problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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8. Given a convex $n(n \geqslant 4)$-sided polygon $M$, color $n-3$ diagonals green and the other $n-3$ diagonals red, such that diagonals of the same color do not intersect inside $M$. Find the maximum number of intersection points between the red and green diagonals inside $M$. | 8. It is easy to obtain that in a convex $n$-gon, there are at most $n-3$ diagonals such that any two diagonals do not intersect inside the convex $n$-gon. These $n-3$ diagonals divide the convex $n$-gon into $n-2$ triangles, and at least two vertices do not have any diagonals drawn from them. Therefore, at least two d... | \left\lceil \frac{3}{4}(n-3)^2 \right\rceil | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,559 |
1. If $x+y=1, x^{3}+y^{3}=\frac{1}{3}$, then the value of $x^{5}+y^{5}$ is ( ).
(A) $\frac{11}{81}$
(B) $\frac{31}{81}$
(C) $\frac{11}{243}$
(D) $\frac{31}{243}$ | -1.A.
Given $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=\frac{1}{3}, x+y$ $=1$, we have $x^{2}-x y+y^{2}=\frac{1}{3}$.
Also, since $x^{2}+2 x y+y^{2}=1$, then $3 x y=\frac{2}{3}, x y=\frac{2}{9}$.
Therefore, $x^{5}+y^{5}=\frac{32}{243}+\frac{1}{243}=\frac{33}{243}=\frac{11}{81}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,560 |
2. Given $2 x-3 \sqrt{x y}-2 y=0(x>0)$. Then the value of $\frac{x^{2}+4 x y-16 y^{2}}{2 x^{2}+x y-9 y^{2}}$ is ( ).
(A) $\frac{2}{3}$
(B) $\frac{4}{9}$
(C) $\frac{16}{25}$
(D) $\frac{16}{27}$ | 2.D.
From the original equation, we get $2\left(\frac{x}{y}\right)-3 \sqrt{\frac{x}{y}}-2=0$.
Let $\sqrt{\frac{x}{y}}=t$, then the equation transforms into $2 t^{2}-3 t-2=0$, which is $(2 t+1)(t-2)=0$.
Solving, we get $t_{1}=2, t_{2}=-\frac{1}{2}$ (discard). Hence $\frac{x}{y}=4$.
Substituting $x=4 y$ into the fractio... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,561 |
Example 6 In isosceles $\triangle A B C$, $A B=A C$, there is a circle that is internally tangent to the circumcircle of $\triangle A B C$, and is tangent to $A B$ and $A C$ at points $P$ and $Q$ respectively. Prove: The midpoint of the line segment connecting $P$ and $Q$ is the incenter of $\triangle A B C$. | As shown in Figure 6, let $D$ be the point of tangency of the two circles. By the symmetry of the circles and the isosceles triangle, we know that $AD$ is the diameter of the circumcircle of $\triangle ABC$. Let $E$ be the intersection of $AD$ and $PQ$. Since $AP = AQ$, we know that $E$ is the midpoint of $PQ$, and $AE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,562 |
3. In a convex polygon, a quadrilateral has two diagonals, a pentagon has 5 diagonals. Observe and explore how many diagonals a convex decagon has ( ).
(A) 29
(B) 32
(C) 35
(D) 38 | 3.C.
Draw and observe to explore the number of diagonals in polygons: for quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, the number of diagonals are: $\begin{array}{lllllll}2 & 5 & 9 & 14 & 20 & 27 & 35\end{array}$ The pattern is: $2+35+49+514+620+727+8$ | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,563 |
4. In $\triangle A B C$, it is known that $A B=A C=8 \sqrt{3}$, and the altitude $A D$ $=8$. Then the radius of the circumcircle of $\triangle A B C$ is $(\quad$.
(A) 8
(B) 9
(C) 10
(D) 12 | 4.D.
As shown in Figure 4, extend $AD$ to intersect the circumcircle at point $E$, then $AE$ is the diameter. Connecting $BE$, we know that $\triangle ABE$ is a right triangle, thus we have
$$
AB^{2}=AD \cdot AE .
$$
Therefore, the radius is 12. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,564 |
5. If a positive integer can be expressed as the difference of squares of two positive integers, it is called a "wise number" (for example, $3=$ $2^{2}-1^{2}, 16=5^{2}-3^{2}$). It is known that wise numbers, arranged in ascending order, form the following sequence:
$$
\begin{array}{l}
3,5,7,8,9,11,12,13,15,16,17,19,20,... | 5.C.
Observing and exploring the pattern, we know that all wise numbers, from smallest to largest, can be grouped in sets of three, and starting from the 2nd group, the first number of each group is a multiple of 4. By induction, the first number of the $n$-th group is $4n$ ($n \geqslant 2$).
Since $2006=3 \times 668... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,565 |
1. As shown in Figure 2, given the vertices of square $A B C D$ are $A(1,1)$, $B(3,1)$, $C(3,3)$, and $D(1,3)$, the line $y=2 x+b$ intersects $A B$ at point $E$ and $C D$ at point $F$. Then the range of the y-intercept $b$ of the line is | $$
\text { II. 1. }-3 \leqslant b \leqslant-1 \text {. }
$$
From the line $y=2 x+b$ moving parallel with different values of $b$, we know that when the line passes through point $A$, we get $b=-1$; when the line passes through point $C$, we get $b=-3$. Therefore, $-3 \leqslant b \leqslant-1$. | -3 \leqslant b \leqslant -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,567 |
2. The function expression of the image $l_{2}$, which is symmetric to the image $l_{1}$ of the linear function $y = a x + b$ with respect to the line $y = -x$, is $\qquad$ . | 2. $y=\frac{1}{a} x+\frac{b}{a}$.
The line $y=a x+b$ intersects the $x$-axis and $y$-axis at points $A_{1}\left(-\frac{b}{a}, 0\right)$ and $B_{1}(0, b)$, respectively. The points $A_{1}$ and $B_{1}$, when reflected over the line $y=-x$, become $A_{2}\left(0, \frac{b}{a}\right)$ and $B_{2}(-b, 0)$. Using the method of... | y=\frac{1}{a} x+\frac{b}{a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,568 |
3. Regardless of the value of $m$, the vertex of the parabola
$$
y=x^{2}+2 m x+m^{2}+m-1
$$
always lies on a straight line. Then the function expression of this straight line is $\qquad$ . | 3. $y=-x-1$.
Transform the quadratic function into $y=(x+m)^{2}+m-1$, knowing that the coordinates of the vertex of the parabola are $\left\{\begin{array}{l}x=-m, \\ y=m-1 .\end{array}\right.$ Eliminating $m$, we get $x+y=-1$. | y=-x-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,569 |
4. When $a<0$, the solution to the equation
$$
x|x|+|x|-x-a=0
$$
is $\qquad$ . | 4. $x=-1-\sqrt{1-a}$.
When $a<1$, the solution is $x=\frac{-2 \pm \sqrt{4-4a}}{2}=-1 \pm \sqrt{1-a}$ (take the negative value). | x=-1-\sqrt{1-a} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,570 |
5. A square is paved with regular hexagonal tiles with a side length of $36 \mathrm{~cm}$. Now, a circular disc with a radius of $6 \sqrt{3} \mathrm{~cm}$ is tossed upwards. The probability that the disc, upon landing, does not intersect with the gaps between the tiles is approximately $\qquad$ | 5. $\frac{4}{9}$.
To ensure the circular disc does not overlap the gaps between the tiles, the center of the disc must fall within a smaller regular hexagon that is centered with the tile and has sides parallel to the tile edges, with a distance of $6 \sqrt{3} \mathrm{~cm}$ (as shown in Figure 5). Construct $O C_{1} \... | \frac{4}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,571 |
6. Place small balls of red, white, and yellow colors into red, white, and yellow boxes, with each box containing balls of the same color. It is known that
(1) The number of balls in the yellow box is more than the number of yellow balls;
(2) The number of balls in the red box is different from the number of white ball... | 6. Yellow, Red, White.
From condition (2), we know that the red box does not contain a white ball, and from condition (3), we know that the white box does not contain a white ball, so the yellow box contains the white ball.
Assume the white box contains the yellow ball. From condition (3), we know that the white ball... | Yellow, Red, White | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,572 |
Example 7 As shown in Figure 7, in $\triangle A B C$, $\angle A=60^{\circ}$, the incircle $\odot I$ of $\triangle A B C$ touches sides $A B$ and $A C$ at points $D$ and $E$, respectively. Line $D E$ intersects lines $B I$ and $C I$ at points $F$ and $G$, respectively. Prove:
$$
F G=\frac{1}{2} B C .
$$ | Explanation: Since $\angle B I G=\frac{1}{2}(\angle B+\angle C)$
$$
\begin{array}{l}
\text { Also, } \angle B D G=\angle A D E=\frac{180^{\circ}-\angle A}{2} \\
=\frac{1}{2}(\angle B+\angle C),
\end{array}
$$
Therefore, $B 、 I 、 D 、 G$ are concyclic.
Hence $\angle B G C=\angle B D I=90^{\circ}$.
