problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
5. Let $a_{n}=(2+\sqrt{7})^{2 n+1}$, $b_{n}$ be the fractional part of $a_{n}$. Then when $n \in \mathbf{N}_{+}$, the value of $a_{n} b_{n}$ ( ).
(A) must be an irrational number
(B) must be an even number
(C) must be an odd number
(D) can be either an irrational number or a rational number | 5. C.
Let $u=2+\sqrt{7}, v=2-\sqrt{7}$. Then $u+v=4, u v=-3$,
knowing that $u, v$ are the two roots of the equation $x^{2}=4 x+3$.
Therefore, $u^{2}=4 u+3, v^{2}=4 v+3$.
Thus, when $n \geqslant 2$,
$$
u^{n}=4 u^{n-1}+3 u^{n-2}, v^{n}=4 v^{n-1}+3 v^{n-2} \text {. }
$$
Let $S_{n}=u^{n}+v^{n}$. Then when $n \geqslant 2$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,430 |
6. Let $n$ be a positive integer, and $3 n+1$ and $5 n-1$ are both perfect squares. For the following two propositions:
甲: $7 n+13$ must be a composite number;
乙: $8\left(17 n^{2}+3 n\right)$ must be the sum of two squares.
Your judgment is ( ).
(A) 甲 is correct, 乙 is incorrect
(B) 甲 is incorrect, 乙 is correct
(C) Both... | 6. C.
Let $3 n+1=a^{2}, 5 n-1=b^{2}\left(a, b \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
7 n+13=9(3 n+1)-4(5 n-1) \\
=(3 a)^{2}-(2 b)^{2} \\
=(3 a-2 b)(3 a+2 b) .
\end{array}
$$
Therefore, $3 a-2 b$ is a positive integer.
If $3 a-2 b=1$, then
$$
27 n+9=(3 a)^{2}=(2 b+1)^{2}=4 b^{2}+4 b+1 \text {, }
$$
whic... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,431 |
7. Draw a line $l$ through the point $P(1,1)$ such that the midpoint of the chord intercepted by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ is exactly $P$. Then the equation of the line $l$ is $\qquad$ . | $=, 7.4 x+9 y=13$.
Let the equation of line $l$ be $y=k(x-1)+1$.
Substituting into the ellipse equation and rearranging, we get
$$
\left(9 k^{2}+4\right) x^{2}+18 k(1-k) x+9 k^{2}-18 k-27=0 \text {. }
$$
Let the two roots be $x_{1} 、 x_{2}$. Then $\frac{x_{1}+x_{2}}{2}=1$, i.e.,
$$
-\frac{18 k(1-k)}{9 k^{2}+4}=2 \Righ... | 4 x+9 y=13 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,432 |
8. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=\sqrt{x^{2}+1}+\sqrt{(x-12)^{2}+16}
$$
has a minimum value of $\qquad$ . | 8. 13 .
As shown in Figure 1, take $A$ as the origin of the number line, $A B=$ 12, and then construct the perpendicular lines $A C$ and $B D$ such that
$$
\begin{array}{l}
A C=1, \\
B D=4 .
\end{array}
$$
On the number line, take point $P$ such that $A P=x$. Then
$$
f(x)=|C P|+|D P| \text {. }
$$
When points $C$, $... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,433 |
9. In tetrahedron $ABCD$, face $ABC$ and face $BCD$ form a dihedral angle of $60^{\circ}$, the projection of vertex $A$ onto face $BCD$ is the orthocenter $H$ of $\triangle BCD$, and $G$ is the centroid of $\triangle ABC$. If $AH=4$ and $AB=AC$, then $GH=$ $\qquad$ | 9. $\frac{4 \sqrt{21}}{9}$.
Let the plane $A H D$ intersect $B C$ at point $F$. Given $A B=A C$, we know point $G$ lies on $A F$, and
$$
G F=\frac{1}{3} A F, \angle A F H=60^{\circ} \text {. }
$$
Then $A F=\frac{A H}{\sin 60^{\circ}}=\frac{8}{\sqrt{3}}, F H=\frac{1}{2} A F=\frac{4}{\sqrt{3}}$,
$$
G F=\frac{8}{3 \sqrt... | \frac{4 \sqrt{21}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,434 |
10. $\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ}=$ | $\begin{array}{l}\text { 10. } \frac{\sqrt{3}}{8} \\ 8 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\ =4\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\ =4 \sin 80^{\circ} \cdot \cos 20^{\circ}-2 \sin 80^{\circ} \\ =2\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-2 \sin 80^{\circ} \\ =2 \s... | \frac{\sqrt{3}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,435 |
Example 3 As shown in Figure 7, in square $A B C D$, it is known that $E$ and $F$ are points on sides $B C$ and $C D$ respectively, satisfying $E F=B E+D F$, and $A E$ and $A F$ intersect the diagonal $B D$ at points $M$ and $N$ respectively. Prove:
(1) $\angle E A F=45^{\circ}$;
(2) $M N^{2}=B M^{2}+D N^{2}$.
(2007, S... | Proof: (1) As shown in Figure 7, extend $CD$ to point $E_1$ such that $DE_1 = BE$, and connect $AE_1$. Then $\triangle ADE_1 \cong \triangle ABE$.
Thus, $\angle DAE_1 = \angle BAE$, and $AE_1 = AE$.
Therefore, $\angle EAE_1 = 90^\circ$.
In $\triangle AEF$ and $\triangle AE_1F$,
$EF = BE + DF = E_1D + DF = E_1F$, then
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,436 |
11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for each $n \in \mathbf{N}_{+}, a_{n} 、 a_{n+1}$ are the two roots of the equation $x^{2}+3 n x+b_{n}=0$. Then $\sum_{k=1}^{20} b_{k}=$ $\qquad$ | 11. 6385.
For each $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=-3 n, \\
a_{n} a_{n+1}=b_{n} .
\end{array}
$$
Rewrite equation (1) as
$$
a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right) \text {. }
$$
Therefore, $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is... | 6385 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,437 |
12. From the set $M=\{1,2, \cdots, 2008\}$ consisting of the first 2008 positive integers, take a $k$-element subset $A$, such that the sum of any two numbers in $A$ cannot be divisible by their difference. Then the maximum value of $k$ is $\qquad$ | 12.670.
First, take 670 yuan set $A=\{1,4,7, \cdots, 2008\}$, where the sum of any two numbers in $A$ cannot be divisible by 3, but their difference is a multiple of 3.
Then, divide the numbers in $M$ from smallest to largest into 670 segments, each containing three numbers:
$$
1,2,3 ; 4,5,6 ; \cdots ; 2005,2006,2007... | 670 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,438 |
13. Given that $A D$ is the altitude on the hypotenuse $B C$ of the right triangle $\triangle A B C$ $(A B<A C)$, $I_{1}$ and $I_{2}$ are the incenters of $\triangle A B D$ and $\triangle A C D$ respectively. The circumcircle of $\triangle A I_{1} I_{2}$, denoted as $\odot O$, intersects $A B$ and $A C$ at points $E$ a... | Three, 13. As shown in Figure 2, connect $O I_{1}$, $O I_{2}$, $I_{1} D$, and $I_{2} D$.
Since $\angle B A C=90^{\circ}$, the circumcenter $O$ of $\triangle A I_{1} I_{2}$ lies on $E F$.
Given that $I_{1}$ and $I_{2}$ are the incenter, $\angle I_{1} A I_{2}=45^{\circ}$. Therefore,
$\angle I_{1} O I_{2}=2 \angle I_{1} A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,439 |
14. Let $x, y, z$ be non-negative real numbers, satisfying $xy + yz + zx = 1$. Prove:
$$
\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geqslant \frac{5}{2} .
$$ | 14. From $x y+y z+z x=1$, we know that at most one of $x, y, z$ is 0.
By symmetry, assume $x \geqslant y \geqslant z \geqslant 0$. Then $x>0, y>0, z \geqslant 0, x y \leqslant 1$.
(1) When $x=y$, the condition becomes
$x^{2}+2 x z=1 \Rightarrow z=\frac{1-x^{2}}{2 x}, x^{2} \leqslant 1$.
And $\frac{1}{x+y}+\frac{1}{y+z}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,440 |
15. For the $2n$-element set $M=\{1,2, \cdots, 2n\}$, if the $n$-element sets
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}
$$
satisfy $A \cup B=M, A \cap B=\varnothing$, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, then $A \cup B$ is called an "equal-sum partition" ... | 15. Solution 1: Without loss of generality, let $12 \in A$. Since when set $A$ is determined, set $B$ is uniquely determined, we only need to consider the number of sets $A$.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}, a_{6}$ be the largest number.
