Vasily Petrov is completing an English language assignment. In this assignment, there are 10 English expressions and their Russian translations in random order. The task is to establish the correct correspondences between the expressions and their translations. 1 point is given for each correctly established correspondence. Thus, one can score from 0 to 10 points. Vasya knows nothing, so he chooses the options at random. Find the probability that he will score exactly 9 points.

Solution:
Solution. It cannot happen that exactly nine correct answers are guessed, because in this case the tenth answer would also be correct. Therefore, the desired probability is zero.

Answer: 0

Answer: 0


Solve the equation $\frac{1}{\sqrt{\log _{5}(5 x)}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$.

Solution: Solution: Using the properties of logarithms, our equation can be rewritten as $\frac{1}{\sqrt{1+\log _{5} x}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$. Let $t=\log _{5} x$. Then $\frac{1}{\sqrt{1+t}+\sqrt{t}}+\sqrt{t}=2$. By multiplying the numerator and denominator of the first fraction by $\sqrt{t+1}-\sqrt{t}$, we arrive at the equation $\sqrt{t+1}=2$. From this, we get that $t=3$, so $x=5^{3}=125$. Answer: 125.

Answer: 125


For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$?

Solution:
Suppose that the $n$th term of the sequence $S_k$ is $2005$. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$.  The [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002)$, $(3,668)$, $(4,501)$, $(6,334)$, $(12,167)$, $(167,12)$,$(334,6)$, $(501,4)$, $(668,3)$, $(1002,2)$ and $(2004,1)$, and each of these gives a possible value of $k$. Thus the requested number of values is $12$, and the answer is $\boxed{012}$.
Alternatively, notice that the formula for the number of [divisors](https://artofproblemsolving.com/wiki/index.php/Divisor) states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^2\cdot 3^1\cdot 167^1$.

Answer: 12


In a school, there are 1200 students, each of whom has five lessons every day. Any teacher in this school gives 4 lessons per day. How many teachers work in the school if there are exactly 30 students in each class?

Solution:
# Solution

Since each student has 5 lessons per day, if there was only one student in the class, the total number of lessons per day would be $5 \times 1200=6000$. Since there are 30 students in the class, the number of lessons conducted in the school each day is $\frac{6000}{30}=200$. Therefore, the number of teachers in the school is $\frac{200}{4}=50$.

Answer: 50.

Answer: 50


Solution:
Solve the 10 factors with the same structure $a^{4}+4 \times 3^{4}$. By Theorem 3,
$$
a^{4}+4 \cdot 3^{4}=\left[(a+3)^{2}+3^{2}\right]\left[(a-3)^{2}+3^{2}\right] \text {, }
$$

Thus the original expression is
$$
\begin{array}{l}
=\frac{\left(10^{4}+4 \times 3^{4}\right)\left(22^{4}+4 \times 3^{4}\right) \cdots\left(58^{4}+4 \times 3^{4}\right)}{\left(4^{4}+4 \times 3^{4}\right)\left(16^{4}+4 \times 3^{4}\right) \cdots\left(52^{4}+4 \times 3^{4}\right)} \\
=\frac{\left[\left(13^{2}+3^{2}\right)\left(25^{2}+3^{2}\right)\left(37^{2}+3^{2}\right)\left(49^{2}+3^{2}\right)\left(61^{2}+3^{2}\right)\right]}{\left[\left(7^{2}+3^{2}\right)\left(19^{2}+3^{2}\right)\left(31^{2}+3^{2}\right)\left(43^{2}+3^{2}\right)\left(55^{2}+3^{2}\right)\right]} \\
=\frac{\left[\left(7^{2}+3^{2}\right)\left(19^{2}+3^{2}\right)\left(31^{2}+3^{2}\right)\left(43^{2}+3^{2}\right)\left(55^{2}+3^{2}\right)\right]}{\left[\left(1^{2}+3^{2}\right)\left(13^{2}+3^{2}\right)\left(25^{2}+3^{2}\right)\left(33^{2}+3^{2}\right)\left(49^{2}+3^{2}\right)\right]} \\
=\frac{61^{2}+3^{2}}{1^{2}+3^{2}}=\frac{3730}{10}=373 .
\end{array}
$$

