qquad$ different ways to climb and jump. | 4.81.
Starting with simple cases:
(1) If there is 1 step, then there is only one way to climb. That is, $a_{1}=1$.
(2) If there are 2 steps, then there are 2 ways to climb:
(1) Climb one step at a time;
(2) Leap two steps at once, i.e., $a_{2}=2$.
(3) If there are 3 steps, then there are 4 ways to climb:
(1) Climb one... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, and $x_{n+1}=$ $\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $x_{1999}-x_{601}=$ $\qquad$ . | 3.0 .
From $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$ we get $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$.
Let $x_{n}=\tan \alpha_{n}$, then
$$
x_{n+1}=\tan \alpha_{n+1}=\tan \left(\alpha_{n}+\frac{\pi}{6}\right) \text {. }
$$
Therefore, $x_{n+6}=x_{n}$,
which means the sequence $\left... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | 4.7.
As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$
which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$
... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed. | 6.5.
First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.
Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.
Since 4 people sit at 3 tables, after th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Arrange all powers of 3 and the sums of distinct powers of 3 in an increasing sequence:
$$
1,3,4,9,10,12,13, \cdots \text {. }
$$
Find the 100th term of this sequence. | Analysis: Write the known sequence in the form of the sum of powers of 3:
$$
\begin{array}{l}
a_{1}=3^{0}, a_{2}=3^{1}, a_{3}=3^{1}+3^{0}, \\
a_{4}=3^{2}, a_{5}=3^{2}+3^{0}, a_{6}=3^{2}+3^{1}, \\
a_{7}=3^{2}+3^{1}+3^{0}, \cdots
\end{array}
$$
It is easy to find that the terms correspond exactly to the binary represent... | 981 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Find the smallest positive integer $n$, such that every $n$-element subset of $S=\{1,2, \cdots, 150\}$ contains 4 pairwise coprime numbers (it is known that $S$ contains a total of 35 prime numbers). | Solution: Consider the number of multiples of 2, 3, or 5 in $S$. We have
$$
\begin{array}{l}
{\left[\frac{150}{2}\right]+\left[\frac{150}{3}\right]+\left[\frac{150}{5}\right]-\left[\frac{150}{2 \times 3}\right]-\left[\frac{150}{2 \times 5}\right]} \\
-\left[\frac{150}{3 \times 5}\right] +\left[\frac{150}{2 \times 3 \ti... | 111 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5. Does there exist a prime number that remains prime when 16 and 20 are added to it? If so, can the number of such primes be determined? | Solution: Testing with prime numbers $2, 3, 5, 7, 11, 13 \cdots$, we find that 3 meets the requirement. Below, we use the elimination method to prove that apart from 3, no other prime number satisfies the requirement.
Divide the natural numbers into three categories: $3n, 3n+1, 3n+2$ $(n \in \mathbb{N})$,
$\because 2n... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 What is the largest even number that cannot be written as the sum of two odd composite numbers?
The largest even number that cannot be written as the sum of two odd composite numbers is 38.
Note: The original text did not provide the answer, this is based on known mathematical results. However, if you need... | Solution: First, prove that even numbers $n$ greater than 38 can all be written as the sum of two odd composite numbers.
Obviously, if the unit digit of $n$ is 0, then $n=5k+15$; if the unit digit of $n$ is 2, then $n=5k+27$; if the unit digit of $n$ is 4, then $n=5k+9$; if the unit digit of $n$ is 6, then $n=5k+21$; ... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $a_{1}=1, a_{2}=\frac{5}{2}, a_{n+1}=$ $\frac{a_{n}+b_{n}}{2}, b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Prove:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .
$$ | Proof: From the characteristic of the simultaneous recurrence relations, we know that $\left\{a_{n}\right\}$ is an arithmetic mean sequence, and $\left\{b_{n}\right\}$ is a harmonic mean sequence. By considering the inequality of means, we can insert the geometric mean between $\left\{a_{n}\right\}$ and $\left\{b_{n}\r... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
3. Given that $x$ and $y$ are positive integers, and $xy + x + y$ $=23, x^2y + xy^2=120$. Then $x^2 + y^2=$ $\qquad$ | 3.34 .
Let $x+y=s, xy=t$. Then $s+t=23, st=120$. We get $s=8, t=15$ or $s=15, t=8$ (discard). $x^{2}+y^{2}=(x+y)^{2}-2xy=s^{2}-2t=34$. | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A positive integer, if added to 100 and 168 respectively, can result in two perfect squares. This positive integer is $\qquad$ | 4. $n=156$.
