problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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5. If $\tan x_{1} \cdot \tan x_{2} \cdots \cdots \tan x_{2000}=$ 1 , then the maximum value of $\sin x_{1} \cdot \sin x_{2} \cdots \cdot \sin x_{2 \text { obo }}$ is $\qquad$ . | 5. $\frac{1}{2^{1000}}$.
From $\tan x_{1} \cdot \tan x_{2} \cdots \cdot \tan x_{2000}=1$, we get
$$
\begin{array}{l}
\sin x_{1} \cdot \sin x_{2} \cdots \cdot \sin x_{2000} \\
=\cos x_{1} \cdot \cos x_{2} \cdots \cdot \cos x_{2000} .
\end{array}
$$
Thus $2^{2000}\left(\sin x_{1} \cdot \sin x_{2} \cdots \cdots \cdot \s... | \frac{1}{2^{1000}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,910 |
6. Given $f(x)=\frac{1+x}{2-x}$, for $n \in \mathbf{N}$, define: $f_{1}(x)=f(x), f_{n+1}(x)=f\left[f_{n}(x)\right]$. If $f_{13}(x)=f_{31}(x)$, then the analytical expression of $f_{16}(x)$ is . $\qquad$ | 6. $\frac{x-1}{x}$.
It is easy to know that the inverse function of $f(x)$ is $f^{-1}(x)=\frac{2 x-1}{x+1}$. Then
$$
\begin{array}{l}
f^{-1}[f(x)]=\frac{2\left(\frac{1+x}{2-x}\right)-1}{\frac{1+x}{2-x}+1}=x . \\
\therefore f^{-1}\left[f_{n+1}(x)\right]=f^{-1}\left\{\left[f_{n}(x)\right]\right\}=f_{n}(x) . \\
\because ... | \frac{x-1}{x} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,911 |
Three, (20 points) Does there exist a set of positive numbers $a$, $b$, $c$, $d$ such that the following three inequalities hold simultaneously? Prove your conclusion.
$$
\begin{array}{l}
a+b<c+d, \\
(a+b)(c+d)<a b+c d, \\
(a+b) c d<a b(c+d) .
\end{array}
$$ | Three, assuming there exist positive numbers $a, b, c, d$, such that inequalities (1), (2), and (3) all hold.
(1) $\times$ (2) gives
$$
(a+b)^{2}0, \\
\therefore 4 c d<(a+b)(c+d).
\end{array}
$$
Combining (2) we get $4 c d<a b+c d, c d<\frac{1}{3} a b$.
From (4) we get $(a+b)^{2}<\frac{4}{3} a b$,
i.e., $a^{2}+b^{2}<-... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 711,912 |
Four. (20 points) Let $p>0$, when $p$ varies, $C_{p}$: $y^{2}=2 p x$ is a family of parabolas. Line $l$ passes through the origin and intersects $C_{p}$ at the origin and point $A_{p}$. Also, $M$ is any point on the $x$-axis other than the origin, and line $M A_{p}$ intersects $C_{p}$ at points $A_{p}$ and $B_{p}$. Pro... | Let $l: y=k x(k \neq 0)$, and by solving the equation of $C_{p}$ together, we get the coordinates of point $A_{p}$ as $\left(\frac{2 p}{k^{2}}, \frac{2 p}{k}\right)$.
Assume the coordinates of point $M$ are $(a, 0)$, then $a \neq 0$. It is easy to get the equation of line $M A_{p}$ as
$$
\frac{2 p}{k}\left(x-\frac{2 p... | y = -ka | Algebra | proof | Yes | Yes | cn_contest | false | 711,913 |
Five. (20 points) For an arithmetic sequence $\left\{a_{n}\right\}$ with a common difference of $d(d \neq 0)$, prove that the sum of any two distinct terms in the sequence is still a term in the sequence if and only if there exists an integer $m \geqslant-1$, such that $a_{1}=m d$.
---
The translation maintains the o... | Necessity. For any two different terms $a_{s}$ and $a_{t}$ ($s \neq t$) in the arithmetic sequence $\left\{a_{n}\right\}$, if there exists $k$ such that $a_{s}+a_{t}=a_{k}$, then
$$
2 a_{1}+(s+t-2) d=a_{1}+(k-1) d.
$$
This gives $a_{1}=(k-s-t+1) d$.
Thus, there exists an integer $m=k-s-t+1$ such that $a_{1}=m d$.
We n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 711,914 |
一、(50 分)As shown in Figure 1, quadrilateral $ABCD$ is inscribed in a circle. The extensions of $AB$ and $DC$ intersect at $E$, and the extensions of $AD$ and $BC$ intersect at $F$. $P$ is any point on the circle, and $PE$ and $PF$ intersect the circle at $R$ and $S$, respectively. If the diagonals $AC$ and $BD$ interse... | Connect $P D$, $A S$,
$R C$, $B R$, $A P$,
$S D$. From
$\triangle E B R \sim \triangle E P A$,
$\triangle F D S \sim \triangle F P A$,
we have $\frac{B R}{P A}=\frac{E B}{E P}$,
$$
\frac{P A}{D S}=\frac{F P}{F D} \text {. }
$$
Multiplying the two equations, we get $\frac{B R}{D S}=\frac{E B \cdot F P}{E P ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,915 |
Three. (50 points) Let $E=\{1,2,3, \cdots, 200\}, G$ $=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{100}\right\} \subset E$, and $G$ has the following two properties:
(1) For any $1 \leqslant i<j \leqslant 100$, it always holds that $a_{i}+a_{j} \neq 201$;
(2) $a_{1}+a_{2}+a_{3}+\cdots+a_{100}=10080$.
Prove: The number of od... | Three, noticing that $200+1=199+2=198+3=\cdots=101+100=201$, and $a_{i}+a_{j} \neq 201$, the set $E$ can be divided into 100 subsets: $\{1,200\},\{2,199\},\{3,198\}$, $\cdots, \{100,101\}$, then the elements of $G$ can only take 1 from each of these 100 subsets.
To discuss the number of odd numbers in $G$, we further ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,917 |
Example 1 Given that the polynomial $3 x^{3}+a x^{2}+b x+1$ can be divided by $x^{2}+1$, and the quotient is $3 x+1$. Then, the value of $(-a)^{i}$ is $\qquad$
（Dinghe Hesheng Junior High School Math Competition） | Solution: According to the polynomial identity, we have
$$
3 x^{3}+a x^{2}+b x+1=\left(x^{2}+1\right)(3 x+1) \text {. }
$$
Taking $x=1$ gives $a+b+4=8$.
Taking $x=-1$ gives $a-b-2=-4$.
Solving these, we get $a=1, b=3$.
$$
\therefore(-a)^{b}=(-1)^{3}=-1 \text {. }
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,918 |
Example 2 The remainder of the polynomial $x^{12}-x^{6}+1$ divided by $x^{2}-1$ is $\qquad$
(A) 1
(B) -1
(C) $x-1$
(D) $x+1$
(1993, National Junior High School Mathematics Competition) | Solution: Let the quotient be $g(x)$. Since the divisor is a quadratic polynomial, the remainder is at most a linear polynomial, so we can set
$$
x^{12}-x^{6}+1=\left(x^{2}-1\right) g(x)+a x+b \text {. }
$$
Taking $x=1$, we get $1=a+b$;
Taking $x=-1$, we get $1=-a+b$.
Solving these, we get $a=0, b=1$.
Therefore, the r... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,919 |
Example 11 Rationalize the denominator:
$$
\frac{3+2 \sqrt{2}-\sqrt{3}-\sqrt{6}}{1+\sqrt{2}-\sqrt{3}}=
$$
$\qquad$
(Fifth National Partial Provinces and Cities Junior High School Mathematics Competition) | Solution: From $\sqrt{6}=\sqrt{2} \times \sqrt{3}$, we know that the numerator of the original expression is the product of two radicals containing $\sqrt{2}$ and $\sqrt{3}$. Therefore, we can set
$$
\begin{aligned}
& 3+2 \sqrt{2}-\sqrt{3}-\sqrt{6} \\
& =(1+\sqrt{2}-\sqrt{3})(a+b \sqrt{2}+c \sqrt{3}) \\
= & (a+2 b-3 c)... | 1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,920 |
Height 97 Find the last three digits of $2^{2^{2} 1100}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | $1000=2^{3} \times 5^{3}$ divided by the remainder.
Obviously, $x==1\left(\bmod ^{1} 2^{1}\right)$.
Next, consider the remainder of $x$ divided by $5^{3}$.
By Euler's theorem, $2^{2 / 5^{3}} \equiv \mathrm{I}\left(\bmod 5^{3}\right)$.
That is, $2^{+5^{2}}=1\left(\operatorname{mxl} 55^{3}\right)$.
