Suppose that $a$ and $b$ are positive real numbers such that $3\log_{101}\left(\frac{1030301-a-b}{3ab}\right) = 3 - 2 \log_{101}(ab)$. Find $101 - \sqrt[3]{a}-  \sqrt[3]{b}$.

Solution:
1. Start with the given equation:
   \[
   3\log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = 3 - 2 \log_{101}(ab)
   \]

2. Divide both sides by 3:
   \[
   \log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = 1 - \frac{2}{3} \log_{101}(ab)
   \]

3. Let \( x = \log_{101}(ab) \). Then the equation becomes:
   \[
   \log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = 1 - \frac{2}{3} x
   \]

4. Rewrite the right-hand side using properties of logarithms:
   \[
   \log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = \log_{101}(101) - \log_{101}((ab)^{2/3})
   \]

5. Since \(\log_{101}(101) = 1\), we have:
   \[
   \log_{101}\left(\frac{1,030,301 - a - b}{3ab}\right) = \log_{101}\left(\frac{101}{(ab)^{2/3}}\right)
   \]

6. Equate the arguments of the logarithms:
   \[
   \frac{1,030,301 - a - b}{3ab} = \frac{101}{(ab)^{2/3}}
   \]

7. Cross-multiply to clear the fraction:
   \[
   (1,030,301 - a - b)(ab)^{2/3} = 303ab
   \]

8. Let \( \sqrt[3]{a} = x \) and \( \sqrt[3]{b} = y \). Then \( ab = x^3 y^3 \) and the equation becomes:
   \[
   (101^3 - x^3 - y^3)(x^2 y^2) = 303x^3 y^3
   \]

9. Simplify the equation:
   \[
   101^3 x^2 y^2 - x^5 y^2 - x^2 y^5 = 303x^3 y^3
   \]

10. Factor out \( x^2 y^2 \):
    \[
    101^3 - x^3 - y^3 = 303xy
    \]

11. Let \( s = x + y \) and \( p = xy \). Then:
    \[
    101^3 - s^3 + 3sp = 303p
    \]

12. Rearrange the equation:
    \[
    101^3 = s^3 - 3sp + 303p
    \]

13. Since \( s = 101 \), substitute \( s \) into the equation:
    \[
    101^3 = 101^3 - 3 \cdot 101 \cdot p + 303p
    \]

14. Simplify:
    \[
    0 = -303p + 303p
    \]

15. This confirms that \( s = 101 \) is correct. Therefore:
    \[
    101 - \sqrt[3]{a} - \sqrt[3]{b} = 101 - s = 101 - 101 = 0
    \]

The final answer is \(\boxed{0}\).

Answer: 0


In $\triangle A B C$, it is known that $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=2, C D=3$, then $S_{\triangle A B C}$ $=$ $\qquad$

Solution:
Solve as shown in Figure 5, with $AB$ as the axis of symmetry, construct the symmetric figure of $\triangle ADB$ as $\triangle AGB$, and with $AC$ as the axis of symmetry, construct the symmetric figure of $\triangle ADC$ as $\triangle AFC$, and extend $GB$ and $FC$ to intersect at point $E$. Then it is easy to know that quadrilateral $AGEF$ is a square.
Assume $AD=h$. Then
$$
\begin{array}{l}
BE=h-2, CE=h-3 . \\
\text { By } BC^{2}=BE^{2}+CE^{2} \\
\Rightarrow(h-2)^{2}+(h-3)^{2}=5^{2} \\
\Rightarrow h^{2}-5 h-6=0 \\
\Rightarrow h=6 \\
\Rightarrow S_{\triangle ABC}=\frac{1}{2} BC \cdot AD=\frac{1}{2} \times 5 \times 6=15 .
\end{array}
$$

Answer: 15


Given positive real numbers $x, y$ satisfy $\frac{(x+2)^{2}}{y}+\frac{(y+2)^{2}}{x}=16$, then $x+y=$

Solution: $16=\frac{(x+2)^{2}}{y}+\frac{(y+2)^{2}}{x} \geqslant \frac{(x+y+4)^{2}}{x+y}=x+y+\frac{16}{x+y}+8 \geqslant 16$, equality holds when $x+y=4$.

Answer: 4


Solution:
Solution: From $(1+x)^{11}=\sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k}$, we have
$$
\begin{array}{l}
\int_{0}^{1}(1+x)^{11} \mathrm{~d} x=\int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x . \\
\text { and } \int_{0}^{1} \sum_{k=0}^{11} \mathrm{C}_{11}^{k} x^{k} \mathrm{~d} x \\
=\left.\left(\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k} x^{k+1}}{k+1}\right)\right|_{0} ^{1}=\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1} \text {, } \\
\int_{0}^{1}(1+x)^{11} \mathrm{~d} x=\left.\frac{(1+x)^{12}}{12}\right|_{0} ^{1}=\frac{1365}{4} \text {. } \\
\end{array}
$$

Therefore, $\sum_{k=0}^{11} \frac{\mathrm{C}_{11}^{k}}{k+1}=\frac{1365}{4}$.
Similarly, if we first differentiate the expansion of $(1+x)^{n}$, then multiply both sides by $x$, and differentiate again, and finally set $x=1$, we can prove Example 1; if we differentiate $(1+x)^{n}$ twice and then set $x=1$, we can prove the following conclusion
$$
\begin{array}{l}
2 \times 1 \times \mathrm{C}_{n}^{2}+3 \times 2 \times \mathrm{C}_{n}^{3}+\cdots+n(n-1) \mathrm{C}_{n}^{n} \\
=n(n-1) 2^{n-2} .
\end{array}
$$

Answer: \frac{1365}{4}


Let $\Gamma$ be the maximum possible value of $a+3b+9c$ among all triples $(a,b,c)$ of positive real numbers such that
\[ \log_{30}(a+b+c) = \log_{8}(3a) = \log_{27} (3b) = \log_{125} (3c) .\]
If $\Gamma = \frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers, then find $p+q$.

Solution:
1. We start by setting all the logarithmic expressions equal to a common variable \( k \):
   \[
   \log_{30}(a+b+c) = k, \quad \log_{8}(3a) = k, \quad \log_{27}(3b) = k, \quad \log_{125}(3c) = k
   \]

2. Converting these logarithmic equations to exponential form, we get:
   \[
   30^k = a + b + c
   \]
   \[
   8^k = 3a \implies 2^{3k} = 3a \implies a = \frac{2^{3k}}{3}
   \]
   \[
   27^k = 3b \implies 3^{3k} = 3b \implies b = 3^{3k-1}
   \]
   \[
   125^k = 3c \implies 5^{3k} = 3c \implies c = \frac{5^{3k}}{3}
   \]

3. We now multiply the expressions for \(a\), \(b\), and \(c\):
   \[
   a \cdot b \cdot c = \left(\frac{2^{3k}}{3}\right) \cdot 3^{3k-1} \cdot \left(\frac{5^{3k}}{3}\right) = \frac{2^{3k} \cdot 3^{3k-1} \cdot 5^{3k}}{9}
   \]

4. Simplifying the product:
   \[
   a \cdot b \cdot c = \frac{2^{3k} \cdot 3^{3k-1} \cdot 5^{3k}}{9} = \frac{(2 \cdot 3 \cdot 5)^{3k}}{9 \cdot 3} = \frac{30^{3k}}{27}
   \]

5. Using the equation \(30^k = a + b + c\), we substitute \(a + b + c\) into the product:
   \[
   (a + b + c)^3 = 30^{3k}
   \]
   \[
   (a + b + c)^3 = 27 \cdot a \cdot b \cdot c
   \]

6. From the AM-GM inequality, we know:
   \[
   \frac{a + b + c}{3} \geq \sqrt[3]{abc}
   \]
   Equality holds when \(a = b = c\). Therefore, we set \(a = b = c\).

7. Substituting \(a = b = c\) into the logarithmic equations, we get:
   \[
   2^{3k} = 3a, \quad 3^{3k} = 3b, \quad 5^{3k} = 3c
   \]
   Since \(a = b = c\), we have:
   \[
   2^{3k} = 3^{3k} = 5^{3k}
   \]
   This implies \(k = 0\).

8. Substituting \(k = 0\) back into the equations, we get:
   \[
   a = b = c = \frac{1}{3}
   \]

9. Finally, we calculate \(a + 3b + 9c\):
   \[
   a + 3b + 9c = \frac{1}{3} + 3 \left(\frac{1}{3}\right) + 9 \left(\frac{1}{3}\right) = \frac{1}{3} + 1 + 3 = \frac{13}{3}
   \]

10. Since \(\Gamma = \frac{13}{3}\), and \(p = 13\) and \(q = 3\) are relatively prime, we find:
   \[
   p + q = 13 + 3 = 16
   \]

The final answer is \(\boxed{16}\).

Answer: 16


Find the $ n $-th term of the sequence of numbers $ a_1, a_2, a_n, \ldots $, where $ a_1 = 1 $, $ a_2 = 3 $, $ a_3 = 6 $, and for every natural number $ k $

Solution:
We can write equality (1) in the form

Let's denote $ a_{i+1} - a_i = r_i $ for $ i = 1, 2, 3, \ldots $; according to the above equality, $ r_{k+2} - 2r_{k+1} + r_k = 0 $, from which $ r_{k+2} - r_{k+1}= r_{k+1} - r_k $.
The sequence of differences $ r_k $ is therefore an arithmetic progression. Since $ r_1 = a_2 - a_1 = 2 $ and $ r_2 = a_3 - a_2 = 3 $, the difference of this progression is $ r_2 - r_1 = 1 $. Hence, for every natural $ k $, we have $ r_k = r_1 + (k - 1) \cdot 1 = k + 1 $, thus

for every natural $ k $. From this, it easily follows (e.g., by induction) that $ a_n $ is the sum of natural numbers from $ 1 $ to $ n $:

Answer: \frac{n(n+1)}{2}


Solution: $\begin{array}{l}\left(x-\frac{1}{x}\right)^{2}=2007 \\ x^{2}-2+\frac{1}{x^{2}}=2007 \\ x^{2}+\frac{1}{x^{2}}=2009 \\ \left(x^{2}+\frac{1}{x^{2}}\right)^{2}=2009^{2}=4036081 \\ x^{4}+2+\frac{1}{x^{4}}=4036081 \\ x^{4}+\frac{1}{x^{4}}=4036079\end{array}$

Answer: 4036079


Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$

where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$

Solution:
10. 112 .

From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$

When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
When $n=6$, $S_{7}-S_{6}<0$.
Therefore, $S_{n}$ is monotonically decreasing in the interval $[1,7]$ and monotonically increasing in the interval $[7,+\infty)$.
Hence, the minimum value of $S_{n}$ is $S_{7}=112$.

Answer: 112


Dividing 1722 by a two-digit number, Xiao Ming mistakenly reversed the tens and units digits of this number, resulting in the incorrect answer of 42. What should the correct result be? $\qquad$ .

Solution:
【Analysis】Based on the dividend and the incorrect quotient, find the wrong divisor, then swap the tens and units digits of the wrong divisor to find the correct divisor, and then find the correct quotient.

【Solution】Solution: $1722 \div 42=41$,
so the correct divisor is 14,
$$
1722 \div 14=123 ;
$$

Answer: The correct result should be 123.
Therefore, the answer is: 123.
【Comment】The key to solving this problem is to first find the wrong divisor, then derive the correct divisor, and finally use the dividend $\div$ divisor = quotient.

Answer: 123


Find all functions $f: \mathbb{Z} \longrightarrow \mathbb{Z}$ such that $f(p)>0$ for all prime $p$ and such that for all prime $p$ and for all $x \in \mathbb{Z}$:

$$
p \mid (f(x) + f(p))^{f(p)} - x
$$

Solution:
We take $x=p$ an odd prime, which gives $p \mid f(p)$ for all odd prime $p$. In particular, $x=0$ gives $p \mid f(0)$ for all odd prime $p$, so $f(0)=0$. With $x=0$ and $p=2$, we get $2 \mid f(2)$.

Our intuition is that the solution will likely be the identity, so the only prime factors of $f(n)$ are those of $n$. Let's now show that $f(q)$ has only $q$ as a prime factor.

Let $p$ be a factor of $f(q)$, take $x=q$. This gives: $p \mid (f(p)+f(q))^{f(p)}-q$, so $p \mid q$. This is obviously impossible. Therefore, $f(q)=q^{\alpha}$.

We suspect that $f$ is the identity, so we will try to show that $f(x)-x$ has infinitely many divisors. We have in particular $f(p) \equiv 0 \pmod{p}$ and $f(x)^{f(p)} \equiv f(x) \pmod{p}$ by Fermat's little theorem, so: $p \mid (f(x)+f(p))^{f(p)}-x$, thus $p \mid f(x)^{f(p)}-x$, hence $p \mid f(x)-x$. Therefore, $f(x)-x$ has infinitely many divisors and is 0.

Answer: f(x)=x


Let $ p,q,r$ be distinct real numbers that satisfy: $ q=p(4-p), \, r=q(4-q), \, p=r(4-r).$ Find all possible values of $ p+q+r$.

Solution:
1. Let \( p, q, r \) be distinct real numbers that satisfy the given system of equations:
   \[
   q = p(4 - p), \quad r = q(4 - q), \quad p = r(4 - r).
   \]
   We introduce new variables \( a, b, c \) such that:
   \[
   a = p - 2, \quad b = q - 2, \quad c = r - 2.
   \]
   This transforms the system into:
   \[
   b = f(a), \quad c = f(b), \quad a = f(c),
   \]
   where \( f(x) = 2 - x^2 \).

2. We need to solve the equation \( a = f(f(f(a))) \). Substituting \( f(x) = 2 - x^2 \), we get:
   \[
   f(f(f(a))) = 2 - (2 - (2 - a^2)^2)^2.
   \]
   Therefore, we need to solve:
   \[
   f(f(f(a))) - a = 0.
   \]

3. To find the roots of this equation, we note that \( f(a) = a \) gives the fixed points of \( f \):
   \[
   2 - a^2 = a \implies a^2 + a - 2 = 0 \implies (a - 1)(a + 2) = 0.
   \]
   Thus, \( a = 1 \) or \( a = -2 \).

