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
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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6. Given the function
$$
y=\left\{\begin{array}{ll}
x^{2}-2 x, & x \leqslant 3 ; \\
x^{2}-10 x+24, & x>3,
\end{array}\right.
$$
and the equation $y=k$ has exactly three solutions for $x$. Then the value of $k$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | 6. A.
As shown in Figure 9.
From the graph, it is easy to see that the line $y=3$ intersects the function graph at exactly three points, i.e., when $k=3$, the $x$ values that satisfy $y=k$ are exactly three. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,693 |
Example 1 Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ . | $$
\begin{array}{l}
\left(x^{4}+2 x^{3}+x^{2}\right)+\left[(2+k) x^{2}+(2+k) x\right]+2 k=0 \\
\Rightarrow\left(x^{2}+x\right)^{2}+(2+k)\left(x^{2}+x\right)+2 k=0 \\
\Rightarrow\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 .
\end{array}
$$
Since the original equation has real roots, and $x^{2}+x+2=0$ has no real root... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,694 |
Example 2 If the equation
$$
x^{3}-5 x^{2}+(4+k) x-k=0
$$
has three roots that can serve as the lengths of the three sides of an isosceles triangle, then the value of the real number $k$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | The original equation can be transformed into
$$
(x-1)\left(x^{2}-4 x+k\right)=0 \text {. }
$$
Given that the three roots of the original equation can serve as the three side lengths of an isosceles triangle, we know that $x=1$ is a root of $x^{2}-4 x+k=0$, or $x^{2}-4 x+k=0$ has two equal roots.
Solving this, we get ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,695 |
Example 11 The equation about $x$
$$
x^{4}+(k-1) x^{3}+k x^{2}+(k-1) x+1=0
$$
has no real roots. Then the range of real number $k$ is $\qquad$ | Solve: Dividing both sides of the original equation by $x^{2}$, we get
$$
x^{2}+\frac{1}{x^{2}}+(k-1)\left(x+\frac{1}{x}\right)+k=0 \text {. }
$$
Let $x+\frac{1}{x}=y \in(-\infty,-2] \cup[2,+\infty)$.
Then $x^{2}+\frac{1}{x^{2}}=y^{2}-2$
$$
\begin{array}{l}
\Rightarrow y^{2}+(k-1) y+k-2=0 \\
\Rightarrow(y+1)[y+(k-2)]=... | 0<k<4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,696 |
8. Regular tetrahedron
$ABCD$ and regular
tetrahedron $A_{1}B_{1}C_{1}D_{1}$,
and a regular octahedron
KEFGHI (as shown in
Figure 1) have equal
edge lengths. When they
are joined together, the
faces $ACD$ and $A_{1}C_{1}D_{1}$
are made to coincide with
the faces $EFK$ and $GHK$, respectively. The number of fa... | 8.6.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,697 |
9. There are 4 colors of light bulbs (with enough of each color), and we need to install a light bulb at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at ... | 9.216 .
We can first install $A$, $B$, and $C$, which has $\mathrm{A}_{4}^{3}$ ways; then select one vertex from $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, which has $\mathrm{C}_{3}^{1}$ ways; finally, there are 3 ways to install the remaining two vertices.
Therefore, there are 216 di... | 216 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,698 |
10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ . | 10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$
Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$
Thus, the smalle... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,699 |
11. Let $A$ and $B$ be the common vertices of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
and the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
Let $P$ and $M$ be two moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\... | 11. -5 .
Let $A(-a, 0), B(a, 0), P\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M})$, we know that points $O, P, M$ are collinear, and we can find that
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_... | -5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,700 |
12. $\triangle A B C$ is an inscribed triangle in $\odot O$, $A B>A C$ $>B C$, point $D$ is on arc $\overparen{B C}$, perpendiculars from $O$ to $A B$ and $A C$ intersect $A D$ at points $E$ and $F$, respectively, and rays $B E$ and $C F$ intersect at point $P$. If $B P=P C+P O$, then $\angle B A C=$ $\qquad$ . | 12. $30^{\circ}$.
Connect $O B, O C$. It is easy to know that $\triangle E A B$ and $\triangle F A C$ are both isosceles triangles.
From the property that the supplement of the vertex angle of an isosceles triangle is twice the base angle, we have
$$
\begin{array}{l}
\angle B P C=\angle A E P+\angle C F D \\
=2(\angl... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,701 |
13. Given the function
$$
f(x)=4 \sin x \cdot \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+\cos 2 x \text {. }
$$
(1) Let $\omega>0$, and be a constant. If the function $y=f(\omega x)$ is increasing on the interval $\left[-\frac{\pi}{2}, \frac{2 \pi}{3}\right]$, find the range of $\omega$;
(2) The sets are
$$
\begin... | Three, 13. (1) Notice,
$$
\begin{array}{l}
f(x)=2 \sin x\left[1-\cos \left(\frac{\pi}{2}+x\right)\right]+\cos 2 x \\
=2 \sin x(1+\sin x)+\cos 2 x \\
=2 \sin x+1 .
\end{array}
$$
Given that $f(\omega x)=2 \sin \omega x+1$ is an increasing function on $\left[-\frac{\pi}{2}, \frac{2 \pi}{3}\right]$, then
$$
\begin{array}... | m \in(1,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,702 |
14. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=\frac{n+2}{n} a_{n}+\frac{1}{n}$. Find the general term formula of the sequence $\left\{a_{n}\right\}$. | 14. From the given, we have
$$
a_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(a_{n}+\frac{1}{2}\right) \text {. }
$$
Let $b_{n}=a_{n}+\frac{1}{2}$. Then
$$
b_{n+1}=\frac{n+2}{n} b_{n} \text {. }
$$
Thus, $\frac{b_{n+1}}{b_{1}}=\frac{b_{n+1}}{b_{n}} \cdot \frac{b_{n}}{b_{n-1}} \cdots \cdots \frac{b_{2}}{b_{1}}$
$$
\begin{arra... | a_{n}=\frac{(n-1)(n+2)}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,703 |
15. Find the maximum value of the positive real number $A$ such that for any real numbers $x, y, z$, the inequality
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4}+x^{2} y z+x y^{2} z+x y z^{2}- \\
A(x y+y z+z x)^{2} \geqslant 0
\end{array}
$$
holds. | $$
a_{2}=1, a_{3}=\frac{5}{2}, a_{4}=\frac{9}{2}, \cdots,
$$
Guess: $a_{n}=\frac{(n-1)(n+2)}{4}$, then prove it using mathematical induction.
15. When $x=y=z$, the inequality transforms to
$$
6 x^{4} \geqslant A \cdot 9 x^{4} (x \in \mathbf{R}),
$$
which means $A \leqslant \frac{2}{3}$.
Next, we prove: the maximum va... | \frac{2}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,704 |
16. Does there exist 2011 different positive integers such that for any two numbers $a$ and $b$, we have
$$
|a-b|=(a, b)
$$
holds? | 16. Existence.
Below is the proof using mathematical induction.
(1) Since $|a-b| \geqslant (a, b)$, we have $|a-b| = (a, b) \Leftrightarrow (a-b) \mid a$.
Clearly, 2 and 3 satisfy the condition.
(2) Assume there exist $k\left(k \in \mathbf{N}_{+}\right)$ positive integers $a_{1}, a_{2}$,
$\cdots, a_{k}$, such that for ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,705 |
1. If $\frac{a^{-1}+b}{a+b^{-1}}=k$, then $\frac{a^{-2}+b^{2}}{a^{2}+b^{-2}}=(\quad)$.
(A) $k$
(B) $\frac{1}{2} k$
(C) $k^{2}$
(D) $\frac{1}{2} k^{2}$ | $$
\begin{array}{l}
k=\frac{a^{-1}+b}{a+b^{-1}}=\frac{b(1+a b)}{a(1+a b)}=\frac{b}{a} . \\
\text { Then } \frac{a^{-2}+b^{2}}{a^{2}+b^{-2}}=\frac{b^{2}\left(1+a^{2} b^{2}\right)}{a^{2}\left(1+a^{2} b^{2}\right)}=\frac{b^{2}}{a^{2}}=k^{2} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,706 |
If the equation
$$
x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0
$$
has exactly one real solution, then the value of $p$ is $\qquad$ | Hint: Use factorization to transform the original equation into $x^{2}+3 x-p=0$ or $x^{2}+3 x-2 p=0$.