Answer: 373


Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}-z^{2}=1, z=0, z=2
$$

Solution:
## Solution

In the section of the given figure by the plane $z=$ const, there is an ellipse:

$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=z^{2}+1
$$

The area of the ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$

Let's find the radii of the ellipse:

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-47.jpg?height=923&width=1108&top_left_y=992&top_left_x=817)

$$
\begin{aligned}
& \frac{x^{2}}{25 \cdot\left(z^{2}+1\right)}+\frac{y^{2}}{9 \cdot\left(z^{2}+1\right)}=1 \rightarrow a=5 \sqrt{z^{2}+1} ; b=3 \sqrt{z^{2}+1} \\
& \Rightarrow S=\pi a b=\pi \cdot 5 \sqrt{z^{2}+1} \cdot 3 \sqrt{z^{2}+1}=15 \pi \cdot\left(z^{2}+1\right) \\
& V=\int_{0}^{2} S(z) d z=15 \pi \int_{0}^{2}\left(z^{2}+1\right) d z=\left.15 \pi\left(\frac{z^{3}}{3}+z\right)\right|_{0} ^{2}= \\
& =15 \pi\left(\frac{2^{3}}{3}+2-\frac{0^{3}}{3}-0\right)=5 \pi\left(3 \cdot \frac{8}{3}+3 \cdot 2-0-0\right)=70 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_

\%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-28$ "

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking

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- Last modified: 11:48, 23 June 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 20-29

## Material from PlusPi

Answer: 70\pi


Find all integer pairs $(x, y)$ for which

$$
(x+2)^{4}-x^{4}=y^{3}
$$

Solution:
$$
(x+2)^{4}-x^{4}=y^{3} \text {. }
$$

Introducing the notations $x+1=u$ and $y / 2=v$, our equation transforms into:

$$
u^{3}+u=v^{3}
$$

If equation (1) holds for integers $x, y$, then from (2) it is clear that $v^{3}$ is an integer, and $v$ is rational (since it is half of an integer), which is only possible if $v$ itself is an integer. Therefore, we will only look for integer solutions to (2).

$$
(u+1)^{3}-\left(u^{3}+u\right)=\frac{(3 u+1)^{2}}{3}+\frac{2}{3}>0
$$

and

$$
\left(u^{3}+u\right)-(u-1)^{3}=\frac{(3 u-1)^{2}}{3}+\frac{2}{3}>0
$$

due to these identities, for every $u$,

$$
(u+1)^{3}>u^{3}+u>(u-1)^{3}
$$

thus $u^{3}+u$ can only be equal to a cube number if $u^{3}+u=u^{3}$, i.e., $u=v=0$. In this case, $x=-1$, and $y=0$, which is the only integer solution to (1).

Answer: (x,y)=(-1,0)


In 2013, there was a "special" sandal trading rule in the Winniebago County market: if you buy one pair of sandals, it is traded at the original price of 50 yuan; if you buy a second pair, it is traded at a 60% discount of the original price; if you buy a third pair, it is half price. It is known that Javier bought three pairs of sandals according to this trading rule. Then he saved ( ).
(A) $25 \%$
(B) $30 \%$
(C) $33 \%$
(D) $40 \%$
(E) $45 \%$

Solution:
12. B.

From the problem, what we need to find is
$$
150-(50+50 \times 60 \%+50 \times 50 \%)=45 \text { (yuan). }
$$

Thus, the savings amount to $\frac{45}{150} \times 100 \%=30 \%$.

Answer: B


If positive numbers $a$, $b$, $c$ satisfy $a+2b+3c=abc$, then the minimum value of $abc$ is $\qquad$.

Solution:
$-1.9 \sqrt{2}$.
From $a b c=a+2 b+3 c \geqslant 3 \sqrt[3]{6 a b c}$ $\Rightarrow a b c \geqslant 9 \sqrt{2}$.
When $a=3 \sqrt{2}, b=\frac{3 \sqrt{2}}{2}, c=\sqrt{2}$, $a b c$ achieves the minimum value $9 \sqrt{2}$.