4. We now consider the polynomial \( f(f(f(a))) - a \):
   \[
   f(f(f(a))) - a = 2 - (2 - (2 - a^2)^2)^2 - a.
   \]
   This polynomial is of degree 8. The roots of this polynomial include \( a = 1 \) and \( a = -2 \). The sum of the roots of a polynomial \( P(x) = 0 \) of degree \( n \) with leading coefficient 1 is given by Vieta's formulas as the negative of the coefficient of \( x^{n-1} \) divided by the leading coefficient. Here, the sum of the roots of \( f(f(f(a))) - a = 0 \) is zero because the coefficient of \( a^7 \) is zero.

5. The polynomial \( f(f(f(a))) - a \) factors as:
   \[
   (a - 1)(a + 2)(a^3 - 3a - 1)(a^3 - a^2 - 2a + 1) = 0.
   \]
   The roots of \( a^3 - 3a - 1 = 0 \) and \( a^3 - a^2 - 2a + 1 = 0 \) are the remaining roots.

6. We need to verify that the roots of \( a^3 - 3a - 1 = 0 \) and \( a^3 - a^2 - 2a + 1 = 0 \) form triples \( \{a_1, f(a_1), f(f(a_1))\} \) and \( \{a_2, f(a_2), f(f(a_2))\} \) respectively. The sum of the roots of each cubic polynomial is zero by Vieta's formulas.

7. Therefore, the sum of the roots of \( f(f(f(a))) - a = 0 \) is zero, and the sum of the roots of the two cubic polynomials is zero. The total sum of \( a, b, c \) is zero.

8. Finally, we need to find \( p + q + r \):
   \[
   p + q + r = (a + 2) + (b + 2) + (c + 2) = a + b + c + 6.
   \]
   Since \( a + b + c = 0 \), we have:
   \[
   p + q + r = 6.
   \]

The final answer is \(\boxed{6}\).

Answer: 6


A certain natural number has only two prime divisors (in some powers), and its square has 35 different divisors. How many different divisors does the cube of this number have?

Solution:
$\triangle$ Let the original number have the form:

$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}
$$

where $p_{1}$ and $p_{2}$ are distinct prime numbers, and $\alpha_{1}$ and $\alpha_{2}$ are natural numbers. Then

$$
a^{2}=p_{1}^{2 \alpha_{1}} \cdot p_{2}^{2 \alpha_{2}}, \quad a^{3}=p_{1}^{3 \alpha_{1}} \cdot p_{2}^{3 \alpha_{2}}
$$

Since $a^{2}$ has 35 distinct divisors, we have

$$
\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right)=35
$$

From this, we obtain:

$$
2 \alpha_{1}+1=5, \quad 2 \alpha_{2}+1=7 ; \quad \alpha_{1}=2, \quad \alpha_{2}=3
$$

Therefore,

$$
\tau\left(a^{3}\right)=\left(3 \alpha_{1}+1\right)\left(3 \alpha_{2}+1\right)=7 \cdot 10=70
$$

Answer: 70.

$1001^{\circ}$. Prove that a natural number has an odd number of distinct divisors if and only if it is a square of a natural number.

$\triangle$ Apply the statement of problem 995 (with the same notation). The product

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)
$$

is odd if and only if all its factors are odd, i.e., when all numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ are even, i.e., when the original number

$$
a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{n}^{\alpha_{n}}
$$

is a square of a natural number.

Answer: 70


Three circles with radii 1, 1, and \(2 \sqrt{\frac{13-6 \sqrt{3}}{13}}\) are arranged such that the triangle formed by their centers is equilateral with a side length of \(\sqrt{3}\). Find the radius of the circumcircle of the triangle, each vertex of which is the point of intersection of two of these circles, farthest from the center of the third circle.

Answer: \(4 \sqrt{3}-6\)

Solution:
# Solution:

Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.

The radii of the first and second circles are equal, so point $A$ will lie on the perpendicular bisector of $O_{1} O_{2}$. The triangle formed by the centers of the circles is equilateral, so point $O_{3}$ will also lie on this perpendicular bisector. Therefore, the entire picture is symmetric with respect to this perpendicular bisector, in particular, points $B$ and $C$ are symmetric to each other.

Triangle $O_{1} A O_{2}$ is isosceles with vertex $A$; moreover, its sides are 1, 1, and $\sqrt{3}$, from which its angles can be easily found: $30^{\circ}, 30^{\circ}$, and $120^{\circ}$.

Let $\angle O_{1} A C$ be $\alpha$ and $\angle C O_{1} O_{3}$ be $\beta$. In triangle $C O_{1} O_{3}$, we know all the sides, so we can find

$$
\cos \beta=\frac{1+3-4 \cdot \frac{13-6 \sqrt{3}}{13}}{2 \cdot 1 \cdot \sqrt{3}}=\frac{12}{13}
$$

Accordingly, $\sin \beta=\frac{5}{13}$. Next, $\angle A O_{1} C=\angle A O_{1} O_{2}+\angle O_{2} O_{1} O_{3}+\angle O_{3} O_{1} C=30^{\circ}+60^{\circ}+\beta=90^{\circ}+\beta$

Triangle $C O_{1} A$ is isosceles with angles $\alpha, 90^{\circ}+\beta$, and $\alpha$, so $\alpha=\frac{90^{\circ}-\beta}{2}$, from which

$$
\cos \alpha=\sqrt{\frac{1+\cos \left(90^{\circ}-\beta\right)}{2}}=\sqrt{\frac{1+\sin \beta}{2}}=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}
$$

Accordingly, $\sin \alpha=\sqrt{\frac{4}{13}}=\frac{2}{\sqrt{13}}$. Moreover, $A C=2 O_{1} A \cos \alpha=2 \cdot 1 \cdot \frac{3}{\sqrt{13}}=\frac{6}{\sqrt{13}}$.

$\angle B A C=120-2 \alpha=30+\beta$, so

$$
\begin{aligned}
& \sin \angle B A C=\frac{1}{2} \cdot \frac{12}{13}+\frac{\sqrt{3}}{2} \cdot \frac{5}{13}=\frac{12+5 \sqrt{3}}{26} \\
& \cos \angle B A C=\frac{\sqrt{3}}{2} \cdot \frac{12}{13}-\frac{1}{2} \cdot \frac{5}{13}=\frac{12 \sqrt{3}-5}{26} \\
& \sin \angle A B C=\cos \frac{\angle B A C}{2}=\sqrt{\frac{1+\cos \angle B A C}{2}}=\sqrt{\frac{21+12 \sqrt{3}}{52}}=\frac{2 \sqrt{3}+3}{2 \sqrt{13}}
\end{aligned}
$$

By the Law of Sines,

$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{\frac{6}{\sqrt{13}}}{2 \cdot \frac{2 \sqrt{3}+3}{2 \sqrt{13}}}=\frac{6}{2 \sqrt{3}+3}=\frac{6(2 \sqrt{3}-3)}{3}=4 \sqrt{3}-6
$$

Answer: 4\sqrt{3}-6


Let $n$ be a natural number. In the plane, there are $2n+2$ points, no three of which lie on the same line. A line in the plane is a separator if it passes through two of the given points and on each side of this line there are exactly $n$ points. Determine the largest number $m$ such that there are always at least $m$ separators in the plane.

Solution:
IV /4. We will show that there are at least $n+1$ lines that pass through two of the given points and are separators. First, let's give an example where the number of separators is exactly $n+1$. Let the points be the vertices of a regular $(2 n+2)$-gon. Denote them by $A_{1}, A_{2}, \ldots, A_{2 n+2}$. For each $1 \leq i \leq n+1$, it is clear that the line $A_{i} A_{i+n+1}$ is a separator. No other line is a separator, so there are exactly $n+1$ separators.

We will also show that there are always at least $n+1$ separators regardless of the position of the points. We will show that through each of the given points, there is at least one separator. Let $A$ be one of the points and $A_{1}, A_{2}, \ldots, A_{2 n+1}$ the remaining points. Consider the lines $A A_{1}, A A_{2}, \ldots, A A_{2 n+1}$. Color the part of the plane that lies on one side of the line $A A_{1}$ red. Let $r_{1}$ be the number of points in the red part (the line $A A_{1}$ is not in the red part). We will rotate the red part of the plane around the point $A$ in the positive direction. In each step, we rotate so that the red part borders the next possible line among $A A_{1}, A A_{2}, \ldots, A A_{2 n+1}$. We can assume that the lines follow in the order $A A_{1}, A A_{2}, \ldots, A A_{2 n+1}$ and then repeat $A A_{1}$. Let $r_{i+1}$ be the number of points in the red part of the plane after we have rotated the red part $i$ times.

Suppose that in some step we had the line $A A_{i}$ and the corresponding red part of the plane, and then we rotated the red part so that it bordered the line $A A_{i+1}$. We distinguish four cases based on the position of the points $A_{i}$ and $A_{i+1}$. If the point $A_{i+1}$ was in the original red part and $A_{i}$ is in the rotated red part, then $r_{i}=r_{i+1}$. If the point $A_{i+1}$ was in the original red part and $A_{i}$ is not in the rotated red part, the number of points in the red part has decreased by 1, so $r_{i+1}=r_{i}-1$. Suppose that $A_{i+1}$ was not in the original red part. If $A_{i}$ is not in the new red part, then $r_{i+1}=r_{i}$, otherwise $r_{i+1}=r_{i}+1$.

In each step, the number of red points changes by at most 1. All numbers are integers. Initially, there were $r_{1}$ points in the red part, and after $2 n+1$ rotations, we return to the situation where the red part borders the line $A A_{1}$, but on the opposite side of the line $A A_{1}$ as it was initially. Therefore, the number of points in the red part at the end is $2 n-r_{1}$. If $r_{1}=n$, then the line $A A_{1}$ is a separator. Otherwise, one of the numbers $r_{1}$ or $2 n-r_{1}$ is greater than $n$, and the other is less than $n$. During the rotation of the red part, the numbers $r_{i}$ change by 1, which means there is a step where $r_{i}=n$. The corresponding line $A A_{i}$ is then a separator.

We have shown that there is a separator through each point. Since there are $2 n+2$ points, we get $2 n+2$ lines, each counted twice (once for each point that lies on it). Thus, there are always at least $\frac{2 n+2}{2}=n+1$ separators.

Answer: n+1


Solution:
Solution (1) Since $f(1)=1, f(2)=2, \frac{f(n+2)}{f(n)}=\frac{f^{2}(n+1)+1}{f^{2}(n)+1}(n \geqslant 1)$, it is easy to see that for all $n \geqslant 1, f(n) \neq 0$.

When $n \geqslant 1$, from $\frac{f(n+2)}{f(n)}=\frac{f^{2}(n+1)+1}{f^{2}(n)+1}$ we can get $\frac{f(n+2) f(n+1)}{f^{2}(n+1)+1}=\frac{f(n+1) f(n)}{f^{2}(n)+1}$, thus $-\frac{f(n+2)}{f(n+1)+\frac{1}{f(n+1)}}=\frac{f(n+1)}{f(n)+\frac{1}{f(n)}}$.

By successively using the above relation, we get $\frac{f(n+1)}{f(n)+\frac{1}{f(n)}}=\frac{f(n)}{f(n-1)+\frac{1}{f(n-1)}}=\frac{f(n-1)}{f(n-2)+\frac{1}{f(n-2)}}=$ $\cdots=\frac{f(2)}{f(1)+\frac{1}{f(1)}}=\frac{2}{1+\frac{1}{1}}=1$, hence $f(n+1)=f(n)+\frac{1}{f(n)}$.
(2) Clearly, from $f(1)=1$ and $f(n+1)=f(n)+\frac{1}{f(n)}(n \geqslant 1)$, we know: for all $n \geqslant 1$, $f(n) \geqslant 1$ holds. Therefore, $06022$, so $63^{2}<f^{2}(2008)<78^{2}$, thus $63<$ $f(2008)<78$.

Answer: 63<f(2008)<78


Let the quadratic function $f(x)$ satisfy
$$
f(x+2)=f(-x+2)
$$

and its graph intersects the $y$-axis at the point $(0,1)$, and the segment it intercepts on the $x$-axis is $2 \sqrt{2}$. Find the analytical expression of $f(x)$.

Solution:
Explanation: Utilizing the symmetry of the graph of a quadratic function, and combining it with the fact that the length of the segment it intercepts on the $x$-axis is $2 \sqrt{2}$, we know that the intersection points of $f(x)$ with the $x$-axis are $(2+\sqrt{2}, 0)$ and $(2-\sqrt{2}, 0)$. Therefore, we can use the zero-product form to solve.
Let $f(x)=a(x-2+\sqrt{2})(x-2-\sqrt{2})$.
Since the graph of the quadratic function passes through the point $(0,1)$, substituting into the undetermined form gives $a=\frac{1}{2}$. Thus,
$$
f(x)=\frac{1}{2}(x-2+\sqrt{2})(x-2-\sqrt{2}),
$$

which simplifies to $f(x)=\frac{1}{2} x^{2}-2 x+1$.

Answer: f(x)=\frac{1}{2} x^{2}-2 x+1


A road is 400 meters long, and on both sides of the road, a trash can is placed every 20 meters, with the start and end points being bus stops where no trash cans are placed. How many trash cans are placed in total?

Solution:
【Solution】Solution: $400 \div 20-1$
$$
\begin{array}{l}
=20-1 \\
=19 \text { (units) } \\
19 \times 2=38 \text { (units) }
\end{array}
$$

Answer: A total of 38 trash bins need to be placed.

Answer: 38


An infinite sequence of integers, $a_{0}, a_{1}, a_{3} \ldots$, with $a_{0}>0$, has the property that for any $n \geq 0, a_{n+1}=a_{n}-b_{n}$, where $b_{n}$ is the number having the same sign as $a_{n}$, but having the digits written in the reverse order. For example if $a_{0}=1210$, $a_{1}=1089$ and $a_{2}=-8712$, etc. Find the smallest value of $a_{0}$ so that $a_{n} \neq 0$ for all $n \geq 1$.

