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6. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$. Then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$ . | 6.4.
Definition: $f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$.
It can be rewritten as $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\right)^{x}+\left(\frac{5}{6}\right)^{x}$.
It is easy to see that the function $f(x)=\left(\frac{3}{6}\right)^{x}+\left(\frac{4}{6}\... | 6.4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,347 |
3. (20 points) As shown in Figure 1, in rectangle $ABCD$, $AB = \sqrt{3}$, $BC = a$. Also, $PA \perp$ plane $ABCD$, $PA = 4$.
(1) If there exists a point $Q$ on side $BC$ such that $PQ \perp QD$, find the range of values for $a$;
(2) When there is exactly one point $Q$ on $BC$ such that $PQ \perp QD$, find the angle be... | (1) Let $B Q=t$, then
$P Q^{2}=19+t^{2}, \varphi()^{2}=3+(a-t)^{2}$,
$P I)^{2}=16+a^{2}$.
If $P Q \perp Q D$, we get $19+t^{2}+3+(a-1)^{2}=16+a^{2}$,
thus $t^{2}-a t+3=0$.
From $\Delta=a^{2}-12 \geqslant 0$, solving gives $a \geqslant 2 \sqrt{3}$.
(2) For the minimum value of $B C$: there exists a unique point $Q$, suc... | \arccos \frac{\sqrt{15}}{5} \text{ or } \arccos \frac{\sqrt{7}}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,348 |
Four. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ and a fixed point $P(t, 0)(t>0)$, a line $l$ with a slope of $\frac{1}{2}$ passes through point $P$ and intersects the ellipse $C$ at two distinct points $A$ and $B$. For any point $M$ on the ellipse, there exists $\theta \in[0,2 \pi]$, such that $O M=\co... | Let $M(x, y)$ and points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. The equation of line $AB$ is $x-2 y-t=0$.
From the cubic equation, we get
$2 x^{2}-2 t x+t^{2}-4=0$ and $8 y^{2}+4 t y+t^{2}-4=0$.
Thus, $x_{1} x_{2}=\frac{t^{2}-4}{2}$ and $y_{1} y_{2}=\frac{t^{2}-4}{8}$.
Given $O M=\cos \theta \cdot O ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,349 |
Five. (20 points) Given that the rhombus $A B C D$ is an inscribed quadrilateral of the ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
(1) Prove that $\frac{1}{O A^{2}}+\frac{1}{O B^{2}}$ is a constant;
(2) Find the maximum and minimum values of the area of the rhombus $A B C D$. | (1) As shown in Figure 3, in the rhombus $\triangle BCD$, let $OA=m, OB=n$, $\angle A Ox=\alpha$. Then the point $A(m \cos \alpha, m \sin \alpha)$, $B(-n \sin \alpha, n \cos \alpha)$. Since points $A$ and $B$ are both on the ellipse $C$, we have $\frac{m^{2} \cos ^{2} \alpha}{a^{2}}+\frac{m^{2} \sin ^{2} \alpha}{b^{2}}... | \frac{4 a^{2} b^{2}}{a^{2}+b^{2}} \text{ and } 2 a b | Geometry | proof | Yes | Yes | cn_contest | false | 717,350 |
Six. (20 points) Let two positive real numbers $x, y, z$. Try to find the maximum value of the algebraic expression $\frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z}$.
| $$
\begin{array}{l}
\text { Six, because } \frac{16 x+9 \sqrt{2 x y}+9 \sqrt[3]{3 x y z}}{x+2 y+z} \\
=\frac{16 x+\frac{9 \sqrt{x \cdot 18 y}}{3}+\frac{3 \sqrt[3]{x \cdot 18 y \cdot 36 z}}{2}}{x+2 y+z} \\
\leqslant \frac{16 x+\frac{3(x+18 y)}{2}+\frac{x+18 y+36 z}{2}}{x+2 y+z}=18,
\end{array}
$$
Therefore, the equalit... | 18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,351 |
1. Let $S=\{(x, y) \mid x y>0\}, T=\{(x, y) \mid$ $x>0$ and $y>0\}$. Then ().
(A) $S \cup T=S$
(B) $S \cup T=T$
(C) $S \cap T=S$
(D) $S \cap T=\varnothing$ | $$
\begin{array}{l}
\text { I.1.A. } \\
\text { Given } S=\{(x, y) \mid x y>0\} \\
=\{(x, y) \mid x>0, y>0\} \cup\{(x, y) \mid x<0, y<0\},
\end{array}
$$
we know that $T \subset S$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,352 |
1. Let $x_{1}, x_{2}$ be the real roots of the equation
$$
2 x^{2}-4 m x+2 m^{2}+3 m-2=0
$$
When $m$ is what value, $x_{1}^{2}+x_{2}^{2}$ has the minimum value? And find this minimum value.
(2001, Jiangsu Province Junior High School Mathematics Competition) | Answer: When $m=\frac{2}{3}$, $x_{1}^{2}+x_{2}^{2}$ attains its minimum value $\frac{8}{9}$. ) | \frac{8}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,353 |
2. If $f(x)=\frac{1}{x}$ has a domain $A$, and $g(x)=f(x+1)-f(x)$ has a domain $B$, then ( ).
(A) $A \cup B=\mathbf{R}$
(B) $A \supsetneqq B$
(C) $A \subseteq B$
(D) $A \cap B=\varnothing$ | 2. B.
From the problem, we know
$$
A=\{x \mid x \neq 0\}, B=\{x \mid x \neq 0 \text { and } x \neq 1\} \text {. }
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,354 |
3. Given $\tan \alpha>1$, and $\sin \alpha+\cos \alpha<0$
(A) $\cos \alpha>0$
(B) $\cos \alpha<0$
(C) $\cos \alpha=0$
(D) The sign of $\cos \alpha$ is uncertain | 3. B.
From $\tan \alpha>0$, we know that $\alpha$ is in the first and third quadrants.
From $\sin \alpha+\cos \alpha<0$, we know that $\alpha$ is in the second, third, and fourth quadrants. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,355 |
4. Let $a>0, a \neq 1$. If the graph of the inverse function of $y=a^{x}$ passes through the point $\left(\frac{\sqrt{2}}{2},-\frac{1}{4}\right)$, then $a=(\quad)$.
(A) 16
(B) 4
(C) 2
(D) $\sqrt{2}$ | 4. B.
From the problem, we know that the graph of the function $y=a^{x}$ passes through the point $\left(-\frac{1}{4}, \frac{\sqrt{2}}{2}\right)$, i.e., $a^{-\frac{1}{4}}=\frac{\sqrt{2}}{2}$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,356 |
5. Given $a \neq 0$. The necessary and sufficient condition for the graph of the function $f(x)=a x^{3}+b x^{2}+c x +d$ to be symmetric with respect to the origin is ( ).
(A) $b=0$
(B) $c=0$
(C) $d=0$
(D) $b=d=0$ | 5.D.
It is known that $f(x)$ is an odd function, so $b=d=0$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,357 |
6. If the three sides of $\triangle A B C$ are $a=\sin \frac{3}{4}$, $b=\cos \frac{3}{4}$, $c=1$, then the size order of $\angle A, \angle B, \angle C$ is ( ).
(A) $\angle A<\angle B<\angle C$
(B) $\angle B<\angle A<\angle C$
(C) $\angle C<\angle B<\angle A$
(D) $\angle C<\angle A<\angle B$ | 6. A.
Since $\frac{3}{4}\cos \frac{3}{4} > \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.
Then $a < b < c$. Hence $\angle A < \angle B < \angle C$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,358 |
7. If the real number $x$ satisfies $\log _{2} x=3+2 \cos \theta$, then $|x-2|+|x-33|$ equals ( ).
(A) $35-2 x$
(B) 31
(C) $2 x-35$
(D) $2 x-35$ or $35-2 x$ | 7. B.
Since $-1 \leqslant \cos \theta \leqslant 1$, we have $1 \leqslant \log _{2} x \leqslant 5$, thus $2 \leqslant x \leqslant 32$. Therefore, $|x-2|+|x-33|=x-2+33-x=31$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,359 |
8. The image set interval of the interval $[0, m]$ under the mapping $f: x \rightarrow 2 x+m$ is $[a, b]$. If the length of the interval $[a, b]$ is 5 units greater than the length of the interval $[0, m]$, then $m=(\quad)$.
(A) 5
(B) 10
(C) 2.5
(D) 1 | 8. A.
The length of the interval $[0, m]$ is $m$. Under the mapping $f$, the corresponding image set interval is $[m, 3m]$, with an interval length of $2m$. Therefore, $m=2m-m=5$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,360 |
9. Let the sequence $\left\{a_{n}\right\}\left(a_{n}>0\right)$ have the sum of its first $n$ terms as $S_{n}$, and the arithmetic mean of $a_{n}$ and 2 equals the geometric mean of $S_{n}$ and 2. Then the general term of $\left\{a_{n}\right\}$ is ( ).
