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3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$ | $3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as t... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,677 |
4. For $\triangle A B C$ with an area of $3 \sqrt{15}$, the lengths of the two medians $A D$ and $B E$ are 3 and 6, respectively. Then the length of the third median $C F$ is $\qquad$ | 4.6 or $6 \sqrt{3}$.
Let the centroid of $\triangle ABC$ be $G$, and $CF=3x$. Extend $GF$ to $P$ such that $FP = GF$.
Then quadrilateral $APBG$ is a parallelogram.
$$
\begin{array}{l}
\frac{1}{2} S_{\triangle APBG} = S_{\triangle ABG} = \frac{1}{3} S_{\triangle ABC} = \sqrt{15}, \\
\frac{1}{2} S_{\square APBG} = S_{\... | 6 \text{ or } 3 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,678 |
一、(满分 20 分) Determine the smallest positive integer $n$, whose last digit is 7, such that if the last digit 7 is moved to the front, the resulting number is 5 times the original number. | Let $n=\overline{a_{k} a_{k-1} \cdots a_{2} a_{1}}, x=\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}}$, where $a_{1}$ $=7, k$ is a natural number greater than 1. Then
$$
\begin{aligned}
n & =\overline{a_{k} a_{k-1} \cdots a_{2}} \times 10+7=10 x+7, \\
5 n & =\overline{7 a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{... | 142857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,679 |
II. (Full marks 25 points) Given that $AO$ is the altitude from the vertex $A$ to the base $EF$ of isosceles $\triangle AEF$, and $AO = EF$. Extend $AE$ to $B$ such that $BE = AE$, and draw a perpendicular from $B$ to $AF$, meeting at point $C$. Prove that point $O$ is the incenter of $\triangle ABC$.
保留源文本的换行和格式，直接输出... | $$
\left.\begin{array}{l}
\left.\begin{array}{c}
A E=A F \\
A O \perp E F
\end{array}\right\} \Rightarrow\left\{\begin{array}{c}
O E=\frac{1}{2} E F \\
\angle O A E=\angle O A F
\end{array}\right. \\
\angle A C B=90^{\circ} \\
\left.\begin{array}{c}
\angle A O B=90^{\circ}
\end{array}\right\} \Rightarrow A, B, D, C \te... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,680 |
Three. (Full marks 25 points) Does there exist 100 different lines in the plane, such that their intersection points total 1998? | Three, there are 1998 intersection points.
First, divide 98 lines into two sets of parallel lines, one set has $k$ lines, and the other set has $(98-k)$ lines. The number of intersection points is $k(98-k)$. When the value of the positive integer $1998-k(98-k)$ is minimized, $k=26$ or 72.
When $k=26$ or 72, $k(98-k)=2... | 1998 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,681 |
1. Let $A=\{1,2\}$, then the number of mappings from $A$ to $A$ that satisfy $f[f(x)]=f(x)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | -、1.C.
The functions are $f, g, h$, such that $f(1)=1, f(2)$ $=2 ; g(1)=g(2)=1 ; h(1)=h(2)=2$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,682 |
2. In the square $R$ (including the boundary) with vertices at $(1997,0)$, $(0,1997)$, $(-1997,0)$, $(0,-1997)$, the number of integer points is ( ) .
(A) 7980011
(B) 7980013
(C) 7980015
(D) 7980017 | 2. B.
In general, consider the square with vertices at $(N, 0)$, $(0, N)$, $(-N, 0)$, and $(0, -N)$.
When $N=1$, the integer points are $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$, and $(0,0)$, totaling 5 points, i.e., $a_{1}=5$.
When $N$ increases to $N+1$, in the first quadrant and on the positive x-axis, the added integ... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 709,683 |
3. Let $M=\{(x, y):|x y|=1, x>0\}, N=$ $\{(x, y): \operatorname{arctg} x+\operatorname{arcctg} y=\pi\}$. Then, ( )
(A) $M \cup N=\{(x, y):|x y|=1\}$
(B) $M \cup N=M$
(C) $M \cup N=N$
(D) $M \cup N=\{(x, y):|x y|=1$ and $x, y$ are not both negative\} | 3. B.
In $M$, $|x y|=1$ is equivalent to $x y=1$ and $x y=-1$. But $x>0$, so it represents the graph of a hyperbolic function in the $\mathrm{I}$ and $\mathrm{N}$ quadrants.
In $N$, from $\operatorname{arctg} x+\operatorname{arcctg} y=\pi$, we get
$$
\operatorname{tg}(\operatorname{arctg} x+\operatorname{arcctg} y)=\f... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,684 |
4. In tetrahedron $ABCD$, faces $ABC$ and $BCI$ are both equilateral triangles with side length $2a$, and $AD = 2\sqrt{2}a$. Points $M$ and $N$ are the midpoints of edges $AB$ and $CD$, respectively. The shortest distance between $M$ and $N$ on the tetrahedron is ( ).
(A) $2a$
(B) $\frac{3}{2}a$
(C) $a$
(D) $\frac{5}{2... | 4. A.
On the surface of the tetrahedron, the possible paths from point $M$ to $N$ are as follows: (1) passing through edge $A C$; (2) passing through edge $A D$; (3) passing through edge $B C$; (4) passing through edge $B D$. When the distances from $M$ to $N$ in the above four cases are not all equal, take the smalle... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,686 |
5. Given three triangles $\triangle, \triangle_{1}, \triangle_{2}$ with perimeters $p, p_{1}, p_{2}$ respectively. If $\triangle \backsim \triangle_{1} \backsim \triangle_{2}$, and the smaller two triangles $\triangle_{1}$ and $\triangle_{2}$ can be placed inside the larger triangle $\triangle$ without overlapping. The... | 5.13.
Let the perimeter of $\triangle$ be $p$, and the area be $S$; the perimeter of $\Delta$ be $p_{1}$, and the area be $S_{1}$; the perimeter of $\triangle_{2}$ be $p_{2}$, and the area be $S_{2}$. From the condition, we have $S \geqslant S_{1} + S_{2}$.
$$
\begin{array}{l}
\because \triangle \backsim \triangle_{1}... | \sqrt{2} p | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,687 |
G. The number of non-overlapping triangles formed with the vertices of a regular $n$-sided polygon is equal to ( ).
(A) $\left[\frac{n^{2}}{10}\right]$
(B) $\left[\frac{n^{2}}{11}\right]$
(C) $\left[\frac{n^{2}}{12}\right]$
(D) None of the above | 6. D.
Let the number of distinct triangles formed by the vertices of a regular $n$-sided polygon be $N$, among which there are $N_{1}$ equilateral triangles, $N_{2}$ isosceles triangles that are not equilateral, and $N_{3}$ scalene triangles. There is 1 equilateral triangle with vertex $A$, 3 isosceles triangles, and ... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 709,688 |
1. Let $p, q \in N$, and $1 \leqslant p<q \leqslant n$, where $n$ is a natural number not less than 3. Then the sum of all fractions of the form $\frac{p}{q}$ is $\qquad$ . | $=1 \cdot \frac{1}{4} n(n-1)$.
Group all fractions of the form $\frac{p}{q}$ as follows:
$$
\left(\frac{1}{2}\right),\left(\frac{1}{3}, \frac{2}{3}\right), \cdots,\left(\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}\right) \text {, }
$$
where the $k$-th group contains $k$ fractions. It is not difficult to find that t... | \frac{1}{4} n(n-1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,689 |
3. If $2 f(1-x)+1=x f(x)$, then $f(x)=$ | 3. $\frac{x-3}{x^{2}-x+4}$.
Let there exist a function $f(x)$ that satisfies the given equation
$$
2 f(1-x)+1=x f(x)
$$
Substitute $1-x$ for $x$ to get
$$
2 f(x)+1=(1-x) f(1-x) .
$$
From (1), we have $f(1-x)=\frac{1}{2}[x f(x)-1]$.