Thus, $16^{35}=1\left... | 136 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,921 |
Example 12 Simplify the radical expression:
$$
\sqrt{2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})} \text {. }
$$
(1997, Taiyuan City Junior High School Mathematics Competition) | Solution: Let $2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})=(a+$ $b \sqrt{3}+c \sqrt{5})^{2}$, that is,
$$
\begin{aligned}
12 & -4 \sqrt{3}-4 \sqrt{5}+2 \sqrt{15} \\
= & \left(a^{2}+3 b^{2}+5 c^{2}\right)+2 a b \sqrt{3}+2 a c \sqrt{5} \\
& +2 b c \sqrt{15} .
\end{aligned}
$$
Comparing the coefficients of the corresponding ter... | -2+\sqrt{3}+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,923 |
Example 13 The solution to the equation $\frac{1}{x^{2}+x}+\frac{1}{x^{2}+3 x+2}+$ $\frac{1}{x^{2}+5 x+6}+\frac{1}{x^{2}+7 x+12}=\frac{4}{21}$ is $\qquad$ -
(Tenth Zu Chongzhi Cup Junior High School Mathematics Competition) | Solution: After transformation, each fraction in the original equation is in the form of $\frac{1}{n(n+1)}$. If we convert it into partial fractions, the original equation can be simplified.
Let $\frac{1}{n(n-1)}=\frac{A}{n}+\frac{B}{n+1}$,
i.e., $1=A(n+1)+B n$.
Taking $n=0$, we get $A=1$;
Taking $n=-1$, we get $B=-1$... | x_{1}=3, x_{2}=-7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,924 |
Example 14 Express $\frac{6 x^{2}-22 x+18}{(x-1)(x-2)(x-3)}$ as partial fractions. | $$
\text { Solution: Let } \begin{aligned}
\frac{6 x^{2}-22 x+18}{(x-1)(x-2)(x-3)} \\
\quad=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3},
\end{aligned}
$$
i.e.,
$$
\begin{array}{l}
6 x^{2}-22 x+18 \\
=A(x-2)(x-3)+B(x-3)(x-1) \\
\quad+C(x-1)(x-2) .
\end{array}
$$
Taking $x=1$, we get $2=2 A$, i.e., $A=1$;
Taking $x=2$, w... | \frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,925 |
Example 15 If $x=\frac{\sqrt{5}-1}{2}$, then $x^{4}+x^{2}+2 x-$
$1=$ $\qquad$
(Sixth National Partial Provinces and Cities Junior High School Mathematics Competition) | Given $x=\frac{\sqrt{5}-1}{2}$, we know $x(x+1)=1$. We can express $x^{4}+x^{2}+2 x-1$ in terms of the powers of $x^{2}+x$. Let
$$
\begin{aligned}
x^{4} & +x^{2}+2 x-1 \\
= & A\left(x^{2}+x\right)^{2}+(B x+C)\left(x^{2}+x\right) \\
& +D x+E . \\
= & A x^{4}+(2 A+B) x^{3}+(A+B+C) x^{2} \\
& +(C+D) x+E .
\end{aligned}
$$... | 3-\sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,926 |
For example, the remainder when $16 x^{5}+x+1$ is divided by $x^{2}+1$ is $\qquad$ | To find the remainder, we can first express $x^{5}+x+1$ using the factor $x^{2}+1$. By the properties of polynomials, we can assume
$$
\begin{array}{l}
x^{5}+x+1 \\
=A x\left(x^{2}+1\right)^{2}+B x\left(x^{2}+1\right)+C x+D \\
=A x^{5}+(2 A+B) x^{3}+(A+B+C) x \\
\quad+D .
\end{array}
$$
By comparing the coefficients o... | 2x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,927 |
1. Find the quotient $q(x)$ and the remainder $r(x)$ when $f(x)=3 x^{4}+x^{3}-4 x^{2}-17 x+5$ is divided by $g(x)=x^{2}+x+1$. | $q(x)=3 x^{2}-2 x-5, r(x)=-10 x+10$ | q(x)=3 x^{2}-2 x-5, r(x)=-10 x+10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,928 |
3. For any real number $x$, the quadratic trinomial $x^{2}+3 m x+m^{2}-$ $m+\frac{1}{4}$ is a perfect square, then $m=$ $\qquad$ | \begin{array}{l}m=-1 \text { or } m= \\ \frac{1}{5} .\end{array} | m=-1 \text{ or } m=\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,930 |
4. Factorize:
$$
2 x^{2}-7 x y+6 y^{2}+2 x-y-12 \text {. }
$$ | $\begin{array}{l}(2 x-3 y-4)(x \\ -2 y+3) .\end{array}$ | (2 x-3 y-4)(x-2 y+3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,931 |
Example 3 Find the values of $a$ and $b$ such that the polynomial
$$
x^{4}+(2 a+1) x^{3}+(a-1)^{2} x^{2}+b x+4
$$
can be factored into the product of two quadratic polynomials $P(x)$ and $Q(x)$ with leading coefficients of 1, and
(1) $P(x)=0$ has two distinct real roots $r$ and $s$;
(2) $\boldsymbol{Q}(r)=s, Q(s)=r$.
... | Solution: From (1), let
$$
\begin{aligned}
P(x) & =x^{2}+u x+v \\
& =x^{2}-(r+s) x+r s .
\end{aligned}
$$
Let $Q(x)=x^{2}+w x+t$, then from (2) we have
$$
\left\{\begin{array}{l}
r^{2}+w r+t=s, \\
s^{2}+w s+t=r .
\end{array}\right.
$$
Solving, we get $\left\{\begin{array}{l}w=-(r+s+1) \\ t=r s+r+s .\end{array}\right.... | a=2, b=-14 \text { or } a=-1, b=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,932 |
7. Let $f^{\prime}(x)=x^{4}+a x^{3}+b x^{2}+c x+d$. Where $a$, $b$, $c$, $d$ are constants. If $f(1)=1, f(2)=2, f(3)=3$, then the value of $\frac{1}{4}[f(4)+f(0)]$ is $(\quad)$.
(A) 1
(B) 4
$\begin{array}{ll}\text { (C) } 7 & \end{array}$
(D) 8 | Select (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 | 711,935 |
8. Given $y=y_{1}+y_{2}, y_{1}$ is directly proportional to $x$, $y_{2}$ is inversely proportional to $x^{2}$, and when $x=2$ and $x=3$, the value of $y$ is 19. Find the functional relationship between $y$ and $x$. | $y=5 x+\frac{36}{x^{2}}$ | y=5 x+\frac{36}{x^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,936 |
9. Given the equation $x^{3}-x^{2}-8 x+12=0$ has two equal roots, solve this equation. | $x_{1}=x_{2}=2, x_{3}=-3$ | x_{1}=x_{2}=2, x_{3}=-3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,937 |
11. Express $\frac{5 x-4}{(x-1)(2 x-1)}$ as partial fractions. | $\frac{1}{x-1}+\frac{3}{2 x-1}$ | \frac{1}{x-1}+\frac{3}{2 x-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,939 |
12. Express $x^{5}-3 x+2$ in terms of the powers of $x^{2}-x-1$.
Use $x^{2}-x-1$'s powers to represent $x^{5}-3 x+2$. | $\begin{array}{l}(x+2)\left(x^{2}-x-1\right)^{2}+(5 x+2)\left(x^{2}-x-1\right) \\ +2 x+2\end{array}$ | (x+2)\left(x^{2}-x-1\right)^{2}+(5 x+2)\left(x^{2}-x-1\right) +2 x+2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,940 |
13. Express $\frac{x^{3}+2}{(x-1)^{4}}$ as partial fractions. | (Hint: Let $x^{3}+2=[(x-1)+1]^{3}+2=(x-1)^{3}$ $+3(x-1)^{2}+3(x-1)+3$. The answer is: $\frac{1}{x-1}+$ $\left.\frac{3}{(x-1)^{2}}+\frac{3}{(x-1)^{3}}+\frac{3}{(x-1)^{4}}.\right)$ | \frac{1}{x-1}+\frac{3}{(x-1)^{2}}+\frac{3}{(x-1)^{3}}+\frac{3}{(x-1)^{4}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,941 |
Example 1 In $\triangle ABC$, points $A_{1}, B_{1}, C_{1}$ satisfy $\overrightarrow{B A_{1}}=g \overrightarrow{B C}, \overrightarrow{C B_{1}}=g \overrightarrow{C A}, \overrightarrow{A C_{1}}=g \overrightarrow{A B}$ (where $g$ can take any value), points $A_{2}, B_{2}, C_{2}$ satisfy $\overrightarrow{C A_{2}}=g \overrig... | Prove: Transform $\triangle A B C$ into an isosceles right triangle with legs of length 1 and with $A$ as the right-angle vertex, as shown in Figure 4. Establish the coordinate system as follows:
$$
A(0,0), B(1,
$$
$0), C(0,1)$.