(A) $a_{n}=n^{2}+n$
(B) $a_{n}=n^{2}-n$
(C) $a_{n}=3 n-1$
(D) $a_{n}... | 9. D.
According to the problem, we have $\frac{a_{n}+2}{2}=\sqrt{2 S_{n}}$. Therefore,
$$
\begin{array}{l}
S_{n}=\frac{\left(a_{n}+2\right)^{2}}{8}, \\
S_{n-1}=\frac{\left(a_{n-1}+2\right)^{2}}{8}(n \geqslant 2) .
\end{array}
$$
Subtracting (2) from (1) and simplifying, we get
$$
\left(a_{n}+a_{n-1}\right)\left(a_{n}-... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,361 |
10. The function $f(x)=-9 x^{2}-6 a x+2 a-a^{2}$ has a maximum value of -3 on the interval $\left[-\frac{1}{3}, \frac{1}{3}\right]$. Then the value of $a$ is ( ).
(A) $-\frac{3}{2}$
(B) $\sqrt{6}+2$ or $-\sqrt{2}$
(C) $\sqrt{6}+2$ or $2-\sqrt{6}$
(D) $2-\sqrt{6}$ or $-\sqrt{2}$ | 10. B.
Since $f(x)=-9\left(x+\frac{a}{3}\right)^{2}+2 a$, therefore, when $-\frac{a}{3}>\frac{1}{3}$, i.e., $a>1$, we have
$$
f(x)_{\max }=f\left(-\frac{1}{3}\right)=-1+4 a-a^{2}=-3 \text {. }
$$
Solving this gives $a=\sqrt{6}+2$ (discard $2-\sqrt{6}$).
When $-\frac{1}{3} \leqslant-\frac{a}{3} \leqslant \frac{1}{3}$,... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,362 |
11. Given the function defined on the set of non-zero natural numbers
$$
f(n)=\left\{\begin{array}{ll}
n+2, & n \leqslant 2005 ; \\
f(f(n-4)), & n>2005 .
\end{array}\right.
$$
then when $n \leqslant 2005$, $n-f(n)=$ $\qquad$
when $2005<n \leqslant 2007$, $n-f(n)=$
$\qquad$ | 11. $-2,0$.
When $n \leqslant 2005$, we have
$$
n-f(n)=n-(n+2)=-2 \text {. }
$$
When $2005<n \leqslant 2007$, we have
$$
2001<n-4 \leqslant 2003,2003<n-2 \leqslant 2005 \text {. }
$$
Thus, $n-f(n)=n-f(f(n-4))$
$$
=n-f(n-2)=n-n=0 \text {. }
$$ | -2,0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,363 |
Example 1 Given $|a b+2|+|a+1|=0$. Find
$$
\begin{array}{l}
\frac{1}{(a-1)(b+1)}+\frac{1}{(a-2)(b+2)}+\cdots+ \\
\frac{1}{(a-2006)(b+2006)}
\end{array}
$$
the value. | Explanation: From the given conditions and the property of non-negative numbers, we have
$$
a b+2=0 \text { and } a+1=0 \text {. }
$$
Solving these, we get $a=-1, b=2$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{1}{-2 \times 3}+\frac{1}{-3 \times 4}+\cdots+\frac{1}{-2007 \times 2008} \\
=\left(\frac{1... | -\frac{1003}{2008} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,364 |
Example 2 Given the quadratic equation in $x$
$x^{2}+n^{2} x+n-1=0(n$ is a natural number, $n \geqslant 2$ ). When $n=2$, the two roots of this equation are denoted as $\alpha_{2}$ and $\beta_{2}$; when $n=3$, the two roots are denoted as $\alpha_{3}$ and $\beta_{3}$; $\cdots \cdots$ when $n=2006$, the two roots are de... | Explanation: From the relationship between roots and coefficients, we have
$$
\begin{array}{l}
\alpha_{n}+\beta_{n}=-n^{2}, \alpha_{n} \beta_{n}=n-1 . \\
\text { Then }\left(\alpha_{n}-1\right)\left(\beta_{n}-1\right) \\
=\alpha_{n} \beta_{n}-\left(\alpha_{n}+\beta_{n}\right)+1 \\
=n-1+n^{2}+1 \\
=n(n+1) .
\end{array}
... | \frac{2005}{4014} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,365 |
Example 11 Let $y=\frac{x}{1+x}=f(x)$, and $f(1)$ represents the value of $y$ when $x=1$, i.e., $f(1)=\frac{1}{1+1}=\frac{1}{2}$; $f\left(\frac{1}{2}\right)$ represents the value of $y$ when $x=\frac{1}{2}$, i.e., $f\left(\frac{1}{2}\right)=$ $\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3} ; \cdots \cdots$. Try to find
... | Explanation: Given $y=\frac{x}{1+x}=f(x)$, we have
$$
\begin{array}{l}
f\left(\frac{1}{n}\right)=\frac{\frac{1}{n}}{1+\frac{1}{n}}=\frac{1}{1+n} \\
f(n)=\frac{n}{1+n} \\
f(0)=\frac{0}{1+0}=0 .
\end{array}
$$
Thus, $f(n)+f\left(\frac{1}{n}\right)=\frac{1}{n+1}+\frac{n}{n+1}=1$.
Therefore, the original expression is
$$
... | 2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,366 |
1. If $a_{n}=\left\{\begin{array}{cc}1, & n \text { is odd, } \\ -1, & n \text { is even, }\end{array} a_{n}\right.$ can be represented by a unified formula, which is $a_{n}=(-1)^{n+1}$. Given $\sin \frac{\pi}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}, \cos \frac{\pi}{8}=\frac{1}{2} \sqrt{2+\sqrt{2}}$. Then these two formulas c... | $\begin{array}{l}\text { II. } 1 \cdot \sin \left[\frac{1+(-1)^{n}}{2} \cdot \frac{\pi}{2}+\frac{\pi}{8}\right] \\ =\frac{1}{2} \sqrt{2+(-1)^{n} \sqrt{2}} .\end{array}$ | \frac{1}{2} \sqrt{2+(-1)^{n} \sqrt{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,367 |
2. Given $A=\{1,2,3,4,5,6\}, f: A \rightarrow A$, the number of mappings $f$ that satisfy $f(f(x)) \neq x$ is $\qquad$ | 2.7360 .
First, there should be $f(x) \neq x$, and the number of mappings satisfying $f(x) \neq x$ is $5^{6}$.
If $x \rightarrow a, a \rightarrow x(a \neq x)$, then it is called a pair of cyclic correspondences.
If there is one pair of cyclic correspondences and $f(x) \neq x$, the number of such mappings is $C_{6}^{2}... | 7360 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,368 |
3. Given that the maximum value of $x$ satisfying the inequality
$$
\left|x^{2}-4 x+p\right|+|x-3| \leqslant 5
$$
is 3. Then the solution set of this inequality is
$\qquad$ | 3. $\{x \mid 2 \leqslant x \leqslant 3\}$.
According to the problem, 3 should be a root of the equation $\left|x^{2}-4 x+p\right|+|x-3|=$ 5, so we can get $p=-2$ or $p=8$.
If $p=-2$, the original inequality becomes
$$
\left|x^{2}-4 x-2\right|+|x-3| \leqslant 5 \text {. }
$$
Obviously, $x=4$ is a solution to this ineq... | \{x \mid 2 \leqslant x \leqslant 3\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,369 |
4. There is a light source at point $P$ on the ceiling of the room, and the projection of $P$ on the ground is $Q$. A square pyramid $S-A B C D$ (with base $A B C D$ touching the ground) is placed on the ground. It is known that the height of the square pyramid $S-A B C D$ is $1 \mathrm{~m}$, the side length of the bas... | 4. $\frac{3 \sqrt{2}-1}{8} \mathrm{~m}^{2}$.
As shown in Figure 2, connect $P S$ and extend it to intersect line $A C$ at point $R$, then connect $R B$ and $R D$.
Thus, the concave quadrilateral $R B A D$ is the shape of the shadow, and its area is the area of the shadow we are looking for.
