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In an acute triangle $\triangle ABC$, let $R$ and $r$ be the radii of its circumcircle and incircle, respectively, and $s$ be its semiperimeter. $\sum$ denotes the cyclic sum.
Reference [1] strengthens the Child inequality from Reference [2]
$$
\sum \sec B \cdot \sec C \geqslant 12
$$
to
$$
\sum \sec B \cdot \sec C \g... | Proof: From the three-color path equation ${ }^{[2]}$
$$
\begin{array}{l}
\sum \cos A=\frac{R+r}{R}, \\
\cos A \cdot \cos B \cdot \cos C=\frac{s^{2}-4 R^{2}-4 R r-r^{2}}{4 R^{2}}
\end{array}
$$
and the Gerretsen inequality ${ }^{[2]}$
$$
s^{2} \leqslant 4 R^{2}+4 R r+3 r^{2}
$$
we get
$$
\begin{array}{l}
\sum \sec B ... | \frac{2 R(R+r)}{r^2} | Inequalities | proof | Yes | Yes | cn_contest | false | 711,390 |
In 1987, D. M. Milosevic proposed and proved the following inequality:
For a triangle $\triangle A B C$ with side lengths $a, b, c$, and corresponding heights $R, r$, we have
$$
\bar{h}_{b}+\frac{a}{h_{c}}+\frac{b}{h_{c}+h_{a}}+\frac{c}{h_{a}+h_{b}} \geqslant \frac{9 R \sqrt{3}}{2(4 R+r)} .
$$
In [1], Milosevic's ine... | Proof: From $h_{a}=\frac{2 \triangle}{a}$ and $a b c=s r$ (where $\triangle$ is the area, $s$ is the semi-perimeter), the left side of equation (2) is
$$
\begin{array}{l}
\frac{a}{h_{b}+h_{c}} \cdot \frac{b}{h_{c}+h_{a}}+\frac{c}{h_{a}+h_{b}} \\
=\frac{(a b c)^{3}}{8 \triangle^{3}(b+c)(c+a)(a+b)} \\
=\frac{s^{3} r^{3}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,391 |
Example 6 Given three real numbers $x_{1}, x_{2}, x_{3}$, any one of these numbers plus five times the product of the other two always equals 6. The number of such triples $\left(x_{1}, x_{2}, x_{3}\right)$ is $\qquad$.
$(1995$, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition) | Solving the system of equations given by the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{3} x_{1}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.
$$
(1) - (2) gives
$$
\left(x_{1}-x_{2}\right)\left(1-5 x_{3}\right)=0 \text {. }
$$
(2) - (3) gives
$$
\left(x_{2}-x_{3}\right)\left(1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,392 |
1. The value of $\sqrt{14+6 \sqrt{5}}-\sqrt{14-6 \sqrt{5}}$ is ( ).
(A) 1
(B) $\sqrt{5}$
(C) $2 \sqrt{5}$
(D) 5 | $-、 1$. (C).
$$
\begin{array}{l}
\because 14+6 \sqrt{5}=(3+\sqrt{5})^{2}, 14-6 \sqrt{5}=(3-\sqrt{5})^{2}, \\
\begin{aligned}
\therefore \text { Original expression } & =\sqrt{(3+\sqrt{5})^{2}}-\sqrt{(3-\sqrt{5})^{2}} \\
& =(3+\sqrt{5})-(3-\sqrt{5})=2 \sqrt{5} .
\end{aligned}
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,394 |
2. If $\frac{x}{3 y}=\frac{y}{2 x-5 y}=\frac{6 x-15 y}{x}$, then the value of $\frac{4 x^{2}-5 x y+6 y^{2}}{x^{2}-2 x y+3 y^{2}}$ is ( ).
(A) $\frac{9}{2}$.
(B) $\frac{9}{4}$
(C) 5
(D) 6 | 2. (A).
From the given conditions, we know that $x \neq 0, y \neq 0$. Transform the given equation and use the ratio to eliminate $y$, we get
$$
\text { - } \begin{aligned}
\frac{25 x}{75 y} & =\frac{15 y}{30 x-75 y}=\frac{6 x-15 y}{x} \\
& =\frac{25 x+15 y+16 x-15}{75}+\frac{1}{x} \\
& =\frac{31 x}{31 x}=1 .
\end{ali... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,395 |
3. Let $a, b$ be any distinct positive numbers, and $x=\frac{b^{2}+1}{a}, y=\frac{a^{2}+1}{b}$. Then $x, y$ must ( ).
(A) both not be greater than 2
(B) both not be less than 2
(C) at least one is greater than 2
(D) at least one is less than 2 | 3. (C).
Let $a=1, b=3$, then $x=10, y=\frac{2}{3}$. This negates (A) and (B). Let $a=3, b=4$, then $x=\frac{17}{3}, y=\frac{10}{4}$. This negates (D), hence only (C) is correct. | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 711,396 |
4. The positive integer $n$ is less than 100, and satisfies the equation $\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=n$, where $[x]$ denotes the greatest integer not exceeding $x$. How many such positive integers $n$ are there?
(A)2
(B) 3
(C) 12
(D) 16 | 4. (D).
From $\frac{n}{2}+\frac{n}{3}+\frac{n}{6}=n$, and if $x$ is not an integer, then $[x]<x$, we know that $2|n, 3| n, 6 \mid n$, which means $n$ is a multiple of 6. Therefore, the number of such positive integers less than 100 is $\left[\frac{100}{6}\right]=16$. | 16 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,397 |
5. Given the lengths of the four sides of a trapezoid are $1, 2, 3, 4$. Then the area of this trapezoid is ( ).
(A) 4
(B) 6
(C) $8 \sqrt{2}$
(D) $\frac{10}{3} \sqrt{2}$ | 5. (D).
Using $1, 2, 3, 4$ as sides to form a trapezoid, there are only six possible combinations:
(1) with 1, 2 as the bases; (2) with 1, 3 as the bases; (3) with 1, 4 as the bases; (4) with 2, 3 as the bases; (5) with 2, 4 as the bases; (6) with 3, 4 as the bases.
It is easy to see that only (3) can form a trapezoi... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,398 |
6. Given that $A B C D$ is a cyclic quadrilateral inscribed in a circle with radius $R$, $A B=12, C D=6$, and the extensions of $A B$ and $D C$ intersect at $P$, with $B P=8$ and $\angle A P D=60^{\circ}$. Then $R$ equals ( . .
(A) 10
(B) $2 \sqrt{21}$
(C) $12 \sqrt{2}$
(D) 14 | 6. (R).
By the secant theorem, we have
$$
P E \cdot P A = P C \cdot P D,
$$
so $8 \times 20$
$$
= P C (P C + 6).
$$
Solving this, we get $P C = 10$.
As shown in Figure 3, connect AC. In
$\triangle P A C$, since $P A =$
$2 P C$ and $\angle A P C = 60^{\circ}$, we have
$\angle P C A = 90^{\circ}$. Therefore, $A D$ is ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,399 |
2. A fruit store is conducting a promotional sale, with the following combinations: Combination A: 2 kg of fruit $A$, 4 kg of fruit $B$; Combination B: 3 kg of fruit $A$, 8 kg of fruit $B$, 1 kg of fruit $C$; Combination C: 2 kg of fruit $A$, 6 kg of fruit $B$, 1 kg of fruit $C$. It is known that fruit $A$ costs 2 yuan... | 2.150.
Let the number of sets of fruit A, B, and C sold on that day be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
2(2 x+3 y+2 z)=116, \\
8.8 x+25.6 y+21.2 z=441.2,
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
2 x+3 y+2 z=58, \\
22 x+64 y+53 ... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,401 |
3. Real numbers $x, y$ satisfy $x \geqslant y \geqslant 1$ and $2 x^{2}-x y$ $-5 x+y+4=0$. Then $x+y=$ $\qquad$ | 3.4 .
From the given equation, we know that $2 x^{2}-5 x+4=y(x-1) \leqslant x(x-1)$, which leads to $x^{2}-4 x+4 \leqslant 0$, or $(x-2)^{2} \leqslant 0$. Therefore, $x=2$.
Substituting $x=2$ into the given equation yields $y=2$, hence $x+y=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,402 |
4. Let the side length of the equilateral $\triangle ABC$ be $2, M$ is the midpoint of side $AB$, $P$ is any point on side $BC$, and $PA + PM$ are denoted as $s$ and $t$ for their maximum and minimum values, respectively. Then $s^{2}-t^{2}=$ $\qquad$. | $4.4 \sqrt{3}$.
First, find $s$.
$$
\begin{array}{l}
\because P A \leqslant A C, P M \leqslant C M, \\
\therefore P A+P M \leqslant C A+C M=2+\sqrt{3} .
\end{array}
$$
When point $P$ is at vertex $C$, the equality holds.
Thus, $s=2+\sqrt{3}$.