From the given information, we have
$B_{2}(0, g)$,
$A_{1}(1-g, g)$.
Then the line $A A_{1}... | \left|\frac{2 g-1}{g-2}\right| | Geometry | proof | Yes | Yes | cn_contest | false | 711,942 |
Example 4 Factorize:
$$
x^{2}+x y-6 y^{2}+x+13 y-6 \text{. }
$$
(10th Jinyun Cup Junior High School Mathematics Invitational Competition) | Solution: Since the original expression is a quadratic in two variables and can only be factored into the product of two linear expressions in two variables, considering
$$
x^{2}+x y-6 y^{2}=(x+3 y)(x-2 y) \text {, }
$$
we can assume
$$
\begin{array}{l}
x^{2}+x y-6 y^{2}+x+13 y-6 \\
=(x+3 y+a)(x-2 y+b) \\
= x^{2}+x y... | (x+3 y-2)(x-2 y+3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,943 |
Example 2 In a convex quadrilateral $ABCD$, take points $E$ and $F$ on sides $AB$ and $BC$ respectively, such that segments $DE$ and $DF$ trisect $AC$. It is known that the areas of $\triangle ADE$ and $\triangle CDF$ are each equal to $\frac{1}{4}$ of the area of quadrilateral $ABCD$. Prove that $ABCD$ is a parallelog... | Prove: The conditions and conclusions of the problem satisfy the aforementioned invariant properties. Transform $\triangle A B C$ into the right-angled triangle shown in Figure 5, with $|A B|=|B C|=$ 3. Then
```
A(3,0), C(0,3),
F(2,1), Q(1,2).
```
Let $D(a, b)$ be the point to be found. Then the line
```
D E: y-1=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,944 |
For example, $3 A B C D$ is a quadrilateral, and $B C \parallel A D, M$ is the midpoint of $C D, P$ is the midpoint of $M A, Q$ is the midpoint of $M B$, and lines $D P, C Q$ intersect at point $N$. Prove that the point $N$ is not outside $\triangle A B M$ if and only if the ratio of the lengths of the upper and lower ... | Proof: The conditions and conclusions of the problem satisfy the invariance property under scaling transformation. Let the midpoint of $AB$ be $R$, and transform $\triangle AMR$ into an isosceles right triangle with $R$ as the right-angle vertex. As shown in Figure 6, we can set up the coordinate system such that:
$$
A... | \frac{1}{3} \leqslant \frac{|AD|}{|BC|} \leqslant 3 | Geometry | proof | Yes | Yes | cn_contest | false | 711,945 |
Example 4 In $\triangle ABC$, $AB=12, AC=16$, $M$ is the midpoint of $BC$, $E, F$ are on $AB, AC$ respectively, $EF$ intersects $AM$ at $G$, and $AE=2AF$. Find: the ratio $\frac{EG}{GF}$.
(29th IMO Preliminary Problem) | Proof: Let $A F=a, A E=2 a$, then
$$
\begin{array}{l}
\frac{A F}{A C}=\frac{a}{16}, \\
\frac{A E}{A \bar{B}}=\frac{2 a}{12}=\frac{a}{6} .
\end{array}
$$
Transform $\triangle A B C$ from Figure 7 to the triangle shown in Figure 8. Let $B(5,0), C(0, 16)$, then $E(a, 0), F(9, a), M(3,8)$.
The equation of $A M$ is:
$$
y=\... | \frac{8}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,946 |
Example 5 In an arbitrary quadrilateral $ABCD$, one pair of opposite sides $BA$ and $CD$ intersect at $M$. A secant through $M$ intersects the lines containing the other pair of opposite sides at $H$ and $L$, and the lines containing the diagonals at $H'$ and $L'$. Prove that: $\frac{1}{MH}+\frac{1}{ML}=\frac{1}{MH'}+\... | Analysis: Although the conclusion does not satisfy the invariance property, we only need to transform the original conclusion into an equivalent one:
$$
\begin{array}{l}
M L\left(\frac{1}{M H}+\frac{1}{M L}\right) \\
=M L\left(\frac{1}{M H^{\prime}}+\frac{1}{M L^{\prime}}\right),
\end{array}
$$
i.e., $\frac{M L}{M H}+... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,947 |
2. Let the points dividing the sides $BC$, $CA$, and $AB$ of $\triangle ABC$ internally in the ratio $t:(1-t)$ be $P$, $Q$, and $R$, respectively. The area of the triangle formed by the segments $AP$, $BQ$, and $CR$ is $K$, and the area of $\triangle ABC$ is $L$. Find $\frac{K}{L}$. | $t^{2}-t+1$ | t^2 - t + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,948 |
First question As shown in Figure 1, on the side $BC$ of the acute triangle $\triangle ABC$, there are two points $E$ and $F$,
satisfying $\angle BAE = \angle CAF$. Draw $FM \perp AB, FN \perp AC$
($M, N$ are the feet of the perpendiculars),
extend $AE$ to intersect the circumcircle of $\triangle ABC$ at point $D... | Proof 1: As shown in Figure 1, connect $BD$, then $\triangle ABD \cong \triangle AFC$, so $AF \cdot AD = AB \cdot AC$.
Let $\angle BAE = \angle CAF = \alpha, \angle EAF = \beta$, then
$S_{\text{quadrilateral AMEN}} = \frac{1}{2} AM \cdot AD \sin \alpha$
$+ \frac{1}{2} AD \cdot AN \sin (\alpha + \beta)$
$= \frac{1}{2} A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 711,950 |
The second question: Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
a_{0}=1, b_{0}=0 \text {, }
$$
and $\left\{\begin{array}{l}a_{n+1}=7 a_{n}+6 b_{n}-3, \\ b_{n+1}=8 a_{n}+7 b_{n}-4,\end{array} n=0,1,2, \cdots\right.$.
Prove: $a_{n}(n=0,1,2, \cdots)$ is a perfect square. | Proof 1: Substituting $b_{n}$ from the first equation into the second and simplifying, we get
$$
a_{n+2}-14 a_{n+1}+a_{n}+6=0,
$$
which can be rewritten as
$$
\left(a_{n+2}-\frac{1}{2}\right)-14\left(a_{n+1}-\frac{1}{2}\right)+\left(a_{n}-\frac{1}{2}\right)=0.
$$
Let $A_{n}=a_{n}-\frac{1}{2}$, then
$$
A_{n+2}-14 A_{n+... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 711,951 |
Third question: There are $n$ people, and it is known that any two of them make at most one phone call. Any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.
---
The translation maintains the original text's format and line... | Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:
If... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,952 |
Example 1 Let $x \geqslant y \geqslant z \geqslant \frac{\pi}{12}$, and $x+y+z=$ $\frac{\pi}{2}$. Find the maximum and minimum values of the product $\cos x \cdot \sin y \cdot \cos z$.
(1997, National High School Mathematics Competition) | Solution: From the given conditions, we have
$$
\begin{array}{l}
x=\frac{\pi}{2}-(y+z) \leqslant \frac{\pi}{2}-\left(\frac{\pi}{12}+\frac{\pi}{12}\right)=\frac{\pi}{3}, \\
\sin (x-y) \geqslant 0, \sin (y-z) \geqslant 0 .
\end{array}
$$
Thus,
$$
\begin{array}{l}
\cos x \cdot \sin y \cdot \cos z \\
=\frac{1}{2} \cos x \... | \frac{1}{8} \text{ and } \frac{2+\sqrt{3}}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,953 |
Example 5 Factorize:
$$
(x+1)(x+2)(x+3)-6 \times 7 \times 8 \text {. }
$$
(1987, Sichuan Province Junior High School Mathematics Competition) | Solution: Let $f(x)=(x+1)(x+2)(x+3)-$ $6 \times 7 \times 8$. Obviously, $f(5)=0$. By the Factor Theorem, $f(x)$ has a factor $(x-5)$. Therefore, we can assume
$$
\begin{array}{l}
(x+1)(x+2)(x+3)-6 \times 7 \times 8 \\
=(x-5)\left(x^{2}+a x+b\right) .
\end{array}
$$
Taking $x=-1$, we get $-6 \times 7 \times 8=-6(1-a+b)... | (x-5)(x^2+11x+66) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,954 |
Example 2 Given a positive integer $n$ and a positive number $M$. For all arithmetic sequences $a_{1}$, $a_{2}, a_{3}, \cdots$ satisfying the condition $a_{1}^{2}+a_{n+1}^{2} \leqslant M$, find the maximum value of $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$.