Given $S O=1, P Q=3, O Q=3... | \frac{3 \sqrt{2}-1}{8} \mathrm{~m}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,370 |
5. The maximum value of the function $y=|\sin x|+|\sin 2 x|$ is $\qquad$ . | 5. $\frac{\sqrt{414+66 \sqrt{33}}}{16}$
Notice that $y=|\sin x|(1+2|\cos x|)$, then
$$
\begin{array}{l}
y^{2}=\sin ^{2} x(1+2|\cos x|)^{2} \\
=(1-|\cos x|)(1+|\cos x|)(1+2|\cos x|)^{2} \\
= \frac{1}{\mu \lambda} \cdot \mu(1-|\cos x|) \cdot \lambda(1+|\cos x|) \cdot \\
(1+2|\cos x|)(1+2|\cos x|)(\lambda, \mu>0) \\
\le... | \frac{\sqrt{414+66 \sqrt{33}}}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,371 |
6. Given the function $f(x)=[x[x]]$, where $[x]$ denotes the greatest integer not exceeding $x$. If $x \in [0, n] (n \in \mathbf{N}_{+})$, the range of $f(x)$ is $A$, and let $a_{n}=\operatorname{card}(A)$, then $a_{n}=$ $\qquad$ | 6. $\frac{1}{2}\left(n^{2}-n+4\right)$.
When $n<x<n+1$, $[x]=n, x[x]=n x \in\left(n^{2}, n^{2}+n\right)$. Therefore, $f(x)$ can take
$$
n^{2}, n^{2}+1, n^{2}+2, \cdots, n^{2}+n-1
$$
When $x=n+1$, $f(x)=(n+1)^{2}$.
Also, when $x \in[0, n]$, it is clear that $f(x) \leqslant n^{2}$. Therefore, $a_{n+1}=a_{n}+n$.
It is a... | \frac{1}{2}\left(n^{2}-n+4\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,372 |
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$ with the general term $a_{n}=$ $n$, and the sum of the first $n$ terms is $S_{n}$. If $S_{n}$ is a perfect square, find $n$.
| Three, according to the problem, we have $S_{n}=\frac{n(n+1)}{2}=y^{2}$, which means $(2 n+1)^{2}-8 y^{2}=1\left(n 、 y \in \mathbf{N}_{+}\right)$.
Thus, the problem is transformed into:
Finding the positive integer solutions of the Pell equation $x^{2}-8 y^{2}=1$.
Obviously, $(3,1)$ is a minimal positive integer soluti... | n=\frac{(\sqrt{2}+1)^{2 m}+(\sqrt{2}-1)^{2 m}-2}{4}\left(m \in \mathbf{N}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,373 |
Four. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$, $F_{1}$ and $F_{2}$ are its left and right foci, $P$ is any point on the ellipse $C$, the centroid of $\triangle F_{1} P F_{2}$ is $G$, the incenter is $I$, and $I G=\lambda F_{1} \boldsymbol{F}_{2}$.
(1) Find the eccentricity... | (1) Let $P(x, y)(y \neq 0)$.
Since $F_{1}(-c, 0), F_{2}(c, 0)$, then $G\left(\frac{x}{3}, \frac{y}{3}\right)$.
Since $IG=\lambda F_{1} F_{2}$, then $IG \parallel F_{1} F_{2}$. Therefore, the y-coordinate of point $I$ is the same as the y-coordinate of point $G$.
Thus, the inradius of $\triangle F_{1} P F_{2}$ is $r=\fr... | \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,374 |
Five. (20 points) In the tetrahedron $P-ABC$, $AB=2$, $PA+PC=4$, $AC=x$, and the maximum volume of $V_{P-ABC}$ is $V$.
(1) Express $V$ in terms of $x$;
(2) If $V_{P-ABC}=\frac{4}{3}$, find the lateral surface area of the tetrahedron $P-ABC$. | (1) As shown in Figure 3, draw $P O \perp$ plane $A B C$ and $P H \perp A C$, with $O$ and $H$ being the feet of the perpendiculars.
$$
\begin{aligned}
& \text { Then } V_{P-A B C} \\
= & \frac{1}{3} S_{\triangle A B C} \cdot P O \\
= & \frac{1}{3} \times \frac{1}{2} A B \cdot A C \sin \angle B A C \cdot P O \\
\leqsla... | 4+2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,375 |
One, (50 points) As shown in Figure 1, in quadrilateral $ABCD$, the two diagonals $AC$ and $BD$ intersect at point $O$. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. $MQ \perp BD$ at $Q$, $NR \perp AC$ at $R$, and the lines $MQ$ and $NR$ intersect at point $P$. Moreover, $OP \perp AD$ at $H$. Prove that ... | As shown in Figure 5, connect $A Q$, $Q R$, and $D R$.
Since $\angle O Q P + \angle O R P = 180^{\circ}$, points $O$, $Q$, $P$, and $R$ are concyclic. Therefore, $\angle 1 = \angle 2$.
Also, $\angle O Q P = \angle P H D = 90^{\circ}$, so points $Q$, $P$, $D$, and $H$ are concyclic.
Hence, $\angle 1 = \angle 3$.
Thus, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,376 |
$$
\begin{array}{l}
\frac{2}{2007}\left[\left(1^{2}+3^{2}+5^{2}+\cdots+2005^{2}\right)-\right. \\
\left.\left(2^{2}+4^{2}+6^{2}+\cdots+2006^{2}\right)\right] .
\end{array}
$$ | Answer: -2006 . | -2006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,377 |
II. (50 points) Given the equations $x^{2}+p x+q=0$ and $y^{2}-p y+r=0$ both have real roots $(p, q, r \in \mathbf{R}, p \neq 0)$, and the roots of the two equations can be appropriately ordered as $x_{1}, x_{2}$ and $y_{1}, y_{2}$. Then the necessary and sufficient condition for $x_{1} y_{1}-x_{2} y_{2}=1$ to hold is
... | (1) Necessity
According to the problem, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=q, \\
y_{1}+y_{2}=p, y_{1} y_{2}=r, \\
x_{i}^{2}+p x_{i}+q=0(i=1,2), \\
y_{j}^{2}-p y_{j}+r=0(i=1,2) .
\end{array}
$$
From $x_{1} y_{1}-x_{2} y_{2}=1$, we get
$$
x_{1}^{2} y_{1}^{2}+x_{2}^{2} y_{2}^{2}-2 x_{1} x_{2} y_{1} y... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,378 |
Three. (50 points) Given $a, b, \lambda \in \mathbf{R}_{+}$. Determine the range of
$$
g(a, b)=\sqrt{\frac{a}{a+\lambda b}}+\sqrt{\frac{b}{b+\lambda a}}
$$ | Three, by replacing $a, b$ with $\frac{a}{a+b}, \frac{b}{a+b}$ respectively, $g(a, b)$ remains unchanged. Therefore, under the condition $a+b=1$, find the range of $g(a, b)$.
$$
\begin{array}{l}
g^{2}(a, b)=\frac{a}{a+\lambda b}+\frac{b}{b+\lambda a}+2 \sqrt{\frac{a b}{(a+\lambda b)(b+\lambda a)}} \\
=\frac{\lambda+2(1... | 1<g(a, b) \leqslant \frac{2}{\sqrt{1+\lambda}} \text{ for } 0<\lambda<2; \quad 1<g(a, b) \leqslant \frac{\lambda}{\sqrt{\lambda^{2}-1}} \text{ for } 2 \leqslant \lambda<3; \quad \frac{2}{\sqrt{1+\lambda}} \le | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,379 |
In $\triangle A B C$, $\angle A B C$ and $\angle A C B$ are both acute angles. Points $D$ and $E$ are on sides $A B$ and $A C$, respectively. $D F \perp B C$ at point $F$, and $E G \perp B C$ at point $G$. Let $B E$ intersect $D F$ at point $M$, and $C D$ intersect $E G$ at point $N$. $B N$ and $C M$ intersect at point... | Proof: As shown in Figure 2, draw $A Q \perp B C$ at $Q$. Extend $C M$ to intersect $A B$ at point $X$, and extend $B N$ to intersect $A C$ at point $Y$. Then we have
$$
D F / / A Q / / E G \text {. }
$$
Considering $\triangle A D C$ being intersected by the line $Y N B$, by Menelaus' theorem, we get
$$
\frac{C Y}{Y A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,380 |
Given a unit square $ABCD$, point $M$ (different from points $B$ and $C$) lies on side $BC$. The perpendicular bisector $l$ of segment $AM$ intersects $AB$, $CD$, and $BD$ at points $E$, $F$, and $K$ respectively.
(1) Which is longer: $AE$ or $DF + BM$? Please explain your reasoning;
(2) Which is longer: $EF$ or $AM$? ... | Solution: (1) As shown in Figure 3, let the line $l$ intersect $AM$ at the midpoint $N$ of $AM$. Denote $BM = x$, then
$$
\begin{array}{l}
MC = 1 - x, \\
AM = \sqrt{1 + x^2}, \\
AN = NM \\
= \frac{\sqrt{1 + x^2}}{2}.
\end{array}
$$
Extend $AM$ and $DC$ to intersect at point $P$.