Answer: 9\sqrt{2}


The function
$$
f(x)=\sqrt{3} \sin ^{2} x+\sin x \cdot \cos x-\frac{\sqrt{3}}{2}\left(x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right]\right)
$$

has the range . $\qquad$

Solution:
2. $\left[-\frac{1}{2}, 1\right]$.

From the problem, we have
$$
\begin{array}{l}
f(x)=\sqrt{3} \times \frac{1-\cos 2 x}{2}+\frac{1}{2} \sin 2 x-\frac{\sqrt{3}}{2} \\
=\frac{1}{2} \sin 2 x-\frac{\sqrt{3}}{2} \cos 2 x=\sin \left(2 x-\frac{\pi}{3}\right) .
\end{array}
$$

Since $x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right]$, we have
$$
\begin{array}{l}
-\frac{\pi}{6} \leqslant 2 x-\frac{\pi}{3} \leqslant \frac{2 \pi}{3} \\
\Rightarrow-\frac{1}{2} \leqslant \sin \left(2 x-\frac{\pi}{3}\right) \leqslant 1 .
\end{array}
$$

Answer: \left[-\frac{1}{2}, 1\right]


At the award ceremony of a district olympiad, the following was announced:

Exactly one ninth of all participants in this district olympiad won a prize. Exactly one tenth of all participants in the district olympiad are members of the district club Young Mathematicians. Of the prize winners, exactly 75 percent are from the district club. Exactly 6 of the students who participated in the district olympiad and are members of the district club did not receive a prize.

Determine the total number of participants in this district olympiad!

Solution:
Let $T$ be the number of all participants in the district olympiad. According to the problem, $\frac{1}{9}$ of the participants are prize winners, and $75\%$, or $\frac{3}{4}$, of the prize winners are club members.

Thus, $\frac{1}{9} \cdot \frac{3}{4} \cdot T = \frac{1}{12} \cdot T$ of the club members receive a prize, meaning that one twelfth of the participants are both prize winners and club members.

The number of club members who receive a prize $\left(\frac{1}{12} \cdot T\right)$ and those who do not receive a prize (6) is equal to $\frac{1}{10}$ of the participants. Therefore,

$$
\frac{1}{12} \cdot T + 6 = \frac{1}{10} \cdot T \quad \text{and thus} \quad T = 360
$$

The number of participants in the district olympiad was 360.

Answer: 360


Determine all pairs $(a,b)$ of real numbers with $a\leqslant b$ that maximize the integral $$\int_a^b e^{\cos (x)}(380-x-x^2) \mathrm{d} x.$$

Solution:
1. **Identify the function to be integrated:**
   The integral to be maximized is:
   \[
   \int_a^b e^{\cos(x)} (380 - x - x^2) \, dx
   \]

2. **Find the roots of the quadratic function:**
   The quadratic function inside the integral is \( 380 - x - x^2 \). To find its roots, solve the equation:
   \[
   380 - x - x^2 = 0
   \]
   Rearrange it to:
   \[
   x^2 + x - 380 = 0
   \]
   Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -380 \):
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-380)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 1520}}{2} = \frac{-1 \pm \sqrt{1521}}{2}
   \]
   Since \( \sqrt{1521} = 39 \), the roots are:
   \[
   x = \frac{-1 + 39}{2} = 19 \quad \text{and} \quad x = \frac{-1 - 39}{2} = -20
   \]

3. **Determine the interval for the integral:**
   The quadratic function \( 380 - x - x^2 \) is positive between its roots \( x = -20 \) and \( x = 19 \). Therefore, to maximize the integral, we should integrate over this interval.

4. **Verify the conditions \( a \leq b \):**
   Given \( a \leq b \), the interval \([-20, 19]\) satisfies this condition.

5. **Conclusion:**
   The integral is maximized when \( a = -20 \) and \( b = 19 \).