Substitute (3) into (2), to get
$$
2 f(x)+1=(1-x) \cdot \frac{1}{2}[x f(x)-1] .
$$
T... | \frac{x-3}{x^{2}-x+4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,691 |
4. In $\triangle A B C$, $D$ is on $B C$, $B D: D C=$ $3: 2, E$ is on $A D$, $A E: E D=5: 6$, extend $B E$ to intersect $A C$ at $F$. Then $B E: E F=$ . $\qquad$ | $4.9: 2$
Solve using the Angle Bisector Theorem. As shown in the figure, we have
$$
\begin{array}{l}
\frac{A F}{F C}=\frac{A B \cdot \sin \angle 1}{B C \cdot \sin \angle 2} \\
=\frac{A B \cdot \sin \angle 1}{\frac{5}{3} B D \cdot \sin \angle 2}
\end{array}
$$
Thus, $A F=\frac{1}{3} A C$.
$$
\text { Also, } \frac{B E}{... | 9: 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,692 |
5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=p, a_{n+1}=a_{n}^{2}+2 a_{n}$. Then the general term $a_{n}=$ $\qquad$ . | $\begin{array}{l}\text { 5. }(p+1)^{2^{n-1}-1}- \\ \because a_{n}=a_{n-1}^{2}+ \\ 2 a_{n-1}, \\ \quad \therefore a_{n}+1 \\ \quad=\left(a_{n-1}+1\right)^{2} \\ \quad=\left(a_{n-2}+1\right)^{2^{2}}=\cdots=\left(a_{1}+1\right)^{2^{n-1}}=(p+1)^{2^{n-1}}, \\ \text { i.e., } a_{n}=(p+1)^{2^{n-1}}-1 .\end{array}$ | a_{n}=(p+1)^{2^{n-1}}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,693 |
6. Given the sets $A=\{1,2,3,4,5,6\}, B=$ $\{6,7,8,9\}$, by selecting 3 elements from $A$ and 2 elements from $B$, we can form $\qquad$ new sets with 5 elements. | 6. Selecting 3 elements from set $A$ has $C_{6}^{3}$ ways, and selecting 2 elements from set $B$ without 6 has $C_{3}^{2}$ ways, resulting in a total of $C_{6}^{3} \cdot C_{3}^{2}$ ways; selecting 3 elements from set $A$ without 6 has $C_{5}^{3}$ ways, and selecting 1 element from set $B$ with 6 has $C_{3}^{1}$ ways, r... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,694 |
Three. (Full marks 23 points) Given that $M$ is a point on the moving chord $AB$ of the parabola $y^{2}=$ $2 p x$, $O$ is the origin, $O A$ $\perp O B, O M \perp A B$. Find the equation of the locus of point $M$.
| Three, let $A B$ intersect the $x$-axis at point $C\left(x_{0}, 0\right)$, and the coordinates of $A, B$ are $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, respectively. Then
$$
\begin{array}{l}
y_{1}^{2}=2 p x_{1}, \\
y_{2}^{2}=2 p x_{2} . \\
\because O A \perp O E, \\
\therefore \frac{y_{1} y_{2}}{x_{1} x_... | (x-p)^{2}+y^{2}=p^{2} .(x \neq 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,695 |
Four, (Full marks 24 points) The board is initially written with the numbers 11 and 13. Now, perform the following operations:
i) Rewrite a number once;
ii) Add the two numbers and write the sum. Prove:
(1) The number 119 will never appear on the board;
(2) Any natural number greater than 119 can appear on the board af... | (1) Proof by contradiction. Assume there exist non-negative integers $m, n$ such that $11 m + 13 n = 119$, i.e.,
$$
11(m+1) + 13 n = 130.
$$
Thus, $13 \mid (m+1), m+1 \geq 13, m \geq 12$.
$$
\text{Also, } 11 m + 13(n+1) = 132,
$$
Thus, $11 \mid (n+1), n+1 \geq 11, n \geq 10$.
This way, $11 m + 13 n \geq 11 \times 12 ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,697 |
Five. (Full marks 25 points) Given $m \geqslant 0, f(x)=x^{2}+$ $\sqrt{m} x+m+1$. Prove that for all $x_{1}, x_{2}, \cdots, x_{n} \in$ $R^{+}$, we have
$$
\begin{array}{l}
f\left(\sqrt[n]{x_{1} x_{2} \cdots x_{n}}\right) \\
\leqslant \sqrt[n]{f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)},
\end{arr... | V. $\because \Delta=(\sqrt{m})^{2}-4(m-1)0$.
First, we prove by induction for $n=2^{k}$, the proposition holds. In fact,
i) When $k=1$, i.e., $n=2$, since
$$
\begin{array}{l}
\sqrt{m} x_{1} x_{2}\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2}+(m+1)\left(x_{1}-x_{2}\right)^{2} \\
+\sqrt{m}(m+1)\left(\sqrt{x_{1}}-\sqrt{x_{2}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,698 |
One, (Full marks 50 points) Given that $ABCD$ is any convex quadrilateral. Construct equilateral triangles $\triangle ADH$ and $\triangle BCF$ outside the quadrilateral on sides $AD$ and $BC$ respectively; construct isosceles triangles $\triangle ABE$ and $\triangle CDG$ outside the quadrilateral on bases $AB$ and $CD$... | Sure, here is the translation:
---
I. The complex number solution and pure proof are left to the reader.
Let the letter represent the complex number of this point. Set
$$
\begin{array}{l}
z_{1}=\frac{E-B}{A-B}, z_{2}=\frac{D-C}{G-C}, \\
z_{3}=\frac{D-H}{A-H}, z_{4}=\frac{B-F}{C-F} .
\end{array}
$$
Thus,
$$
\begin{arr... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,699 |
II. (Full marks 50 points) If the next student participates in the mathematics competition, among any $m(m \geqslant 3)$ students there is a unique common friend (if A is a friend of B, then B is also a friend of A). How many students are participating in the mathematics competition? | Based on the known conditions, each student has friends.
"If there are $k(\leqslant m)$ students who are friends with each other, then according to the known conditions, they have a common friend, and we get $k+1$ students who are friends with each other. By analogy, we conclude that there are $m+1$ students $A_{1}$, $... | m+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,700 |
Three. (Full marks 50 points) $\alpha$ is a cyclic decimal, $f_{x}(m)$ represents the product of the digits in the $m$ consecutive positions starting from the $k$-th position after the decimal point of $\alpha$. Prove that there exist natural numbers $p, q$, such that for any $s, t$, we have
$$
\left[f_{p}(s)\right]^{\... | Let $\alpha$ be a pure repeating decimal, $\alpha=0 . \overline{a_{1} a_{2} \cdots a_{n}}, \alpha$'s repeating segment length is $n$, i.e.,
$$
a_{i}=a_{n+i}, i=1,2, \cdots \text {. }
$$
If some $a_{k}=0$, we can take $p=q=k$, so we also assume $a_{i} \neq 0, 1 \leqslant i \leqslant n, G=\sqrt[n]{a_{1} a_{2} \cdots a_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,701 |
Let $n=19^{97}, a_{1}, a_{2}, \cdots, a_{n}$ be any permutation of $1,2, \cdots, n$. Let $a_{i}^{\prime}=\left|a_{19 i-18}-a_{19 i-17}+a_{19 i-16}-\cdots+a_{19 i}\right|, i$ $=1,2, \cdots, 19^{96} ; a_{j}^{\prime \prime}=\mid a^{\prime}{ }_{19 j-18}-a_{19 j-17}^{\prime}+a_{19 j-16}^{\prime}-\cdots$ $+a^{\prime}{ }_{19 ... | Proof: Obviously,
$$
\begin{array}{l}
a_{i}^{\prime} \equiv a_{19 i-18}-a_{19 i-17}+\cdots+a_{19 i} \\
\equiv a_{19 i-18}+a_{19 i-17}+\cdots+a_{19 j}(\bmod 2), \\
\therefore a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{196}^{\prime} \equiv a_{1}+a_{2}+\cdots+a_{n}(\bmod 2),
\end{array}
$$
Min Tian, $a_{1}^{\prime \prime}+a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 709,703 |
Let $P, Q, R$ be points on the sides $BC, CA, AB$ of $\triangle ABC$ respectively, such that $\frac{BP}{PC}=\frac{CQ}{QA}=\frac{AR}{RB} = k$. If the sides of $\triangle ABC$ are denoted as $a, b, c$, then:
$$
A P^{4}+B Q^{4}+C R^{4} \geqslant \frac{9}{16}\left(a^{4}+b^{4}+c^{4}\right) .
$$ | Proof: $\because \frac{B P}{P C}=\frac{C Q}{Q A}=\frac{A R}{R B}$,
$$
\therefore \frac{B P}{B C}=\frac{C Q}{C A}=\frac{A R}{A B} \text {. }
$$
Let the ratio be $\lambda$, then we have $B P=\lambda a$.
In $\triangle A B C$, by the cosine rule, we have
$$
\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a} .
$$
In $\triangle A B P$... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,704 |
Given $a 、 \beta 、 \gamma$ are the interior angles of an acute $\triangle A B C$. Prove:
$$
\begin{array}{l}
\left(\alpha^{2}-\frac{\pi^{2}}{9}\right) \sin \frac{\alpha^{2}}{2}+\left(\beta^{2}-\frac{\pi^{2}}{9}\right) \sin \frac{\beta^{2}}{2} \\
+\left(\gamma^{2}-\frac{\pi^{2}}{9}\right) \sin \frac{\gamma^{2}}{2} \geqs... | Proof: It is known that $\frac{\alpha^{2}}{2} 、 \frac{\pi^{2}}{18} \in\left(0, \frac{\pi}{2}\right)$.