Let this number be $n$, and $n+168=a^{2}, n+100=b^{2}$. Then $a^{2}-b^{2}=68=2^{2} \times 17$,
which means $(a+b)(a-b)=2^{2} \times 17$.
However, $a+b$ and $a-b$ have the same parity,
so $a+b=34, a-b=2$.
Thus, $a=18$, and hence $n=156$. | 156 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radiu... | Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $m$ and $n$ are integers, the equation
$$
x^{2}+(n-2) \sqrt{n-1} x+m+18=0
$$
has two distinct real roots, and the equation
$$
x^{2}-(n-6) \sqrt{n-1} x+m-37=0
$$
has two equal real roots. Find the minimum value of $n$, and explain the reasoning. | 1. $\left\{\begin{array}{l}n \geqslant 1, \\ (n-2)^{2}(n-1)-4(m+18)>0, \\ (n-6)^{2}(n-1)-4(m-37)=0 .\end{array}\right.$
(2) - (3), and rearrange to get
$$
(n-4)(n-1)>27.5 \text {, }
$$
then $n \geqslant 8$.
When $n=8$, $m=44$, which satisfies the given conditions. Therefore, the minimum value of $x$ is 8. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $f(x)=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x$ is an even function, $g(x)=2^{x}+\frac{a+b}{2^{x}}$ is an odd function, where $a 、 b \in \mathbf{C}$. Then $a+a^{2}+a^{3}+\cdots+a^{2000}+b+b^{2}$ $+b^{3}+\cdots+b^{2000}=$ | 3. -2 .
From the known $f(-x)=f(x), g(-x)=-g(x)$ we get
$$
\begin{array}{l}
\log _{\frac{1}{3}}\left(3^{-x}+1\right)-\frac{1}{2} a b x \\
=\log _{\frac{1}{3}}\left(3^{x}+1\right)+\frac{1}{2} a b x . \\
2^{-x}+\frac{a+b}{2^{-x}}=-\left(2^{x}+\frac{a+b}{2^{x}}\right) .
\end{array}
$$
Simplifying, from equation (1) we g... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the smallest positive integer $n$ that has exactly 144 different positive divisors, and among them, there are 10 consecutive integers.
(26th IMO Shortlist) | Analysis: According to the problem, $n$ is a multiple of the least common multiple of 10 consecutive integers, and thus must be divisible by $2, 3, \cdots, 10$. Since $8=2^{3} ; 9=3^{2}, 10=2 \times 5$, its standard factorization must contain at least the factors $2^{3} \times 3^{2} \times 5 \times 7$. Therefore, we ca... | 110880 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, if
the three sides of $\triangle A B C$ are
$n+x, n+2 x, n$
$+3 x$, and the height $A D$ from $B C$ is $n$, where $n$
is a positive integer, and $0<x \leqslant$
1. Then the number of triangles that satisfy the above conditions is $\qquad$. | 6.12 .
Let $a=n+x, b=n+2 x, c=n+3 x, s=\frac{1}{2}(a+b+c)$. By Heron's formula and the general area formula, we have
$$
\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} n(n+2 x) .
$$
Simplifying, we get $12 x=n$.
And $0<x=\frac{n}{12} \leqslant 1$,
so $0<n \leqslant 12$,
which means $n$ has exactly 12 possibilities (correspondin... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. If the sum of positive integers $a$ and $b$ is $n$, then $n$ can be transformed into $a b$. Can this method be used several times to change 22 into 2001? | 1. Reverse calculation, $2001=3 \times 667$ from $3+667=670$; $670=10 \times 67$ from $10+67=77$; $77=7 \times 11$ from $7+11=18$. From any $n=1+(n-1)$, we can get $n-1=1 \times(n-1)$. Therefore, starting from 22, we can sequentially get 21, $20,19,18,77,670$ and 2001. | 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Three-digit number $\overline{a b c}=a^{2}+1+(\overline{b c})^{2}$. Then $\overline{a b c}=$ | 3.726 .
From $100 a+\overline{b c}=a^{2}+(\overline{b c})^{2}+1$, we have $(\overline{b c}-1) \overline{b c}=100 a-a^{2}-1$.
Obviously, the left side is even,
so the right side $a$ must be odd to make both sides equal.