Next, find $t$. As shown in Figure 4, construct the equilateral $\triangle A... | 4 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,404 |
Three. (Full marks 20 points) Let $p$ be a real number, and the graph of the quadratic function $y=x^{2}-2 p x-p$ intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
(1) Prove that $2 p x_{1}+x_{2}^{2}+3 p>0$;
(2) If the distance between points $A$ and $B$ does not exceed $|... | $$
\begin{array}{l}
\Delta=(-2 p)^{2}-4(-p)=4 p^{2}+4 p>0, \\
x_{2}^{2}-2 p x_{2}-p=0 . \\
\begin{array}{l}
\therefore 2 p x_{1}+x_{2}^{2}+3 p \\
\quad=2 p x_{1}+2 p x_{2}+p+3 p \\
\quad= 2 p\left(x_{1}+x_{2}\right)+4 p \\
\quad=4 p^{2}+4 p>0 .
\end{array}
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } \because A B... | \frac{9}{16} | Algebra | proof | Yes | Yes | cn_contest | false | 711,405 |
Four, (Full score 25 points) As shown in Figure 1, EFGH is an inscribed quadrilateral of square $ABCD$, $\angle BEG$ and $\angle CFH$ are both acute angles. Given that $EG=3, FH=4$, and the area of quadrilateral $EFGH$ is 5. Find the area of square $ABCD$.
保留源文本的换行和格式，直接输出翻译结果如下：
```
Four, (Full score 25 points) As s... | In square $ABCD$, perpendicular lines are drawn from $E, F, G, H$ to the opposite sides, forming rectangle $PQRT$. Let the side length of $ABCD$ be $a$, $PQ=b$, and $QR=c$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
b=\sqrt{3^{2}-a^{2}}, \\
c=\sqrt{4^{2}-a^{2}} .
\end{array}
$$
From $S_{\triangle AEH}=S_{... | \frac{44}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,406 |
Five. (Full marks 25 points) Let the quadratic equation in $x$ be $\left(k^{2}-6 k+8\right) x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}=4$ with both roots being integers. Find all real values of $k$ that satisfy the condition.
---
Translate the above text into English, please retain the original text's line breaks and f... | $$
\begin{array}{l}
(k-4)(k-2) x^{2}+\left(2 k^{2}-6 k-4\right) x \\
+(k-2)(k+2)=0, \\
(k-4) x+(k-2)][(k-2) x+(k+2)]=0, \\
\because(k-4)(k-2) \neq 0, \\
\therefore x_{1}=-\frac{k-2}{k-4}=-1-\frac{2}{k-4}, \\
x_{2}=-\frac{k+2}{k-2}=-1-\frac{4}{k-2} . \\
\therefore k-4=-\frac{2}{x_{1}+1}, k-2=-\frac{2}{x_{2}+1} . \\
\lef... | k=6,3, \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,407 |
1. Does there exist an odd number of 2000 digits, which is a multiple of the integer
$(1999$, Ukrainian Mathematical Olympiad) | Sol: If a 2000-digit integer is entirely composed of 5s, it is clearly not a perfect square.
If 1999 digits are "5" and only one digit is not 5, let the 2000-digit integer be $n$.
(i) If the non-5 digit is not the unit digit, then there exists $k$ (an odd number) such that
$$
n=(5 k)^{2}=25 k^{2}=25(4 m+1)=100 m+25 .
... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,408 |
3. Find all positive integer solutions to $(x+1)^{y+1}+1=(x+2)^{z+1}$.
(1999, Taiwan Mathematical Olympiad) | Let $x+1=a, y+1=b, z+1=c$. Then $a, b, c \geqslant 2$ and $a, b, c \in \mathbf{N}$. The original equation becomes
$a^{b}+1=(a+1)^{c}$.
For equation (1), take $\bmod (a+1)$, then we have $(-1)^{3}+1 \equiv 0(\bmod (a+1))$.
Thus, $b$ must be odd.
In fact, if $b$ is even, then $a=1$ leads to a contradiction.
Equation (1) ... | x=1, y=2, z=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,410 |
5. Let $p$ be a prime, $p>2$, and let $a, b, c, d$ be integers not divisible by $p$. Suppose that for any integer $r$ not divisible by $p$, we have
$$
\left\{\frac{r a}{p}\right\}+\left\{\frac{r b}{p}\right\}+\left\{\frac{r c}{p}\right\}+\left\{\frac{r d}{p}\right\}=2 \text {. }
$$
Prove that at least two of the six n... | Proof: Since $p \backslash d$, then $0, d, 2d, \cdots, (p-1)d$ form a complete residue system modulo $p$. Among these, there must be an $R \in \{1, 2, \cdots, p-1\}$ such that $Rd \equiv p-1 \pmod{p}$. Let
$$
\begin{array}{l}
Ra \equiv A \pmod{p}, \quad Rb \equiv B \pmod{p}, \\
Rc \equiv C \pmod{p}, \quad Ra' \equiv C ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,412 |
1 . Find real numbers $x, y$ such that
$$
\left\{\begin{array}{l}
4^{-x}+27^{-y}=\frac{5}{6}, \\
\log _{27} y-\log _{4} x \geqslant \frac{1}{6}, \\
27^{y}-4^{x} \leqslant 1
\end{array}\right.
$$
holds.
(1999, Romanian Mathematical Olympiad) | Solution: Let $4^{x}=a>0, 27^{y}=b>0$. Then
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=\frac{5}{6} \\
b-a \leqslant 1
\end{array}
$$
Substituting $b \leqslant a+1$ into (1), we have
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{a+1} \leqslant \frac{5}{6} \\
5 a^{2}-7 a-6 \geqslant 0
\end{array}
$$
Solving this, we get $a... | x=\frac{1}{2}, y=\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,413 |
Example 8 Solve the equation
$$
\frac{13 x-x^{2}}{x+1}\left(x+\frac{13-x}{x+1}\right)=42 \text {. }
$$
(9th Zu Chongzhi Cup Mathematics Invitational Competition) | Let $y=\frac{13-x}{x+1}$, then $x+y=\frac{13+x^{2}}{x+1}$, $x y=\frac{13 x-x^{2}}{x+1}$. Thus, the original equation can be transformed into the system of equations
$$
\left\{\begin{array}{l}
x y(x+y)=42, \\
x y+(x+y)=13 .
\end{array}\right.
$$
Let $x+y=u, x y=v$, then the above system of equations becomes
$$
\left\{\... | x_{1}=1, x_{2}=6, x_{3}=3+\sqrt{2}, x_{4}=3-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,414 |
2. Solve the system of equations
$$
\left\{\begin{array}{l}
\left(1+4^{2 x-y}\right) 5^{1-2 x+y}=1+2^{2 x-y+1}, \\
y^{3}+4 x+1+\ln \left(y^{2}+2 x\right)=0 .
\end{array}\right.
$$
(1999, Vietnam Mathematical Olympiad) | Solution: In (1), let $u=2 x-y$. Then
$$
f(u)=5 \times\left(\frac{1}{5}\right)^{u}+5 \times\left(\frac{4}{5}\right)^{u}-2 \times 2^{u}-1=0 \text {. }
$$
Since $f(u)$ is monotonically decreasing on $\mathbf{R}$, and $f(1)=0$, we have $u=1$, i.e., $2 x=y+1$. Substituting into equation (2), we get
$$
\begin{aligned}
g(y)... | x=0, y=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,415 |
3. There is a sequence, the first term is a positive integer, Tang constructs all the even terms, Xia constructs the subsequent odd terms. Tang's construction method is to subtract any digit of the previous term from the previous term; Xia's construction method is to add any digit of the previous term to the previous t... | Proof: If the first term is a single-digit number, then all subsequent terms are 0, which is obviously true.
If the first term is an $n$-digit number, we will prove that no term in the sequence can carry over.
If a carry over can occur, it must happen at an odd term, let this term be the $(2k+1)$-th term. Then the $(2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,416 |
5. Given the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots, n \geqslant 1, a_{i} \in \mathbf{N}$, and $a_{a_{n}}+a_{n}=2 n$. Prove: $a_{n}=n$.
(1999, Ukrainian Mathematical Olympiad) | Proof: Let $a_{1}=t \geqslant 1$, then $a_{t}+t=2$. Since $a_{t} \geqslant 1$, it follows that $t=1, a_{1}=1$.
If for $n \leqslant k$, we have $a_{n}=n$. For $n=k+1$, let $a_{k+1}=t$, then $a_{t}+t=2(k+1)$.
If $t \leqslant k$, then $a_{t}+t=2 t \leqslant 2 k<2(k+1)$, so $k+1 \leqslant t \leqslant 2 k+1, t \in \mathbf... | a_{n}=n | Algebra | proof | Yes | Yes | cn_contest | false | 711,418 |
6. For all real numbers $x$ satisfying $|x| \neq 1$, the function $f(x)$ satisfies
$$
f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x .
$$
Find all possible $f(x)$.