$\because y=\sin x$ is an increasing function on $\left(0, \frac{\pi}{2}\right)$.
$$
\begin{array}{l}
\therefore\left(\frac{\alpha^{2}}{2}-\frac{\pi^{2}}{18}\right)\left(\sin \frac{\alpha^{2}}{2}-\sin \frac{\pi^{2}}{18... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,705 |
Example 1 Fill the numbers $1,2,3, \cdots, n$ into $n$ cells labeled $1,2,3, \cdots, n$, with one number per cell. How many ways are there to fill the cells so that the number in each cell matches its label? | Let $a_{n}$ be the number of valid ways. Add the number $n+1$ and the cell labeled $n+1$.
For each valid way in $a_{n}$, we move the number in the $k$-th cell to the $n+1$-th cell, and fill $n+1$ into the $k$-th cell, obtaining $n a_{n}$ valid ways;
For the $n$ numbers, if the $k$-th cell is filled with $k (1 \leqsla... | a_{n}=n!\sum_{i=1}^{n}(-1)^{i} \frac{1}{i!} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,706 |
Example 2 Let $s_{n}$ be the number of sequences $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, where $a_{i} \in\{0,1\}$ and no segment of 6 consecutive numbers is identical. Prove that as $n \rightarrow \infty$, $s_{n} \rightarrow \infty$. | Proof: Let $B_{n}$ be the set of sequences that satisfy the requirements of the problem, then $s_{n}=\left|B_{n}\right|$.
For each sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ in $B_{n}$, we perform the following operations:
\[
\left\{\begin{array}{l}
\text { (1) } a_{n}=0 \text {, add } a_{n+1}=1 ; \\
\text { ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,707 |
Example 3 Divide a circle into $n(n \geqslant 2)$ sectors, sequentially denoted as $S_{1}, S_{2}, \cdots, S_{n}$ (as shown in the figure). Each sector can be painted with any of the two colors: red, white, or blue, with the requirement that adjacent sectors must have different colors. How many coloring methods are ther... | Let the number of coloring methods be $a_{n}$, and add a sector $S_{n+1}$. We first color $S_{1}$, which has 3 options; coloring $S_{2}$ has 2 options; $\cdots$; coloring $S_{n}$ has 2 options; coloring $S_{n+1}$ is temporarily only required to be different from $S_{n}$. There are a total of $3 \times 2^{n}$ coloring m... | a_{n}=2^{n}+2(-1)^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,708 |
Example 4 Suppose
$$
x=\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}+\frac{1+1}{1+\cdots}}}
$$
with 1000 horizontal fraction bars. Is $x^{2}+x>1$ always true? (1990, Wisconsin Talent Search) | Let $x_{n}$ represent a number in the form of $x$ with $n$ fraction lines. Then, we have $x_{n+1}=\frac{1}{1+x_{n}}$.
It is easy to find $x_{1}=\frac{1}{2}, x_{2}=\frac{2}{3}, x_{3}=\frac{3}{5}, x_{4}=\frac{5}{8}$,
$\cdots$. Let $y_{n}=x_{n}^{2}+x_{n}$, then
$$
y_{1}=\frac{3}{4}1, y_{3}=\frac{24}{25}1$. Conjecture: whe... | x^{2}+x>1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,709 |
Example 3 In the plane where square $ABCD$ is located, find a point $P$ such that $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ are all isosceles triangles. Then the number of points $P$ with this property is ( ).
(A) 1
(B) 5
(C) 9
(D) 17 | Analysis: For this problem, it is easy to think that there is only 1 point (1 is the intersection of the perpendicular bisectors of the sides, i.e., the center of the regular polygon), this point is the vertex with $P$ as the apex, and $A$, $B$, $C$, $D$ as the base angles of the isosceles triangles. Does there exist a... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,710 |
Example 5 Prove: For any natural number $n$,
$$
\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}<2 .
$$ | Prove: Construct the sequence $\left\{a_{n}\right\}$:
$$
\begin{array}{l}
a_{0}=2, a_{1}=a_{0}^{2}-1, \\
a_{2}=\left(a_{0}^{2}-1\right)^{2}-2, \\
a_{3}=\left[\left(a_{0}^{2}-1\right)^{2}-2\right]^{2}-3, \\
a_{4}=\left\{\left[\left(a_{0}^{2}-1\right)^{2}-2\right]^{2}-3\right\}^{2}-4, \\
\cdots \cdots .
\end{array}
$$
T... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,711 |
Example 7 For a natural number $k, g(k)$ represents the greatest odd divisor of $k$ (for example, $g(3)=3, g(20)=5$), find $g(1)+$ $g(2)+g(3)+\cdots+g\left(2^{n}\right.$ ) (where $n \in N$ ). | Let $S_{n}=g(1)+g(2)+g(3)+\cdots+g\left(2^{n}\right)$. It is easy to see that $S_{1}=g(1)+g\left(2^{1}\right)=2$.
From the definition of $g(k)$: when $k$ is odd, $g(k)=k$; when $k$ is even, i.e., $k=2 m(m \in N)$, $g(k)=g(m)$.
Thus,
$$
\begin{aligned}
S_{n}= & {\left[g(1)+g(3)+\cdots+g\left(2^{n}-1\right)\right]+} \\
... | \frac{4^{n}+2}{3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,713 |
Example 8 Let $P_{n}(k)$ be the number of permutations of the set $\{1,2, \cdots, n\}$ that keep $k$ elements fixed. Prove that:
$$
\sum_{k=0}^{n} k P_{n}(k)=n!.
$$ | Proof: The distribution of permutations of the set $\{1,2, \cdots, n\}$ that keep $k$ elements fixed is as follows: First, choose $k$ elements from $n$ elements to remain fixed, which can be done in $C_{n}^{k}$ ways. The remaining $n-k$ elements are then permuted such that all elements change their positions, which can... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,714 |
Example 9 Let $f(n)$ represent the number of arrangements of length $n$ (such as 00101, 10100 are both of length 5) consisting of 0s and 1s where no two 1s are adjacent, with the convention that $f(0) = 1$. Prove that $f(4m-2)$ is divisible by 3, $m \in \mathbb{N}$. | Proof: Permutations of length 1 are only 0,1, i.e., $f(1) = 2$. Permutations of length 2 are $00, 01, 10, 11$, i.e., $f(2) = 3$. When $n > 2$, permutations of length $n$ can be divided into two categories: those ending in 0 and those ending in 01.
(1) The number of permutations ending in 0 with no two 1s adjacent is $f... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 709,715 |
Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers 1, 2, $\cdots, n$. Let $f(n)$ be the number of such permutations that satisfy: (1) $a_{1}=1$; (2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1$, $2, \cdots, n-1$. Determine whether $f(1996)$ is divisible by 3. | Solution: It is easy to find that $f(1)=f(2)=1, f(3)=2$.
When $n \geqslant 4$, we have $a_{1}=1, a_{2}=2$ or 3.
(a) When $a_{2}=2$, we proceed as follows: delete $a_{1}$, $b_{1}=a_{2}-1, b_{2}=a_{3}-1, \cdots, b_{n-1}=a_{n}-1$, to get a permutation $b_{1}, \cdots, b_{n-1}$ that meets the conditions, and the number of s... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,716 |
Conclusion 1 As shown in Figure 1, for the squares $A E D B, A C F G$ with vertex $A$, then
(1) $B G$ and $C E$ are perpendicular and equal;
(2) $B G, C E, D F$ are concurrent;
(3) Let $B G, C E, D F$ intersect at $K$, then $A K \perp$ $D F$. | Proof: As shown in Figure 1,
(1) Rotate $\triangle A B G$ 90 degrees clockwise around point $A$, it will coincide with $\triangle A E C$. This proves the conclusion.
(2) Let $B G$ and $C E$ intersect at $K$, connect $A K$, $K D$, and $K F$. Since $\angle A G K = \angle A C K$, points $A$, $K$, $C$, and $G$ are concycli... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,717 |
Conclusion 2 As shown in Figure 2, if squares $A E D B$ and $A C F G$ share vertex $A$, and if we take the midpoints $O_{1}$, $M$, and $O_{2}$ of $A D$, $B C$, and $A F$ respectively, then $\triangle O_{1} M O_{2}$ is an isosceles right triangle, with $M$ as the right-angle vertex. | Prove: Connect $B E$ and $C G$, then $O_{1}$ must be on $B E$, and $O_{2}$ must be on $C G$. Connect $B G$ and $C E$, then $O_{1} M$ II $\frac{1}{2} C E, O_{2} M$ II $\frac{1}{2} B G$. By conclusion (1), it is proved. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,718 |
Conclusion 3 As shown in Figure 3, if the squares $A E D B$ and $A C F G$ share the vertex $A$, and the midpoint of $E G$ is $M$. Then
(1) $A M=\frac{1}{2} B C$;
(2) Extending $M A$ to intersect $B C$ at $H$, then $A H \perp B C$;
(3) The lines $A H, C D, B F$ are concurrent. | Proof: (1) Extend $A M$ to $P$, such that $M P = A M$, then $P G = A E = A B$. Therefore, $\triangle P G A \cong \triangle B A C$, hence $P A = B C$. From this, $A M = \frac{1}{2} B C_{y}$?