Similarly, $\angle C F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,573 |
Three, (16 points) Remove the middle digit of a three-digit number $\overline{a b c}$, making it a two-digit number $\overline{a c}$, and satisfy $\overline{a b c}=$ $9 \overline{a c}+4 \bar{c}$ (for example, $155=9 \times 15+4 \times 5$). Try to find all such three-digit numbers. | Three, since $\overline{a b c}=100 a+10 b+c, \overline{a c}=10 a+c$, from the problem we get $100 a+10 b+c=9(10 a+c)+4 c$. Simplifying, we get
$$
5(a+b)=6 c \text {. }
$$
Here $0 \leqslant a, b, c \leqslant 9$, and $a \neq 0$.
Since 5 is a prime number, we have $\left\{\begin{array}{l}c=5, \\ a+b=6 \text {. }\end{arra... | 155,245,335,425,515,605 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,574 |
Four. (16 points) Given the quadratic function
$$
y=a x^{2}+4 a x+4 a-1
$$
the graph of which is $C_{1}$.
(1) Find the function expression of the image $C_{2}$ that is centrally symmetric to $C_{1}$ about the point $R(1,0)$;
(2) Let the intersection points of curves $C_{1}$ and $C_{2}$ with the $y$-axis be $A$ and $B$... | (1) From $y=$ $a(x+2)^{2}-1$, we know that the vertex of parabola $C_{1}$ is $M(-2,-1)$.
From Figure 6, we know that the point $M(-2,-1)$ is centrally symmetric to point $N(4,1)$ with respect to point $R(1,0)$. With $N(4,1)$ as the vertex, the image $C_{2}$, which is centrally symmetric to parabola $C_{1}$ with respec... | a=1 \text{ or } a=-\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,575 |
Five, (17 points) Find all positive integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=$ 2006.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Five, factoring both sides of the equation, we get
$$
(2 x+y)(x+2 y)=2 \times 17 \times 59 \text {. }
$$
Assume without loss of generality that $x \geqslant y \geqslant 1$, then we have
$$
2 x+y \geqslant x+2 y>x+y>1 \text {. }
$$
Thus, there are only three cases:
$$
\left\{\begin{array}{l}
2 x+y=59, \\
x+2 y=34,
\en... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,576 |
Six. (17 points) As shown in Figure 3, given that $AB$ is a chord of $\odot O$, $M$ is the midpoint of $AB$, and $P$ is any point on $\odot O$. A circle is drawn with $P$ as the center and $2MO$ as the radius, intersecting $\odot O$ at points $C$ and $D$. $AC$ and $BD$ intersect at point $Q$. Please answer:
(1) What is... | Six, Analysis: When point $P$ is on the perpendicular bisector $MO$ of chord $AB$, point $Q$ is also on line $MO$. At this time, $PQ \perp AB$, so consider $Q$ as the orthocenter of $\triangle PAB$.
(1) As shown in Figure 7, draw the diameter $BE$ of $\odot O$, and connect $PD$, $DE$, and $EA$.
Since $\angle BAE = 90^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,577 |
2. Given that $f(x)$ is a positive function defined on $\mathbf{R}$, and satisfies $f(x+1) f(x-1)=f(x)$, then it is a periodic function. One period of such a function is ( ).
(A) 2
(B) 3
(C) 4
(D) 6 | 2.D.
Substitute $x-1$ for $x$ in the equation, then we have
$$
f(x+1)=\frac{1}{f(x-2)} \text {. }
$$
Therefore, $f(x+3)=\frac{1}{f(x)}$.
Thus, $f(x+6)=\frac{1}{f(x+3)}=f(x)$.
Note: Here 6 is not necessarily the smallest positive period. For example, the constant function $f(x)=1(x \in \mathbf{R})$ also satisfies the ... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,579 |
3. When three dice are thrown simultaneously, the probability that the sum of the numbers facing up is no more than 7 is ( ).
(A) $\frac{35}{6^{3}}$
(B) $\frac{34}{6^{3}}$
(C) $\frac{31}{6^{3}}$
(D) $\frac{30}{6^{3}}$ | 3.A.
Let the numbers rolled by the three dice be $x, y, z$, then the number of positive integer triples $(x, y, z)$ that satisfy $x+y+z \leqslant 7$ is
$$
C_{6}^{2}+C_{5}^{2}+C_{4}^{2}+C_{3}^{2}+C_{2}^{2}=35 \text { (sets) } \text {. }
$$
Therefore, the probability of the situation described in the problem is $\frac{... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,580 |
4. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B$ $=2, A A_{1}=A D=1$, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. Then, the volume of the tetrahedron $B_{1} E F G$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{3}{8}$
(C) $\frac{1}{2}$
(D) $\frac{5}{8}... | 4.B.
Take a point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{4}$. It is easy to prove that $H E / / B_{1} G, H E / /$ plane $B_{1} F G$. Therefore,
$$
V_{B_{1}-E F C}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H} .
$$
And $S_{\triangle B_{1} F H}=\frac{9}{8}$, the distance from point $G$ to p... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,581 |
5. Let $n (n<150)$ be a positive integer, and $n^{3}+23$ is divisible by 24. How many such $n$ are there?
(A)6
(B) 7
(C) 11
(D) 12 | 5.B.
From the given, we have $24 \mid\left(n^{3}-1\right)$. Therefore,
$$
\begin{array}{l}
n^{3}-1=(n-1)\left(n^{2}+n+1\right) \\
=(n-1)[n(n+1)+1] .
\end{array}
$$
Since $n(n+1)+1$ is odd, it must be that $2^{3} \mid (n-1)$.
If $3 \mid (n-1)$, then $3 \mid n(n+1)$.
Thus, $3 \mid [n(n+1)+1]$.
Therefore, it must be tha... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,582 |
6. Given positive integers $x_{1}, x_{2}, \cdots, x_{9}$ satisfying $x_{1}<x_{2}$ $<\cdots<x_{9}$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{9}^{2} \leqslant 675$. Then the maximum value of $x_{7}-$ $x_{4}$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 6. C.
Optimal solution:
$$
\begin{array}{l}
x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=5, x_{6}=6, \\
x_{9}=15, x_{8}=14, x_{7}=13 .
\end{array}
$$
However, at this point $x_{1}^{2}+x_{2}^{2}+\cdots+x_{9}^{2}=681$ (which does not meet the requirement), so $x_{7}-x_{4}<9$.
And $1^{2}+2^{2}+\cdots+6^{2}+12^{2}+14^{2}+1... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,583 |
Example 8 As shown in Fig. $8(\mathrm{a}), \odot O$ is the circumcircle of $\triangle A B C$, $A M$ and $A T$ are the median and angle bisector of $\triangle A B C$ respectively, the tangents to $\odot O$ at points $B$ and $C$ intersect at point $P$, connect $A P$, which intersects $B C$ and $\odot O$ at points $D$ and... | Explanation: First, prove that $A T$ is the angle bisector of $\angle M A E$, which means proving $\angle B A M=\angle C A P$.
As shown in Figure 8(a), draw $C F \perp A B$, with the foot of the perpendicular at $F$, and connect $M F$. Then $F M=\frac{1}{2} B C=M C$.
Since $\angle B A C=\angle B C P$, we have,
$$
\beg... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,584 |
7. Given $\sin ^{2}\left(x+\frac{\pi}{8}\right)-\cos ^{2}\left(x-\frac{\pi}{8}\right)=$ $\frac{1}{4}$, and $x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value of $\tan x$ is $\qquad$ . | $$
=7 . \frac{2 \sqrt{14}+\sqrt{7}}{7} \text {. }
$$
From the problem, we know
$$
\frac{1}{2}\left[1-\cos \left(2 x+\frac{\pi}{4}\right)\right]-\frac{1}{2}\left[1+\cos \left(2 x-\frac{\pi}{4}\right)\right]=\frac{1}{4} \text {, }
$$
which means $\cos \left(2 x+\frac{\pi}{4}\right)+\cos \left(2 x-\frac{\pi}{4}\right)=-... | \frac{2 \sqrt{14}+\sqrt{7}}{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,585 |
8. In a regular quadrilateral pyramid $S-A B C D$, the angle between a lateral face and the base is $\frac{\pi}{3}$. Then the ratio of the radius $R$ of the circumscribed sphere to the radius $r$ of the inscribed sphere $\left(\frac{R}{r}\right)$ is $\qquad$ . | 8. $\frac{5}{2}$.
Let the side length of the base be $a$, then $r=\frac{a}{2} \tan \frac{\pi}{6}$. Draw $\mathrm{SO}_{1} \perp$ plane $A B C D$, with $O_{1}$ as the foot of the perpendicular. Let $S O_{1}=h$, and the center of the circumscribed sphere be $O$.
In the right triangle $\triangle O O_{1} C$, we have
$(h-R)... | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,586 |
9. If the solution set of the inequality with respect to $x$
$$
a x^{2}-|x+1|+2 a<0
$$
is an empty set, then the range of values for $a$ is $\qquad$. | 9. $\left[\frac{\sqrt{3}+1}{4},+\infty\right)$.
From the problem, for all real numbers $x$, we have
$$
a x^{2}-|x+1|+2 a \geqslant 0 \text {, }
$$
which means $a \geqslant f(x)=\frac{|x+1|}{x^{2}+2}$.
Therefore, $a \geqslant \max f(x)$.
Let $t=x+1$, then
$$
f(x)=\frac{|t|}{(t-1)^{2}+2}=\frac{|t|}{t^{2}-2 t+3} \text {... | \left[\frac{\sqrt{3}+1}{4},+\infty\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,587 |
10. Let $a_{n}>0, a_{1}=2$, and for $n \geqslant 2$, we have
$$
a_{n}+a_{n-1}=\frac{n}{a_{n}-a_{n-1}}+2 \text {. }
$$
Then the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ | 10. $\sqrt{\frac{n(n+1)}{2}}+1$.