(12th Korean Mathematical Olympiad) | Solve: Let $x$ be $\frac{x-3}{x+1}$ and substitute into the original equation, we get
$$
f\left(\frac{3+x}{1-f}\right)+f(x)=\frac{x-3}{x+1} \text {. }
$$
Let $x$ be $\frac{3+x}{1-x}$ and substitute into the original equation, we have
$$
f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{3+x}{1-x} \text {. }
$$
(1) + (2), and su... | f(x)=\frac{x^{3}+7 x}{2-2 x^{2}},|x| \neq 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,419 |
7. Let $T$ be the set of all non-negative integers not exceeding 1999, and $N$ be the set of all non-negative integers. Find all functions $f: N \rightarrow T$ that satisfy the following conditions:
(i) For all $t \in T$, $f(t)=t$;
(ii) For all $m, n \in N$,
$$
f(m+n)=f(f(m)+f(n)).
$$
(1999, Vietnam Mathematical Olympi... | Solution: Let $f(2000)=u, u \in T$, and denote $p=2000-u$. For all $k \geqslant 2000$, we prove by induction that
$$
f(k)=\max \{l \in T \mid k \equiv l(\bmod p)\} .
$$
First, when $k=2000$, $2000=u(\bmod p)$ and $u+p>\max T$, i.e., $2000 \notin T$, so
$f(2000)=u=\max \{l \in T \mid l \equiv 2000(\bmod p)\}$.
Thus, eq... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,420 |
8. Let the function $f(x)$ satisfy for all rational numbers $m, n$,
$$
|f(m+n)-f(m)| \leqslant \frac{n}{m}.
$$
Prove: For all positive integers $k$, we have
$$
\sum_{i=1}^{k}\left|f\left(2^{k}\right)-f\left(2^{i}\right)\right| \leqslant \frac{k(k-1)}{2} \text {. }
$$
(12th Korean Mathematical Olympiad) | Proof: Induction on $k$.
When $k=1$, left $=0=$ right.
When $k=2$,
$$
\begin{aligned}
\text { left } & =|f(4)-f(2)| \\
& =|f(2+2)-f(2)| \leq 1=\text { right }
\end{aligned}
$$
$$
\begin{array}{l}
\left|f\left(2^{n}\right)-f(2)\right|+\left|f\left(2^{n}\right)-f(4)\right|+\cdots \\
+\left|f\left(2^{n}\right)-f\left(2^{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,421 |
9. Let $a, b, c$ be positive real numbers, and $a b c + a + c = b$. Determine the maximum value of
$$
p=\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1}
$$
(1999, Vietnam Mathematical Olympiad) | Given the condition $a+c=(1-a c) b$, it is clear that $1-a c \neq 0$, hence $b=\frac{a+c}{1-a c}$.
Let $\alpha=\arctan a, \beta=\arctan b, \gamma=\arctan c, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then
$$
\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \tan \gamma}=\tan (\alpha+\gamma).
... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,422 |
10. Let $a_{1}, a_{2}, \cdots, a_{1999}$ be non-negative real numbers, and satisfy:
(1) $a_{1}+a_{2}+\cdots+a_{1999}=2$;
(2) $a_{1} a_{2}+a_{2} a_{5}+\cdots+a_{1998} a_{1999}+a_{1999} a_{1}=1$.
Let $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{1999}^{2}$, find the maximum and minimum values of $S$.
(12th Korean Mathematical Olympi... | Solution: For $a_{1}+a_{2}+\cdots+a_{1999}=2$, the following calculation is made:
$$
\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{1999}\right) \\
=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{1998} a_{1999}+a_{1999} a_{1}
\end{array}
$$
Find the maximum value.
Lemma: There exists an $i \in\{1,2, \cdots, 1999\}$, such that $a... | S_{\min }=1 \frac{1}{2}, S_{\max }=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,423 |
Example 9 Solve the equation
$$
x=\left(x^{2}+3 x-2\right)^{2}+3\left(x^{2}+3 x-2\right)-2
$$
(1996, Sichuan Province Junior High School: Mathematics Competition) | Let $y=x^{2}+3 x-2$, then the original equation can be transformed into the system of equations
$$
\left\{\begin{array}{l}
x=y^{2}+3 y-2, \\
y=x^{2}+3 x-2 .
\end{array}\right.
$$
(1) - (2), factorizing, we get
$$
\begin{array}{l}
(x-y)(x+y+4)=0 . \\
\therefore x-y=0 \text { or } x+y+4=0 .
\end{array}
$$
From $x-y=0$ w... | x_{1}=-1+\sqrt{3}, x_{2}=-1-\sqrt{3}, x_{3}=-2+\sqrt{2}, x_{4}=-2-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,425 |
2. Given that $n^{2}+5 n+13$ is: a perfect square. Then the natural number $n$ ( )
(A) does not exist
(B) has only one
(C) has more than one, but a finite number
(D) has infinitely many | 2. (B).
$$
\begin{array}{l}
\because(n+2)^{2}<n^{2}+5 n+13<(n+4)^{2}, \\
\therefore n^{2}+5 n+13=(n+3)^{2} .
\end{array}
$$
Solving gives $n=4$. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,428 |
3. If the equations $x^{2}+2 a x+b^{2}=0$ and $x^{2}+2 c x$ $-b^{2}=0$ have a common root, and $a, b, c$ are the sides of a triangle, then this triangle must be ( ).
(A) right triangle
(B) isosceles triangle
(C) equilateral triangle
(D) isosceles right triangle | 3. (A).
Let the common root of the two equations be $m$, then
$$
\begin{array}{l}
m^{2}+2 a m+b^{2}=0, \\
m^{2}+2 c m-b^{2}=0 .
\end{array}
$$
Subtracting the two equations gives
$$
(c-a) m=b^{2} \text {. }
$$
When $c=a$, $b=0$, which is not in line with the problem, so,
$$
c \neq a, m=\frac{b^{2}}{c-a} \text {. }
$... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,429 |
4. Let $n$ be a natural number, and $a_{n}=\sqrt[3]{n^{2}+2 n+1}$ $+\sqrt[3]{n^{2}-1}+\sqrt[3]{n^{2}-2 n+1}$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{3}}+\frac{1}{a_{5}}$ $+\cdots+\frac{1}{a_{997}}+\frac{1}{a_{999}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. (C).
$$
\begin{array}{l}
a_{n}=(\sqrt[3]{n+1})^{2}+\sqrt[3]{n+1} \cdot \sqrt[3]{n-1}+(\sqrt[3]{n-1})^{2}, \\
\frac{1}{a_{n}}=\frac{1}{2}(\sqrt[3]{n+1}-\sqrt[3]{n-1}) . \\
\therefore \sum_{i=1}^{300} \frac{1}{a_{2 i}-1}=\frac{1}{2}[(\sqrt[3]{2}-0)+(\sqrt[3]{4}-\sqrt[3]{2}) \\
\quad+\cdots+(\sqrt[3]{1000}-\sqrt[3]{998... | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,430 |
5. As shown in Figure 1, in the pentagon $A B C D E$ with all sides equal, $\angle A B C$ $=2 \angle D B E$. Then, $\angle A B C$ is ().
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$ | 5. (B).
Rotate $\triangle B A E$ clockwise around point $B$ by $\angle A B C$ to get $\triangle B C F$, as shown in Figure 4.
$$
\because \angle D B F=\angle D B C+\angle A B E=\frac{1}{2} \angle A B C=
$$
$\angle D B E$,
$\therefore \triangle D B E \cong$
$\triangle D B F(S A S)$.
Therefore, $\triangle C D F$ is an
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,431 |
6. Given that $n$ is an integer greater than 1, and $2^{2^{n}}+1$ is a prime number. Then $2^{2^{n}}-1$ must be ( ).
(A) a prime number
(B) a composite number with a factor of 3
(C) a composite number with a factor of 5
(D) a composite number with a factor of 15 | 6. (D).
$\because n$ is an integer greater than 1,
$$
\therefore N=2^{2^{n}}-1=16^{2^{n-2}}-1 \text {. }
$$
Therefore, the unit digit of $N$ is 5, so $5 \mid N$; $\because 2^{2^{n}}-1, 2^{2^{n}}, 2^{2^{n}}+1$ are three consecutive natural numbers, $\therefore$ one of them must be divisible by 3.
$\because 3 \times 2^{... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,432 |
2. Given $x=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then $x^{6}-2 \sqrt{2} x^{5}-x^{4}$ $+x^{3}-2 \sqrt{3} x^{2}+2 x-\sqrt{2}$ is $\qquad$. | 2. $\sqrt{3}$.
$$
\begin{array}{l}
\text { Given } x=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2} \text {, we get } x^{2}-2 \sqrt{2} x-1=0 \text {, } \\
x^{2}-2 \sqrt{3} x+1=0 \text {. } \\
\therefore \text { the original expression }=x^{4}\left(x^{2}-2 \sqrt{2} x-1\right)+x\left(x^{2}-2 \sqrt{3} x\right. \\
+1)+x-\sq... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,434 |
4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\frac{16}{3}$. Then the maximum value of $a$ is $\qquad$ . | 4. 2 .