The final answer is \( \boxed{ (a, b) = (-20, 19) } \)

Answer: (a, b) = (-20, 19)


In daily life, decimal is often used to represent numbers, using 10 digits: $0,1,2,3,4,5,6,7,8,9$. In electronic computers, binary is used, which only requires two digits: 0 and 1. Just as in decimal addition, "carry over when reaching ten," in binary, "carry over when reaching 2." Thus, we can obtain the following table comparing the decimal and binary representations of natural numbers:
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline Decimal & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & $\cdots$ \\
\hline Binary & 0 & 1 & 10 & 11 & 100 & 101 & 110 & 111 & 1000 & $\cdots$ \\
\hline
\end{tabular}

The decimal 0 is still 0 in binary, the decimal 1 is still 1 in binary, the decimal 2 becomes $1+1=10$ in binary, the decimal 3 becomes $10+1=11$, and so on. Knowing that the product of ten 2s in decimal is 1024, i.e., $2^{10}=1024$, in binary it is 10000000000. So, what is the decimal representation of the binary number 10110?

Answer: 22


Insert a digit into the middle of a two-digit number to form a three-digit number. Some two-digit numbers, when a certain digit is inserted in the middle, become a three-digit number that is $k$ times the original two-digit number ($k$ is a natural number). The maximum value of $k$ is . $\qquad$

Solution: answer: 19

Answer: 19


Solution:
$$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$

Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$

Answer: 2


Find the smallest positive integer $n$ (where $n>1$) such that the quadratic mean of the first $n$ natural numbers is an integer.
(Note: The quadratic mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{n}$ is $\left.\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}}.\right)$

Solution:
[Solution] Let $\sqrt{\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}}=m, \quad m \in \mathbb{Z}$, then
$$
\begin{array}{l}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2}, \\
\frac{(n+1)(2 n+1)}{6}=m^{2}, \\
(n+1)(2 n+1)=6 m^{2} .
\end{array}
$$

Since $6 m^{2}$ is even and $2 n+1$ is odd, $n+1$ must be even, hence $n$ is odd.
Let $n=6 p+3$ or $n=6 p-1$ or $n=6 p+1$.
(1) When $n=6 p+3$, then
$$
\begin{array}{c}
(6 p+4)(12 p+7)=6 m^{2}, \\
72 p^{2}+90 p+28=6 m^{2} .
\end{array}
$$

Since
$$
6172,6190,6 \times 28 \text {, }
$$

there is no solution in this case.
(2) When $n=6 p-1$, then
$$
\begin{array}{c}
6 p(12 p-1)=6 m^{2}, \\
m^{2}=p(12 p-1) .
\end{array}
$$

Since $p$ and $12 p-1$ are coprime, for the equation to hold, both $p$ and $12 p-1$ must be perfect squares.
Let
$$
p=s^{2}, \quad 12 p-1=t^{2} \text { . }
$$

Eliminating $p$ gives
$$
t^{2}=12 s^{2}-1=4\left(3 s^{2}-1\right)+3 .
$$

Since a square number modulo 4 can only be 0 or 1, the above equation cannot hold. Thus, there is no solution in this case.
(3) When $n=6 p+1$, then
$$
\begin{array}{c}
(6 p+2)(12 p+3)=6 m^{2} \\
m^{2}=(3 p+1)(4 p+1)
\end{array}
$$

Since $3 p+1$ and $4 p+1$ are coprime, both $3 p+1$ and $4 p+1$ must be perfect squares.
Let
$$
3 p+1=u^{2}, \quad 4 p+1=v^{2} \text { . }
$$

Eliminating $p$ gives
$$
4 u^{2}-3 v^{2}=1 \text { . }
$$

Clearly, $u=v=1$ is one solution, but in this case
$$
p=0, \quad n=1
$$

which contradicts $n>1$.
From $4 u^{2}-3 v^{2}=1$, we know that $v$ must be odd.
Let $v=2 q+1$, then the equation becomes
$$
u^{2}-3 q(q+1)-1=0 \text { . }
$$