From the conditions, we have
$$
\left\{\begin{array} { l }
{ \Delta _ { 1 } = 9 + 4 p = 0 , } \\
{ \Delta _ { 2 } = 9 + 8 p < 0 }
\end{array} \text { or } \left\{\begin{array}{l}
\Delta_{1}=9+4 p<0, \\
\Delta_{2}=9+8 ... | -\frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,707 |
2. Given the radius of the base of a cylinder is $r$, and the height is $h$. If the volume of the cylinder is 1, and the surface area is 12, then $\frac{1}{r}+\frac{1}{h}=$ ( ).
(A) 2
(B) 3
(C) 6
(D) 12 | 2. C.
From the given equations, we have
$$
\pi r^{2} h=1,2 \pi r^{2}+2 \pi r h=12 \text {. }
$$
Dividing the two equations yields $\frac{1}{r}+\frac{1}{h}=6$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,708 |
3. Given
$$
\begin{array}{l}
\left(\sqrt{x^{2}+201 \cdot 1}-x\right)\left(\sqrt{y^{2}+2011}-y\right) \\
=2011 . \\
\text { Then } \frac{x+2012 y}{2012 x+y}=(\quad) .
\end{array}
$$
(A) 2012
(B) 2011
(C) 1
(D) -1 | 3. D.
From the given, we have
$$
\sqrt{x^{2}+2011}-x=\sqrt{y^{2}+2011}+y \text {. }
$$
Similarly, $\sqrt{y^{2}+2011}-y=\sqrt{x^{2}+2011}+x$.
Adding the two equations yields $x+y=0$.
$$
\text { Then } \frac{x+2012 y}{2012 x+y}=\frac{2011 y}{2011 x}=\frac{y}{x}=-1 \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,709 |
4. Given the equation of line $A B$ is: $y=k x+m$ passing through points $A(a, a), B(b, 8 b)(a>0, b>0)$. When $\frac{b}{a}$ is an integer, the number of integer $k$ that satisfies the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 4. B.
When $a=b$, since $a=b=0$ does not meet the problem's conditions, we have $a \neq b$.
Thus, $k=\frac{a-8 b}{a-b}=\frac{1-8 \times \frac{b}{a}}{1-\frac{b}{a}}$.
Let $\frac{b}{a}=t$. Then,
$$
k=\frac{1-8 t}{1-t}=8-\frac{7}{1-t}=8+\frac{7}{t-1} .
$$
Since $\frac{b}{a}=t$ is a positive integer not equal to 1, and $... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,710 |
5. As shown in Figure 1, in the Cartesian coordinate system, there is a rectangle $A O B C$. The inverse proportion function $y=\frac{k}{x}$ passes through the intersection point $G$ of the diagonals of rectangle $A O B C$. A circle with radius $4-2 \sqrt{2}$ is inscribed in $\triangle A B C$. Then the minimum value of... | 5. A.
Let $G(a, b)$, and denote the radius of the circle as $r$. Then
$$
\begin{aligned}
& A C=2 a, B C=2 b, \\
& r=\frac{1}{2}(A C+B C-A B) \\
= & a+b-\sqrt{a^{2}+b^{2}} \\
= & \frac{2 a b}{a+b+\sqrt{a^{2}+b^{2}}} \\
\leqslant & \frac{2 a b}{2 \sqrt{a b}+\sqrt{2 a b}} \\
= & (2-\sqrt{2}) \sqrt{a b} .
\end{aligned}
$$... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,711 |
1. In $\triangle A B C$, $A B=A C$. If $\sin A=\frac{3}{5}$, then $\frac{B C}{A B}=$ . $\qquad$ | $=.1 . \frac{\sqrt{10}}{5}$.
As shown in Figure 7, draw $BH \perp AC$, with $H$ as the foot of the perpendicular.
$$
\begin{array}{l}
\text { Let } BH=3k, \\
AB=AC=5k .
\end{array}
$$
Then $AH=4k$,
$$
\begin{array}{c}
CH=k, \\
BC=\sqrt{BH^{2}+CH^{2}} \\
=\sqrt{10} k .
\end{array}
$$
Therefore, $\frac{BC}{AB}=\frac{\s... | \frac{\sqrt{10}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,713 |
3. As shown in Figure 4, the parabola $y=a x^{2}+b x-3$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$, and $O B=O C=3 O A$. The line $y=-\frac{1}{3} x+1$ intersects the $y$-axis at point $D$, and $E$ is the vertex of the parabola. If $\angle D B C=\alpha, \angle C B E=\beta$, t... | 3.45.
It is known that $D(0,1), C(0,-3), A(-1,0)$, $B(3,0)$.
Then $y=a(x+1)(x-3) \Rightarrow a=1$.
Thus, $y=x^{2}-2x-3$, and its vertex is $E(1,-4)$.
Therefore, $BC=3\sqrt{2}, CE=\sqrt{2}, BE=2\sqrt{5}$.
Since $BC^{2}+CE^{2}=BE^{2}$, $\triangle BCE$ is a right triangle.
Thus, $\tan \beta=\frac{CE}{CB}=\frac{1}{3}$.
Al... | 45^{\circ} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,715 |
4. As shown in Figure 5, given $\triangle A B C$ is inscribed in $\odot O$, chord $A F \perp B C$ at point $H, G$ is the midpoint of $B F$. Then $\frac{A C}{O G}=$ | 4. 2 .
Connect $F O$. Then
$$
\begin{array}{l}
\text { Rt } \triangle A B H \backsim \text { Rt } \triangle O \\
\Rightarrow \frac{A H}{B H}=\frac{O G}{F G} . \\
\text { Also } \triangle A C H \backsim \triangle B F H \\
\Rightarrow \frac{A H}{B H}=\frac{A C}{B F} .
\end{array}
$$
Therefore $A C=2 O G \Rightarrow \fr... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,716 |
One. (20 points) In $\triangle ABC$, it is known that $DE \parallel BC$, the incircle of $\triangle ABC$ touches $DE$ and $BC$ at points $M$ and $N$ respectively, and $BE$ intersects $CD$ at point $P$. Prove: $M$, $N$, and $P$ are collinear. | As shown in Figure 8, let $B E$ intersect $M N$ at point $P'$.
Since $D E \parallel B C$,
we have,
$$
\begin{array}{l}
\frac{B P}{P E}=\frac{B C}{D E}, \\
\frac{B P^{\prime}}{P^{\prime} E}=\frac{B N}{E M} .
\end{array}
$$
Therefore, we only need to
prove:
$$
\frac{B C}{D E}=\frac{B N}{E M} .
$$
Clearly,
$$
\begin{ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,717 |
2. The equation
$$
3 x^{3}+2 \sqrt{2} x^{2}-(17-9 \sqrt{2}) x-(6-5 \sqrt{2})=0
$$
has solutions
$$
x_{1}=
$$
$\qquad$
$$
, x_{2}=
$$
$\qquad$
$$
, x_{3}=
$$
$\qquad$ . | Answer: $\frac{\sqrt{2}}{3}, \sqrt{2}-1,1-2 \sqrt{2}$. | \frac{\sqrt{2}}{3}, \sqrt{2}-1,1-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,718 |
II. (25 points) Let $x^{5}+y^{5}=2$. Prove: $x+y \leqslant 2$.
保留了原文的换行和格式，如上所示。 | Given that $x, y$ at least one is greater than 0, without loss of generality, assume $x>0$.
First, we prove: $x+y>0$.
If $x+y=0$, then $y=-x$, and we have
$$
2=x^{5}+y^{5}=x^{5}+(-x)^{5}=0 \text {, }
$$
which is not true;
If $x+y<x$, then
$$
x^{5}-2=-y^{5}=(-y)^{5}>x^{5} \text {, }
$$
which is not true.
Therefore, $x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,719 |
$$
\begin{array}{l}
\text { Three. (25 points) Let } \\
x=a+b-c, y=c+a-b, z=b+c-a,
\end{array}
$$
where $a$, $b$, and $c$ are prime numbers, and satisfy $x^{2}=y$, $\sqrt{z}-\sqrt{y}$ $=2$. Question: Can $a$, $b$, and $c$ form the three sides of a triangle? If they can, find the area of the triangle; if not, please ex... | Three, cannot.
From the given, we have
$$
a=\frac{1}{2}(x+y), b=\frac{1}{2}(x+z), c=\frac{1}{2}(y+z).
$$
Since $y=x^{2}$, we have
$$
a=\frac{1}{2}\left(x+x^{2}\right)=\frac{x(x+1)}{2}.
$$
Given that $x$ is an integer and $a$ is a prime number, then $x=2$ or $-3, a=3$.