Construct the function
$$
y=3 x^{2}-2(b+c+d) x+\left(b^{2}+c^{2}+d^{2}\right) \text {. }
$$
Since $y(x-b)^{2}+(x-c)^{2}+(x-d)^{2} \geqslant 0$, and the graph
is a parabola opening upwards, we have
$$
\Delta=4(b+c+d)^{2}-12\left(b^{2}+c^{2}+d^{2}\right) \leqslant 0,
$$
which simplifies to $(4-a)^{2}-3\left(\fr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,437 |
One. (20 points) Given the equation about $x$: $x^{3}-a x^{2}$ $-2 a x+a^{2}-1=0$ has only one real root. Find the range of real number $a$.
| Consider the original equation as a quadratic equation in $a$, i.e.,
$$
a^{2}-\left(x^{2}+2 x\right) a+x^{3}-1=0 \text {. }
$$
Factoring, we get
$$
[a-(x-1)]\left[a-\left(x^{2}+x+1\right)\right]=0 .
$$
Thus, $x=a+1$ or $x^{2}+x+1-a=0$.
Since $x=a+1$ is not a root of equation (1), the original equation has exactly one... | a<\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,438 |
II. (25 points) As shown in Figure 3, in the right triangle $\triangle ABC$, $\angle C=90^{\circ}, CD \perp AB$ at $D$, the incenter of $\triangle ADC$ and $\triangle CDB$ are $I_{1}$ and $I_{2}$ respectively, and $I_{1} I_{2}$ intersects $CD$ at $K$. Prove that: $\frac{1}{AC}+\frac{1}{BC}=\frac{1}{CK}$. | As shown in Figure 6, let $D I_{1}$ and $D I_{2}$ intersect $A C$ and $B C$ at $E$ and $F$, respectively, and connect $I_{1} C$ and $I_{2} B$. Clearly, $\triangle D I_{1} C \sim \triangle D I_{2} B$.
From this, we know
$\triangle D I_{1} I_{2} \sim \triangle D C B$.
Thus, $\angle D I_{1} K = \angle D C B = \angle A$.
$... | \frac{1}{AC} + \frac{1}{BC} = \frac{1}{CK} | Geometry | proof | Yes | Yes | cn_contest | false | 711,439 |
1. Given $x, y, z \in \mathbf{R}^{+}$ and satisfy $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is ( ).
(A) 2
(B) 3
(C) $\frac{4 \sqrt{3}}{3}$
(D) 1 | \begin{array}{l}-1 \cdot(\mathrm{A}) . \\ (x+y)(y+z)=x y+x z+y^{2}+y z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 \sqrt{\frac{1}{x z} \cdot x z} \\ =2\left(x, y, z \in \mathbf{R}^{+}\right) .\end{array} | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,441 |
2. For a finite sequence $P=\left(p_{1}, p_{2}, \cdots\right.$, $p_{n}$ ), the Cesàro sum of $P$ (Cesàro being a mathematician's name) is defined as $\frac{s_{1}+s_{2}+\cdots+s_{n}}{n}$, where $s_{k}=p_{1}+p_{2}+$ $\cdots+p_{k}(1 \leqslant k \leqslant n)$. If a 99-term sequence $\left(p_{1}\right.$, $p_{2}, \cdots, p_{... | 2. (B).
$$
\begin{array}{l}
\because \frac{s_{1}+s_{2}+\cdots+s_{99}}{99}=1000, \\
\therefore s_{1}+s_{2}+\cdots+s_{99}=99000 .
\end{array}
$$
Therefore, $\frac{1+\left(1+s_{1}\right)+\left(1+s_{2}\right)+\cdots+\left(1+s_{99}\right)}{100}$
$$
\begin{array}{l}
=\frac{100+\left(s_{1}+s_{2}+\cdots+s_{99}\right)}{100} \\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,442 |
3. Let $x^{2}-\frac{1}{x^{2}}$ and $x^{3}-\frac{1}{x^{3}}$ both be integers for the positive number $x(\quad$.
(A) does not exist
(B) there is only one
(C) there are only two
(D) there are infinitely many | 3. (B).
$\because x^{3}-\frac{1}{x}$ is an integer,
and $\left(x^{3}-\frac{1}{x^{3}}\right)^{2}=x^{6}-2+\frac{1}{x^{6}}$,
$\therefore x^{6}+\frac{1}{x^{6}}$ is also an integer.
$\because x^{2}-\frac{1}{x^{2}}$ is an integer,
and $\left(x^{2}-\frac{1}{x^{2}}\right)^{3}=x^{6}-3 x^{2}+\frac{3}{x^{2}}-\frac{1}{x^{6}}$
$$
=... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,443 |
4. In the permutation $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of $1,2,3,4,5$, the permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that satisfy $a_{1}a_{3}, a_{3}a_{5}$ are ( ) kinds.
(A) 8
(B) 16
(C) 24
(D) 10 | 4. (B).
In the permutations $a_{1}, a_{2}, \cdots, a_{5}$ that satisfy the conditions, 5 can only be in the second or fourth position, i.e., $5=a_{2}$ or $a_{4}$.
When $a_{2}=5$, $a_{1}a_{3}$ naturally satisfies the condition, $a_{1}$ can be any number from $1,2,3,4$, and $a_{3}, a_{4}, a_{5}$ are the remaining three... | 16 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 711,444 |
5. Put four small balls with a radius of 1 into a large ball, then the minimum radius of this large ball is ( ).
(A) $1+\frac{\sqrt{6}}{2}$
(B) $1+\frac{\sqrt{3}}{2}$
(C) $2+\sqrt{6}$
(D) $1+\frac{2 \sqrt{3}}{3}$ | 5. (A).
Let the radius of the large sphere be $R$, and the small spheres inside the large sphere must be tangent to each other in pairs. The centers of the small spheres form a regular tetrahedron with an edge length of 2, and the center of the tetrahedron $O$ is the center of the large sphere. Let the distance betwee... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,445 |
6. The function $F(x)=\cos ^{2} x+2 \sin 2 \cdot \cos x$ $-\sin ^{2} x+A x+B \mid$ has a maximum value $M$ on $0 \leqslant x \leqslant \frac{3 \pi}{2}$ that depends on the parameters $A$ and $B$. When $M$ is minimized, the values of $A$ and $B$ are ( . . .
(A) $A=0, B=1$
(B) $A=1, B=0$
(C) $A=B=0$
(D) $A=B=1$ | 6. (C).
When $A=B=0$, $F(x)=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right|$ has three points, $x=\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}$, in $\left[0, \frac{3 \pi}{2}\right]$, where $F(x)$ reaches its maximum value $\sqrt{2}$.
When $A$ and $B$ are not both zero, $\max _{1 \leqslant x \leqslant \frac... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,446 |
Example 1 Given $x+y=3, x^{2}+y^{2}-x y=4$. Then the value of $x^{4}+y^{4}+x^{3} y+x y^{3}$ is $\qquad$
(13th Junior High School Mathematics Competition of Jiangsu Province) | $$
\begin{array}{l}
\text { Solution: } x^{4}+y^{4}+x^{3} y+x y^{3} \\
=(x+y)^{2}\left(x^{2}+y^{2}-x y\right)
\end{array}
$$
Substitute $x+y=3, x^{2}+y^{2}-x y=4$, we get
$$
x^{4}+y^{4}+x^{3} y+x y^{3}=3^{2} \times 4=36 .
$$ | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,447 |
Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
$(1995$, National Junior High School Mathematics League) | Solution: According to the problem, we have $a^{2}+a=\frac{1}{4}$.
$$
\therefore \text { the original expression }=\frac{(a-1)\left(a^{2}+a+1\right)}{(a-1)\left(a^{2}+a\right)^{2}}=20 \text {. }
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,448 |
Example 11 Given $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$. Then $m^{5}+n^{5}=$ $\qquad$ .
(From the riverbed Jiangsu Province Junior High School Mathematics Competition) | Solution: Construct a quadratic equation $x^{2}-x-1=0$. From the given conditions, we know that $m$ and $n$ are the two roots of the equation $x^{2}-x-1=0$, so $m+n=1, m n=-1$. Then
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,449 |
Three, (Full marks 16 points) In the cyclic quadrilateral $ABCD$, $\angle A = 60^{\circ}, \angle B = 90^{\circ}, AD = 3, CD = 2$.
(1) Find the length of $BC$;
(2) Find the area of quadrilateral $ABCD$;
(3) Compare $AB + BC$ with $AD + CD$.