Since $q(q+1)$ is even, $u$ must also be odd.
Let $u=2 j+1$, then the equation becomes
$$
4 j(j+1)=3 q(q+1) \text { . }
$$

Clearly, the left side of the equation is a multiple of 8. To make the right side a multiple of 8 and to minimize $n$, let $q+1=8$.
In this case, $q=7, j=6, j+1=7$.
Thus, $u=2 j+1=13, \quad v=2 q+1=15$,
$$
u^{2}=3 p+1=169, \quad p=56 \text { . }
$$

Therefore,
$$
n=6 p+1=337 \text { . }
$$

Answer: 337


Two bus stations, $A$ and $B$, continuously send out a bus at the same intervals, with each bus traveling at the same speed. A cyclist on the road between $A$ and $B$ notices that a bus passes from behind every $a$ minutes and a bus passes from the front every $b$ minutes. How often do stations $A$ and $B$ send out a bus?

Solution:
Let $A, B$ two stations send out a bus every $x$ minutes. According to the problem, we have $\left(\frac{1}{a}+\frac{1}{b}\right) x=2$.
Solving for $x$ gives $x=\frac{2 a b}{a+b}$.
Answer: $A, B$ two stations send out a bus every $\frac{2 a b}{a+b}$ minutes.

Answer: \frac{2 a b}{a+b}


Solution:
Answer: $x=1$.

Solution. Notice that $\sqrt{8088}=2 \sqrt{2022}$. This observation suggests that $x=1$ is a root of the equation. We will show that there are no other roots. Indeed, the left side represents the difference of two exponential functions: the base of the first is greater than one, and the base of the second is less than one (it can be noted that the product of these bases equals one). Therefore, the left side is the difference of an increasing and a decreasing function over the entire real line, i.e., it is a strictly increasing function, and thus it takes the value equal to the number on the right side at the unique point $x=1$.

Answer: 1


Let $f$ be a function defined on the set of positive integers, and with values in the same set, which satisfies:
$\bullet$ $f (n + f (n)) = 1$ for all $n\ge 1$.
$\bullet$ $f (1998) = 2$
Find the lowest possible value of the sum $f (1) + f (2) +... + f (1999)$, and find the formula of $f$ for which this minimum is satisfied.

Solution:
1. **Understanding the given conditions:**
   - We are given a function \( f \) defined on the set of positive integers with values in the same set.
   - The function satisfies \( f(n + f(n)) = 1 \) for all \( n \geq 1 \).
   - We are also given \( f(1998) = 2 \).

2. **Analyzing the function:**
   - From the condition \( f(n + f(n)) = 1 \), we know that for any \( n \), \( n + f(n) \) must be such that \( f(n + f(n)) = 1 \).
   - Given \( f(1998) = 2 \), we have \( 1998 + f(1998) = 2000 \), and thus \( f(2000) = 1 \).

3. **Inductive reasoning:**
   - We need to find the minimum possible value of the sum \( f(1) + f(2) + \cdots + f(1999) \).
   - We start by noting that \( f(1997 + f(1997)) = 1 \). Since \( f(1998) = 2 \), it implies \( f(1997) \geq 2 \).
   - By induction, we can show that \( f(n) \geq 2 \) for all \( n \leq 1998 \).

4. **Establishing a lower bound:**
   - We need to ensure that \( f(n) \geq 1999 - n \) for all \( n \leq 1997 \). If \( f(n) < 1999 - n \), then \( n + f(n) \leq 1998 \), which would contradict \( f(n + f(n)) = 1 \).