When $a=1$, $(\overline{b c}-1) \overline{b c}=98=2 \times 7^{2}$;
When $a=3$, $(\overline{b c}-1) \... | 726 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $x, y \in \mathbf{R}$. If $2 x, 1, y-1$ form an arithmetic sequence, and $y+3,|x+1|+|x-1|, \cos (\arccos x)$ form a geometric sequence, then the value of $(x+1)(y+1)$ is $\qquad$ . | 2.4.
Given that $2 x, 1, y-1$ form an arithmetic sequence, we have
$$
y=3-2 x \text{. }
$$
From $\cos (\arccos x)=x$ and $-1 \leqslant x \leqslant 1$, it follows that
$$
|x+1|+|x-1|=2 \text{. }
$$
Thus, $y+3, 2, x$ form a geometric sequence, which gives
$$
x(y+3)=4 \text{. }
$$
Substituting (1) into (2), we get $2 ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R$, where $O$ is the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\q... | 5.2.
Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter ).
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta}$.
Then $|A Q|=\frac{2 a b^{2}|\cos \theta|}{b^{2... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Draw 63 lines on the coordinate plane: $y=b, y=\sqrt{3} x + 2b, y=-\sqrt{3} x+2b$, where $b=-10,-9,-8, \ldots, 8,9,10$. These lines divide the plane into several equilateral triangles. The number of equilateral triangles with side length $\frac{2}{\sqrt{3}}$ is $\qquad$ $-$ | 6.660.
The six outermost straight lines determine a regular hexagon with a side length of $\frac{20}{\sqrt{3}}$. The three straight lines passing through the origin $O$ divide this hexagon into six equilateral triangles with a side length of $\frac{20}{\sqrt{3}}$. Since the side length of each large triangle is 10 tim... | 660 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Given the equation about $x$
Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1}=\frac... | 15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given 997 points in the plane, the midpoints of the line segments connecting every pair of points are colored red. Prove that there are at least 1991 red points. Can you find 997 points that result in exactly 1991 red points? | (遈示: Let $A B$ be the longest line connecting two points among 997 points. Point $A$ is connected to the other 996 points, and the midpoints of these lines are all within the circle centered at $A$ with a radius of $\frac{A B}{2}$. Point $B$ is connected to the other 996 points, and the midpoints of these lines are all... | 1991 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Four, (20 points) Given that $x$ and $y$ are real numbers, and satisfy
$$
\begin{array}{l}
x y + x + y = 17, \\
x^{2} y + x y^{2} = 66 .
\end{array}
$$
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$. | Given the conditions $x y + x + y = 17$ and $x y (x + y) = 66$, we know that $x y$ and $x + y$ are the two real roots of the equation
$$
t^{2} - 17 t + 66 = 0
$$
Solving this equation, we get
$$
t_{1} = 6, \quad t_{2} = 11.
$$
Thus, $x y = 6$ and $x + y = 11$; or $x y = 11$ and $x + y = 6$.
When $x y = 6$ and $x + y ... | 12499 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Find all four-digit numbers that satisfy the following conditions: they are divisible by 111, and the quotient obtained is equal to the sum of the digits of the four-digit number. | $$
\overline{a b c d}=a \times 10^{3}+b \times 10^{2}+c \times 10+d
$$
satisfies the condition. Then
$$
\begin{array}{l}
\frac{a \times 10^{3}+b \times 10^{2}+c \times 10+d}{11 i} \\
=9 a+b+\frac{a-11 b+10 c+d}{11 i} . \\
\because-98 \leqslant a-11 b+10 c+d \leqslant 108, \text { and } \overline{a b c d} \text { is di... | 2997 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) (1) In a $4 \times 4$ grid paper, some small squares are painted red, and then 2 rows and 2 columns are crossed out. If no matter how they are crossed out, at least one red small square remains uncrossed, how many small squares must be painted red at least? Prove your conclusion.
(2) If the “$4 \time... | Three, (1) At least 7 cells need to be colored.
If the number of colored cells $\leqslant 4$, then by appropriately crossing out 2 rows and 2 columns, all the colored cells can be crossed out.
If the number of colored cells is 5, then at least one row has 2 cells colored. By crossing out this row, the remaining colore... | 5 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 1: 20 teams participate in the national football championship. To ensure that in any three teams that have already played, there are two teams that have already played against each other, what is the minimum number of matches that need to be played?