Since $a_{n}+a_{n-1}=\frac{n}{a_{n}-a_{n-1}}+2$, we have
$$
a_{n}^{2}-a_{n-1}^{2}=2\left(a_{n}-a_{n-1}\right)+n .
$$
Therefore, $\left(a_{n}-1\right)^{2}-\left(a_{n-1}-1\right)^{2}=n$.
Hence, $\left(a_{n}-1\right)^{2}$
$$
\begin{array}{l}
=\sum_{k=2}^{n}\left[\left(a_{k}-1\right)^{2}-... | a_{n}=\sqrt{\frac{n(n+1)}{2}}+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,588 |
11. Given that $A$ and $B$ are two moving points on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $O$ is the origin, and $O A \perp O B$. Then the minimum value of $S_{\triangle A O B}$ is | 11. $\frac{36}{13}$.
Let $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right), B\left(-r_{2} \sin \theta, r_{2} \cos \theta\right)$. Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\begin{array}{l}
\frac{r_{1}^{2} \cos ^{2} \theta}{9}+\frac{r_{1}^{2} \sin ^{2} \theta}{4}=1, \\
\f... | \frac{36}{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,589 |
13. Let \( f(x, y) = \frac{a x^{2} + x y + y^{2}}{x^{2} + y^{2}} \), satisfying
\[
\max _{x^{2}+y^{2}+0} f(x, y) - \min _{x^{2}+y^{2}+0} f(x, y) = 2 \text{. }
\]
Find \( a \). | $$
\begin{array}{l}
=\max _{0 \leqslant \theta<2 \pi}\left(\frac{a-1}{2} \cos 2 \theta+\frac{1}{2} \sin 2 \theta+\frac{a+1}{2}\right) \\
=\max _{0 \leqslant \theta<2 \pi}\left(\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}} \sin (2 \theta+a)+\frac{a+1}{2}\right) \\
=\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}}+\... | a=1 \pm \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,591 |
14. Let $A=\{1,2, \cdots, 10\}, f$ be a one-to-one mapping from $A$ to $A$. Define
$$
f_{1}(x)=f(x), f_{n+1}(x)=f\left(f_{n}(x)\right) \text {. }
$$
The mapping $f$ also satisfies the conditions:
(1) For all $x \in A, f(x) \neq x$;
(2) For all $x \in A, f_{21}(x)=x$.
How many mappings $f$ satisfy the above conditions... | 14. Since $f$ is a one-to-one mapping from $A$ to $A$, and $A$ is a finite set, for every $x \in A$, there is a positive integer $k(1 \leqslant k \leqslant 10)$ such that $f_{k}(x)=x$.
From condition (1), it follows that for any $x \in A$, $k \neq 1$;
From condition (2), it follows that for $x \in A$, $k=7$ or 3.
Since... | 14400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,592 |
15. A football team participates in the World Cup, and in a certain match, 11 players start the game, using a 4-4-2 formation (4 defenders, 4 midfielders, 2 forwards). There are 7 substitute players (2 defenders, 2 midfielders, 2 forwards, and 1 goalkeeper). The rules allow for a maximum of three substitutions. If the ... | 15. (1) Let the total number of substitutions in the match be $t$, where $t=0,1,2,3$.
When $t=0$, 11 players play the entire match, which is only 1 way.
When $t=1$, the substituted player can be a midfielder or a defender, resulting in $\mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} + \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{1} = 16... | 189 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,593 |
1. Let $a>0$, and the function $f:(0,+\infty) \rightarrow \mathbf{R}$ satisfies $f(a)=1$. If for any positive real numbers $x, y$, we have
$$
f(x) f(y) + f\left(\frac{a}{x}\right) f\left(\frac{a}{y}\right) = 2 f(x y),
$$
prove that $f(x)$ is a constant.
(Zhu Huawei) | 1. In equation (1), let $x=y=1$, we get
$$
f^{2}(1)+f^{2}(a)=2 f(1),(f(1)-1)^{2}=0 \text {. }
$$
Then $f(1)=1$.
In equation (1), let $y=1$, we get
$$
\begin{array}{l}
f(x) f(1)+f\left(\frac{a}{x}\right) f(a)=2 f(x), \\
f(x)=f\left(\frac{a}{x}\right)(x>0) .
\end{array}
$$
In equation (1), take $y=\frac{a}{x}$, we get
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,594 |
2. Let the diagonals of convex quadrilateral $ABCD$ intersect at point $O$. The circumcircles of $\triangle OAD$ and $\triangle OBC$ intersect at points $O$ and $M$. Line $OM$ intersects the circumcircles of $\triangle OAB$ and $\triangle OCD$ at points $T$ and $S$, respectively. Prove that $M$ is the midpoint of segme... | Proof 1: As shown in Figure 1, connect $B T, C S, M A, M B, M C, M D$.
Since $\angle B T O = \angle B A O, \angle B C O = \angle B M O$, we have
$\triangle B T M \sim \triangle B A C$, thus
$$
\frac{T M}{A C} = \frac{B M}{B C}.
$$
Similarly, $\triangle C M S \sim \triangle C B D$, thus
$$
\frac{M S}{B D} = \frac{C M}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,596 |
3. Prove that for $i=1,2,3$, there are infinitely many positive integers $n$ such that among $n, n+2, n+28$, exactly $i$ of them can be expressed as the sum of three positive cubes. | 3. The sum of the cubes of three integers cannot have a remainder of 4 or 5 when divided by 9, because integers can be written as $3k$ or $3k \pm 1 (k \in \mathbf{Z})$, and
$$
(3k)^{3}=9 \times 3k^{3}, (3k \pm 1)^{3}=9(3k^{3} \pm 3k^{2}+k) \pm 1.
$$
For $i=1$, let $n=3(3m-1)^{3}-2 (m \in \mathbf{Z}_{+})$, then the rem... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,597 |
4.8 Eight people attend a party.
(1) If among any 5 of them, there are 3 people who all know each other, prove that it is possible to find 4 people who all know each other;
(2) Ask: If among any 6 of them, there are 3 people who all know each other, can we definitely find 4 people who all know each other?
(Su Chun prov... | 4. (1) Case analysis:
Case 1: If there exist 3 people who are all mutually unfamiliar, then the remaining 5 people must all be mutually familiar. Otherwise, if there are 2 people among them who are unfamiliar, then in the group of 5 people including these 2 and the original 3, it would be impossible to find 3 people w... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,598 |
5. The set of integer points on the plane
$$
S=\{(a, b) \mid 1 \leqslant a, b \leqslant 5(a, b \in \mathbf{Z})\},
$$
$T$ is a set of integer points on the plane, such that for any point $P$ in $S$, there exists a point $Q$ in $T$ different from $P$, such that the line segment $P Q$ contains no other integer points exce... | 5. The minimum number is 2.
First, we prove that $T$ cannot consist of only one point.
Otherwise, suppose
$$
T=\left\{Q\left(x_{0}, y_{0}\right)\right\}.
$$
In $S$, take a point $P\left(x_{1}, y_{1}\right)$ such that $\left(x_{1}, y_{1}\right) \neq \left(x_{0}, y_{0}\right)$, and $x_{1}$ has the same parity as $x_{0}... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,599 |
6. Let the set
$$
\begin{array}{l}
M=\{1,2, \cdots, 19\}, \\
A=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\} \subseteq M .
\end{array}
$$
Find the smallest $k$, such that for any $b \in M$, there exist $a_{i}, a_{j} \in A$, satisfying $b=a_{i}$ or $b=a_{i} \pm a_{j}\left(a_{i}, a_{j}\right.$ can be the same).
(Supplied b... | 6. By the hypothesis, in $A$, there are $k(k+1)$ possible combinations. Thus, $k(k+1) \geqslant 19$, i.e., $k \geqslant 4$.
When $k=4$, we have $k(k+1)=20$. Suppose $a_{1}<a_{2}<a_{3}<a_{4}$. Then $a_{4} \geqslant 10$.
(1) When $a_{4}=10$, we have $a_{3}=9$. At this time, $a_{2}=8$ or 7.
If $a_{2}=8$, then $20,10-9=1,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,600 |
7. Let $x_{i}>0(i=1,2, \cdots, n), k \geqslant 1$. Prove:
$$
\sum_{i=1}^{n} \frac{1}{1+x_{i}} \cdot \sum_{i=1}^{n} x_{i} \leqslant \sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}} \cdot \sum_{i=1}^{n} \frac{1}{x_{i}^{k}} \text {. }
$$ | Proof 1: The original inequality is equivalent to
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}} \cdot \sum_{i=1}^{n} \frac{1}{x_{i}^{k}}-\sum_{i=1}^{n} \frac{1}{1+x_{i}} \cdot \sum_{i=1}^{n} x_{i} \geqslant 0 . \\
\text { The left side of the above equation }=\sum_{i \neq j} \frac{x_{i}^{k+1}}{1+x_{i}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,601 |
8. Let $p$ be a prime number greater than 3. Prove that there exist several integers $a_{1}, a_{2}, \cdots, a_{i}$ satisfying the condition
$$
-\frac{p}{2}<a_{1}<a_{2}<\cdots<a_{i}<\frac{p}{2},
$$
such that the product
$$
\frac{p-a_{1}}{\left|a_{1}\right|} \cdot \frac{p-a_{2}}{\left|a_{2}\right|} \cdots \cdots \frac{p... | 8. Proof 1: By the division algorithm, there exist unique integers $q, r$ such that $p=3q+r (00)$.