By the similarity of right triangles $\... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,381 |
183 Let the points $0, 1, 2, 3, 4$ on the number line be $O, A, B, C, D$ respectively. A particle starts at point $O$, and each time it jumps one unit to the left or to the right, and the particle jumps back and forth among these five points. How many ways are there to jump $m$ times to $O, A, B, C, D$ respectively? | Solution: Let the number of ways a particle can jump $m$ times from point $O$ to points $O, A, B, C, D$ be $f_{1}(m), f_{2}(m), f_{3}(m), f_{4}(m), f_{5}(m)$, respectively.
When $m=1$, the particle must jump from point $O$ to point $A$, so we have
$$
f_{2}(1)=1, f_{1}(1)=f_{3}(1)=f_{4}(1)=f_{5}(1)=0 \text {. }
$$
When... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,382 |
Clothing 184 Let $a, b, c$ be positive numbers, and $a+b+c \geqslant a b c$. Prove:
$$
a^{2}+b^{2}+c^{2} \geqslant \sqrt{3} a b c .
$$ | Prove: Transform the condition $a+b+c \geqslant a b c$ into
$$
\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b} \geqslant 1,
$$
Transform the conclusion $a^{2}+b^{2}+c^{2} \geqslant \sqrt{3} a b c$ into
$$
\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b} \geqslant \sqrt{3} \text {. }
$$
In fact, we have
$$
\frac{a}{b c}+\frac{b}{c... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,383 |
$$
\begin{array}{l}
\frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\cdots+ \\
\left(\frac{1}{2006}+\frac{2}{2006}+\cdots+\frac{2005}{2006}\right)
\end{array}
$$ | (Hint: Use the holistic idea, set the original expression as equation (1). Rearrange the terms inside the parentheses in reverse order to get equation (2). (1) + (2) yields the result. Answer:
$$
\left.\frac{2011015}{2} .\right)
$$ | \frac{2011015}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,384 |
3. Calculate
$$
\begin{array}{c}
\frac{\frac{1}{2}}{1+\frac{1}{2}}+\frac{\frac{1}{3}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)}+ \\
\frac{\frac{1}{4}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)}+\cdots+ \\
\frac{\frac{1}{2006}}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3... | (Hint: Use the general term to find the pattern. Answer: $\frac{2005}{2007}$.)
| \frac{2005}{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,385 |
4. Calculate $\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2006}}$. | (Hint: Use the method of borrowing and returning "numbers", that is, add $\frac{1}{2^{2006}}$ to the original expression, then subtract $\frac{1}{2^{2006}}$. Answer: $1-\frac{1}{2^{2006}}$.) | 1-\frac{1}{2^{2006}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,386 |
5. Let the linear function $y=-\frac{n}{n+1} x+\frac{\sqrt{2}}{n+1}(n$ be a positive integer) intersect the $x$-axis and $y$-axis to form a triangle with area $S_{n}$. Find the value of $S_{1}+S_{2}+\cdots+S_{2006}$. | (Hint: $S_{n}=\frac{1}{n}-\frac{1}{n+1}$. Answer: $\frac{2006}{2007}$.)
| \frac{2006}{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,387 |
Example 1 Given three fixed points on a line in sequence as $A, B, C$, and $\Gamma$ is a circle passing through $A, C$ with its center not on $AC$. Draw tangents to the circle $\Gamma$ through points $A, C$ respectively, and let them intersect at point $P$. $PB$ intersects the circle $\Gamma$ at point $Q$. Prove: The a... | As shown in Figure 1, let the angle bisector of $\angle A Q C$ intersect $A C$ at point $R$, extend $Q R$ to intersect the circle $\Gamma$ at point $G$, and connect $A G$ and $C G$. It is easy to see that
$$
\begin{array}{l}
A G=C G, \\
P A=P C . \\
\text { Let } \angle A P Q=\gamma, \\
\angle C P Q=\theta,
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,388 |
Example 2 Given that $A B$ is the diameter of a semicircle $\odot O$, $Q$ is a fixed point on $O B$, $P Q \perp A B$, point $P$ is on the semicircle $\odot O$, and moving points $M, N$ are both on the semicircle $\odot O$, satisfying $\angle P Q M = \angle P Q N$. Let the line $M N \cap A B = S$. Prove: Regardless of h... | As shown in Figure 2, extend the semicircle $\odot O$ to form the full circle $\odot O$, extend $N Q$ to intersect $\odot O$ at point $M^{\prime}$, and connect $O M$ and $O M^{\prime}$.
Since $\angle P Q M = \angle P Q N$ and $P Q \perp A B$, then
$$
\angle A Q M = \angle B Q N = \angle A Q M^{\prime}.
$$
It is easy ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,389 |
Example 3 Given an acute triangle $\triangle ABC$ is a fixed triangle, two moving points $D, E$ are on sides $AB, AC$ respectively, and $DF \perp BC, EG \perp BC$, with $F, G$ being the feet of the perpendiculars. Let $BE \cap CD=O, FE \cap GD=P$. Prove: the line $OP$ always passes through a certain fixed point.
---
... | Explanation: As shown in Figure 3, draw the altitude $A Q$, then the foot of the perpendicular $Q$ is a fixed point.
Below is the proof: The line $O P$ passes through point $Q$, i.e., points $O$, $P$, and $Q$ are collinear.
Consider $\triangle A B E$ being intersected by the line $D O C$. By Menelaus' theorem, we hav... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,390 |
Example 4 Given a fixed line $l$ and a fixed circle $\odot O$ that do not intersect, $O D$ $\perp l$ at $D$. Through a moving point $P$ on $l$, draw $P A$ and $P B$ tangent to $\odot O$ at points $A$ and $B$, respectively. Draw $D M \perp P A$ at $M$, and $D N \perp P B$ at $N$. Prove: the line $M N$ always passes thro... | Explanation: As shown in the figure,
4, connect $O P$, $O A$,
$O B$, $D B$, $A B$ intersects
$O D$ at point $Q$, draw
$D L \perp A B$ at $L$.
Since $\angle O A P$
$$
\begin{array}{l}
=\angle O B P \\
=\angle O D P \\
=90^{\circ},
\end{array}
$$
Therefore, points $P$, $A$, $O$, $B$, $D$ are concyclic, meaning point $D$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,391 |
$$
\begin{array}{l}
\frac{1}{2}-\left(\frac{1}{3+\sqrt{3}}+\frac{1}{5 \sqrt{3}+3 \sqrt{5}}+\frac{1}{7 \sqrt{5}+5 \sqrt{7}}+\cdots+ \\
\frac{1}{2007 \sqrt{2005}+2005 \sqrt{2007}}\right) .
\end{array}
$$ | Explanation: By finding the pattern, we know that the above radicals can all be expressed as $\frac{1}{(n+2) \sqrt{n}+\sqrt{n+2} \cdot n}(n \geqslant 1$ and $n$ is an odd number $)$. Simplifying the above radical, we get
$$
\begin{array}{l}
\frac{1}{\sqrt{n} \cdot \sqrt{n+2}(\sqrt{n+2}+\sqrt{n})} \\
=\frac{\sqrt{n+2}-\... | \frac{\sqrt{2007}}{4014} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,392 |
Example 5 Given a convex quadrilateral $A B C D, B C=A D$, and $B C$ is not parallel to $A D$. Let points $E$ and $F$ be on the interiors of sides $B C$ and $A D$, respectively, such that $B E=D F$. Line $A C$ and $B D$ intersect at point $P$, line $B D$ and $E F$ intersect at point $Q$, and line $E F$ and $A C$ inters... | This is an interesting and thought-provoking competition problem. Its fixed point is the intersection $S$ of the perpendicular bisectors of $AC$ and $BD$.
As shown in Figure 5, construct $BX \parallel AC \parallel DY$, with points $X$ and $Y$ both on line $EF$. Connect $SA$, $SB$, $SC$, $SD$, $SQ$, and $SR$.
It is eas... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,393 |
Example 6 In $\triangle A B C$, $D$ is the midpoint of $A B$, point $E$ is on side $B C$, $A C=B E-E C$. Points $M, N$ are taken on the extensions of $C A, C B$ respectively, such that $A M=B N$. Line $E D$ intersects $M N$ at point $F$. Prove: the line through $F$ and perpendicular to $M N$ passes through a fixed poin... | Explanation: This is another thought-provoking and interesting geometry problem. The fixed point is the intersection of the perpendicular bisector of $AB$ and the circumcircle of $\triangle ABC$, which is the midpoint $G$ of the major arc $\overparen{ACB}$.