(3rd All-Union Mathematical Olympiad) | Solution: Suppose that in any three teams that have already played, there are two teams that have played against each other. Let team $A$ be the team with the fewest matches played, having played $k$ matches. Thus, the $k$ teams that have played against team $A$, as well as team $A$ itself, have all played no fewer tha... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$. Try to find the maximum value of $n$.
The difference between any two adjacent interior angles of a convex $n$-sided polygon is $18^{\circ}$. Try to find the maximum value of $n$. | Solution: Complete in three steps.
(1) Prove that $n$ is even. Let the convex $n$-gon be $A_{1} A_{2} A_{3} \cdots$
$A_{n}$, then $A_{2}=A_{1} \pm 18^{\circ}$, (where “ + ” or “-” is chosen)
$A_{3}=A_{2} \pm 18^{\circ}=A_{1} \pm 18^{\circ} \pm 18^{\circ}$, (choose “ + ” or “ - ” as before) $\qquad$
$$
\begin{array}{l}
... | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 In a football invitational tournament, 16 cities participate, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. After several days of the tournament, it was found that except... | Solution: Consider the general case: Suppose there are $n$ cities participating, and let the number of matches played by Team B from City A be $a_{n}$. Clearly, $a_{1}=$ 0.
Since there are $n$ cities participating, according to the competition rules, each team can play at most $2(n-1)$ matches. Given the conditions, t... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 There is a type of sports competition with $M$ events, and athletes $A$, $B$, and $C$ participate. In each event, the first, second, and third places receive $p_{1}$, $p_{2}$, and $p_{3}$ points respectively, where $p_{1}$, $p_{2}$, $p_{3} \in \mathbf{Z}^{+}$, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 2... | Solution: Consider the total score of three people, we have
$$
\begin{array}{l}
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40 . \\
\because p_{1}, p_{2}, p_{3} \in \mathbf{Z}^{+}, \text { and } p_{1}>p_{2}>p_{3}, \\
\therefore p_{1}+p_{2}+p_{3} \geqslant 3+2+1=6 .
\end{array}
$$
Thus, $6 M \leqslant 40, M \leqslant 6$.
Si... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In a certain competition, each player plays exactly one game against each of the other players. A win earns the player 1 point, a loss 0 points, and a draw 0.5 points for each player. After the competition, it is found that exactly half of each player's points were earned in games against the 10 lowest-scorin... | Solution: Let there be a total of $n$ players, then they collectively score $\mathrm{C}_{n}^{2}$ points. The 10 players with the lowest scores collectively score $\mathrm{C}_{10}^{2}=45$ points through their matches against each other, which is half of their total score, so these 10 players collectively score 90 points... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 After 20 figure skaters have performed, 9 judges each assign them ranks from 1 to 20. It is known that for each skater, the difference between any two ranks does not exceed 3. If the sum of the ranks each skater receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant ... | If 9 judges all give a certain athlete the first place, then $C_{1}=9$. If two athletes are both judged as the first place, then one of them gets no less than 5 first places, and the other 4 ranks no higher than fourth. Therefore, $C_{1}=5 \times 1+$ $4 \times 4=21$.
If 3 athletes all get the first place, then their o... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. In the final stage of a professional bowling tournament, the top five players compete as follows: First, the fifth-place player competes with the fourth-place player, the loser gets fifth place, the winner competes with the third-place player; the loser gets third place, the winner competes with the first-place play... | (Due to determining the top five rankings, 4 races were held, and each race has two possible outcomes. By the multiplication principle, there are $2^{4}=16$ possible ranking orders for the top five winners.) | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In a cross-country race, there are two teams participating, each with 5 members. When the athletes reach the finish line, their ranks are recorded, and the team of the $n$-th arriving athlete scores $n$ points, with the team having the lower total score winning. Now, assuming no two team members arrive at the finish... | (The hint: The sum of the scores of 10 athletes is $1+2+\cdots+10$ $=55$, thus the total score of the winning team does not exceed $\left[\frac{55}{2}\right]=27$. Also, the total score of the winning team is no less than $1+2+\cdots+5=15$, it is easy to see that the score of the winning team can take all values from $1... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example $1\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}=$ $\qquad$
(1999, National Junior High School Mathematics League Wuhan Selection Competition) | Let $3^{999}=a, 5^{099}=b, 7^{999}=c$, then
$$
\begin{array}{l}
\text { Original expression }=\frac{c}{a} \sqrt{\frac{a^{2}+(a b)^{2}}{c^{2}+(b c)^{2}}} \\
=\frac{c}{a} \sqrt{\frac{a^{2}\left(1+b^{2}\right)}{c^{2}\left(1+b^{2}\right)}} . \\
=\frac{c}{a} \cdot \frac{a}{c}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let the parabola
$$
y=x^{2}+(2 a+1) x+2 a+\frac{5}{4}
$$
intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+323 a^{-6}$.