(1999, National High School Mathematics Competition) | Solution: Let the common difference be $d$, $a_{n+1}=\alpha$, then
$$
\begin{array}{l}
S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1} \\
=(n+1) \alpha+\frac{n(n+1)}{2} d .
\end{array}
$$
Thus, $\alpha+\frac{n d}{2}=\frac{S}{n+1}$.
Then
$$
\begin{aligned}
M & \geqslant a_{1}^{2}+a_{n+1}^{2}=(\alpha-n d)^{2}+\alpha^{2} \\
& =\frac{... | \frac{\sqrt{10}}{2}(n+1) \sqrt{M} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,955 |
Example 3 Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2} = 5$, let $s: x^{2}+y^{2}$. Then the value of $\frac{1}{s_{\text {max }}}+\frac{1}{s_{\text {min }}}$ is $\qquad$ (1993, National High School Mathematics Competition) | Solution: From the given, we know that $(x y)^{2}=\left(\frac{4}{5} s-1\right)^{2}$.
Also, $x^{2}+y^{2}=s$,
$\therefore x^{2}, y^{2}$ are the two real roots of the equation $t^{2}-s t+\left(\frac{4}{5} s-1\right)^{2}=0$.
$$
\begin{aligned}
\therefore \Delta & =s^{2}-4\left(\frac{4}{5} s-1\right)^{2} \\
& =-\frac{39}{25... | \frac{8}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,956 |
Example 4 Given the family of curves $2(2 \sin \theta-\cos \theta+3)$. $x^{2}-(8 \sin \theta+\cos \theta+1) y=0, \theta$ is a parameter. Find the maximum value of the length of the chord intercepted by the line $y=2 x$ on this family of curves.
(1995, National High School Mathematics Competition) | Solution: Clearly, the family of curves always passes through the origin, and the line $y$ $=2 x$ also passes through the origin, so the length of the chord intercepted by the family of curves on $y=2 x$ depends only on the coordinates of the other intersection point of the family of curves with $y=2 x$.
Substituting $... | 8 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,957 |
Example 5 Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy $a+b=2$. Then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$
(1990, National High School Mathematics Competition) | Solution: $\because a, b>0$,
$$
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1 \text {. }
$$
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}}$ $\geqslant 1$.
When $a=b=1$, the above expression $=1$, hence the minimum value is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 711,958 |
Example 6 Suppose $0<\theta<\pi$, then the maximum value of $\sin \frac{\theta}{2}(1+\cos \theta)$ is $\qquad$
(1994, National High School Mathematics Competition) | Solution: From $00$.
$$
\begin{array}{l}
\sin \frac{\theta}{2}(1+\cos \theta)=2 \sin \frac{\theta}{2} \cos ^{2} \frac{\theta}{2} \\
=\sqrt{2} \sqrt{2 \sin ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2}} \\
\leqslant \sqrt{2} \sqrt{\left(\frac{2 \sin ^{2} \frac{\theta}{2}+\cos ^{... | \frac{4 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,959 |
Example 7 Let $a=\lg z+\lg \left[x(y z)^{-1}+1\right]$, $b=\lg x^{-1}+\lg (x y z+1), c=\lg y+$ $\lg \left[(x y z)^{-1}+1\right]$. Let $M$ be the maximum of $a$, $b$, and $c$. Then the minimum value of $M$ is $\qquad$
(1997, National High School Mathematics Competition) | Given the conditions, we have
$$
\begin{array}{l}
a=\lg \left(x y^{-1}+z\right), b=\lg \left(y z+x^{-1}\right), \\
c=\lg \left[(x z)^{-1}+y\right] .
\end{array}
$$
Let the maximum of \(x y^{-1}+z, y z+x^{-1}, (x z)^{-1}+y\) be \(u\), then \(M=\lg u\).
From the given conditions, \(x, y, z \in \mathbf{R}^{+}\), thus
$$
... | \lg 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,960 |
Example 8 When $s$ and $t$ take all real numbers, then the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is $\qquad$
(1989, National High School Mathematics Competition) | Solution: As shown in Figure 1, the distance squared between any point on the line
$$
\left\{\begin{array}{l}
x=s+5, \\
y=s
\end{array}\right.
$$
and any point on the elliptical arc
$$
\left\{\begin{array}{l}
x=3|\cos t|, \\
y=2|\sin t|
\end{array}\right.
$$
is what we are looking for.
Indeed, the shortest distance ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,961 |
Example 9 Function
$$
\begin{aligned}
f(x)= & \sqrt{x^{4}-3 x^{2}-6 x+13} \\
& -\sqrt{x^{4}-x^{2}+1}
\end{aligned}
$$
The maximum value is $\qquad$
$(1992$, National High School Mathematics Competition) | Solution: Transform the function, we get
$$
\begin{aligned}
f(x)= & \sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}} \\
& -\sqrt{(x-0)^{2}+\left(x^{2}-1\right)^{2}} .
\end{aligned}
$$
It can be known that the geometric meaning of the function $y=f(x)$ is:
As shown in Figure 2, for a point $P(x, x^{2})$ on the parabola $y=x^{... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,962 |
Given $a, \sqrt{2}<a<2$. A convex quadrilateral $ABCD$ is inscribed in the unit circle $\Gamma$ and satisfies the following conditions:
(1) The center of the circle is inside this convex quadrilateral;
(2) The longest side is $a$, and the shortest side is $\sqrt{4-a^{2}}$. Tangents to the circle $\Gamma$ at points $A, ... | Let the center of $\odot \Gamma$ be $O$, and denote $\angle A O B=2 \theta_{1}$, $\angle B O C=2 \theta_{2}$, $\angle C O D=2 \theta_{3}$, $\angle D O A=2 \theta_{4}$. Thus, $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}$ are all acute angles and $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=\pi$. It is not difficult t... | \frac{8}{a^{2}\left(4-a^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,963 |
Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$, such that for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
（Supplied by Zhang Zhusheng） | To divide $X$ into the following 5 subsets for examination:
$$
\begin{array}{l}
2001=1024+977 \geqslant x \geqslant 1024-977=47, \\
46=32+14 \geqslant x \geqslant 32-14=18, \\
17=16+1 \geqslant x \geqslant 16-1=15, \\
14=8+6 \geqslant x \geqslant 8-6=2, \\
x=1 .
\end{array}
$$
To construct an example that does not sat... | 999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,964 |
Example 6: There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs 3.15 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs 4.20 yuan. Now, if you buy 1 piece each of A, B, and C, it will cost $\qquad$ yuan.
(1985. National Junior High Scho... | Solution: Let the cost of 1 item of A be $x$ yuan, 1 item of B be $y$ yuan, and 1 item of C be $z$ yuan, then the cost of 1 item each of A, B, and C is $(x+y+z)$ yuan. From the given conditions, we have
$$
\begin{array}{l}
3 x+7 y+z=3.15, \\
4 x+10 y+z=4.20 .
\end{array}
$$
Let
$$
\begin{array}{l}
x+y+z \\
=a(3 x+7 y+... | 1.05 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,965 |
Three, on each vertex of a regular $n$-sided polygon, there is a magpie. When startled, all the magpies fly away. After some time, they all return to these vertices, with one magpie on each vertex, but not necessarily to their original vertices. Find all positive integers $n$, such that there must exist 3 magpies, for ... | Three, Solution 1: When $n=3,4$, the conclusion is obviously true.
When $n=5$, in a regular pentagon with 5 diagonals, there are exactly 5 obtuse triangles and 5 acute triangles. When 5 magpies $A$, $B$, $C$, $D$, and $E$ land in the states shown in Figure 1, acute triangles and obtuse triangles transform into each oth... | n \geqslant 3, n \neq 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,966 |
Let $a$, $b$, $c$, $a+b-c$, $a+c-b$, $b+c-a$, and $a+b+c$ be 7 distinct prime numbers, and suppose that the sum of two of $a$, $b$, and $c$ is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$.
(Liang Darong, problem contributor) | $$
\begin{array}{l}
\text { Let's assume } a<b<c<d \text {, and } a, b, c, d \text { are all prime numbers. } \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, and $b+c$ is 800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \times 47$ and 798 are not prime numbers, but 797 is a prime num... | 1594 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,967 |
Five, divide a circle with a circumference of 24 into 24 equal segments, and select 8 points from the 24 points; such that the arc length between any two points is not equal to 3 and 8. How many different ways are there to select such 8-point groups? Explain your reasoning.
（Supplied by Li Chengzhang） | Solution 1: Number the 24 points as $1,2, \cdots, 24$, and arrange them in a $3 \times 8$ number table according to their "bad relationship":
$1,4,7,10,13,16,19,22$,
$9,12,15,18,21,24,3, \quad 6$,
$17,20,23, \quad 2, \quad 5, \quad 8, \quad 11, \quad 14$.