Let $b_{i}, b_{i+1}, \cdots, b_{j-1}$ be arranged in ascending order, then the original proposition is proved.
Proof 2: Consider two cases.
(1) The case $p=6k+1$:
$$
\begin{array}{l}
\frac{p-1}{1} \cdot \frac{p+2}{2} \cdo... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,602 |
1. Given that $m$ and $n$ are constants, the sum of the three monomials $4 x^{2} y$, $m x^{3-n^{2}} y$, and $8 x^{3} y$ is still a monomial. Then the number of possible values for $m+n$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4 | -、1.C.
Since $4 x^{2} y$ and $8 x^{3} y$ are not like terms, therefore, the sum of $4 x^{2} y$ and $m x^{3-n^{2}} y$ is 0 or the sum of $m x^{3-n^{2}} y$ and $8 x^{3} y$ is 0.
Thus, we have $\left\{\begin{array}{l}m=-4, \\ 3-n^{2}=2\end{array}\right.$ or $\left\{\begin{array}{l}m=-8, \\ 3-n^{2}=3 .\end{array}\right.$
S... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,603 |
2. Given in a convex quadrilateral $A B C D$, $A B=5 \sqrt{2}$, $B C=\sqrt{13}$, $C D=3 \sqrt{13}$, $A D=7$, $\angle B C D=$ $120^{\circ}$. Then the degree of $\angle B A D$ is $(\quad)$.
(A) $120^{\circ}$
(B) $105^{\circ}$
(C) $135^{\circ}$
(D) $150^{\circ}$ | 2.C.
As shown in Figure 5, draw $B E \perp$ $A D$ and $B F \perp C D$ at points $E$ and $F$ respectively. Connect $B D$.
Let $B E=x$,
$$
A E=y \text{.}
$$
Since $\angle B C D=120^{\circ}$, we have
$$
\angle B C F=180^{\circ}-\angle B C D=60^{\circ} \text{.}
$$
In the right triangle $\triangle B C F$, we have
$$
\beg... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,604 |
3. Given real numbers $a, b$ are constants, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
x+a^{2} \leqslant 2 a, \\
x+6 \geqslant \sqrt{b}
\end{array}\right.
$$
has only 8 integer solutions. Then the value of $a+b$ is ( ).
(A) 1
(B) 0
(C) -1
(D) 2 | 3.A.
The solution set of the system of inequalities is $\sqrt{b}-6 \leqslant x \leqslant 2 a-a^{2}$. From the problem, we have $\left(2 a-a^{2}\right)-(\sqrt{b}-6)+1 \geqslant 8$, which simplifies to $(a-1)^{2}+\sqrt{b} \leqslant 0$.
Solving this, we get $a=1, b=0$.
It is easy to see that quadrilateral $B C D F$ is a ... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,605 |
2. If a straight line bisects both the area and the perimeter of a triangle, then the line must pass through the ( ) of the triangle.
(A) incenter
(B) circumcenter
(C) centroid
(D) orthocenter | (提示: Let line $D E$ bisect the perimeter and area of $\triangle A B C$, with $D 、 E$ on sides $A B 、 A C$ respectively. Draw the angle bisector of $\angle A$ intersecting $D E$ at point $P$, and let the distances from point $P$ to $A B 、 A C$ be $r$, and the distance from point $P$ to $B C$ be $r_{1}$. From
$$
\left\{\... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,606 |
4. As shown in Figure 1, in the right trapezoid $A B C D$, $A D / / B C$, $C D \perp A D, E$ is a point on $C D$, $\angle A B E=45^{\circ}, B C=C D$ $=6, A E=5$. Then the value of $\tan \angle A E B$ is ( ).
(A) 1.5
(B) 2
(C) 3
(D) None of the above | 4.D.
As shown in Figure 6, draw $B F \perp A D$ at point $F$, extend $A F$ to point $G$ such that $F G=C E$, and connect $B G$.
Therefore, $\triangle B F G \cong \triangle B C E$.
We have $\angle A B G=\angle A B E=45^{\circ}, B E=B G$.
Also, $A B=A B$, so $\triangle A B G \cong \triangle A B E$, which gives
$$
A G=A ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,607 |
5. Given that $a$ and $b$ are real numbers, and satisfy
$$
\sqrt{a+1}+\sqrt{a^{2} b-4 a^{2}}+|6-2 b|=2 \text {. }
$$
Then the number of real number pairs $(a, b)$ that satisfy the condition is ( ).
(A) 4
(B) 3
(C) 2
(D) 1 | 5. B.
(1) When $a=0$, the original equation becomes
$$
1+|6-2 b|=2 \text {. }
$$
Solving, we get $b=\frac{5}{2}$ or $\frac{7}{2}$.
At this point, $(a, b)$ is $\left(0, \frac{5}{2}\right)$ or $\left(0, \frac{7}{2}\right)$.
(2) When $a \neq 0$, $a^{2}>0$.
From the problem, we know $a^{2} b-4 a^{2} \geqslant 0$.
Solving... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,608 |
6. Let the lengths of the three sides of a triangle be $a$, $b$, and $c$ satisfying the system of equations
$$
\left\{\begin{array}{l}
2 a b+c^{2}-a=91, \\
2 b c-a^{2}+b=24, \\
2 a c-b^{2}-c=61 .
\end{array}\right.
$$
Then the value of $a+b-c$ is ( ).
(A) -2 or 4
(B) 3
(C) -2 or 3
(D) 4 | 6. B.
From (1)-(2) - (3), we get
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 a c-a-b+c=6 \text {, } \\
\text { i.e., }(a+b-c)^{2}-(a+b-c)-6=0 \text {. } \\
\end{array}
$$
Solving, we get $a+b-c=3$ or -2.
Since $a, b, c$ are the lengths of the three sides of a triangle, then
$$
a+b-c>0 \text {. }
$$
Therefore... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,609 |
1. Given in the Cartesian coordinate system, $O(0,0)$, $A(6,0)$, $\angle O B A=60^{\circ}$. When the area of $\triangle O A B$ is maximized, the coordinates of point $B$ are $\qquad$ | 2. $(3,3 \sqrt{3})$ or $(3,-3 \sqrt{3})$.
Since $O$ and $A$ are two fixed points, $\angle O B A=60^{\circ}$, therefore, point $B$ lies on the two major arcs with $O A$ as the chord, and the central angle subtended by chord $O A$ is $120^{\circ}$ (as shown in Figure 7).
Draw $O_{1} C \perp O A$ at point $C$, intersecti... | (3,3 \sqrt{3}) \text{ or } (3,-3 \sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,610 |
2. Given in $\triangle A B C$, $\angle A: \angle B: \angle C=1$ : $2: 5$. Then $\frac{A B}{B C}=$ $\qquad$ . | $2.1+\sqrt{2}$.
Since $\angle A: \angle B: \angle C=1: 2: 5$, and $\angle A+\angle B+\angle C=180^{\circ}$, therefore, $\angle A=22.5^{\circ}, \angle B=45^{\circ}$.
As shown in Figure 8, draw $C D \perp A B$ at point $D$, and intercept $D E=B D$ on $A D$, connect $C E$.
We have $B D=C D$, then
$$
B C=\sqrt{2} B D \tex... | 1+\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,611 |
3. Given that the equation with respect to $x$
$$
|1-| x||=m-2008
$$
has exactly two real solutions for $x$. Then the range of real values for $m$ is | 3. $m=2008$ or $m>2009$.
Let $y=|1-| x||=m-2008$.
Sketch the graph of the function $y=$
$|1-| x||$
(see Figure 9).
When $y=0$ or $y$
$>1$, the line $y=$
$m-2008$ intersects the broken line
$y=|1-| x||$ at exactly two points, meaning there are exactly two real numbers $x$ that satisfy the condition.
Thus, $m-2008=0$ o... | m=2008 \text{ or } m>2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,612 |
4. As shown in Figure 2, it is known that ray $O M$ and ray $O N$ are perpendicular to each other, $A$ is a point on a semicircular iron sheet with diameter $P Q$ of $6 \mathrm{~cm}$, and the degree of $\overparen{A Q}$ is $60^{\circ}$. A moving point $P$ starts sliding from point $O$ along ray $O M$, while point $Q$ s... | 4. $(9-3 \sqrt{3}) \mathrm{cm}$.
In $\angle M O N$, draw ray $O B$ through point $O$ such that $\angle B O N = 30^{\circ}$. Connect $A P$ and $A Q$.
Since $P Q$ is the diameter of the semicircle, and $\overparen{A Q^{\circ}} = 60^{\circ}$, we have $\angle P A Q = 90^{\circ}$ and $\angle A P Q = 30^{\circ}$.
Therefore,... | 9 - 3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,613 |
One. (20 points) Place a rectangular iron block of size $a \mathrm{~cm} \times b \mathrm{~cm} \times c \mathrm{~cm}$ (where $a<b<c$) (as shown in Figure 3(a)) into a rectangular water tank (as shown in Figure 3(b)). The iron block does not touch the four walls of the water tank. Now, water is uniformly injected into th... | (i) From the three images, the maximum value of $h$ is 10, so the depth of the water tank is $10 \, \text{cm}$.
