problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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9. Arrange the sequence of numbers
$$
\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, \cdots, 10 \sqrt{2}
$$
in the following manner:
\begin{tabular}{ccccc}
$\sqrt{2}$ & 2 & $\sqrt{6}$ & $2 \sqrt{2}$ & $\sqrt{10}$ \\
$2 \sqrt{3}$ & $\sqrt{14}$ & 4 & $3 \sqrt{2}$ & $2 \sqrt{5}$ \\
$\sqrt{22}$ & $2 \sqrt{6}$ & $\cdots$ & ... | 9. $(20,3)$.
The sequence is arranged according to the square roots of even numbers, $\sqrt{2}, \sqrt{4}, \sqrt{6}, \cdots$, $\sqrt{200}$, totaling 100. Each row contains 5 numbers, making 20 rows in total, with the largest rational number being $\sqrt{196}=14$. | (20,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,450 |
10. Let $0 . \overline{a_{1} a_{2} a_{3} 1}$ be a four-digit decimal pure fraction, where $a_{i}(i=1,2,3)$ can only take 0 or 2. Let $T$ be the number of all such four-digit fractions, and $S$ be the sum of all such four-digit fractions. | 10.0.111 1
$a_{1} a_{2} a_{3}$ can be respectively
$$
000,200,020,002,022,202,220,222 \text {. }
$$
a total of 8 kinds. Then $S=0.8888$. | 0.8888 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,451 |
11. Given positive numbers $a, b, c, d, e, f$, which simultaneously satisfy
$$
\begin{array}{l}
\frac{b c d e f}{a}=\frac{1}{2}, \frac{a c d e f}{b}=\frac{1}{4}, \frac{a b d e f}{c}=\frac{1}{8}, \\
\frac{a b c e f}{d}=2, \frac{a b c d f}{e}=4, \frac{a b c d e}{f}=8 .
\end{array}
$$
Then the value of $a+b+c+d+e+f$ is $... | 11. $\frac{5}{2}+\frac{15 \sqrt{2}}{4}$.
Multiplying both sides of the 6 equations yields $(a b c d e f)^{4}=1$. Thus, $a b c d e f=1$, $a^{2}=2, b^{2}=4, c^{2}=8, d^{2}=\frac{1}{2}, e^{2}=\frac{1}{4}, f^{2}=\frac{1}{8}$.
Therefore, $a+b+c+d+e+f$
$$
\begin{array}{l}
=\sqrt{2}+\sqrt{4}+\sqrt{8}+\sqrt{\frac{1}{2}}+\sqrt{... | \frac{5}{2}+\frac{15 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,452 |
13. As shown in Figure 4, the area of isosceles right $\triangle ABC$ is $98$, and $D$ is a point on the hypotenuse $BC$ such that $BD: DC = 2: 5$. Then the area of the square $ADEF$ with side $AD$ is $\qquad$ | 13.116.
Draw $D G \perp A B, D H \perp A C$, with the feet of the perpendiculars at $G$ and $H$ respectively. It is easy to see that $A B=A C=14$.
Also, $B D: D C=G D: H D=2: 5$, so $D G=4, D H=10$. Therefore, $A D=\sqrt{4^{2}+10^{2}}$. Hence, $S_{\text {square ADEF }}=116$. | 116 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,454 |
14. As shown in Figure 5, given $\square A B C D, \angle A C B=30^{\circ}$, fold $\triangle A B C$ along $A C$ so that point $B$ lands at point $E$ on the plane of $\square A B C D$, and connect $D E$. Then the value of $\frac{A C+D E}{A D}$ is $\qquad$ | 14. $\sqrt{3}$.
Point $B$ rotates $180^{\circ}$ around $A C$ and lands on point $E$. Points $A, B, C, D, E$ are on the same plane. $E C = B C = A D, A E = A B = C D$, so quadrilateral $A C D E$ is an isosceles trapezoid, and $\angle D A C = 30^{\circ}$. Extend $A C$ to point $F$ such that $C F = E D$, then $\triangle ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,455 |
15. (16 points) In an activity class, the teacher asked each student to make a rectangular box with a lid as shown in Figure 6. The length, width, and height of the rectangular box are $x \mathrm{~cm}, y \mathrm{~cm}$, and $z \mathrm{~cm}$, respectively. When Xiao Yang presented his box, he told the class: “The length,... | Three, 15. Since $x y=x z+3$, therefore, $x(y-z)=3$.
Also, $x, y, z$ are all positive integers, then
$$
\left\{\begin{array}{l}
x = 3, \\
y - z = 1
\end{array} \text { or } \left\{\begin{array}{l}
x=1, \\
y-z=3 .
\end{array}\right.\right.
$$
(1) When $\left\{\begin{array}{l}x=3, \\ y=z+1\end{array}\right.$, by $y z=x y... | 22 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,456 |
16. (16 points) As shown in Figure 7, in rectangle $ABCD$, points $E$ and $F$ are on $AB$ and $AD$ respectively. Lines $EH$ and $FG$ are drawn parallel to each other, intersecting $BC$ and $CD$ at points $H$ and $G$ respectively. Let the lengths of $AB$ and $AD$ be $a$ and $b$ ($a < b$).
(1) If $AE = AF = 1$ and $EH \p... | 16. (1) Does not exist.
Since $A B=a, A D=b, A E=A F=1$, we have
$$
B E=a-1, D F=b-1 \text {. }
$$
Also, since $E H / / A C$, then $\frac{B H}{B C}=\frac{B E}{A B}$, which means
$$
\frac{b-C H}{b}=\frac{a-1}{a} \Rightarrow C H=\frac{b}{a} \text {. }
$$
Similarly, we get $C G=\frac{a}{b}$.
Since $a<b$, then $C G$ is ... | S_{\triangle A H G}=S_{\triangle C E F} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,457 |
Example 9 Calculate
$$
(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) .
$$ | $$
\begin{array}{l}
=1 \times(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdots\left(2^{2^{n}}+1\right) \\
=(2-1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) \\
=\left(2^{2}-1\right)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) \\
=\cdots \c... | 2^{2^{n+1}}-1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,458 |
17. (18 points) Among 200 small balls numbered $1, 2, \cdots, 200$, any $k$ balls are drawn such that there must be two balls with numbers $m$ and $n$ satisfying
$$
\frac{2}{5} \leqslant \frac{n}{m} \leqslant \frac{5}{2} \text {. }
$$
Determine the minimum value of $k$ and explain the reasoning. | 17. Divide 200 balls numbered $1 \sim 200$ into 6 groups, such that the ratio of the numbers of any two balls in each group is no less than $\frac{2}{5}$ and no more than $\frac{5}{2}$.
Grouping as follows:
Group 1 $(1,2)$
Group 2 $\quad(3,4,5,6,7)$
Group 3 $(8,9,10, \cdots, 20)$
Group 4 $\quad(21,22,23, \cdots, 52)$
G... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,459 |
2. According to the state regulations, the tax calculation method for personal income from publishing the same work in the form of books or periodicals is as follows:
(1) No tax is levied on royalties not exceeding 800 yuan;
(2) For royalties exceeding 800 yuan but not exceeding 4000 yuan, a tax of 14% is levied on the... | 2.C.
Since $4000 \times 11 \% = 440 > 420$, the manuscript fee is between 800 and 4000 yuan. Let the manuscript fee be $x$ yuan. According to the problem,
$$
(x-800) \times 14 \% = 420 \text{.}
$$
Solving for $x$ gives $x = 3800$ (yuan). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,461 |
3. Given two equations about $x$: $x^{2}+2 b x+a$ $=0$ and $x^{2}+a x+2 b=0$ have exactly one common root. Then the minimum value of $a^{2}+b^{2}$ is $(\quad)$.
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{3}{5}$
(D) $\frac{4}{5}$ | 3. A.
Let the common root of the two equations be $x=t$, then we have $t^{2}+2 b t+a=0, t^{2}+a t+2 b=0$.
Subtracting the two equations yields
$$
(2 b-a) t+a-2 b=0 .
$$
Given that $a-2 b \neq 0$, we have $t=1$.
Substituting $t=1$ into the equation gives $1+2 b+a=0$.
Thus, $a^{2}+b^{2}=(-1-2 b)^{2}+b^{2}$
$$
=5 b^{2}+... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,462 |
4. As shown in Figure 1, in the rectangular paper piece $A B C D$, $A B=3 \mathrm{~cm}, B C$ $=4 \mathrm{~cm}$. Now the paper is folded and pressed flat so that $A$ coincides with $C$, and the crease is set as $E F$. Then the area of the overlapping part $\triangle A E F$ is equal to ( ).
(A) $\frac{73}{8}$
(B) $\frac{... | 4.D.
Let $CE=x$, then $AE=AF=x, BE=4-x$. In the right triangle $\triangle ABE$, by $AE^2=AB^2+BE^2$, we get $x^2=3^2+(4-x)^2$.
Solving for $x$ gives $x=\frac{25}{8}$.
Therefore, $S_{\triangle AEF}=\frac{1}{2} AF \cdot AB=\frac{1}{2} \times \frac{25}{8} \times 3=\frac{75}{16}$. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,463 |
5. If a natural number $n$ allows the vertical addition
$$
n+(n+1)+(n+2)
$$
to not produce any carry, then $n$ is called a "continuous number". For example, 12 is a continuous number, because $12+13+14$ does not produce a carry; but 13 is not a continuous number. Then, the number of continuous numbers less than 1000 i... | 5.C.