(2) From $\angle M A = \angle 4 C 4$, we have $\angle H C A + \angle H A C = \angle M A G + \angle H A C = 90^{\circ}$. Therefore,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,719 |
Conclusion 4 As shown in Figure 6, if the squares $A G B M$ and $A N C H$ share the vertex $A$, and $BC$ intersects $AM$ at $D$, and intersects $AN$ at $E$. Then $\frac{B D \cdot B E}{C D \cdot C E}=\frac{A B^{2}}{A C^{2}}$. | Prove: Connect $A B$,
$A C$, draw $A K \perp B C$ at
$K$, by $\frac{B M}{A K}=\frac{B D}{A D}$
and $\frac{C N}{A K}=\frac{C E}{A E}$, we get
$\frac{B D}{A D} \cdot \frac{A E}{C E}=\frac{B M}{C N}$.
Since $\frac{A B}{A C}=\frac{B M}{C N}$,
thus $\frac{B D}{A D} \cdot \frac{A E}{C E}=\frac{A B}{A C}$.
Also, using $\frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,720 |
Example 4 As shown in Figure 4, on a plane: Given isosceles $\triangle A B C$, where $A B=A C$. Try to find all points $M$ on the plane that satisfy the following conditions, such that $\triangle A B M$ and $\triangle A C M$ are both isosceles triangles. | Solution: (1) All points on the circle with center $A$ and radius $A B$, except for points $B$, $C$, and the points symmetric to $B$, $C$ about $A$, i.e., $B^{\prime}$, $C^{\prime}$;
(2) In addition, there are 6 points (see Figure 4): the intersection point $M_{1}$ of the perpendicular bisectors of $A B$ and $A C$, the... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,721 |
Conclusion 7. As shown in Figure 9, $A C H K$ share vertices $A, B, C$, and $J$ is outside $\triangle A B C$.
(1) Let $A B=a, A C=b$, then the maximum value of $S_{\triangle A K D}+S_{\triangle B E F}+S_{\triangle C G H}$ is $\frac{3}{2} a b$;
(2) $D K, E F, G H$ are twice the lengths of the medians from $B C, A C, A B... | Proof: (1) omitted.
(2) Draw $C M \underline{=} B A$, connect $B M$ intersecting $A C$ at $N$, then $B N$ is the median of side $A C$. From $\triangle E B F \cong \triangle M C B$, we have $E F=2 B N$.
Similarly, the other conclusions can be proven.
(3) $D, E, F, G, H, K$ are concyclic $\Leftrightarrow$ the perpendicul... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,724 |
Conclusion 9 As shown in Figure 11, squares $A B C D$, $D E F G$, and $F H L K$ share vertices $D$, $F$, and $P$ is the midpoint of $A K$, then $P E \perp C H$, and $P E=\frac{1}{2} C H$. | Prove: Connect $G H, G C, E K, A E$, extend $A E$ to intersect $C H$ at $M$, extend $E P$ to $N$, making $P N=E P$. Connect $A N, K N$, then $E K N A$ is a parallelogram. From conclusion 1 (1), we have $A E=C G$ and $A E \perp C G, E K=G H$ and $E K \perp G H$.
Also, $\angle N K E$ is supplementary to $\angle A E K$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,726 |
Conclusion 12
As shown in Fig. 14,
the squares
$A P Q R$,
$PBTV$,
$TCRS$ share vertices
$P, T, R$,
with their centers
being $O_{4}, O_{5}$,
$O_{6}$, respectively. Let the midpoints of $B C$,
$C A$, and $A B$ be $X, Y, Z$. Then $O_{4} X, O_{5} Y,$
$O_{6} Z$ are concurrent. | Prove: Connect $A O_{4}, A O_{5}, O_{4} O_{5}, C O_{6}, C O_{5}$, and $O_{5} O_{6}$. Then, in $\triangle A O_{5} O_{4}$ and $\triangle O_{5} C O_{6}$, we have $A O_{1}=$ $\frac{1}{2} P R=O_{5} \Theta_{6}, O_{4} O_{5}=\frac{1}{2} R T=O_{6} C, \angle A O_{1} O_{5}$ $=90^{\circ}+\angle P R T=\angle O_{5} O_{6} C$. Therefo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,728 |
Conclusion 13: Squares $BAFE$, $CBHG$, $JIDC$, and $DLKA$ share vertices $A$, $B$, $C$, and $D$, and are located outside quadrilateral $ABCD$, as shown in Figure 15.
(1) Let $\mathrm{O}_{1}$, $\mathrm{O}_{2}$, $\mathrm{O}_{3}$, and $\mathrm{O}_{4}$ be the centers of the respective squares. Then $\mathrm{O}_{1}\mathrm{O... | Figure 15
Proof: (1) Let the midpoint of $BD$ be $O$, then $OO_{3} \underline{I} \frac{1}{2} BJ, OO_{2} \underline{I} \frac{1}{2} GD$. From Conclusion 1(1), $GD$ and $BJ$ are perpendicular and equal, so $OO_{3}$ and $OO_{2}$ are also perpendicular and equal. Similarly, $OO_{4}$ and $OO_{1}$ are perpendicular and equal,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,729 |
Conclusion 14 As shown in Figure 16, squares $D C A B$, $Q D F E$, $Q R S P$, and $S G H C$ share vertices $C$, $D$, $Q$, and $S$, and $P R \| \frac{1}{2} B D$, then points $E$, $R$, and $G$ are collinear, and $R$ is the midpoint of $E G$. | Prove: Take the midpoint $N$ of $PD$, and the midpoint $M$ of $PC$, then $MN \parallel \frac{1}{2} DC$. Connect $QN$ and $SM$ and extend them so that $NL = QN, MK = SM$. From $PR \parallel \frac{1}{2} BD$, we know that $QSMN$ is a parallelogram. Then, by conclusions $3(1), (2)$, we know that $QL \parallel SK$. This pro... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,730 |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers $(n \geqslant 3)$, and let $p=\sum_{i=1}^{n} x_{i}, q=\sum_{i \leqslant i<j \leqslant n} x_{i} x_{j}$. Find to make:
(i) $\frac{n-1}{n} p^{2}-2 q \geqslant 0$;
(ii) $\left|x_{i}-\frac{p}{n}\right| \leqslant \frac{n-1}{n} \sqrt{p^{2}-\frac{2 n}{n-1} q}$ $(i=1,2... | Proof: (i) By Zhao Da's theorem, the real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are the $n$ roots of the real-coefficient $n$-degree polynomial $x^{n}-p x^{n-1}+q x^{n-2}+\cdots +a_{1} x+a_{0}=0$. Thus, by the theorem (1), we have
$$
\Delta_{1}=(n-1)(-p)^{2}-2 n q \geqslant 0 \text {. }
$$
which simplifies to $\frac{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 709,731 |
Example 5 In triangles with side lengths being consecutive natural numbers and a perimeter not exceeding 100, the number of acute triangles is $\qquad$ | Analysis: Let the three sides be $n-1$, $n$, and $n+1$. They should satisfy the following system of inequalities:
$$
\left\{\begin{array}{l}
(n-1)+n+(n+1) \leqslant 100, \\
(n-1)+n>n+1, \\
(n-1)^{2}+n^{2}>(n+1)^{2}.
\end{array}\right.
$$
From (1), we get $n \leqslant 33 \frac{1}{3}$.
From (2), we get $n>2$.
From (3), ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,732 |
Example 2 Let $n \geqslant 4, \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} ; \beta_{1}, \beta_{2}, \cdots, \beta_{n}$ be two sets of real numbers, satisfying
$$
\sum_{i=1}^{n} \alpha_{i}^{2}<1, \sum_{i=1}^{n} \beta_{i}^{2}<1 .
$$
Let $A^{2}=1-\sum_{i=1}^{n} \alpha_{i}^{2}, B^{2}=1-\sum_{i=1}^{n} \beta_{i}^{2}$,
$$
W=\fr... | When $\lambda=\hat{*}$, the equation has only the real root 0.
When $\lambda \neq \neq$ (),
$$
\begin{aligned}
\Delta_{i} & ==(n-1)(\lambda A B)^{2}-2 n(\lambda W) \lambda \\
& =\lambda^{2}\left[(n-1) A^{2} B^{2}-2 n W\right] .