(1998, National Junior High School Mathematics Competition) | Solution: (1) From the given condition, the equation
$$
x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0
$$
has two equal real roots, hence
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$
which simplifies to $a^{2}-a-1=0$.
Solving for $a$, we get $a=\frac{1 \pm \sqrt{5}}{2}$.
(2) From (1), we know $a^{2}-a-1=0$, thus
$$
\... | 5796 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $a^{2}+a+1=0$. Then, $a^{1992}+$ $a^{322}+a^{2}=$ $\qquad$
(Harbin 15th Junior High School Mathematics Competition) | Given the condition $a \neq 1$, from $(a-1)(a^2 + a + 1) = 0$, we get $a^3 - 1 = 0$, hence $a^3 = 1$.
Substituting 1 for $a^3$ and 0 for $a^2 + a + 1$, we get
$$
\begin{aligned}
\text { Original expression } & =\left(a^{3}\right)^{664}+\left(a^{3}\right)^{107} a+a^{2} \\
& =1^{664}+1^{107} a+a^{2} \\
& =1+a+a^{2}=0 .
\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
(Jiangsu Province 14th Junior High School Mathematics Competition) | Solution: From the conditions, we have
$$
\begin{array}{l}
x y+(x+y)=71, \\
x y(x+y)=880,
\end{array}
$$
Therefore, $x y$ and $x+y$ are the two roots of the equation
$$
z^{2}-71 z+880=0
$$
Solving this equation, we get
$$
x+y=16, x y=55,
$$
or $x+y=55, x y=16$.
When $x+y=55, x y=16$, $x$ and $y$ are the two roots of... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $f(x)=\sum_{0}^{5}\left[2^{i} x\right]$. Find $f(3.14)$. | Given: $\because x=3.14, \frac{4}{2^{5}} \leqslant 0.14<\frac{5}{2^{5}}$, so take $p=3, k=4$.
$$
\begin{aligned}
\therefore f(3.14) & =\sum_{0}^{5} 2^{i} \times 3+\sum_{0}^{5}\left[\frac{4}{2^{i}}\right] \\
& =3 \times 63+7=196 .
\end{aligned}
$$ | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In a $3 \times 3$ square grid, fill in the numbers as shown in the table below. The operation on the table is as follows: each operation involves adding a number to two adjacent numbers in the grid (adjacent means two small squares that share a common edge).
\begin{tabular}{|l|l|l|}
\hline 0 & 3 & 2 \\
\hline 6 & 7 ... | (Tip: (1) After 5 operations, all cells can be 0.
(2) As shown in the table below, the invariant $S=a-$ $b+c-d+e-f+g-h+k$.
\begin{tabular}{|l|l|l|}
\hline$a$ & $b$ & $c$ \\
\hline$d$ & $e$ & $f$ \\
\hline$g$ & $h$ & $k$ \\
\hline
\end{tabular}
The initial state is
$$
S=0-3+2-6+7-0+4-9+5=0,
$$
The target state is
$$
S... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
1. A bus starts a 100-kilometer journey at 12:20 PM. There is a computer on the bus that at 1:00 PM, 2:00 PM, 3:00 PM, 4:00 PM, 5:00 PM, and 6:00 PM, says: “If the average speed in the future is the same as the average speed in the past, then it will take one more hour to reach the destination.” Is this possible? If so... | 1. The computer is correct.
At $k: 00$ PM, where $1 \leqslant k \leqslant 6$, the distance the car has traveled is $\frac{60 k-20}{60 k+40}$ of the total distance, and the average speed of the car up to this point is $\frac{1}{60 k+40}$. If it continues to travel the remaining $\frac{60}{60 k+40}$ at the same speed, i... | 85 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. The cube of an $n$-digit number is an $m$-digit number, can $n+m=$ 2001? | 2. Notice that $10^{500}$ has 501 digits, and $\left(10^{500}\right)^{3}$ has 1501 digits, totaling 2002 digits.