Solution:
4. If $a_{0}$ has a single digit, then $a_{1}=0$. Thus $a_{0}$ has at least 2 digits. If $a_{0}=\overline{a b}=10 a+b$, then $a_{1}=9(a-b)$ which is divisible by 9 . it follows that all subsequent terms are divisible by 9. Checking all 2-digit multiples of 9 shows that eventually 9 appears (Note that $\overline{a b}$ and $\overline{b a}$ give rise to the same sequence, but with opposite signs):
$$
81 \rightarrow 63 \rightarrow 27 \rightarrow 45 \rightarrow 9 .
$$

If $a_{0}=\overline{a b c}$, then $a_{1}=99(a-c)$. Thus if suffices to investigate 3-digit multiples of 99 , i.e., $198, \ldots, 990$. Here we find that 99 will eventually appear:
$$
990 \rightarrow 891 \rightarrow 693 \rightarrow 297 \rightarrow-495 \rightarrow 99 \text {. }
$$

If $a_{0}=\overline{a b c d}$, then $a_{1}=999(a-d)+90(b-c)$. If $b, c$ are both 0 , then $a_{1}$ and all subsequent terms are multiples of 999 . However, if such numbers appear in the sequence, eventually 999 will appear:
$$
9990 \rightarrow 8991 \rightarrow 6993 \rightarrow 2997 \rightarrow-4995 \rightarrow 999 .
$$

For 1010, we get 909 and for 1011 we get -90. For 1012, we get
$$
1012 \rightarrow-1089 \rightarrow-8712 \rightarrow 6534 \rightarrow 2178 \rightarrow-6534
$$
and the sequence becomes periodic thereafter. Thus the smallest $a_{0}=1012$.

Answer: 1012


Find the real number $k$ such that $a$, $b$, $c$, and $d$ are real numbers that satisfy the system of equations
\begin{align*}
abcd &= 2007,\\
a &= \sqrt{55 + \sqrt{k+a}},\\
b &= \sqrt{55 - \sqrt{k+b}},\\
c &= \sqrt{55 + \sqrt{k-c}},\\
d &= \sqrt{55 - \sqrt{k-d}}.
\end{align*}

Solution:
1. **Substitute \( c' = -c \) and \( d' = -d \):**
   Let \( c' = -c \) and \( d' = -d \). Then, for \( x \in \{a, b\} \), we have \( x = \sqrt{55 \pm \sqrt{k+x}} \). For \( x \in \{c', d'\} \), we have \( -x = \sqrt{55 \pm \sqrt{k+x}} \). In either case, \( a, b, c', d' \) satisfy \( x^2 = 55 \pm \sqrt{k+x} \).

2. **Square both sides of the equation:**
   \[
   x^2 = 55 \pm \sqrt{k+x}
   \]
   Squaring both sides, we get:
   \[
   (x^2 - 55)^2 = (\sqrt{k+x})^2
   \]
   Simplifying, we obtain:
   \[
   (x^2 - 55)^2 = k + x
   \]

3. **Expand and rearrange the equation:**
   \[
   (x^2 - 55)^2 = x^4 - 110x^2 + 3025
   \]
   Therefore, the equation becomes:
   \[
   x^4 - 110x^2 + 3025 = k + x
   \]
   Rearranging, we get:
   \[
   x^4 - 110x^2 - x + (3025 - k) = 0
   \]

4. **Determine the product of the roots:**
   The product of the roots of the polynomial \( x^4 - 110x^2 - x + (3025 - k) = 0 \) is given by the constant term divided by the leading coefficient (which is 1 in this case). Thus, the product of the roots is:
   \[
   3025 - k = ab c' d' = ab cd = 2007
   \]

5. **Solve for \( k \):**
   \[
   3025 - k = 2007
   \]
   Solving for \( k \), we get:
   \[
   k = 3025 - 2007 = 1018
   \]

The final answer is \( \boxed{1018} \).

Answer: 1018


For a regular tetrahedron $ABCD$ with edge length $1$, $E$ is a point inside $\triangle ABC$. The sum of the distances from point $E$ to the sides $AB$, $BC$, and $CA$ is $x$, and the sum of the distances from point $E$ to the planes $DAB$, $DBC$, and $DCA$ is $y$. Then $x^{2}+y^{2}$ equals ( ).
(A) 1
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{5}{3}$
(D) $\frac{17}{12}$

Solution:
3.D.

Point $E$ to the distances of sides $A B, B C, C A$ sum up to the height of $\triangle A B C$, which is $\frac{\sqrt{3}}{2}$, hence $x=\frac{\sqrt{3}}{2}$.
$$
\begin{array}{l}
\frac{1}{3} S_{\triangle A B C} \cdot h \\
=\frac{1}{3} S_{\triangle \operatorname{MAB}} \cdot y_{1}+\frac{1}{3} S_{\triangle H C C} \cdot y_{2}+\frac{1}{3} S_{\triangle N C A} \cdot y_{3} \text {, } \\
\end{array}
$$

Here $h$ is the height of the regular tetrahedron, and $y_{1}, y_{2}, y_{3}$ are the distances from point $E$ to the planes $D A B$, $D B C$, $D C A$ respectively. Thus, $h=\frac{\sqrt{6}}{3}$. Therefore, $y_{1}+y_{2}+y_{3}=y$. So, $y=\frac{\sqrt{6}}{3}$.

Answer: D


Find all monic polynomials $P(x)=x^{2023}+a_{2022}x^{2022}+\ldots+a_1x+a_0$ with real coefficients such that $a_{2022}=0$, $P(1)=1$ and all roots of $P$ are real and less than $1$.

Solution:
1. We are given a monic polynomial \( P(x) = x^{2023} + a_{2022}x^{2022} + \ldots + a_1x + a_0 \) with real coefficients such that \( a_{2022} = 0 \), \( P(1) = 1 \), and all roots of \( P \) are real and less than \( 1 \).

2. Since \( P(x) \) is monic and has real coefficients, we can write it as:
   \[
   P(x) = (x - r_1)(x - r_2) \cdots (x - r_{2023})
   \]
   where \( r_i \) are the roots of \( P(x) \).

3. Given that all roots are real and less than \( 1 \), we have \( r_i < 1 \) for all \( i \).

4. We are also given that \( a_{2022} = 0 \). By Vieta's formulas, the sum of the roots of \( P(x) \) is zero:
   \[
   r_1 + r_2 + \cdots + r_{2023} = 0
   \]

5. Additionally, we know that \( P(1) = 1 \). Substituting \( x = 1 \) into the polynomial, we get:
   \[
   P(1) = (1 - r_1)(1 - r_2) \cdots (1 - r_{2023}) = 1
   \]

6. Since \( r_i < 1 \) for all \( i \), each term \( (1 - r_i) \) is positive. Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   1 \leq \left( \frac{(1 - r_1) + (1 - r_2) + \cdots + (1 - r_{2023})}{2023} \right)^{2023}
   \]

7. Simplifying the expression inside the AM-GM inequality:
   \[
   1 \leq \left( 1 - \frac{r_1 + r_2 + \cdots + r_{2023}}{2023} \right)^{2023}
   \]

8. Since \( r_1 + r_2 + \cdots + r_{2023} = 0 \), we have:
   \[
   1 \leq 1
   \]

9. The equality condition in the AM-GM inequality holds if and only if all the terms are equal. Therefore, we must have:
   \[
   1 - r_1 = 1 - r_2 = \cdots = 1 - r_{2023}
   \]

10. This implies that \( r_1 = r_2 = \cdots = r_{2023} = 0 \).

11. Hence, the polynomial \( P(x) \) must be:
    \[
    P(x) = x^{2023}
    \]

The final answer is \( \boxed{ P(x) = x^{2023} } \).

Answer: P(x) = x^{2023}


Given that $\overline{A B C A}$ represents a four-digit number. If the two-digit number $\overline{A B}$ is a prime number, $\overline{B C}$ is a perfect square not equal to 1, and $\overline{C A}$ is the product of a prime number and a perfect square not equal to 1, then the number of four-digit numbers that satisfy these conditions is ( ).
(A) 0
(B) 1
(C) 2
(D) 10

Solution:
-,1.C.
Since $\overline{A B}$ is a prime number, $B$ can only be $1,3,5, 7,9$. Also, since $\overline{B C}$ is a perfect square not equal to 1, $\overline{B C}$ can only be 16 or 36. Since $\overline{C A}$ is the product of a prime number and a perfect square not equal to 1, $A$ must be 3 or 8.
Therefore, the four-digit number $\overline{A B C A}$ is 3163 or 8368.

Answer: C


Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other nodes, then this line segment is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments.

Solution:
Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
Thus, $e=v+2015$.
Note that, the points with degree 2 in graph $G$ are exactly 4, the rest of the points have a degree of 3 or 4. Points with degree 2 are the vertices of one rectangle, points with degree 3 are the vertices of two rectangles, and points with degree 4 are the vertices of four rectangles.

On one hand, the total number of vertices of all small rectangles is $4 \times 2016$.

On the other hand, the total number of vertices of all small rectangles is no less than $2(v-4)+4=2 v-4$.
Thus, $2 v-4 \leqslant 4 \times 2016 \Rightarrow v \leqslant 4034$.
Therefore, from equation (1), we get $e \leqslant 6049$.
When the rectangle $R$ is divided into $1 \times 2016$ small rectangles, the above equality holds.
Thus, the maximum value of $e$ is $e_{\text {max }}=6049$.
Next, consider the minimum value of $e$.
Suppose the rectangle $R$ is divided using $a$ horizontal lines and $b$ vertical lines, excluding the boundaries.

Since $a$ horizontal lines and $b$ vertical lines can divide the area into at most $(a+1)(b+1)$ regions, we have
$$
(a+1)(b+1) \geqslant 2016 \text {. }
$$

Let the number of points with degree 3 be $x$, and the number of points with degree 4 be $y$. Then,
$$
v=4+x+y, 4+2 x+4 y=4 \times 2016 .
$$

Each horizontal and vertical line has its endpoints as the vertices of two rectangles, thus they are all points with degree 3, and these points are distinct.
Hence, $x \geqslant 2 a+2 b$.
Combining with equation (2), we get
$$
\begin{array}{l}
x \geqslant 2(a+1)+2(b+1)-4 \\
\geqslant 4 \sqrt{(a+1)(b+1)}-4 \\
\geqslant 4 \sqrt{2016}-4>175.59 \\
\Rightarrow x \geqslant 176 .
\end{array}
$$

Thus, from equations (3) and (4), we get
$$
v=2019+\frac{1}{2} x \geqslant 2017 \text {. }
$$

From equation (1), we immediately get $e \geqslant 4122$, where the equality holds in the $42 \times 48$ division.
Thus, $e_{\min }=4122$.

Answer: 4122


Given that $a, b, c$ are not all zero, then the maximum value of $\frac{a b+2 b c}{a^{2}+b^{2}+c^{2}}$ is

Solution: 4. $\frac{\sqrt{5}}{2}$ Detailed Explanation: Let $x, y \in \mathbf{R}_{+}$, then $x^{2} a^{2}+b^{2} \geqslant 2 x a b, y^{2} b^{2}+c^{2} \geqslant 2 y c b$, thus $\frac{x}{2} a^{2}+\frac{b^{2}}{2 x} \geqslant a b, y b^{2}+\frac{1}{y} c^{2} \geqslant 2 b c$. Let $\frac{x}{2}=\frac{1}{2 x}+y=\frac{1}{y} \Rightarrow x=\sqrt{5}, y=\frac{2 \sqrt{5}}{5}$, substituting into the above equations gives $\frac{\sqrt{5}}{2}\left(a^{2}+b^{2}+c^{2}\right) \geqslant a b+2 b c, \therefore \frac{a b+2 b c}{a^{2}+b^{2}+c^{2}} \leqslant \frac{\sqrt{5}}{2}$, equality is achieved when $\sqrt{5} a=b=\frac{\sqrt{5}}{2} c$.

Answer: \frac{\sqrt{5}}{2}


The sequence $\left\{x_{n}\right\}$ satisfies $x_{0}=0$ and $x_{n+1}=3 x_{n}+\sqrt{8 x_{n}^{2}+1}$, find the general term formula.

Solution:
From $x_{n+1}=3 x_{n}+\sqrt{8 x_{n}^{2}+1}$, we get $\left(x_{n+1}-x_{n}\right)^{2}=8 x_{n}^{2}+1$, which means
$$
x_{n+1}^{2}-6 x_{n+1} x_{n}+x_{n}^{2}-1=0 \text {, }
$$

By replacing the subscript $n$ with $n-1$ in the above equation, we get $\left(x_{n+1}-x_{n}\right)^{2}=8 x_{n}^{2}+1$, which means
$$
x_{n-1}^{2}-6 x_{n+1} x_{n}+x_{n}^{2}-1=0,
$$

From these two equations, we know that $x_{n+1}, x_{n-1}$ are the two roots of the quadratic equation $t^{2}-6 x_{n} t+x_{n}^{2}-1=0$. By Vieta's formulas, we have $x_{n+1}+x_{n-1}=6 x_{n}$. This is a second-order linear recurrence sequence, whose characteristic equation is $\lambda^{2}-6 \lambda+1=0$, with characteristic roots $\lambda_{1}=3+2 \sqrt{2}, \lambda_{2}=3-2 \sqrt{2}$. ….. 10 points Therefore, we can assume its general term to be $x_{n}=C_{1}(3+2 \sqrt{2})^{n}+C_{2}(3-2 \sqrt{2})^{n}$. From $x_{0}=0$ and $x_{n+1}=x_{n}+\sqrt{8 x_{n}^{2}+1}$, we get $x_{1}=1$, so $C_{1}$ and $C_{2}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
C_{1}+C_{2}=0 \\
(3+2 \sqrt{2}) C_{1}+(3-2 \sqrt{2}) C_{2}=1 .
\end{array}\right.
$$

Solving this, we get $C_{1}=\frac{\sqrt{2}}{8}, C_{2}=-\frac{\sqrt{2}}{8}, x_{n}=\frac{\sqrt{2}}{8}(3+2 \sqrt{2})^{n}-\frac{\sqrt{2}}{8}(3-2 \sqrt{2})^{n} \cdots \cdots 20$ points

Answer: x_{n}=\frac{\sqrt{2}}{8}(3+2\sqrt{2})^{n}-\frac{\sqrt{2}}{8}(3-2\sqrt{2})^{n}


In trapezoid $ABCD$ shown, $AD$ is parallel to $BC$, and $AB=6, BC=7, CD=8, AD=17$. If sides $AB$ and $CD$ are extended to meet at $E$, find the resulting angle at $E$ (in degrees).

Solution:
Solution: 90
Choose point $F$ on $A D$ so that $B C D F$ is a parallelogram. Then $B F=C D=8$, and $A F=A D-D F=A D-B C=10$, so $\triangle A B F$ is a 6-8-10 right triangle. The required angle is equal to $\angle A B F=90^{\circ}$.

Answer: 90


On the board, there are 5 integers. By adding them in pairs, the following set of 10 numbers was obtained: $-1,4,6,9,10,11,15,16,20,22$. Determine which numbers are written on the board. In your answer, write their product.