By $1+2+\cdots+12=78$, we know
$a_{1}+a_{2}+\cdots+a_{6}=39, a_{... | 29 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,441 |
2. If $a, b, c$ represent three distinct digits among $0,1, \cdots, 9$, and satisfy the equation
$$
\underbrace{a a \cdots a b b \cdots b}_{n \uparrow}+1=(\underbrace{c c \cdots c}_{n \uparrow}+1)^{2} \text {, }
$$
then the number of valid triples $(a, b, c)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 721,443 |
3. Let distinct non-zero real numbers $a, b, c, x, y, z$ satisfy the relation
$$
\frac{y z}{b z+c y}=\frac{z x}{c x+a z}=\frac{x y}{a y+b x}=\frac{x^{2}+y^{2}+z^{2}}{a^{2}+b^{2}+c^{2}} \text {. }
$$
Then the algebraic expression $\frac{a+b+c}{x+y+z}=(\quad$.
(A) $\frac{1}{2}$
(B) $\frac{1}{4}$
(C) 2
(D) 4 | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,444 |
4. Given an isosceles $\triangle A B C$ with the vertex angle $\angle A=72^{\circ}$, the altitude from $A$ to side $B C$ is of length $h$, the longer of the four equal angle bisectors of $\angle A$ is $m$, and the shorter of the three equal angle trisectors of the exterior angle of $\angle A$ is $n$. Then ( ).
(A) $\fr... | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,445 |
Example 4 As shown in Figure 8, given that the side length of square $ABCD$ is 1, points $M$ and $N$ are on $BC$ and $CD$ respectively, such that the perimeter of $\triangle CMN$ is 2. Find:
(1) the size of $\angle MAN$;
(2) the minimum value of the area of $\triangle MAN$.
(2003, (Yu Zhen Cup) Shanghai Junior High Sch... | Solution: (1) As shown in Figure 8, extend $CB$ to point $L$ such that $BL = DN$. Then, Rt $\triangle ABL \cong \text{Rt} \triangle ADN$.
Thus, $AL = AN$, $\angle LAB = \angle NAD$,
$\angle NAL = \angle DAB = 90^\circ$.
Also, $MN = 2 - CN - CM$
$= 1 - CN + 1 - CM = DN + BM$
$= BL + BM = ML$,
and $AM$ is a common side, ... | 45^\circ, \sqrt{2} - 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,447 |
6. Given that $O$, $G$, and $H$ are the circumcenter, centroid, and orthocenter of $\triangle ABC$ respectively, $BC=a$, $CA=b$, $AB=c$, and the radius of the circumcircle of $\triangle ABC$ is $R$. Then among the following conclusions:
(1) $O G^{2}=R^{2}-\frac{1}{9}\left(a^{2}+b^{2}+c^{2}\right)$,
(2) $O G^{2}=\frac{1... | 6. A | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,448 |
1. Given positive real numbers $x, y, z, w$ satisfy $2007 x^{2}=2008 y^{2}=2009 z^{2}=2010 w^{2}$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}=1$.
Then $\sqrt{2007 x+2008 y+2009 z+2010 w}$ $=$ . $\qquad$ | $$
\text { II. 1. } \sqrt{2007}+\sqrt{2008}+\sqrt{2009}+\sqrt{2010} \text {. }
$$
According to the problem, let
$$
\begin{array}{l}
2007 x^{2}=2008 y^{2}=2009 z^{2}=2010 w^{2}=k, \\
x>0, y>0, z>0, w>0 . \\
\text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w} \\
=\frac{1}{\sqrt{k}}(\sqrt{2007}+\sqrt{2008}+\sq... | \sqrt{2007}+\sqrt{2008}+\sqrt{2009}+\sqrt{2010} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,449 |
2. As shown in Figure 1, given that $P A$ and $P B$ are tangent to $\odot O$ at points $A$ and $B$, $P C$ satisfies $A B \cdot P B - A C \cdot P C = A B \cdot P C - A C \cdot P B$, and $A P \perp P C$, $\angle P A B = 2 \angle B P C$. Then $\angle A C B =$ $\qquad$ | 2. $30^{\circ}$.
As shown in Figure 3.
$$
\begin{array}{l}
\text { From } A B \cdot P B- \\
A C \cdot P C \\
= A D \cdot P C- \\
A C \cdot P B,
\end{array}
$$
we get
$$
\begin{array}{l}
(A B+A C) \cdot \\
(P B-P C)=0 .
\end{array}
$$
Thus, $P B=P C$.
Also, $P A=P B$, so
$$
P A=P B=P C \text {. }
$$
Therefore, poin... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,450 |
3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$
Then $x_{1}+x_{2}+x_{3}=$ | 3. 8 .
Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\l... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,451 |
4. Given that $a, b, c$ are the lengths of the three sides of $\triangle ABC$, and satisfy the conditions
$$
\frac{2 a^{2}}{1+a^{2}}=b, \frac{2 b^{2}}{1+b^{2}}=c, \frac{2 c^{2}}{1+c^{2}}=a \text {. }
$$
Then the area of $\triangle ABC$ is . $\qquad$ | 4. $\frac{\sqrt{3}}{4}$.
From the problem, we have
$$
\begin{array}{l}
\frac{2 a^{2}}{1+a^{2}}=b \Rightarrow 1+\frac{1}{a^{2}}=\frac{2}{b}, \\
\frac{2 b^{2}}{1+b^{2}}=c \Rightarrow 1+\frac{1}{b^{2}}=\frac{2}{c}, \\
\frac{2 c^{2}}{1+c^{2}}=a \Rightarrow 1+\frac{1}{c^{2}}=\frac{2}{a} .
\end{array}
$$
(1) + (2) + (3) giv... | \frac{\sqrt{3}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,452 |
(20 points) (1) If the quadratic function $y=a x^{2}-$ $b x+c$, when $x$ takes the values $0, 1, 2$, $y$ is an integer, prove: for any integer $x$, the value of $y$ is an integer;
(2) If for any integer $x$, the value of the quadratic function $y=a x^{2}-$ $b x+c$ is an integer, is it necessary that $a, b, c$ are integ... | (1) From the problem, we know that $c$, $a-b+c$, and $4a-2b+c$ are all integers. Therefore,
$$
a-b=(a-b+c)-c
$$
and $4a-2b=(4a-2b+c)-c$
are both integers. Furthermore,
$$
2a=(4a-2b)-2(a-b)
$$
and $2b=(4a-2b)-4(a-b)$
are both integers.
Thus, when $x$ is even (let's say $x=2k$), $y=4ak^2-2bk+c$ is an integer;
when $x$ ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,453 |
II. (25 points) As shown in Figure 2, through the vertices $B$ and $C$ of $\triangle ABC$, draw the tangents $BP$ and $CP$ to the circumcircle of $\triangle ABC$, intersecting at point $P$. Connect $AP$ to intersect the side $BC$ and the circumcircle of $\triangle ABC$ at points $D$ and $Q$, respectively. Prove that $\... | II. Since $PB$ and $PC$ are tangents to $\odot O$, we have $\triangle PBQ \sim \triangle PAB$, $\triangle PCQ \sim \triangle PAC$. Therefore, $\frac{BQ}{AB} = \frac{PB}{PA}$, $\frac{CQ}{AC} = \frac{CP}{PA}$.
Since $PB = PC$, it follows that $\frac{BQ}{AB} = \frac{CQ}{AC}$, which means $\frac{AC}{AB} = \frac{CQ}{BQ}$.
G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,454 |
Three. (25 points) Pentagram
The five edges of a pentagram intersect pairwise at ten points. Fill in the numbers $2001, 2002, \cdots, 2010$ at these ten points so that the sum of the four numbers on each edge is equal. Can you do it? If yes, please fill them out; if not, please explain why. | Three, it cannot be done. The reasons are as follows.
When filling the ten numbers at the intersection points and calculating the sum of the numbers on each edge, because each number is added twice, to make the sum of the four numbers on each edge equal, the sum of the numbers on each edge should be
$$
\frac{1}{5}(200... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,455 |
2. Given positive numbers $a, b$, for $x>0, y>0$, and $x^{20 \times 9}$ $+y^{2009}=1$. Then the minimum value of $f(x, y)=\frac{a}{x^{336}}+\frac{b}{y^{336}}$ is . $\qquad$ | 2. $\left(a^{\frac{2009}{2345}}+b^{\frac{2009}{2345}}\right)^{\frac{2345}{2009}}$.