\end{aligned}
$$
By $1-\sqrt{2 \bar{W}}=\sum_{i=1}^{n} \alpha_{i} \beta_{i} \leqslant \sum_... | \lambda=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,733 |
Example 3 Find the largest integer $n$, such that all non-zero solutions of the equation $(z+1)^{n}=z^{n}+1$ lie on the unit circle. | $$
\begin{array}{l}
\text { Solution: Using the binomial theorem, the equation can be transformed into } \\
z\left(C_{n}^{1} z^{n-2}+C_{n}^{2} z^{n-3}+C_{n}^{3} z^{n-4}+\cdots+C_{n}^{n-2} z+\right. \\
\left.C_{n}^{n-1}\right)=0(n>3) .
\end{array}
$$
Let the non-zero solutions of the equation be $z_{i}(i=1,2, \cdots, n... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,734 |
As shown in the figure, let $a, b, c, p, R, r$ represent the sides, semi-perimeter, circumradius, and inradius of $\triangle ABC$, respectively, and let $I$ be the incenter. Then we have
$$
\begin{array}{c}
\frac{\triangle_{D E F}}{\triangle}= \\
\frac{2 a b c}{(a+b)(b+c)(c+a)} .
\end{array}
$$
(1958, Shangyou High Sch... | $$
\begin{array}{l}
\frac{\triangle}{\triangle_{D E F}}=\frac{(a+b)(b+c)(c+a)}{2 a b c} \\
=\frac{(a+b+c)(a b+b c+c a)-a b c}{2 a b c} \\
=\frac{2 p\left(p^{2}+4 R r+r^{2}\right)-4 p R r}{2 \cdot 4 p R r} \\
=\frac{p^{2}+2 R r+r^{2}}{4 R r} \\
\left\{\begin{array}{l}
\geqslant \frac{16 R r-5 r^{2}+2 R r+r^{2}}{4 R}=\fr... | \frac{9}{2}-\frac{r}{R} \leqslant \frac{\triangle}{\triangle_{D E F}} \leqslant \frac{3}{2}+\frac{R}{r}+\frac{r}{R} | Geometry | proof | Yes | Yes | cn_contest | false | 709,735 |
36 IMO-5: Let $A B C D E F$ be a convex hexagon, $A B=B C=C D, D E=E F=F A, \angle B C D=$ $\angle E F A=60^{\circ}, G, H$ are two points inside the hexagon such that $\angle A G B=\angle D H E=120^{\circ}$. Prove:
$$
A G+G B+G H+D H+H E \geqslant C F . \quad(*)
$$ | Proof: As shown in the figure, construct equilateral triangles $\triangle A B M$ and $\triangle D E N$ outside the hexagon with $A B$ and $D E$ as sides. Rotate $\triangle A G B$ counterclockwise by $60^{\circ}$ around $A$ to $\triangle A G^{\prime} M$, then $\triangle A G G^{\prime}$ is an equilateral triangle. Theref... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,736 |
Example 1 In $\triangle ABC$, $AB=12, AC=16$, $M$ is the midpoint of $BC$, $E, F$ are on $AB, AC$ respectively, $EF$ intersects $AM$ at $G, AE=2AF$. Find the ratio $\frac{EG}{GF}$. | Solution: Connect $C E$ intersecting $A M$ at $O, \triangle C B E$ and $\triangle C E F$ are intersected by the same...
line $A M$, using Menelaus Theorem,
we get (as shown in the figure):
$$
\frac{B M}{M C} \cdot \frac{C O}{O E} \cdot \frac{E A}{A B}=1, \frac{E O}{O C} \cdot \frac{C A}{A F} \cdot \frac{F G}{G E}=1 .
$... | \frac{8}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,737 |
Example 2 In any quadrilateral $ABCD$, the pair of opposite sides $BA$ and $CD$ intersect at $M$. Through $M$, a secant line intersects the lines containing another pair of opposite sides at $H$ and $L$, and intersects the lines containing the diagonals at $H'$ and $L'$. Prove:
$$
\frac{1}{M H}+\frac{1}{M L}=\frac{1}{M... | Proof: Let $A D \times X C$, and intersect at $O, \triangle B M L$ and $\triangle C M L$ are intersected by the line $A O$, then
$$
\begin{array}{l}
\frac{B A}{A M}=\frac{H L}{M H} \cdot \\
\frac{O B}{\overline{L O}}, \\
\frac{C D}{D M}=\frac{H L}{M H} \cdot \frac{O C}{L O} . \\
(1) \times L C+(2) \times B L, \text { w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,738 |
Let $\triangle A B C$ and its medial triangle's sides, semi-perimeter, area, circumradius, and inradius be denoted as $a$, $b$, $c$, $a^{\prime}$, $b^{\prime}$, $c^{\prime}$; $p$, $p^{\prime}$, $\triangle$, $\triangle^{\prime}$; $R$, $R^{\prime}$, $r$, $r^{\prime}$. Then the theorem states $2 R^{\prime} \geqslant R \ge... | Proof: First, there are the following facts: (1) $A B^{\prime}=A^{\prime} B$ $=p-c$, etc.;
(2) $\frac{a^{\prime} b^{\prime} c^{\prime}}{a b c} \geqslant$
$\frac{r}{4 R}$; (3) $\frac{\triangle^{\prime}}{\triangle}=$
$\frac{r}{2 R}$; (4) $2 p^{\prime} \geqslant$
$p$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,740 |
1. On the right
side, in a column of squares, each square except for 9.7 represents a number. It is known that the sum of any 3 consecutive squares is 19. Then $A+H+M+O$ equals ( ).
(A) 21
(B) 23
(C) 25
(D) 26 | -1. D.
From the given, we easily know that $A+H=10$.
$$
\begin{array}{l}
\because M+O+X=O+X+7, \\
\therefore M=7 .
\end{array}
$$
Similarly, $O=9$.
$$
\therefore A+H+M+O:=26 \text {. }
$$ | 26 | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 709,741 |
Example 6 A square piece of paper contains 100 points. Together with the vertices of the square, there are 104 points, and no three points among these 104 points are collinear. Now, the square piece of paper is entirely cut into triangles, with each vertex of these triangles being one of the 104 points, and each of the... | Analysis: If there is only one point $P_{0}$ inside the square, connecting $P_{0}$ to the four vertices results in four triangles, which clearly require the first 4 cuts. If another point $P_{1}$ is added, this point must fall within one of the triangles, let's assume it falls within $\triangle P_{0} C D$ (as shown in ... | 301 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,743 |
4. On the hypotenuse $AB$ of the isosceles right triangle $ABC$, take two points $M$ and $N$ such that $\angle MCN = 45^{\circ}$. Let $AM = m$, $MN = x$, and $BN = n$. The shape of the triangle with side lengths $x$, $m$, and $n$ is ( ).
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Varies with the chan... | 4. B.
$$
\begin{array}{l}
\because \angle M C N=\angle A=\angle B=45^{\circ}, \\
\therefore \triangle C M N \backsim \triangle A C N \backsim \triangle B M C \\
\therefore \frac{B C}{x+n}=\frac{m+x}{A C} . \\
\because B C=A C, \\
\therefore B C^{2}=(x+n)(x+m) . \\
\text { Also } 2 B C^{2}=A B^{2}, \\
\therefore 2(x+m)(... | x^2 = m^2 + n^2 | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,745 |
5. Given that $p$ and $q$ are rational numbers, and $x=\frac{\sqrt{5}-1}{2}$ satisfies $x^{3}+p x+q=0$. Then the value of $p+q$ is ( ).
(A) -1
(B) 1
(C) -3
(D) 3 | 5. A.
Substitute $x=\frac{\sqrt{5}-1}{2}$ into $x^{3}+p x+q=0$, and simplify to get
$$
4 \sqrt{5}(2+p)+4(2 q-p-4)=0 \text {. }
$$
Since $p, q$ are rational numbers, we have
$$
\left\{\begin{array}{l}
2+p=0, \\
2 q-p-4=0 .
\end{array}\right.
$$
Solving these, we get $p=-2, q=1$.
Therefore, $p+q=-1$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,746 |
1. As shown in the figure, in $\odot O$, $\overparen{A D B} = 90^{\circ}$, chord $A B = a$, and a circular arc with center $B$ and radius $B A$ intersects $\odot O$ at another point $C$. Then the area $S$ of the crescent shape (the shaded part in the figure) formed by the two circular arcs is $S=$ . $\qquad$ | ii. $1 \cdot \frac{1}{2} a^{2}$.
It is known that $AC$ is the diameter of $\odot O$, connect $BC$ (figure not shown). Then $\triangle ABC$ is an isosceles right triangle:
$\therefore OA=\frac{1}{2} AC=\frac{1}{2} \sqrt{2} a$.