Solution:
Answer: -4914 (numbers on the board: $-3,2,7,9,13$).

Solution. The sum of the numbers in the obtained set is 112. Each of the original five numbers appears 4 times in this sum. Therefore, the sum of the required numbers is $112: 4=28$. The sum of the two smallest numbers is -1, and the sum of the two largest numbers is 22. Therefore, the middle number (the third largest of the five) is $28-22-(-1)=7$. In the set given in the problem, the second number is equal to the sum of the first and third required numbers, from which the first number is $4-7=-3$, and the second is 2. Similarly, we obtain that the fourth number is 9, and the fifth is 13. Thus, the numbers on the board are $-3,2,7,9,13$, and their product is -4914.

Answer: -4914


Solution:
## Solution.

Domain of definition: $3-4 x^{2}>0 \Leftrightarrow-\frac{\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.

From the condition

$$
5^{\log _{5}\left(3-4 x^{2}\right)}+1.5 x \log _{2^{-3}} 2^{2}=0 \Leftrightarrow 3-4 x^{2}-x=0 \Leftrightarrow 4 x^{2}+x-3=0
$$

from which $x_{1}=-1, x_{2}=\frac{3}{4} ; x_{1}=-1$ does not satisfy the domain of definition.

Answer: $\frac{3}{4}$.

Answer: \frac{3}{4}


Find all ordered pairs $(a, b)$ such that $a$ and $b$ are integers and $3^{a}+7^{b}$ is a perfect square.

Solution:
It is obvious that $a$ and $b$ must be non-negative.

Suppose that $3^{a}+7^{b}=n^{2}$. We can assume that $n$ is positive. We first work modulo 4. Since $3^{a}+7^{b}=n^{2}$, it follows that

$$
n^{2} \equiv (-1)^{a} + (-1)^{b} \quad (\bmod 4)
$$

Since no square can be congruent to 2 modulo 4, it follows that we have either (i) $a$ is odd and $b$ is even or (ii) $a$ is even and $b$ is odd.

Case (i): Let $b=2c$. Then

$$
3^{a} = (n - 7^{c})(n + 7^{c}).
$$

It cannot be the case that 3 divides both $n - 7^{c}$ and $n + 7^{c}$. But each of these is a power of 3. It follows that $n - 7^{c} = 1$, and therefore

$$
3^{a} = 2 \cdot 7^{c} + 1.
$$

If $c=0$, then $a=1$, and we obtain the solution $a=1, b=0$. So suppose that $c \geq 1$. Then $3^{a} \equiv 1 (\bmod 7)$. This is impossible, since the smallest positive value of $a$ such that $3^{a} \equiv 1 (\bmod 7)$ is given by $a=6$, and therefore all $a$ such that $3^{a} \equiv 1 (\bmod 7)$ are even, contradicting the fact that $a$ is odd.

Case (ii): Let $a=2c$. Then

$$
7^{b} = (n - 3^{c})(n + 3^{c}).
$$

Thus each of $n - 3^{c}$ and $n + 3^{c}$ is a power of 7. Since 7 cannot divide both of these, it follows that $n - 3^{c} = 1$, and therefore

$$
7^{b} = 2 \cdot 3^{c} + 1.
$$

Look first at the case $c=1$. Then $b=1$, and we obtain the solution $a=2, b=1$. So from now on we may assume that $c > 1$. Then $7^{b} \equiv 1 (\bmod 9)$. The smallest positive integer $b$ such that $7^{b} \equiv 1 (\bmod 9)$ is given by $b=3$. It follows that $b$ must be a multiple of 3. Let $b=3d$. Note that $d$ is odd, so in particular $d \geq 1$.

Let $y = 7^{d}$. Then $y^{3} - 1 = 2 \cdot 3^{c}$, and therefore

$$
2 \cdot 3^{c} = (y - 1)(y^{2} + y + 1).
$$

It follows that $y - 1 = 2 \cdot 3^{u}$ for some positive $u$, and that $y^{2} + y + 1 = 3^{v}$ for some $v \geq 2$. But since

$$
3y = (y^{2} + y + 1) - (y - 1)^{2},
$$

it follows that $3 \mid y$, which is impossible since $3 \mid (y - 1)$.

Answer: a=1, b=0 \text{ and } a=2, b=1


Consider an equilateral triangle $ABC$, where $AB=BC=CA=2011$. Let $P$ be a point inside $\triangle ABC$. Draw line segments passing through $P$ such that $DE\|BC, FG\|CA$ and $HI \| AB$. Suppose $DE: FG: HI=8: 7: 10$. Find $DE+FG+HI$.

Solution:
34. Answer: 4022.
Set $D P=G P=a, I P=F P=b, E P=H P=c$. Then
$$
D E+F G+H I=(a+c)+(a+b)+(b+c)=2(a+b+c)=2 \times 2011=4022
$$

Answer: 4022


A student participates in military training and engages in target shooting, which must be done 10 times. In the 6th, 7th, 8th, and 9th shots, he scored 9.0 points, 8.4 points, 8.1 points, and 9.3 points, respectively. The average score of his first 9 shots is higher than the average score of his first 5 shots. If he wants the average score of 10 shots to exceed 8.8 points, how many points does he need to score at least in the 10th shot? (The points scored in each shot are accurate to 0.1 points)

Solution:
Solution: From the given, the average score of the first 5 shots is less than
$$
\frac{9.0+8.4+8.1+9.3}{4}=8.7 \text{. }
$$

The total score of the first 9 shots is at most
$$
8.7 \times 9-0.1=78.2 \text{. }
$$

Therefore, the score of the tenth shot must be at least
$$
8.8 \times 10+0.1-78.2=9.9 \text{ (points). }
$$

The above solution, if we set the total score of the first 5 shots as $x$, and the score of the tenth shot as $y$, can be represented by the system of inequalities as
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{5}<8.7 \\
x<43.5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x<53.2
\end{array} \text{ and the scores are accurate to } 0.1 \text{ points }
\end{array}\right\}
$$

Starting point: Assuming this student scores 9.9 points on the tenth shot, can his average score over 10 shots definitely exceed 8.8 points? The answer is not certain. For example, if the total score of the first 5 shots is 40 points (satisfying the given condition), then the total score of 10 shots is
$$
40+(9.0+8.4+8.1+9.3)+9.9=84.7<88
$$

The system of inequalities
$$
\left\{\begin{array}{l}
x<A \\
x+y<B
\end{array}(B>A)\right.
$$

is not solved by $\left\{\begin{array}{l}x<A \\ y<B-A\end{array}\right.$. Its solution should be, when $x$ takes real values $(x<A)$, $y$ takes values $(y<B-x$. The solution set is
$$
G=\{(x, y) \mid x<A, y<B-x\} .
$$

The graphical solution is as follows:
As shown in Figure 1, draw the line
$$
L_{1}: x+y=B . x+y
$$
$>B$ is the set of points above the line $L_{1}$; draw the line $L_{2}$: $x=A, x<A$, and $y<B-A$ is only the range of $y$ values.

From the above analysis, to find the minimum value of $y$, we must first determine the value of $x$ within its range, for example, $x=C(C<A)$, then $y<B-C$.

For this problem, after solving $\left\{\begin{array}{l}x<53.2\end{array}\right.$, how do we determine the minimum value of $y$? This requires first determining the value of $x$. Since the maximum score for each shot is 10 points, we have
$$
x+(9.0+8.4+8.1+9.3)+10>88 \text{. }
$$

Therefore, $x>43.2$.
From $x<53.2 \Rightarrow y>9.9 \Rightarrow y \geqslant 10 \text{; }$
$$
(iii) \text{ When } x=43.4 \text{, }
$$
$$
43.4+y>53.2 \Rightarrow y>9.8 \Rightarrow y \geqslant 9.9 \text{. }
$$

In summary, to make the average score of 10 shots exceed 8.8 points, we must analyze the total score of the first 5 shots $(x)$. If $x \leqslant 43.2$ points, it is impossible for him to make the average score of 10 shots exceed 8.8 points; only when $x=43.3$ points, 43.4 points, the tenth shot must score at least 10 points and 9.9 points, respectively.

Answer: 9.9


Let $n$ be a three-digit positive integer without the digit 0. If the digits of $n$ in the units, tens, and hundreds places are permuted arbitrarily, the resulting three-digit number is never a multiple of 4. Find the number of such $n$.

Solution: Hint: Classify by the number of even digits (i.e., $2,4,6,8$) appearing in the three-digit code of $n$. It is known from the discussion that the number of $n$ satisfying the condition is $125+150+0+8=283$.

Answer: 283


In a park, there are 10,000 trees planted in a square grid (100 rows of 100 trees each). What is the maximum number of trees that can be cut down so that the following condition is met: if you stand on any stump, you will not see any other stump? (The trees can be considered sufficiently thin.)

Solution:
Let's divide the trees into 2500 quadruples, as shown in the figure. In each such quadruple, no more than one tree can be cut down. On the other hand, all the trees growing in the upper left corners of the squares formed by our quadruples of trees can be cut down. Therefore, the maximum number of trees that can be cut down is 2500.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e668e47ffca00611739g-28.jpg?height=706&width=992&top_left_y=1259&top_left_x=539)

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Answer: 2500


Find all real numbers $s$ for which the equation

$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=0
$$

has four distinct real roots and the product of two of these roots is -2 .

Solution:
Solution. Assume that $s$ is a number as above, and denote the four roots of the equation by $x_{1}, x_{2}, x_{3}$ and $x_{4}$ in such a way that

$$
x_{1} x_{2}=-2 .
$$

From the factorization

$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=4\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$

we obtain, upon multiplying out the brackets and comparing the coefficients at like powers of $x$ on both sides, the familiar Vièta's relations

$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+x_{4} & =5 \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4} & =\frac{s}{4}, \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4} & =-\frac{11}{2}, \\
x_{1} x_{2} x_{3} x_{4} & =-\frac{1}{2} .
\end{aligned}
$$

From the equalities (0) and (4) it follows immediately that

$$
x_{3} x_{4}=\frac{1}{4} \text {. }
$$

Rewriting (3) as

$$
\left(x_{1}+x_{2}\right) x_{3} x_{4}+\left(x_{3}+x_{4}\right) x_{1} x_{2}=-\frac{11}{2}
$$

and substituting the known values for $x_{1} x_{2}$ and $x_{3} x_{4}$ we obtain

$$
\frac{1}{4}\left(x_{1}+x_{2}\right)-2\left(x_{3}+x_{4}\right)=-\frac{11}{2},
$$

which together with the equation (1) forms a system of two linear equations for the unknown sums $x_{1}+x_{2}$ and $x_{3}+x_{4}$. An easy calculation shows that its solution is given by

$$
x_{1}+x_{2}=2 \quad \text { and } \quad x_{3}+x_{4}=3
$$

Inserting all this into the equality (2) rewritten in the form

$$
x_{1} x_{2}+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+x_{3} x_{4}=\frac{s}{4}
$$

we find that necessarily $s=17$.

Conversely, from the equalities

$$
x_{1}+x_{2}=2 \quad \text { and } \quad x_{1} x_{2}=-2
$$

it follows that the numbers $x_{1,2}$ are the roots of the quadratic equation

$$
x^{2}-2 x-2=0, \quad \text { or } \quad x_{1,2}=1 \pm \sqrt{3}
$$

and from the equalities

$$
x_{3}+x_{4}=3 \quad \text { and } \quad x_{3} x_{4}=\frac{1}{4}
$$

it follows that the numbers $x_{3,4}$ are the roots of the quadratic equation

$$
x^{2}-3 x+\frac{1}{4}=0, \quad \text { or } \quad x_{3,4}=\frac{3}{2} \pm \sqrt{2}
$$

We see that $x_{1,2,3,4}$ are indeed four mutually different real numbers which satisfy the system (1)-(4) for the value $s=17$, hence are the roots of the original equation from the statement of the problem.

There is thus only one such number $s$, namely $s=17$.

Answer: 17


Find $a$ such that the sum of the squares of the real roots of the equation $x^{4}+a x^{2}-2017=0$ is 4.

Solution:
Answer: $1006.5$.

Solution: Let's make the substitution $t=x^{2}$. The equation $t^{2}+a t-2017=0$ has two roots, the product of which is -2017 according to Vieta's theorem. Therefore, one of them is negative. Let $t_{1}>0$, then the roots of the original equation are $\pm \sqrt{t_{1}}$. The sum of their squares is $2 t_{1}$, so $t_{1}=2$. Substituting into the equation $t^{2}+a t-2017=0$, we get $4+2 a-2017=0$, from which $a=1006.5$.

## Variant $2-$ -

Answer: 1006.5


In how many ways can we color exactly $k$ vertices of an $n$-gon in red such that any $2$ consecutive vertices are not both red. (Vertices are considered to be labeled)

Solution:
1. **Define the problem for a straight line:**
   Let \( f(n, k) \) denote the number of ways to color exactly \( k \) points red on a straight line of \( n \) points such that no two consecutive points are both red.

2. **Establish the recurrence relation:**
   - If the \( n \)-th point is red, then the \((n-1)\)-th point cannot be red. The remaining \( n-2 \) points must have \( k-1 \) red points, which can be done in \( f(n-2, k-1) \) ways.
   - If the \( n \)-th point is not red, then the remaining \( n-1 \) points must have \( k \) red points, which can be done in \( f(n-1, k) \) ways.
   - Therefore, the recurrence relation is:
     \[
     f(n, k) = f(n-1, k) + f(n-2, k-1)
     \]

3. **Prove the closed form by induction:**
   - Base case: For \( k = 0 \), \( f(n, 0) = 1 \) if \( n \geq 0 \) (since there is exactly one way to color zero points red).
   - Inductive step: Assume \( f(n, k) = \binom{n-k+1}{k} \) holds for all \( n \leq m \). We need to show it holds for \( n = m+1 \).
     \[
     f(m+1, k) = f(m, k) + f(m-1, k-1)
     \]
     By the inductive hypothesis:
     \[
     f(m, k) = \binom{m-k+1}{k}, \quad f(m-1, k-1) = \binom{m-k+2}{k-1}
     \]
     Using the Pascal's identity:
     \[
     \binom{m-k+1}{k} + \binom{m-k+2}{k-1} = \binom{m-k+2}{k}
     \]
     Therefore:
     \[
     f(m+1, k) = \binom{m-k+2}{k}
     \]
     This completes the induction.