By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(\frac{a}{x^{336}}+\frac{b}{y^{336}}\right)^{\frac{2009}{2345}}\left(x^{2009}+y^{2009}\right)^{\frac{336}{2345}} \\
\geqslant\left(\frac{a}{x^{336}}\right)^{\frac{2009}{2345}}\lef... | \left(a^{\frac{2009}{2345}}+b^{\frac{2009}{2345}}\right)^{\frac{2345}{2009}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,457 |
Example 5 As shown in Figure 9, given a square $A B C D$ divided into four smaller rectangles by two line segments $E F$ and $G H$ parallel to the sides, $P$ is the intersection of $E F$ and $G H$.
If the area of rectangle $P F C H$ is exactly twice the area of rectangle $A G P E$, determine the size of $\angle H A F$ ... | Solution: Guess $\angle F A H=45^{\circ}$. Proof as follows:
Let $A G=a, B G=b, A E=x, E D=y$. Then $a+b=x+y, 2 a x=b y$.
Thus, $a-x=y-b$, which means $a^{2}-2 a x+x^{2}=y^{2}-2 b y+b^{2}=y^{2}-4 a x+b^{2}$. Therefore, $(a+x)^{2}=b^{2}+y^{2}$, which means $a+x=\sqrt{b^{2}+y^{2}}=F H$.
Hence, $D H+B F=F H$.
Extend $C B$... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,458 |
3. Let the tetrahedron $ABCD$ be a regular tetrahedron, and $E$ be the midpoint of edge $BC$. Then the distance $d$ between the skew lines $AE$ and $CD$ is $\qquad$ . | 3. $\frac{4 \sqrt{22}}{11}$.
As shown in Figure 2, let $F$ be the midpoint of side $BD$, and connect $AF$, $EF$, and $CF$. Then $d$ is the distance from point $C$ to the plane $AEF$.
Given that the edge length of the regular tetrahedron $a=4$, its volume is
$$
V=\frac{\sqrt{2}}{12} a^{3}=\frac{16 \sqrt{2}}{3} .
$$
T... | \frac{4 \sqrt{22}}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,459 |
4. Let $a, b$ be real numbers, and satisfy
$$
\begin{array}{l}
\tan (2009 a+2008)+\arcsin (2009 a+2008) \\
=-2007, \\
\tan (2009 b+2008)+\arcsin (2009 b+2008) \\
=2007 .
\end{array}
$$
Then $a+b=$ . $\qquad$ | 4. $-\frac{4016}{2009}$.
Let $f(x)=\tan x+\arcsin x$. Then $f(x)$ is an odd and monotonically increasing function on $[-1,1]$.
From the condition, we have $-1 \leqslant 2009 a+2008 \leqslant 1$, which implies $-1 \leqslant a \leqslant-\frac{2007}{2009}$.
Similarly, $-1 \leqslant b \leqslant-\frac{2007}{2009}$.
Thus, $... | -\frac{4016}{2009} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,460 |
5. Given that $f(x)$ is a monotonically increasing function on $\mathbf{R}$, and for any $x \in \mathbf{R}$, $f(f(f(x)))=x$. Then $f(2009)=$ $\qquad$ | 5.2 009.
If $f(2009)2009$, then we have
$$
f(f(f(2009)))>2009,
$$
contradiction. | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,461 |
6. $A, B$ two people are playing a table tennis match, with the rule that the first to win 3 games more than the opponent wins. After 13 games, $A$ finally wins with a record of 8 wins and 5 losses. Then the number of all different possible outcomes of these 13 games is $\qquad$ | 6. 243.
As shown in Figure 3, use the grid point $(m, n)$ to represent the moment when $m+n$ games have been played (with $A$ winning $m$ games and $B$ winning $n$ games). The problem then transforms into finding the number of grid paths from $(0,0)$ to $(8,5)$, where the points $(x, y)$ passed through satisfy $|x-y|<... | 243 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,462 |
7. Let the side length of the equilateral triangle $\triangle A B C$ be $1, t$ be any real number. Then the minimum value of $|\overrightarrow{A B}+t \overrightarrow{A C}|$ is $\qquad$ | 7. $\frac{\sqrt{3}}{2}$.
Let $a=\overrightarrow{A B}, b=\overrightarrow{A C}$. Then $|a|=|b|=1$, and the angle between $a$ and $b$ is $60^{\circ}$. Therefore,
$$
\begin{array}{l}
|\overrightarrow{A B}+t \overrightarrow{A C}|^{2}=|a+t b|^{2} \\
=a^{2}+t^{2} b^{2}+2 t a \cdot b=1+t^{2}+2 t \cos 60^{\circ} \\
=1+t^{2}+t=... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,463 |
8. Two people agree: to go to place $A$ together on a certain day, meeting at place $B$ between 7:00 AM and 8:00 AM. The one who arrives at $B$ first will wait for a maximum of $5 \mathrm{~min}$, and if the other does not show up, they will go ahead. Assume the times at which the two people arrive at $B$ are random, in... | 8. $\frac{23}{144}$.
Let the times when the two people arrive at location $A$ be 7:00 plus $m$ minutes and 7:00 plus $n$ minutes $(0 \leqslant m, n \leqslant 60)$. Represent the times when the two people arrive at location $A$ with the pair $(m, n)$. In a Cartesian coordinate system, the domain of point $(m, n)$ is a ... | \frac{23}{144} | Other | math-word-problem | Yes | Yes | cn_contest | false | 721,464 |
9. (14 points) Let there exist $n$ integers $a_{1}, a_{2}, \cdots, a_{n}$, such that $\prod_{i=1}^{n} a_{i}=n$, and $\sum_{i=1}^{n} a_{i}=0$. Find all possible values of the positive integer $n$.
| 9. First prove $4 \mid n$.
If $n$ is odd, then from $\prod_{i=1}^{n} a_{i}=n$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers.
Thus, $\sum_{i=1}^{n} a_{i}$ is the sum of an odd number of odd numbers, which cannot be 0, contradicting $\sum_{i=1}^{n} a_{i}=0$.
Therefore, $n$ is even.
From $\sum_{i=1}^{n}... | n=4k\left(k \in \mathbf{N}_{+}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,465 |
10. (15 points) As shown in Figure 1, in the pyramid $P-ABCD$, the base $ABCD$ is a rhombus, $\frac{AB}{AC}=\frac{2}{3}$, $PA \perp$ plane $ABCD$, and $M$ is a point on edge $PC$.
(I) If the angles formed by $PA$, $PB$, $PC$, $PD$ with plane $ABCD$ are $\alpha$, $\beta$, $\gamma$, $\delta$ respectively, and $\alpha=\be... | 10. (1) As shown in Figure 5, since $P A \perp$ plane $A B C D$, we know $\alpha=90^{\circ}$, and $A B, A C, A D$ are the projections of $P B, P C, P D$ in plane $A B C D$, respectively. Therefore,
$$
\angle P B A=\beta, \angle P C A=\gamma, \angle P D A=\delta.
$$
Since quadrilateral $A B C D$ is a rhombus, we have $... | \pi-\arccos \frac{2 \sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,466 |
11. (15 points) Let $A$ and $B$ be the two intersection points of the line $y=x+b$ and the ellipse $3 x^{2}+y^{2}=1$, and $O$ be the origin. If $\triangle O A B$ is an acute triangle, find the range of $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the ... | 11. Substituting $y=x+b$ into $3 x^{2}+y^{2}=1$ yields $3 x^{2}+(x+b)^{2}=1$, which simplifies to $4 x^{2}+2 b x+b^{2}-1=0$.
From $\Delta>0$, we get $-\frac{2 \sqrt{3}}{3}<b<\frac{2 \sqrt{3}}{3}$.
In $\triangle O A B$, as $|O A|>0$, when $b (b>0)$ increases to $b^{\prime}$, points $A$ and $B$ become $A^{\prime}$ and $B... | \left(-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,467 |
One. (50 points) Three circles intersect at a point $O$, and intersect pairwise at points $A, B, C$. A circle $\omega$ with center $O$ intersects the above three circles at points $D, D', E, E', F, F'$, where points $D, D'$ lie on the circle not containing point $A$, and so on. Let the circumcircles of $\triangle AEF, ... | Given a circle $\omega$ as the base circle for inversion, the images of each point after inversion are denoted by the same letters (all letters appearing below are the images of the points after inversion). Then, the inverted shapes of the three circles are three lines $B C$, $C A$, and $A B$ (as shown in Figure 7). We... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,468 |
Example 2 If the function
$$
y=\frac{1}{2}\left(x^{2}-100 x+196+\left|x^{2}-100 x+196\right|\right),
$$
then when the independent variable $x$ takes the natural numbers $1,2, \cdots, 100$, the sum of the function values is ( ).