There are $0,1,2$, a total of 3 single-digit numbers;
There are $\overline{a 0}, \overline{a 1}, \overline{a 2}, a=1,2,3$, a total of $3 \times 3=9$ two-digit numbers;
There are $\overline{a b 0}, \overline{a b 1}, \overline{a b 2}, a=1,2,3, b=0,1,2,3$, a total of $3 \times 4 \times 3=36$ three-digit numbers.
The... | 48 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,464 |
6. For natural numbers $a, b, c$, define the operation * such that it satisfies
$$
\begin{array}{l}
(a * b) * c=a *(b c), \\
(a * b)(a * c)=a *(b+c) .
\end{array}
$$
Then the value of $3 * 4$ is ( ).
(A) 12
(B) 64
(C) 81
(D) None of the above | 6.C.
In $(a * b) * c=a *(b c)$, let $b=c=1$, we get $(a * 1) * 1=a * 1$,
so, $a * 1=a$.
In $(a * b)(a * c)=a *(b+c)$,
let $b=c=1$, we get $(a * 1)^{2}=a * 2$, so, $a * 2=a^{2} ;$
let $b=2, c=1$, we get $(a * 2)(a * 1)=a * 3$, so, $a * 3=a^{2} \cdot a=a^{3} ;$
let $b=3, c=1$, we get $(a * 3)(a * 1)=a * 4$, so, $a * 4=a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,465 |
2. Let natural numbers $x, y$ satisfy
$$
x<y, x^{3}+19 y=y^{3}+19 x
$$
Then $x+y=$ $\qquad$ | 2.5.
From $x^{3}-y^{3}=19(x-y), x<y$, we get $x^{2}+x y+y^{2}=19$.
Thus, $3 x^{2}<x^{2}+x y+y^{2}=19$.
Since $x$ is a natural number, we have $x=2$.
Therefore, $y=3$, so $x+y=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 717,467 |
3. The 2008 Olympic Games will be held in Beijing. Figure 3 is the Olympic five-ring logo, where $a, b, c, d, e, g, h, i$ are filled with different numbers from $1 \sim 9$. If the sum of the numbers in each circle is equal, then the maximum value of this sum is $\qquad$
| 3.14 .
Let the sum of the numbers in each circle be $N$, then
$$
(1+2+\cdots+9)+(b+d+f+h)=5 N .
$$
Therefore, $N=\frac{b+d+f+h}{5}+9$.
To maximize $N$, $b+d+f+h$ must be the largest possible value and a multiple of 5.
Since $b+d+f+h=6+7+8+9=30$ is the maximum, and $N=15$, at this point, no matter how you fill in the... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,468 |
Example 10 Calculation
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2006}\right)\left(1+\frac{1}{2}+\cdots+\frac{1}{2005}\right) \\
\left(1+\frac{1}{2}+\cdots+\frac{1}{2006}\right)\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2005}\right)
\end{array}
$$ | Explanation: Let $1+\frac{1}{2}+\cdots+\frac{1}{2005}=a$,
$$
\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2005}=b,
$$
then $a-b=1$.
Therefore, the original expression is
$$
\begin{array}{l}
\quad=\left(b+\frac{1}{2006}\right) a-\left(a+\frac{1}{2006}\right) b \\
=\frac{a-b}{2006}=\frac{1}{2006} .
\end{array}
$$ | \frac{1}{2006} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,469 |
(20 points) Read the material and answer the questions:
As shown in Figure 5, the shapes of rhombuses, rectangles, and squares differ. We refer to the degree of similarity between these shapes as "closeness."
When studying closeness,
the closeness of similar figures
should be equal.
(1) Let the two adjacent interior an... | (1)(i)20 (ii)0 (2)Unreasonable
In rectangles with side lengths of $a, b$ and $2a, 2b$, it is clear that these two rectangles are similar, but $|2a-2b| \neq |a-b|$.
The proximity of a rectangle can be defined as $\frac{a}{b}(a \leqslant b)$, then the closer $\frac{a}{b}$ is to 1, the closer the rectangle is to a square... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,471 |
II. (25 points) $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are two isosceles right triangles, $\angle A=\angle A_{1}=90^{\circ}$, and the vertex $B_{1}$ of $\triangle A_{1} B_{1} C_{1}$ is located at the midpoint of side $B C$.
(1) As shown in Figure 6(a), let $A_{1} B_{1}$ intersect $A B$ at point $E$, and $B... | (1) In $\triangle B B_{1} E$ and $\triangle C F B_{1}$, since $\angle B B_{1} E + \angle B E B_{1} = \angle B B_{1} E + \angle C B_{1} F = 135^{\circ}$, therefore, $\angle B E B_{1} = \angle C B_{1} F$.
Also, $\angle B_{1} B E = \angle F C B_{1} = 45^{\circ}$, so, $\triangle B B_{1} E \backsim \triangle C F B_{1}$.
(2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,472 |
$$
\begin{array}{l}
\text { Three. (25 points) Given the quadratic function } \\
y=a x^{2}+b x+c(a<b)
\end{array}
$$
the graph is always above or on the $x$-axis, and $m<\frac{a+b+c}{b-a}$ always holds. Find the range of values for $m$. | Three, it is known that $y=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a} \geqslant 0$ always holds.
Therefore, $a>0, c \geqslant \frac{b^{2}}{4 a}$.
From $m1$. Thus,
$$
\begin{array}{l}
\frac{a+b+c}{b-a} \geqslant \frac{a+b+\frac{b^{2}}{4 a}}{b-a}=\frac{4 a^{2}+4 a b+b^{2}}{4 a(b-a)} \\
=\frac{\left(\frac{b}{a}... | m<3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,473 |
1. Given the set
$$
\begin{aligned}
M & =\left\{f(x) \mid f^{2}(x)-f^{2}(y)\right. \\
& =f(x+y) \cdot f(x-y), x, y \in \mathbf{R}\} .
\end{aligned}
$$
The following statements are given:
(1) If $f_{1}(x)=\left\{\begin{array}{cc}1, & x \geqslant 0, \\ -1, & x<0,\end{array}\right.$ then $f_{1}(x) \in$
$M$;
(2) If $f_{2}... | - 1.C.
It is easy to verify that (1) and (4) are incorrect, while (2) and (3) are correct.
Let $x=y=0$, we get $f(0)=0$;
Then let $x=0$, we have $f^{2}(0)-f^{2}(y)=f(y) f(-y)$, which simplifies to $f(y)(f(y)+f(-y))=0$.
Therefore, $f(y)=0$ or $f(y)+f(-y)=0$, so we always have
$$
f(-y)=-f(y)
$$
Thus, $f(x)$ is an odd f... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,474 |
2. Given the sequence $\left\{a_{n}\right\}$ consists of positive integers and is an increasing sequence, and satisfies $a_{n+2}=a_{n+1}+2 a_{n}\left(n \in \mathbf{N}_{+}\right)$. If $a_{5}$ $=52$, then $a_{7}=(\quad)$.
(A) 102
(B) 152
(C) 212
(D) Insufficient conditions, $a_{7}$ cannot be uniquely determined | 2.C.
$$
\begin{array}{l}
a_{3}=a_{2}+2 a_{1}, \\
a_{4}=a_{3}+2 a_{2}=3 a_{2}+2 a_{1}, \\
a_{5}=a_{4}+2 a_{3}=5 a_{2}+6 a_{1} .
\end{array}
$$
The positive integer solutions to the equation $5 a_{2}+6 a_{1}=52$ are $a_{1}=2, a_{2}=8$ or $a_{1}=7, a_{2}=2$.
Since $a_{2}>a_{1}$, we have $a_{1}=2, a_{2}=8$.
Thus, $a_{7}=a... | 212 | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,475 |
3. The curve represented by the equation $\frac{x^{2}}{\sin \left(19^{2007}\right)^{\circ}}+\frac{y^{2}}{\cos \left(19^{2007}\right)^{\circ}}=1$ is ( ).
(A) Hyperbola
(B) Ellipse with foci on the $x$-axis
(C) Ellipse with foci on the $y$-axis
(D) None of the above answers is correct | 3.C.
$$
\begin{array}{l}
19^{2007}=19 \times\left(19^{2}\right)^{1003}=19(360+1)^{1003} \\
=19(360 n+1)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $\sin \left(19^{2007}\right)^{\circ}=\sin (360 \times 19 n+19)^{\circ}=\sin 19^{\circ}$,
$$
\cos \left(19^{2007}\right)^{\circ}=\cos (360 \times 19 n+19)^{\circ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,476 |
4. The equation $8 x\left(2 x^{2}-1\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ has ( ) real roots.
(A) 1
(B) 3
(C) 5
(D) 7 | 4.D.
If $|x| \geqslant 1$, then $2 x^{2}-1 \geqslant 1, 8 x^{4}-8 x^{2}+1 \geqslant 1$.
Therefore, $\left|8 x\left(2 x^{2}-1\right)\left(8 x^{4}-2 x^{2}+1\right)\right| \geqslant 8$.
Hence $|x|<1$.
Let $x=\cos \theta, \theta \in(0, \pi)$.