From Figure 3(c), after $21 \, \text{s}$ of water injection, the water depth is $6 \, \text{cm}$. At this point, the water surface is at the same level as the top surface of the iron block or just covers the... | 240 \, \text{cm}^2, 30 \, \text{cm}^3/\text{s}, 15, 53 \, \text{s}, 53 \, \text{s} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,614 |
II. (25 points) As shown in Figure 4, given that $E$ is the midpoint of side $AB$ of quadrilateral $ABCD$, $BC=CD=CE$, $\angle B=75^{\circ}$, $\angle D=90^{\circ}$. Find the degree measure of $\angle DAB$. | As shown in Figure 11, draw $E F \perp B C$ at point $F$, and draw $A G \perp B C$ at point $G$. Let $B C = C D = C E = 2 m$.
Since $A G \perp B C$ and $E F \perp B C$,
we have $A G \parallel E F$, and $\angle A G B = 90^{\circ}$.
Also, since $E$ is the midpoint of $A B$, we know that $F$ is the midpoint of $B G$, thus... | 105^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,615 |
Three, (25 points) Find all integer pairs $(x, y)$ such that
$$
\begin{array}{l}
x^{3}-x^{2} y+x y^{2}-y^{3} \\
=4 x^{2}-4 x y+4 y^{2}+47 .
\end{array}
$$ | $$
m-2= \pm 1, \pm 5, \pm 11, \pm 55 .
$$
Thus, $m=3,1,7,-3,13,-9,57,-53$.
(1) When $m=3$, from equation (2) we get $n=28$.
Solving simultaneously, we get $(x, y)=(7,4),(-4,-7)$.
(2) When $m=7$, from equation (2) we get $n=-10$.
Solving simultaneously, we get $(x, y)=(5,-2),(2,-5)$.
(3) When $m=1,-3,13,-9,57,-53$, f... | (x, y)=(7,4),(-4,-7),(5,-2),(2,-5) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,616 |
3. As shown in Figure $9, A B$ is the diameter of a semicircle, and $C$ is a point on the semicircular arc. One side $D G$ of the square $D E F G$ lies on the diameter $A B$,
and the other side $D E$ passes through the incenter $I$ of $\triangle A B C$, with point
$E$ on the semicircular arc. If the area of the squar... | (Given: $A D=\frac{b+c-a}{2}, B D=\frac{c+a-b}{2}$. From $D E^{2}$ $=A D \cdot D B$, we get $100=\frac{b+c-a}{2} \cdot \frac{c+a-b}{2}=\frac{a b}{2}$, hence $\left.S_{\triangle B B C}=100.\right)$ | 100 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,617 |
1. Given point $P(1,2)$ is inside (including the boundary) the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and outside (including the boundary) the circle $x^{2}+y^{2}=\frac{a^{2}+2 b^{2}}{3}$. If $a, b \in \mathbf{R}_{+}$. Then the minimum value of $a+b$ is ( ).
(A) $\sqrt{5}+2$
(B) $\sqrt{6}+\sqrt{3}$
(C) $2... | - 1.B.
From the given, we have $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1$ and $a^{2}+2 b^{2} \leqslant 15$. Therefore,
$$
\frac{b^{2}}{b^{2}-4} \leqslant a^{2} \leqslant 15-2 b^{2} \text {. }
$$
Simplifying, we get $\left(b^{2}-5\right)\left(b^{2}-6\right) \leqslant 0$.
Thus, $5 \leqslant b^{2} \leqslant 6$.
Hence... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,618 |
2. Divide the set of positive odd numbers in ascending order into groups, with the $n$-th group containing $n^{2}$ odd numbers. Then 2005 is in the ( )-th group.
(A) 14
(B) 15
(C) 16
(D) 17 | 2. A.
Consider the sequence of groups formed by adding 1 to each odd number and then dividing by 2:
$$
\{1\},\{2,3,4,5\}, \cdots \text {. }
$$
Thus, 2005 corresponds to 1003 at this point, so we only need to consider the group where 1003 is located.
Assume 1003 is in the $n$-th group, then we have
$$
\begin{array}{l}... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,619 |
3. Given that $\alpha, \beta$ satisfy
$$
\csc (\alpha-2 \beta), \csc \alpha, \csc (\alpha+2 \beta)
$$
form an arithmetic sequence with a non-zero common difference. Then the value of $\frac{\sin \alpha}{\cos \beta}$ could be ( ).
(A) $\pm 1$
(B) $\pm \sqrt{2}$
(C) $\pm \sqrt{3}$
(D) None of the above | 3. B.
From the given, we have $\frac{2}{\sin \alpha}=\frac{1}{\sin (\alpha-2 \beta)}+\frac{1}{\sin (\alpha+2 \beta)}$.
Thus, $2 \sin (\alpha-2 \beta) \cdot \sin (\alpha+2 \beta)$
$$
=[\sin (\alpha-2 \beta)+\sin (\alpha+2 \beta)] \sin \alpha \text {, }
$$
which means $2 \sin (\alpha-2 \beta) \cdot \sin (\alpha+2 \beta... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,620 |
4. Given $2 n$ positive real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, b_{1}$ $\geqslant b_{2} \geqslant \cdots \geqslant b_{n}$ and $0 \leqslant \lambda \leqslant 2$, let
$$
\begin{array}{l}
M=\sum_{i=1}^{n} \sqrt{a_{i}^{2}+b_{i}^{2}-\lambda a_{i} b_{i}}, \\
N=\sum_{i=1}^{n} \sqrt{a_{i+1}^{2}+b_... | 4. B.
Construct $\angle A O B$, with the included angle $\theta=\arccos \frac{\lambda}{2} \in[0, \pi]$.
As shown in Figure 1, take $n$ points $A_{1}$,
$A_{2}, \cdots, A_{n}$ and $B_{1}$,
$B_{2}, \cdots, B_{n}$ on the rays
$O A$ and $O B$ respectively, such that
$O A_{i}=a_{i}, O B_{i}=$
$b_{i}$. By the cosine rule,
we... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 717,621 |
5. In space, there are 5 points, no 4 of which are coplanar. If several line segments are drawn and there is no tetrahedron in the graph, then the graph can have at most ( ) triangles.
(A) 3
(B) 4
(C) 5
(D) 6 | 5.B.
First, construct Graph 2, which is easily known to meet the conditions and has exactly 4 triangles.
If there exists a situation where the number of triangles is no less than 5, we will prove that this leads to a contradiction.
If only two line segments are not connected, then these two unconnected line segments... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,622 |
6. The sequence $\left\{a_{n}\right\}$ is called a difference-ratio sequence if and only if the sequence satisfies $a_{0}=0,\left\{a_{n+1}-q a_{n}\right\}$ forms a geometric sequence with a common ratio of $q$. $q$ is called the difference-ratio of this difference-ratio sequence. Then, when the difference-ratio is grea... | 6.C.
Let the $n+1$ terms of the geometric sequence composed of natural numbers within 100 be
$$
a_{k}=k a_{1} q^{k-1} \quad(k=0,1,2, \cdots, n) \text { and } q>1 \text {. }
$$
From $\frac{a_{k+1}}{a_{k}}=\frac{k+1}{k} \cdot q$, we know that $q$ is a positive rational number, so we can let $q=$ $\frac{p}{r}$ (where $p... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,623 |
1. Simplify $\arctan \sqrt{5}-\frac{1}{2} \arcsin \frac{2}{3}=$ | $=1 \cdot \frac{\pi}{4}$.
Let $z_{1}=1+\sqrt{5} \mathrm{i}, z_{2}=-\sqrt{5}+2 \mathrm{i}$, then we have $\arg z_{1}+\frac{1}{2} \arg z_{2}=\frac{1}{2} \arg \left(z_{1}^{2} z_{2}\right)$
$$
\begin{array}{l}
=\frac{1}{2} \arg [(-4+2 \sqrt{5} \mathrm{i})(-\sqrt{5}+2 \mathrm{i})] \\
=\frac{1}{2} \arg (-18 \mathrm{i})=\frac... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,624 |
2. Let $c(X)$ denote the number of subsets of set $X$. If $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ with different numbers of elements satisfy:
$$
\left(\left|A_{1}\right|+1\right)\left(\left|A_{2}\right|+1\right) \cdots\left(\left|A_{n}\right|+1\right)=2006
$$
and
$$
\begin{array}{l}
c\left(A_{1}\right)+c\left(A_{2}\rig... | 2.58.
Assume $\left|A_{1}\right|>\left|A_{2}\right|>\cdots>\left|A_{n}\right| \geqslant 0$.
When $n \geqslant 3$, by $c(X)=2^{|X|}$ and the condition, we know that $2^{\left|A_{1}\right|}+2^{\left|A_{2}\right|}+\cdots+2^{\left|A_{n}\right|}=2^{\left|A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right|}+2^{\left|A_{1} \cap A... | 58 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,625 |
3. Given that the ellipse $C_{1}$ and the hyperbola $C_{2}$ share foci $F_{1}(3,0), F_{2}(-3,0)$, and have coincident minor axes. Then the number of lattice points inside the region enclosed by the intersection points of $C_{1}$ and $C_{2}$ is $\qquad$
Translate the above text into English, please retain the original ... | 3.25.
From the problem, we can let
$$
C_{1}: \frac{x^{2}}{m+9}+\frac{y^{2}}{m}=1, C_{2}: \frac{x^{2}}{9-m}-\frac{y^{2}}{m}=1,
$$
where $0<m<9$.