(A) 540
(B) 390
(C) 194
(D) 97
(1999, National Junior High School Mathematics Competition) | Solution: Since $x^{2}-100 x+196=(x-2)(x-98)$, therefore, when $2 \leqslant x \leqslant 98$,
$$
\left|x^{2}-100 x+196\right|=-\left(x^{2}-100 x+196\right) \text {. }
$$
Thus, when the independent variable $x$ takes $2,3, \cdots, 98$, the function value is 0; and when $x$ takes $1,99,100$, the sum of the function value... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,470 |
Example 11 Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition) | Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Therefore,
$$
\begin{array}{l}
\Delta=[-(2... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,471 |
2. As shown in Figure 3, in $\triangle A B C$, $B D$ bisects $\angle A B C$, $C D \perp B D$ at point $D$, $\angle B A C=\angle A C D, B C=$ $5, B A=11$. Then the area of $\triangle A B C$ is | 2. $\frac{132}{5}$.
As shown in Figure 8, extend $CD$ to intersect $AB$ at point $E$. It is easy to see that
$\triangle B E D \cong \triangle B C D$.
Therefore, $B E=B C=5$,
$E D=C D$.
Thus, $A E=A B-B E$
$$
=11-5=6 \text {. }
$$
Also, $\angle B A C=\angle A C D$, so $C E=A E=6$.
Therefore, $C D=\frac{1}{2} C E=3$.
B... | \frac{132}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,472 |
3. For any real numbers $x, y$, the inequality
$$
|x-1|+|x-3|+|x-5| \geqslant k(2-|y-9|)
$$
always holds. Then the maximum value of the real number $k$ is $\qquad$ | 3. 2 .
From the geometric meaning of absolute value, we know that the minimum value of $|x-1|+|x-3|+|x-5|$ is 4 (at this time $x=3$), and the maximum value of $2-|y-9|$ is 2 (at this time $y=9$). According to the condition, we get $4 \geqslant k \cdot 2$, so, $k \leqslant 2$. Therefore, the maximum value of $k$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 721,473 |
One, (20 points) Given the quadratic function $y=x^{2}-2 k x+$ $3 k+1$, the independent variable $x$ and real numbers $a, b$ satisfy $4 a^{2}+9 b^{2}=$ $x+6 a b=2, y_{\text {minimum }}=1$: Find the value of the constant $k$. | Given:
\[4 a^{2}-6 a b+9 b^{2}=x\].
Also,
\[4 a^{2}+9 b^{2}=2\],
Subtracting (1) from (2) gives:
\[2 a \cdot 3 b=2-x\].
From equations (2) and (3), we get:
\[
\begin{array}{l}
(2 a+3 b)^{2}=4 a^{2}+9 b^{2}+12 a b \\
=2+2(2-x)=6-2 x .
\end{array}
\]
Thus,
\[2 a+3 b= \pm \sqrt{6-2 x}\].
Therefore, \(2 a\) and \(3 b\) ... | k=-1 \text{ and } k=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,475 |
II. (25 points) As shown in Figure 4, $AB$ is the diameter of $\odot O$, $C$ is a point outside $\odot O$, $CD \perp AB$ at point $D$, intersecting $\odot O$ at point $E$. $EF$ is a non-diameter chord, and
$$
CF = CE. FG \perp AB
$$
at point $G$, $CH \perp EF$ at point $H$, points $D$ and $G$ are on opposite sides of ... | II. As shown in Figure 10, connect $O H, O E, O F$.
Given $C E=C F, C H \perp E F$, we have $E H=F H$.
By the perpendicular chord theorem, $O H \perp E F$.
Also, $F G \perp A B \Rightarrow \angle F H O+\angle F G O=180^{\circ}$
$\Rightarrow O, G, F, H$ are concyclic $\Rightarrow \angle G F O=\angle G H O$.
Similarly, $... | (2-\sqrt{2}):4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,476 |
$$
\begin{array}{l}
\text { Three. (25 points) Given the equation in } x \text { : } \\
x^{2}+\sqrt{a-2009} x+\frac{a-2061}{2}=0
\end{array}
$$
has two integer roots. Find all real values of $a$ that satisfy the condition. | Three, let the two integer roots of the equation be \(x_{1}\) and \(x_{2}\).
By Vieta's formulas, we have \(x_{1} + x_{2} = -\sqrt{a-2009}\) is an integer, and \(x_{1} x_{2} = \frac{a-2061}{2}\) is an integer.
Thus, the original equation is a quadratic equation with integer coefficients, and \(a-2009\) is a perfect sq... | 2013, 2109 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,477 |
1. Choose 5 numbers (repetition allowed) from $\{1,2, \cdots, 100\}$. Then the expected number of composite numbers taken is | $$
-1 . \frac{37}{10} \text {. }
$$
In the set $\{1,2, \cdots, 100\}$, there are 74 composite numbers. Let $\xi$ be the number of composite numbers drawn. Then
$$
P(\xi=i)=\mathrm{C}_{5}^{i}\left(\frac{74}{100}\right)^{i}\left(\frac{26}{100}\right)^{5-i}(0 \leqslant i \leqslant 5) .
$$
Therefore, $\xi$ follows a binom... | \frac{37}{10} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,478 |
3. The maximum value of the area of an inscribed triangle in a circle with radius $R$ is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. $\frac{3 \sqrt{3}}{4} R^{2}$.
Let the inscribed triangle of $\odot O$ be $\triangle A B C$.
Obviously, when $\triangle A B C$ is an acute or right triangle, the area can reach its maximum value (because if $\triangle A B C$ is an obtuse triangle, the side opposite to the obtuse angle (let's assume it is $A$) can be... | \frac{3 \sqrt{3}}{4} R^{2} | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 721,480 |
4. Given $a_{1}=1, a_{n+1}=2 a_{n}+\frac{n^{3}-2 n-2}{n^{2}+n}$. Then the general term formula of $\left\{a_{n}\right\}$ is $\qquad$ . | 4. $a_{n}=2^{n-1}+\frac{1-n^{2}}{n}$.
From the given condition,
$$
a_{n+1}+\frac{n(n+2)}{n+1}=2\left[a_{n}+\frac{(n+1)(n-1)}{n}\right] \text {. }
$$
Let $b_{n}=a_{n}+\frac{(n+1)(n-1)}{n}$. Then $b_{1}=1$.
Therefore, $\left\{b_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2.
Thus, $b_{n}=... | a_{n}=2^{n-1}+\frac{1-n^{2}}{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,481 |
5. $f(\theta)=\frac{1-\sqrt{2} \sin \theta}{\sqrt{7-\left[6 \sin \left(\theta+\frac{\pi}{4}\right)+2 \cos \left(\theta+\frac{\pi}{4}\right)\right]}}$ The range of values is . $\qquad$ | 5. $\left[\frac{\sqrt{3}-\sqrt{8}}{5}, \frac{\sqrt{3}+\sqrt{8}}{5}\right]$.
Notice
$$
\begin{array}{l}
f(\theta)=\frac{1-\sqrt{2} \sin \theta}{\sqrt{7-\left[6 \sin \left(\theta+\frac{\pi}{4}\right)+2 \cos \left(\theta+\frac{\pi}{4}\right)\right]}} \\
=\frac{1-\sqrt{2} \sin \theta}{\sqrt{(-2+\sqrt{2} \cos \theta)^{2}+(... | \left[\frac{\sqrt{3}-\sqrt{8}}{5}, \frac{\sqrt{3}+\sqrt{8}}{5}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,483 |
6. A function is called a "good function" if and only if it satisfies:
(1) It is defined on $\mathbf{R}$;
(2) There exist $a<b$, such that it is monotonically increasing on $(-\infty, a)$ and $(b, +\infty)$, and monotonically decreasing on $(a, b)$.
Then the following functions that are good functions are $\qquad$
(1) ... | 6. (1)(2)(3).
(1) $y=x | x-2$ | is monotonically increasing on $(-\infty, 1)$ and $(2,+\infty)$, and monotonically decreasing on $(1,2)$, satisfying the definition.
(2) $y=x^{3}-x+1$.
Notice that $y^{\prime}=3 x^{2}-1>0$
$\Leftrightarrow x \in\left(-\infty,-\frac{\sqrt{3}}{3}\right) \cup\left(\frac{\sqrt{3}}{3},+\infty... | (1)(2)(3) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,484 |
7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$. | 7.5.
Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other on... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,485 |
8. A classroom has desks arranged in 6 rows and 7 columns, with 40 students. Two positions in the last row are left empty, and the rest of the students are seated based on their height and vision. There are 24 students who are tall, 18 students who have good vision, and 6 students who have both conditions. It is known ... | 8. 35 .
The number of students who are short and have poor eyesight is 4, the number of students who are tall and have good eyesight is 6, and the remaining students number 30 people.
The number of ways to leave two seats empty in the last row is $C_{7}^{2}$, where the power of 2 is 0;
The number of ways to arrange ... | 35 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,486 |
9. (14 points) Let $x, y, z$ be non-negative real numbers, satisfying $xy + yz + zx = 1$.
Prove: $\frac{z}{1-xy} + \frac{y}{1-yx} + \frac{x}{1-yz} \geqslant \frac{5}{2}$. | $$
\begin{array}{l}
\sum \frac{z}{1-x y}=\sum \frac{1}{x+y}=\sum \frac{1}{\cot A+\cot B} \\
=\sum \frac{\sin A \cdot \sin B}{\sin C} \\
=\left[\sin C\left(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}-2\right)+\frac{\sin A \cdot \sin B-\sin ^{2} t}{\sin C}\right]+ \\
\left(2 \sin C+\frac{\sin ^{2} t}{\sin C}\right) \\
=\... | \frac{5}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 721,487 |
10. (15 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $M(m, 0)(0<m<a)$ is a moving point. Let $f(m)$ be the radius of the smallest circle centered at point $M$ that is tangent to the ellipse $C$. Find $f(m)$. | 10. Since $\odot M$ is the circle with the smallest radius that is tangent to the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, the radius should be the minimum distance from a point on the ellipse to $M$. Let any point on the ellipse be $P\left(x_{0}, y_{0}\right)$. Then $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^... | f(m)=\left\{\begin{array}{ll}\frac{b}{c} \sqrt{c^{2}-m^{2}}, & 0<m \leqslant \frac{c^{2}}{a} ; \\ a-m, & \frac{c^{2}}{a}<m<a .\end{array}\right.} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,488 |
11. (15 points) Given the equation $8 t^{3}-4 t^{2}-4 t+1=$ 0 has a root $x$ in $\left(0, \frac{\pi}{13}\right)$. Find $x$.
| 11. Notice
$$
\begin{array}{l}
8 x^{3}-4 x^{2}-4 x+1=0\left(x \in\left(0, \frac{\pi}{13}\right)\right) \\
\Rightarrow(x+1)\left(8 x^{3}-4 x^{2}-4 x+1\right)=0 \\
\Rightarrow 8 x^{4}+4 x^{3}-8 x^{2}-3 x+1=0 \\
\Rightarrow 3 x-4 x^{3}=8 x^{4}-8 x+1 .
\end{array}
$$
For any $\alpha \in \mathbf{R}$, we have
$$
\begin{arra... | x=\sin \frac{\pi}{14} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,489 |
One, (50 points) In the acute isosceles triangle $\triangle ABC (AB = AC)$, the midpoint of the minor arc $\overparen{BC}$ on the circumcircle $\Gamma$ is $D$. Points $E$ and $F$ are on the minor arcs $\overparen{BD}$ and $\overparen{AB}$, respectively. The incenter of $\triangle ABE$ and $\triangle ACE$ are $P$ and $Q... | As shown in Figure 1, let $I$ be the incenter of $\triangle ABC$, and connect the auxiliary lines as shown.
It is easy to see that $\angle PAQ = \frac{1}{2} \angle BAC = \angle BAD = \angle BFD = \angle PFQ$.
Therefore, points $P, Q, A, F$ are concyclic (denoted as $\Gamma_{1}$).
Also, $\angle APB = \frac{\pi}{2} + \f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,490 |
```
Given real numbers $a, b, c, t$, satisfying $a, b, c \in (t, t+1) (t > 0)$. Prove:
\[
\begin{array}{l}
\frac{1}{\sqrt{(a-b+1)(b-a+2)}}+ \\
\frac{1}{\sqrt{(b-c+1)(c-b+2)}}+ \\
\frac{1}{\sqrt{(c-a+1)(a-c+2)}} \geqslant \frac{3}{\sqrt{2}} .
\end{array}
\]
``` | Let $x=a-b+1, y=b-c+1, z=c-a+1$. Given $a, b, c \in (t, t+1)$, we know $x, y, z > 0$.
Then the inequality to be proved can be transformed into
$$
\frac{1}{\sqrt{x(y+z)}}+\frac{1}{\sqrt{y(x+z)}}+\frac{1}{\sqrt{z(x+y)}} \geqslant \frac{3}{\sqrt{2}},
$$
where $x+y+z=3$.
By the Cauchy-Schwarz inequality, we have
$$
\begin... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,491 |
Three, (50 points) On a plane, there are 32 points, with no three points being collinear. Prove: Among these 32 points, there are at least 2135 sets of four points that form the vertices of a convex quadrilateral. | Three, let $n=32$.
Since no three points among the $n$ points are collinear, $\mathrm{C}_{n}^{3}$ triangles can be formed by any three points among the $n$ points, among which the one with the largest area is denoted as $\triangle A B C$.
Draw lines parallel to the opposite sides through points $A, B, C$, intersecting... | 2135 | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,492 |
2. If
$$
\begin{array}{l}
a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5,
\end{array}
$$
then, the value of $a+b+c$ is $\qquad$ | (Tip: Refer to Example 3. Answer: 20. ) | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,493 |
Initially 255, set 2009 can be decomposed into the sum of squares of four positive integers, among which, the ratio of two numbers is $\frac{5}{14}$, and the ratio of the other two numbers is $\frac{1}{2}$. Write down this decomposition. | Solution: Let the required decomposition be
$$
2009=(14 x)^{2}+(5 x)^{2}+(2 y)^{2}+y^{2} \text {. }
$$
Obviously, $221 x^{2}+5 y^{2}=2009$, hence
$$
x^{2}=\frac{2009-5 y^{2}}{221}<10 \text {. }
$$
Thus, $x=1, 2$ or 3.
(1) If $x=1, y^{2}=\frac{2009-221 x^{2}}{5}=\frac{1788}{5}$,
no solution;
(2) If $x=2, y^{2}=\frac{... | 2009=30^{2}+28^{2}+15^{2}+10^{2}, \\ 2009=42^{2}+15^{2}+4^{2}+2^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,495 |
As shown in Figure 3, from a point $P$ outside circle $\odot O$, two secants $PAB$ and $PCD$ are drawn, intersecting $\odot O$ at points $A, B$ and $C, D$ respectively. Chords $AD$ and $BC$ intersect at point $Q$, and $PQ$ intersects $\odot O$ at point $E$. If $PE = \frac{12}{5}$, the radius of $\odot O$ is 3, and the ... | Solution: As shown in Figure 3, draw $PT$ tangent to $\odot O$ at point $T$, and connect $OP, OT$. Then
$$
OP=5, OT=3, PT \perp OT \text{. }
$$
Thus, $PT=\sqrt{OP^2-OT^2}=\sqrt{5^2-3^2}=4$.
Extend $PQ$ to intersect $\odot O$ at point $F$.
Then $PT^2=PE \cdot PF \Rightarrow 4^2=\frac{12}{5} PF \Rightarrow PF=\frac{20}{... | \frac{96}{85} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,496 |
255 As shown in Figure 4, let the incenter of $\triangle ABC$ be $I$, and the extensions of $AI$, $BI$, and $CI$ intersect the circumcircle of $\triangle ABC$ at points $A_1$, $B_1$, and $C_1$, respectively. Given that $B_1C_1$ intersects $AI$, $C_1A_1$ intersects $BI$, and $A_1B_1$ intersects $CI$ at points $A_2$, $B_... | Proof: Let $\angle A, \angle B, \angle C$ and $\angle A_{1}, \angle B_{1}, \angle C_{1}$ represent the interior angles of $\triangle ABC$ and $\triangle A_{1}B_{1}C_{1}$, respectively.