It is easy to see that the arc length between two points represe... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,968 |
4. Given two circles $C_{1}: x^{2}+y^{2}=1$ and $C_{2}$ : $(x-2)^{2}+y^{2}=16$. Then the locus of the center of the circle that is externally tangent to $C_{1}$ and internally tangent to $C_{2}$ is $\qquad$
$\qquad$ | 4. $\frac{(x-1)^{2}}{\frac{25}{4}}+\frac{y^{2}}{\frac{21}{4}}=1$ | \frac{(x-1)^{2}}{\frac{25}{4}}+\frac{y^{2}}{\frac{21}{4}}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,973 |
6. If $[x]$ denotes the greatest integer not exceeding the real number $x$, then the solution set of the equation $[\cot x]=2 \cos ^{2} x$ is $\qquad$
$\qquad$ | 6. $\left\{x \left\lvert\, x=k \pi+\frac{\pi}{2}\right.\right.$ or $\left.x=l \pi+\frac{\pi}{4}, k 、 l \in \mathbf{Z}\right\}$ | \left\{x \left\lvert\, x=k \pi+\frac{\pi}{2}\right.\right. \text{ or } \left.x=l \pi+\frac{\pi}{4}, k 、 l \in \mathbf{Z}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,975 |
Example 7 Let $x_{1}, x_{2}, \cdots, x_{7}$ all be integers, and
$$
\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6} \\
+49 x_{7}=1, \\
4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6} \\
+64 x_{7}=12, \\
9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6} \\
+81 x_{7}=123 .
\end{array}
$$
Find $1... | Solution: Since the coefficients of the same letters in the four equations are the squares of four consecutive natural numbers, i.e., $n^{2}, (n+1)^{2}, (n+2)^{2}, (n+3)^{2}$, to find the value of (4), it is necessary to express (4) using (1), (2), and (3), i.e., to express $(n+3)^{2}$ using $n^{2}, (n+1)^{2}, (n+2)^{2... | 334 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,976 |
II. (Problem 16) The permutation $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ of $1,2,3,4,5$ has the following property: for $1 \leqslant i \leqslant 4$, $a_{1}$, $a_{2} \cdots, a_{i}$ does not form a permutation of $1,2, \cdots, i$. Find the number of such permutations. | Obviously, $a_{1} \neq 1$.
When $a_{1}=5$, all $4!$ permutations meet the requirements.
When $a_{1}=4$, the $3!$ permutations where $a_{5}=5$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!)$.
When $a_{1}=3$, permutations in the form of $3 \times \times \times 5$ and $... | 71 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 711,980 |
Three, (This question is worth 16 points) How many ordered pairs of positive integers $(x, y)$ have the following properties:
$y<x \leqslant 100$, and $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are both integers? | Three, let $\frac{x+1}{y+1}=m(m \in \mathbf{N}, m>1)$, then $x=m y+(m-1)$.
Since $y \mid x$, it follows that $y \mid(m-1)$.
Let $m-1=k y(k \in \mathbf{N})$, then $m=k y+1$,
Substituting into (1), we get
$$
\begin{array}{l}
x=(k y+1) y+k y=k y(y+1)+y \leqslant 100, \\
k \leqslant \frac{100-y}{y(y+1)} .
\end{array}
$$
T... | 85 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,981 |
Four. (This question is worth 18 points) Let $p_{1}, p_{2}, \cdots, p_{n}$ be $n$ distinct prime numbers. Using these primes as terms (repetition allowed), form any sequence such that the product of no adjacent terms is a perfect square. Prove: The number of terms in such a sequence has a maximum value (denoted as $L(n... | Let $a_{1}, a_{2}, \cdots, a_{m}$ be a sequence with terms $p_{1}, p_{2}, \cdots, p_{n}$, where $m \geqslant 2^{n}$. Consider the sequence
$$
b_{0}=1, b_{1}=a_{1}, b_{2}=a_{1} a_{2}, \cdots, b_{m}=a_{1} a_{2} \cdots a_{m},
$$
then each $b_{i}$ can be written as $p_{1}^{(i)} p_{2}^{(i)} \cdots p_{n}^{a_{n}^{(i)}}$, whe... | L(n)=2^{n}-1 | Number Theory | proof | Yes | Yes | cn_contest | false | 711,982 |
1. Among the following four equations, the incorrect one is ( ).
(A) $1-a-b+a b=(1-a)(1-b)$
(B) $1-a+b+a b=(1-a)(1+b)$
(C) $1+a+b+a b=(1+a)(1+b)$
(D) $1+a-b-a b=(1+a)(1-b)$ | -、1.(B).
By swapping the shadows of $a$ and $b$, it can be found that (B) and (D) cannot be converted into each other, so one of them must be wrong. Also, by expanding $(1-a)(1+b)$, there should be two minus signs $-a-ab$, indicating that (B) does not hold. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,984 |
2. Given that $a$ is a non-negative integer, if the equation $2 x-a \sqrt{1-x}-a+4=0$ has at least one integer root. Then the number of possible values for $a$ is ( ).
(A) 1
(i) 2
(C) 3
(D) 4 | 2. (C).
From the meaningful radicand, we know $x \leqslant 1$.
Also, $2 x+4=a(\sqrt{1-x}+1) \geqslant 0$,
so $x \geqslant-2$.
When $x=1,0,-2$, $a$ takes the values $6,2,0$ respectively.
When $x=-1$, $a$ is not an integer.
Therefore, $a$ has a total of 3 possible values. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,985 |
3. In quadrilateral $A B C D$, $B C=8, C D=1$, $\angle A B C=30^{\circ}, \angle B C D=60^{\circ}$. If the area of quadrilateral $A B C D$ is $\frac{13 \sqrt{3}}{2}$, then the value of $A B$ is ( ).
(A) $\sqrt{3}$
(B) $2 \sqrt{3}$
(C) $3 \sqrt{3}$
(D) $4 \sqrt{3}$ | 3. (C).
As shown in Figure 3, since $\angle A B C$ and $\angle B C D$ are complementary, extending $B A$ and $C D$ to intersect at $E$, we have $\angle B E C=90^{\circ}$, and
$$
\begin{array}{l}
C E=\frac{1}{2} B C=4, \\
B E=\frac{\sqrt{3}}{2} B C=4 \sqrt{3} .
\end{array}
$$
We have $S_{\triangle A D E}=S_{\triangle ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,986 |
Example 8 Try to find a 3rd degree polynomial $y(x)$, such that it satisfies $y(0)=y(1)=0, y(2)=1, y(3)=5$.
(Guangdong Province Olympiad 1987 Winter Training) | Solution: Let $y(x)=a x^{3}+b x^{2}+c x+d$. From the given information, we have
$$
\left\{\begin{array}{l}
d=0, \\
a+b+c+d=0, \\
8 a+4 b+2 c+d=1, \\
27 a+9 b+3 c+d=5 .
\end{array}\right.
$$
Solving, we get $a=\frac{1}{3}, b=-\frac{1}{2}, c=\frac{1}{6}, d=0$.
$$
\therefore y(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}+\frac... | y(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}+\frac{1}{6} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,987 |
4. Among the following 3 geometric propositions,
(1) Two similar triangles, if their perimeters are equal, then these two triangles are congruent;
(2) Two similar triangles, if two sets of sides are equal, then these two triangles are congruent;
(3) Two similar triangles, the angles (not obtuse) formed by their corresp... | 4. (B).
The first proposition is true. Let the side lengths of two similar triangles be $a, b, c$ and $a_{1}, b_{1}, c_{1}$, respectively. Then, by similarity,
$$
\frac{a}{a_{1}}=\frac{b}{b_{1}}=\frac{c}{c_{1}}=\frac{a+b+c}{a_{1}+b_{1}+c_{1}}=1,
$$
which implies $a=a_{1}, b=b_{1}, c=c_{1}$.
Therefore, the two triangl... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,988 |
5. The ground floor of a hotel has 5 fewer rooms than the second floor. A tour group has 48 people. If all are arranged to stay on the ground floor, each room can accommodate 4 people, but there are not enough rooms; if each room accommodates 5 people, some rooms are not fully occupied. If all are arranged to stay on t... | 5. (B).
Let the number of guest rooms on the ground floor be $x$, then the number of rooms on the second floor is $x+5$. According to the problem, we have the system of inequalities
$$
\left\{\begin{array}{l}
\frac{48}{5}<x<\frac{48}{4}, \\
\frac{48}{4}<x+5<\frac{48}{3} .
\end{array}\right.
$$
which simplifies to $\l... | 10 | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 711,989 |
6. As shown in Figure 1, in the acute triangle $\triangle ABC$, the three altitudes $AD$, $BE$, and $CF$ intersect at $H$. Connect $DE$, $EF$, and $FD$. Then the number of triangles in the figure is ( ).