When $x=2$,
$$
y=x^{2}=4 \Rightarrow z=(\sqrt{y}+... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,720 |
1. For the system of inequalities about real numbers $x, y$
$$
\left\{\begin{array}{l}
y \geqslant x^{2}+2 a, \\
x \geqslant y^{2}+2 a
\end{array}\right.
$$
to have a unique solution, the values of the parameter $a$ are | 1. $\frac{1}{8}$.
From the problem, we know that the parabolas
$$
y=x^{2}+2 a \text { and } x=y^{2}+2 a
$$
are tangent to each other, and $y=x$ is their common tangent line.
Therefore, $a=\frac{1}{8}$. | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,721 |
2. The number of real numbers $a$ that make the equation
$$
x^{2}+a x+8 a=0
$$
have only integer solutions is $\qquad$ . | 2. 8 .
Let the equation (1) have integer solutions $m, n (m \leqslant n)$. Then $m+n=-a, mn=8a$.
Thus, $(m+8)(n+8)=64$.
Solving for $(m, n)$
$$
\begin{array}{c}
=(-72,-9),(-40,-10),(-24,-12), \\
(-16,-16),(-7,56),(-6,24), \\
(-4,8),(0,0) .
\end{array}
$$
Correspondingly,
$$
\begin{array}{l}
a=-(m+n) \\
=81,50,36,32,-... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,722 |
3. Let the function $f(x, y)$ satisfy:
(1) $f(x, x)=x$;
(2) $f(k x, k y)=k f(x, y)$;
(3) $f\left(x_{1}+x_{2}, y_{1}+y_{2}\right)=f\left(x_{1}, y_{1}\right)+f\left(x_{2}, y_{2}\right)$;
(4) $f(x, y)=f\left(y, \frac{x+y}{2}\right)$.
Then $f(x, y)=$ . $\qquad$ | 3. $\frac{x}{3}+\frac{2 y}{3}$.
Notice that,
$$
\begin{array}{l}
f(x, 0)+f(0, x) \stackrel{(3)}{=} f(x, x)=x, \\
f(x, 0) \stackrel{(4)}{=} f\left(0, \frac{x}{2}\right) \stackrel{(2)}{=} \frac{1}{2} f(0, x) .
\end{array}
$$
Subtracting the two equations gives $f(0, x)=\frac{2 x}{3}$.
Similarly, $f(0, y)=\frac{2 y}{3}$... | \frac{x}{3}+\frac{2 y}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,723 |
5. All prime numbers $p$ that make $2 p^{4}-p^{2}+36$ a perfect square are $\qquad$ . | 5.2.
When $p=2$,
$$
2 p^{4}-p^{2}+36=64
$$
is a perfect square, thus, $p=2$ is the solution.
When $p=3$,
$$
2 p^{4}-p^{2}+36=189
$$
is not a perfect square.
When $p$ is an odd prime greater than 3, let
$2 p^{4}-p^{2}+36=k^{2}$ ( $k$ is an odd positive integer).
Then $p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6)$
$\Rightar... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,725 |
6. Let $\alpha, \beta, \gamma$ be the angles formed by the body diagonal of a rectangular parallelepiped with the three faces meeting at a vertex. Then
$$
f(\alpha, \beta, \gamma)=\sum \frac{\sin ^{4} \alpha+\sin ^{2} \beta}{\sin ^{2} \beta+\sin ^{2} \gamma}
$$
the range of values for $f(\alpha, \beta, \gamma)$ is $\q... | 6. $[2,+\infty)$.
From the problem, we have
$$
\begin{array}{l}
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1 . \\
\text { Then } \sum \frac{\sin ^{4} \alpha+\sin ^{2} \beta}{\sin ^{2} \beta+\sin ^{2} \gamma}+1 \\
=\sum\left(\frac{\sin ^{4} \alpha+\sin ^{2} \beta}{\sin ^{2} \beta+\sin ^{2} \gamma}+\sin ^{2} \alp... | [2,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,726 |
7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the last four digits of $a_{2 \text { on }}$ are $\qquad$ | 7.8124.
Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$
subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$
Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^... | 8124 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,727 |
8. Given $x \leqslant-\frac{1}{2}$. Then the minimum value of the bivariate function
$$
\begin{array}{l}
f(x, y) \\
=\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}-2 x+2}+\sqrt{y^{2}-2 y+5}
\end{array}
$$
is . $\qquad$ | $\begin{array}{l}\text { 8. } \frac{\sqrt{13}+\sqrt{29}}{2} . \\ \text { Let } A(x, 0), B(0, y), C(1,-1), D(2,1) \text {. } \\ \text { Then } f(x, y)=|A B|+|A C|+|B D| \\ \geqslant|A C|+|A D| \\ \geqslant \sqrt{\left(\frac{3}{2}\right)^{2}+1}+\sqrt{\left(\frac{5}{2}\right)^{2}+1} \\ =\frac{1}{2}(\sqrt{13}+\sqrt{29}) \t... | \frac{\sqrt{13}+\sqrt{29}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,728 |
3. Solve the equation:
$$
(6 x+7)^{2}(3 x+4)(x+1)=6 \text {. }
$$ | $$
\begin{array}{l}
(6 x+7)^{2}(6 x+8)(6 x+6)=72, \\
\text { i.e. }\left(36 x^{2}+84 x+49\right)\left(36 x^{2}+84 x+48\right)=72 .
\end{array}
$$
Answer: $x_{1}=-\frac{2}{3}, x_{2}=-\frac{5}{3}$. | x_{1}=-\frac{2}{3}, x_{2}=-\frac{5}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,729 |
9. (16 points) As shown in Figure 1, let $P$ be a point outside the plane of square $A B C D$. Points $M$ and $N$ are on $P A$ and $B D$ respectively, and $\frac{P M}{M A}=\frac{B N}{N D}=\frac{5}{8}$. Prove: Line $M N \parallel$ plane $P B C$.
保留源文本的换行和格式，直接输出翻译结果如下：
9. (16 points) As shown in Figure 1, let $P$ be a... | $$
\begin{array}{l}
\overrightarrow{M N}=\overrightarrow{M P}+\overrightarrow{P B}+\overrightarrow{B N} \\
=\frac{5}{13}(\overrightarrow{A B}-\overrightarrow{P B})+\overrightarrow{P B}+\frac{5}{13}(\overrightarrow{B C}+\overrightarrow{C D}) \\
=\frac{8}{13} \overrightarrow{P B}+\frac{5}{13} \overrightarrow{B C} .
\end{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,730 |
10. (20 points) Let \( a, b, c, d \in \mathbf{R}_{+} \). Prove:
$$
\sqrt[3]{a b}+\sqrt[3]{c d}<\sqrt[3]{(a+b+c)(b+c+d)} .
$$ | 10. Notice that, when $x, y, z \in \mathbf{R}_{+}$ and are distinct, we have
$$
\sqrt[3]{x y z}<\frac{x+y+z}{3} .
$$
Let $x=\frac{a}{a+b+c}, y=\frac{b+c}{b+c+d}, z=\frac{b}{b+c}$. Then
$$
\begin{array}{c}
\sqrt[3]{\frac{a b}{(a+b+c)(b+c+d)}} \\
<\frac{1}{3}\left(\frac{a}{a+b+c}+\frac{b+c}{b+c+d}+\frac{b}{b+c}\right) .... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,731 |
11. (20 points) If $N(m, 0)$ is a point on the internal axis of symmetry (the $x$-axis) of the quadratic curve $\Gamma$:
$$
A x^{2}+B y^{2}+D x+F=0
$$
$\left(A^{2}+B^{2} \neq 0,2 A m+D \neq 0\right)$, and a moving line through $N$ intersects the quadratic curve $\Gamma$ at points $P$ and $Q$. Let $M$ be a point on the ... | 11. Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)\left(x_{1}<m<x_{2}\right)$,
$$
\begin{array}{l}
M\left(x_{0}, 0\right), l_{P Q}: x=k y+m . \\
\quad \text { From }\left\{\begin{array}{l}
x=k y+m, \\
A x^{2}+B y+D x+F=0,
\end{array} \text { eliminate } x\right. \text { to get } \\
\quad\left(A k^{2}+B\rig... | M\left(-\frac{D m+2 F}{2 A m+D}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,732 |
One. (40 points) As shown in Figure 2, let $M$ and $N$ be the midpoints of the minor arcs $\overparen{BC}$ and $\overparen{CA}$ of the circumcircle of $\triangle ABC$, respectively. Draw $PC \parallel NM$ intersecting the circumcircle at point $P$, and let $I$ be the incenter of $\triangle ABC$. The line $PI$ intersect... | As shown in Figure 3, from the given conditions, we know that points $A, I, M$ and points $B, I, N$ are collinear respectively. Connect $C M, C I, C N$.