Let the intersection point be $P\left(x_{0}, y_{0}\right)$. Then, since point $P$ lies on $C_{1}$ and $C_{2}$, we have $\frac{x_{0}^{2}}{m+9}+\frac{x_{0}^{2}}{9-m}=2$, which ... | 25 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 717,626 |
4. Given a unit cube $A B C D-E F G H$ with moving points $Q 、 P$ on edge $A D$ and line $B C$ respectively. If the line segment $M N$ obtained by the intersection of $\triangle P Q G$ and $\triangle B D E$ has a length of $y$, and $A Q$ $=x(0 \leqslant x \leqslant 1)$, then the minimum value of $y$ expressed as a func... | 4. $y=\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3-x}$.
When $A Q=x$, let $G Q$ intersect the plane $B D E$ at point $N$, draw $N M$ $\perp B D$ at point $M$, connect $Q M$ to intersect line $B C$ at point $P^{\prime}$, and take point $P^{\prime}$ as point $P$. It is known that at this time $y=|M N|$ is minimized.
Establish ... | y=\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3-x} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,627 |
For example, if $1 A E$ is the angle bisector of $\triangle A B C$, and the circumcircle of $\triangle A B E$ intersects $A C$ at point $D$. Prove:
$$
A B \cdot C D=B E^{2}+B E \cdot E C .
$$ | Proof: As shown in Figure 1, connect $B D$. Since $A E$ bisects $\angle B A C$, we have
$$
\frac{B E}{A B}=\frac{C E}{C A} \text {. }
$$
Obviously, $C E \cdot C B$
$$
\begin{array}{l}
=C D \cdot C A \text {, that is } \\
\qquad \frac{C D}{C B}=\frac{C E}{C A} .
\end{array}
$$
Therefore, $\frac{C D}{C B}=\frac{B E}{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,628 |
Example 2 As shown in Figure 2, draw the radii $O A$ and $O B$ of $\odot O$, $E$ is the projection of point $B$ on line $O A$, $P$ is the projection of point $E$ on line $A B$, and $R$ is the radius of $\odot O$. Prove:
$$
O P^{2}+E P^{2}=R^{2} \text {. }
$$ | Proof: As shown in Figure 2, let the line $O P$ intersect $\odot O$ at points $F$ and $G$. Clearly,
$$
\begin{array}{l}
E P^{2}=P A \cdot P B=P G \cdot P F \\
=(O G-O P)(O F+O P)=R^{2}-O P^{2} .
\end{array}
$$
Therefore, $O P^{2}+E P^{2}=R^{2}$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,629 |
Example 1 Given that $a, b$ are positive real numbers, the equations
$$
x^{2}+a x+2 b=0 \text{ and } x^{2}+2 b x+a=0
$$
both have real roots. Find the minimum value of $a^{2}+b^{2}$. | Explanation: Since the given equations all have real roots, we obtain the constraint conditions
$$
\left\{\begin{array}{l}
a^{2}-8 b \geqslant 0 \\
b^{2}-a \geqslant 0 \\
a>0, b>0
\end{array}\right.
$$
To find the minimum value of $a^{2}+b^{2}$, using algebraic methods is clearly challenging, so we consider a numerica... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,630 |
5. As shown in Figure 1, in the convex quadrilateral $A B C D$, $A B=4, B C$ $=3, C D=\frac{5}{2}$, and $\angle A D C=$ $\angle A B C=90^{\circ}$. Then $B C \cdot A D$ equals ( ).
(A) $3 \sqrt{3}+\frac{25}{4}$
(B) $3 \sqrt{3}+\frac{27}{4}$
(C) $3 \sqrt{3}+8$
(D) $3 \sqrt{3}+\frac{29}{4}$ | 5.B.
As shown in Figure 4, by the Pythagorean theorem, we have
$$
\begin{array}{l}
A C=\sqrt{4^{2}+3^{2}}=5 \\
=2 \times \frac{5}{2}=2 C D,
\end{array}
$$
and $\angle A D C=90^{\circ}$, thus
$$
\angle C A D=30^{\circ} \text {. }
$$
Since $\angle A D C=\angle A B C$
$=90^{\circ}$, therefore, points $A, B, C, D$ are c... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,631 |
6. As shown in Figure 2, let $A-BCD$ be a regular tetrahedron (the base $BCD$ is an equilateral triangle), and construct $AO \perp$ the base $BCD$, with $O$ being the foot of the perpendicular. $P$ is a point on the height $AO$ such that $PA = \frac{1}{m} AO (m > 1)$. A plane through point $P$ parallel to the base $BCD... | 6.D.
Let $A O=a$, then
$$
A P=\frac{a}{m}, V_{A-B_{1}} c_{1} D_{1}=\frac{1}{m^{3}} V_{A-B C D} .
$$
Thus, $V_{B C D-B_{1} c_{1} D_{1}}=\left(1-\frac{1}{m^{3}}\right) V_{A-B C D}$.
Suppose a plane parallel to the base $B C D$ is drawn through point $Q$, intersecting the three edges $A B, A C, A D$ at points $B_{2}, C_... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,632 |
1. If $\theta$ is an acute angle, and
$$
\cos \theta \cdot \cos \left(\theta-\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{10},
$$
then the value of $\sin \theta$ is $\qquad$ . | $=1 . \frac{2 \sqrt{5}}{5}$.
From the problem, we have
$$
\cos \theta\left(\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta\right)=\frac{3 \sqrt{2}}{10},
$$
which simplifies to $\cos ^{2} \theta+\cos \theta \cdot \sin \theta=\frac{3}{5}$.
Thus, $5 \cos ^{2} \theta+5 \cos \theta \cdot \sin \theta=3 \cos ^{... | \frac{2 \sqrt{5}}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,633 |
2. Let $a, b, c \in \mathbf{R}$, and satisfy the system of equations
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+c^{2}-10 a-11=0, \\
a^{2}-b c-4 a-5=0 .
\end{array}\right.
$$
Then the range of values for $a b+b c+c a$ is | $2 \cdot[-40,72]$. From the given, we have
$$
b c=a^{2}-4 a-5, b^{2}+c^{2}=-a^{2}+10 a+11 \text {. }
$$
Then $b+c= \pm \sqrt{(b+c)^{2}}= \pm \sqrt{b^{2}+c^{2}+2 b c}$
$$
= \pm \sqrt{a^{2}+2 a+1}= \pm(a+1) \text {. }
$$
By the relationship between roots and coefficients, $b$ and $c$ are the two real roots of the quadr... | [-40,72] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,634 |
3. Let the real number $a$ be such that the quadratic equation $5 x^{2}-5 a x+66 a-1715=0$ has two integer roots. Then all such $a$ are $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.870.
Let the two integer roots be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have $a=x_{1}+x_{2}$, thus, $a$ is an integer. From the original equation, we get
$$
\begin{array}{l}
a=\frac{5 x^{2}-1715}{5 x-66}=x+13+\frac{x-857}{5 x-66} \in \mathbf{Z} \\
\L... | 870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,635 |
4. As shown in Figure 3, let $\odot O$ be the circumcircle of the unit square $ABCD$, and point $P$ lies on the minor arc $AB$ (excluding $A$ and $B$). $DP$ intersects $AO$ at point $Q$. Then the range of $k=\frac{S_{\triangle AQD}}{S_{\triangle OPC}}$ is | 4. $[2 \sqrt{2}-2,1)$.
Let the radius of $\odot O$ be $r$. By the Pythagorean theorem, it is easy to get $r=\frac{1}{\sqrt{2}}$.
Let $\angle O P C=\angle O C P=\angle A C P=\angle A D P=\theta\left(0^{\circ}<\theta<45^{\circ}\right)$. Draw $Q T \perp A D$, with $T$ as the foot of the perpendicular, and let $Q T=h$. Th... | [2 \sqrt{2}-2,1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,636 |
5. If $a$ is a real constant, the function $f(x)$ for any non-zero real number $x$ satisfies $f\left(\frac{1}{x}\right)=a f(x)-x-1$, and $f(1)=1$, then the range of the function $F(x)=f(x)(x \in D=\{x \mid x$ $\in \mathbf{R}, x \neq 0, f(x) \geqslant x\})$ is $\qquad$ | 5. $\left(-\infty, \frac{1}{2}-\frac{\sqrt{3}}{4}\right] \cup\left[\frac{1}{2}+\frac{\sqrt{3}}{4},+\infty\right)$.
Let $x=1$, we get $f(1)=a f(1)-2$.
Then, by $f(1)=1$, we get $a=3$.
Thus, $f\left(\frac{1}{x}\right)=3 f(x)-x-1$.
Substitute $x$ with $\frac{1}{x}$ in equation (1), we get
$$
f(x)=3 f\left(\frac{1}{x}\rig... | \left(-\infty, \frac{1}{2}-\frac{\sqrt{3}}{4}\right] \cup\left[\frac{1}{2}+\frac{\sqrt{3}}{4},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,637 |
6. If from the set $S=\{1,2, \cdots, 20\}$, we take a three-element subset $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, such that it simultaneously satisfies: $a_{2}-a_{1} \geqslant 5,4 \leqslant a_{3}-a_{2} \leqslant 9$, then the number of all such subsets $A$ is $\qquad$ (answer with a specific number). | 6.251.