Then $8 \cos \theta\left(2 \cos ^{2} \theta-1\right)\left(8 \cos ^{4} \theta-8 \... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 717,477 |
5. Reverse the order of the three digits of a three-digit number, and add the resulting number to the original number. If none of the digits in the sum is even, then the number is called a "odd-sum number". Therefore, among all three-digit numbers, the number of odd-sum numbers is ( ).
(A) 100
(B) 120
(C) 160
(D) 200 | 5.A.
Let the three-digit number be $\overline{a_{1} a_{2} a_{3}}$. Then
$$
\begin{array}{l}
\overline{a_{1} a_{2} a_{3}}+\overline{a_{3} a_{2} a_{1}} \\
=100\left(a_{1}+a_{3}\right)+10\left(a_{2}+a_{2}\right)+\left(a_{1}+a_{3}\right)
\end{array}
$$
If $a_{1}+a_{3}$ does not carry over, then the tens digit of the sum ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 717,478 |
6. If the arithmetic mean of the focal distance and the length of the major axis of an ellipse equals the geometric mean of the length of the major axis and the length of the minor axis, then the value of the eccentricity of the ellipse should be in the interval ( ).
(A) $\left(0, \frac{1}{2}\right)$
(B) $\left(\frac{1... | 6.C.
According to the problem, we have
$$
\begin{array}{l}
a+c=2 \sqrt{a b} \\
\Rightarrow 1+e=2 \sqrt{\frac{b}{a}}=2 \sqrt[4]{\frac{a^{2}-c^{2}}{a^{2}}}=2 \sqrt[4]{1-e^{2}} \\
\Rightarrow(1+e)^{4}=16\left(1-e^{2}\right) \\
\Rightarrow(1+e)^{3}=16(1-e) \\
\Rightarrow(1+e)^{3}+16(e-1)=0 . \\
\text { Let } f(x)=(x+1)^{3... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 717,479 |
Example 1 Find all real parameters $a$ such that the equations
$$
x^{2}-(2 a+1) x+a=0
$$
and $x^{2}+(a-4) x+a-1=0$
have roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$, respectively, and satisfy
$$
\frac{x_{1}}{x_{3}}+\frac{x_{4}}{x_{2}}=\frac{x_{1} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)}{a} .
$$
(2005, Bulgarian Winter Ma... | Explanation: By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=2 a+1, x_{1} x_{2}=a, \\
x_{3}+x_{4}=4-a, x_{3} x_{4}=a-1 .
\end{array}
$$
Transform the given equations into
$$
\begin{array}{l}
a\left(x_{1} x_{2}+x_{3} x_{4}\right) \\
=x_{1} x_{2} x_{3} x_{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right) .
\end{array}... | a=-1 \pm \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,480 |
Example 2 For any real numbers $p, q, r, s (q \neq -1, s \neq -1)$, prove: the quadratic equations
$$
x^{2} + p x + q = 0 \text{ and } x^{2} + r x + s = 0
$$
have a common root and their other roots are reciprocals of each other if and only if the coefficients $p, q, r, s$ satisfy
$$
\begin{array}{l}
p r = (q + 1)(s +... | Let's consider: If the two roots of $x^{2}+p x+q=0$ are $u$ and $v$, and the two roots of $x^{2}+r x+s=0$ are $u$ and $\frac{1}{v}$, then by Vieta's formulas, we have
$$
\begin{array}{l}
p=-(u+v), q=u v, \\
r=-\left(u+\frac{1}{v}\right), s=\frac{u}{v} .
\end{array}
$$
Since $p r=\frac{(u+v)(u v+1)}{v}$,
$$
\begin{arra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 717,481 |
Example 4 On the line segment $A_{1} A_{n}(n \geqslant 2)$, there are $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. Now, some of these points (or the empty set) are painted red, so that the colored points are not adjacent. Find the number of ways to color. | Let the number of coloring methods that satisfy the conditions be $a_{n}$.
When $n=2$, there are (red, none), (none, red), (none, none), a total of 3 cases, $a_{2}=3$;
When $n=3$, there are (red, none, none), (none, red, none), (none, none, red), (red, none, red), (none, none, none), a total of 5 cases, $a_{3}=5$.
Whe... | a_{n} = a_{n-1} + a_{n-2} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,482 |
Example 5 "Tossing a coin continuously until two consecutive heads appear" occurs in all possible ways on the $n(n \geqslant 2)$-th toss.
How many such ways are there? | Explanation: Let the number of methods that satisfy the conditions be $a_{n}$.
When $n=2$, there is only 1 case of "positive positive", $a_{2}=1$;
When $n=3$, there is only 1 case of "negative positive positive", $a_{3}=1$.
For $n \geqslant 4$, the number of methods $a_{n}$ where "positive positive" appears exactly a... | 233 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,483 |
Example 6 Prove: the inequality $1+\frac{1}{2}+\cdots+\frac{1}{2^{n}}<2$.
| Explanation: The inequality can be directly seen from Figure 1: a part is less than the whole.
If we use mathematical induction to prove it, we can set
$$
S_{n}=1+\frac{1}{2}+\cdots+\frac{1}{2^{n}} .
$$
Verify for $n=1$, we have $S_{1}=1+\frac{1}{2}<2$.
The proposition holds.
Assume $S_{k}<2$, then
$$
S_{k+1}=S_{k}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,484 |
Example 7 Prove the inequality:
$$
1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}<2 .
$$ | Explanation: If mathematical induction is used to prove, when $n=1$, it is obviously true. The second step, if the recursive relation
$$
S_{k+1}=S_{k}+\frac{1}{(k+1)!}<2+\frac{1}{(k+1)!},
$$
is used, it will fall into the same dilemma as Example 6. However, by using the recursive relation
$$
\begin{array}{l}
S_{k+1}=1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,485 |
Example 8 A point whose both coordinates are integers is called an integer point. If $n$ is a non-negative integer, then, in the region
$$
\{(x, y) \mid | x|+| y | \leqslant n\}
$$
how many integer points are there? | Explanation: As shown in Figure 2, it is evident that the region is a square.
Let there be $a_{\mathrm{n}}$ integer points in it, and obviously $a_{0}=1$.
When $a_{n}$ increases to $a_{n+1}$, the points in the first quadrant and on the positive half-axis of the $x$-axis that will be added are those with the sum of th... | 2 n^{2}+2 n+1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,486 |
Example 9 As shown in Figure 3, a brass plate is equipped with three diamond fine columns, one of which is fitted with 64 rings of different sizes, with the larger ones at the bottom and the smaller ones on top. These rings can be transferred from one column to another one at a time, but it is not allowed to place a la... | Explanation: This problem is about the Tower of Hanoi.
Let $a_{n}$ represent the minimum number of moves required for $n$ disks, with $a_{1}=1$.
For $n>1$, we can first move the $n-1$ disks from column $A$ to column $C$ as required, which takes $a_{n-1}$ moves; then move the largest disk from column $A$ to column $B$,... | 2^{64}-1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 717,487 |
Example 1 In $\triangle ABC$, $AB=AC$, $BE$ is a median, point $O$ is the circumcenter of $\triangle ABC$, and point $G$ is the centroid of $\triangle ABE$. Prove: $OG \perp BE$. | To prove: The auxiliary line is added as shown in Figure 2.
To prove $O G \perp B E$, it is sufficient to prove
$$
\begin{array}{l}
O B^{2}+E G^{2} \\
=O E^{2}+B G^{2} .
\end{array}
$$
It is easy to see that $O E \perp A C$, thus, we have
$$
O B^{2}-O E^{2}=O A^{2}-O E^{2}=A E^{2} \text {. }
$$
By the property of th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,488 |
Example 2 In a convex pentagon $A B C D E$, $A B=A E$, $\angle A B C=\angle A E D=90^{\circ}$, point $P$ is inside $\triangle A C D$, satisfying $P B \perp A C, P E \perp A D$. Prove: $A P \perp C D$.
| Proof: As shown in Figure 3, connect $PC$ and $PD$. Clearly, quadrilateral $ACPD$ is a concave quadrilateral.
To prove $AP \perp CD$, it suffices to show that
$$
PC^{2} + AD^{2} = PD^{2} + AC^{2}.
$$
In quadrilateral $ABCP$,
since $PB \perp AC$, we have
$$
PC^{2} + AB^{2} = AP^{2} + BC^{2}.
$$
In quadrilateral $AEDP... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,489 |
Example 3 In $\triangle A B C$, $\angle A C B=90^{\circ}, C D$ bisects $\angle A C B, D E \perp B C$ at point $E, D F \perp A C$ at point $F$, $A E \cap B F=H$. Prove: $C H \perp A B$.