5. **Summing the values:**
   - We calculate the sum \( f(1) + f(2) + \cdots + f(1999) \) under the assumption that \( f(n) = 1999 - n \) for \( n \leq 1997 \), \( f(1998) = 2 \), and \( f(1999) = 1 \).
   - The sum is:
     \[
     \sum_{k=1}^{1997} (1999 - k) + f(1998) + f(1999)
     \]
   - This can be simplified as:
     \[
     \sum_{k=1}^{1997} 1999 - \sum_{k=1}^{1997} k + 2 + 1
     \]
   - The first sum is:
     \[
     1999 \times 1997 - \sum_{k=1}^{1997} k
     \]
   - The sum of the first 1997 positive integers is:
     \[
     \sum_{k=1}^{1997} k = \frac{1997 \times 1998}{2}
     \]
   - Therefore, the total sum is:
     \[
     1999 \times 1997 - \frac{1997 \times 1998}{2} + 3
     \]
   - Simplifying further:
     \[
     1999 \times 1997 - 998501 + 3 = 1999 \times 1997 - 998501 + 3 = 1999 \times 1997 - 998498
     \]
   - Calculating \( 1999 \times 1997 \):
     \[
     1999 \times 1997 = 1999 \times (2000 - 3) = 1999 \times 2000 - 1999 \times 3 = 3998000 - 5997 = 3992003
     \]
   - Therefore, the sum is:
     \[
     3992003 - 998498 = 1997003
     \]

6. **Verifying the function:**
   - The function \( f(n) = 1999 - n \) for \( n \leq 1997 \), \( f(1998) = 2 \), and \( f(1999) = 1 \) satisfies the given conditions and achieves the minimum sum.

The final answer is \( \boxed{1997003} \).

Answer: 1997003


Find the range of $k$ such that the equation
$$
x^{2}-k x-k+3=0
$$

has two roots in the open intervals $(0,1)$ and $(1,2)$, respectively.

Solution:
Solution: The original equation can be transformed into $x^{2}+3=k(x+1)$. Let $y_{1}=x^{2}+3, y_{2}=k(x+1)$.

Draw the graphs of $y_{1}$ and $y_{2}$, as shown in Figure 1. The intersection points of the parabola and the family of lines are on $\overparen{A B}$ and $\overparen{B C}$ (excluding points $A, B, C$) when the two roots of the equation are in the intervals $(0,1)$ and $(1,2)$. Thus, we have
$$
k_{D B}=2, k_{D C}=\frac{7}{3} .
$$

Clearly, $2<k<\frac{7}{3}$.

Answer: 2<k<\frac{7}{3}


On the extension of side $AB$ of rhombus $ABCD$ beyond point $B$, a point $M$ is taken such that $MD=MC$ and $\angle MDC=\arctan 8/5$. Find the ratio of segments $MA$ and $MB$.

Solution:
Let the side of the rhombus be denoted as $10 x$. Let $P$ be the projection of point $M$ on $D C$, and $K$ be the projection of point $D$ on $A B$. Since triangle $D M C$ is isosceles, $D P = 5 x$. Then, $M P = D P \operatorname{tg} \angle M D C = 8 x, D K = M P = 8 x, A K^{2} = A D^{2} - K D^{2} = 36 x^{2}, M K = D P = 5 x, A M = A K + M K = 11 x, M B = A M - A B = x, A M: M B = 11: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-03.jpg?height=358&width=784&top_left_y=1233&top_left_x=640)

## Answer

$11: 1$.

Answer: 11:1


In triangle $ABC$, $AB = 3 \cdot BC$, point $M$ is the midpoint of side $AB$, and $BD$ is the angle bisector. Find the angle $MDB$.

Solution:
4. Answer. $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_73882b07206f9b28bd67g-2.jpg?height=234&width=532&top_left_y=310&top_left_x=206)

Let $X$ be the midpoint of segment $M B$, and draw segment $D X$. By the property of the angle bisector, we have

$$
A D: D C = A B: B C = 3: 1
$$

On the other hand, $A X: X B = 3: 1$, so $A D: D C = A X: X B$. Therefore, $X D \parallel B C$. Then $\angle X B D = \angle D B C = \angle X D B$, which means that $X D = X B = X M$. From this, we get that $\angle M D B = 90^{\circ}$.

Answer: 90


Given $7 \sin \alpha+24 \cos \alpha=25$, find the value of $\tan \alpha$.