As shown in Figure 6, connect $GA$, $GM$, $GB$, and $GN$. It i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,394 |
Example 7 As shown in Figure 7, $\triangle A B C$ is a fixed triangle, and a moving point $P$ is on side $B C$. Construct $P M / / A C, P N / / A B$, with points $M, N$ on sides $A B, A C$ respectively. There is a fixed point $Q$ such that points $A, M, Q, N$ are concyclic. Determine the geometric position of point $Q$... | Explanation: This example is both ingenious and challenging. Its "ingenuity" and "difficulty" lie in how to find the fixed point $Q$.
After repeated trials and explorations, the fixed point $Q$ is determined as follows: Draw $\odot O_{1}$ through points $A$ and $B$ such that $AC$ is tangent to $\odot O_{1}$ (how to do... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,395 |
1. $A B$ is the diameter of $\odot O$, $\odot C$ is a moving circle, it is internally tangent to $\odot O$ at point $P$ and tangent to $A B$ at point $Q$. Prove: the line $P Q$ always passes through a fixed point. | (Prompt: Let $P Q$ intersect $\odot O$ at point $R$. It is easy to see that $O, C, P$ are collinear. Connect $O R, C Q$. Try to prove $O R \parallel C Q$.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,396 |
2. In $\triangle A B C$, $A B=A C, \odot O$ is tangent to $A B$ and $A C$, and $X$ is the point of tangency on side $A B$. Draw $C Y$ tangent to $\odot O$ at point $Y$ (where $Y$ is inside the shape). Prove that regardless of the size of $\odot O$, line $X Y$ always passes through a certain point. | (It is easy to prove $\triangle O B X \cong \triangle O C Y$. Let $X Y$ intersect side $B C$ at point $M$, and then it can be proved that $O, X, B, M$ are concyclic.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,397 |
3. On a fixed line segment $AB$, there is a fixed point $D$, with $AD > DB$. Line $l$ is perpendicular to $AB$, with $D$ as the foot of the perpendicular. Point $C$ moves on $l$, and $AE \perp CB$ at $E$, $BF \perp CA$ at $F$. Prove: Line $FE$ always passes through a fixed point. | (Tip: Take the midpoint $M$ of $A B$, let $A D=a, D B=b, a>b$. Let $F E \cap A B=P, B P=x$. Note, $M 、 D 、 E 、 F$ are concyclic - defined by the nine-point circle.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,398 |
5. Given that $\odot A$ and $\odot B$ are equal and intersect. Two moving circles $\odot C$ and $\odot D$ are externally tangent at point $K$, and they are internally tangent to $\odot A$ and externally tangent to $\odot B$. Prove: regardless of how $\odot C$ and $\odot D$ move, their common internal tangent always pas... | (Tip: Take the midpoint $O$ of $A B$, and try to prove that $O K \perp C D$, i.e., $O K$ is the inner common tangent of $\odot C$ and $\odot D$. Since $A B$ is a fixed line segment, its midpoint $O$ must be a fixed point.) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,400 |
Example 1 Given that in the right-angled $\triangle ABC$, the two legs are $AC=2, CB=3$, and $CP$ is the angle bisector of $\angle ACB$ (point $P$ is on the hypotenuse $AB$). Fold the right-angled triangle along $CP$ to form a dihedral angle $A-CP-B$. When $AB=2\sqrt{2}$, the size of the dihedral angle $A-CP-B$ is $\qq... | Solution: As shown in Figure 2, we know
$$
\begin{array}{l}
\alpha=\beta=45^{\circ}, \\
\cos \gamma \\
=\cos \angle A C B \\
=\frac{2^{2}+3^{2}-(2 \sqrt{2})^{2}}{2 \times 3 \times 2} \\
=\frac{5}{12} .
\end{array}
$$
By the formula for the cosine of the dihedral angle, we get
$$
\cos \theta=\frac{\frac{5}{12}-\frac{\s... | \arccos \left(-\frac{1}{6}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,401 |
Example 2 As shown in Figure 3, let $E$, $F$, and $G$ be the midpoints of the edges $AB$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\operatorname{ar... | Solution: As shown in Figure 4, let the edge length of the regular tetrahedron be $2a$, $CE$ is the height of the equilateral $\triangle ABC$ with side length $2a$, i.e., $CE=\sqrt{3}a$, $EG$ is the height of the isosceles $\triangle ECD$ with legs of length $\sqrt{3}a$ and base of length $2a$, i.e., $EG=\sqrt{2}a$.
Fr... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,402 |
Example 4 Calculate
$$
\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots \cdot\left(1-\frac{1}{2006^{2}}\right) .
$$ | $$
\begin{aligned}
= & \left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right) \cdots \\
& \left(1-\frac{1}{2006}\right)\left(1+\frac{1}{2006}\right) \\
= & \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \cdots \times \frac{2005}{2006} \times... | \frac{2007}{4012} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,403 |
Example 3 As shown in Figure 5, in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ and $F$ are the midpoints of edges $A B$ and $B C$ respectively. Find the size of the dihedral angle $B-F B_{1}-E$. | Solution: As shown in Figure 6, by the Pythagorean theorem, we have
$$
\begin{array}{l}
B_{1} E=B_{1} F=\frac{\sqrt{5}}{2} a, \\
E F=\frac{\sqrt{2}}{2} a .
\end{array}
$$
In the right triangle $\triangle E B B_{1}$, we have
$$
\cos \gamma=\frac{2}{\sqrt{5}}.
$$
Since $\triangle E B_{1} B \cong \triangle F B_{1} B$, w... | \arccos \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,404 |
Example 4 As shown in Figure 7, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the degree of the dihedral angle $A-B D_{1}-A_{1}$ is | Solution: As shown in Figure 8, let the edge length of the cube be 1, then $B A_{1}=\sqrt{2}, B D_{1}=\sqrt{3}$.
By the theorem of three perpendiculars,
$B A_{1} \perp A_{1} D_{1}, \cos \alpha=\frac{\sqrt{2}}{\sqrt{3}}$.
Also, $B A \perp A D_{1}, \cos \beta=\frac{1}{\sqrt{3}}$, so,
$\sin \alpha=\frac{1}{\sqrt{3}}, \sin... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,405 |
Example 5 As shown in Figure 9, in the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$, respectively. The sine value of the plane angle formed by the dihedral angle between plane $C E B_{1}$ and plane $D_{1} F B_{1}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) ... | Solution: Let the edge length of the cube be 1. Then
$$
E C=E B_{1}=\frac{\sqrt{5}}{2} \text {. }
$$
Since $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$ respectively, it follows that $D_{1} F$, $D A$, and $C E$ must intersect at a point $G$.
Connecting $B_{1} G$. By the theorem of parallel lines cutting inters... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,406 |
Question: Let a regular $n$-sided polygon with side length 1 in the plane, whose vertices are $P_{1}, P_{2}, \cdots, P_{n}$. If two different points $P_{n+1} 、 P_{n+2}$ are placed arbitrarily within the shape or on its boundary, try to find: $\min _{\leqslant i<j \leqslant n+2} P_{i} P_{j}$'s maximum value. | Solution: Take points $P_{n+1}$ and $P_{n+2}$ on the perpendicular bisectors of $P_{1} P_{2}$ and $P_{\left[\frac{n}{2}\right]+1} P_{\left[\frac{n}{2}\right]+2}$, respectively, such that $P_{n+1}$ and $P_{n+2}$ are inside the $n$-sided polygon, and
$P_{1} P_{n+1}=P_{n+1} P_{n+2}=P_{n+2} P_{\left[\frac{n}{2}\right]+1}$
... | \frac{\sqrt{3+\cos ^{2} \frac{\pi}{n}}-\cos \frac{\pi}{n}}{3 \sin \frac{\pi}{n}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,407 |
1. Let $I$ be the incenter of $\triangle ABC$, and $P$ be a point inside $\triangle ABC$ such that
$$
\angle PBA + \angle PCA = \angle PBC + \angle PCB.
$$
Prove: $AP \geqslant AI$, and state the necessary and sufficient condition for equality, which is $P=I$. | 1. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$.
Since $\angle P B A+\angle P C A+\angle P B C+\angle P C B=\beta+$ $\gamma$, by assumption we have
$$
\angle P B C+\angle P C B=\frac{\beta+\gamma}{2} .
$$
Since points $P$ and $I$ are on the same side of side $B C$, points $B, C, I, P$ are concyclic, meaning... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,408 |
2. Let $P$ be a regular 2006-gon. If an end of a diagonal of $P$ divides the boundary of $P$ into two parts, each containing an odd number of sides of $P$, then the diagonal is called a "good edge". It is stipulated that each side of $P$ is a good edge.