(2003, Romanian Mathematical Olympiad) | To prove: As shown in Figure 4, extend $CH$ to intersect $AB$ at point $G$. To prove $CH \perp AB$, i.e., $CG \perp AB$, it suffices to prove
$$
AG^{2} + BC^{2} = BG^{2} + AC^{2}.
$$
Let $BC = a$, $AC = b$, and the side length of the square $C E D F$ be $x$. From $\triangle A F D \sim \triangle A C B$, we have
$$
\fra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,490 |
Example 4 In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. The centroids of $\triangle OAB$ and $\triangle OCD$ are $M_{1}$ and $M_{2}$, respectively, and the orthocenters of $\triangle OBC$ and $\triangle ODA$ are $H_{1}$ and $H_{2}$, respectively. Prove that $M_{1} M_{2} \perp H_{... | Proof: As shown in Figure 5, take the midpoints $E$ and $F$ of $AB$ and $CD$ respectively. Clearly, $O, M_{1}, E$ and $O, M_{2}, F$ are collinear, and $M_{1} M_{2} \parallel E F$. Other auxiliary lines are added as shown in Figure 5. To prove $M_{1} M_{2} \perp H_{1} H_{2}$, it suffices to prove $E F \perp H_{1} H_{2}$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,491 |
Example 3 Given non-zero real numbers $x, y, z$ satisfy
$$
x+y+z=x y z, x^{2}=y z \text {. }
$$
Prove: $x^{2} \geqslant 3$. | Explanation: Since $y z=x^{2}, y+z=x^{3}-x$, therefore, $y, z$ are the two real roots of the quadratic equation
$$
t^{2}-\left(x^{3}-x\right) t+x^{2}=0
$$
Thus, $\Delta=\left[-\left(x^{3}-x\right)\right]^{2}-4 x^{2} \geqslant 0$, which means
$$
x^{2}\left(x^{2}-3\right)\left(x^{2}+1\right) \geqslant 0 \text {. }
$$
S... | x^{2} \geqslant 3 | Algebra | proof | Yes | Yes | cn_contest | false | 717,492 |
Example 5 As shown in Figure 6, the convex quadrilateral $ABCD$ is inscribed in $\odot O$, the rays $AB, DC$ intersect at point $E$, the rays $BC, AD$ intersect at point $F$, and $AC, BD$ intersect at point $P$. Prove: $OP \perp EF$. | Prove: Take a point $K$ on ray $E P$ such that points $K$, $D$, $C$, and $P$ are concyclic, hence points $E$, $D$, $K$, and $B$ are also concyclic.
Let the radius of $\odot O$ be $R$, and the length of the tangent from point $E$ to $\odot O$ be $t$. Thus,
$$
\begin{array}{l}
E P \cdot E K=E C \cdot E D=t^{2}=O E^{2}-R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,493 |
Example 6 As shown in Figure 7, $A D$ is a common chord of $\odot O_{1}$ and $\odot O_{2}$, and a line through point $D$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $B$ and $C$. Point $E$ is on $A D$, and ray $C E$ intersects $\odot O_{1}$ at points $P$ and $Q$, while ray $B E$ intersects $\odot O_{2}$ at poin... | Proof: Proving that points $P, M, Q, N$ are concyclic is straightforward and is omitted here.
As shown in Figure 7, connect $O_{3} B, O_{3} C$. Let the radius of $\odot O_{3}$ be $r$, and the lengths of the tangents drawn from points $B, C$ to $\odot O_{3}$ be $m, n$ respectively. Thus,
$$
\begin{array}{l}
O_{3} B^{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,494 |
Example 1 In the convex quadrilateral $ABCD$, diagonal $AC$ bisects $\angle BAD$, $E$ is a point on side $CD$, $BE$ intersects $AC$ at point $G$, $DG$ intersects $BC$ at point $F$. Prove: $\angle FAC = \angle EAC$.
(1999, National High School Mathematics Competition) | Proof: As shown in Figure 1, since point $A$ lies on the line of diagonal $CG$ of the complete quadrilateral CFBGDE, and satisfies $\angle BAC = \angle DAC$, it follows from Property 10 of the complete quadrilateral that $\angle FAC = \angle EAC$.
Note: In Figure 11 of [1], Property 10 only points out the following tw... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,495 |
Example 2 Property 7, that is, the midpoints of the three diagonals of a complete quadrilateral are collinear.
This article presents another proof method.
保留了原文的换行和格式，但请注意，原文中的“本文”被翻译为了“This article”以保持语义的准确。 | Proof: As shown in Figure 2, let \( M, N, P \) be the midpoints of the three diagonals \( AD, BF, CE \) of the complete quadrilateral \( ABCDEF \). \( AD \) intersects \( BF \) at point \( K \). Since the lines containing sides \( AB \) and \( FD \) intersect at point \( C \), and the lines containing sides \( AF \) an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,496 |
Example 3 As shown in Figure 3, the extensions of the two pairs of opposite sides of quadrilateral $ABCD$ meet at points $E$ and $F$, the diagonal $BD \parallel EF$, and the extension of $AC$ intersects $EF$ at point $G$. Prove: $EG=GF$. (1978, National High School Mathematics Competition) | In the complete quadrilateral $A B E C F D$, by property 2 (or Ceva's theorem), we have
$$
\frac{A B}{B E} \cdot \frac{E G}{G F} \cdot \frac{F D}{D A}=1 \text {. }
$$
Thus, $\frac{E G}{G F}=1$. Therefore, $E G=G F$.
Proof 2: Let $\frac{A C}{C G}=p_{1}, \frac{E C}{C D}=p_{2}, \frac{F C}{C B}=p_{3}$.
By property 5, we k... | E G=G F | Geometry | proof | Yes | Yes | cn_contest | false | 717,497 |
Example 4 As shown in Figure 4, for any pentagram $A_{1} A_{2} A_{3} A_{4} A_{5}-$ $C_{1} C_{2} C_{3} C_{4} C_{5}$, the circumcircles of the five small triangles intersect at five points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ outside the star. Prove that $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ are concyclic. | Proof: Note that a pentagram can be seen as composed of five complete quadrilaterals. By property 3, each complete quadrilateral has a Miquel point, so this problem reduces to proving that the five Miquel points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ are concyclic.
Let the extension of $B_{2} A_{2}$ intersect the extensi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,498 |
Example 5 In a complete quadrilateral $A B C D E F$, if $O_{1} 、 O_{2} 、 O_{3} 、 O_{4}$ are the circumcenters of $\triangle A C F 、 \triangle A B E 、$ $\triangle D E F 、 \triangle B C D$ respectively, then $O_{1} 、 O_{2}$ 、 $\mathrm{O}_{3} 、 \mathrm{O}_{4}$ are concyclic. | Proof: As shown in Figure 5, by Property 3, we know that $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}$ intersect at point $M$.
Connecting $O_{1} O_{4}, C O_{4}, O_{4} M, M O_{3}, O_{1} O_{3}$, then
$$
\begin{array}{l}
\angle O_{1} O_{4} M=180^{\circ}-\frac{1}{2} \angle C O_{4} M \\
=180^{\circ}-\angle C D M .
\e... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,499 |
Example 6 In a complete quadrilateral $A B C D E F$, if the diagonal $C E$ intersects the extension of diagonal $A D$ at point $G$, and $B G, F G, B F$ are connected, then $S_{\triangle G F B} \leqslant \frac{1}{4} S_{\triangle A B C}$. | Proof: As shown in the figure
$$
\begin{array}{l}
6, \text { let } \frac{A D}{D G}=p_{1}, \\
\frac{C D}{D F}=p_{2}, \frac{E D}{D B}= \\
p_{3}, \frac{C G}{G E}=\lambda_{1}, \frac{E F}{F A} \\
=\lambda_{2}, \frac{A B}{B C}=\lambda_{3} .
\end{array}
$$
By property 5, we know
$$
\begin{array}{l}
p_{1} p_{2} p_{3}=p_{1}+p_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,500 |
Example 7 In a complete quadrilateral ABECFD, diagonals $A C$ and $B D$ intersect at point $P$. Draw $P O$ perpendicular to diagonal $E F$ at point $O$, and connect $B O$, $C O$, $A O$, $D O$. Prove: $\angle B O C=\angle A O D$. | To prove: As shown in Figure 7, let the extension of diagonal $AC$ intersect $EF$ at point $Q$. To prove $\angle BOC = \angle AOD$, it is sufficient to prove
$$
\angle POC = \angle POA \text{ and } \angle POB = \angle POD.
$$
To prove $\angle POC = \angle POA$, it is sufficient to prove
$$
\angle COE = \angle AOF.
$$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,501 |
Example 8 As shown in Figure 8, in the complete quadrilateral $A B C D E F$, point $H$ is on the diagonal $A D$, and point $G$ is on the extension of $A D$. Line $H F$ intersects line $G E$ at point $Q$, and line $H B$ intersects line $G C$ at point $P$.
(1) Prove that lines $P F$, $B Q$, and $A D$ are concurrent;
(2) ... | Proof: (1) Let $A D$ and $B F$ intersect at point $K$. For the complete quadrilateral $A B C D E F$, by property 2, equation (5) (or applying Ceva's theorem to $\triangle A B F$ and point $D$), we have
$$
\frac{B C}{C A} \cdot \frac{A E}{E F} \cdot \frac{F K}{K B}=1.
$$
Furthermore, applying property 1 to the complete... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,502 |
For example, as shown in Figure 9, in the complete quadrilateral $A B C D E F$, $P, Q, R, S$ are points on $A B, B D, D F, F A$ respectively. Prove that the necessary and sufficient condition for the lines $P Q, A D, S R$ to be parallel or concurrent is
$$
\frac{A P}{P B} \cdot \frac{B Q}{Q D} \cdot \frac{D R}{R F} \cd... | Proof: Necessity.
If the lines $P Q, A D, S R$ are parallel to each other, then
$$
\frac{A P}{P B}=\frac{D Q}{Q B}, \frac{F S}{S A}=\frac{F R}{R D} \text {. }
$$
20
Intermediate Mathematics
Thus, $\frac{A P}{P B} \cdot \frac{B Q}{Q D} \cdot \frac{D R}{R F} \cdot \frac{F S}{S A}=1$.