Solution:
Analysis: From the known equation and the Cauchy-Schwarz inequality, we get
\[1 \cdot \left(7^{2} + 24^{2}\right) = \left(\sin ^{2} \alpha + \cos ^{2} \alpha\right) \left(7^{2} + 24^{2}\right) \geqslant (7 \sin \alpha + 24 \cos \alpha)^{2} = 25^{2}.\]
The equality in the above equation holds if and only if \(\frac{\sin \alpha}{7} = \frac{\cos \alpha}{24}\), which implies \(\tan \alpha = \frac{7}{24}\).

Answer: \frac{7}{24}


Lisa writes a positive whole number in the decimal system on the blackboard and now makes in each turn the following:
The last digit is deleted from the number on the board and then the remaining shorter number (or 0 if the number was one digit) becomes four times the number deleted number added. The number on the board is now replaced by the result of this calculation.
Lisa repeats this until she gets a number for the first time was on the board.
(a) Show that the sequence of moves always ends.
(b) If Lisa begins with the number $53^{2022} - 1$, what is the last number on the board?

Example: If Lisa starts with the number $2022$, she gets $202 + 4\cdot 2 = 210$ in the first move and overall the result $$2022 \to 210 \to 21 \to 6 \to 24 \to 18 \to 33 \to 15 \to 21$$.
Since Lisa gets $21$ for the second time, the turn order ends.

Solution:
1. **Show that the sequence of moves always ends.**

   Let's denote the number on the board as \( X \). In each turn, Lisa performs the following operation:
   \[
   X = 10a + b \quad \text{(where \( a \) is the number formed by all digits except the last one, and \( b \) is the last digit)}
   \]
   The new number \( Y \) is given by:
   \[
   Y = a + 4b
   \]

   We need to show that this sequence always ends. First, observe that if \( X \) is a single-digit number, the sequence will eventually repeat because there are only 10 possible single-digit numbers (0 through 9).

   Next, consider the case when \( X \) has more than one digit. If \( X \geq 40 \), then \( a \geq 4 \). We can show that \( X \) strictly decreases:
   \[
   X = 10a + b \quad \text{and} \quad Y = a + 4b
   \]
   Since \( a \geq 4 \), we have:
   \[
   3a \geq 12 > b \implies 10a + b > a + 4b \implies X > Y
   \]
   Therefore, the number on the board always decreases strictly until it falls below 40. Once it falls below 40, it will eventually repeat because all numbers from 0 to 39 are represented in the sequences provided.

   Hence, the sequence of moves always ends.

2. **If Lisa begins with the number \( 53^{2022} - 1 \), what is the last number on the board?**

   We need to determine the final number on the board when Lisa starts with \( 53^{2022} - 1 \).

   First, note that:
   \[
   X \to Y \implies 4X \equiv Y \pmod{13}
   \]
   because:
   \[
   40a + 4b \equiv a + 4b \pmod{13}
   \]
   Each operation multiplies the number on the board by 4 modulo 13. Since \( 53 \equiv 1 \pmod{13} \), we have:
   \[
   53^{2022} \equiv 1^{2022} \equiv 1 \pmod{13}
   \]
   Therefore:
   \[
   53^{2022} - 1 \equiv 1 - 1 \equiv 0 \pmod{13}
   \]
   The initial number is divisible by 13, so the final number will also be divisible by 13. The possible numbers are 13, 26, or 39.

   Next, we consider the modulo 3 condition. The number on the board does not change modulo 3:
   \[
   53 \equiv 2 \pmod{3} \implies 53^{2022} \equiv 2^{2022} \pmod{3}
   \]
   Since \( 2^2 \equiv 1 \pmod{3} \), we have:
   \[
   2^{2022} \equiv (2^2)^{1011} \equiv 1^{1011} \equiv 1 \pmod{3}
   \]
   Therefore:
   \[
   53^{2022} - 1 \equiv 1 - 1 \equiv 0 \pmod{3}
   \]
   The final number must also be divisible by 3. The only number that is divisible by both 13 and 3 is 39.

   Therefore, the last number on the board is 39.