$a_{2}-a_{1} \geqslant 5 \Leftrightarrow\left(a_{2}-4\right)-a_{1} \geqslant 1$,
$4 \leqslant a_{3}-a_{2} \leqslant 9 \Leftrightarrow 1 \leqslant\left(a_{3}-7\right)-\left(a_{2}-4\right) \leqslant 6$.
Let $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-4, a_{3}^{\prime}=a_{3}-7$, then $a_{1}^{\prime}, a_{2}^{\prime}... | 251 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,638 |
Three, (20 points) Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x, y \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(4 x-y-1)+4 f(x+y)-16 f(x) \\
=(6 x-2)^{2}+45 y^{2}
\end{array}
$$ | Three, taking $y=0$ in equation (1), we get
$$
f(4 x-1)-12 f(x)=(6 x-2)^{2} \text {. }
$$
Taking $y=3 x-1$ in equation (1), we get
$$
\begin{array}{l}
4 f(4 x-1)-15 f(x) \\
=(6 x-2)^{2}+45(3 x-1)^{2} .
\end{array}
$$
(3) $-4 \times$ (2) yields $f(x)=(3 x-1)^{2}$.
Upon verification, it is known that $f(x)=(3 x-1)^{2}$... | f(x)=(3 x-1)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,639 |
In the Cartesian coordinate system $x 0 y$, there are points $A(3,3)$, $B(-2,1)$, $C(1,-2)$, and $T$ represents the set of all points inside and on the sides (including vertices) of $\triangle ABC$. Try to find the range of values for the bivariate function
$$
f(x, y)=\max \{2 x+y, \min \{3 x-y, 4 x+y\}\}
$$
(points $(... | As shown in Figure 5.
$l_{B C}: y=-x-1$,
$l_{A B}: y=\frac{2}{5} x+\frac{9}{5}$,
$l_{A C}: y=\frac{5}{2} x-\frac{9}{2}$.
We will discuss the following cases:
(1) If $4 x+y \geqslant 3 x-y$, then $x+2 y \geqslant 0$.
In this case, $f(x, y)=\max \{2 x+y, 3 x-y\}($ point $(x, y) \in$ $T$, and $x+2 y \geqslant 0$ ).
Let t... | [-3,9] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,640 |
Example 2 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right) \cdot (x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $A=\{(x, y) \mid x>0, y>0\}$. | Explanation: If $x+y \leqslant 1$, then
$$
\begin{array}{l}
x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4} \text {. } \\
\text { Hence } f(x, y) \\
=6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y- \\
4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
=6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1) \text {. } \\
\left(x-\frac{1}... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,641 |
Five. (20 points) Given the parabola $\Gamma_{1}: y=x^{2}$ and the circle $\Gamma_{2}:(x-3)^{2}+y^{2}=1$, the focus of $\Gamma_{1}$ is $F$, and point $A$ is on $\Gamma_{2}$. The line $F A$ intersects $\Gamma_{1}$ at points $B$ and $C$ (point $B$ is between point $F$ and point $A$). The tangents to $\Gamma_{1}$ at point... | Clearly, the focus of $\Gamma_{1}$ is $F\left(0, \frac{1}{4}\right)$. If the slope of line $F A$ is $k$, then the equation of line $F A$ is
$$
y=\frac{1}{4}+k x \text {. }
$$
Next, we find the range of $k$.
Since point $A(x, y)$ satisfies both equation (1) and $\Gamma_{2}$, substituting equation (1) into $\Gamma_{2}$ ... | y=-\frac{1}{4}\left(\frac{-3-\sqrt{129}}{64} \leqslant x \leqslant \frac{-3+\sqrt{129}}{64}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,642 |
(50 points) In $\triangle ABC$, $I$ is the incenter, and lines $BI$, $CI$ intersect sides $AC$, $AB$ at points $D$, $E$ respectively.
(1) Find the necessary and sufficient condition for $\frac{IC}{IB} = \frac{CD}{BE}$ to hold;
(2) If $\frac{IC}{IB} = \frac{CD}{BE} \neq 1$, find the range of $k = \frac{S_{\triangle BIC}... | (1) From $\frac{I C}{C D}=\frac{\sin \left(A+\frac{B}{2}\right)}{\sin \frac{B+C}{2}}$,
$$
\begin{array}{l}
\frac{I B}{B E}=\frac{\sin \left(A+\frac{C}{2}\right)}{\sin \frac{B+C}{2}}, \\
\text { then } \frac{I C}{I B}=\frac{C D}{B E} \Leftrightarrow \sin \left(A+\frac{B}{2}\right)=\sin \left(A+\frac{C}{2}\right) \\
\Lef... | \left(\frac{1}{3}, \frac{1}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,643 |
II. (50 points) Try to find the smallest positive integer $m$, such that the following conditions are satisfied simultaneously:
(1) $\left[\frac{2}{1977} m^{2}\right] \geqslant m+2006$ ( $[x]$ denotes the greatest integer not exceeding $x$);
(2) $99^{m}$ leaves a remainder of 11 when divided by 190. | From condition (1), we know that $m > 1977$.
Otherwise, if $m \leqslant 1977$, then we have
$$
\begin{array}{l}
{\left[\frac{2}{1977} m^{2}\right] \leqslant \frac{2}{1977} m^{2} \leqslant \frac{2}{1977} \times 1977 m} \\
= 2m \cdot 1977.
$$
Let $m = 1977 + k$ ($k$ is a positive integer), and substitute it into the ineq... | 2004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,644 |
Three. (50 points) Let $m, n$ be two positive integers (allowing $m$ to equal $n$), and $A, B$ be two sets composed of several real numbers, with $|A|=m, |B|=n$ (allowing $A=B$). The set $A$ satisfies:
If $x, y, u, v \in A$, and $x+y=u+v$, then either $x=y$ and $u=v$, or $\{x, y\}=\{u, v\} (x \neq y$, and $u \neq v)$.... | Let $A=\left\{a_{1}<a_{2}<\cdots<a_{m}\right\}$, $B=\left\{b_{1}<b_{2}<\cdots<b_{n}\right\}$. The table is listed as follows (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
$A$ \\
$B$
\end{tabular} & $a_{1}$ & $a_{2}$ & $\cdots$ & $a_{m-1}$ & $a_{m}$ \\
\hline$b_{1}$ & $a_{1}+b_{1}$ & $... | |A+B|_{\min }=\left\{\begin{array}{l}
m n-C_{n}^{2}=n\left(m-\frac{n-1}{2}\right), m \geqslant n ; \\
n+C_{m}^{2}=n+\frac{m(m-1)}{2}, m<n .
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,645 |
In $\triangle A B C$, $A B: B C$
$$
: C A=3: 5: 4, \odot O_{1} \text { and }
$$
$\odot O_{2}$ are two externally
tangent equal circles, both tangent to side
$B C$, where $\odot O_{1}$
is also tangent to side $A B$,
and $\odot O_{2}$ is also tangent to side $A C$. Given that line $O_{1} O_{2}$ intersects the two circles... | Proof: As shown in Figure 4, the internal common tangents of the two circles intersect the lines $BC$, $BA$, and $AC$ at points $D$, $E$, and $F$ respectively.
Given $AB: BC: CA = 35: 4$, it is clear that $\triangle ABC$ is a right triangle. Clearly, Rt $\triangle AEF \sim \text{Rt} \triangle ACB$.
Let $AB = 3x$, $AF ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,646 |
For positive integers $x, y$, let $m_{a}, m_{b}$ be the arithmetic mean and geometric mean of $x, y$ respectively.
(1) If $m_{a}+m_{b}=y-x$, find $\frac{x}{y}$;
(2) Prove: there exists a unique pair of positive integers $(x, y) (x<y)$ such that $m_{a}+m_{b}=40$. | (1) Solution: From the given, we have $\frac{x+y}{2}+\sqrt{x y}=y-x$, which leads to $\sqrt{x y}=\frac{y-3 x}{2}$.
From this, we get $9 x^{2}-10 x y+y^{2}=0$, which is
$(9 x-y)(x-y)=0$.
Solving this, we get $\frac{x}{y}=\frac{1}{9}$ or 1 (discard the latter).
(2) Proof: From the given, we have $\frac{x+y}{2}+\sqrt{x y... | \frac{x}{y}=\frac{1}{9}, (x, y) = (5, 45) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,647 |
Given 190 Let $\left\{a_{n}\right\}$ be a positive arithmetic sequence $\left(n \in \mathbf{N}_{+}\right.$ and $n>$
1). Prove:
$$
\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \ldots\left(1+\frac{1}{a_{n}}\right) \leq\left(1+\frac{a_{1}+a_{n}}{2 a_{1} a_{n}}\right)^{n} .
$$ | Proof: Let $\{a_{n}\}$ have a common difference of $d$, and $i \in \mathbf{N}_{+}$ with $i \leq n$.
Notice that
$$
\begin{array}{l}
a_{i}+a_{n+1-i}=a_{1}+(i-1) d+a_{1}+(n-i) d \\
=a_{1}+\left[a_{1}+(n-1) d\right]=a_{1}+a_{n}, \\
a_{i} a_{n+1-i}=\left[a_{1}+(i-1) d\right]\left[a_{1}+(n-i) d\right] \\
=a_{1}^{2}+(n-1) a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,648 |
Example 3 Let $0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant 1$. Try to find the minimum value of the function
$$
f(x, y)=(2 y-1) \sin x+(1-y) \sin (1-y) x
$$ | It is known that for all $0 \leqslant x \leqslant \pi$, we have
$$
\sin x \geqslant 0, \sin (1-y) x \geqslant 0 \text {. }
$$
Thus, when $\frac{1}{2} \leqslant y \leqslant 1$, $f(x, y) \geqslant 0$, and the equality holds when $x=0$.