Given 2003 non-intersecting diagonals inside $P$ that partition $... | 2. If an isosceles triangle has two good sides, it is briefly referred to as a "good triangle". Let $\triangle A B C$ be a good triangle, and $A B, B C$ be the good sides. Then, there are an odd number of edges between points $A$ and $B$; the same applies to $B$ and $C$. We say these edges belong to the good $\triangle... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,409 |
3. Find the smallest real number $M$, such that for all real numbers $a, b, c$, we have
$$
\begin{array}{l}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
\leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{array}
$$ | 3. First consider
$$
P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right) .
$$
It is easy to see that $P(b)=P(c)=P(-c-b)=0$. Therefore, we have
$$
\begin{array}{l}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
=|P(a)|=|(b-c)(a-b)... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,410 |
4. Find all integer pairs $(x, y)$ such that
$$
1+2^{x}+2^{2 x+1}=y^{2} .
$$ | 4. If $(x, y)$ is a solution, then $x \geqslant 0, (x, -y)$ is also a solution.
When $x=0$, the solutions are $(0,2), (0,-2)$.
Assume $(x, y)$ is a solution, $x>0$. Without loss of generality, let $y>0$.
Thus, the original equation is equivalent to
$$
2^{x}\left(1+2^{x+1}\right)=(y-1)(y+1) \text {. }
$$
Therefore, $y... | (0,2),(0,-2),(4,23),(4,-23) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,411 |
5. Let $P(x)$ be an integer-coefficient polynomial of degree $n(n>1)$, and $k$ be a positive integer. Consider the polynomial
$$
Q(x)=P(P(\cdots P(P(x)) \cdots)),
$$
where $P$ appears $k$ times. Prove: There are at most $n$ integers $t$ such that $Q(t)=t$. | 5. If every integer fixed point of $Q$ is also a fixed point of $P$, then the conclusion holds.
Assume an integer $x_{0}$ satisfies $Q\left(x_{0}\right)=x_{0}$, but $P\left(x_{0}\right) \neq x_{0}$.
Define $x_{i+1}=P\left(x_{i}\right)(i=0,1,2, \cdots)$.
Then $x_{k}=x_{0}$.
Clearly, for different $u, v$, we have
$$
(u-v... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,412 |
6. For any side $b$ of a convex polygon $P$, construct a triangle with the largest area inside $P$ using $b$ as one of its sides. Prove: For each side of $P$, the sum of the areas of the triangles obtained by the above method is at least twice the area of $P$. | 6. First, we prove a lemma.
Lemma: For every convex $2n$-gon with area $S$, there exists a triangle formed by its sides and vertices, whose area is not less than $\frac{S}{n}$.
Proof of the lemma: A main diagonal of a $2n$-gon is a diagonal that divides the $2n$-gon into two $(n+1)$-gons. For any side $b$ of the $2n$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,413 |
Example 5 Find the value of $2-2^{2}-2^{3}-\cdots-2^{2005}+2^{2000}$. | Explanation: Utilizing the characteristic $2^{n+1}-2^{n}=2^{n}$, rearrange the terms of the original expression in reverse order.
Therefore, the original expression
$$
\begin{array}{l}
=2^{2006}-2^{2005}-2^{2004}-\cdots-2^{3}-2^{2}+2 \\
=2^{2006}(2-1)-2^{2005}-\cdots-2^{3}-2^{2}+2 \\
=2^{2005}(2-1)-2^{2004}-\cdots-2^{3... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,414 |
1. Find all monic integer polynomials $P(x)$ of degree 2 such that there exists an integer polynomial $Q(x)$ for which all coefficients of $P(x) Q(x)$ are $\pm 1$. | 1. $P(x)=x^{2} \pm x \pm 1, x^{2} \pm 1, x^{2} \pm 2 x+1$.
Let $F(x)$ be any polynomial of degree $n$ with coefficients all being $\pm 1$. If $z$ is a root of $F(x)$, and $|z|>1$, then
$$
\begin{array}{l}
|z|^{n}=\left|z^{n}\right|=\left| \pm z^{n-1} \pm z^{n-2} \pm \cdots \pm 1\right| \\
\leqslant|z|^{n-1}+|z|^{n-2}+... | P(x)=x^{2} \pm x \pm 1, x^{2} \pm 1, x^{2} \pm 2 x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,415 |
2. Let $\mathbf{R}_{+}$ denote the set of positive real numbers. Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ such that for all positive real numbers $x, y$, we have $f(x) f(y)=2 f(x+y f(x))$. | 2. $f(x)=2$.
First, we prove: the function $f(x)$ is non-decreasing.
If there exist positive real numbers $x>z$, such that $f(x)0$, then $x+y f(x)=z+y f(z)$. Thus, we have
$$
\begin{array}{l}
f(x) f(y)=2 f(x+y f(x)) \\
=2 f(z+y f(z))=f(z) f(y),
\end{array}
$$
which implies $f(x)=f(z)$, a contradiction.
If $f(x)$ is n... | f(x)=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,416 |
3. Given real numbers $p, q, r, s$ satisfy
$$
p+q+r+s=9, \quad p^{2}+q^{2}+r^{2}+s^{2}=21 \text {. }
$$
Prove: there exists a permutation $(a, b, c, d)$ of $(p, q, r, s)$, such that $a b - c d \geqslant 2$. | 3. Assume $p \geqslant q \geqslant r \geqslant s$.
If $p+q \geqslant 5$, then
$$
\begin{array}{l}
p^{2}+q^{2}+2 p q \geqslant 25=4+\left(p^{2}+q^{2}+r^{2}+s^{2}\right) \\
\geqslant 4+p^{2}+q^{2}+2 r s,
\end{array}
$$
i.e., $p q-r s \geqslant 2$.
If $p+q<5$, then $4<r+s \leqslant p+q<5$.
Notice that
$$
\begin{array}{l... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,417 |
4. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x, y$, the following holds:
$$
f(x+y)+f(x) f(y)=f(x y)+2 x y+1 .
$$ | 4. $f(x)=2 x-1, f(x)=-x-1, f(x)=x^{2}-1$.
In the original equation, let $y=1$, and let $a=1-f(1)$, then we have $f(x+1)=a f(x)+2 x+1$.
In the original equation, change $y$ to $y+1$, we get
$$
\begin{array}{l}
f(x+y+1)+f(x) f(y+1) \\
=f(x(y+1))+2 x(y+1)+1 .
\end{array}
$$
Substitute $f(x+y+1)=a f(x+y)+2(x+y)+1$,
$$
f(... | f(x)=2x-1, f(x)=-x-1, f(x)=x^2-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,418 |
1. Given $\triangle A B C$ satisfies $A B+B C=3 A C, I$ is the incenter of $\triangle A B C$, the incircle touches sides $A B, B C$ at points $D, E$ respectively. The points symmetric to $D, E$ with respect to $I$ are $K, L$ respectively. Prove: $A, C, K, L$ are concyclic. | 1. As shown in Figure 1, let $BI$ intersect the circumcircle of $\triangle ABC$ at point $P$, and let $M$ be the midpoint of side $AC$. The projection of point $P$ on $IK$ is $N$.
Since $AB + BC$
$$
= 3AC \text{, }
$$
thus, $BD = BE$
$$
\begin{array}{l}
= AC = 2CM. \\
\text{Also, } \angle ABP = \angle ACP,
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,419 |
3. Given a parallelogram $A B C D$, a moving line $l$ through point $A$ intersects the rays $B C$ and $D C$ at points $X$ and $Y$ respectively. In $\triangle A B X$, the excenter of $\angle B A X$ is $K$, and in $\triangle A D Y$, the excenter of $\angle D A Y$ is $L$. Prove: $\angle K C L$ is a constant value. | 3. As shown in Figure 2.
Let $\angle B A X=2 \alpha, \angle D A Y=2 \beta$, and let the points on the extensions of line segments $A B$ and $A D$ be $B^{\prime}$ and $D^{\prime}$, respectively. Then we have
$$
\begin{array}{l}
\angle K A B=\angle K A X=\alpha, \\
\angle L A D=\angle L A Y=\beta, \\
\angle K B B^{\prim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,420 |
5. In an acute triangle $\triangle ABC$, $AB \neq AC$, $H$ is the orthocenter, $M$ is the midpoint of $BC$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and satisfy $AE = AD$, and points $D$, $H$, and $E$ are collinear. Prove: $HM$ is perpendicular to the common chord of the circumcircles of $\triangle A... | 5. As shown in Figure 3, let the circumcenters of $\triangle ABC$ and $\triangle ADE$ be points $O$ and $O_1$, respectively.
Since $O_1O$ is perpendicular to the common chord of the two circles, it suffices to prove that $O_1O \parallel HM$.