$\therefore$ The area of sector $ABC$ $S_{1}=\frac{\pi}{4} \cdot AB^{2}=\frac{\pi}{4} a^{2}$,
... | \frac{1}{2} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,747 |
2. Given $S=1^{2}-2^{2}+3^{2}-1^{2}+\cdots+99^{2}-$ $100^{2}+101^{2}$. Then the remainder when $S$ is divided by 103 is $\qquad$ | 2. 1 .
$$
\begin{aligned}
S= & 1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(99^{2}-98^{2}\right) \\
& +\left(101^{2}-100^{2}\right) \\
= & 1+2+3+\cdots+100+101 \\
= & \frac{101 \times 102}{2}=5151=103 \times 50+1 .
\end{aligned}
$$
Therefore, the required remainder is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,748 |
3. Locations A and B are respectively upstream and downstream on a lake. Each day, there is a boat that departs on time from both locations and travels at a constant speed towards each other, usually meeting at 11:00 AM on the way. ... Due to a delay, the boat from location A left 40 minutes late, and as a result, the ... | 3. 6 .
As shown in the figure, the two ships usually meet at point $A$. On the day when ship B is late, they meet at point $B$. According to the problem, ship A takes $15^{\prime}$ to navigate segment $AB$, and ship C takes
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,749 |
4. Given that $a$ is an integer, the two real roots of the equation $x^{2}+(2 a-1) x+$ $a^{2}=0$ are $x_{1}$ and $x_{2}$. Then $\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=$ | 4. 1.
From the problem, the discriminant $\Delta=-4 a+1 \geqslant 0$, so $a \leqslant 0$.
$$
\begin{array}{l}
\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2}=\left(x_{1}+x_{2}\right)-2 \sqrt{x_{1} x_{2}} \\
=1-2 a+2 a=1 \\
\therefore\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,750 |
5. Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers, and $x_{1}<x_{2}$ $<\cdots<x_{6}<x_{7}$, also $x_{1}+x_{2}+\cdots+x_{7}=159$. Then the maximum value of $x_{1}+x_{2}+x_{3}$ is $\qquad$ | 5. 61.
Given that $x_{7} \geqslant x_{6}+1 \geqslant x_{5}+2 \geqslant x_{4}+3 \geqslant x_{3}+4 \geqslant$ $x_{2}+5 \geqslant x_{1}+6$.
Similarly, $x_{6} \geqslant x_{1}+5, x_{5} \geqslant x_{1}+4, x_{4} \geqslant x_{1}+3, x_{3} \geqslant x_{1}+2, x_{2} \geqslant x_{1}+1$.
$$
\begin{array}{l}
\therefore 159=x_{1}+x_... | 61 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,751 |
One. (20 points)
As shown in the figure, the square $E F G H$ is inscribed in $\triangle A B C$. Let $B C=\overline{a b}$ (this is a... digit number), $E F=c$, and the altitude $A D=d$. It is known that $a, b, c, d$ are exactly four consecutive positive integers in ascending order. Try to find the area of $\triangle A ... | From the figure, we know that
$S_{\triangle A B C}=S_{\triangle A E F}+S_{\triangle B E N}+S_{\triangle F C C}+S_{\text {square HEEFGH }}$, that is, $\frac{1}{2} \overline{a b} \times d=\frac{1}{2} c(d-c)+\frac{1}{2}(\overline{a b}-c) c+c^{2}$.
$$
\begin{array}{l}
\overline{a b} \times d=(\overline{a b}+d) c, \\
\there... | 24 \text{ or } 224 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,752 |
Example 7 If the perimeter of a hexagon is 20, each side length is an integer, and no triangle can be formed with any three sides, then such a hexagon ( ).
(A) does not exist
(B) exists uniquely
(C) has a finite number, but more than one
(D) has infinitely many | Analysis: Observing the four options, due to the instability of an $n$-sided polygon $(n \geqslant 4)$, if there exists such a hexagon, there must be infinitely many. Therefore, we need to find whether there exist six positive integers $a_{1}, a_{2}, \cdots, a_{6}$ (without loss of generality, assume $a_{1} \leqslant a... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,754 |
1. Try to design a method to divide a square into 8 smaller squares without repetition or omission (the sizes of the smaller squares can be different); also, how to divide a square into 31 smaller squares under the same requirements? | Three, 1. It is easy to divide a square into $4^{2}=16$ smaller squares, then combine 9 of them located at one corner into one square, resulting in a total of $16-9+1=8$ squares.
Divide into 16 smaller squares, then further divide any 5 of them into 4 smaller squares each, resulting in a total of $16-5+5 \times 4=31$ ... | 31 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,755 |
1. If $4 a-b \sqrt{b-1}=4 a^{2}-\sqrt{b-1}+1$, then among the following four equations:
(1) $2 a+2 b-3=0$,
(2) $(a-b)^{2}+3 a^{2}-b=0$,
(3) $a^{2}+3 a-2 b=0$,
(4) $4 a^{2}+b_{1}^{2}-2 j=0$
which of the following are correct?
(A) 1
(B) 2
(C) 3
(D) 4 | -.1.C.
Transform $4 a-b \sqrt{b-1}=4 a^{2}-\sqrt{b-1}+1$ into $(2 a-1)^{2}+(b-1)^{\frac{3}{2}}=0$.
It is found that $a=\frac{1}{2}, b=1$. Substituting these values into the four equations listed, all but (3) hold true. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,757 |
2. Let the altitudes of $\triangle A B C$ be $A D, B E, C F$, and let the orthocenter $H$ not coincide with any vertex, then the number of circles that can be determined by some four points among $A, B, C, D, E, F, H$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 2. D.
As shown in the figure, the following sets of four points are concyclic:
$$
A F H E, B F H D, C D H E, A F D C, B F E C, C D F A \text {. }
$$
If $\triangle A B C$ is an obtuse triangle, the same conclusion applies. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,758 |
3. If $[a]$ denotes the integer part of the real number $a$, then $\left[\frac{1}{\sqrt{16-6 \sqrt{7}}}\right]$ equals ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. B.
$$
\begin{array}{l}
\because 16-6 \sqrt{7}=4+7-2 \sqrt{63} \\
\quad=(\sqrt{9}-\sqrt{7})^{2}, \\
\therefore \frac{1}{\sqrt{16-6 \sqrt{7}}}=\frac{1}{\sqrt{9}-\sqrt{7}}=\frac{3+\sqrt{7}}{2} .
\end{array}
$$
And $2<\sqrt{7}<3$,
$$
\therefore \frac{5}{2}<\frac{3+\sqrt{7}}{2}<3 \text {. }
$$
Thus $\left[\frac{1}{\sqr... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,759 |
4. In $\triangle A B C$, if $\angle A=2 \angle B$, side $b=4, c=5$, then side $a$ equals ( ).
(A) 6
(B) 7
(C) $3 \sqrt{5}$
(D) 5 | 4. A.
As shown in the figure, draw the angle bisector $AD$ of $\angle A$. By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
CD=\frac{4}{9} a, BD=\frac{5}{9} a . \\
\chi \angle CDA=2 \angle DAB=
\end{array}
$$
$\angle A$,
$$
\begin{array}{l}
\therefore \triangle ADC \sim \triangle ABC . \\
\text { Hence } \fra... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,760 |
5. Let $(x+\alpha)^{4}=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$. If $a_{1}$ $+a_{2}+a_{3}=64$, then $\alpha$ equals ( ).
(A) -4
(B) 4
(C) -2
(D) 2 | 5. D.
Let $x=1$, then
$$
(1+\alpha)^{4}=1+a_{1}+a_{2}+a_{4}+a_{4}=65+a_{4} \text {. }
$$
Let $x=0$, then $\alpha^{4}=a_{4}$.
$$
\therefore(1+\alpha)^{4}-\alpha^{4}=65 \text {. }
$$
This leads to $2 \alpha+3 \alpha^{2}+2 \alpha^{3}-32=0$,
which is $(\alpha-2)\left(2 \alpha^{2}+7 \alpha+16\right)=0$.
ifij $2 \alpha^{2... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,761 |
6. Given the quadratic equation $a x^{2}+b x+c=0(a c \neq 0)$ has two distinct real roots $m$ and $n$, and $m<|m|$. Then, the root situation of the quadratic equation $c x^{2}+(m-n) a x-a=0$ is ( ).
(A) It has two negative roots
(B) It has one positive root
(C) It has two roots of opposite signs
(D) It has no real root... | 6. A.
$\because m, n$ have opposite signs and $m0 . \\
\text { Therefore, } m n=\frac{c}{a}0 . \\
\text { The discriminant of the equation } c x^{2}+(m-n) a x-a=0 \text { is } \\
\begin{array}{l}
\Delta=a^{2}(m-n)^{2}+4 a c \\
=a^{2}\left[(m-n)^{2}+4 \cdot \frac{c}{a}\right]=a^{2}(m+n)^{2} \geqslant 0,
\end{array}
\end... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,762 |
7. On June 9, 1898, Britain forced the Qing government to sign a treaty leasing 975.1 square kilometers of Hong Kong land to Britain for 99 years. On July 1, 1997, it was known that it was a Tuesday. Therefore, June 9, 1898, was a ( ).