(a)
(b)
Figure 9
If the lines $P Q, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,504 |
(2) Let $k$ be any positive integer, then there exists a smallest highly composite positive integer $n$, such that $\alpha_{p}(n) \geqslant k$, and except for finitely many highly composite integers, all highly composite integers $n^{\prime}$ satisfy
$$
\alpha_{p}\left(n^{\prime}\right) \geqslant k .
$$ | Below is the proof: $\frac{n}{p}$ is also a highly composite integer.
If $\frac{n}{p}$ is not a highly composite integer, then there exists a highly composite integer $m<\frac{n}{p}$, such that $d(m) \geqslant d\left(\frac{n}{p}\right)$.
Since $\alpha_{p}(m)<\alpha_{p}(n)$, we have
$$
\begin{array}{l}
d(m p)=d(m) \frac... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,506 |
Given $A$ is a point outside $\odot O$, and $O A$ intersects $\odot O$ at point $B$. Question: Does there exist a point $Q$ on $O A$ such that for any point $P$ on $\odot O$, $\angle A P Q$ is bisected by $P B$? Please provide the conclusion and prove it.
Conclusion: There exists a unique point $Q$ on $O A$ such that ... | Proof: (i) Determine the number and position of point $Q$.
As shown in Figure 5, draw the tangent line $AM$ from point $A$ to $\odot O$, with $M$ being the point of tangency, and connect $OM$.
Let $Q$ be the point on $OA$ that satisfies the given condition. When point $P$ is at point $M$, since $PB$ must bisect $\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,507 |
Height 185 Through the centroid $G$ of $\triangle A B C$, draw a line intersecting $A C$ and $A B$ at points $D$ and $E$ respectively, $B D$ and $C E$ intersect at point $P$. Let $k$ be the positive real number that for any $\triangle A B C, \triangle D E P$ can satisfy the inequality $S_{\triangle D E P} \leqslant k S... | Solution: Let $\frac{A E}{A B}=x, \frac{A D}{A C}=y$. First, we prove $\frac{1}{x}+\frac{1}{y}=3$.
As shown in Figure 7, draw $B M / / D E$, and draw $C N / / D E$ intersecting line $A G$ at points $M$ and $N$ respectively, then
$$
\begin{array}{l}
\frac{A B}{A E}=\frac{A M}{A G}, \\
\frac{A C}{A D}=\frac{A N}{A G} .
\... | \frac{4}{45} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,508 |
Example 5 Given the function $f(x)=\frac{a x+b}{x^{2}+2}$ has a maximum value of 1 and a minimum value of -2. Find the values of the real numbers $a, i$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: It seems there i... | Let $f(x)=y$, then we have
$$
y x^{2}-a x+(2 y-b)=0 .
$$
Considering this as a quadratic equation in $x$, since $x$ can be any real number, the equation must have real roots. Therefore, we have
$$
\Delta=(-a)^{2}-4 y(2 y-b) \geqslant 0,
$$
which simplifies to $y^{2}-\frac{b}{2} y-\frac{a^{2}}{8} \leqslant 0$.
Given t... | a= \pm 4, b=-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,510 |
Example 6 Let $k$ be any positive integer, and the equations determined by the integer pairs $(a, b)$ are
$$
x^{2}-2 a x+b=0 \text { and } y^{2}+2 a y+b=0
$$
with real roots $x_{1}, x_{2}$ and $y_{1}, y_{2}$, respectively, and
$$
x_{1} y_{1}-x_{2} y_{2}=4 k \text {. }
$$
(1) For a given $k$, find the maximum value of ... | Explanation: The necessary and sufficient condition for two equations to have real roots is $a^{2}-b \geqslant 0$, and
$$
\begin{array}{l}
\left\{x_{1}, x_{2}\right\}=\left\{a+\sqrt{a^{2}-b}, a-\sqrt{a^{2}-b}\right\}, \\
\left\{y_{1}, y_{2}\right\}=\left\{-a+\sqrt{a^{2}-b},-a-\sqrt{a^{2}-b}\right\} .
\end{array}
$$
To... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,511 |
Example 7 Given the inequality
$$
\left|x^{2}-5 x+6\right| \leqslant x+a,
$$
where $a$ is a real parameter. If the inequality has exactly 3 integer solutions, find all values of $a$ that satisfy the condition.
(2005, Bulgarian Winter Mathematical Competition) | Explanation: Since we are considering integer roots, and all integers $x$ satisfy $-\infty<x \leqslant 2$ or $3 \leqslant x<+\infty$, we have
$$
x^{2}-5 x+6 \geqslant 0 \text {. }
$$
Thus, the original inequality becomes
$$
x^{2}-5 x+6 \leqslant x+a \text {, }
$$
which simplifies to $x^{2}-6 x+6-a \leqslant 0$.
Since... | -2 \leqslant a < 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,512 |
Example 2 Find $\sum_{k=0}^{1003} \mathrm{C}_{2006-k}^{k}$. | Explanation: Transform this problem into:
Climbing a 2006-step staircase, one can take 1 step at a time or 2 steps at a time. How many different ways are there to climb it?
On one hand, classify by the number of times taking 2 steps at a time, there are 0, $1, \cdots, 1003$ for a total of 1004 cases (as shown in Table... | a_{2006}=\sum_{k=0}^{1003} \mathrm{C}_{2006-k}^{k} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,514 |
Example 3 In the set $S_{n}=\{1,2, \cdots, n\}$, a subset that does not contain two consecutive natural numbers is called a "good subset". How many good subsets are there in $S_{n}$? | Let $S_{n}$ have $a_{n}$ good subsets.
When $n=1$, $S_{n}$ has 2 subsets, $\varnothing$ and $\{1\}$, both of which are good subsets, $a_{1}=2$;
When $n=2$, $S_{n}$ has 4 subsets, $\varnothing$, $\{1\}$, $\{2\}$, $\{1,2\}$, except for $\{1,2\}$, all are good subsets, $a_{2}=3$.
For $n \geqslant 3$, let $M$ be a good su... | a_{n}=a_{n-1}+a_{n-2}, \\ a_{1}=2, a_{2}=3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,515 |
Example 2 In the circumcircle of $\triangle A B C$, take points $K$ and $L$ on the arcs $\overparen{A B}$ (not containing point $C$) and $\overparen{B C}$ (not containing point $A$), respectively, such that the line $K L$ is parallel to the line $A C$. Prove: The distances from the incenter of $\triangle A B K$ and the... | If $A B=B C$, then the conclusion is obviously true.
Assume $A B<B C$ as shown in Figure 2. Let $I_{1}$ and $I_{2}$ represent the incenters of $\triangle A K B$ and $\triangle C L B$, respectively. Connect $B I_{1}$ and $B I_{2}$ and extend them to intersect the circumcircle of $\triangle A B C$ at points $P$ and $Q$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,516 |
Example 3 Let $I$ be the incenter of $\triangle ABC$, and $P$ be a point inside $\triangle ABC$ such that
$$
\angle PBA + \angle PCA = \angle PBC + \angle PCB.
$$
Prove that $AP \geqslant AI$, and state the necessary and sufficient condition for equality, which is $P=I$. | Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Since $\angle P B A+\angle P C A+\angle P B C+\angle P C B=\beta+\gamma$, by the given condition, we have $\angle P B C+\angle P C B=\frac{\beta+\gamma}{2}$.
Therefore, $\angle B P C=\pi-\frac{\beta+\gamma}{2}=\angle B I C$.
Since points $P$ and $I$ are on the sam... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,517 |
4. As shown in Figure 10, in the acute triangle $\triangle ABC$, $AM$ is the median on side $BC$, and $X$ is the intersection of the lines passing through points $B$ and $C$ and tangent to the circumcircle $\odot O$ of $\triangle ABC$. Prove that: $\frac{AM}{AX}=\cos \angle BAC$. | (提示: Set $A X$ intersects $\odot O$ at point $A_{1}$, connect $O B$, $O C$, $O A_{1}$, $O X$. First prove $\triangle X M A$ $\backsim \triangle X A_{1} O$, then $\frac{A M}{A X}=\frac{O A_{1}}{O X}=\frac{O B}{O X}$. It is also easy to know that $\angle B O X=$ $\angle B A C$, then
$$
\left.\frac{A M}{A X}=\frac{O B}{O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,518 |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2, a_{2}=6$, and $\frac{a_{n+2}+a_{n}}{a_{n+1}+1}=2$. Then $\lim _{n \rightarrow+\infty} \frac{\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}}}{n^{2}}$ $=$ . $\qquad$ | 5. $\frac{1}{2}$.
From the given, we have $a_{n+2}-a_{n+1}=a_{n+1}-a_{n}+2$, thus,
$$
a_{n+1}-a_{n}=\left(a_{2}-a_{1}\right)+2(n-1)=2(n+1) \text {. }
$$
Therefore, $a_{n}=\left(a_{n}-a_{{n}-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+$
$$
\begin{array}{c}
\left(a_{2}-a_{1}\right)+a_{1} \\
=2(n+n-1+\cdots+2)+2=n(n+1... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,519 |
6. Let $a, b, c$ be non-negative real numbers. Then the algebraic expression
$$
\begin{aligned}
S= & \sqrt{\frac{a b}{(b+c)(c+a)}}+\sqrt{\frac{b c}{(c+a)(a+b)}} \\
& +\sqrt{\frac{c a}{(a+b)(b+c)}}
\end{aligned}
$$
has a minimum value of $\qquad$ and a maximum value of $\qquad$. | $6.1, \frac{3}{2}$.