When $0 \leqslant yx>\sin x$,
and $\sin (x+\delta)=\sin x \cdot \cos \delta+\cos x \... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,649 |
Example 4 Let $x_{1}, x_{2}, \cdots, x_{n}$ take values in some interval of length 1. Let $x=\frac{1}{n} \sum_{j=1}^{n} x_{j}, y=\frac{1}{n} \sum_{j=1}^{n} x_{j}^{2}$. Find the maximum value of the function $f=y-x^{2}$: | Let $x_{1}, \cdots, x_{n} \in [a, a+1] (a \in \mathbf{R})$.
When $n=1$, $f=0$, obviously, $f_{\text{max}}=0$.
When $n>1$, if we fix $x_{2}, x_{3}, \cdots, x_{n}$, then we have
$$
\begin{array}{l}
f=y-x^{2}=\frac{1}{n} \sum_{j=1}^{n} x_{j}^{2}-\left(\frac{1}{n} \sum_{j=1}^{n} x_{j}\right)^{2} \\
=\left(\frac{1}{n}-\frac... | \frac{1}{4}, \text{ if } n \text{ is even; } \frac{n^{2}-1}{4 n^{2}}, \text{ if } n \text{ is odd.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,650 |
Example 5 Let $a, b, c$ be positive real numbers. Find the minimum value of
$$
\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}
$$
(2004, Girls' Mathematical Olympiad) | Let $a+2b+c=x, a+b+2c=y, a+b+3c=z$, then
$$
\begin{array}{l}
a+3c=2y-x, b=x+z-2y, c=z-y. \\
\text{Therefore, } \frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c} \\
=\frac{2y-x}{x}+\frac{4(x+z-2y)}{y}-\frac{8(z-y)}{z} \\
=-17+\frac{2y}{x}+\frac{4x}{y}+\frac{4z}{y}+\frac{8y}{z} \\
\geqslant-17+2\sqrt{8}+2\sqrt{32}=... | -17+12\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,651 |
Example 6 Let $a, b, c$ be positive real numbers, and $abc + a + c = b$. Try to determine the maximum value of $P=\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1}$.
(1999, Vietnam Mathematical Olympiad) | Given that $a+c=(1-a c) b$.
It is easy to see that $1-a c \neq 0$. Therefore, we have $b=\frac{a+c}{1-a c}$.
This structure reminds us of the tangent addition formula, hence we make a tangent substitution.
Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$. Then
$$... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,652 |
Example 7 Let positive numbers $a, b, c, x, y, z$ satisfy
$$
c y+b z=a, a z+c x=b, b x+a y=c .
$$
Find the minimum value of the function $f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}$.
(2005, National High School Mathematics Competition) | Given the conditions:
$$
\begin{array}{l}
b(a z+c x)+c(b x+a y)-a(c y+b z) \\
=b^{2}+c^{2}-a^{2},
\end{array}
$$
we have \( x=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \).
Similarly, \( y=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \), \( z=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \).
Using \(\sum\) to denote cyclic summation, we have:
$$
\begin{ar... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,653 |
Example 8 For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
(40th IMO Chinese National Team Selection Exam) | Explanation: (1) For $x, y > 0$,
$$
\begin{array}{l}
(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\
= x y\left(4 x^{2}+6 x y+4 y^{2}\right)- \\
x y\left(5 x^{3}+10 x y^{2}+10 x^{2} y+5 y^{3}\right) \\
= \frac{1}{2} x y\left(7 x^{2}+12 x y+7 y^{2}+x^{2}+y^{2}\right)- \\
5 x y\left(x^{3}+2 x y^{2... | \frac{1}{12} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,654 |
Example 9 Let $a, b, c$ be positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary | Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By Vieta's formulas, we know $x_{1}x_{2} = \frac{c}{a} = 9$. Therefore,
\[ b^{2} \geqslant 4ac = 4 \times \frac{a}{c} \times c^{2} > 4 \times 9 \times 1^{2} = 36. \]
Thus, $b \geqslant 7$.
Also, $\frac{b}{a} = (-x_{1}) + (-x_{2}) \Rightarrow \frac{3}{2} b \... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,655 |
Example 10 Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
\sum_{i=1}^{n} x_{i}^{2}=1 \quad(n \geqslant 3) \text {. }
$$
Find $\min _{1 \leqslant i<j \leqslant n}\left\{\left|x_{i}-x_{j}\right|\right\}$'s maximum value. | Explanation: By symmetry, without loss of generality, assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, and let $\min _{1 \leqslant i<j \leqslant n}\left\{\left|x_{i}-x_{j}\right|\right\}=m$.
By the minimality of $m$, we have
$$
x_{j}-x_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant m) \text {. }
$$
The... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,656 |
Example 3 Given a cyclic quadrilateral $ABCD$, $AD$ and $BC$ intersect at point $P$, $AC$ and $BD$ intersect at point $M$. Prove:
$$
PM^2 = PA \cdot PD - AM \cdot MC \text{.}
$$ | Proof: As shown in Figure 3, let the circumcircle of $\triangle A M D$ intersect line $P M$ at point $H$, and connect $H A$.
Since $\angle P M D = \angle P A H$ and $\angle P B D = \angle C A D$, we have $\angle P M D - \angle P B D = \angle P A H - \angle C A D$,
thus $\angle M P B = \angle M A H$.
Therefore, points $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,657 |
Example 1 Let any real numbers $x, y$ satisfy $|x|<1$, $|y|<1$. Prove:
$$
\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geqslant \frac{2}{1-x y} \text {. }
$$
(19th Moscow Mathematical Olympiad) | Proof: Since $|x|<1,|y|<1$, therefore, $x^{2}<1$, $y^{2}<1$. Thus, we have
$$
\begin{array}{l}
\frac{1}{1-x^{2}}=1+x^{2}+x^{4}+\cdots, \\
\frac{1}{1-y^{2}}=1+y^{2}+y^{4}+\cdots .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } \frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \\
=\left(1+x^{2}+x^{4}+\cdots\right)+\left(1+y^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,659 |
Example 2 Let $a_{1}, a_{2}, \cdots, a_{n}>0(n \geqslant 2)$, and $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that:
$$
\frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} \geqslant \frac{n}{2 n-1} .
$$
(1984, Balkan Mathematical Olympiad) | Proof: Given $0 < a_{1}, a_{2}, \cdots, a_{n} < 1$, we have
$$
\begin{array}{l}
\frac{a_{1}}{2-a_{1}}=\frac{\frac{a_{1}}{2}}{1-\frac{a_{1}}{2}}=\sum_{k=1}^{\infty}\left(\frac{a_{1}}{2}\right)^{k}, \\
\frac{a_{2}}{2-a_{2}}=\frac{\frac{a_{2}}{2}}{1-\frac{a_{2}}{2}}=\sum_{k=1}^{\infty}\left(\frac{a_{2}}{2}\right)^{k}, \\
... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,660 |
Example 3 Let $a, b, c$ be positive real numbers, and satisfy $abc = 1$. Try to prove:
$$
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }
$$
(36th IMO) | Analysis: Transform each term, convert to $\frac{a_{1}}{1-q}(|q|<1)$, since $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>0$, so, $0<\frac{1}{t a} 、 \frac{1}{t b} 、 \frac{1}{t c}<1$.
Also, $a b c=1$, then
$$
\begin{array}{l}
\frac{1}{a^{3}(b+c)}=\frac{a b c}{a^{3}(b+c)}=\frac{b c}{a^{2}(b+c)} \\
=\frac{\frac{1}{a^{2}}}{\frac{... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,661 |
Example 4 Given $x, y \in(-2,2)$, and $x y=-1$. Find the minimum value of the function $u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}$. (2003, National High School Mathematics Competition) | Solution: Since $x, y \in (-2,2)$, we have
$$
\left(\frac{x}{2}\right)^{2}<1, \left(\frac{y}{3}\right)^{2}<1.
$$
Given that $xy = -1$, we have
$$
\begin{array}{l}
u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}} \\
=\frac{1}{1-\left(\frac{x}{2}\right)^{2}}+\frac{1}{1-\left(\frac{y}{3}\right)^{2}} \\
=\sum_{k=0}^{\infty}\left(\fr... | \frac{12}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,662 |
Question As shown in Figure 1, the ellipse with foci at $B_{0}$ and $B_{1}$ intersects the sides $A B_{i}$ of $\triangle A B_{0} B_{1}$ at points $C_{i}(i=0,1)$. On the extension of $A B_{0}$, take any point $P_{0}$, and draw the arc $\overparen{P_{0} Q_{0}}$ with $B_{0}$ as the center and $B_{0} P_{0}$ as the radius, ... | Solution: As shown in Figure 1, the intersection point $O$ of the angle bisector of $\angle Q_{0} B_{0} P_{0}$ and the angle bisector of $\angle A C_{1} B_{0}$ is the rotation center of the rotation transformation from point $P_{0}$ to point $P_{1}$, with the rotation angle being $\angle P_{0} B_{0} Q_{0}+\angle Q_{0} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,663 |
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