4. **Return to the original problem for an \( n \)-gon:**
   - Fix a vertex \( P \). If \( P \) is red, the two adjacent vertices cannot be red. The remaining \( n-3 \) vertices must have \( k-1 \) red vertices, which can be done in \( f(n-3, k-1) \) ways.
   - If \( P \) is not red, the remaining \( n-1 \) vertices must have \( k \) red vertices, which can be done in \( f(n-1, k) \) ways.
   - Therefore, the total number of ways is:
     \[
     f(n-3, k-1) + f(n-1, k)
     \]

5. **Substitute the closed form:**
   \[
   f(n-3, k-1) = \binom{(n-3)-(k-1)+1}{k-1} = \binom{n-k-1}{k-1}
   \]
   \[
   f(n-1, k) = \binom{(n-1)-k+1}{k} = \binom{n-k}{k}
   \]
   Therefore, the required answer is:
   \[
   \binom{n-k-1}{k-1} + \binom{n-k}{k}
   \]

The final answer is \( \boxed{ \binom{n-k-1}{k-1} + \binom{n-k}{k} } \)

Answer: \binom{n-k-1}{k-1} + \binom{n-k}{k}


The sum of all irreducible proper fractions with a denominator of 1994 is ( ).
(A) 497.5
(B) 498
(C) 498.5
(D) 997

Solution:
$-1 . B$.
From $1994=2 \times 997$, and knowing that 997 is a prime number, the sum of the reduced proper fractions in the problem is
$$
\begin{aligned}
N= & \frac{1}{1994}+\frac{3}{1994}+\cdots+\frac{995}{1994}+\frac{999}{1994}+\cdots \\
& +\frac{1993}{1994} \\
= & \frac{1+1993}{1994}+\frac{3+1991}{1994}+\cdots+\frac{995+999}{1994} \\
= & 1+1+\cdots+1 \quad\left(\frac{995+1}{2}=498 \uparrow 1\right) \\
= & 498 .
\end{aligned}
$$

Answer: B


At the ball, $n$ family couples arrived. In each couple, the husband and wife are of exactly the same height, but there are no two couples of the same height. A waltz begins, and all the attendees randomly form pairs: each gentleman dances with a randomly chosen lady. Find the mathematical expectation of the random variable $X$ "The number of gentlemen who are shorter than their partner."

Solution:
Solution. Let's assign the cavaliers numbers from 1 to $n$. Let $I_{j}$ be the indicator of the event "cavalier number $j$ is shorter than his lady."

$$
\mathrm{P}\left(I_{j}=1\right)=\frac{j-1}{n} . \text { Then } \mathrm{E} X=\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{n}=\frac{1}{n}(0+1+\ldots+n-1)=\frac{n(n-1)}{2 n}=\frac{n-1}{2}
$$

Answer: $\mathrm{E} X=\frac{n-1}{2}$.

Answer: \frac{n-1}{2}


On a strip, numbers from 1 to 1598 were written in order and the strip was cut into several parts. It turned out that the arithmetic mean of all numbers in the first part is some natural number $n$, the second part - the number $2 n$, the third part - the number $3 n$, and so on. Explain for which $n$ this is possible.

Solution:
Solution. Consider several consecutive natural numbers. If their quantity is even, then they can be paired with equal odd sums, hence their arithmetic mean is not an integer. If their quantity is odd, then their arithmetic mean is the middle number in this set. Therefore, all segments of the strip consist of an odd number of numbers.

The arithmetic mean of the numbers from 1 to $k$ is a natural number $n$, hence $k=2 n-1$. Thus, the first strip consists of numbers from 1 to $2 n-1$, the second strip consists of the single number $2 n$, the first number of the third strip is $2 n+1$, and the last number is $4 n-1$ (the sum of the extreme numbers must equal $6 n$), then the next number is $4 n$, followed by numbers from $4 n+1$ to $6 n-1$, and so on.

Thus, the last numbers of the strips alternate: $2 n-1, 2 n, 4 n-1, 4 n, 6 n-1, 6 n, \ldots, 2 k n-1, 2 k n$. Since the number 1598 is even and is the last number in some strip, it equals some $2 k n$. Hence $k n=799=17 \cdot 47$, so $n$ can be one of the divisors of this number: $1, 17, 47, 799$. It is clear from the solution that corresponding examples for each such $n$ can be found.

Answer: 1,17,47,799


The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find $AB$, if $CA=9, CB=4$.

Solution:
Answer: 6.

Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio of the adjacent sides, we have $\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\mathrm{CD} / \mathrm{BD}$ $=\mathrm{AC} / \mathrm{AB} . \quad$ Similarly, $\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right)$ $=A E / E C=A B / B C$. Since $S_{1}=S_{2}$, then $\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\quad\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right) \quad$ from which $A B / B C=A C / A B . \quad A B / 4=9 / A B . \quad A B^{2}=36$,

![](https://cdn.mathpix.com/cropped/2024_05_06_209f947ed7e329235966g-2.jpg?height=508&width=458&top_left_y=436&top_left_x=942)
$\mathrm{AB}=6$ (since the length of a segment is a positive number). It is not difficult to verify that such a triangle exists $(4+6>9)$.

Criteria. The length of the side is found correctly, but there is no check for the existence of the triangle: 6 points.

Answer: 6


Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its left focus at point $F$, and let a tangent line be drawn from a point $A$ on the ellipse $C$ intersecting the $y$-axis at point $Q$. If $\angle Q F O=45^{\circ}, \angle Q F A=30^{\circ}$, then the eccentricity of the ellipse is $\qquad$.

Solution:
7. $\frac{\sqrt{6}}{3}$.

As shown in Figure 2, let $A\left(x_{0}, y_{0}\right)$, then the equation of the tangent line passing through point $A$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$. Let the tangent line intersect the $x$-axis at point $P$, then $P\left(\frac{a^{2}}{x_{0}}, 0\right), C\left(0, \frac{b^{2}}{y_{0}}\right)$, so
$$
\begin{array}{l}
\overrightarrow{F Q} \cdot \overrightarrow{F O}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot(c, 0)=c^{2}, \\
\overrightarrow{F Q} \cdot \overrightarrow{F A}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot\left(x_{0}+c, y_{0}\right)=c x_{0}+a^{2} .
\end{array}
$$

Thus,
$$
\begin{aligned}
\frac{\cos \angle Q F O}{\cos \angle Q F A} & =\frac{\overrightarrow{F Q} \cdot \overrightarrow{F A}}{\overrightarrow{F Q} \cdot \overrightarrow{F O}} \cdot \frac{|\overrightarrow{F Q}| \cdot|\overrightarrow{F O}|}{|\overrightarrow{F Q}| \cdot|\overrightarrow{F A}|} \\
& =\frac{c^{2}}{c x_{0}+a^{2}} \cdot \frac{e x_{0}+a}{c} \\
& =\frac{c}{a} \cdot \frac{e x_{0}+a}{e x_{0}+a}=e,
\end{aligned}
$$

Therefore, the eccentricity of the ellipse
$$
e=\frac{\cos \angle Q F O}{\cos \angle Q F A}=\frac{\cos 45^{\circ}}{\cos 30^{\circ}}=\frac{\sqrt{6}}{3} .
$$

Answer: \frac{\sqrt{6}}{3}


The average height of boys in Class 5-1 is 149 cm, and the average height of girls is 144 cm. The average height of the entire class is 147 cm. Therefore, the number of boys in the class is $\qquad$ times the number of girls.

Solution:
【Solution】Solve: $149 x+144 y=147 \times(x+y)$,
$$
\begin{aligned}
149 x+144 y & =147 x+147 y, \\
2 x & =3 y, \\
\frac{\mathrm{x}}{\mathrm{y}} & =\frac{3}{2} ;
\end{aligned}
$$

Answer: The number of boys in the class is $\frac{3}{2}$ times the number of girls. Therefore, the answer is: $\frac{3}{2}$.

Answer: \frac{3}{2}


Given that the three vertices $A, B, C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis, then the altitude $CD$ from $C$ to $AB$ equals $\qquad$.

Solution:
Let $A\left(-s, s^{2}\right), B\left(s, s^{2}\right), C\left(t, t^{2}\right)$, then $\overrightarrow{A C} \cdot \overrightarrow{B C}=(t+s)(t-s)+\left(t^{2}-s^{2}\right)\left(t^{2}-s^{2}\right)$ $=\left(t^{2}-s^{2}\right)\left(t^{2}-s^{2}+1\right)=0 \Rightarrow t^{2}=s^{2}-1 \Rightarrow y_{C}=y_{A}-1$.
Therefore, the altitude $C D$ from vertex $C$ to the hypotenuse $A B$ equals 1.

Answer: 1


$8 n^{2}(n \geqslant 4)$ positive numbers are arranged in several rows and columns:
\begin{tabular}{llllll}
$a_{11}$ & $a_{12}$ & $a_{13}$ & $a_{14}$ & $\ldots$ & $a_{1 n}$, \\
$a_{21}$ & $a_{22}$ & $a_{23}$ & $a_{24}$ & $\ldots$ & $a_{2 n}$, \\
$a_{31}$ & $a_{32}$ & $a_{33}$ & $a_{34}$ & $\ldots$ & $a_{3 n}$, \\
$\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ \\
$a_{n 1}$ & $a_{n 2}$ & $a_{n 3}$ & $a_{n 4}$ & $\ldots$ & $a_{n n}$,
\end{tabular}

where the numbers in each row form an arithmetic sequence, and the numbers in each column form a geometric sequence, and all the common ratios are equal. Given $a_{24}=1, a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, find $a_{11}+a_{22}+a_{33}+\cdots+a_{n n}$.

Solution:
Given that an arithmetic sequence can be uniquely determined by its first term and common difference, and a geometric sequence can be uniquely determined by its first term and common ratio. If we set \(a_{11}=a\), the common difference of the first row is \(d\), and the common ratio of the first column is \(q\), it is easy to calculate that \(a_{st}=[a+(t-1)] q^{-1}\). From the given conditions, we can establish a system of equations to solve for \(n\), \(d\), and \(q\).

Solution: Let the common difference of the first row be \(d\), and the common ratio of each column be \(q\). Then the common difference of the fourth row is \(d q^{2}\). Thus, we can obtain the system of equations:
\[
\left\{\begin{array}{l}
a_{24}=\left(a_{11}+3 d\right) q=1 \\
a_{42}=\left(a_{11}+d\right) q^{3}=\frac{1}{8} \text { Solving this system of equations, we get } a_{11}=d=q= \pm \frac{1}{2} . \text { Since all } n^{2} \text { numbers } \\
a_{43}=a_{42}+d q^{3}=\frac{3}{16},
\end{array}\right.
\]

are positive, we must have \(q>0\), hence \(a_{11}=d=q=\frac{1}{2}\). Therefore, for any \(1 \leqslant k \leqslant n\), we have \(a_{k k}=a_{1 k} q^{k-1}=\left[a_{11}+(k-1)\right] q^{k-1}=\frac{k}{2^{k}}\). Thus, \(S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}\). Also, \(\frac{1}{2} S=\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\cdots+\frac{n}{2^{n+1}}\). Subtracting these two equations, we get \(\frac{1}{2} S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}-\frac{n}{2^{n+1}}\), so \(S=2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}\).

Answer: 2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}


Determine the force of water pressure on a vertical parabolic segment, the base of which is $4 \mathrm{m}$ and is located on the water surface, while the vertex is at a depth of $4 \mathrm{m}$.

Solution:
Solution. We have $|B A|=2 x=4$ (m). Point $A$ in the chosen coordinate system has coordinates $(2 ; 4)$. The equation of the parabola relative to this system is $y = a x^{2}$ or $4=a \cdot 2^{2}$, from which $a=1$, i.e., $y=x^{2}$.

Consider an elementary area $d S$ at a distance $y$ from the origin. The length of this area is $2 x$, and its area is $d S=2 x d y$, where $d y$ is the width of the area. This area will experience a pressure force

$$
d P=9.81 \gamma(4-y) d S=9.81 \gamma(4-y) 2 x d y
$$

Summing all the elementary forces of pressure as $y$ changes from 0 to 4 and taking the limit as $\Delta y \rightarrow 0$, the force of pressure on the entire segment will be expressed by the following integral:

$$
P=9.81 \gamma \int_{0}^{4}(4-y) 2 x d y
$$

To find this integral, express $x$ in terms of $y$; since $y=x^{2}$, then $x=\sqrt{y}$. Then we get

$$
\begin{aligned}
& P=19.62 \cdot 1000 \int_{0}^{4}(4-y) \sqrt{y} d y=19620\left(\int_{0}^{4} 4 \sqrt{y} d y-\int_{0}^{4} y \sqrt{y} d y\right)= \\
&=19620\left(4 \int_{0}^{4} y^{1 / 2} d y-\int_{0}^{4} y^{3 / 2} d y\right)=\left.19620\left(\frac{4 y^{3 / 2}}{3 / 2}-\frac{y^{5 / 2}}{5 / 2}\right)\right|_{0} ^{4}= \\
&=\left.19620\left(\frac{8}{3} y^{3 / 2}-\frac{2}{5} y^{5 / 2}\right)\right|_{0} ^{4}=19620\left(\frac{8 \cdot 2}{3}-\frac{2 \cdot 32}{5}\right)= \\
&=\frac{19620 \cdot 128}{15}=167424 \text { (N). }
\end{aligned}
$$

Answer: 167424


How many triples of non-negative integers $(x, y, z)$ satisfying the equation
$$
x y z+x y+y z+z x+x+y+z=2012 ?
$$

Solution:
31. Answer: 27.
$$
(x+1)(y+1)(z+1)=2013=3 \times 11 \times 61 .
$$

If all $x, y, z$ are positive, there are $3!=6$ solutions.
If exactly one of $x, y, z$ is 0 , there are $3 \times 6=18$ solutions.
If exactly two of $x, y, z$ are 0 , there are 3 solutions.
$$
6+18+3=27 \text {. }
$$

Answer: 27


To steal a precious jewel, a thief must discover the code that allows him to open the safe's door. The information he has managed to gather is as follows:

- the code is a number
- any consecutive subsequence of digits of the code (thus each digit taken individually, as well as each pair of digits, etc., up to the entire number) represents a prime number (for example, 217 is not good, because 1 is not a prime and 21 is not a prime)
- the code is the largest number that has this property.

What is the secret code to open the safe?

Solution:
13. The answer is 373.

Let's find all the numbers that satisfy the conditions of the text.