Let $\frac{a}{b+c}=u^{2}, \frac{b}{c+a}=v^{2}, \frac{c}{a+b}=w^{2}$, and set $a+b+c=1$, then $a=\frac{u^{2}}{1+u^{2}}, b=\frac{v^{2}}{1+v^{2}}, c=\frac{w^{2}}{1+w^{2}}$.
Thus, $\frac{u^{2}}{1+u^{2}}+\frac{v^{2}}{1+v^{2}}+\frac{w^{2}}{1+w^{2}}=1$,
which means $(w v)^{2}+(v w)^{2}+(w u)^{2}+2(w w)(v ... | 1, \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,520 |
Three. (20 points) Let real numbers $p, q, r$ satisfy: there exists $a$ which is one of $p, q, r$, and the other two are exactly the two real roots of the equation
$$
x^{2}+(a-3) x+a^{2}-3 a=0
$$
Find the minimum possible value of $p^{3}+q^{3}+r^{3}$. | Three, let $a=p$, then $q, r$ are exactly the two roots of the equation. By Vieta's formulas, we know $p+q+r=3$ and $qr=p^2-3p$.
From $\Delta=(p-3)^2-4(p^2-3p) \geqslant 0$, we know $-1 \leqslant p \leqslant 3$.
$$
\begin{array}{l}
\text { Then } p^2+q^2+r^2=p^2+(q+r)^2-2qr \\
=p^2+(3-p)^2-2(p^2-3p)=9 \text {. }
\end{a... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,521 |
Four. (20 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with an inscribed $\triangle A B C$ where sides $A B$ and $A C$ pass through the left and right foci $F_{1}$ and $F_{2}$ respectively, and the left and right vertices of the ellipse are $D$ and $E$ respectively. The lines $D B$ and $C E$ intersect ... | As shown in Figure 5, let \( A(2 \cos \theta, \sqrt{3} \sin \theta), B(2 \cos \alpha, \sqrt{3} \sin \alpha), C(2 \cos \beta, \sqrt{3} \sin \beta) \), where \(\alpha, \beta, \theta\) are distinct. The line passing through points \(A\) and \(B\) is given by
\[
l_{A E}: y=\frac{\sqrt{3}(\sin \theta-\sin \alpha)}{2(\cos \t... | \frac{x^{2}}{4}+\frac{y^{2}}{27}=1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 717,522 |
Five, color the numbers in $S=\{0,1,2, \cdots, n\}$ with two colors arbitrarily. Find the smallest positive integer $n$, such that there must exist $x, y, z \in S$ of the same color, satisfying $x+y=2 z$. | Let $A=\{0,2,5,7\}$ and $B=\{1,3,4,6\}$. Then neither $A$ nor $B$ contains a 3-term arithmetic progression. By coloring the elements of $A$ red and the elements of $B$ yellow, we can see that $n>7$.
When $n \geqslant 8$, we will prove that there must be a color for which the numbers form a 3-term arithmetic progressio... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,523 |
One, (50 points) Given that $\odot O_{1}$ and $\odot O_{2}$ intersect at two distinct points $A$ and $B$, points $P$ and $E$ are on $\odot O_{1}$, points $Q$ and $F$ are on $\odot O_{2}$, and it is satisfied that $EF$ is a common tangent of the two circles, $PQ \parallel EF$, and $PE$ intersects $QF$ at point $R$. Prov... | As shown in Figure 6, let the radii of $\odot O_{1}$ and $\odot O_{2}$ be $r_{1}$ and $r_{2}$, respectively. Draw a line through point $B$ parallel to $EF$ intersecting $\odot O_{1}$ and $\odot O_{2}$ at points $G$ and $H$, respectively.
Let $B R \cap P Q=Z, B G \cap E P=X, B H \cap F Q=Y$,
$P Q \cap E O_{1}=N_{1}, P Q... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,524 |
$\begin{array}{l}\text { II. (50 points) Given } a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}_{+} \text {, and } \\ \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=\frac{n-1}{2}\left(\sum_{i=1}^{n} a_{i}\right) . \\ \text { Prove: } \sum_{i=1}^{n} \frac{1}{n-1+a_{i}} \leqslant 1 .\end{array}$ | Let $M=\sum_{i=1}^{n} a_{i}, N=\sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}$. Then we have
$$
N=\frac{n-1}{2} M \text {. }
$$
Substituting into the original inequality, we get
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{1}{\frac{2 N}{M}+a_{i}}=\sum_{i=1}^{n} \frac{M}{2 N+a_{i} M} \leqslant 1 \\
\Leftrightarrow \sum_{i=... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,525 |
Three. (20 points) Given a sequence $\left\{a_{n}\right\}$ with all terms no less than 1, satisfying: $a_{1}=1, a_{2}=1+\frac{\sqrt{2}}{2},\left(\frac{a_{n}}{a_{n+1}-1}\right)^{2}+$ $\left(\frac{a_{n}-1}{a_{n-1}}\right)^{2}=2$. Try to find:
(1) The general term formula of the sequence $\left\{a_{n}\right\}$;
(2) The va... | (1) Let $b_{n}=\left(\frac{a_{n+1}-1}{a_{n}}\right)^{2}$.
Then $\frac{1}{b_{n}}+b_{n-1}=2$, and $b_{1}=\frac{1}{2}$.
From this, $b_{2}=\frac{2}{3}, b_{3}=\frac{3}{4}, \cdots$.
Observing, we find $b_{n}=\frac{n}{n+1}$.
We will prove this by mathematical induction.
For $n=1,2,3$, the conclusion is obviously true.
Assume ... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,526 |
As 187 As shown in Figure 2, from a point $P$ outside the circle $\odot O$, two secants $PAB$ and $PCD$ intersect the circle at points $A, B, C, D$ respectively, and the chords $AD$ and $BC$ intersect at point $G$. A secant $PEF$ is drawn through point $G$, intersecting the circle $\odot O$ at points $E$ and $F$, and i... | Proof: As shown in Figure 3, it is easy to see that $\angle PGC > \angle PBC = \angle PDA$.
Take a point $R$ on $PF$, connect $RD$, such that $\angle RDP = \angle PGC$. Then, points $C, D, R, G$ are concyclic. Therefore,
$$
PG \cdot PR = PC \cdot PD = PA \cdot PB.
$$
Thus, points $A, B, R, G$ are concyclic.
Extend $DR... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,527 |
5. As shown in Figure 11, in $\triangle ABC$, let $AB > AC$, and draw the tangent line $l$ to the circumcircle of $\triangle ABC$ at point $A$. Also, draw a circle with center $A$ and radius $AC$ intersecting line segment $AB$ at point $D$ and intersecting line $l$ at points $E$ and $F$. Prove that line $DE$ passes thr... | (提示: As shown in Figure 11, connect $DE$, $DC$, and draw the angle bisector of $\angle BAC$, intersecting $DE$ and $DC$ at points $I$ and $G$ respectively, and connect $IC$. It is easy to see that $ID=IC$. It is easy to prove that points $A$, $I$, $C$, and $E$ are concyclic, so $\angle CIE = \angle CAE = \angle ABC$. S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,529 |
Height 187 As shown in Figure 5, in $\triangle A B C$, $\angle A C B=90^{\circ}$, $\angle A B C=30^{\circ}$, point $D$ is on side $A B$ (excluding endpoints), and point $E$ is on the extension of $C A$, such that $C E+2 B D=\sqrt{3} C B$. Connect $C D$ and $B E$. Prove: $C D=\frac{1}{2} B E$. | Proof: First, we prove the following lemma.
Lemma: Given three rays $PA, PB, PC$ from point $P$, let $\angle APB = \alpha, \angle CPB = \beta, \angle APC = \alpha + \beta < 180^\circ$. If
\[ PC \sin \alpha + PA \sin \beta = PB \sin (\alpha + \beta), \]
then $P, A, B, C$ are concyclic.
Proof of the lemma: As shown in F... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,530 |
Given $n \in \mathbf{N}$, and $n \geqslant 2$. Prove:
$$
\frac{2 n}{n+3}<\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}<\frac{2 n}{\sqrt{3(2 n+3)}} \text {. }
$$ | Proof: From the inequality $\frac{1}{x}+\frac{1}{y} \geqslant \frac{4}{x+y}$ (with equality if and only if $x=y$), we can derive:
$$
\begin{array}{l}
\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1} \\
=\frac{1}{2}\left[\left(\frac{1}{2}+\frac{1}{n+1}\right)+\left(\frac{1}{3}+\frac{1}{n}\right)+\cdots+\left(\frac{1}{n+1}+\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,531 |
6. Given a convex quadrilateral $A B C D$, $I_{1}$ and $I_{2}$ are the incenters of $\triangle A B C$ and $\triangle D B C$ respectively. The line through points $I_{1}$ and $I_{2}$ intersects $A B$ and $D C$ at points $E$ and $F$ respectively. Extend $A B$ and $D C$ to intersect at point $P$, and $P E = P F$. Prove th... | (It is easy to know that $\angle P E F=\angle P F E$. Then, from $\angle P E F=\angle I_{2} I_{1} B-\angle E B I_{1}=\angle I_{2} I_{1} B-\angle I_{1} B C, \angle P F E=\angle I_{1} I_{2} C-\angle F C I_{2}=\angle I_{1} I_{2} C-\angle I_{2} C B$, we get $\angle I_{2} I_{1} B + \angle I_{2} C B=\pi$. Therefore, $I_{1}, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,532 |
Example 10 The sports meet lasted for $n(n>1)$ consecutive days, and a total of $m$ medals were awarded. On the first day, 1 medal and $\frac{1}{7}$ of the remaining $(m-1)$ medals were awarded. On the second day, 2 medals and $\frac{1}{7}$ of the remaining medals were awarded, and this pattern continued for subsequent... | Explanation: Suppose after $k$ days of the sports meet, there are still $a_{k}$ medals left. The number of medals issued on the $k$-th day is
$$
k+\frac{1}{7}\left(a_{k-1}-k\right),
$$
Therefore, $a_{k}=a_{k-1}-\left[k+\frac{1}{7}\left(a_{k-1}-k\right)\right]$.