保留源文本的换行和格式，直接输出翻译结果。 | Three 1$) \because \angle B$ is a right angle, we know $AC$ is the diameter,
$$
\therefore \angle D=90^{\circ} \text {. }
$$
As shown in Figure 5, extend $AB$ and
$DC$ to intersect at $E$, with $\angle E=$ $30^{\circ}$, then
$$
\begin{array}{l}
AE=2AD=6, \\
DE=3\sqrt{3}, \\
EC=3\sqrt{3}-2 .
\end{array}
$$
Thus, $BC=\... | AB + BC < AD + CD | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,452 |
Four, (Full marks 16 points) Given that $x$ and $y$ are positive integers, and they satisfy the conditions $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation r... | $$
\begin{array}{l}
(x+1)(y+1)=72, \\
x y(x+y)=2^{4} \times 5 \times 11 .
\end{array}
$$
From equation (1), we know that at least one of $x$ and $y$ is odd. Without loss of generality, let $x$ be odd.
From equation (2), $x$ can only be 5, 11, or 55.
However, when $x=55$, $x+y \geqslant 56, x y(x+y) \geqslant 55 \times... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,453 |
Five. (Full marks 16 points) Xiao Ming has 5 RMB notes, with a total face value of 20 yuan.
(1) The face values of Xiao Ming's 5 RMB notes are $\qquad$ yuan, $\qquad$ yuan, $\qquad$ yuan, $\qquad$ yuan.
(2) Xiao Ming went to a fruit shop and weighed $x$ pounds of apples ($x$ is an integer), and according to the marked ... | (1) Xiaoming's 5 RMB notes have denominations of 1 yuan, 2 yuan, 2 yuan, 5 yuan, and 10 yuan.
It is easy to see that, among the 5 RMB notes, the smallest denomination cannot be less than "1 yuan." For example, if the smallest denomination is "1 jiao," at least 3 more notes would be needed to make up a whole 1 yuan, an... | x=4, y=6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,454 |
Six. (Full marks 16 points)
As shown in Figure 4, for the pentagon
$A B C D E$, each side is
translated 4 units outward
along the direction perpendicular to that side,
resulting in a new pentagon
$A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$.
(1) Can the 5 shaded
parts in the figure be assembled into a penta... | Six, (1) The 5 shaded parts in Figure 4 can form a small pentagon.
$$
\begin{array}{c}
\because B F=A G=A H=E I=E K=D L=D M \\
=C N=C O=B P=4, \\
\angle B F B^{\prime}=\angle A G A^{\prime}=90^{\circ}, \\
\angle C O C^{\prime}=\angle B P B^{\prime}=90^{\circ}, \\
\angle D M D^{\prime}=\angle C N C^{\prime}=90^{\circ}, ... | 25 | Geometry | proof | Yes | Yes | cn_contest | false | 711,455 |
1. Given the proposition "A quadrilateral with one pair of opposite sides parallel and the other pair of opposite sides equal is a parallelogram." Then ( ).
(A) This proposition and its negation are both true propositions
(B) This proposition and its negation are both false propositions
(C) This proposition is true, an... | 1. 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 | 711,456 |
2. For a project, Team A can complete it alone in $a$ days, and Team B can complete it alone in $b$ days. If the two teams work together, the time required to complete the project is ( ).
(A) $\frac{1}{a} \frac{1}{b}$
(B) $\left(\frac{1}{a}+\frac{1}{b}\right)$ days
(C) $\frac{a b}{a+b}$ days
(D) $\frac{1}{a b}$ days | 2. C
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 | 711,457 |
3. As shown in Figure 1, $\angle C G E=$ $\alpha$. Then $\angle A+\angle B+\angle C$ $+\angle D+\angle E+\angle F=$ ( ).
(A) $360^{\circ}-\alpha$
(B) $270^{\circ}-\alpha$
(C) $180^{\circ}+\alpha$
(D) $2 \alpha$ | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,458 |
4. If $|x|+|i x|-1 \mid=1$. Then ( ).
(A) $(x+1)(x-1)>0$
(B) $(x+1)(x-1)<0$
(C) $(x+1)(x-1) \geqslant 0$
(D) $(x+1)(x-1) \leqslant 0$ | 4. 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 | 711,459 |
Example 12 Given positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
z^{2}+x z+x^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 x z$.
(18th All-Union Mathematical Olympiad for Secondary Schools) | The original system of equations is equivalent to
$$
\left\{\begin{array}{l}
x^{2}-2 x \cdot \frac{y}{\sqrt{3}} \cos 150^{\circ}+\left(\frac{y}{\sqrt{3}}\right)^{2}=5^{2}, \\
\left(\frac{y}{\sqrt{3}}\right)^{2}+z^{2}=3^{2}, \\
z^{2}-2 x \cos 120^{\circ}+x^{2}=4^{2} .
\end{array}\right.
$$
Notice that $3^{2}+4^{2}=5^{2... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,460 |
2. Given $a, b, c$ are non-zero real numbers, and $a+b+c$ $=0$. Then the value of $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ is . $\qquad$ | 2. -3 | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,466 |
4. As shown in Figure 4, on the side $AB$ of rectangle $ABCD$, there is a point $E$, and $\frac{AE}{EB}=\frac{3}{2}$. On the side $DA$, there is a point $F$, and $EF=18$. When the rectangle is folded along $EF$, point $A$ lands on point $G$ on side $BC$. Then $AB=$ $\qquad$ | $4.5 \sqrt{6}$ | 4.5 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,468 |
Three. (Full marks 20 points) As shown in Figure 5, $AD$ is the altitude from the right angle $A$ to the hypotenuse $BC$ of the right triangle $\triangle ABC$. $P$ is the midpoint of $AD$, and $BP$ is extended to intersect $AC$ at $E$. Given that $AC: AB = k$. Find $AE: EC$.
---
The translation maintains the original... | From $\mathrm{R} \triangle A C O \cap R \triangle B D A \sim S \mathrm{Rt} \triangle B A C$,
we get $\frac{D C}{A D}=\frac{A D}{B D}:=\frac{A C}{A B}=k$.
Let $B D=a$, then $A D=k a, D C=k^{2} a$. Therefore, we have
$$
\begin{array}{l}
\frac{D C}{B D}=\frac{k^{2} a}{a}=k^{2}, \\
\frac{B C}{B D}=\frac{D C+B D}{B D}=\frac... | \frac{1}{1+k^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,469 |
Four, (Full marks 20 points) Given the equations $x^{2}+a_{1} x+a_{2} a_{3}=0$ and $x^{2}+a_{2} x+a_{1} a_{3}=0$ have exactly one common root. Prove: The other two roots (excluding the common root) of these two equations are the roots of the equation $x^{2}+a_{3} x+a_{1} a_{2}=0$.
---
Please note that the format and ... | Let the roots of the equation $x^{2}+a_{1} x+a_{2} a_{3}=0$ be $\alpha, \beta$, and the roots of the equation $x^{2}+a_{2} x+a_{1} a_{3}=0$ be $\alpha, \gamma$, where $\alpha$ is the common root of the two equations. According to the problem, we have
$$
\begin{array}{l}
\alpha^{2}+a_{1} \alpha+a_{2} a_{3}=0, \\
\alpha^... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,470 |
Five. (Full marks 30 points) There are several weights of 9 grams and 13 grams each. To weigh an object of 3 grams on a balance, what is the minimum number of such weights needed? Prove your conclusion.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the... | On the three pans of a balance, assuming that when the balance is level, 9 grams are used, and the number is an integer. Similarly, assuming that 13 grams of weights are used $|y|$ times. Therefore, when the balance is level and measures a 3-gram object, there should be
$$
9 x+13 y=3 \text {. }
$$
The problem becomes ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,472 |
4. A certain commodity is priced at a percentage $p$ higher than its cost. When the commodity is sold at a discount, to avoid incurring a loss, the discount (i.e., the percentage reduction in price) must not exceed $d \%$. Then $d$ can be expressed in terms of $p$ as ( ).
(A) $\frac{p}{100+p}$
(B) $p$
(C) $\frac{100 p}... | 4.C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems there was a misunderstanding in your request. The text "4.C" does not require translation as it is already in a form that is the same in both Chinese and ... | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,476 |
11. (10 points) If the equation with respect to $x$
$$
\frac{2 k}{x-1}-\frac{x}{x^{2}-x}=\frac{k x+1}{x}
$$
has only one solution, find the value of $k$. | When $k=0$, equation (1) has a unique solution $x=\frac{1}{2}$.
When $k \neq 0$, the discriminant of equation (1) is positive,
but the original equation has only one root, so one of the roots must be an extraneous root of the original equation. The extraneous root can only be 0 or 1. Clearly, 0 is not a root of (1), s... | k=0 \text{ or } \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,484 |
12. (12 points) Given the line $y=2 x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Points $P$ and $Q$ have coordinates $P(-2,0)$ and $Q(0, k)$, where $k<6$. A circle is drawn with $Q$ as the center and $PQ$ as the radius. Then
(1) For what value of $k$ does the circle $\odot Q$ touch the ... | 12. Let $x=0$, get $y=6$; let $y=0$, get $x=-3$.