By the properties of the incenter, we have
$$
N I = N C, \quad M I = M C.
$$
Thus, quadrilateral $N I M C$ is a kite, meaning that $M N$ is the perpendicular bisecto... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,733 |
Sure, here is the translated text:
```
II. (40 points) Let positive real numbers $a, b$ satisfy $a+b=1$.
Find the minimum value of the function
$$
f(a, b)=\left(\frac{1}{a^{5}}+a^{5}-2\right)\left(\frac{1}{b^{5}}+b^{5}-2\right)
$$
``` | Let $a b=t$. Then $t \in\left(0, \frac{1}{4}\right]$.
From $a+b=1$, we know
$$
\begin{array}{l}
a^{5}+b^{5}=a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4} \\
=\left(a^{2}+b^{2}\right)^{2}+a^{2} b^{2}-a b(a+b)^{2} \\
=(1-2 t)^{2}+t^{2}-t \\
=5 t^{2}-5 t+1 .
\end{array}
$$
Then $f(a, b)=\frac{\left(a^{5}-1\right)^{2}}{a^{5}} \... | \frac{31^4}{32^2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,734 |
Three, (50 points) Given any prime $p$. Prove: There exists an integer $x_{0}$ such that $p \mid \left(x_{0}^{2}-x_{0}+1\right)$ if and only if there exists $y_{0}$ such that $p \mid \left(y_{0}^{2}-y_{0}+7\right)$. | Three, it is known that $x_{0}^{2}-x_{0}+1$ and $y_{0}^{2}-y_{0}+7$ are both odd numbers. Let $p$ be an odd prime. Then
$$
\begin{array}{l}
p \mid\left(x_{0}^{2}-x_{0}+1\right) \\
\Leftrightarrow p \mid 4\left(x_{0}^{2}-x_{0}+1\right) \\
\Leftrightarrow p \mid\left(2 x_{0}-1\right)^{2}+3, \\
p \mid\left(y_{0}^{2}-y_{0}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,735 |
Four. (50 points) From $1,2, \cdots, 2011$, what is the minimum number of different numbers that must be selected to ensure that among the selected numbers, there must be three different numbers that can form the lengths of the three sides of a triangle.
Translate the above text into English, please retain the origina... | $$
\begin{array}{l}
\text { Let the required smallest positive integer be } n. \\
\text { Consider the problem from the opposite angle. How many numbers can there be at most without any three different numbers forming the sides of a triangle? } \\
\text { When } 0<a_{1} . \\
\text { Hence } a_{2} \geqslant 2, a_{3} \ge... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,736 |
Let $a, b \in \mathbf{R}_{+}$, and
$$
\frac{1}{a^{2}+2}+\frac{1}{b^{2}+2}=\frac{1}{3} \text {. }
$$
Prove: $\frac{1}{a}+\frac{1}{b} \geqslant 1$. | Prove that,
$$
\begin{aligned}
\frac{1}{a^{2}+2} & +\frac{1}{b^{2}+2}=\frac{1}{3} \\
& \Leftrightarrow 3\left(a^{2}+b^{2}+4\right)=\left(a^{2}+2\right)\left(b^{2}+2\right) \\
& \Leftrightarrow a^{2}+b^{2}+8=a^{2} b^{2} .
\end{aligned}
$$
From equation (1) and $a^{2}+b^{2} \geqslant 2 a b\left(a, b \in \mathbf{R}_{+}\r... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,737 |
Given 314 As shown in Figure $2, \odot O$ is the incircle of $\triangle A B C$, and $D, E, M$ are the points of tangency. Connect $M O$ and extend it to intersect $D E$ and $\odot O$ at points $K$ and $F$ respectively. Connect $A F$ and $A K$ and extend them to intersect $B C$ at points $N$ and $L$ respectively. Prove:... | Prove as shown in Figure 2, draw a line through point $K$ parallel to $BC$, intersecting $AB$ and $AC$ at points $R$ and $S$ respectively, and connect $OD$, $OE$, $OR$, and $OS$. Draw a tangent line to $\odot O$ through point $F$, intersecting $AB$ and $AC$ at points $P$ and $Q$ respectively.
Since $\angle OKS = \angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,738 |
Given 313 As shown in Figure $3, \triangle A B C$ has an incircle $\odot I$ that is tangent to $B C, C A, A B$ at points $D, E, F$ respectively. Connect $A D$ to intersect $\odot I$ at point $P$, and intersect $E F$ at point $T$. Connect $B P, C P$ to intersect $\odot I$ at points $H, G$ respectively. $H F$ intersects ... | Prove that in Figure 4, draw the tangent line $l$ of $\odot I$ through point $P$, intersecting $A B$ and $A C$ at points $R$ and $S$ respectively.
For the circumscribed quadrilateral $BCSR$, by Newton's theorem, points $C$, $T$, and $R$ are collinear.
Let $H F$ intersect the tangent line $l$ at point $K^{\prime}$.
To ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,739 |
4. The solution to the equation $x=\left(x^{2}-2\right)^{2}-2$ is | Answer: $x_{1}=2, x_{2}=-1, x_{3,4}=\frac{-1 \pm \sqrt{5}}{2}$. | x_{1}=2, x_{2}=-1, x_{3,4}=\frac{-1 \pm \sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,740 |
314 Assuming that a smart person answers at least 9 out of 10 questions correctly, and a confused person answers at least 9 out of 10 questions incorrectly. Among 2011 people, there are exactly 7 confused people, and the rest are smart people. To determine the identity (smart or confused) of 28 people, the examiner des... | Solution: First, number the questions as: $T_{1}, T_{2} \cdots, T_{26}$, and the personnel as: $P_{1}, P_{2} \cdots, P_{28}$.
Arrange the answering of questions according to Table 1 and Table 2. Among them, $P_{14}$ answers the questions
$$
T_{1}, T_{2}, \cdots, T_{6}, T_{20}, T_{21}, \cdots, T_{26} ;
$$
$P_{28}$ answ... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 724,741 |
5. To make the equation
$$
x^{4}+(m-4) x^{2}+2(1-m)=0
$$
have exactly one real root that is not less than 2, the range of values for $m$ is $\qquad$. | Let $x^{2}=y$. Then the original equation becomes
$$
y^{2}+(m-4) y+2(1-m)=0 \text {. }
$$
By the problem statement, this equation has exactly one real root that is not less than 4.
Answer: $m \leqslant-1$. | m \leqslant -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,742 |
6. If the two roots of the equation $x^{2}-2 x+\frac{\sqrt{3}}{2}=0$ are $\alpha, \beta$, and they are also the roots of the equation $x^{4}+p x^{2}+q=0$, then $p=$ $\qquad$ . | $\sqrt{3}-4$ | \sqrt{3}-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,743 |
Example 1 Proof: For all positive integers $n$ and all real numbers $x$,
$$
\begin{array}{l}
\sum_{k=0}^{n}(-1)^{k} \mathrm{C}_{n}^{k} \frac{x}{x+k} \\
=\frac{n!}{(x+1)(x+2) \cdots(x+n)} .
\end{array}
$$ | Prove that when $n=1$,
left side $=1-\frac{x}{x+1}=\frac{1}{x+1}=$ right side.
Assume that when $n=m$, for all real numbers,
$$
\sum_{k=0}^{m}(-1)^{k} \mathrm{C}_{m}^{k} \frac{x}{x+k}=\frac{m!}{(x+1)(x+2) \cdots(x+m)} .
$$
Then when $n=m+1$, for all real numbers $x$,
$$
\begin{aligned}
& \sum_{k=0}^{m+1}(-1)^{k} \math... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,744 |
Example 2 Let the set $S_{n}=\{1,2, \cdots, n\}, X$ be a subset of $S_{n}$. The sum of all numbers in $X$ is called the "capacity" of $X$ (the capacity of the empty set is defined as 0). If the capacity of $X$ is odd (even), then $X$ is called an "odd (even) subset" of $S_{n}$. Prove:
(1) The number of odd subsets of $... | Proof: Let the number of odd subsets of $S_{n}$ be $a_{n}$, and the sum of their sizes be $A_{n}$; the number of even subsets be $b_{n}$, and the sum of their sizes be $B_{n}$.
(1) Prove by mathematical induction: $a_{n}=b_{n}$.