Let the diameter of $\odot O$ be $AP$, and let $BH$ intersect $AC$ at point ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,421 |
6. Given $\triangle A B C$ with median $A M$ intersecting its incircle $\Gamma$ at points $K$ and $L$. Lines through $K$ and $L$ parallel to $B C$ intersect the circle $\Gamma$ at points $X$ and $Y$, respectively. $A X$ and $A Y$ intersect $B C$ at points $P$ and $Q$. Prove: $B P = C Q$. | 6. As shown in Figure 4, let the incenter of $\triangle ABC$ be $I$, and the points where the incircle touches sides $BC$, $CA$, and $AB$ be $D$, $E$, and $F$ respectively. $EF$ intersects $DI$ at point $T$, and a line parallel to $BC$ through $T$ intersects $AB$ and $AC$ at points $U$ and $V$ respectively.
Since $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,422 |
7. In an acute triangle $\triangle ABC$, the projections of points $A, B, C$ onto sides $BC, CA, AB$ are $D, E, F$ respectively. The projections of points $A, B, C$ onto sides $EF, FD, DE$ are $P, Q, R$ respectively. Let the perimeters of $\triangle ABC, \triangle PQR, \triangle DEF$ be $p_{1}, p_{2}, p_{3}$ respective... | 7. As shown in Figure 5, since $\triangle ABC$ is an acute triangle, $P, Q, R$ are points inside $EF, FD, DE$ respectively. Let the projections of points $E, F$ on $AB, AC$ be $K, L$ respectively. Then,
$$
\begin{array}{l}
\angle AKL = \angle AEF \\
= \angle ABC,
\end{array}
$$
Therefore, $KL \parallel BC$.
Since $\tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,423 |
1. Simplify $\left(\frac{2 x}{x+y}-\frac{4 x}{y-x}\right) \div \frac{8 x}{x^{2}-y^{2}}$, we get ( ).
(A) $\frac{x+3 y}{4}$
(B) $-\frac{x+3 y}{4}$
(C) $-\frac{3 x+y}{4}$
(D) $\frac{3 x+y}{4}$ | $\begin{array}{l}\text {-.1.D. } \\ \left(\frac{2 x}{x+y}-\frac{4 x}{y-x}\right) \div \frac{8 x}{x^{2}-y^{2}} \\ =\left(\frac{2 x}{x+y}+\frac{4 x}{x-y}\right) \times \frac{(x+y)(x-y)}{8 x} \\ =\frac{x-y}{4}+\frac{x+y}{2}=\frac{x-y+2 x+2 y}{4}=\frac{3 x+y}{4} .\end{array}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,424 |
Example 6 Calculate
$$
1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+2006} \text {. }
$$ | Explanation: Using the formula
$$
1+2+\cdots+n=\frac{n(n+1)}{2} \text {, }
$$
we know that $\frac{1}{1+2+\cdots+n}=\frac{2}{n(n+1)}$.
Therefore, the original expression is
$$
\begin{aligned}
& =\frac{2}{1 \times 2}+\frac{2}{2 \times 3}+\cdots+\frac{2}{2006 \times 2007} \\
& =2\left(\frac{1}{1 \times 2}+\frac{1}{2 \tim... | \frac{4012}{2007} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,425 |
2. The number of all integers that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
\frac{2 x-1}{3}+1 \geqslant x-\frac{5-3 x}{2} \\
\frac{x}{5}<3+\frac{x-1}{3}
\end{array}\right.
$$
is ( ).
(A) 1
(B) 2
(C) 21
(D) 22 | 2.C.
Solve the inequality $\frac{2 x-1}{3}+1 \geqslant x-\frac{5-3 x}{2}$, we get $x \leqslant \frac{19}{11}$.
Solve the inequality $\frac{x}{5}-20$.
Therefore, the solution set of the original system of inequalities is $-20<x \leqslant \frac{19}{11}$. Then all integers that satisfy the original system of inequalities... | C | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,426 |
3. Two similar triangles, their perimeters are 36 and 12, respectively. The largest side of the triangle with the larger perimeter is 15, and the smallest side of the triangle with the smaller perimeter is 3. Then the area of the triangle with the larger perimeter is ( ).
(A) 52
(B) 54
(C) 56
(D) 58 | 3. B.
Let the smallest side of the larger perimeter triangle be $x$. Then $\frac{x}{3}=$ $\frac{36}{12}$. Solving for $x$ gives $x=9$.
The other side should be $36-15-9=12$.
Since $15^{2}=12^{2}+9^{2}$, both triangles are right triangles, and the area of the larger perimeter triangle is $\frac{1}{2} \times 9 \times 12... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,427 |
4. If the two roots of the quadratic equation $x^{2}+p x+q=0$ are $p$ and $q$, then $p q$ equals ( ).
(A) 0
(B) 1
(C) 0 or -2
(D) 0 or 1 | 4. C.
According to the relationship between the roots and coefficients of a quadratic equation, we have
$$
\left\{\begin{array}{l}
p+q=-p, \\
p q=q .
\end{array}\right.
$$
When $q=0$, we have $p=q=0$, so $p q=0$;
When $q \neq 0$, we have $p=1, q=-2$, so $p q=-2$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,428 |
5. As shown in Figure 1, in $\triangle A B C$, $\angle B=$ $45^{\circ}, A C$'s perpendicular bisector intersects $A C$ at point $D$, and intersects $B C$ at point $E$, and $\angle E A B: \angle C A E=3: 1$. Then $\angle C$ equals ( ).
(A) $27^{\circ}$
(B) $25^{\circ}$
(C) $22.5^{\circ}$
(D) $20^{\circ}$ | 5.A.
Since $E D \perp$ bisects $A C$, therefore,
$$
\begin{array}{l}
\angle C=\angle E A C, \\
\angle C+\angle C A B=180^{\circ}-45^{\circ}=135^{\circ} .
\end{array}
$$
And $\angle C A B=\angle C A E+\angle E A B=4 \angle C A E=4 \angle C$, then $\angle C+\angle C A B=\angle C+4 \angle C=5 \angle C=135^{\circ}$. Thus... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,429 |
6. $70 \%$ of the students in the class participate in the biology group, $75 \%$ participate in the chemistry group, $85 \%$ participate in the physics group, and $90 \%$ participate in the mathematics group. The percentage of students who participate in all four groups is at least ( ).
(A) $10 \%$
(B) $15 \%$
(C) $20... | In fact,
$70 \%+75 \%=145 \%$, indicating that at least $45 \%$ of the students participated in both the biology and chemistry groups;
$45 \%+85 \%=130 \%$, indicating that at least $30 \%$ of the students participated in the biology, chemistry, and physics groups;
$30 \%+90 \%=120 \%$, indicating that at least 20\% of... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,430 |
7. There is a barrel of pure pesticide. After pouring out $20 \mathrm{~L}$ and then topping up with water, another $10 \mathrm{~L}$ is poured out and topped up with water again. At this point, the volume ratio of pure pesticide to water in the barrel is $3: 5$. Then the capacity of the barrel is $(\quad) L$.
(A) 30
(B)... | 7.B.
Let the capacity of the bucket be $x \mathrm{~L}$.
The first addition of water is $20 \mathrm{~L}$, and when $10 \mathrm{~L}$ is poured out, the $10 \mathrm{~L}$ contains $\frac{20}{x} \times 10 \mathrm{~L}$ of water.
Finally, adding another $10 \mathrm{~L}$ of water, so the amount of water in the bucket is $20-\... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,431 |
8. The triangle formed by the intersection of the three external angle bisectors of a triangle ( ).
(A) must be an acute triangle
(B) must be an obtuse triangle
(C) must be a right triangle
(D) is similar to the original triangle | 8. A.
According to the theorem that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles (as shown in Figure 5), we have
$$
\begin{array}{l}
\angle C B A^{\prime}+\angle A^{\prime} C B \\
= \frac{1}{2}(\angle A B C+ \\
\angle B C A+2 \angle C A B) \\
= \frac{1}{2} \times 180^{\c... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,432 |
9. As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $A D \perp B C$, $C G / / A B$, $B G$ intersects $A D$ and $A C$ at points $E$ and $F$ respectively. If $\frac{E F}{B E}=\frac{a}{b}$, then $\frac{G E}{B E}$ equals $\qquad$ | Two $9 \cdot \frac{b}{a}$.
As shown in Figure 6, connect $C E$ and extend it to intersect $A B$ at point $H$. Clearly, $E H=E F, B E=E C$. Since $C G / / A B$, we know $\triangle E H B \backsim \triangle E C G$. Therefore, we have
$$
\begin{array}{l}
\frac{G E}{E B}=\frac{E C}{E H} \\
=\frac{B E}{E F}=\frac{b}{a} .