(1) From the given conditions, we have
$$
\begin{array}{l}
\angle A_{1}=\frac{1}{2}(\angle B+\angle C), \\
\angle B_{1}=\frac{1}{2}(\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,497 |
Let $a 、 b 、 c$ be distinct real numbers. Prove:
$$
\left(\frac{a}{a-b}+3\right)^{2}+\left(\frac{b}{b-c}+3\right)^{2}+\left(\frac{c}{c-a}+3\right)^{2} \geqslant 25 .
$$ | Prove: Let $x=\frac{a}{a-b}, y=\frac{b}{b-c}, z=\frac{c}{c-a}$.
$$
\begin{array}{l}
\text { Then }(x-1)(y-1)(z-1) \\
=\frac{b}{a-b} \cdot \frac{c}{b-c} \cdot \frac{a}{c-a} \\
=\frac{a}{a-b} \cdot \frac{b}{b-c} \cdot \frac{c}{c-a}=x y z .
\end{array}
$$
Transforming, we get $x+y+z=x y+y z+z x+1$.
$$
\begin{array}{l}
\t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 721,498 |
3. If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ . | (Tip: Refer to Example 3. Answer: 4. ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,499 |
4. Given that $a$ and $b$ are positive numbers, $a+b=3$. For what values of $a$ and $b$ is $\sqrt{a^{2}+4}+\sqrt{b^{2}+25}$ minimized, and what is the minimum value? | ( Hint: Refer to Example 5. Answer: The minimum value is $\sqrt{58}$, $\left.a==\frac{6}{7}, b=\frac{15}{7}.\right)$ | \sqrt{58} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,500 |
5. Find the maximum value of the function $y=\sqrt{2 x^{2}+3 x+1}+\sqrt{7-2 x^{2}-3 x}$. | ( Hint: Refer to Example 8. Answer: The maximum value is 4 . ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,501 |
6. Given that the roots of the equation $x^{2}-6 x-4 n^{2}-32 n=0$ are integers. Find the integer value of $n$.
(2004, National Junior High School Mathematics League) | (Tip: $\Delta=4\left(4 n^{2}+32 n+9\right)$. Since the roots of the equation are all integers, $4 n^{2}+32 n+9$ is a perfect square. Let $4 n^{2}+32 n+9=m^{2}$ (where $m$ is a natural number). Then
$$
\begin{array}{l}
(2 n+8+m)(2 n+8-m)=55 . \\
\text { Also } 55=( \pm 1) \times( \pm 55)=( \pm 5) \times( \pm 11), \\
2 n... | n=-18,-8,0,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,502 |
Example 1
(1) In a convex hexagon where each side is greater than 1, can we always find a diagonal that is longer than 2?
(2) In a convex hexagon $A B C D E F$, if the diagonals $A D, B E, C F$ are all longer than 2, can we always find a side that is longer than 1?
(Eighth All-Union Mathematical Olympiad) | (1) Consider a regular hexagon, the length of the diagonal passing through the center of symmetry is exactly twice the side length. However, with a slight adjustment, a hexagon can be obtained where each diagonal is no longer than 2 and each side is greater than 1.
In fact, as shown in Figure 1, construct hexagon $A B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,503 |
Example 2 Does there exist a point \( P \) in the plane of the convex quadrilateral \( ABCD \) such that the angle bisectors of \( \angle APB \), \( \angle BPC \), \( \angle CPD \), and \( \angle DPA \) intersect \( AB \), \( BC \), \( CD \), and \( DA \) at points \( K \), \( L \), \( M \), and \( N \) respectively, a... | Explanation: As shown in Figure 3, consider when $P$ is the intersection of the perpendicular bisector of $A C$ and the perpendicular bisector of $B D$. By the Angle Bisector Theorem, we have
$$
\frac{A K}{K B}=\frac{A P}{B P}=\frac{C P}{B P}=\frac{C L}{L B} \text {. }
$$
Therefore, $K L \parallel A C$.
Similarly, $M ... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,504 |
Example 3 Given an acute triangle $\triangle ABC$.
(1) Find the locus of the centers of the inscribed rectangles of $\triangle ABC$;
(2) Does there exist a point that is the center of three different inscribed rectangles of $\triangle ABC$?
(2007, Italian National Team Selection Exam) | Explanation: (1) Consider the case where one side of the inscribed rectangle is parallel to $BC$. When the height of the rectangle approaches the height from $A$ to $BC$, the center of the rectangle approaches the midpoint $N$ of the height; when the height of the rectangle approaches 0, the center of the rectangle app... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,505 |
Example 3 Real numbers $x, y, z$ satisfy
$$
x+y+z=4(\sqrt{x-5}+\sqrt{y-4}+\sqrt{z-3}) \text {. }
$$
Then $x=$ $\qquad$ ,$y=$ $\qquad$ ,$z=$ $\qquad$ .
(10th Five Sheep Cup Junior High School Mathematics Competition) | Solution: From the original equation, we have
$$
\begin{array}{l}
(\sqrt{x-5}-2)^{2}+(\sqrt{y-4}-2)^{2}+(\sqrt{z-3}-2)^{2}=0 . \\
\text { Also, } (a-b)^{2} \geqslant 0, \text { so } \\
\sqrt{x-5}-2=0, \sqrt{y-4}-2=0, \\
\sqrt{z-3}-2=0 .
\end{array}
$$
$$
\text { Therefore, } x=9, y=8, z=7 \text {. }
$$ | x=9, y=8, z=7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,506 |
Example 4 On a plane, there is a convex quadrilateral $A B C D$.
(1) If there exists a point $P$ on the plane such that the areas of $\triangle A B P$, $\triangle B C P$, $\triangle C D P$, and $\triangle D A P$ are equal, what condition must the quadrilateral $A B C D$ satisfy?
(2) How many such points $P$ can there b... | Explanation: If point $P$ is inside the quadrilateral and not on side $A C$ (as shown in Figure 6), by the equality of areas, the distances from points $A$ and $C$ to line $B P$ are equal. Therefore, line $B P$ passes through the midpoint $M$ of $A C$. Similarly, $D P$ passes through the midpoint $M$ of $A C$. Thus, po... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,507 |
Example 5 Let $M$ be the intersection of the two diagonals of a convex quadrilateral $ABCD$. Find all such convex quadrilaterals for which there exists a line $l$ through point $M$ such that $l$ intersects $BC$ and $CD$ at points $P$ and $Q$ respectively, and $\triangle APM$, $\triangle BPM$, $\triangle CQM$, $\triangl... | Explanation: As shown in Figure 8, consider two similar triangles $\triangle A P M$ and $\triangle B P M$.
Notice that the corresponding angle of $\angle A P M$ can only be $\angle B P M$, thus,
$\angle A P M=\angle B P M=90^{\circ}$.
Therefore, $P Q \perp A B$.
Similarly, $P Q \perp C D$.
Hence, $A B / / C D$.
Now co... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,508 |
Example $6 R$ is a rectangle, which is composed of several smaller rectangles $R_{i}(1 \leqslant i \leqslant n)$, satisfying:
(1) The sides of $R_{i}$ are parallel to the sides of $R$;
(2) The $R_{i}$ do not overlap;
(3) Each $R_{i}$ has at least one side of integer length. Prove: $R$ has at least one side of integer l... | Explanation: Let the four vertices of $R$ be $A$, $B$, $C$, and $D$. As shown in Figure 9, establish a Cartesian coordinate system with $A$ as the origin, $AB$ as the positive $x$-axis, and $AD$ as the $y$-axis. Thus, the problem is reduced to showing that one of $B$, $C$, and $D$ must be an integer point.
If not, ass... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,509 |
Example 7 Given two intersecting circles $\odot O_{1}$ and $\odot O_{2}$ in the plane, and $A$ is one of the intersection points. There are two moving points $M_{1}$ and $M_{2}$ starting from point $A$ simultaneously, moving at constant speeds along $\odot O_{1}$ and $\odot O_{2}$ in the same direction, and each return... | Explanation: As shown in Figure 10, imagine two congruent triangles with vertices at points $P, O_{1}, M_{1}$ and $P, O_{2}, M_{2}$, respectively. Thus, point $P$ can be directly constructed. Let the perpendicular bisector of line segment $O_{1} O_{2}$ be $l$, and construct point $P$ as the reflection of point $A$ abou... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,510 |
Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such th... | Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.
When $n=5$,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,511 |
Example 9 Prove or disprove: In the plane of convex quadrilateral $ABCD$, there exists a point $E$ such that
$$
\triangle ABE \backsim \triangle CDE \text{. }
$$
(2007, Taiwan Mathematical Olympiad Selection Exam) | Consider the plane containing the convex quadrilateral $ABCD$ as the platinum plane, and agree to use $A, B, C, D$ to represent the complex numbers corresponding to each vertex. Let the complex number $E$ satisfy
$$
\frac{C-E}{A-E}=\frac{D-E}{B-E},
$$
which implies $[(A+D)-(B+C)] E=A D-B C$.