$\therefore$ The coordinates of points $A$ and $B$ are $A(-3,0)$, $B(0,6)$.
$$
\begin{array}{l}
\therefore B Q=6-k, P Q=\sqrt{k^{2}+4}, \\
A B=\sqrt{36+9}=3 \sqrt{5} .
\end{array}
$$
As shown in Figure 5, draw $Q Q_{1} \perp A B$, with the foot of the perpendicular bein... | k=1 \text{ or } k=-4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,485 |
13. (13 points) As shown in Figure 3, given that $\odot O$ is the circumcircle of $\triangle ABC$, $D$ is the midpoint of the minor arc $\overparen{BC}$, and $H$ is the midpoint of the minor arc $\overparen{AB}$. Connect $C, H$, intersecting $AB$ at $E$: Connect $A, D$, intersecting $CH$ at point $G$, extend $CH$ to po... | 13. (1) As shown in Figure 6, connect $A$ and $H$, then $\angle 1 = \angle 3$.
Since $H$ is the midpoint of $\overparen{A B}$,
$$
\begin{array}{l}
\therefore \angle 2 = \angle 3. \\
\therefore \angle 1 = \angle 2.
\end{array}
$$
Since $D$ is the midpoint of $\overparen{B C}$,
$$
\begin{array}{l}
\therefore \angle 4 =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,486 |
14. (15 points) As shown in Figure 4,
there are two mounds $A$ and
$B$, a depression $E$,
and a pond $F$. The volumes of the two mounds are 781 cubic meters and
1584 cubic meters, respectively. The depression $E$ needs
to be filled with 1025 cubic meters of soil, and the pond $F$
can be filled with 1390 cubic meters o... | $\vdots 1$. "cubic meter・meter" as the unit of labor cost for transporting soil, let the amount of soil transported from $A$ to $E$ be $x_{1}$, and to $F$ be $y_{1}$; the amount of soil transported from $B$ to $E$ be $x_{2}$, and to $F$ be $y_{2}$. The total "cubic meter・meter" of soil transported is $W$. According to ... | 207170 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 711,487 |
1. Given $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ are the midpoints of the arcs $\overparen{B C} 、 \overparen{C A}$ and $\overparen{A B}$ on the circumcircle of $\triangle A B C$, not containing $A 、 B 、 C$ respectively. $B C$ intersects $C^{\prime} A^{\prime} 、 A^{\prime} B^{\prime}$ at points $M 、 N$ respectively, $C A... | Prove: As shown in Figure 1, connect $C^{\prime} A$, $C^{\prime} B$, $C^{\prime} C$, $B B^{\prime}$, and $A A^{\prime}$. Then
$$
\begin{array}{l}
\angle A R Q=\angle B^{\prime} C^{\prime} A \\
+\angle B A C^{\prime}= \\
\angle A B B^{\prime}+\angle B C C^{\prime} \\
=\frac{1}{2}(\angle A B C+ \\
\angle A C B)
\end{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,488 |
2. Given $I$ as the incenter of $\triangle A B C$, connect $A I, B I, C I$. If one of the triangles $\triangle B I C, \triangle C I A, \triangle A I B$ is similar to $\triangle A B C$, find the sizes of the angles of $\triangle A B C$. | Solution: Assume without loss of generality that $\triangle B I C$ is similar to $\triangle A B C$, then $\angle B I C \neq \angle A$.
In fact, if $\angle B I C=\frac{1}{2} \angle A+90^{\circ}=\angle A$, then $\angle A$ $=180^{\circ}$, which is a contradiction.
Assume without loss of generality that $\angle I B C=\ang... | \frac{\pi}{7}, \frac{2 \pi}{7}, \frac{4 \pi}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,489 |
3. $\triangle A B C$ is an acute triangle, $B^{\prime}$ is the midpoint of $C A$, $A^{\prime} 、 C^{\prime}$ are any points on $B C 、 A B$ respectively. Prove:
(i) The area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is not greater than $\cdots$ half of the area of $\triangle A B C$;
(ii) The area of $\triangle A^{\... | Proof: (i) Let $A_{1}$ and $C_{1}$ be the midpoints of $BC$ and $AB$ respectively. If $C^{\prime}$ is on $BC_{1}$ and $A^{\prime}$ is on $BA_{1}$, then
$S_{\triangle A^{\prime} B^{\prime} C} \leqslant S_{\text {quadrilateral } A_{1} I C_{1} B^{\prime}}=\frac{1}{2} S_{\triangle A B C} ;$
If $C^{\prime}$ is on $AC_{1}$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,490 |
4. In an isosceles trapezoid $A B C D$, $A B / / C D$, the incircle of $\triangle B C D$ touches $C D$ at $E$, $F$ is a point on the angle bisector of $\angle D A C$, and $E F \perp C D$. The circumcircle of $\triangle A C F$ intersects $C D$ at $G$. Prove that $\triangle A F G$ is an isosceles triangle. | Proof: Taking $E$ as the origin and $CD$ as the $x$-axis to establish a Cartesian coordinate system, let the radius of the incircle $I$ of $\triangle BCD$ be $1$, and the coordinates of points $C$, $D$, and $F$ be $(c, 0)$, $(-d, 0)$, and $(0, -f)$, respectively, where $c$, $d$, and $f > 0$.
$$
\begin{array}{l}
\becaus... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,491 |
5. In $\triangle A B C$, it is known that the excircle opposite to $\angle B$ is tangent to $C A$ at $D$, and the excircle opposite to $\angle C$ is tangent to $A B$ at $E$. A line is drawn through the midpoints $M$ and $N$ of $D E$ and $B C$ respectively. Prove that the line $M N$ bisects the perimeter of $\triangle A... | Proof: As shown in Figure 2, draw $D^{\prime} E^{\prime} \parallel D E$ such that $N$ is the midpoint of $D^{\prime} E^{\prime}$. Therefore,
$\therefore \triangle B E E^{\prime} \cong \triangle C D D^{\prime}$, which implies $\angle B E E^{\prime} = \angle C D D^{\prime}$.
Also, $\because \angle B E E^{\prime} + \angle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,492 |
6. Let $R, r$ be the circumradius and inradius of $\triangle A B C$, and $R^{\prime}, r^{\prime}$ be the circumradius and inradius of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Prove: If $\angle C=\angle C^{\prime}, R r^{\prime}=R^{\prime} r$, then $\triangle A B C \sim \triangle A^{\prime} B^{\prime} C^{\prime}$. | Proof: $\because \angle C=\angle C^{\prime}, R=\frac{c}{2 \sin C}, R^{\prime}=\frac{c^{\prime}}{2 \sin C}$, $\therefore c r^{\prime}=c^{\prime} r$, thus $\frac{c}{r}=\frac{c^{\prime}}{r^{\prime}}$,
i.e., $\cot \frac{A}{2}+\cot \frac{B}{2}=\cot \frac{A^{\prime}}{2}+\cot \frac{B^{\prime}}{2}$.
$$
\begin{aligned}
\quad &... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,494 |
8. Given that quadrilateral $ABCD$ is a cyclic quadrilateral. Prove:
$$
|AB-CD|+|AD-BC| \geqslant 2|AC-BD| .
$$ | Proof: Let $O$ be the circumcenter of quadrilateral $ABCD$, and let the radius of circle $O$ be $1$, $\angle AOB = 2\alpha$, $\angle BOC = 2\beta$, $\angle COD = 2\gamma$, $\angle DOA = 2\delta$. Then $\alpha + \beta + \gamma + \delta = \pi$, and without loss of generality, assume $\alpha \geq \gamma$, $\beta \geq 0$.
... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,496 |
9. Let $O x, O y, O z, O t$ be four different rays in three-dimensional space, and the angle between any two of them is the same.
(i) Find this angle;
(ii) Let $O r$ be a ray different from the above four, and let $\alpha, \beta, \gamma, \delta$ be the angles formed between $O r$ and $O x, O y, O z, O t$, respectively,... | (i) Solution: On the rays $Ox$, $Oy$, $Oz$, and $Ot$, intercept $OX = OY = OZ = OT = 1$. Let the angle between any two of these rays be $\theta$. By the cosine rule, we have
$$
\begin{aligned}
XY & = YZ = ZX = TX = TY = TZ \\
& = \sqrt{2 - 2 \cos \theta},
\end{aligned}
$$
which means the distances between any two of t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,497 |
10. A polyhedron has $10 n$ faces. Prove: there are $n$ faces with the same number of edges.
untranslated text remains in its original format and line breaks are preserved. | Proof: Let this convex polyhedron have $x$ vertices, and each face has $a_{1}, a_{2}, \cdots, a_{10 n}$ vertices. Thus, each face has $a_{1}, u_{2}, \cdots, a_{10 n}$ edges, so the number of edges of the convex polyhedron is $\frac{1}{2} \sum_{-1}^{1 \omega_{1}} a_{1}$. By Euler's theorem, we get
$$
\begin{array}{l}
10... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,498 |
1. If $n$ is a positive integer, then the value of $\frac{\left(1-(-1)^{n+1}\right)\left(n^{2}-1\right)}{2}$ is ( ).
(A) an even number
(B) zero or an odd number
(C) not necessarily an odd number
(D) definitely zero | $-1 .(B)$
When $n$ is odd, the expression $=0$; when $n$ is even, the original expression is an odd number. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,499 |
2. In an acute triangle $\triangle ABC$, two altitudes $BY'$ and $CC'$ intersect at $H$. If $AH + AB' + AC' = BC + BB' + CC'$, then $\angle BAC = (\quad)$.