(A) 63
(B) 47
(C) 45
(D) 40 | 6. (B).
The number of triangles with $A$ as a vertex is 13 (excluding $\triangle A B C$); similarly, the number of triangles with $B, C$ as vertices is also 13 each. In total, there are 39.
In $\triangle D E F$, the number of triangles with $D$ as a vertex is 8 (excluding $\triangle D E F$); similarly, the number of ... | 47 | Geometry | MCQ | Yes | Yes | cn_contest | false | 711,990 |
$$
\begin{array}{l}
\text { 1. Let } \\
A=\frac{8}{3} \sqrt{2 \sqrt{13}-5}+2 \sqrt{2 \sqrt{13}-2}, \\
B=\frac{2}{3} \sqrt{2 \sqrt{13}-5}+2 \sqrt{2 \sqrt{17}+7} .
\end{array}
$$
Then $A-B$ can be simplified to | $$
\begin{array}{l}
\text { II, 1.0. } \\
\begin{aligned}
A-B= & 2 \sqrt{2 \sqrt{13}-5}+2 \sqrt{2 \sqrt{13}-2} \\
& -2 \sqrt{2 \sqrt{13}+7} .
\end{aligned}
\end{array}
$$
Below we prove
$$
\sqrt{2 \sqrt{13}-5}+\sqrt{2 \sqrt{13}-2}=\sqrt{2 \sqrt{13}+7} \text {. }
$$
Squaring, we get $\sqrt{62-14 \sqrt{13}}=7-\sqrt{13}... | A-B=0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,991 |
2. If the integer point $(n, m)$ in the first quadrant lies on the parabola $y=19 x^{2}-98 x$, then the minimum value of $m+n$ is $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.102.
From $m=(19 n-98) n$, we know that there exists a positive integer $k$, such that $k=19 n-98$.
Taking $n=1,2, \cdots, 6$, we find that when $n$ takes the minimum value 6, $k$ takes the minimum positive integer value 16, thus
$$
m+n=n k+n=6 \times 17=102 .
$$ | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 711,992 |
3. Given the radii of $\odot O_{1}$ and $\odot O_{2}$ are 3 and 5, respectively, and $O_{1} O_{2}=10$. Then the area of the triangle formed by the two inner common tangents and one outer common tangent of the two circles is | 3. $\frac{45}{4}$.
As shown in Figure 5, let the internal common tangent $AB$ and $EF$ intersect at $O$, and extend $AB$ and $EF$ to intersect the external common tangent $CD$ at $N$ and $M$. Connect $O_{1}A$ and $O_{2}B$. From $\mathrm{Rt} \triangle A O_{1} O \sim \mathrm{Rt} \triangle \mathrm{BO}_{2} \mathrm{O}$, we... | \frac{45}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,993 |
4. Given "two triangles, with two sides and one angle corresponding equal", please add conditions to make these two triangles congruent: $\qquad$
$\qquad$ (Example: If the equal angle is the included angle of the equal sides, then these two triangles are congruent.) | 4. The answer is not unique.
Options available:
(1) If this angle is a right angle, then the two triangles are congruent;
(2) If these two sides are equal, then the two triangles are congruent;
(3) If the opposite side of this angle is the larger of the two sides, then the two triangles are congruent;
(4) If this angl... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 711,994 |
One, (Full marks 20 points) Given positive numbers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}+c^{2}=4^{2}, \\
b^{2}+c^{2}=5^{2} .
\end{array}\right.
$$
Find the range of $a^{2}+b^{2}$. | Let $a^{2}+b^{2}=k$. From the given information, we have
$$
b^{2}-a^{2}=\left(b^{2}+c^{2}\right)-\left(a^{2}+c^{2}\right)=5^{2}-4^{2}=9 .
$$
Solving this, we get $a^{2}=\frac{k}{2}-9$
$$
\begin{array}{l}
i^{2}=\frac{k+9}{2}>9, \\
c^{2}=\frac{41-k}{2}>0 .
\end{array}
$$
Thus, $9<k<41$.
That is, the value of $a^{2}+b^{... | 9<a^{2}+b^{2}<41 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,995 |
II. (Full marks 25 points) As shown in Figure 2, $PA$ and $PB$ are two tangents to $\odot O$, $PEC$ is a secant, and $D$ is the intersection of $AB$ and $PC$.
1. When $PEC$ passes through the center, prove that $PE \cdot CD = PC \cdot DE$.
2. When $PEC$ does not pass through the center, does $PE \cdot CD = PC - DE$ hol... | II. 1. As shown in Figure 8, by the power of a point theorem, we have
$$
\begin{array}{l}
P E \cdot P C = P A^{2}, \\
D E \cdot C D = A D^{2},
\end{array}
$$
Subtracting $P E \cdot P C - D E \cdot C D$,
$$
\begin{array}{l}
= P A^{2} - A D^{2} = P D^{2} \\
= (P C - C D)(P E + D E) \\
= P C \cdot P E + P C \cdot D E - C... | x = \frac{\sqrt{17} - 3}{2} | Geometry | proof | Yes | Yes | cn_contest | false | 711,996 |
Three. (Full marks 25 points) Take any 2001 distinct positive numbers. Prove that it is always possible to split one of them into two smaller positive numbers, such that when these 2002 numbers are evenly divided into two groups, the sum of the 1001 positive numbers in each group is equal, and the two new numbers are i... | Proof 1: Arrange these 2001 numbers in ascending order, and denote the $i$-th number as $a_{i}$, then
$$
a_{1}0 . \\
y= & \frac{a_{2001}-B+A}{2} \\
= & \frac{1}{2}\left[\left(a_{2}+a_{4}+\cdots+a_{50}\right)+\left(a_{51}+a_{53}+\cdots+\right.\right. \\
& \left.a_{1999}+a_{2001}\right)-\left(a_{1}+a_{3}+\cdots+a_{49}\ri... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 711,997 |
Example 9 If the equation $\left(x^{2}-1\right)\left(x^{2}-4\right)=k$ has 4 non-zero real roots, and the 4 points corresponding to them on the number line are equally spaced, then $k=$ $\qquad$ | Solution: Since the given equation is a biquadratic equation, its 4 non-zero real roots correspond to 4 equidistant points on the number line that are symmetric about the origin. Therefore, we can set
$$
\begin{array}{l}
\left(x^{2}-1\right)\left(x^{2}-4\right)-k \\
=(x+3 a)(x+a)(x-a)(x-3 a),
\end{array}
$$
which is
$... | \frac{7}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 711,998 |
1. If real numbers $a, x$ satisfy $a>x>1$. And
$$
\begin{array}{l}
A=\log _{a}\left(\log _{a} x\right), \\
B=\left(\log _{a} x \right)^{2}, C=\log _{a} x^{2},
\end{array}
$$
then the following correct relationship is ( ).
(A) $A>C>B$
(B) $C>B>A$
(C) $B>C>A$
(D) $C>A>B$ | -1 (B).
$$
\begin{array}{l}
\because A=\log _{a}\left(\log _{a} x\right)=\log _{a}\left(\frac{\lg x}{\lg a}\right), 1\left(\log _{a} x\right)^{2}=B>0, \\
\therefore C>B>A .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 711,999 |
2. In the complex plane, there are 7 points corresponding to the 7 roots of the equation $x^{7}=$ $-1+\sqrt{3} i$. Among the four quadrants where these 7 points are located, only 1 point is in ( ).
(A) the I quadrant
(B) the II quadrant
(C) the III quadrant
(D) the IV quadrant | 2. (C).
$$
\begin{array}{l}
\because x^{7}=-1+\sqrt{3} \mathrm{i}=2\left(\cos 120^{\circ} + i \sin 120^{\circ}\right), \\
\therefore x_{n}=2^{\frac{1}{7}}\left(\cos \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}+\sin \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}\right) .
\end{array}
$$
where $n=0,1,2, \cdots, 6$.
After calcu... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,000 |
3. Given two propositions:
(I ) The dihedral angles formed by adjacent lateral faces of the pyramid $V-A B C D$ are all equal;
(II ) The angles formed by adjacent lateral edges of the pyramid $V-A B C D$ are all equal.
For the above two propositions ().
(A)( I ) is a sufficient but not necessary condition for ( II )
(B... | 3. (D).
Construct two counterexamples: As shown in Figure 2, the quadrilateral pyramid $P-ABCD$ has a rectangular base $ABCD$ (not a square), which satisfies (I) but not (II); As shown in Figure 3, the quadrilateral pyramid $P-ABCD$ has a rhombic base $ABCD$ (not a square), which satisfies (II) but not (I). | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,001 |
4. Given the sequence $\left\{a_{n}\right\}(n \geqslant 1)$ satisfies $a_{n+2}=$ $a_{n+1}-a_{n}$ and $a_{2}=1$. If the sum of the first 1999 terms of this sequence is 2000, then the sum of the first 2000 terms equals ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002 | 4. (C).