Thus, $a_{k}=\frac{6}{7} a_{k-1}-\frac{6}{7} k$.
This giv... | n=6, m=36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,533 |
Example 11 In a convex $n$-sided polygon, by non-intersecting diagonals inside the $n$-sided polygon, divide the $n$-sided polygon into non-overlapping triangles. Please express the number of different methods using a recursive formula. | For a polygon $A_{1} A_{2} \cdots A_{n-1} A_{n}$, let the number of valid constructions be $a_{n}$.
Now consider the recursive relationship based on the line segment $A_{1} A_{n}$.
Construct $\triangle A_{1} A_{k} A_{n}(1<k<n)$, then the original convex $n$-gon
is divided into three parts, namely
$\triangle A_{1} A_{k}... | a_{n}=\sum_{k=2}^{n-1} a_{k} a_{n-k+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,534 |
Example 12 If $n$ letters are to be put into $n$ envelopes addressed to $n$ different people, and each letter is placed in the wrong envelope, how many different ways are there to do this? | Explanation: This problem is about the wrong envelope problem, which can be solved using the principle of inclusion-exclusion, bipartite graphs (bipartite graphs), and other methods. Here, we use the recursive method.
Let $a_{n}$ be the number of ways to place the letters incorrectly, then $a_{1}=0, a_{2}=1$.
We will f... | a_{n}=n!\left[1-\frac{1}{1!}+\frac{1}{2!}-\cdots+(-1)^{n} \frac{1}{n!}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,535 |
Example 13 Try to prove that the equation $x^{2}-2 y^{2}=1$ has infinitely many positive integer solutions. | Explanation: This problem is the Pell's equation, and the proof here uses a recursive method.
First, $(3,2)$ is a positive integer solution to the equation, substituting we get $3^{2}-2 \times 2^{2}=1$, i.e.,
$$
(3+2 \sqrt{2})(3-2 \sqrt{2})=1 \text {. }
$$
Squaring it, we get
$$
(17+12 \sqrt{2})(17-12 \sqrt{2})=1 \te... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 717,536 |
Example 14 Prove: There exists a unique function $f$ from the set of positive real numbers $\mathbf{R}_{+}$ to $\mathbf{R}_{+}$, such that when $x>0$, $f(x)>0$ and
$$
f(f(x))=6 x-f(x) .
$$ | For $x>0$, introduce a recursive sequence
$$
\left\{\begin{array}{l}
a_{0}=x, \\
a_{n+1}=f\left(a_{n}\right) .
\end{array}\right.
$$
From the given, we have
$$
a_{n+2}=6 a_{n}-a_{n+1},
$$
which can be rewritten as
$$
a_{n+2}=3 \times 2 a_{n}-(3-2) a_{n+1} \text {. }
$$
Thus, we have
$$
a_{n+2}+3 a_{n+1}=2\left(a_{n+... | f(x)=2x | Algebra | proof | Yes | Yes | cn_contest | false | 717,537 |
Example 15 In a $2 \times n$ strip of squares, some squares are colored, but no $2 \times 2$ square is completely colored (at most 3 squares are colored), and let $P_{n}$ denote the number of different coloring methods that satisfy the condition. Prove: $3 \mid P_{n}$.
---
The translation maintains the original text'... | Explanation: The number of coloring methods that satisfy the conditions can be divided into two categories based on the last column:
(1) Both cells in the last column are colored, denoted as $a_{n}$ such methods;
(2) At most one cell in the last column is colored (in three ways), denoted as $b_{n}$ such methods. Theref... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,538 |
Example 1 Let $x, y \in \mathbf{R}_{+}, A=\min \left\{x, \frac{y}{x^{2}+y^{2}}\right\}$. Try to find $A_{\max }$. | Solution: We discuss in two cases.
(1) If $x \geqslant \frac{y}{x^{2}+y^{2}}$, then $A=\frac{y}{x^{2}+y^{2}}$. Thus,
$$
\begin{array}{l}
A^{2}=\left(\frac{y}{x^{2}+y^{2}}\right)^{2} \leqslant \frac{y}{x^{2}+y^{2}} \cdot x \\
=\frac{x y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2} .
\end{array}
$$
Hence, $A \l... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,539 |
Example 4 Let $I$ and $O$ be the incenter and circumcenter of $\triangle ABC$ respectively. Prove that $\angle AIO \leqslant 90^{\circ}$ if and only if
$$
2 BC \leqslant AB + AC \text{. }
$$ | Explanation: As shown in Figure 4, extend $A I$ to intersect the circumcircle at point $D$, and connect $B D$, $C D$, and $O D$. Then,
$$
\begin{array}{l}
\angle A I O \leqslant 90^{\circ} \\
\Leftrightarrow A I \geqslant I D \\
\Leftrightarrow 2 \leqslant \frac{A D}{D I} .
\end{array}
$$
By property 3 of the incenter... | 2 BC \leqslant AB + AC | Geometry | proof | Yes | Yes | cn_contest | false | 717,540 |
Example 2 Let $x, y \in \mathbf{R}, M=\max || x+y |$, $|x-y|,|1-x|,|1-y|\}$. Try to find the minimum value of $M$. | Solution: (1) If $xy \geqslant 0$, then
$$
|x-y| \leqslant |x| + |y| = |x + y|.
$$
Thus,
$$
\begin{array}{l}
M = \max \{|x + y|, |x - y|, |1 - x|, |1 - y|\} \\
= \max \{|x + y|, |1 - x|, |1 - y|\}.
\end{array}
$$
Since $M$ is the maximum of $|x + y|, |1 - x|, |1 - y|$, $M$ is not less than their arithmetic mean, i.e.... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,541 |
Example 3 Let $x_{i} \geqslant 0(i=1,2, \cdots, 7)$, and satisfy $x_{1}+x_{2}+\cdots+x_{7}=a$ (a constant), denote
$$
A=\max \left\{x_{1}+x_{2}+x_{3}, x_{2}+x_{3}+x_{4}, \cdots, x_{5}+x_{6}+x_{7}\right\} \text {. }
$$
Try to find $A_{\min }$. | Solution: Notice that
$$
\begin{array}{l}
A=\max \left\{x_{1}+x_{2}+x_{3}, x_{2}+x_{3}+x_{4}, \cdots, x_{5}+x_{6}+x_{7}\right\} \\
\geqslant \max \left\{x_{1}+x_{2}+x_{3}, x_{4}+x_{5}+x_{6}, x_{5}+x_{6}+x_{7}\right\} .
\end{array}
$$
Since $x_{i} \geqslant 0(i=1,2, \cdots, 7)$, we have
$$
x_{5}+x_{6}+x_{7} \geqslant x... | \frac{a}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 717,542 |
Example 4 Try to find the maximum value of the function
$$
y(x)=|\cos x+\alpha \cos 2x+\beta \cos 3x|
$$
among all $\alpha$ and $\beta$.
(49th Moscow Mathematical Olympiad) | Solution: For all $\alpha$ and $\beta$, we have
$$
\begin{array}{l}
\max y(x) \geqslant \max \left\{y\left(\frac{\pi}{6}\right), y\left(\frac{5 \pi}{6}\right)\right\} \\
=\max \left\{\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|,\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|\right\} \\
\geqslant \frac{1}{2}\left(... | \frac{\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,543 |
\begin{aligned} & \text { Example } 5 \max _{a, b, c \in \mathbf{R}_{+}} \min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\} \\ = & \end{aligned} | Solution: Let $t=\min \left\{\frac{1}{a}, \frac{1}{b^{2}}, \frac{1}{c^{3}}, a+b^{2}+c^{3}\right\}$. Then $0<t \leqslant \frac{1}{a}, 0<t \leqslant \frac{1}{b^{2}}, 0<t \leqslant \frac{1}{c^{3}}$,
which means $a \leqslant \frac{1}{t}, b^{2} \leqslant \frac{1}{t}, c^{3} \leqslant \frac{1}{t}$.
Therefore, $t \leqslant a+... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,544 |
Example 6 Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ be $n$ distinct real numbers, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=S$. Try to find $\max \min _{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2}$. | Solution: Without loss of generality, let $a_{1}>a_{2}>\cdots>a_{n}$. Denote
$$
M=\min _{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right)^{2} \quad (a_{i}>a_{j} \text{ if } i<j).
$$
Thus, $\sum_{i<j}\left(a_{i}-a_{j}\right)^{2} \geqslant M \sum_{i<j}(j-i)^{2}$
$$
\begin{array}{l}
=M \sum_{k=1}^{n-1}\left(1^{2}+2^{2... | \frac{12 S}{n\left(n^{2}-1\right)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,545 |
Example 7 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}(n \geqslant 2)$, and satisfy $\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$. Try to find $\max _{1 \leq k \leq n}^{\max }\left|x_{k}\right|$. | Solution: First, find the maximum value of $\left|x_{k}\right|$ for a given $k$.