Since the midpoints of $AD... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,512 |
Example 10 Prove: there exists a convex 1990-gon with the following properties:
(1) All interior angles of the polygon are equal;
(2) The lengths of the sides of the polygon are a permutation of $1^{2}, 2^{2}, \cdots, 1989^{2}, 1990^{2}$.
(31st IMO) | Consider on the complex plane, the problem is equivalent to the existence of a permutation $a_{1}, a_{2}, \cdots, a_{1989}, a_{1990}$ of $1^{2}, 2^{2}, \cdots, 1989^{2}, 1990^{2}$, such that
$$
\begin{array}{l}
\sum_{n=1}^{1990} a_{n} e^{n \theta \mathrm{i}}=0\left(\theta=\frac{\pi}{995}\right). \\
\text { Let }\left\{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,513 |
Example 11 In $\triangle ABC$, draw the median $BD$ and the angle bisector $BE$. Is there such a case where $BD$ is the angle bisector in $\triangle BCE$, and at the same time, $BE$ is the median in $\triangle ABD$? | Assume such a triangle exists.
Let $C D=2 D E=2 A E=2 a, \angle C B D=\alpha$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{B A}{B C}=\frac{A E}{E C}=\frac{1}{3}, \frac{B E}{B C}=\frac{E D}{D C}=\frac{1}{2} . \\
\text { Let } B C=6 b, B E=3 b, B A=2 b .
\end{array}
$$
In $\triangle B C E$ and $\tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,514 |
Example $12 P$ is a point on side $AB$ of $\triangle ABC$, $PQ$ $/ / AC, PR / / BC$, points $Q, R$ are on $BC, CA$ respectively. Question: Is there a fixed point $M$ other than point $C$, such that when $P$ moves on $AB$, $C, Q, M, R$ are always concyclic? | Explanation: As shown in Figure 12, if point $M$ satisfies:
When $P \rightarrow B$,
$Q \rightarrow B, R \rightarrow C$;
When $P \rightarrow A$,
$$
Q \rightarrow C, R \rightarrow A \text {. }
$$
Based on the fact that points $C, Q, M, R$ are concyclic, we conjecture
$$
\angle M A C=\angle M C B, \angle M C A=\angle M B... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,515 |
1. Given an $n \times n$ grid. Question: Does there exist a coloring scheme such that each grid point is colored with one of four colors, and on every line connecting two points of the same color, there is at least one point of a different color. | (Tip: Establish a Cartesian coordinate system, taking the side length of the small square grid as the unit length, so that each intersection of the grid lines is an integer point. Thus, each integer point can be classified into four categories according to the parity of its horizontal and vertical coordinates. Color th... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,516 |
Example 4 If real numbers $a, b, c$ satisfy $a+b+c=0$, $abc=2, c>0$, then ( ).
(A) $ab<0$
(B) $|a|+|b| \geqslant 2$
(C) $|a|+|b| \geqslant 4$
(D) $0<|a|+|b| \leqslant 1$ | Solution: It is easy to know that $a+b=-c$.
Since $abc=2, c>0$, therefore, $a, b$ are both negative, i.e., $|a|+|b|=|a+b|=|c|$, and $|ab|=\frac{2}{|c|}$.
From $|a|+|b| \geqslant 2 \sqrt{|a| \cdot|b|}$, we get $|c| \geqslant 2 \sqrt{\frac{2}{|c|}}$
Solving this, we get $|c| \geqslant 2$.
Therefore, $|a|+|b|=|c| \geqslan... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,517 |
4. Color all points on a circle either black or white. Ask: Is it necessarily true that there exists a triangle with vertices on the circle and all three vertices of the same color:
(1) Isosceles triangle; (2) Equilateral triangle;
(3) Rectangle; (4) Trapezoid?
Prove your conclusion.
(57th Belarusian Mathematical Olymp... | (提示: If half of the circle is colored black and the other half is colored white, then
it is known that (2) and (3) do not hold. Considering a regular pentagon inscribed in the circle, by the pigeonhole principle, there must be three vertices of the same color, thus (1) holds. Considering each inscribed equilateral tria... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,518 |
5 . Find all rhombi with side length $2 a$ such that there exists a circle intersecting each side of the rhombus, and the length of the chord of each side inside the circle is equal to $a$.
$(2007$, Austrian Mathematical Olympiad (Second Round)) | (Tip: By symmetry, the center of the circle $M$ is the center of the rhombus $ABCD$. Let the midpoint of $AB$ be $H$. Then point $H$ lies on a chord of length $a$ (including endpoints). Thus, $BM \geqslant MH$.
Therefore, $\angle BHM \geqslant 60^{\circ}, \angle BAM \geqslant 30^{\circ}$.
Consequently, $\angle BAD \geq... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,519 |
Example 1 Suppose the lengths of the three sides of a triangle are integers $l$, $m$, and $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$
where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceedin... | Solution: Note that, when $a \equiv b(\bmod m)(0 \leqslant b < m)$, we get $500+2 n>1000+n$.
So, $n>500$.
Thus, $n \geqslant 501, m \geqslant 1001, l \geqslant 1501$.
Therefore, $m+n+l \geqslant 3003$, which means the minimum perimeter of the triangle is 3003. | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,520 |
Let $\triangle ABC$ be an acute-angled triangle with incircle $\omega$ and circumcircle $\Omega$, and let the circumradius be $R$. The circle $\omega_A$ is internally tangent to $\Omega$ at $A$ and externally tangent to $\omega$; the circle $\Omega_A$ is internally tangent to $\Omega$ at $A$ and internally tangent to $... | The paper [1] provides the proof of the problem. Inspired by it, this paper obtains the following relationships between $P_{A} Q_{A}$, $P_{B} Q_{B}$, $P_{C} Q_{C}$, and $R$, $r$ (where $r$ represents the inradius of $\triangle A B C$):
$$
\begin{array}{l}
3 r \leqslant P_{A} Q_{A}+P_{B} Q_{B}+P_{C} Q_{C} \leqslant \fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,522 |
1. Let $n$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{k}(k \geqslant 2)$ be distinct integers in the set $\{1,2, \cdots, n\}$, such that for $i=1,2, \cdots, k-1$, we have $n \mid a_{i}\left(a_{i+1}-1\right)$. Prove: $n \times a_{k}\left(a_{1}-1\right)$ (Provided by Australia) | 1. First, use mathematical induction to prove: for any integer $i(2 \leqslant i \leqslant k)$, we have $n \mid a_{1}\left(a_{i}-1\right)$.
In fact, when $i=2$, by the given condition, $n \mid a_{1}\left(a_{2}-1\right)$, the conclusion holds.
Assume that $n \mid a_{1}\left(a_{i}-1\right)(2 \leqslant i \leqslant k-1)$, ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 721,523 |
2. Let $O$ be the circumcenter of $\triangle ABC$, and let points $P$ and $Q$ be on sides $CA$ and $AB$, respectively. Let $K$, $L$, and $M$ be the midpoints of segments $BP$, $CQ$, and $PQ$, respectively, and let $\Gamma$ be the circle passing through points $K$, $L$, and $M$. If line $PQ$ is tangent to circle $\Gamma... | 2. Obviously, the line $P Q$ is tangent to the circle $\Gamma$ at point $M$. By the tangent-chord angle theorem, we know $\angle Q M K = \angle M L K$.
Since points $K$ and $M$ are the midpoints of segments $B P$ and $P Q$ respectively, we have
$K M \parallel B Q \Rightarrow \angle Q M K = \angle A Q P$.
Therefore, $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 721,524 |
3. Let $s_{1}, s_{2}, \cdots$ be a strictly increasing sequence of positive integers, such that the two subsequences
$$
s_{s_{1}}, s_{s_{2}}, \cdots \text { and } s_{s_{1}+1}, s_{s_{2}+1}, \cdots
$$
are both arithmetic sequences. Prove: The sequence $s_{1}, s_{2}, \cdots$ itself is also an arithmetic sequence.
(Provid... | 3. It is easy to see from the conditions that $s_{s_{1}}, s_{s_{2}}, \cdots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \cdots$ are both strictly increasing sequences of positive integers.
Let $s_{s_{k}}=a+(k-1) d_{1}, s_{s_{k}+1}=b+(k-1) d_{2}$ $\left(k=1,2, \cdots, a, b, d_{1}, d_{2} \in \mathbf{N}_{+}\right)$.