First, we notice that if we have found all acceptable numbers of $n$ digits, then an acceptable number of $n+1$ digits must necessarily contain one of the numbers of $n$ digits. In particular, if we discover that there is no number of $k$ digits, then there will be no acceptable number with more than $k$ digits.

The acceptable numbers of one digit are the one-digit primes, i.e., $2, 3, 5,$ and $7$. We note, however, that the digits 2 and 5 can only be at the beginning of an acceptable number (otherwise, the number would have a subsequence divisible by 2 or 5, respectively).

The acceptable numbers of two digits are therefore $23, 37, 53,$ and $73$.

Now let's look at the numbers of three digits. None can end in 23 or 53. Then 237, 537, 273, and 573 are not acceptable as they are multiples of 3. No number can start with the same pair of digits, otherwise it would have a subsequence that is a multiple of 11 (for example, in 773, the subsequence 77 is a multiple of 11). The only ones left to check are 373, which turns out to be prime (and therefore acceptable), and 737, which is a multiple of 11.

Now let's see if there can be numbers of four digits. We only need to check 4: 2373, 3373, 5373, and 7373. The first and the third contain a subsequence that is a multiple of 3. The second contains a subsequence that is a multiple of 11, and the last one is a multiple of 101. Therefore, there are no acceptable numbers of four digits, and, as stated at the beginning, there will be no acceptable numbers with more than four digits. The code we are looking for is thus 373.

Answer: 373


Add appropriate parentheses to $1 \div 2 \div 3 \div 4 \div 5$ to form a complete expression. Then, the number of different values that can be obtained is $(\quad)$.
(A) 6
(B) 8
(C) 9
(D) 10

The denominator must be below the fraction line. By appropriately placing parentheses, the numbers $3, 4, 5$ can be placed either above or below the fraction line, resulting in $2^{3}=8$ different values.
For example, the following operations can yield:
$$
\begin{array}{l}
\{[(1 \div 2) \div 3] \div 4\} \div 5=\frac{1}{2 \times 3 \times 4 \times 5}=\frac{1}{120} . \\
1 \div\{[(2 \div 3) \div 4] \div 5\}=\frac{3 \times 4 \times 5}{2}=30 .
\end{array}
$$

From equation (1),
$$
\begin{array}{l}
{[(1 \div 2) \div 3] \div(4 \div 5)=\frac{5}{2 \times 3 \times 4}=\frac{5}{24} ;} \\
{[(1 \div 2) \div(3 \div 4)] \div 5=\frac{4}{2 \times 3 \times 5}=\frac{2}{15} ;} \\
\{[1 \div(2 \div 3)] \div 4\} \div 5=\frac{3}{2 \times 4 \times 5}=\frac{3}{40} .
\end{array}
$$

From equation (2),
$$
\begin{array}{l}
1 \div[(2 \div 3) \div(4 \div 5)]=\frac{3 \times 4}{2 \times 5}=\frac{6}{5} ; \\
1 \div\{[2 \div(3 \div 4)] \div 5\}=\frac{3 \times 5}{2 \times 4}=\frac{15}{8} ; \\
(1 \div 2) \div[(3 \div 4) \div 5]=\frac{4 \times 5}{2 \times 3}=\frac{10}{3} .
\end{array}
$$

This gives all 8 possible values, and they are all distinct.

Solution:
6. B.

No matter how parentheses are added, 1 must be in the numerator, 2 must

Answer: B


Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}$

Solution:
## Solution

$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{\lim _{x \rightarrow 0} 1+x}=$

$=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{1}=\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=$

Using the substitution of equivalent infinitesimals:

$\sin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

We get:

$$
=\lim _{x \rightarrow 0} \frac{2 x}{x}=\lim _{x \rightarrow 0} \frac{2}{1}=2
$$

## Problem Kuznetsov Limits 18-1

Answer: 2


The equation

$x^{10}+(13x-1)^{10}=0\,$

has 10 [complex](https://artofproblemsolving.com/wiki/index.php/Complex_number) [roots](https://artofproblemsolving.com/wiki/index.php/Root) $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation.  Find the value of

$\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.$

Solution:
Let $t = 1/x$. After multiplying the equation by $t^{10}$, $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$.
Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$.
$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)$.
Since $\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)$, $t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product.
The expression to find is $\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$.
But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$.

Answer: 850


Consider all the real sequences $x_0,x_1,\cdots,x_{100}$ satisfying the following two requirements:
(1) $x_0=0$;
(2) For any integer $i,1\leq i\leq 100$, we have $1\leq x_i-x_{i-1}\leq 2$.
Find the greatest positive integer $k\leq 100$, so that for any sequence $x_0,x_1,\cdots,x_{100}$ like this, we have
\[x_k+x_{k+1}+\cdots+x_{100}\geq x_0+x_1+\cdots+x_{k-1}.\]

Solution:
1. We start by considering the given conditions for the sequence \( x_0, x_1, \ldots, x_{100} \):
   - \( x_0 = 0 \)
   - For any integer \( i \) where \( 1 \leq i \leq 100 \), we have \( 1 \leq x_i - x_{i-1} \leq 2 \).

2. We need to find the greatest positive integer \( k \leq 100 \) such that:
   \[
   x_k + x_{k+1} + \cdots + x_{100} \geq x_0 + x_1 + \cdots + x_{k-1}.
   \]

3. Let's generalize the problem by replacing 100 with \( n \). We need to find \( k \) such that:
   \[
   x_k + x_{k+1} + \cdots + x_n \geq x_0 + x_1 + \cdots + x_{k-1}.
   \]

4. By the given conditions, we know that \( x_{j+d} \geq x_j + d \) for all \( j, d \geq 1 \). This implies that the sequence grows at least linearly.

5. We can write:
   \[
   x_k + x_{k+1} + \cdots + x_n \geq x_{k-1} + x_{k-2} + \cdots + x_{2k-1-n} + (n+1-k)^2.
   \]

6. We need to ensure that:
   \[
   x_1 + x_2 + \cdots + x_{2k-2-n} \leq (n+1-k)^2.
   \]

7. The left-hand side (LHS) can be at most:
   \[
   2 + 4 + \cdots + 2(2k-2-n) = (2k-2-n)(2k-1-n).
   \]

8. We need to check if:
   \[
   (2k-2-n)(2k-1-n) \leq (n+1-k)^2.
   \]

9. Simplifying the inequality:
   \[
   (2k-2-n)(2k-1-n) \leq (n+1-k)^2.
   \]

10. Expanding both sides:
    \[
    (2k-2-n)(2k-1-n) = 4k^2 - 4k - 2kn + n^2 - 2k + n.
    \]
    \[
    (n+1-k)^2 = n^2 + 2n + 1 - 2nk - 2k + k^2.
    \]

11. Comparing the coefficients, we need:
    \[
    4k^2 - 4k - 2kn + n^2 - 2k + n \leq n^2 + 2n + 1 - 2nk - 2k + k^2.
    \]

12. Simplifying further, we get:
    \[
    3k^2 - 2kn - 2k \leq 2n + 1.
    \]

13. Solving for \( k \):
    \[
    3k^2 - 2kn - 2k - 2n - 1 \leq 0.
    \]

14. This inequality is satisfied if:
    \[
    k \leq \frac{2n+2}{3}.
    \]

15. Substituting \( n = 100 \):
    \[
    k \leq \frac{2 \cdot 100 + 2}{3} = \frac{202}{3} \approx 67.33.
    \]

16. Therefore, the greatest integer \( k \) is \( 67 \).

The final answer is \( \boxed{67} \).

Answer: 67


Given that $m, n$ are integers greater than 7, consider a rectangular array of $m \times n$ points. Color $k$ of these points red, such that the three vertices of any right-angled triangle with its two legs parallel to the sides of the rectangle are not all red points. Find the maximum value of $k$.

Solution:
Assume there are $m$ rows and $n (n \leqslant m)$ columns, then the maximum value is at least $m+n-2$, because by selecting any row and column, and coloring all points in these row and column except their common point in red, it satisfies the condition.

Below, we use mathematical induction to prove that in an $m \times n$ grid, the number of red points $\leqslant m+n-2$.
When $n=2$, if the number of red points is $m+n-1=m+1$, then there must be a row with 2 red points, which means the other points in these two columns cannot be red, leading to a contradiction!

When $n=3$, if the number of red points is $m+n-1=m+2$, then there must be a row with 2 red points, which means the other points in these two columns cannot be red, and the other column can have at most $m-1$ red points, leading to a contradiction.

If for $n \leqslant k$, in an $m \times n$ grid, the number of red points $\leqslant m+n-2$, assume for $n=k+1$, in an $m \times n$ grid, there are $m+n-1=m+k$ red points. By the pigeonhole principle, there must be a row with 2 red points, and in the columns of these 2 red points, there are no other red points. Removing these two columns, we get an $m \times (k-1)$ grid with $m-k-2$ red points, which contradicts the induction hypothesis.

Therefore, in an $m \times n$ grid, the number of red points $\leqslant m+n-2$.
In conclusion, the maximum value of $k$ is $m+n-2$.

Answer: m+n-2


On a square $ABCD$ a line segment $BE$ is drawn such that the point $E$ lies on the side $CD$. The perimeter of triangle $BCE$ is three-quarters of the perimeter of the square $ABCD$. The ratio of lengths $CE: CD$ is $\lambda: 1$.
What is the value of $960 \times \lambda$ ?

Solution:
SolUtion
720

Let the square's side have length 1 . The triangle's hypotenuse has length $\sqrt{1+\lambda^{2}}$, so the triangle's perimeter is $1+\lambda+\sqrt{1+\lambda^{2}}=3$. Therefore, $\sqrt{1+\lambda^{2}}=2-\lambda$ from which we obtain the equation $1+\lambda^{2}=4-4 \lambda+\lambda^{2}$.
Solving this equation we find $\lambda=\frac{3}{4}$, so $960 \lambda=720$.

Answer: 720


Find the denominator $q$ of an infinite geometric progression $(|q|<1)$, where each term is four times the sum of all its subsequent terms.

Solution:
## Solution.

From the condition, we have $b_{1}=4\left(s-b_{1}\right)$.

By formula (4.12), we get

$$
b_{1}=4\left(\frac{b_{1}}{1-q}-b_{1}\right), b_{1}=\frac{4 b_{1}(1-1+q)}{1-q}, 1-q=4 q, q=\frac{1}{5}
$$

Answer: $\frac{1}{5}$.

Answer: \frac{1}{5}


Let $a_{1}, a_{2}, a_{3}, \ldots$ be the sequence of all positive integers that are relatively prime to 75, where $a_{1}<a_{2}<a_{3}<\cdots$. (The first five terms of the sequence are: $a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=7, a_{5}=8$.) Find the value of $a_{2008}$.

Solution:
35. Answer: 3764
Let $U=\{1,2,3, \ldots, 74,75\}$ be the set of integers from 1 to 75 . We first find the number of integers in $U$ that are relatively prime to $75=3 \times 5^{2}$. Let $A=\{n \in U: 3$ is a factor of $n\}$ and $B=\{n \in U: 5$ is a factor of $n\}$. Then $A \cup B$ is the set of integers in $U$ that are not relatively prime to 75 . Note that $|A|=25,|B|=15$ and $A \cap B|=|\{n \in U: 15$ is a factor of $n\} \mid=5$. By the principle of inclusion-exclusion, $|A \cup B|=|A|+|B|-|A \cap B|=35$. Therefore the number of integers in $U$ that are relatively prime to 75 is $|U|-|A \cup B|=75-35=40$. Thus
$$
a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=7, a_{5}=8, \ldots, a_{40}=74 \text {. }
$$

Now by division algorithm, any non-negative integer can be written uniquely in the form $75 k+r$ for some integers $k$ and $r$, where $k \geq 0$ and $0 \leq r \leq 74$. Note that $\operatorname{gcd}(75 k+r, 75)=1$ if and only if $\operatorname{gcd}(r, 75)=1$. Therefore the sequence $\left\{a_{n}\right\} n \geq 1$
can be written in the form $a_{n}=75 k+a_{i}$, where $k \geq 0$ and $1 \leq i \leq 40$. Indeed, $k$ is given by $k=\left\lfloor\frac{n}{40}\right\rfloor$ and $i$ is the remainder when $n$ is divided by 40 . Thus for $n=$ 2008, $k=\left\lfloor\frac{2008}{40}\right\rfloor=50$ and $i=8$. Hence
$$
a_{2008}=75 \times 50+a_{8}=3750+14=3764 \text {. }
$$

Answer: 3764


Find the $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x-f(y))=1-x-y$.

Solution: We perform the substitution $(x, y) \leftarrow(x+f(0), 0)$, to find $f(x)=1-x-f(0)$. By evaluating at 0, we then find $f(0)=1-f(0)=\frac{1}{2}$. Conversely, we have $\frac{1}{2}-\left(x-\left(\frac{1}{2}-y\right)\right)=$ $1-x-y:$ the function $f: x \mapsto \frac{1}{2}-x$ is the unique solution.

Answer: f(x)=\frac{1}{2}-x


In a Cartesian coordinate system, given points $A(2,2)$, $B(-2,-3)$, point $P$ lies on the coordinate axes, and $\triangle A B P$ is a right-angled triangle. Then the number of points $P$ is ( ).
(A) 3
(B) 4
(C) 6
(D) 8

Solution:
7. D.

Draw a circle with $AB$ as the diameter, and draw perpendicular lines to $AB$ through points $A$ and $B$. The circle and the perpendicular lines intersect the coordinate axes at 8 points.

Answer: D


The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares.  The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution:
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$. Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.
For $s$, there are $8^2$ [unit squares](https://artofproblemsolving.com/wiki/index.php/Unit_square), $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$.
Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$, and $m+n=\boxed{125}$.

Answer: 125


Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as:
$u_1=6,u_{n+1} = \frac{2n+a}{n} + \sqrt{\frac{n+a}{n}u_n+4}, \forall n \ge 1$
a) With $a=0$, prove that there exist a finite limit of $(u_n)$ and find that limit
b) With $a \ge 0$, prove that there exist a finite limit of $(u_n)$

Solution:
### Part (a)
Given \( a = 0 \), we need to prove that there exists a finite limit of \( (u_n) \) and find that limit.