In fact, the given equation is
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+\cdots+ \\
\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2}=2 .
\end{array}
$$
Notice the identity
$$
\begin{aligned}
\left|x_{... | \sqrt{\frac{2}{n+1}\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,546 |
Example 1 Given $x, y, z > 0$, and $x+y+z=1$, prove: $\left(\frac{1}{x^{2}}+x\right)\left(\frac{1}{y^{2}}+y\right)\left(\frac{1}{z^{2}}+z\right) \geqslant\left(\frac{28}{3}\right)^{3}$. | Prove: Let $g=\left(\frac{1}{x^{2}}+x\right)\left(\frac{1}{y^{2}}+y\right)\left(\frac{1}{z^{2}}+z\right)$, then $\ln g=\ln \left(\frac{1}{x^{2}}+x\right)+\ln \left(\frac{1}{y^{2}}+y\right)+\ln \left(\frac{1}{z^{2}}+z\right)$.
Let $f(x)=\ln \left(\frac{1}{x^{2}}+x\right)(0<x<1)$.
Thus, $f(x)$ is convex on $(0,1)$.
There... | \left(\frac{28}{3}\right)^{3} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,547 |
Example 2: Prove that among the inscribed triangles of a fixed circle, the one with the maximum area is the equilateral triangle inscribed in the circle.
| Proof: Let the radius of a circle be $R$, and the sides of the inscribed $\triangle ABC$ be $a$, $b$, and $c$, with the area being $S$. Then,
$$
S=\frac{1}{2} a b \sin C=2 R^{2} \sin A \cdot \sin B \cdot \sin C \text{. }
$$
From the above equation, we have
$\ln S=\ln 2 R^{2}+\ln \sin A+\ln \sin B+\ln \sin C$.
Since $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,548 |
Example 3 Given $x_{i}>0(i=1,2, \cdots, n), x_{1}+$ $x_{2}+\cdots+x_{n} \geqslant x_{1} x_{2} \cdots x_{n}(n \geqslant 2)$, and $1 \leqslant \alpha \leqslant n$. Prove: $\frac{x_{1}^{\alpha}+x_{2}^{\alpha}+\cdots+x_{n}^{\alpha}}{x_{1} x_{2} \cdots x_{n}} \geqslant n^{\frac{\alpha-1}{n-1}}$. | Prove: Given $x>0, n x=x^{n}$, then $x=n^{\frac{1}{n-1}}$.
(1) When $x_{1} x_{2} \cdots x_{n} \geqslant n^{\frac{n}{n-1}}$, since $x^{\alpha}(\alpha \geqslant 1)$ is a convex function on $(0,+\infty)$, thus, by equation (1) we have
$$
\begin{array}{l}
\frac{x_{1}^{\alpha}+x_{2}^{\alpha}+\cdots+x_{n}^{\alpha}}{x_{1} x_{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 717,549 |
Example 4 Given $x, y, z \geqslant 0$, and $x+y+z=\frac{1}{2}$. Prove: $\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1} \leqslant \frac{3 \sqrt{6}}{10}$. | Prove: Let $f(x)=\frac{\sqrt{x}}{4 x+1}$. Then, we have
$$
\begin{array}{l}
2 f^{\prime}(x)=\frac{1-4 x}{\sqrt{x}(4 x+1)^{2}} \\
\Rightarrow 2 f^{\prime \prime}(x)=\frac{24(x-a)(x+b)}{[\sqrt{x}(4 x+1)]^{3}} .
\end{array}
$$
where, $00 \text {. }
$$
Thus, we know $f^{\prime \prime}(x)<0$, and $f(x)$ is a concave funct... | \frac{3 \sqrt{6}}{10} | Inequalities | proof | Yes | Yes | cn_contest | false | 717,550 |
Example 5 Let the incenter of $\triangle ABC$ be $I$, and connect $AI$ with the circumcircle $\odot O$ of $\triangle ABC$ intersecting at another point $E$, and $AE$ intersects $BC$ at point $D$. Let $R$ and $r$ be the circumradius and inradius of $\triangle ABC$ respectively. Prove:
(1) $E$ is the circumcenter of $\tr... | Explanation: (1) As shown in Figure 5, connect $B E$ and $C E$.
Since $B E = C E = E I$, therefore, $E$ is the circumcenter of $\triangle B C I$.
(2) Since $\triangle A B D \backsim \triangle A E C$,
we have
$$
\frac{A D}{A C} = \frac{A B}{A E},
$$
which means
$$
\begin{array}{l}
A D \cdot A E \\
= A B \cdot A C .
\en... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 717,551 |
Given $x, y \in \mathbf{R}$, and satisfy $\sqrt{x+2}+\sqrt{y-5}$ $=6$. Find the minimum and maximum values of $x+2y$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution 1: Since $\frac{\sqrt{x+2}}{6}+\frac{\sqrt{y-5}}{6}=1$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sqrt{x+2}=6 \cos ^{2} \alpha, \\
\sqrt{y-5}=6 \sin ^{2} \alpha .
\end{array}\right. \\
\text { Solving, we get }\left\{\begin{array}{l}
x=36 \cos ^{4} \alpha-2, \\
y=36 \sin ^{4} \alpha+5 .
\end{array}\... | 32 \text{ and } 80 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 717,552 |
1. A house has an even number of lamps distributed across several rooms, with each room having at least 3 lamps. Each lamp shares a switch with exactly one other lamp (not necessarily in the same room). Each time a switch is toggled, the two lamps connected to that switch change their states simultaneously. Prove: For ... | 1. We refer to two lights controlled by the same switch as "twins". If in a room some lights are on and some are off, we call this room "normal". For this, design a sequence of operations: to increase the number of normal rooms in this building. Thus, after a finite number of operations, all rooms can be made normal.
... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,553 |
2. Let $k$ be a fixed positive integer. A company sells hats using the following method: each customer who buys a hat can persuade two people to buy hats, where these two people have not been persuaded by anyone else. Each new customer can also persuade two other people to buy hats. If a person persuades two people, an... | 2. First consider: If $w$ people received the learning machines, what is the minimum value of the number of people $n$ who bought hats?
When $w=1$, the minimum value of $n$ is
$$
2 k+3=1 \times(k+2)+(k+1) ;
$$
When $w=2$, the minimum value of $n$ is
$$
3 k+5=2(k+2)+(k+1) \text {. }
$$
We will prove by mathematical in... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,554 |
3. Given an $m \times n$ grid. If two unit squares share a common edge, they are called "adjacent". If several unit squares can be arranged in a sequence such that any two consecutively arranged squares in this sequence are adjacent, then this sequence of unit squares is called a "path". Each square in the grid is colo... | 3. First, prove a more general conclusion.
Assume the top and bottom sides of each unit square are the objects of consideration, some unit squares are transparent, and others are opaque. A transparent unit square is colored on only one side, and it looks the same from above and below. An opaque unit square must be col... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,555 |
4. Given a positive integer $n(n \geqslant 3)$. Label each side and each diagonal of the regular $n$-gon $P_{1} P_{2} \cdots P_{n}$ with a positive integer not exceeding $r$, and satisfy:
(1) Each positive integer $1,2, \cdots, r$ appears in the labels of the sides or diagonals;
(2) In each $\triangle P_{i} P_{j} P_{k}... | 4. A labeling that satisfies condition (2) is called "good", and a labeling that satisfies both conditions (1) and (2) is called "very good". The diagonals and sides of the polygon are collectively referred to as edges, and the numbers labeled on the edges are called the distances between two points.
Let the edge labe... | f(n) = \frac{n!(n-1)!}{2^{n-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 717,556 |
5. There are $n$ bookmarks, each with one side white and the other side black. Arrange them in a row, with all bookmarks showing the white side up. Each operation (if possible) involves removing a white-side-up bookmark (not at the edge) and flipping the two adjacent bookmarks to the other side. Prove: The necessary an... | 5. If the bookmark with the white side up is called a "white bookmark" and the opposite is called a "black bookmark," then the parity of the number of black bookmarks remains unchanged. Therefore, if only two bookmarks are left, these two bookmarks must be the same color.
If a white bookmark has $t$ black bookmarks to... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,557 |
7. Given a positive integer $n(n>1)$, the integer sequence $a_{1}, a_{2}$, $\cdots, a_{n}$ satisfies $n!\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. Prove: There exist two permutations $\sigma, \tau$ of $1,2, \cdots, n$, such that for all $i=1,2, \cdots, n$, we have
$$
\sigma(i)+\tau(i) \equiv a_{i}(\bmod n) .
$$ | 7. Suppose there exist permutations $\sigma, \tau$ that satisfy the conditions. If the integer sequence $b_{1}$, $b_{2}, \cdots, b_{n}$ satisfies $n \mid\left(b_{1}+b_{2}+\cdots+b_{n}\right)$, and $b_{1}, b_{2}, \cdots, b_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ differ in only two indices $i_{1}, i_{2}$ modulo $n$, then ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 717,558 |
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