By $s_{k}s_{k}$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 721,525 |
4. In $\triangle A B C$, $A B=A C$, the internal angle bisectors of $\angle C A B$ and $\angle A B C$ intersect the sides $B C$ and $C A$ at points $D$ and $E$, respectively. Let $K$ be the incenter of $\triangle A D C$. If $\angle B E K=45^{\circ}$, find all possible values of $\angle C A B$. (Belgium provided) | 4. Segments $A D$ and $B E$ are the angle bisectors of $\triangle A B C$, intersecting at the incenter $I$ of $\triangle A B C$. Connecting $C I$. Then $C I$ bisects $\angle A C B$.
Since $K$ is the incenter of $\triangle A D C$, point $K$ lies on segment $C I$.
Let $\angle B A C=\alpha$.
Since $A B=A C$, we have $A D... | 60^{\circ} \text{ and } 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 721,526 |
5. Find all functions $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that for all positive integers $a, b$, there exists a non-degenerate triangle with side lengths $a, f(b), f(b+f(a)-1)$ (a triangle is called non-degenerate if its three vertices are not collinear).
(France) | 5. The function $f$ that satisfies the requirements can only be $f(n)=n\left(n \in \mathbf{N}_{+}\right)$.
From the conditions and the discreteness of integers, for any positive integers $a, b$, we have
$$
\begin{array}{l}
f(b)+f(b+f(a)-1)-1 \geqslant a, \\
f(b)+a-1 \geqslant f(b+f(a)-1), \\
f(b+f(a)-1)+a-1 \geqslant ... | f(n) = n | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 721,527 |
Example 5 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$. | Solution: As shown in Figure 2, construct Rt $\triangle P A C$ and Rt $\triangle P B D$ such that $A C=1, B D=2, C P=x, P D=4-x$. Thus, the original problem is transformed into: finding a point $P$ on line $l$ such that the value of $P A + P B$ is minimized.
Since $y = P A + P B \geqslant A B$, the minimum value of $y... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 721,528 |
6. Let $a_{1}, a_{2}, \cdots, a_{n}$ be distinct positive integers. $M$ is a set of $n-1$ positive integers that does not contain the number $s=a_{1}+a_{2}+\cdots+a_{n}$. A grasshopper starts at the origin 0 on the real number line and makes $n$ jumps to the right, with the jump distances being a permutation of $a_{1},... | 6. First, use mathematical induction on $n$.
When $n=1$, $M$ contains no elements, and the conclusion is obviously true.
When $n=2$, $M$ does not contain at least one of $a_{1}$ or $a_{2}$. The grasshopper can jump to this number first, then to the other number.
Assume that for $n \in \mathbf{N}_{+}, n<m$, the conclu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,529 |
1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest ... | 1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.
Below is the proof: 6 is the maximum value.
Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,530 |
2. For each positive integer $n$, find the number of permutations $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of the set $\{1,2, \cdots, n\}$, where $a_{1}, a_{2}, \cdots, a_{n}$ satisfy that for $k=1,2, \cdots, n, 2\left(a_{1}+a_{2}+\cdots\right.$ $\left.+a_{k}\right)$ is divisible by $k$. | 2. For each positive integer $n$, let $F_{n}$ be the number of permutations of the set $\{1,2, \cdots, n\}$ that satisfy the condition, and call these permutations "good". Then for $n=1,2,3$, every permutation is good. Therefore, $F_{1}=1, F_{2}=2, F_{3}=6$.
For $n>3$, consider a good permutation $\left(a_{1}, a_{2}, ... | F_{n}=3 \times 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,531 |
3. Consider the set $S$ of all integer points (points with integer coordinates) on the coordinate plane. For each positive integer $k$, if there exists a point $C \in S$ such that the area of $\triangle A B C$ is $k$, then two distinct points $A, B \in S$ are called “$k$-friends”; if any two points in $T$ are $k$-frien... | 3. First, consider the point $B \in S$ that is a $k$-friend of the point $(0,0)$. Let $B(u, v)$. Then $B$ is a $k$-friend of $(0,0)$ if and only if there exists a point $C(x, y) \in S$ such that $\frac{1}{2}|u y - v x| = k$ (here we use the area formula of $\triangle A B C$, where $A$ is the origin).
There exist integ... | 180180 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 721,532 |
5. Let the set of real numbers with $k+l$ elements be
$$
S=\left\{x_{1}, x_{2}, \cdots, x_{k+l}\right\}
$$
satisfying $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, k+l, k, l$ are positive integers). If it satisfies
$$
\left|\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{l} \sum_{x_{j} \in S \backslash A} x_{j}\right| \... | 5. For a $k$-element subset $A$ of $S$, let $f(A)=\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{l} \sum_{x_{j} \in S \backslash A} x_{j}, \frac{k+l}{2 k l}=d$.
By definition, if $|f(A)| \leqslant d$, then the subset $A$ is good.
For the set $S=\left\{x_{1}, x_{2}, \cdots, x_{k+1}\right\}$ and each permutation $\left(y... | \frac{2}{k+l} \mathrm{C}_{k+l}^{k} | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,533 |
6. For positive integers $n \geqslant 2$, let $S_{1}, S_{2}, \cdots, S_{2^{n}}$ be $2^{n}$ subsets of the set $A=\left\{1,2,3, \cdots, 2^{n+1}\right\}$, and satisfy the following property: there do not exist indices $a, b (a<b)$ and $x, y, z \in A (x<y<z)$ such that $y, z \in S_{a}$ and $x, z \in S_{b}$. Prove: $S_{1},... | 6. We prove: There exists a set $S_{a}$ with at most $3 n+2$ elements.
Given a $k \in\{1,2, \cdots, n\}$. If $z \in S_{a}$, and $S_{a}$ contains two other elements $x, y$, satisfying $x2\left(u_{m-1}-u_{1}\right)$.
In fact, assume for some $m \geqslant 3$, there is $u_{m}-u_{1}$ $\leqslant 2\left(u_{m-1}-u_{1}\right)$... | 3 n+2 | Combinatorics | proof | Yes | Yes | cn_contest | false | 721,534 |
1. If $-\frac{1}{2} \leqslant x \leqslant 1$, then
$$
\begin{array}{l}
\sqrt{x^{2}-2 x+1}+\sqrt{x^{2}-6 x+9}+\sqrt{4 x^{2}+4 x+1} \\
=(\quad) .
\end{array}
$$
(A) $-4 x+3$
(B) 5
(C) $2 x+3$
(D) $4 x+3$ | $\begin{array}{l}\text { i.1. B. } \\ \text { Original expression }=|x-1|+|x-3|+|2 x+1| \\ =1-x+3-x+2 x+1=5 .\end{array}$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,535 |
2. Use three regular polygon tiles with equal side lengths to pave the ground, with their vertices fitting together, just enough to completely cover the ground. Given the number of sides of the regular polygons are $x, y, z$. Then the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $(\quad)$.
(A) 1
(B) $\frac{2}{3}$
... | 2. C.
According to the problem, we have
$$
\begin{array}{l}
\frac{x-2}{x} \times 180^{\circ}+\frac{y-2}{y} \times 180^{\circ}+\frac{z-2}{z} \times 180^{\circ}=360^{\circ} \\
\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,536 |
3. Given that $a$ is a non-negative integer, the equation with respect to $x$
$$
2 x-a \sqrt{1-x}-a+4=0
$$
has at least one integer root. Then the number of possible values of $a$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 3. B.
Obviously, $x \leqslant 1$.
Also, $2x + 4 = a(\sqrt{1 - x} + 1) \geqslant 0$, so $x \geqslant -2$. When $x = 1, 0, -2$, $a$ is $6, 2, 0$ respectively; when $x = -1$, $a$ is not an integer. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 721,537 |
4. As shown in Figure 1, let $\triangle A B C$ and $\triangle C D E$ both be equilateral triangles, and $\angle E B D=62^{\circ}$. Then the degree measure of $\angle A E B$ is ( ).
(A) $124^{\circ}$
(B) $122^{\circ}$
(C) $120^{\circ}$
(D) $118^{\circ}$ | 4. B.
Obviously, $\triangle B C D \cong \triangle A C E$, so $\angle D C B=\angle E C A$.
Let $\angle B A E=\alpha$. Then
$\angle C B D=\angle C A E=60^{\circ}-\alpha$,
$\angle E B C=\angle E B D-\angle C B D=2^{\circ}+\alpha$.
Thus, $\angle E B A=60^{\circ}-\left(2^{\circ}+\alpha\right)=58^{\circ}-\alpha$.
Therefore,... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 721,538 |
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