1. **Initial Condition and Recurrence Relation:**
   \[
   u_1 = 6, \quad u_{n+1} = \frac{2n}{n} + \sqrt{\frac{n}{n}u_n + 4} = 2 + \sqrt{u_n + 4}
   \]

2. **Establishing a Lower Bound:**
   \[
   u_n > 5 \quad \forall n \ge 1
   \]
   This is because \( u_1 = 6 \) and the recurrence relation ensures that \( u_n \) remains above 5.

3. **Analyzing the Difference:**
   \[
   |u_{n+1} - 5| = \left| 2 + \sqrt{u_n + 4} - 5 \right| = \left| \sqrt{u_n + 4} - 3 \right|
   \]

4. **Bounding the Difference:**
   \[
   \sqrt{u_n + 4} - 3 = \frac{u_n - 5}{\sqrt{u_n + 4} + 3}
   \]
   Since \( u_n > 5 \), we have \( \sqrt{u_n + 4} + 3 > 5 \). Therefore,
   \[
   |u_{n+1} - 5| = \frac{|u_n - 5|}{\sqrt{u_n + 4} + 3} < \frac{1}{5} |u_n - 5|
   \]

5. **Iterative Inequality:**
   \[
   |u_{n+1} - 5| < \frac{1}{5} |u_n - 5| < \left( \frac{1}{5} \right)^n |u_1 - 5|
   \]
   Since \( |u_1 - 5| = 1 \),
   \[
   \lim_{n \to \infty} |u_n - 5| = 0 \implies \lim_{n \to \infty} u_n = 5
   \]

### Part (b)
Given \( a \ge 0 \), we need to prove that there exists a finite limit of \( (u_n) \).

1. **Initial Condition and Recurrence Relation:**
   \[
   u_1 = 6, \quad u_{n+1} = \frac{2n + a}{n} + \sqrt{\frac{n + a}{n}u_n + 4}
   \]

2. **Defining a New Sequence:**
   Let \( v_n = \frac{u_n}{n} \). Then,
   \[
   v_1 = 6, \quad v_{n+1} = \frac{\frac{2n + a}{n} + \sqrt{(n + a)v_n + 4}}{n + 1}
   \]

3. **Simplifying the Recurrence Relation:**
   \[
   v_{n+1} < \frac{2n + a}{n^2 + n} + \frac{\sqrt{(n + a)v_n}}{n + 1} + \frac{2}{n + 1}
   \]
   Using the inequality \( \sqrt{a + b} < \sqrt{a} + \sqrt{b} \),
   \[
   v_{n+1} \le \frac{2n + a}{n^2 + n} + \frac{2}{n + 1} + \frac{1}{2} \left( \frac{1}{n + 1} + \frac{(n + a)v_n}{n + 1} \right)
   \]

4. **Applying the Arithmetic Mean-Geometric Mean Inequality (AM-GM):**
   \[
   v_{n+1} \le \frac{2n + a}{n^2 + n} + \frac{5}{2(n + 1)} + \frac{1}{2} \cdot \frac{n + a}{n + 1} v_n
   \]

5. **Bounding the Terms:**
   Since \( \lim_{n \to \infty} \frac{n + a}{n + 1} = 1 \), there exists \( N \) such that for all \( n \ge N \),
   \[
   \frac{n + a}{n + 1} < \frac{3}{2}
   \]
   Therefore, for \( n \ge N \),
   \[
   v_{n+1} < \left( \frac{2n + a}{n^2 + n} + \frac{5}{2(n + 1)} \right) + \frac{3}{4} v_n
   \]

6. **Applying the Lemma:**
   By the well-known lemma, if \( z_n \) is a positive sequence and \( z_{n+1} \le cz_n + q_n \) with \( \lim q_n = 0 \) and \( c \in (0, 1) \), then \( \lim z_n = 0 \). Here, \( v_n \) satisfies this condition, so
   \[
   \lim_{n \to \infty} \frac{u_n}{n} = 0
   \]

7. **Analyzing the Difference Again:**
   \[
   |u_{n+1} - 5| = \left| \frac{u_n - 5 + \frac{au_n}{n}}{\sqrt{\frac{n + a}{n} u_n + 4} + 3} + \frac{a}{n} \right| < \frac{1}{5} |u_n - 5| + \left( \frac{a}{5} \cdot \frac{u_n}{n} + \frac{a}{n} \right)
   \]

8. **Applying the Lemma Again:**
   Since \( \lim_{n \to \infty} \frac{u_n}{n} = 0 \) and \( \frac{a}{n} \to 0 \), applying the lemma, we have
   \[
   \lim_{n \to \infty} |u_n - 5| = 0 \implies \lim_{n \to \infty} u_n = 5
   \]

The final answer is \( \boxed{ 5 } \)

Answer: 5


Given positive real numbers $a, b$ satisfy $a^{b}=(8 a)^{9 b}$, then the value of $\log _{a}\left(4 a^{2}\right)$ is $\qquad$

Solution:
$$
\begin{array}{l}
a^{b}=\left[(8 a)^{9}\right]^{b} \Rightarrow a=(8 a)^{9}=8^{9} a^{9} \Rightarrow a^{8}=8^{-9}=2^{-27} \Rightarrow 8=-27 \log _{a} 2 \\
\Rightarrow \log _{a} 2=-\frac{8}{27} \Rightarrow \log _{a} 4=-\frac{16}{27}, \text { so } \log _{a}\left(4 a^{2}\right)=2+\log _{a} 4=\frac{38}{27} .
\end{array}
$$

Answer: \frac{38}{27}


In a school, $60 \%$ of the students are male, $90 \%$ are minors, and $60 \%$ have brown hair. Which of the following statements is necessarily true?

(A) There is at least one female who is of age.

(B) There is at least one female with brown hair.

(C) There is at least one minor male with brown hair.

(D) There are no males of age with brown hair.

(E) There is at least one blonde male.

Solution:
9) The answer is (C)

The girls are $40 \%$, the non-brown-haired $40 \%$, and the adults $10 \%$. Even if there were no overlap, the total would reach $90 \%$, leaving $10 \%$ for male minors with brown hair. On the other hand, statements (A), (B), (D), and (E) can be false, as can be seen from the following example of a school with 1000 students:

if in the school there are 400 girls, all minors and blonde, and 600 boys all with brown hair, of which 500 minors and 100 adults, all the statements in the text are true, but (A), (B), (D), (E) are false.

Answer: C


In the fictional country of Mahishmati, there are $50$ cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city $A$, an ordered list of cities $C_1,\ldots, C_{50}$ is called an antitour from $A$ if
[list]
[*] every city (including $A$) appears in the list exactly once, and
[*] for each $k\in \{1,2,\ldots, 50\}$, it is impossible to go from $A$ to $C_k$ by a sequence of exactly $k$ (not necessarily distinct) flights.
[/list]
Baahubali notices that there is an antitour from $A$ for any city $A$. Further, he can take a sequence of flights, starting from the capital and passing through each city exactly once. Find the least possible total number of antitours from the capital city.

Proposed by Sutanay Bhattacharya

Solution:
1. Consider a graph with 50 vertices, where the vertices represent the cities and the edges represent the flights between the cities.

2. Define a graph as *good* if there are antitours from each city. We need to show that a good graph cannot have any odd cycle.

3. **Proof by Contradiction:**
   - Assume the contrary, that there is an odd cycle in the graph. Let the odd cycle be \( v_1, v_2, v_3, \ldots, v_{2k+1} \).
   - Consider the antitour from \( v_1 \). Let the antitour be \( \{ C_1, C_2, C_3, \ldots, C_{50} \} \).
   - Since there are \( 2k+1 \) vertices in the odd cycle, at least one of them must have an index \( \geq 2k+1 \) in the antitour, i.e., \( v_t = C_n \) where \( n \geq 2k+1 \) and \( v_t \) is in the odd cycle.
   - We consider different cases based on the parity of \( t \) and \( n \):
     - If \( t \) and \( n \) are both odd, consider the path \( \{ v_1, v_{2k+1}, v_{2k}, \ldots, v_t, v_{t-1}, v_t, v_{t-1}, \ldots \} \) with \( n \) terms.
     - If \( t \) is even and \( n \) is odd, consider the path \( \{ v_1, v_2, v_3, \ldots, v_t, v_{t+1}, v_t, v_{t+1}, \ldots \} \) with \( n \) terms.
     - If \( t \) is odd and \( n \) is even, consider the path \( \{ v_1, v_{2k+1}, v_{2k}, \ldots, v_t, v_{t-1}, v_t, v_{t-1}, \ldots \} \) with \( n \) terms.
     - If \( t \) and \( n \) are both even, consider the path \( \{ v_1, v_{2k+1}, v_{2k}, \ldots, v_t, v_{t-1}, v_t, v_{t-1}, \ldots \} \) with \( n \) terms.
   - In all cases, we find a path of exactly \( n \) flights from \( v_1 \) to \( v_t \), which contradicts the definition of an antitour. Hence, our assumption is false, and a good graph cannot have any odd cycle.

4. Now, let the capital city be \( v_1 \). As given, there is a path from \( v_1 \) to \( v_{50} \) (WLOG) in the graph. Since there are no odd cycles, there is no edge \( \{v_i, v_j\} \) if \( i \) and \( j \) have the same parity.

5. **Claim:** The minimum number of antitours from the capital city is \( (25!)^2 \).

6. **Proof:**
   - Since there is no edge \( \{v_i, v_j\} \) for vertices \( i \) and \( j \) with the same parity, each edge takes us from an odd-indexed vertex to an even-indexed vertex and vice versa.
   - Assign a random odd number to all \( v_i \) if \( i \) is odd, and an even number to \( v_i \) if \( i \) is even, forming the collection \( \{C_1, C_2, \ldots, C_{50} \} \).
   - This is clearly an antitour from \( v_1 \) because an odd number of edges taken from \( v_1 \) take us to an even-numbered vertex, and vice versa.
   - The number of ways to assign odd numbers is \( 25! \), and the number of ways to assign even numbers is \( 25! \). Hence, there are at least \( (25!)^2 \) possible antitours from \( v_1 \).

7. For the construction, join \( v_1 \) to all vertices \( v_i \) if \( i \) is even, and form the \( v_1, v_2, \ldots, v_{50} \) cycle. Clearly, this works.

The final answer is \( \boxed{ (25!)^2 } \).

Answer: (25!)^2


Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ pass through the point $(0,1)$, and have an eccentricity of $\frac{\sqrt{3}}{2}$.
(1) Find the equation of the ellipse $C$.
(2) Let the line $l: x=my+1$ intersect the ellipse $C$ at points $A$ and $B$, and let the point $A$ be symmetric to point $A^{\prime}$ with respect to the $x$-axis (point $A^{\prime}$ does not coincide with point $B$). Is the line $A^{\prime} B$ intersecting the $x$-axis at a fixed point? If so, write down the coordinates of the fixed point and prove your conclusion; if not, explain your reasoning.

Solution:
21. (1) According to the problem, we have
$$
\left\{\begin{array}{l}
b=1, \\
\frac{c}{a}=\frac{\sqrt{3}}{2}, \\
a^{2}=b^{2}+c^{2}
\end{array} \Rightarrow a=2 .\right.
$$

Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+y^{2}=1$.
(2) From $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ x=m y+1\end{array}\right.$
$$
\Rightarrow\left(m^{2}+4\right) y^{2}+2 m y-3=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $A^{\prime}\left(x_{1},-y_{1}\right)$, and
$$
y_{1}+y_{2}=-\frac{2 m}{m^{2}+4}, y_{1} y_{2}=-\frac{3}{m^{2}+4} \text {. }
$$

Let the equation of the line passing through points $A^{\prime}\left(x_{1},-y_{1}\right), B\left(x_{2}, y_{2}\right)$ be
$$
\begin{array}{l}
y+y_{1}=\frac{y_{2}+y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right) . \\
\text { Let } y=0 . \text { Then } \\
x=\frac{y_{1} x_{2}+y_{2} x_{1}}{y_{2}+y_{1}} . \\
=\frac{y_{1}\left(m y_{2}+1\right)+y_{2}\left(m y_{1}+1\right)}{y_{2}+y_{1}} \\
=\frac{2 m y_{1} y_{2}+y_{1}+y_{2}}{y_{2}+y_{1}}=4,
\end{array}
$$

Thus, the line $A^{\prime} B$ intersects the $x$-axis at the fixed point $(4,0)$.

Answer: (4,0)


Represent the number 2017 as the sum of some number of natural numbers so that their product is the largest.

Answer. $2017=3+3+\ldots+3+2+2$ (671 threes and two twos).

Solution:
Solution If the sought representation contains at least one addend greater than 4, then the product is not maximal. Indeed, if $n \geq 5$ is replaced by the sum $2+(n-2)$, then the product $2 \cdot(\mathrm{n}-2)$ turns out to be greater than $\mathrm{n}$. Also, note that if the representation contains an addend 4, it can be replaced by $2+2$, and the product will remain unchanged.

Thus, the maximum product is obtained when the number is broken down into a sum of twos and threes (it is not difficult to understand that the replacements $2 \rightarrow 1+1$ and $3 \rightarrow 2+1, 2+2 \rightarrow 3+1$ only decrease the product). The replacement $2+2+2 \rightarrow 3+3$ does not change the sum but increases the product, so the number of twos should be as few as possible.

From this, we get the decomposition $2015=671 \cdot 3+2+2$, i.e., the number 2015 should be represented as a sum of 673 addends, two of which are equal to two, and the rest are equal to three.

Answer: 2017=3+3+\ldots+3+2+2


A 6-digit number starts with 1. If we move this digit 1 from the first position to the last on the right, we get a new 6-digit number that is three times the original number. What is this number?

Solution:
The problem is to determine the digits $a, b, c, d$ and $e$ such that the number $a b c d e 1$ is three times $1 a b c d e$:

$$
\frac{\times 3}{a b c d e 1}
$$

Initially, we see that $e=7$, and from there we can start discovering each of the digits:

| $1 a b c d 7$ | $1 a b c 57$ | $1 a b 857$ | $1 a 2857$ |
| :---: | :---: | :---: | :---: |
| $\times 3$ | $\times 3$ | $\times 3$ | $\times 3$ |
| $a b c d 71$ | $a b c 571$ | $a b 8571$ | $a 28571$ |

Therefore, $a=4$ and the starting number is 142857.

Answer: 142857


For the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{m}=1$, the circle with the eccentricity as the radius and the right focus as the center is tangent to the asymptotes of the hyperbola. Then $m=(\quad)$.
(A) $\frac{3}{2}$
(B) $\frac{4}{3}$
(C) $\frac{5}{4}$
(D) $\frac{6}{5}$

Solution:
4. B.