When $n=1$, $S_{1}$ has one odd subset $\{1\}$, and one even subset $\varnothing$, so $a_{1... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,745 |
Example 3 㷵 On a board, there is a convex 2011-gon, and Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? [3] | To prove by induction: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. The following $2n-6$ diagonals can be drawn sequentially:
$A_{2} A_{4}, A_{3} A_{5}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
When $n=3$, the... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,746 |
Example 3 Given that $a$ is a positive integer. If the roots of the equation
$$
x^{3}+(a+17) x^{2}+(38-a) x-56=0
$$
are all integers, find the value of $a$ and the integer roots of the equation. | Solve the left side of the equation by factoring to get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Since the equation about $x$
$$
x^{2}+(a+18) x+56=0
$$
has integer roots, the discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}\left(k \in \mathbf{N}_{+}\right)$. The... | a=12, \text{ roots } 1, -2, -28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,747 |
Example 4 The vertices of a finite graph can be colored black or white. Initially, all vertices are black. Each operation selects a vertex $P$, and changes the color of $P$ and its adjacent vertices. Can all vertices be changed from black to white through a series of operations? ${ }^{[4]}$ | Solution: It is possible.
We use induction to prove for a graph with $n$ vertices.
When $n=1$, it is obviously true.
When the graph has $n-1(n \geqslant 2)$ vertices, this operation can be achieved.
Let $X$ be a graph with $n$ vertices, the vertices being $P_{1}, P_{2}, \cdots, P_{n}$.
Let $P_{i}$ be the base point, ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,748 |
For example, let $5 M$ be a set of a finite number of integer points in the plane. Prove: It is possible to color the points in $M$ red or blue, such that on every line parallel to the coordinate axes, the number of points of the two colors is equal or differs by $1 .{ }^{[5]}$ | Prove by induction on the number of points $n$ in the set $M$.
When $n=1$, the conclusion is obvious.
Assume the proposition holds for point numbers less than $n$.
Consider $n$ points.
If there is only one point in $M$ on a line parallel to the coordinate axis, then the remaining $n-1$ points can be colored with two co... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,749 |
Example 6 As shown in Figure 1, an $\mathrm{L}$-shape is composed of three unit squares. Prove: for any positive integer $k$, a similar $\mathrm{L}$-shape that is $k$ times larger can be divided into several unit $\mathrm{L}$-shapes. ${ }^{[4]}$ | Prove that by placing a large L-shape as shown in Figure 2, with its longer sides intersecting at the top-left corner. Starting from the top-left corner, cover the large L-shape with \( k \) unit L-shapes in the same direction along the diagonal. The remaining part consists of two identical "staircase" shapes.
It is su... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,750 |
Example 7 In an $n \times n$ chessboard, the common part of any $k$ rows and any $l$ columns is called its "sub-region," and $k+1$ is referred to as its semi-perimeter. Suppose several sub-regions with semi-perimeters not less than $n$ together cover the entire main diagonal of the chessboard. Prove: the number of squa... | Prove for $n$ using mathematical induction.
When $n=1,2$, the conclusion is obviously true.
When $n>2$, assume the conclusion holds for all $m \times m (m<n)$ chessboards.
We only need to prove that the conclusion also holds for an $n \times n$ chessboard.
In fact, for any set of sub-rectangles that meet the requiremen... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,751 |
1. Let $k$ be a positive integer. Prove: The set $\left\{0,1,2,3, \cdots, 2^{k+1}-1\right\}$ can be divided into two disjoint subsets $\left\{x_{1}, x_{2}, \cdots, x_{2^{k}}\right\}$ and $\left\{y_{1}, y_{2}, \cdots, y_{2^{k}}\right\}$, such that
$$
\sum_{i=1}^{2 k} x_{i}^{m}=\sum_{i=1}^{2 k} y_{i}^{m}
$$
for any $m \... | Use the second principle of mathematical induction on $k$.
When $k=1$, it holds. Assume the proposition holds for all values less than or equal to $k$, and consider the case for $k+1$. By the assumption, we have
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{2^{k}}, 2^{k+1}+y_{1}, 2^{k+1}+y_{2}, \cdots, 2^{k+1}+y_... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,752 |
3. Prove: In a square chessboard composed of $2^{n} \times 2^{n}$ unit squares, after arbitrarily removing one square, it is always possible to cover the remaining area with L-shaped blocks consisting of three unit squares, without any overlap.
| Hint: Use the first principle of mathematical induction. When $n=1$, the conclusion is obvious. Assume the proposition holds for $n-1$, consider the case for $n$.
Divide this chessboard into 4 equal parts, each part being a $2^{n-1} \times 2^{n-1}$ chessboard. Suppose the removed small square is in the first part. Fir... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,753 |
Question 1 In $\triangle A B C$, $A B=A C, \angle C A B$ and $\angle A B C$'s internal angle bisectors intersect the sides $B C$ and $C A$ at points $D$ and $E$ respectively. Let $K$ be the incenter of $\triangle A D C$. If $\angle B E K=45^{\circ}$, find all possible values of $\angle C A B$ ${ }^{[1,2]}$. | As shown in Figure 1, if $\angle C E K=45^{\circ}$, then because $\angle B E K=45^{\circ}$, we have $\angle B E C=90^{\circ}$.
Also, $\angle A B E=\angle C B E$, so $A B=B C$.
Therefore, $A B=A C=B C$.
Thus, $\angle B A C=60^{\circ}$.
As shown in Figure 2, if $\angle C E K \neq 45^{\circ}$, then $\angle B E C \neq 90^{... | 60^{\circ} \text{ or } 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,754 |
Question 3 In isosceles $\triangle ABC$, $M$ is the midpoint of base $AB$, and there is a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC$. Prove: $\angle APM + \angle BPC = 180^{\circ}$.
[3] | Prove: As shown in Figure 3, construct the circumcircle $\odot O$ of $\triangle A B P$. Then, by $\angle P A B=\angle P B C$, we know that $B C$ is a tangent to $\odot O$.
Similarly, $A C$ is also a tangent to $\odot O$.
Extend $C P$ to intersect $A B$ at point $E$, and $\odot O$ at point $D$. Connect $A D$, $D M$, a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,755 |
Question 5 In $\triangle ABC$, $\angle ACB=90^{\circ}$, $D$ is a point on side $AC$, $K$ is a point on side $BD$, and $\angle ABC = \angle KAD = \angle AKD$. Prove: $BK=2DC.^{[4]}$ | Prove that in Figure 4, extend \( AC \) to point \( E \) such that \( CE = CD \). Connect \( BE \). Then
\[
\begin{array}{l}
BE = BD, \\
\angle CBE = \angle CBD.
\end{array}
\]
Let \(\angle ABC\).
\[
\begin{array}{l}
= \angle KAD \\
= \angle AKD = \alpha.
\end{array}
\]
Then \(\angle BAC = 90^\circ - \angle ABC = 90^... | BK=2DC | Geometry | proof | Yes | Yes | cn_contest | false | 724,756 |
Example 2 Let $A B C D E F$ be a complete quadrilateral, $A C$ intersects $B D$ at point $P$, and a line through $P$ parallel to $E F$ intersects $B C, A D, A B, C D$ at points $M, N, X, Y$ respectively. Prove:
$$
P M=P N, P X=P Y .
$$ | Prove that, as shown in Figure 2, extend $AC$ to intersect $EF$ at point $Q$.
By the proposition, we have
$$
\frac{A P}{P C}=\frac{A Q}{Q C} \text { or } \frac{A P}{A Q}=\frac{P C}{Q C} .
$$
Using the properties of parallel lines, we get
$$
\frac{P N}{Q F}=\frac{A P}{A Q}=\frac{P C}{Q C}=\frac{P M}{Q F} \text {. }
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,757 |
Example 4 Given that $a$, $b$, and $2$ are the side lengths of a triangle, and $a$, $b$ are the roots of the equation
$$
\left(3 x^{2}-4 x-1\right)\left(3 x^{2}-4 x-5\right)=12
$$
Then the perimeter of the triangle can only be ( ).
(A) $\frac{10}{3}$ or $\frac{8}{3}$
(B) $\frac{14}{3}$ or $\frac{10}{3}$
(C) $\frac{16}{... | Let $3 x^{2}-4 x-1=y$.
Then the original equation becomes $y(y-4)=12$.
Solving this equation yields $y=-2$ or 6.
Thus, $3 x^{2}-4 x-1=-2$ or $3 x^{2}-4 x-1=6$.
Solving these gives $x_{1}=1, x_{2}=\frac{1}{3}, x_{3}=\frac{7}{3}, x_{4}=-1$.
If $a \neq b$, from $a+b>2,|a-b|2$, we know $a=b=\frac{7}{3}$.