(A) $75^{\circ}$
(B) $60^{\circ}$.
(C) $45^{\circ}$
(D) $30^{\circ}$ | 2. (C).
It is easy to know $\triangle A H B^{\prime} \backsim \triangle B C B^{\prime}, \triangle A B B^{\prime} \backsim \triangle A C C^{\prime}$
$$
\begin{array}{l}
\Rightarrow \frac{A H}{B C}=\frac{A B^{\prime}}{B B^{\prime}}=\frac{A C^{\prime}}{C C^{\prime}}=\frac{A H+A B^{\prime}+A C^{\prime}}{B C+B B^{\prime}+C... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,500 |
3. Given $p q<0, p^{2}+q^{2}=p^{2} q^{2}$. Simplify $p \sqrt{1-\frac{1}{p^{2}}}-q \sqrt{1-\frac{1}{q^{2}}}$ and the result is ( ).
(A) $p q$ or $-p q$
(B) $p q$
(C) $-p q$
(D) 2 | 3. (A)
It is easy to know that $\frac{1}{p^{2}}+\frac{1}{q^{2}}=1$.
$$
\begin{aligned}
\text { Original expression } & =\frac{p}{|q|}-\frac{q}{|p|}=\frac{p|p|-q|q|}{|p q|} \\
& =-\frac{p|p|-q|q|}{p q} .
\end{aligned}
$$
When $p0>q$, the original expression $=-p q$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,501 |
4. When $m$ and $n$ are both odd numbers, the equation
$$
x^{2}+4 m x-3 n x+4 m^{2}+n^{2}+9=0
$$
has two roots that are prime numbers. These two roots are ( ).
(A) 2 and 11
(B) 3 and 7
(C) 2 and 5
(D) 3 and 11 | 4. (A).
Let the two roots be $x_{1}, x_{2}$. Then $x_{1}+x_{2}=3 n-4 m$.
Since $m, n$ are odd numbers, $3 n-4 m$ is odd, and $x_{1}, x_{2}$ are prime numbers,
Therefore, one of $x_{1}, x_{2}$ must be the even number 2.
Substituting 2 and rearranging, we get
$4(m+1)^{2}+(n-3)^{2}=0$, solving this gives $m=-1, n=3$.
Sub... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,502 |
4. Real numbers $a$ and $b$ satisfy the equation $\frac{a^{2} b^{2}}{a^{4}-2 b^{4}}=1$. Then $\frac{a^{2}-b^{2}}{19 a^{2}+96 b^{2}}=$ $\qquad$ - | (Answer: $\frac{1}{134}$ ) | \frac{1}{134} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,504 |
6. The parabola $y=a x^{2}$ intersects with the lines $x=1$, $x=2$, $y=1$, and $y=2$ to form a square. The range of values for $a$ is ( ).
(A) $\frac{1}{4} \leqslant a \leqslant 1$
(B) $\frac{1}{2} \leqslant a \leqslant 2$
(C) $-\frac{1}{2} \leqslant a \leqslant 1$
(D) $-\frac{1}{4} \leqslant a \leqslant 2$ | 6. (D).
From the problem, we know $a>0$. When $a>0$, the larger the value of $a$, the smaller the opening of the parabola, as shown in Figure 6. The square has vertices at $A, B, C, D$. When the parabola passes through $A(1,2)$ and $C(2,1)$, $a$ reaches its maximum and minimum values. At this time, $a=2$ and $a$
$$
=\... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,505 |
1. In the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=1$, $a$, $b$, $c$, $d$, and $e$ are all natural numbers and are distinct. Then $a=$
,$b=$ $\qquad$ ,$c=$ $\qquad$
$\qquad$ ,$e=$ $\qquad$ ـ. | $\begin{array}{l}=.1 \cdot 2,5,6,12,20 . \\ \because\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right) \\ \quad=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}, \\ \therefore \frac{1}{2}+\frac{1}{5}+\frac{1}{6}+\frac{1}{12}+\frac{1... | a=2, b=5, c=6, d=12, e=20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,506 |
3. The integer part of $\frac{1}{3-\sqrt{7}}$ is $a$, and the fractional part is $b$. Then $a^{2}+(1+\sqrt{7}) a b=$ $\qquad$ . | 3.10 . From $\frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{2}$, we know that $2<\frac{3+\sqrt{7}}{2}<3$, thus $a=2, b=\frac{1}{3-\sqrt{7}}-2=\frac{\sqrt{7}-1}{2}$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,508 |
One, (Total 20 points) $m$ and $n$ are real numbers, and $m^{3}+$ $n^{3}+3 m n=1$. Find the value of $m+n$.
| $$
\begin{array}{l}
\left\{\left(m^{3}+n^{3}\right)-\left(m^{2}-m n+n^{2}\right)\right\}+(m+n)^{2}-1=0, \\
\text { i.e., }\left(m^{2}-m n+n^{2}\right)(m+n-1)+(m+n+1)(m+n-1) \\
=0, \\
\quad(m+n-1)\left(m^{2}-m n+n^{2}+m+n+1\right)=0, \\
\frac{1}{2}(m+n-1)\left[(m-n)^{2}+(m+1)^{2}+(n+1)^{2}\right]=0 .
\end{array}
$$
Whe... | 1 \text{ or } -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,510 |
II. (Total 25 points) As shown in Figure 4, $P$ is the midpoint of line segment $AB$, $\angle BPC = \angle BPD$, and $PB$ is the mean proportional between $PC$ and $PD$. If points $A$, $B$, and $C$ are all on circle $\odot K$, then is point $D$ inside $\odot K$, on $\odot K$, or outside $\odot K$? Prove your conclusion... | II. Conclusion: Point $D$ is on
$\odot K$, i.e., points $A$, $B$, $C$,
$D$ are concyclic. Auxiliary lines see Figure 8.
It is easy to know that $\triangle P A E \cong$
$\triangle P B C$, then $A B C E$ is
an isosceles trapezoid, which must have a circumcircle, i.e., points $A$, $B$, $C$, $E$ are concyclic.
Also, $\beca... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,511 |
Three. (Total 25 points) Given that $a, b, c$ are two unequal rational numbers, and $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is also a rational number. Prove: $\sqrt{a} 、 \sqrt{b} 、 \sqrt{c}$ are all rational numbers. | Three, let $r=\sqrt{a}+\sqrt{b}+\sqrt{c}$ be a rational number.
Then $r^{2}$ must be a rational number, so $s=\sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ is a rational number.
Then $s^{2}$ must be a rational number, so $t=b \sqrt{a c}+c \sqrt{a b}+a \sqrt{b c}$ is a rational number.
Let $u=t-b s=(a-b) \sqrt{b c}+(c-b) \sqrt{a ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,512 |
1. Distinct complex numbers $a, b, c$ satisfy $abc=1$, $ab+a+1 \neq 0$, and let
$$
\begin{array}{l}
M=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1} \\
N=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}, \\
P=\frac{ab}{ab+a+1}+\frac{bc}{bc+b+1}+\frac{ca}{ca+c+1} .
\end{array}
$$
Then the relationship between $M,... | - $1 .(\mathrm{D})$.
$$
\begin{array}{l}
\text { From } \frac{1}{a b+a+1}=\frac{b c}{b c(a b+a+1)}=\frac{b c}{b+1+b c}, \\
\frac{1}{a b+a+1}=\frac{c}{c(a b+a+1)}=\frac{c}{1+a c+c},
\end{array}
$$
we get $\frac{1}{a b+a+1}=\frac{b c}{b c+b+1}=\frac{c}{a c+c+1}$.
Similarly, $\frac{a}{a b+a+1}=\frac{1}{b c+b+1}=\frac{a c... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,513 |
2. As shown in Figure 1, construct squares $A B E F$ and $A C G H$ outside the acute triangle $\triangle A B C$, and draw the altitude $A D$ from $A$ to the base $B C$, intersecting $F H$ at $K$. The relationship between $B C$ and $A K$ is ().
(A) $B C>2 A K$
(B) $B C=2 A K$
(C) $B C<2 A K$
(D) Cannot be determined | 2. (B)
As shown in Figure 5, $\triangle A B C$ is rotated $90^{\circ}$ clockwise around point $A$ to get $\triangle A F M$. Given $A C = A H$, and $H, A, M$ are collinear, with $A$ being the midpoint of $MH$.