For any positive number $a>10^{500}, a$ and $a^{3}$ have at least 2002 digits combined.
If $a<10^{500}$, it has at most 500 digits, and $a^{3}<10^{1500}$ has at most 1500 digits. Therefore, the sum of the ... | 2001 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. According to the "Personal Income Tax Law of the People's Republic of China," citizens do not need to pay tax on the portion of their monthly salary and wages that does not exceed 800 yuan. The portion exceeding 800 yuan is the monthly taxable income, which is taxed according to the following table in segments:
\be... | $=、 11.1350$ | 1350 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$ | 3.3.
Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.
From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discar... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) A chemical plant, starting from January this year, if it does not improve its production environment and continues to produce as it is, will earn 700,000 yuan per month. At the same time, it will receive penalties from the environmental protection department, with the first month's penalty being 30,0... | Three, let the cumulative income for $n$ months without modifying the equipment, under the original conditions, be $a(n)$, and let the cumulative income for $n$ months after modifying the equipment be $b(n)$. From the given conditions, we have $a(n)=70 n$, and we can assume
$$
b(n)=a n^{2}+b n+c, \quad n \leqslant 5.
$... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ has $n$ different points $P_{1}, P_{2}, \ldots, P_{n}, F$ is the right focus, $\left\{\left|P_{i} F\right|\right.$ \} forms an arithmetic sequence with a common difference $d>\frac{1}{100}$. Then the maximum value of $n$ is ( ).
(A ) Fang data $) 200$
( C) 99
( D ) 100 | 5.( B ).
$$
\begin{array}{l}
\because\left|P_{n} F\right|=\left|P_{1} F\right|+(n-1) d, \\
\therefore n=1+\frac{\left|P_{n} F\right|-\left|P_{1} F\right|}{d} .
\end{array}
$$
In an ellipse, the maximum value of the focal radius $\left|P_{i} F\right|$ is $a+c$, and the minimum value is $a-c$, so
$$
\begin{array}{l}
\le... | 200 | Geometry | MCQ | Yes | Yes | cn_contest | false |
3. Given $x, y, z \in \mathbf{R}, x y+y z+z x=-1$. Then the minimum value of $x^{2}+5 y^{2}+8 z^{2}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.4 .
$$
\begin{array}{l}
\text { Given }(x+2 y+2 z)^{2}+(y-2 z)^{2} \geqslant 0 , \\
x^{2}+5 y^{2}+8 z^{2} \geqslant-4(x y+y z+z x)=4 ,
\end{array}
$$
we have $x=\frac{3}{2} , y=-\frac{1}{2} , z=-\frac{1}{4}$ ,
or $x=-\frac{3}{2}, y=\frac{1}{2}, z=\frac{1}{4}$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $S_{n}$ be the sum of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \cdots, \frac{1}{2^{n-1}}\right\}$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=$ $\qquad$ . | 6.1 .
For any element in set $A$, it appears in a subset containing 3 elements $\mathrm{C}_{n-1}^{2}$ times, then
$$
\begin{array}{l}
S_{n}=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{n-1}}\right) \mathrm{C}_{n-1}^{2} . \\
\text { Therefore, } \lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2}}=\lim _{n \rightarrow \infty} \f... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=$ $b_{0}=1, a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$, $(n=1,2 \ldots)$. Find the value of $a_{2001}^{2}-2 b_{2001}^{2}$. | $$
\begin{array}{l}
a_{0}^{2}-2 b_{0}^{2}=-1, \\
a_{1}^{2}-2 b_{1}^{2}=1, \\
\ldots \ldots
\end{array}
$$
Conjecture $a_{n}^{2}-2 b_{n}^{2}=(-1)^{n+1}(n=0,1 \ldots)$.
We will prove this by mathematical induction.
Assume the proposition holds for $n=k$, that is,
$$
a_{k}^{2}-2 b_{k}^{2}=(-1)^{k+1} \text {. }
$$
Then f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.
---
Translate the above text into English, please retain the original text's line bre... | As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length form... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 A class participated in a math competition, with a total of $a$, $b$, and $c$ three questions. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are worth 25 points each. After the competition, every student answered at least one question corr... | Solution: Let $x, y, z$ represent the number of correct answers for questions $a$, $b$, and $c$, respectively. Then we have
$$
\left\{\begin{array} { l }
{ x + y = 2 9 , } \\
{ x + z = 2 5 , } \\
{ y + z = 2 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=17, \\
y=12, \\
z=8 .