(A) Tuesday
(B) Wednesday
(C) Thursday
(D) Friday
(Note: In the Gregorian calendar, ... | 7.C.
Solution 1: $\because 365 \equiv 1(\bmod 7)$,
$\therefore$ If $S$ represents the total number of days from the beginning of the year to the required day, $x$ represents the year, and $y$ represents the day of the year, then we have
$$
\begin{array}{l}
S \equiv(x-1)+\left[\frac{x-1}{4}\right]-\left[\frac{x-1}{100... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 709,763 |
8. For the quadratic function $y=a x^{2}+b x+c(a \neq 0)$, the vertex of its graph is in the first quadrant, and it passes through the points $(0,1)$ and $(-1,0)$. Then the range of values for $S=a+b+c$ is ( ).
(A) $0<S<1$
(B) $0<S<2$
(C) $1<S<2$
(D) $-1<S<1$ | 8. B.
Solution one: Let $x=0, y=1$ and $x=-1, y=0$, we get
$c=1, a=b-1$.
$\therefore S=a+b+c=2b$.
From the problem, we know that $-\frac{b}{2a}>0$, and $a<0$.
Also, from $b=a+1$ and $a<0$, it leads to a contradiction. Therefore, (D) can be excluded.
Solution three: Exclude (A) and (C) as above. If $a+b+c=0$, then whe... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,764 |
Example 8 - 8 smaller cubes are combined to form a larger cube with an edge length of 10 units, which is then painted and subsequently divided into the original smaller cubes. The number of smaller cubes that have at least one face painted is ( ).
(A) 600
(B) 512
(C) 488
(D) 480 | Very complicated. We can consider the opposite situation, that is, the $8 \times 8 \times 8$ cube in the very center of the cube with each side measuring 10 units, then the answer is not hard to find: $1000-512=488$ units. | 488 | Geometry | MCQ | Yes | Yes | cn_contest | false | 709,765 |
1.
$$
\frac{3 \sqrt{15}-\sqrt{10}-2 \sqrt{6}+3 \sqrt{3}-\sqrt{2}+18}{\sqrt{5}+2 \sqrt{3}+1}
$$
$=$ | $\begin{array}{l}=1 .3 \sqrt{3}-\sqrt{2} \text {. } \\ \text { Solution 1: Expression }=\frac{3 \sqrt{3} \cdot \sqrt{5}-\sqrt{2} \cdot \sqrt{5}-2 \sqrt{2} \cdot \sqrt{3}+3 \sqrt{3}-\sqrt{2}+3 \sqrt{3} \cdot 2 \sqrt{3}}{\sqrt{5}+2 \sqrt{3}+1} \\ =\frac{3 \sqrt{3}(\sqrt{5}+2 \sqrt{3}+1)-\sqrt{2}(\sqrt{5}+2 \sqrt{3}+1)}{\... | 3 \sqrt{3}-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,766 |
3: In the sequence of natural numbers $1,2,3, \cdots, n, \cdots$, the sum of numbers from the $m$-th $(m$ $<n$ ) to the $n$-th number is 1997. Then $m$ equals $\mathrm{f}$ | 3. 998 .
$(m+n)(n-m+1)=2 \cdot 1997, 1997$ is a prime number.
(1) $\left\{\begin{array}{l}m+n=1997, \\ n-m+1=2 ;\end{array}\right.$
(2) $\left\{\begin{array}{l}m+n=2 \cdot 1997, \\ n-m+1=1\end{array}\right.$
| (1), $n=999, m=998$.
| (2), $m=n=1997$ (discard). | 998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,768 |
4. Given that the radius of $\odot O$ is $R$, $C, D$ are two points on the circumference of the circle on the same side of the diameter $A B$, the degree measure of $\overparen{A C}$ is $96^{\circ}$. The degree measure of $\overparen{B D}$ is $36^{\circ}$, and a moving point $P$ is on $A B$. Then the minimum value of $... | 4. $\sqrt{3} R$.
If $D^{\prime}$ is the point symmetric to $D$ with respect to line $A B$, and $C D^{\prime}$ intersects $A B$ at point $P$, then point $P$ minimizes $(P+P D)$ (proof omitted).
Given $36^{\circ}$, thus the measure of $\overparen{C D}$ is $180^{\circ}-96^{\circ}-36^{\circ}=48^{\circ}$. Therefore, the m... | \sqrt{3} R | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,769 |
One, (20 points) Solve the equation:
$$
\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{a}(a)
$$
$0)$.
Discuss the solution of this equation for the value of the positive number $a$.
| Given $\sqrt{x-\sqrt{2 x-1}}=\frac{|x-1|}{\sqrt{x+\sqrt{2 x-1}}}$, squaring both sides yields
$$
2 x+2|x-1|=a \text {. }
$$
When $x>1$, the solution is $x=\frac{2+a}{4}$.
When $x \leqslant 1$, we get $2=a$, i.e., when $a=2$, $\frac{1}{2} \leqslant x \leqslant 1$.
Conclusion (1) When $a>2$, there is one root $x=\frac{2... | (1) \text{ When } a>2, x=\frac{2+a}{4}; (2) \text{ When } a=2, \frac{1}{2} \leqslant x \leqslant 1; (3) \text{ When } a<2, \text{ no solution} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,770 |
II. (This question is worth 20 points) As shown in the figure, in $\triangle ABC$, $TD \perp AB$, and $TE \perp AC$. Prove:
$$
\text{(1) } \angle AHD =
$$
$\angle AHE$;
(2) $\frac{BH}{BD} = \frac{CH}{CE}$. | Ni, (1) $\because T D \perp A B, T E \perp A C, A H \perp B C$,
$\therefore$ points $D, E, H$ are all on the circle with $A T$ as the diameter:
Therefore, $\angle A H D=\angle A T D, \angle A H E=\angle A T E$.
Also, $\because A T$ is the angle bisector, it is easy to know that $\angle A T D=\angle A T E$,
$\therefore ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 709,771 |
Equal to their greatest common divisor and least common multiple profit. Find such natural numbers.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let the required natural numbers be $x$ and $y$, and let $m$ be their greatest common divisor. We can set $x = ma$ and $y = mb$, where $m, a, b$ are natural numbers and $a, b$ are coprime.
Then their least common multiple must be $mab$. Therefore, according to the problem statement, we have
$$
ma \times mb - (ma + mb)... | 4 \text{ and } 6, \text{ or } 3 \text{ and } 6, \text{ or } 4 \text{ and } 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 709,772 |
3. For $a b \neq 0, a^{2} \neq b^{2}$, the minimum value of the quadratic function $y=(x-a)(x-b)$ is ( ).
(A) $\left(\frac{a+b}{2}\right)^{2}$
(B) $-\left(\frac{a+b}{2}\right)^{2}$
(C) $\left(\frac{a-b}{2}\right)^{2}$
(D) $-\left(\frac{a-b}{2}\right)^{2}$ | 3. 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 | 709,775 |
Example 9 On the square $A B C D$: there are 10 points, 8 of which are inside $\triangle A B C$, and 2 points are on the side $\mathrm{I}$ of the square (not at the vertices). [L These 10 points, together with points $A, B, C, D$, are not collinear. Find how many small triangles these 10 points, along with the 4 vertic... | Solution: Since adding a point on the plane is equivalent to adding a full angle, the total angle of the figure formed by 14 vertices is $360^{\circ} \times 14=5040^{\circ}$. We have $\}^{\circ}, B, C, D$ as the vertices of the square, so $\angle A+\angle B+\angle C+\angle D=$ $90^{\circ} \times 4=360^{\circ}$. Two of ... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,776 |
4. As shown in the figure, the diagonals $A C$ and $B D$ of rhombus $A B C D$ intersect at $O$, $\angle A B C \neq 90^{\circ}$. Then the number of pairs of congruent triangles in the figure is ( ).
(A) 4 pairs
(B) 6 pairs
(C) 8 pairs
(D) 12 pairs | 4. C
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 | 709,777 |
7. As shown in the figure, in the isosceles right $\triangle ABC$, $AD$ is the altitude on the hypotenuse. Two perpendicular rays are drawn from point $D$ intersecting the two legs at $E$ and $F$. Connecting $EF$ intersects $AD$ at $G$. The relationship between $\angle AED$ and $\angle AGF$ is ( ).
(A) $\angle AED > \a... | 7. 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 | 709,780 |
9. As shown in the figure, there is a point $P$ $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ inside a circle with the center at the origin and a radius of 2. A chord $A B$ is drawn through $P$, and the minor arc $A B$ forms a lens shape. The minimum value of the area of this lens shape is ( . .