\en... | \frac{b}{a} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,433 |
10. The solution to the equation ||$x-3|+3 x|=1$ is | 10. -2 or -1.
Square both sides of the equation, we get
$$
(x-3)^{2}+6 x|x-3|+9 x^{2}=1 \text {. }
$$
When $x \geqslant 3$, we have
$$
x^{2}-6 x+9+6 x^{2}-18 x+9 x^{2}=1 \text {. }
$$
Simplifying, we get $16 x^{2}-24 x+8=0$.
Solving, we get $x=\frac{1}{2}$ or $x=1$ (which does not satisfy $x \geqslant 3$, so we disc... | -2 \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,434 |
11. $A D 、 B E 、 C F$ are the three medians of $\triangle A B C$. If $B C=a, C A=b, A B=c$, then $A D^{2}+B E^{2}+$ $C F^{2}=$ $\qquad$ . | 11. $\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$.
As shown in Figure 7, draw
$A P \perp B C$ at point $P$. Then we have
$$
\begin{array}{l}
A P^{2}=A C^{2}-P C^{2} \\
=A B^{2}-B P^{2}, \\
2 A P^{2}=A C^{2}+A B^{2}-P C^{2}-B P^{2} \\
=b^{2}+c^{2}-\left(\frac{a}{2}-D P\right)^{2}-\left(\frac{a}{2}+D P\right)^{2} \\
=b^{2... | \frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,435 |
Example 7 Simplify
$$
\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}+\sqrt{3}}+\cdots+\frac{1}{\sqrt{2005}+\sqrt{2006}}
$$ | $$
\begin{aligned}
= & \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\cdots+ \\
& \frac{\sqrt{2006}-\sqrt{2005}}{(\sqrt{2006}+\sqrt{2005})(\sqrt{2006}-\sqrt{2005})} \\
= & (\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{2006}-\sqrt{2005}) \\
= & \sqrt{2006}-... | \sqrt{2006}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,436 |
12. There are two two-digit numbers, their difference is 58, and the last two digits of their square numbers are the same. Then these two two-digit numbers are $\qquad$ . | 12.79 and 21.
Let these two two-digit numbers be $m, n (m<n)$.
Since $n-m=58, n^{2}$ and $m^{2}$ have the same last two digits, then
$$
n^{2}-m^{2}=(n+m)(n-m)=58(n+m)
$$
is a multiple of 100.
Also, $(58,100)=2$, so, we have
$n+m=50$ or $n+m=100$ or $n+m=150$.
Thus, we get the system of equations
Solving them respecti... | 79 \text{ and } 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,437 |
13. As shown in Figure 3, in $\triangle A B C$, $A B=1, A C=2$,
$D$ is the midpoint of $B C$, $A E$
bisects $\angle B A C$ and intersects $B C$ at
point $E$, and $D F / / A E$. Find the length of $C F$. | Three, 13. As shown in Figure 8, draw $E H \perp A B$, intersecting $A B$ at point $H$, and $E G \perp A C$, intersecting $A C$ at point $G$. Since $A E$ bisects $\angle B A C$, we have
$$
E H=E G .
$$
Thus, we have $\frac{B E}{C E}=\frac{S_{\triangle A B E}}{S_{\triangle A E C}}=\frac{A B}{A C}=\frac{1}{2}$.
Furtherm... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,438 |
14. A construction company
has contracted two projects, each to be constructed by two different teams. According to the progress of the projects, the construction company can adjust the number of people in the two teams at any time. If 70 people are transferred from Team A to Team B, then the number of people in Team ... | 14. Let team A have $x$ people, then team B has $[2(x-70)-70]$ people, which means team B has $(2 x-210)$ people. Suppose $y$ people are transferred from team B to team A, making the number of people in team A three times that of team B. Then
$$
3(2 x-210-y)=x+y,
$$
which simplifies to $x=126+\frac{4}{5} y$.
Given $y>... | 130 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,439 |
15. Place the numbers 1, 2, $3, \cdots, 9$ into the 9 circles in Figure 4, such that the sum of the numbers in the three circles on each side of $\triangle ABC$ and $\triangle DEF$ is 18.
(1) Provide one valid arrangement;
(2) How many different arrangements are there? Prove your conclusion. | 15. (1) Figure 9 gives one arrangement that meets the requirements.
(2) There are 6 different ways to fill in the numbers.
Let the sum of the three numbers in circles $A, B, C$ be $x$; the sum of the three numbers in circles $D, E, F$ be $y$; and the sum of the remaining three circles be $z$. Clearly, we have
$$
x+y+z... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,440 |
1. Given that the real number $a$ satisfies
$$
|2004-a|+\sqrt{a-2005}=a \text {. }
$$
Then, the value of $a-2004^{2}$ is ( ).
(A) 2003
(B) 2004
(C) 2005
(D) 2006 | -1.C.
Since $a-2005 \geqslant 0$, then, $a-2004+\sqrt{a-2005}=a$. Therefore, $a-2004^{2}=2005$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,441 |
2. A store sells a certain product and makes a profit of $m$ yuan per item, with a profit margin of $20\%$ (profit margin $=\frac{\text{selling price} - \text{cost price}}{\text{cost price}}$). If the cost price of this product increases by $25\%$, and the store raises the selling price so that it still makes a profit ... | 2.C.
Let the original purchase price be $a$ yuan, and the profit margin after the price increase be $x \%$, then $m=a \cdot 20 \% = a(1+25 \%) \cdot x \%$.
Solving for $x \%$ gives $x \% = 16 \%$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,442 |
4. If $x_{0}$ is a root of the quadratic equation $a x^{2}+b x+c=0(a \neq 0)$, set
$$
M=\left(2 a x_{0}+b\right)^{2}, N=b^{2}-4 a c .
$$
Then the relationship between $M$ and $N$ is ( ).
(A) $M<N$
(B) $M=N$
(C) $M>N$
(D) $N<M<2 N$ | 4.B.
$x_{0}=\frac{-b \pm \sqrt{\Delta}}{2 a}$. Then $\left(2 a x_{0}+b\right)^{2}=b^{2}-4 a c$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,444 |
5. As shown in Figure 3, a square piece of paper is cut into a triangle and a trapezoid. If the area ratio of the triangle to the trapezoid is $3: 5$, then their perimeter ratio is ( ).
(A) $3: 5$
(B) $4: 5$
(C)5:6
(D) $6: 7$ | 5.D.
Let the lower base of the trapezoid be $a$, the upper base be $b(a>b)$, the perimeter be $p$, and the two legs of the right triangle be $a$ and $a-b$, with the perimeter being $p_{1}$. Then we have
$$
a(a+b): a(a-b)=5: 3 .
$$
Thus, $b=\frac{1}{4} a$.
Let $a=4, b=1$, then
$$
p_{1}: p=(3+4+5):(1+4+4+5)=6: 7 \text ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,445 |
6. Given the origin $O$ and point $A(2,-2)$, $B$ is a point on the coordinate axis. If $\triangle A O B$ is an isosceles triangle, then the number of such points $B$ is ( ) .
(A) 4
(B) 5
(C) 6
(D) 8 | 6.D.
The points $B$ that satisfy the condition are located as follows: there are 3 on the positive direction of the $x$-axis, 3 on the negative direction of the $y$-axis, and 1 on each of the other two half-axes, making a total of 8 points. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,446 |
Example 8 Calculate
$$
\begin{array}{l}
1 \frac{1}{2048}+2 \frac{1}{1024}+4 \frac{1}{512}+\cdots+512 \frac{1}{4}+ \\
1024 \frac{1}{2} .
\end{array}
$$ | Explanation: Since each number added to itself yields the next number, we can add $1 \frac{1}{2048}$. Therefore, the original expression is:
$$
\begin{aligned}
= & 1 \frac{1}{2048}+1 \frac{1}{2048}+2 \frac{1}{1024}+4 \frac{1}{512}+ \\
& \cdots+512 \frac{1}{4}+1024 \frac{1}{2}-1 \frac{1}{2048} \\
= & 2 \frac{1}{1024}+2 ... | 2047 \frac{2047}{2048} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,447 |
7. In the quadratic equation $x^{2}+m x+n=0$, the coefficients $m, n$ can take values from $1,2,3,4,5,6$. Then, among the different equations obtained, the number of equations with real roots is ( ).
(A) 20
(B) 19
(C) 16
(D) 10 | 7.B.
$\Delta=m^{2}-4 n \geqslant 0$, then $m^{2} \geqslant 4 n$. When $n=1$, $m=2,3,4,5,6$, a total of 5; when $n=2$, $m=3,4,5,6$, a total of 4; when $n=3$, $m=4,5,6$, a total of 3; when $n=4$, $m=4,5,6$, a total of 3; when $n=5$, $m=5,6$, a total of 2; when $n=6$, $m=5,6$, a total of 2. In total, 19. | 19 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,448 |
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