Therefore, the per... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,758 |
Example 3 Given a moving point $O$ inside $\triangle A B C$, rays $A O, B O, C O$ intersect the opposite sides at points $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Let $B^{\prime} C^{\prime}$ intersect $A O$ at $D$, $C^{\prime} A^{\prime}$ intersect $B O$ at $E$, and $A^{\prime} B^{\prime}$ intersect $C O$ at $... | Prove that, as shown in Figure 3, there exist three complete quadrilaterals.
From the complete quadrilateral \( A C^{\prime} O B^{\prime} B C \), the complete quadrilateral \( B A^{\prime} O C^{\prime} C A \), and the complete quadrilateral \( C B^{\prime} O A^{\prime} A B \), we get
\[
\begin{array}{l}
\frac{O D}{A D}... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 724,759 |
Prove: For any positive real numbers $a, b, c, d$, we have
\[
\begin{array}{l}
\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+ \\
\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geqslant 0,
\end{array}
\]
and determine the conditions under which equality holds. | Proof: Let $x_{1}=a+b+c, x_{2}=b+c+d$,
$$
x_{3}=c+d+a, x_{4}=d+a+b \text {. }
$$
Obviously, $x_{i}>0$, and
$$
\left|\frac{x_{i}-x_{j}}{x_{k}}\right|<1(i \neq j \neq k \in\{1,2,3,4\}) .
$$
Then equation (1) is equivalent to
$$
\begin{array}{l}
\frac{\left(x_{2}-x_{3}\right)\left(x_{2}-x_{4}\right)}{x_{1}}+\frac{\left(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,760 |
Given $x$ is an integer greater than 3, and $n=$ $x^{6}-1$. Let $p^{k}\left(p\right.$ be a prime, $\left.k \in \mathbf{N}_{+}\right)$ be a factor of $n$. Prove: $p^{3 k}<8 n$. ${ }^{[1]}$ | Prove that given
$$
n=(x-1)(x+1)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right) \text {, }
$$
and $p^{k} \mid n$.
(1) $p=3$.
Since $x^{2} \equiv 0,1,4,7(\bmod 9)$, therefore,
$$
\begin{array}{l}
\left(x^{2}-x+1\right)\left(x^{2}+x+1\right) \\
=x^{4}+x^{2}+1 \equiv 1 \text { or } 3(\bmod 9) .
\end{array}
$$
Hence $p^{k-1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,761 |
1. As shown in Figure 1, given $\angle A O B=90^{\circ}, \angle B=30^{\circ}$, $\triangle A^{\prime} O B^{\prime}$ can be considered as
the result of rotating $\triangle A O B$ around point $O$
clockwise by an angle $\alpha$.
If point $A^{\prime}$ lies on
$A B$, then the size of $\alpha$ is ( ).
(A) $30^{\circ}$
(B) $4... | - 1. C.
It is known that $\angle A=60^{\circ}$.
Since $O A^{\prime}=O A$, we have $\angle A^{\prime} O A=\angle O A^{\prime} A=\angle A=60^{\circ}$. Therefore, $\alpha=60^{\circ}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,762 |
2. Given the seven-digit number $\overline{13 a b 45 c}$ is divisible by 792. Then the value of $\frac{b}{a+c}$ is ( ).
(A) 0
(B) 1
(C) greater than 1
(D) greater than 0 and less than 1 | 2. A.
From $792=8 \times 9 \times 11$, we know that 45c can be divided by 8.
Thus, $c=6$.
Since $\overline{13 a b 45 c}$ can be divided by 9, therefore,
$a+b=17$ or 8;
Also, since $\overline{13 a b 45 c}$ can be divided by 11, therefore, $a-b=8$ or -3.
Given that $a, b$ are integers between $0 \sim 9$, we know $a=8, b... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,763 |
3. Given that the equilateral $\triangle A B C$ is inscribed in $\odot O, S_{\triangle B B C}=a$, and the quadrilateral $D E F G$ is an inscribed square in the semicircle $\odot O$, $S_{\text {quadrilateral } D E F G}=b$. Then the value of $\frac{a}{b}$ is $($.
(A) 2
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{3 \sqrt{3}}{5}$
... | 3. D.
Let the radius of $\odot O$ be $r$. Then
$$
a=\frac{3 \sqrt{3} r^{2}}{4}, b=\frac{4 r^{2}}{5} .
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,764 |
4. Let prime numbers $a, b$ satisfy $a^{2}-9 b-4=0$. Then the median of the data $a, b, 2, 3$ is ().
(A) 4
(B) 7
(C) 4 or 7
(D) 4.5 or 6.5 | 4. C.
From $(a+2)(a-2)=9 b$ (where $a$ and $b$ are prime numbers), we know that $(a+2, a-2)$ $=(9, b)$ or $(b, 9)$ or $(9 b, 1)$ or $(3 b, 3)$.
Solving, we get $(a, b)=(7,5)$ or $(11,13)$.
Therefore, the median of $2, 3, 5, 7$ is $4$; the median of $2, 3, 11, 13$ is $7$. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,765 |
5. A moving point $P$ starts from the origin on the number line and moves in the positive direction of the number line according to the procedure of advancing 5 units and then retreating 3 units. It is known that $P$ advances or retreats 1 unit per second, and let $x_{n}$ represent the number corresponding to the posit... | 5. B.
Notice that,
$$
2011=8 \times 251+3,2 \times 251+3=505 \text {. }
$$
Therefore, the answer is B. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,766 |
6. As shown in Figure 2, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and point $E$ is on $A C$ such that $\angle A E D=90^{\circ}+\frac{1}{2} \angle C$. Then $B C+2 A E$ equals ( ).
(A) $A B$
(B) $A C$
(C) $\frac{3}{2} A B$
(D) $\frac{3}{2} A C$ | 6. B.
Draw $B F / / D E$ intersecting $A C$ at point $F$. Then $\angle B F C=\angle D E F$.
From $D$ being the midpoint of $A B$, we know $E F=A E$.
Notice,
$$
\begin{array}{l}
\angle B F C=\angle D E F=180^{\circ}-\angle A E D \\
=180^{\circ}-\left(90^{\circ}+\frac{1}{2} \angle C\right)=90^{\circ}-\frac{1}{2} \angle ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,767 |
7. Let positive integers $a$, $m$, $n$ satisfy
$$
\sqrt{a^{2}-4 \sqrt{2}}=\sqrt{m}-\sqrt{n} \text {. }
$$
Then the values of $a$, $m$, $n$ ( ).
(A) have one set
(B) have two sets
(C) have more than two sets
(D) do not exist | 7. A.
Squaring the given equation, we get
$$
\begin{array}{l}
a^{2}-4 \sqrt{2}=m+n-2 \sqrt{m n} . \\
\text { Then }\left\{\begin{array}{l}
m n=8, \\
a^{2}=m+n .
\end{array}\right.
\end{array}
$$
Given $m \geqslant n$, we know $(m, n)=(8,1)$ or $(4,2)$.
Upon verification, the solution that meets the conditions is
$$
(... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,768 |
Example $\mathbf{5}$ equation
$$
\begin{array}{l}
\left(x^{3}-3 x^{2}+x-2\right)\left(x^{3}-x^{2}-4 x+7\right)+ \\
6 x^{2}-15 x+18=0
\end{array}
$$
all distinct real roots are $\qquad$ | Let $x^{3}-3 x^{2}+x-2=A$, $x^{3}-x^{2}-4 x+7=B$.
It is easy to see that $6 x^{2}-15 x+18=3(B-A)-9$. Therefore, the original equation can be rewritten as $A B+3(B-A)-9=0$, which simplifies to $(A+3)(B-3)=0$.
If $A+3=0$, then
$$
\begin{array}{l}
x^{3}-3 x^{2}+x+1=0 \\
\Rightarrow(x-1)\left(x^{2}-2 x-1\right)=0 \\
\Right... | 1, \pm 2, 1 \pm \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,769 |
8. Line $l$ :
$$
m(2 x-y-5)+(3 x-8 y-14)=0
$$
is intercepted by circle $\odot A$ with center $A(1,0)$ and radius 2, the length of the shortest chord is ( ).
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $2 \sqrt{2}$
(D) $2 \sqrt{3}$ | 8. C.
Given that the line $l$ passes through the fixed point $B(2,-1)$, and $A B=\sqrt{2}$ $<2$, we know that point $B$ is inside $\odot A$.
When the line $l$ is perpendicular to $A B$, the chord intercepted by the line $l$ on $\odot A$ is the shortest, and the length of the shortest chord is $2 \sqrt{2}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,770 |
9. Given that $F(x)$ represents a fifth-degree polynomial in $x$. If
$$
\begin{array}{l}
F(-2)=F(-1)=F(0)=F(1)=0, \\
F(2)=24, F(3)=360,
\end{array}
$$
then the value of $F(4)$ is (.