(A) $\pi-1$
(B)... | 9. D
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 | 709,782 |
12. The minimum value of the function $y=x^{4}+4 x+4$ is ( ).
(A) does not exist
(B) 0
(C) 1
(D) 2 | 1. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 709,785 |
6. The coefficients of the expansion of $(a+b)^{n}$, when $n=1,2,3, \cdots$, can be arranged in the form of "Pascal's Triangle" (as shown below). Please use "Pascal's Triangle" to find the value of $1.01^{9}$ $\qquad$ . (accurate to three decimal places) | $6.1 .01^{9}=1.093685272684360901 \approx 1.096$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.1 .01^{9}=1.093685272684360901 \approx 1.096$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,795 |
7. As shown in the left figure, in the right-angled sector $O A B$, $O A=O B=1$. Take any point $C$ on $\overparen{A B}$, and draw $C D \perp O B$ at $D$. Then the maximum value of $O D+$ $D C$ is $\qquad$ . | 7. $\sqrt{2}$ | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,796 |
8. Liu Shang and Luo In $\triangle A B C$, take a point $D, E, F$ on the three sides $A B, B C, C A$ respectively, such that $\frac{A D}{D B}=\frac{B E}{E C}=\frac{C F}{F A}=r>0$. Then the ratio of the area of $\triangle D E F$ to the area of $\triangle A B C$ is . $\qquad$ | 8. $\frac{1+2 r-2 r^{2}}{1+2 r+r^{2}}$ | \frac{1+2 r-2 r^{2}}{1+2 r+r^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,797 |
Example 1 Let $y=a x^{2}+b x+c$. Given: when $x=1$, $y=0$; when $x=-1$, $y$ is an even number. Prove: the two roots of the equation $x^{2}-$ $(a+b+c) x+b a+b c=0$ are integer roots. (1993, Sichuan Province Junior High School Mathematics Competition) | Prove: The equation $x^{2}-(a+b+c) x+b a+b c$ $=0$ can be factored into $(x-b)(x-a-c)=0$.
Thus, $x_{1}=b, x_{2}=a+c$.
From $x=1, y=0$ we get $a+b+c=0$;
From $x=-1, y=2 n$ (where $n$ is an integer) we get $a-b+c=2 n$.
Solving these, we get $b=-n, a+c=n$, both of which are integers.
Therefore, the two roots of the equati... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 709,798 |
For what integer $2 m$ does the equation $\left(m^{2}-1\right) x^{2}$ $-6(3 m-1) x+72=0$ have two distinct positive integer roots? | Solution: Clearly, $m \neq \pm 1$. The original equation can be factored as
$$
[(m-1) x-6][(m+1) x-12]=0 \text {. }
$$
Thus, $x_{1}=\frac{6}{m-1}, x_{2}=\frac{12}{m+1}$.
Since $x_{1}$ and $x_{2}$ are positive integers,
$$
\therefore m-1=1,2,3,6 \text { and } m+1=1,2,3,4 \text {, }
$$
6,12.
Solving, we get $m=2$ or $m=... | m=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,799 |
Evaluate the equation
$$
x^{2}+10 a x+5 b+3=0
$$
and $x^{2}+10 a x+5 b-3=0$
both do not have integer roots.
(1978, Northern Vietnam Student Mathematics Competition) | Proof: (Proof by Contradiction) Let $\alpha$ be an integer root of the original equation. Then $a^{2}+10 a \alpha+5 b \pm 3=0$ can be rearranged to $\alpha^{2}=-5(2 a \alpha+b) \mp 3$.
However, the units digit of the left side can only be $0,1,4,5,6$, $\neq$ the right side.
The assumption is false, hence the original ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,800 |
2. Given tetrahedron $ABCD$ with volume $V, E$ is a point on edge $AD$, extend $AB$ to $F$ such that $BF=AB$, let the plane through points $C, E, F$ intersect $BD$ at $G$. Then the volume of tetrahedron $CDGE$ is $\qquad$ . | 2. $\frac{1}{3} V$.
In $\triangle A F D$, $E$ is the midpoint of $A D$, and $B$ is the midpoint of $A F$.
$$
\begin{array}{l}
\therefore D G=\frac{2}{3} B D . \\
\therefore S_{Z_{2} E D G}=-\frac{1}{3} S_{\triangle A D B} .
\end{array}
$$
$\because$ The pyramids $C-E D G$ and $C-A D B$ have the same height,
$$
\theref... | \frac{1}{3} V | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,801 |
3. Satisfy $\sin (x+\sin x)=\cos (x-\cos x)$ then $x=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $x=\frac{\pi}{4}$.
The original equation is equivalent to
$$
\cos \left(\frac{\pi}{2}-x-\sin x\right)=\cos (x-\cos x).
$$
Then $x-\cos x=2 k \pi+\left(\frac{\pi}{2}-x-\sin x\right)(k \in \mathbb{Z})$ or $x-\cos x=2 k \pi-\left(\frac{\pi}{2}-x-\sin x\right)(k \in \mathbb{Z})$.
From (1) we get $2 x+\sin x-\cos x=2 k... | x=\frac{\pi}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,802 |
4. Let $S_{n}$ be the sum of the products of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \frac{1}{4}, \cdots, \frac{1}{2^{n-1}}\right\}$, and $\lim _{n \rightarrow \infty} \frac{4 S_{n}}{n^{2}}=a$. Then the line described by the polar equation $\rho=\frac{1}{2-a \cos \theta}$ is $\qquad$ ... | 4. The right branch of a hyperbola.
It is known that $S_{n}=C_{n-1}^{2}\left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right)$
$$
\begin{aligned}
& =(n-1)(n-2)\left(1-\frac{1}{2^{n}}\right) . \\
\therefore a & =\lim _{n \rightarrow \infty} \frac{4 S_{n}}{n^{2}}=4 \lim _{n \rightarrow \infty} \frac{(n-1)(n-2)\... | \frac{\frac{1}{2}}{1-2 \cos \theta} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 709,803 |
$B C=a, C A=b, A B=c$. Then $y=\frac{c}{a+b}+\frac{b}{c}$ 的最小值 is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$B C=a, C A=b, A B=c$. Then $y=\frac{c}{a+b}+\frac{b}{c}$ the minimum value is | 5. $\sqrt{2}-\frac{1}{2}$.
Given $b+c \geqslant a$.
$$
\begin{aligned}
\therefore y & =\frac{c}{a+b}+\frac{b}{c}=\frac{c}{a+b}+\frac{b+c}{c}-1 \\
& \geqslant \frac{c}{a+b}+\frac{a+b+c}{2 c}-1 \\
& =\frac{c}{a+b}+\frac{a+b}{2 c}-\frac{1}{2} \\
& \geqslant 2 \sqrt{\frac{c}{a+b} \cdot \frac{a+b}{2 c}}-\frac{1}{2}=\sqrt{2... | \sqrt{2}-\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 709,804 |
6. Then, in: the snake $A_{1} A_{2} B_{2} B_{1}$ is small, and it is known that $A_{1} A_{2} = 2a, A_{1} B_{2} < \sqrt{2} a$. Taking $A_{1} A_{2}$ as the major axis, construct an ellipse $C$ such that the focal distance of $C$ is $\mathrm{f} = \sqrt{2} A_{1} B_{1}$. On $C$, take a point $P$ (not coinciding with the end... | 6. $4 a^{2}$.
Let the line through $A_{1}, A_{2}$ be the $x$-axis, and the perpendicular bisector of $A_{1} A_{2}$ be the $y$-axis to establish a rectangular coordinate system. Suppose $A_{1} B_{1}=2 b$, then the equation of the ellipse $C$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{2 b^{2}}=1$.
Let $P(a \cos \theta, \sqrt... | 4 a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 709,805 |
Three, (Total 20 points) Let $n$ be an integer. Find the value of:
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}-\frac{1}{2}.
$$ | $$
\begin{aligned}
\equiv, & \because 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n} \\
= & \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2 n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right. \\
& \left.+\cdots+\frac{1}{2 n}\right) \\
= & \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\fra... | \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}<\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,806 |
Four, (Full marks 20 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}=$ $2 \sqrt{3}, a_{n-1}=\frac{8 a_{n}}{4-a_{n}^{2}}(n \geqslant 2)$. Find the expression for $a_{n}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text { Let } \frac{a_{n}}{2}=\tan \theta_{n}^{\prime}\left(-\frac{\pi}{2}<\theta_{n}<\frac{\pi}{2}\right), \text { then } \tan \theta_{n-1}=\tan 2 \theta_{n} . \\
\therefore \theta_{n-1}=2 \theta_{n}, \text { and } \theta_{1}=\arctan \sqrt{3}=\frac{\pi}{3}, \\
\therefore \theta_{n}=\frac{\pi}{3} \c... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 709,807 |
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