From the given $a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}$ $=-a_{n}$, i.e., $a_{n+3}+a_{n}=0$, therefore $a_{n}+a_{n+1}+a_{n+2}+$ $a_{n+3}+a_{n+4}+a_{n+5}=0$, which means the sum of any 6 consecutive terms of the sequence is 0. Noting that $1999=6 \times 333+1,2000=6 \times 333+$ 2, we can con... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 712,002 |
5. Given real numbers $a, b$ satisfy $a>b>0, c$ satisfies that a triangle can be formed with $a, b, c$ as its side lengths. If the largest angle among the three interior angles of the triangle is to have the minimum value, then $c$ should be equal to ( ).
(A) $a$
(B) $\frac{a+b}{2}$
(C) $\sqrt{a b}$
(D) $b$ | 5. (A).
As shown in Figure 4, let the sides opposite to the three interior angles of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Assuming points $A$ and $C$ are fixed, it is evident that vertex $B$ moves on the circumference of a circle centered at $C$ with radius $a$. Since in two triangles with two correspond... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,003 |
6. In space, there are 4 non-coplanar fixed points. The number of parallelepipeds that can be formed with these 4 points as vertices is ( ).
(A) 20
(B) 32
(C) 25
(D) 29 | 6. (D).
Each parallelepiped is uniquely determined by the designated 1 vertex and 3 median planes (each of these median planes is equidistant from all vertices of the parallelepiped). For any given 4 non-coplanar points, there exist 7 planes equidistant to these 4 points. Choosing any 3 out of these 7, there are $\math... | 29 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 712,004 |
2. In $\triangle A B C$, $\angle A$ and $\angle B$ satisfy $3 \sin A+4 \cos B=6, 4 \sin B+3 \cos A=1$. Then, the size of $\angle C$ is $\qquad$ | 2. $\frac{\pi}{6}$.
From the given, we have
$$
\begin{array}{l}
9 \sin ^{2} A+16 \cos ^{2} B+24 \sin A \cdot \cos B=36, \\
16 \sin ^{2} B+9 \cos ^{2} A+24 \sin B \cdot \cos A=1 .
\end{array}
$$
Adding the two equations, we get
$$
25+24(\sin A \cdot \cos B+\sin B \cdot \cos A)=37,
$$
which simplifies to $24 \sin (A+B... | \frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,006 |
3. Let $P$ and $Q$ be two points inside the square pyramid $S-ABCD$, with the lateral edge length being $\frac{\sqrt{3}}{2} a$ and the base edge length being $a$. Then the range of $\angle PSQ$ is $\qquad$ | 3. $\left(0, \pi-\arccos \frac{1}{3}\right)$.
As shown in Figure 5, it is clear that
$\angle P S Q \geqslant 0$, when $P$ and $Q$ coincide with the diagonal vertices of the base, $\angle P S Q$ reaches its maximum value, $B D=\sqrt{2} a, S B=$ $S D=\frac{\sqrt{3}}{2} a$; then
$$
\cos \angle B S D=\frac{S B^{2}+S D^{2}... | \left(0, \pi-\arccos \frac{1}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,007 |
Example 10 has the equation
$$
x^{4}-6 x^{3}+8 x^{2}+8 x-16=0
$$
has a root of $1+\sqrt{5}$. Solve this equation. | Solution: From the fact that irrational roots of a rational coefficient equation come in conjugate pairs, we know that $1-\sqrt{5}$ is also a root of the original equation. Therefore, the 4th-degree polynomial on the left side of the original equation contains the factor
$$
(x-1+\sqrt{5})(x-1-\sqrt{5})=x^{2}-2 x-4 \tex... | x_{1,2}=1 \pm \sqrt{5}, x_{3,4}=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,009 |
5. The elements of set $A$ are positive integers, with the property: if $a \in A$, then $42-a \in A$. Therefore, the number of such sets $A$ is $\qquad$.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
5. The elements of set $A$ are positive intege... | $5.2^{21}-1$
It is easy to know that $A$ is a subset of $\{1,2, \cdots, 41\}$, and let $C=$ $\{(1,41),(2,39),(3,38), \cdots,(20,22)\}$. If the number of elements in $A$ is even, we only need to take several pairs from $C$, the number of ways to take is $2^{20}$ -1; if the number of elements in $A$ is odd, then $21 \i... | 2^{21}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,010 |
6. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a>b>$ $0$), point $P$ is a point on it, $F_{1} 、 F_{2}$ are the foci of the ellipse, and the external angle bisector of $\angle F_{1} P F_{2}$ is $l$. Construct $F_{1} 、 F_{2}$ perpendicular to $l$ at $R 、 S$. When $P$ traverses the entire ellipse... | 6. $x^{2}+y^{2}=a^{2}$.
As shown in Figure 7, construct the symmetric point $F^{\prime}$ of $F_{2}$ with respect to $l$, $\angle F^{\prime} P S = \angle F_{2} P S$, and $l$ is the external angle bisector of $\angle F_{1} P F_{2}$. Then, $F^{\prime}$, $P$, and $F_{1}$ are collinear.
Let $S\left(x_{0}, y_{0}\right)$, $... | x^{2} + y^{2} = a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 712,011 |
Four. (20 points) Let $p, q$ be complex numbers $(q \neq 0)$. If the roots of the equation $x^{2} + p x + q^{2} = 0$ have equal modulus, prove: $\frac{p}{q}$ is a real number. | Let $z_{1}, z_{2}$ be the roots of $x^{2} + px + q^{2} = 0$.
Then $z_{1} + z_{2} = -p, z_{1} z_{2} = q^{2}$.
Given that $\left|z_{1}\right| = \left|z_{2}\right|$, then $\left|z_{1}\right|^{2} = \left|z_{2}\right|^{2}$, so $z_{1} \bar{z}_{1} = z_{2} \bar{z}_{2}$.
Thus, $\frac{p^{2}}{q^{2}} = \frac{\left(z_{1} + z_{2}\ri... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,013 |
Five. (20 points) Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, a line $l$ is drawn through one of its foci, intersecting $C$ at points $P$ and $Q$. $A_{1}$ and $A_{2}$ are the endpoints of the real axis of $C$. Prove that for any line $l$ satisfying the conditions, the intersection of ... | After exploration, it is conjectured that the intersection points of $P A_{1}$ and $Q A_{2}$ are only on the corresponding directrix of $F$, which is $x=-\frac{a^{2}}{c}$.
Let $F(\cdots c, 0)$, $P\left(x_{1}, y_{1}\right)$, $Q\left(x_{2}, y_{2}\right)$, $A_{1}(-a, 0)$, and $A_{2}(a, 0)$.
At the same time, let $P A_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,014 |
一、(5/ Insult)
As shown in Figure 1, in rhombus $ABCD$, $\angle A=120^{\circ}$, $\odot O$ is the circle passing through points $A$, $B$, and $C$. $M$ is a point outside the rhombus, connecting $MC$ intersects $AB$ at $E$, and $AM$ intersects the extension of $CB$ at $F$. Prove that the necessary and sufficient condition... | One, Sufficiency.
$\because M$ is on $\odot O$, connect $A C$, then
$$
\angle A M C=\angle A B C=60^{\circ} \text {, }
$$
$\therefore \angle A M C=\angle A C B$, thus $\triangle A M C \sim \triangle A C F$.
$$
\therefore \frac{M C}{M A}=\frac{C F}{C A}=\frac{C F}{C D} \text {. }
$$
Also, $\because \angle A M C=\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,015 |
Sure, here is the translated text:
```
II. (50 points) Let $x_{1}, x_{2}, \cdots, x_{2000}$ be 2000 real numbers, satisfying $x_{i} \in [0,1] (i=1,2, \cdots, 2000)$. Define
$$
F_{i}=\frac{x_{i}^{2000}}{\sum_{j=1}^{2000} x_{j}^{3999} - x_{i}^{3999} + 2000}.
$$
Find the maximum value of $\sum_{i=1}^{2000} F_{i}$, and p... | Two, the maximum value of $\sum_{i=1}^{2000} F_{i}$ is $\frac{2000}{3999}$.