(A) 1800
(B) 2011
(C) 4020
(D) cannot be determined | 9. A.
Let $F(x)=x(x+2)(x+1)(x-1)(a x+b)$.
Then $\left\{\begin{array}{l}24(2 a+b)=24, \\ 120(3 a+b)=360 .\end{array}\right.$
Solving, we get $a=2, b=-3$.
Thus, $F(4)=1800$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,771 |
10. Given that $x$, $y$, and $z$ are all real numbers greater than 0 and less than 1. Then the value of $x(1-y)+y(1-z)+z(1-x)$ is ( ).
(A) greater than 1
(B) equal to 1
(C) less than 1
(D) greater than or equal to 1 | 10. C.
Given an equilateral triangle $\triangle A B C$ with side length 1 (as shown in Figure 7), $D, E, F$ are points on sides $A C, A B, B C$ respectively.
$$
\begin{array}{l}
\text { Let } B F=x, A E=y, \\
C D=z .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } S_{\triangle B E F}=\frac{\sqrt{3}}{4} x(1-y), \\
S_... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,772 |
$$
\begin{array}{l}
\text { 11. If } a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5 \text {, }
\end{array}
$$
then $a+b+c=$ . $\qquad$ | Ni, 11.20.
Notice,
$$
(\sqrt{a-1}-1)^{2}+(\sqrt{b-2}-2)^{2}+\frac{1}{2}(\sqrt{c-3}-3)^{2}=0 \text {. }
$$
Therefore, $a=2, b=6, c=12$.
So $a+b+c=20$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,773 |
12. Given real numbers $a, b$ satisfy
$$
6^{a}=2010,335^{b}=2010 \text {. }
$$
Then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$ | 12. 1 .
Notice, $6^{a b}=2010^{b}, 335^{a b}=2010^{a}$.
Then $(6 \times 335)^{a b}=2010^{a+b}$.
Thus $a b=a+b \Rightarrow \frac{1}{a}+\frac{1}{b}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,774 |
14. As shown in Figure 4, point $A$ is on $B G$, quadrilaterals $A B C D$ and $D E F G$ are both squares, with areas of $7 \mathrm{~cm}^{2}$ and $11 \mathrm{~cm}^{2}$, respectively. Then the area of $\triangle C D E$ is $\qquad$ | 14. $\sqrt{7} \mathrm{~cm}^{2}$.
Draw $E H \perp C D$, intersecting the extension of $C D$ at point $H$.
Since $\angle E D H=\angle G D A, D E=D G$, therefore,
Rt $\triangle E D H \cong$ Rt $\triangle G D A$.
Thus, $E H=A G, D H=A D$.
From the areas of square $A B C D$ and square $D E F G$ being $7$ and $11$ respectiv... | \sqrt{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,776 |
15. As shown in Figure 5, given that quadrilateral $ABCD$ is a parallelogram, $AC$ intersects $BD$ at point $O$, the angle bisector of $\angle ABC$ intersects $AD$ at point $E$ and the extension of $CD$ at point $F$, $\odot O'$ is the circumcircle of $\triangle DEF$, and $G$ is a point on $\odot O'$ such that $AG = CD$... | Three, 15. As shown in Figure 8, connect $O^{\prime} G$, $O^{\prime} A$, $O^{\prime} C$, $O^{\prime} D$, and $O^{\prime} E$.
Since $B E$ bisects $\angle A B C$, and quadrilateral $A B C D$ is a parallelogram, we have
$$
\begin{array}{l}
\angle A B E=\angle E B C=\angle A E B \\
=\angle F E D=\angle E F D .
\end{array}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,777 |
16. As shown in Figure 6, given that $B D$ and $C E$ are the angle bisectors of $\triangle A B C$, $F$ is the midpoint of side $D E$, and the distances from $F$ to $A C$, $A B$, and $B C$ are $F G = a$, $F H = b$, and $F M = c$ respectively. If $c^{2}-c-2 a b+\frac{1}{2} m^{2}-2 m+\frac{5}{2}=0$,
(1) Find the values of... | 16. (1) As shown in Figure 9, draw $E Q \perp A C$ at point $Q$, $E N \perp B C$ at point $N$, and $D K \perp B C$ at point $K$.
Since $C E$ is the angle bisector of $\angle A C B$, we have
$$
E Q=E N .
$$
In $\triangle D E Q$, because $F$ is the midpoint of $D E$ and $F G \parallel E Q$, we have
$$
E N=E Q=2 F G=2 a... | a = \frac{1}{2}, b = \frac{1}{2}, c = 1, m = 2, D G = \frac{1}{4}(B C - C D) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,778 |
1. Let the function $f(x)$ be an even function, and $f(-3)=$ -2. Then the value of $f^{2}(3)-5 f(3)+2$ is ( ).
(A) -12
(B) 16
(C) 17
(D) 8 | - 1. B.
From $f(-x)=f(x)$, we get $f(3)=f(-3)=-2$. Therefore, $f^{2}(3)-5 f(3)+2=16$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,779 |
Example 6 Solve the equation:
$$
x=\left(x^{2}+3 x-2\right)^{2}+3\left(x^{2}+3 x-2\right)-2 \text{. }
$$ | Let $y=x^{2}+3 x-2$.
Then the original equation becomes
$$
x=y^{2}+3 y-2 \text {. }
$$
(1) - (2) gives $(x-y)(x+y+4)=0$.
Therefore, $x-y=0$ or $x+y+4=0$.
From $x-y=0 \Rightarrow y=x$.
Substituting into (1) yields
$$
\begin{array}{l}
x_{1}=-1+\sqrt{3}, x_{2}=-1-\sqrt{3} . \\
\text { From } x+y+4=0 \Rightarrow y=-x-4 .
... | -1 \pm \sqrt{3}, -2 \pm \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,780 |
2. If the graph of the function $y=x^{2}+a x+a$ given in Figure 1 intersects the $x$-axis at only one point, then $a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4 | 2. D.
From the function $y=x^{2}+a x+a$ having only one zero, we know
$$
\begin{array}{l}
\Delta=a^{2}-4 \times 1 \times a=0 \\
\Rightarrow a(a-4)=0 .
\end{array}
$$
Thus, $a=0$ or 4.
But from the axis of symmetry $-\frac{a}{2} \neq 0$, we get $a \neq 0$.
Therefore, $a=4$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,781 |
3. The function
$$
f(x)=\log _{\frac{1}{16}} x-\left(\frac{1}{16}\right)^{x}
$$
has the number of zeros ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 3. D.
The required is the number of intersection points of the graph of the function $y=\log _{16} x$ and the graph of the function $y=\left(\frac{1}{16}\right)^{x}$ for $x>0$.
It is easy to know that the graph of the function $y=\log _{\frac{1}{16}} x$ and the graph of the function $y=\left(\frac{1}{16}\right)^{x}$ ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,782 |
4. The function $f$ defined on the set of real numbers $\mathbf{R}$ satisfies, for every $x \in \mathbf{R}$ and constant $a>0$,
$$
f(x+a)=\frac{1}{2}+\sqrt{f(x)-f^{2}(x)} .
$$
If the range of the function $f$ is denoted by $M$, then ().
(A) $\frac{\pi}{7} \in M$
(B) $\frac{\sqrt{2}}{3} \in M$
(C) $\frac{\sqrt{2}}{2} \... | 4. C.
Notice,
$$
f(x+a)=\frac{1}{2}+\sqrt{f(x)-f^{2}(x)} \geqslant \frac{1}{2},
$$
i.e., for any $x \in \mathbf{R}$, $f(x) \geqslant \frac{1}{2}$.
$$
\begin{array}{l}
\text { By } f(x)-f^{2}(x) \geqslant 0 \\
\Rightarrow f(x)(1-f(x)) \geqslant 0 .
\end{array}
$$
$$
\text { By } f(x) \geq \frac{1}{2}>0
$$
$$
\begin{ar... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,783 |
5. Let $P$ be a point inside square $A B C D$, with $P A=1$, $P B=2, P C=3$. Then the area of $\triangle P B C$ is ( ).
(A) $2+\frac{\sqrt{2}}{2}$
(B) $2-\frac{\sqrt{2}}{2}$
(C) $2+\sqrt{2}$
(D) $2-\sqrt{2}$ | 5. A.
As shown in Figure 3, $\triangle A P B$ is rotated $90^{\circ}$ clockwise around point $B$ to get $\triangle C Q B$.