From $B C \perp A D$, we know that $F M \parallel A D$.
In $\triangle H M F$, since $A$ is the midpoint, and $... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,514 |
3. The system of equations $\left\{\begin{array}{l}y=m(x+2)-4, \\ x^{2}+y^{2}=4(x \geqslant 0)\end{array}\right.$ has real solutions. Then the range of values for $m$ is ( ....).
(A) $[1,3]$
(B) $\left[\frac{3}{4}, 3\right]$
(C) $\left[\frac{3}{4},+\infty\right)$
(D) $(-\infty, 0] \cup\left[\frac{3}{4},+\infty\right)$ | 3. (B).
As shown in Figure 6, the first equation represents a family of lines passing through the fixed point \( P(-2, -4) \), and the second equation represents the right semicircle. To determine the slopes of the lines in the family that intersect the right semicircle, these lines should lie between the tangent \( P... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,516 |
4. As shown in Figure 2, the axial section of the cone is an isosceles right $\triangle S A B, O$ is the center of the base circle, $C$ is the trisection point of the arc $\overparen{A B}$ on the circumference of the base circle, $\overparen{A C}=$ $2 \overparen{C B}$. Then the angle between $S A$ and $O C$ is ( ).
(A)... | 4. (C).
As shown in Figure 7, take another trisection point $D$ on the arc $\overparen{A B}$, such that $\overparen{A D}=\overparen{D C}=\overparen{C B}$. By the inscribed angle theorem, we know
$$
\angle B O C=\angle B A D \text {, }
$$
thus, $O C \parallel A D$.
Therefore, the angle between $O C$ and $S A$ is the s... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,517 |
$5 . n$ tennis players $(n \geqslant 4)$, any two of them can form a doubles team. The number of ways to select two teams from them is ( ).
(A) $\mathrm{C}_{n}^{4}$ ways
(B) $6 \mathrm{C}_{n}^{4}$ ways
(C) $3 C_{n+1}^{4}$ ways
(D) $\left(6 C_{n}^{4}+3 C_{n}^{3}\right)$ ways | 5. (C).
Solution 1: For $n$ athletes, the number of pairs is $\frac{n(n-1)}{2}$. Selecting 2 pairs from these, we have
$$
\begin{aligned}
\frac{C_{n(n-1)}^{2}}{2} & =\frac{1}{2} \cdot \frac{n(n-1)}{2} \cdot\left(\frac{n(n-1)}{2}-1\right) \\
& =\frac{1}{8}(n+1) n(n-1)(n-2) \\
& =3 C_{n+1}^{+} .
\end{aligned}
$$
Soluti... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 711,518 |
6. Given that the natural number $n$ is not a multiple of 5. Then when 5 divides $n^{2}+1, n^{2}-1$, we have ( ).
(A) $5 \mid n^{2}+1$ or $5 \mid n^{2}-1$
(B) $5 \mid n^{2}+1$ and $5 \mid n^{2}-1$
(C) $5 \times n^{2}+1$ and $5 \times n^{2}-1$
(D) Cannot determine, any of the above may occur | 6. (A).
Since $n$ is not a multiple of 5, we have $n \equiv \pm 1$ or $\pm 2(\bmod 5)$,
thus $n^{2} \equiv 1$ or $4(\bmod 5)$,
which gives $n^{2}-1 \equiv 0(\bmod 5)$ or $n^{2}+1 \equiv 0(\bmod 5)$. However, $\left(n^{2}-1\right)+\left(n^{2}+1\right) \neq 0(\bmod 5)$, so option $(B)$ cannot be selected. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 711,519 |
1. The coefficient of $x^{3}$ in the algebraic expression $1+x+x(1+x)+x(1+x)^{2}$ $+\cdots+x(1+x)^{1999}$ is $\qquad$
(Answer with a number) | Ni.1.1331334000.
Starting from the 4th term, there is $x^{3}$, the sum of its coefficients is
$$
\begin{array}{l}
C_{2}^{2}+C_{3}^{2}+\cdots+C_{1999}^{2}=C_{2000}^{3} \\
=\frac{2000 \times 1999 \times 1998}{6}=2000 \times 1999 \times 333 \\
=666000 \times(2000-1)=1331334000 .
\end{array}
$$ | 1331334000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,520 |
3. The base area of the rectangular prism is 4, and the length of the diagonal is
4. Then the maximum value of the lateral surface area of the rectangular prism is $\qquad$ | $3.16 \sqrt{2}$.
As shown in Figure 10, let the length of the rectangular prism be $x$, the width be $y$, and the height be $z$. According to the problem, we have
$$
\left\{\begin{array}{l}
x y=4 \\
x^{2}+y^{2}+z^{2}=16 .
\end{array}\right.
$$
Thus, $z^{2}=16-\left(x^{2}+y^{2}\right) \leqslant 16-2 x y=8$.
This gives ... | 16 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,522 |
5. In flood control and rescue operations, a depression near the river dike has experienced a pipe burst, with $x$ cubic meters of river water already rushing in, and water continues to flow in at a rate of $y$ cubic meters per minute. Now, a water extraction and plugging project needs to be carried out. If 1 water pum... | 5.4
Let at least $n$ water pumps be required. According to the problem, we have
$$
\left\{\begin{array}{l}
x+30 y=30 z, \\
x+10 y=20 z, \\
x+5 y=n \cdot 5 z .
\end{array}\right.
$$
From equations (1) and (2), we can solve for $x=15 z, y=0.5 z$.
Substituting into equation (3), we get
$15 z+0.5 z \leqslant 5 n z$, henc... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,524 |
6. Choose any two non-adjacent numbers from $1,2, \cdots, 10$ and multiply them. The sum of all such products is $\qquad$ | 6.990.
The sum of the products of any two different numbers taken is
$$
\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \text {. }
$$
The sum of the products of any two adjacent numbers is $\sum_{k=1}^{9} k(k+1)$.
The sum that satisfies the condition is
$$
\begin{array}{l}
\frac{1}{2} \sum_{k=1}^{10} k... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,525 |
Three. (Total 20 points) Does there exist $\theta \in$ $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that
$$
x^{2}+6 x+7=(x-\tan \theta)(x-\tan 3 \theta)
$$
holds for all complex numbers $x$? | Three, the $\theta$ for which the identity holds for all $x$ does not exist. We prove this by contradiction as follows.
Assume there exists a $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that the identity holds. Then, taking $x=\mathrm{i}$, we have
$$
\begin{aligned}
6+6 \mathrm{i} & =(\mathrm{i}-\tan \... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,527 |
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
保留源文本的换行和格式，翻译结果如下：
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what ... | Four, prove in three steps that there are at most 6 pairs of mutually perpendicular lines.
(1) 6 pairs can be achieved.
When $V A \perp$ plane $A B C$ and $A B \perp A C$, by the property of line perpendicular to a plane, we have $V A \perp A B, V A \perp B C, V A \perp C A$.
By the theorem of three perpendiculars, we ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,528 |
Five. (Total 20 points) As shown in the figure,
$3, \triangle ABC$ is an isosceles triangle with the vertex angle $\frac{\pi}{5}$,
denoted as $\triangle_{1}$, and its vertex angle is denoted as $\alpha_{1}$ $\left(\alpha_{1}=\angle BAC\right)$. Using one of the base angles of $\triangle_{1}$ $(\angle ABC)$ as the verte... | (1) According to the problem, given $\alpha_{1}=\frac{\pi}{5}$, we have
$$
\begin{array}{l}
\alpha_{2}=\frac{1}{2}\left(\pi-\alpha_{1}\right)=\frac{1}{2}\left(\pi-\frac{\pi}{5}\right)=\frac{2 \pi}{5}, \\
\alpha_{3}=\frac{1}{2}\left(\pi-\alpha_{2}\right)=\frac{1}{2}\left(\pi-\frac{2 \pi}{5}\right)=\frac{3 \pi}{10}, \\
\... | \frac{\pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,529 |
One. (Total 50 points) As shown in Figure 4, in parallelogram $ABCD$, $E$ is any point on $AD$. Draw $EF$ to intersect the extension of $AB$ at $F$, and connect $CE$ and $CF$. Let the circumcenter of $\triangle CDE$ be $O_{1}$, the circumcenter of $\triangle EAF$ be $O_{2}$, and the circumradius of $\triangle CBF$ be $... | As shown in Figure 12, draw $O_{1} G \perp E C$ at $G$, $O_{2} H \perp E F$ at $H$, and connect $G H$. By the midline theorem, we know $G H=\frac{1}{2} C F$.
Let the circumradius of $\triangle CIDE$ be $R_{1}$, and the circumradius of $\triangle E A F$ be $R_{2}$. Connect $O_{1} E$ and $O_{2} E$. We have $O_{1} E=R_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,530 |
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