Actually, since $x_{i} \in[0,1]$, first
$$
\sum_{i=1}^{2000} F_{i} \leqslant \frac{\sum_{i=1}^{2000} x_{i}^{2000}}{\sum_{i=1}^{2000} x_{i}^{3999}+1999},
$$
Let the right side of the inequality be $S$, we only need to prove $S \leqslant \frac{2... | \frac{2000}{3999} | Algebra | proof | Yes | Yes | cn_contest | false | 712,016 |
Three, (50 points) A conference was attended by $12 k$ people $(k \in$ $\mathbf{N}$ ), where each person has greeted exactly $3 k+6$ other people. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference? | Three, connecting the points of people and their acquaintances, each point forms $\mathrm{C}_{3 k+6}^{2}$ angles, each simple connection corresponds to a vertex, so the graph has a total of $12 k \cdot C_{3 k+6}^{2}$ angles. On the other hand, suppose for any two people, there are $n$ people who know both of them, thus... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,017 |
Given positive real numbers $a \geqslant b \geqslant c$. Prove:
$$
\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+a b c \geqslant a+b+c+1 .
$$ | Proof: Consider the following cases.
$1^{\circ}$ When $1 \geqslant a \geqslant b \geqslant c>0$,
$$
\begin{array}{l}
\because(1-a)(1-b) \geqslant 0, \\
\therefore 1-a b \leqslant 2-a-b . \\
\therefore c(1-a b) \leqslant 1-a b \leqslant 2-a-b . \\
\therefore a b \geqslant a+b+c-2 .
\end{array}
$$
$\therefore$ The left s... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 712,018 |
Initial 98
In $\triangle ABC$,
$BM$ and $CN$
are medians, $D$
is any point on
side $BC$,
draw $IE \parallel BM$,
$DF \parallel CN$, intersecting $AC$ and $AB$ at points $E$ and $F$ respectively, and line segment $EF$ intersects medians $BM$ and $CN$ at points $P$ and $Q$ respectively. Prove: $FP = PQ = QE$. | Prove: Connect $R S$.
From $I E / / B M$ we get $\frac{D S}{S E}=\frac{B G}{G M}=2$.
Similarly, we can get $\frac{D R}{R F}=2$.
Therefore, $R S // E F$.
$$
\begin{array}{l}
\because \triangle S E Q \sim \triangle D E F, \\
\therefore \frac{E Q}{E F}=\frac{E S}{E D}=\frac{M G}{M B}=\frac{1}{3} .
\end{array}
$$
That is,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,019 |
Example 1 Let $P$ and $Q$ be two points on line segment $BC$, and $BP = CQ$. $A$ is a moving point outside $BC$ (as shown in Figure 1). When point $A$ moves to
make $\angle BAP = \angle CAQ$,
what type of triangle is $\triangle ABC$? Prove your conclusion.
(1986, National Junior High School Mathematics League) | When point $A$ moves to make $\angle B A P=\angle C A Q$, $\triangle A B C$ is an isosceles triangle.
Proof: As shown in Figure 1, draw lines through points $P$ and $B$ parallel to $A C$ and $A Q$ respectively, and let their intersection be point $D$. Connect $D A$.
In $\triangle D B P$ and $\triangle A Q C$, it is cl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,020 |
Example 2 As shown in Figure 2, quadrilateral $A B C D$ is a parallelogram, $\angle B A F=\angle B C E$. Prove: $\angle E B A=$ $\angle A D E$.
(1979, Qingdao City Mathematics Competition) | Proof: As shown in Figure 2, draw lines through points $A$ and $B$ parallel to $ED$ and $EC$, respectively, and let their intersection be point $P$. Connect $PE$.
Given $AB \cong CD$, it is easy to see that $\triangle PBA \cong \triangle ECD$. Therefore, $PA = ED$ and $PB = EC$.
Clearly, quadrilaterals $PBCE$ and $PA... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,021 |
Example 2 When $x$ is any integer, prove that $x^{9}-$ $6 x^{7}+9 x^{5}-4 x^{3}$ can be divided by 8640. | Analysis: Noting that $8640=2^{6} \times 3^{3} \times 5$, it is only necessary to prove that the given polynomial can be divided by $2^{6}$, $3^{3}$, and 5 respectively. Since these two factors have a coprime relationship, it can be concluded that the conclusion holds. Therefore, the known polynomial should be decompos... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 712,022 |
6. Point $P$ is in the same plane as the square $A B C D$ with side length $\sqrt{2}$, and $P A^{2}+P B^{2}=P C^{2}$. Then the maximum value of $P D$ is ( ).
(A) 4
(B) $2+2 \sqrt{2}$
(C) 6
(D) $2+3 \sqrt{2}$ | 6. (B).
To maximize $PD$, point $P$ and side $CD$ should be on opposite sides of $AB$. Draw $PE \perp AB$ at $E$ (as shown in Figure 3). Let $PE=x$ and $AE=y$. Then
$$
\begin{array}{l}
BE=\sqrt{2}-y, \\
PA^{2}=PE^{2}+AE^{2}=x^{2}+y^{2}, \\
PB^{2}=PE^{2}+BE^{2}=x^{2}+(\sqrt{2}-y)^{2}, \\
PC^{2}=BE^{2}+(PE+BC)^{2} \\
=(... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 712,024 |
1. Cut a wire of length $143 \mathrm{~cm}$ into $n$ small segments $(n \geqslant 3)$, with each segment no less than $1 \mathrm{~cm}$. If no three segments can form a triangle, the maximum value of $n$ is | 2.1.10.
Each segment should be as small as possible, and the sum of any two segments
should not exceed the third segment. A $143 \mathrm{~cm}$ wire should be divided into 10
such segments:
$1 \mathrm{~cm}, 1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}$,
$13 \mathrm{~cm}, 21 \mathrm{~cm}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,025 |
3. Given that $a$, $b$, $c$, $d$ are the thousands, hundreds, tens, and units digits of a four-digit number, respectively, and the digits in lower positions are not less than those in higher positions. When $|a-b|+|b-c|+|c-d|+|d-a|$ takes the maximum value, the maximum value of this four-digit number is $\qquad$ . | $$
\begin{array}{l}
\text { 3. } 1999 . \\
\because a \leqslant b \leqslant c \leqslant d, \\
\therefore a-b \leqslant 0, b-c \leqslant 0, c-d \leqslant 0, d-a \geqslant 0 . \\
\therefore|a-b|+|b-c|+|c-d|+|d-a| \\
=b-a+c-b+d-c+d-a \\
=2(d-a) .
\end{array}
$$
When $d=9, a=1$,
$$
|a-b|+|b-c|+|c-d|+|d-a|
$$
has a maximu... | 1999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 712,027 |
4. A monkey is climbing an 8-rung ladder, each time it can climb one rung or jump two rungs, and at most jump three rungs. From the ground to the top rung, there are $\qquad$ different ways to climb and jump. | 4.81.
Starting with simple cases:
(1) If there is 1 step, then there is only one way to climb. That is, $a_{1}=1$.
(2) If there are 2 steps, then there are 2 ways to climb:
(1) Climb one step at a time;
(2) Leap two steps at once, i.e., $a_{2}=2$.
(3) If there are 3 steps, then there are 4 ways to climb:
(1) Climb one... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 712,028 |
One, (20 points) In the range of positive real numbers, there is only one number that is a solution to the equation $\frac{x^{2}+k x+3}{x-1}=3 x+k$ with respect to $x$. Find the range of real values for $k$.
---
The above text has been translated into English, maintaining the original text's line breaks and format. | One, the original equation is transformed into
$$
2 x^{2}-3 x-(k+3)=0 \text {. }
$$
(1) When $\Delta=0$ for equation (1), that is
$$
(-3)^{2}-4 \times 2 \times[-(k+3)]=0 \text {. }
$$
Solving gives $k=-\frac{33}{8}$.
At this time, the two roots of equation (1) are $x_{1}=x_{2}=\frac{3}{4}$.
Upon verification, $x=\frac... | k=-\frac{33}{8} \text{ or } -4 \text{ or } k \geqslant -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 712,029 |
II. (25 points) As shown in Figure 1, it is known that $\odot O_{1}$ passes through two vertices $A$ and $B$ of trapezoid $ABCD$ and is tangent to the side $CD$ at $N$; $\odot O_{2}$ passes through points $C$ and $D$ and is tangent to the side $AB$ at point $M$. Prove:
$$
A M \cdot M B = C N \cdot N D
$$ | $$
\begin{array}{l}
\text{Since } AD \parallel BC, \\
\therefore \frac{OB}{OA}=\frac{OC}{OD}=t^{2} .
\end{array}
$$
Thus, $OB = OA \cdot t^{2} = a t^{2}$,
$$
OC = OD \cdot t^{2} = b t^{2} \text{. }
$$
Since $ON$ is tangent to $\odot \bigcirc_{1}$,
$$
\begin{array}{l}
\text{we have } ON^{2} = OA \cdot OB \\
= a^{2} t^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 712,030 |
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