Connect $P Q$.
Since $\angle P B Q=90^{\circ}$,
$$
P B=Q B=2 \text{, }
$$
then $\angle P Q B$
$$
\begin{array}{l}
=\angle Q P B=45^{\circ}, \\
P Q=2 \sqrt{2} .
\end{array}
$$
In $\triangle P Q C... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,784 |
6. The function $f(x)$ is an odd function on $\mathbf{R}$, and $g(x)$ is a periodic function on $\mathbf{R}$ with a period of 4. It is known that
$$
f(-2)=g(-2)=6 \text {, }
$$
and $\frac{f(f(2)+g(2))+g(f(-2)+g(-2))}{g^{2}(20 f(2))}=\frac{1}{2}$.
Then the value of $g(0)$ is ( ).
(A) 2
(B) 1
(C) 0
(D) -1 | 6. A.
From $f(x)$ being an odd function, we get
$$
f(2)=-f(-2)=-6, f(0)=0 .
$$
From $g(x)$ being a periodic function with a period of 4, we get
$$
\begin{array}{l}
g(2)=g(-2)=6 . \\
\text { Also, } g(20 f(2))=g(-120) \\
=g(4 \times(-30))=g(0),
\end{array}
$$
Then, from the given equation, we have
$$
\frac{f(-6+6)+g(... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,785 |
1. Calculate $\tan 22.5^{\circ}=$ | 2. $\sqrt{2}-1$.
As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
\angle B A C=45^{\circ}, \\
\angle C=90^{\circ}, \\
A C=B C=1, \\
A B=\sqrt{2} .
\end{array}
$$
Construct the angle bisector $A D$ of $\angle B A C$. Then
$$
\begin{array}{c}
\angle D A C=22.5^{\circ} . \\
\text { In right } \triangle A ... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,786 |
2. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
2 f\left(x^{2}+x\right)+f\left(x^{2}-3 x+2\right) \\
=9 x^{2}-3 x-6 .
\end{array}
$$
Then the value of $f(60)$ is . $\qquad$ | 2. 176.
Substituting $1-x$ for $x$ in equation (1) yields
$$
\begin{array}{l}
2 f\left(x^{2}-3 x+2\right)+f\left(x^{2}+x\right) \\
=9 x^{2}-15 x .
\end{array}
$$
Combining equations (1) and (2) gives
$$
\begin{array}{l}
f\left(x^{2}+x\right)=3 x^{2}+3 x-4 \\
=3\left(x^{2}+x\right)-4 .
\end{array}
$$
Therefore, $f(60... | 176 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,787 |
3. $[x]$ represents the greatest integer not exceeding the real number $x$. Then, in the Cartesian coordinate system $x O y$, the area of the figure formed by all points $(x, y)$ that satisfy $[x][y]=$ 2011 is $\qquad$ | 3.4.
Let $[x]=a,[y]=b$, meaning all such points $(x, y)$ form the region
$$
a \leqslant x<a+1, b \leqslant y<b+1
$$
bounded by these inequalities, with an area of 1.
Since 2011 is a prime number, the points $(x, y)$ that satisfy
$$
[x][y]=2011
$$
form 4 regions, each with an area of 1, making the total area 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,788 |
4. As shown in Figure 2, the radii of two concentric circles are $6$ and $10$, respectively. The sides $AB$ and $CD$ of rectangle $ABCD$ are chords of the two circles. When the area of the rectangle is maximized, its perimeter is $\qquad$ | 4. $\frac{128 \sqrt{34}}{17}$.
Let the centers of these two circles be $O$. Then the area of $\triangle O A D$ is $\frac{1}{4}$ of the area of rectangle $A B C D$. Therefore, when the area of $\triangle O A D$ is maximized, the area of rectangle $A B C D$ also reaches its maximum value.
And $O A, O D$ are constants, ... | \frac{128 \sqrt{34}}{17} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,789 |
5. Given $f(x)=\frac{1}{1+x^{2}}$. Then
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(2011)+ \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{2011}\right) \\
=
\end{array}
$$ | 5.2 010.5.
Notice that,
$$
\begin{array}{l}
f(x)+f\left(\frac{1}{x}\right)=\frac{1}{1+x^{2}}+\frac{1}{1+\frac{1}{x^{2}}} \\
=\frac{1}{1+x^{2}}+\frac{x^{2}}{1+x^{2}}=1 .
\end{array}
$$
Then the original expression is
$$
\begin{array}{l}
=f(1)+\left[f(2)+f\left(\frac{1}{2}\right)\right]+\left[f(3)+f\left(\frac{1}{3}\ri... | 2010.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,790 |
Example 7 The equation in terms of $x$
$$
x^{3}-a x^{2}-2 a x+a^{2}-1=0
$$
has only one real root. Then the range of values for $a$ is | Solve for $a$ as the main variable, to get the equation in terms of $a$
$$
\begin{array}{l}
a^{2}-\left(x^{2}+2 x\right) a+x^{3}-1=0 \\
\Rightarrow[a-(x-1)]\left[a-\left(x^{2}+x+1\right)\right]=0 \\
\Rightarrow x=a+1 \text { or } x^{2}+x+(1-a)=0 .
\end{array}
$$
Since the original equation has only one real root, we h... | a<\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,791 |
6. Given that the lengths of the two legs of a right triangle are the two roots of the quadratic equation $a x^{2}+b x+c=0$. Then the area of the circumcircle of the right triangle is $\qquad$ (the result should be expressed using $a, b, c$ and $\pi$). | 6. $\frac{\pi\left(b^{2}-2 a c\right)}{4 a^{2}}$.
Let the lengths of the two legs of the right triangle be $x_{1}$ and $x_{2}$. Then
$$
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} .
$$
The hypotenuse of the right triangle is
$$
\begin{array}{l}
\sqrt{x_{1}^{2}+x_{2}^{2}}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-2 x_{... | \frac{\pi\left(b^{2}-2 a c\right)}{4 a^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,792 |
7. If the quadratic function $f(x)=a x^{2}-2 x-a$ satisfies
$$
f(2)<f(1)<f(3)<f(0) \text {, }
$$
then the range of the real number $a$ is $\qquad$. | 7. $\left(\frac{1}{2}, \frac{2}{3}\right)$.
From $f(2)<f(1)<f(3)$, we know that the quadratic function opens upwards, and the axis of symmetry is between $[1,2]$.
From $f(2)<f(1)$, we know that the axis of symmetry is in $\left(\frac{3}{2}, 2\right)$, i.e., the x-coordinate of the vertex $\frac{1}{a} \in\left(\frac{3... | \left(\frac{1}{2}, \frac{2}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,793 |
8. Let $D$ be a point inside $\triangle A B C$ such that $\angle B A D=$ $\angle B C D$, and $\angle B D C=90^{\circ}$. Given $A B=5, B C=$ $6, M$ is the midpoint of $A C$. Then $D M=$ $\qquad$ | 8. $\frac{\sqrt{11}}{2}$.
As shown in Figure 5, extend $CD$ to point $E$ such that $DE = DC$, and connect $BE$.
Then, $\triangle BDE \cong \triangle BDC$.
Therefore, $BE = BC = 6$, $\angle BED = \angle BCD = \angle BAD$.
Thus, points $A$, $D$, $B$, and $E$ are concyclic.
Hence, $\angle EAB = \angle EDB = 90^{\circ}$.
... | \frac{\sqrt{11}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,794 |
1. Given sets
$$
\begin{array}{l}
M=\{x \mid(x-1)(x-3)(x-5)<0, x \in \mathbf{R}\}, \\
N=\{x \mid(x-2)(x-4)(x-6)<0, x \in \mathbf{R}\} .
\end{array}
$$
Then $M \cap N=(\quad)$.
(A) $(2,3)$
(B) $(3,4)$
(C) $(4,5)$
(D) $(5,6)$ | -、1. B.
Notice,
$$
\begin{array}{l}
M=(-\infty, 1) \cup(3,5), \\
N=(2,4) \cup(6,+\infty) .
\end{array}
$$
Therefore, $M \cap N=(3,4)$. | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 724,795 |
2. Given $z=(\sqrt{3}-3 \mathrm{i})^{n}$. If $z$ is a real number, then the smallest positive integer $n$ is ().
(A) 3
(B) 4
(C) 5
(D) 6 | 2. A.
Notice,
$$
\begin{array}{l}
z=(\sqrt{3}-3 \mathrm{i})^{n} \\
=(-2 \sqrt{3})^{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{n} .
\end{array}
$$
Therefore, $n=3$ is the smallest positive integer that makes $z$ a real number. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,796 |
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