Notice that $a=2, b=\sqrt{m}$,
$$
c=\sqrt{4+m}, e=\frac{\sqrt{4+m}}{2} \text {. }
$$

The equations of the asymptotes are $\frac{x}{2} \pm \frac{y}{\sqrt{m}}=0$, i.e.,
$$
\sqrt{m} x \pm 2 y=0 \text {. }
$$

The distance from the right focus $(\sqrt{4+m}, 0)$ to the asymptote is $d=b=\sqrt{m}$.
Thus, $\sqrt{m}=\frac{\sqrt{4+m}}{2} \Rightarrow m=\frac{4}{3}$.

Answer: B


How many natural numbers $n$ less than 2022 are there for which $\left(\frac{-\sqrt{2}+i \sqrt{2}}{2 \cos \frac{\pi}{8}-2 i \sin \frac{\pi}{8}}\right)^{n}$ is a real number?

Solution:
## Solution.

Let $z=\frac{-\sqrt{2}+i \sqrt{2}}{2 \cos \frac{\pi}{8}-2 i \sin \frac{\pi}{8}}$.

The trigonometric form of the numerator of $z$ is

$-\sqrt{2}+i \sqrt{2}=2\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

and the denominator

$2 \cos \frac{\pi}{8}-2 i \sin \frac{\pi}{8}=2\left(\cos \left(-\frac{\pi}{8}\right)+i \sin \left(-\frac{\pi}{8}\right)\right)=2\left(\cos \frac{15 \pi}{8}+i \sin \frac{15 \pi}{8}\right)$.

Then we have

$$
\begin{aligned}
z & =\cos \left(\frac{3 \pi}{4}-\frac{15 \pi}{8}\right)+i \sin \left(\frac{3 \pi}{4}-\frac{15 \pi}{8}\right)=\cos \left(-\frac{9 \pi}{8}\right)+i \sin \left(-\frac{9 \pi}{8}\right) \\
& =\cos \left(\frac{7 \pi}{8}\right)+i \sin \left(\frac{7 \pi}{8}\right)
\end{aligned}
$$

Then $z^{n}=\cos \left(\frac{7 \pi n}{8}\right)+i \sin \left(\frac{7 \pi n}{8}\right)$.

This number is real if $\sin \left(\frac{7 \pi n}{8}\right)=0$, that is, if $\frac{7 \pi n}{8}=k \pi, k \in \mathbb{Z}$.

It follows that $n=\frac{8 k}{7}, k \in \mathbb{Z}$.

Since $n$ is a natural number, $k$ must be a positive multiple of 7, and $n$ a multiple of 8, i.e., $n \in\{8,16,24, \ldots, 2016\}$.

Finally, there are 252 such natural numbers $n$.

Answer: 252


Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$

Solution:
Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$

We know $\Delta=s^{2}+\frac{4}{s}\left(s^{2}-\frac{2}{s}\right) \geqslant 0$.
Simplifying, we get $s^{2} \leqslant 8$.
Therefore, $s \leqslant 2$.
Taking $p=q=1$, the maximum value of $p+q$ is 2.
4 Using the definition of absolute value
That is, $|x|= \pm x,|x|^{2}=x^{2}$.

Answer: 2


Given the ellipse $C: \frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, the upper and lower foci are $F_{1}, F_{2}$, and the right vertex is $D$. A line perpendicular to $D F_{2}$ through $F_{1}$ intersects the ellipse $C$ at points $A, B$, and $|B D|-\left|A F_{1}\right|=\frac{8 \sqrt{3}}{39}$.
(1) Find the value of $|A D|+|B D|$.
(2) Construct the tangents to the ellipse $C$ at points $A, B$ which intersect at point $E$. If $F_{1} E$ intersects the $x$-axis at $P$ and $F_{2} E$ intersects the $x$-axis at $Q$, find the value of $|P Q|$.

Solution:
(1) $a=2 c, b=\sqrt{3} c$, connecting $F_{1} D$, then
$\triangle F_{1} F_{2} D$ is an equilateral triangle $\Rightarrow|O M|=\frac{\sqrt{3} c}{3}$
Thus, $A B: x=-\frac{\sqrt{3}}{3} y+\frac{\sqrt{3} c}{3}$,
Solving $\left\{\begin{array}{l}\sqrt{3} x=-y+c, \\ 3 y^{2}+4 x^{2}=12 c^{2}\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow 13 y^{2}-8 c y-32 c^{2}=0 \\
\Rightarrow|A B|=\sqrt{1+\frac{1}{3}} \cdot \frac{\sqrt{64 c^{2}+4 \cdot 13 \cdot 32 c^{2}}}{13} \\
=\frac{2 \sqrt{3}}{3} \cdot \frac{24 \sqrt{3} c}{13}=\frac{48 c}{13} .
\end{array}
$$

Then, $|B D|-\left|A F_{1}\right|=\left|B F_{2}\right|-\left|A F_{1}\right|=2 a-\left(\left|B F_{1}\right|+\left|A F_{1}\right|\right)=4 c-|A B|=\frac{4 c}{13}=\frac{8 \sqrt{3}}{39}$
$\Rightarrow c=\frac{2 \sqrt{3}}{3} \Rightarrow a=\frac{4 \sqrt{3}}{3}, b=2$. Therefore, $|A D|+|B D|=\left|A F_{2}\right|+\left|B F_{2}\right|=4 a-\left(\left|A F_{1}\right|+\left|B F_{1}\right|\right)$
$$
=4 a-|A B|=\frac{8 \sqrt{3}}{3}+\frac{8 \sqrt{3}}{39}=\frac{112 \sqrt{3}}{39} \text {. }
$$
(2) Let $E\left(x_{0}, y_{0}\right), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, from (1) we know the equation of the ellipse $C$ is $3 y^{2}+4 x^{2}=16$,

Then $A E: 3 y_{1} y+4 x_{1} x=16, B E: 3 y_{2} y+4 x_{2} x=16 \Rightarrow A B: 3 y_{0} y+4 x_{0} x=16$.
And $A B: x=-\frac{\sqrt{3}}{3} y+\frac{2}{3} \Rightarrow 3 x+\sqrt{3} y=2 \Rightarrow 8 \sqrt{3} y+24 x=16$,
Thus $y_{0}=\frac{8 \sqrt{3}}{3}, x_{0}=6 \Rightarrow E\left(6, \frac{8 \sqrt{3}}{3}\right)$. Also, $F_{1}\left(0, \frac{2 \sqrt{3}}{3}\right), F_{2}\left(0,-\frac{2 \sqrt{3}}{3}\right)$,
Therefore $k_{E F_{1}}=\frac{\frac{8 \sqrt{3}}{3}-\frac{2 \sqrt{3}}{3}}{6}=\frac{\sqrt{3}}{3} \Rightarrow E F_{1}: y=\frac{\sqrt{3}}{3} x+\frac{2 \sqrt{3}}{3} \Rightarrow P(-2,0)$,
$k_{E F_{2}}=\frac{\frac{8 \sqrt{3}}{3}+\frac{2 \sqrt{3}}{3}}{6}=\frac{5 \sqrt{3}}{9} \Rightarrow E F_{2}: y=\frac{5 \sqrt{3}}{9} x-\frac{2 \sqrt{3}}{3} \Rightarrow Q\left(\frac{6}{5}, 0\right)$, so $|P Q|=\frac{6}{5}+2=\frac{16}{5}$.

Answer: \frac{16}{5}


For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$

and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$

Solution:
2. 2 .

From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$

we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad \Rightarrow a+b \leqslant 2\right. \text {. }
$$

For the fixed point $B(1,1)$, for any $A(x, y) \in D$, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=(1,1) \cdot(x, y)=x+y \leqslant 1 \text {. }
$$

When $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$, the equality holds.
Thus, the point $B(1,1)$ satisfies the conditions of the problem, and in this case, $a+b=2$.
Therefore, $(a+b)_{\max }=2$.

Answer: 2


Find all functions \( f \) defined on the set of all real numbers with real values, such that \( f(x^2 + f(y)) = y + f(x)^2 \) for all \( x, y \).

Solution: The first step is to establish that f(0) = 0. Putting x = y = 0, and f(0) = t, we get f(t) = t 2 . Also, f(x 2 +t) = f(x) 2 , and f(f(x)) = x + t 2 . We now evaluate f(t 2 +f(1) 2 ) two ways. First, it is f(f(1) 2 + f(t)) = t + f(f(1)) 2 = t + (1 + t 2 ) 2 = 1 + t + 2t 2 + t 4 . Second, it is f(t 2 + f(1 + t)) = 1 + t + f(t) 2 = 1 + t + t 4 . So t = 0, as required. It follows immediately that f(f(x)) = x, and f(x 2 ) = f(x) 2 . Given any y, let z = f(y). Then y = f(z), so f(x 2 + y) = z + f(x) 2 = f(y) + f(x) 2 . Now given any positive x, take z so that x = z 2 . Then f(x + y) = f(z 2 + y) = f(y) + f(z) 2 = f(y) + f(z 2 ) = f(x) + f(y). Putting y = -x, we get 0 = f(0) = f(x + -x) = f(x) + f(-x). Hence f(-x) = - f(x). It follows that f(x + y) = f(x) + f(y) and f(x - y) = f(x) - f(y) hold for all x, y. Take any x. Let f(x) = y. If y > x, then let z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z. If y  0 with f(z) = -z = 0. Contradiction. So we must have f(x) = x. s are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 33rd IMO 1992 © John Scholes jscholes@kalva.demon.co.uk 13 Nov 1998

Answer: f(x)=x


In a football tournament, 10 football teams participated, and each team played exactly one match against every other team. For each win, 3 points are awarded, for a draw, 1 point, and for a loss, 0 points. At the end, a final table is given showing the total points of each team. If the total points in the final table amounted to 120, then how many matches in the football tournament ended in a draw?

Solution:
Solution. The total number of football matches played in the tournament is $9+8+7+\ldots+2+1=45$. If we denote the number of draws by $x$, then $45-x$ is the number of matches that ended with a winner. From this, we get the equation

$$
3(45-x)+2 x=120
$$

whose solution is $x=15$.

Thus, 15 matches in the football tournament ended in a draw.

Answer: 15


Let $A B C$ be an acute triangle, $\mathcal{K}$ a circle with diameter $A B$, $E$ and $F$ the intersections of the circle with sides $A C$ and $B C$, and $P$ the intersection of the tangents to the circle $\mathcal{K}$ at points $E$ and $F$. Calculate the ratio of the radii of the circumcircles of triangles $A B C$ and $E F P$.

Solution:
IV/4. Let's denote the lengths of the sides and the sizes of the angles of the triangle as usual. The radius of the circumcircle of triangle $ABC$ is $\frac{c}{2 \sin \gamma}$, and the radius of the other circle is $\frac{|EF|}{2 \sin \angle EPF}$.

Triangle $AOE$ is isosceles with vertex $O$, so $\angle AEO = \alpha$, and the exterior angle $\angle BOE$ is $2\alpha$. Triangle $OBF$ is isosceles with vertex $O$, so $\angle OFB = \beta$, and the exterior angle $\angle FOA$ is $2\beta$. Thus, we get $\angle FOE = 2\alpha + 2\beta - \pi$ and $\angle EPF = \pi - (2\alpha + 2\beta - \pi) = 2(\pi - \alpha - \beta) = 2\gamma$.

Triangle $OEF$ is isosceles, with $|OF| = |OE| = \frac{c}{2}$ and the angle at the vertex $2\alpha + 2\beta - \pi$. Thus, $\frac{|EF|}{2} = \frac{c}{2} \sin \left(\alpha + \beta - \frac{\pi}{2}\right)$, from which we get $|EF| = c \sin \left(\alpha + \beta - \frac{\pi}{2}\right) = -c \cos (\alpha + \beta) = c \cos \gamma$.

Finally, we can calculate the radius of the circumcircle of triangle $EPF$: $\frac{|EF|}{2 \sin \angle EPF} = \frac{c \cos \gamma}{2 \sin (2\gamma)} = \frac{c \cos \gamma}{4 \sin \gamma \cos \gamma} = \frac{c}{4 \sin \gamma}$. The desired ratio is therefore 2.
Writing the radii $\frac{c}{2 \sin \gamma}$ and $\frac{|EF|}{2 \sin \angle EPF}$ ..... 1 point
Angles $\angle FOE = 2\alpha + 2\beta - \pi$ and $\angle EPF = 2\gamma$ ..... 2 points
Writing $|EF| = c \cos \gamma$ ..... 2 points
Radius of the circumcircle of triangle $EPF$ and the calculated ratio ..... 2 points

Answer: 2


Calculate the sum of three consecutive numbers, the smallest of which is 1209?

Answer: 3630


Given $x^{2}+y^{2}+z^{2}=35$, then the maximum value of $x+3 y+5 z$ is greater than the minimum value by

Answer: 70


A certain duplicator can print 3600 sheets of paper per hour, so printing 240 sheets of paper requires $\qquad$ minutes.

Solution:
【Analysis】Convert 1 hour $=60$ minutes, first based on work efficiency $=$ total work $\div$ working time, find out the work efficiency of the duplicator, then according to working time $=$ total work $\div$ work efficiency to solve the problem.

【Solution】Solution: 1 hour $=60$ minutes,
$$
\begin{array}{l}
240 \div(3600 \div 60), \\
=240 \div 60, \\
=4 \text { (minutes), }
\end{array}
$$

Answer: It takes 4 minutes to print 240 sheets of paper.
Therefore, the answer is: 4.
【Comment】This question mainly examines students' ability to solve problems based on the quantitative relationship between working time, work efficiency, and total work.

Answer: 4


$BK$ is the bisector of triangle $ABC$. It is known that $\angle AKB: \angle CKB=4: 5$. Find the difference between angles $A$ and $C$ of triangle $ABC$.

Solution:
Find the angles $A K B$ and $C K B$ and use the theorem about the exterior angle of a triangle.

## Solution

Since $\angle A K B+\angle C K B=180^{\circ}$, then $\angle A K B=80^{\circ}, \angle C K B=100^{\circ}$.

Let $\angle A B K=\angle C B K=\beta$. By the theorem about the exterior angle of a triangle $80^{\circ}=\angle C+\beta, 100^{\circ}=\angle A+\beta, \angle A-\angle C$ $=20^{\circ}$.

## Answer

$20^{\circ}$.

Answer: 20