\end{array}\right.\... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ is a natural number, there exists a linear polynomial with integer coefficients and $a$ as the leading coefficient, which has two distinct positive roots less than 1. Then, the minimum value of $a$ is $\qquad$ . | 2.5.
Let $f(x)=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, $00$. Then from $f(0)$ and $f(1)$ being positive integers, we get $f(0) f(1) \geqslant 1$, that is, $a^{2} x_{1} x_{2}\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1$.
Also, $x(1-x) \leqslant \frac{1}{4}$, with equality when $x=\frac{1}{2}$.... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$. | Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$
Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$
Then $a_{k+1}=3^{a_... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 On a circle with a circumference of $300 \mathrm{~cm}$, there are two balls, A and B, moving in uniform circular motion at different speeds. Ball A starts from point A and moves in a clockwise direction, while ball B starts from point B at the same time and moves in a counterclockwise direction. The two balls... | Solution: Let the length of $\overparen{A C P}$ be $x \mathrm{~cm}$, and the original speeds of A and B be $v_{1}$ and $v_{2}\left(v_{1} \neq v_{2}\right)$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{40}{v_{1}}=\frac{x-40}{v_{2}}, \\
\frac{300-20-(x-40)}{2 v_{1}}=\frac{x-40+20}{\fr... | 120 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 A mall installs an escalator between the first and second floors, which travels upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor (the escalator itself is also moving). If the boy and the girl both move at a uniform speed, and the boy walks twice as many steps per... | Solution: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, and the stairs have $s$ steps, then the boy's speed is $2x$ steps/min. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
Dividing... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. In a regular hexagon divided into six areas for观赏 plants, 加园: $\mathrm{i}$, it is required that the same type of plant be in the same area, and adjacent areas must have different plants. There are 4 different types of plants available, resulting in $\qquad$ planting schemes.
Note: The term "观赏 plants" is directly ... | 12.732.
Consider planting the same type of plant at $A$, $C$, and $E$, there are $4 \times 3 \times 3 \times 3=108$ methods.
Consider planting two types of plants at $A$, $C$, and $E$, there are $3 \times 4 \times 3 \times 3 \times 2 \times 2=432$ methods.
Consider planting three types of plants at $A$, $C$, and $E$, ... | 732 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$
Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left... | Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)}... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. A university has no more than 5000 registered students, of which $\frac{1}{3}$ are freshmen, $\frac{2}{7}$ are sophomores, $\frac{1}{5}$ are juniors, and the remainder are seniors. In the mathematics department's student list, freshmen account for $\frac{1}{40}$ of all freshmen in the university, sophomores account ... | 4. (C).
Let the total number of students in the school be $n$, and the number of students in the mathematics department be $c$. Then $\frac{1}{3} n$ are first-year students, $\frac{2}{7} n$ are second-year students, and $\frac{1}{5} n$ are third-year students. The number of first-year students in the mathematics depar... | 183 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$ | Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, in
Rt $\triangle A B C$,
$\angle C=90^{\circ}$, point $M$
is the intersection of the three
medians of the triangle. Perpendiculars are drawn from $M$
to $A B$, $B C$, and $A C$,
with the feet of the perpendiculars being $D$, $E$, and $F$, respectively. If $A C=3$, $B C=12$, then the area of $\... | 4.4.
$$
\begin{array}{l}
AB = \sqrt{3^2 + 12^2} = \sqrt{153} = 3 \sqrt{17}, \\
S_{\triangle ABC} = \frac{1}{2} \times 3 \times 12 = 18.
\end{array}
$$
Draw a perpendicular from $C$ to $AB$, with the foot of the perpendicular at $H$, and let it intersect $EF$ at $G$. We have
$$
\begin{array}{l}
S_{\triangle ABC} = \fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) In a square array of 16 rows and 16 columns composed of red and blue dots, adjacent dots of the same color are connected by a line segment of the same color, and adjacent dots of different colors are connected by a yellow line segment. It is known that there are 133 red dots, of which 32 are on the b... | Three, each row has 15 line segments, 16 rows have a total of $15 \times 16=240$ horizontal line segments; 16 columns similarly have 240 vertical line segments; in total, there are 480 line segments. It is known that 195 of these line segments are yellow, so the remaining 284 line segments are red and blue.
There are ... | 134 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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