The final answer is \(\boxed{39}\)

Answer: 39


Two cars, A and B, depart from locations $A$ and $B$ respectively at the same time, and travel back and forth between $A$ and $B$ at a constant speed. If after the first meeting, car A continues to travel for 4 hours to reach $B$, while car B only travels for 1 hour to reach $A$, then when the two cars meet for the 15th time (excluding meetings at $A$ and $B$), they have traveled $\qquad$ hours.

Answer: 86


How many sequences of $0 \mathrm{~s}$ and $1 \mathrm{~s}$ are there of length 10 such that there are no three 0s or 1s consecutively anywhere in the sequence?

Solution: Solution: We can have blocks of either 1 or $20 \mathrm{~s}$ and $1 \mathrm{~s}$, and these blocks must be alternating between 0s and 1s. The number of ways of arranging blocks to form a sequence of length $n$ is the same as the number of omino tilings of a 1-by- $n$ rectangle, and we may start each sequence with a 0 or a 1 , making $2 F_{n}$ or, in this case, 178 sequences.

Answer: 178


Find the derivative of the specified order.

$$
y=\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}, y^{IV}=?
$$

Solution:
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}\right)^{\prime}=(-1-2 x) e^{\frac{x-1}{2}}+\frac{1}{2} \cdot\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}= \\
& =\frac{1}{2} \cdot\left(-2-4 x+1-x-x^{2}\right) e^{\frac{x-1}{2}}=-\frac{1}{2} \cdot\left(1+5 x+x^{2}\right) e^{\frac{x-1}{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{1}{2} \cdot\left(1+5 x+x^{2}\right) e^{\frac{x-1}{2}}\right)^{\prime}= \\
& =-\frac{1}{2} \cdot(5+2 x) e^{\frac{x-1}{2}}-\frac{1}{4} \cdot\left(1+5 x+x^{2}\right) e^{\frac{x-1}{2}}= \\
& =-\frac{1}{4} \cdot\left(11+9 x+x^{2}\right) e^{\frac{x-1}{2}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{1}{4} \cdot\left(11+9 x+x^{2}\right) e^{\frac{x-1}{2}}\right)^{\prime}= \\
& =-\frac{1}{4} \cdot(9+2 x) e^{\frac{x-1}{2}}-\frac{1}{8} \cdot\left(11+9 x+x^{2}\right) e^{\frac{x-1}{2}}= \\
& =-\frac{1}{8} \cdot\left(29+13 x+x^{2}\right) e^{\frac{x-1}{2}} \\
& y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-\frac{1}{8} \cdot\left(29+13 x+x^{2}\right) e^{\frac{x-1}{2}}\right)^{\prime}= \\
& =-\frac{1}{8} \cdot(13+2 x) e^{\frac{x-1}{2}}-\frac{1}{16} \cdot\left(29+13 x+x^{2}\right) e^{\frac{x-1}{2}}= \\
& =-\frac{1}{16} \cdot\left(55+17 x+x^{2}\right) e^{\frac{x-1}{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-17

Answer: -\frac{1}{16}\cdot(55+17x+x^{2})e^{\frac{x-1}{2}}


Let $p(x)$ be a polynomial with integer coefficients such that $p(0) = 0$ and $0 \le p(1) \le 10^7$. Suppose that there exist positive integers $a,b$ such that $p(a) = 1999$ and $p(b) = 2001$. Determine all possible values of $p(1)$.

(Note: $1999$ is a prime number.)

Solution:
1. Given that \( p(0) = 0 \), we can write \( p(x) = xq(x) \) where \( q(x) \) is a polynomial with integer coefficients. This implies that \( p(a) = 1999 \) and \( p(b) = 2001 \) for some positive integers \( a \) and \( b \).

2. Since \( p(a) = 1999 \) and \( 1999 \) is a prime number, \( a \) must be a divisor of \( 1999 \). The possible values of \( a \) are \( 1 \) and \( 1999 \).

3. Similarly, since \( p(b) = 2001 \) and \( 2001 = 3 \times 23 \times 29 \), the possible values of \( b \) are \( 1, 3, 23, 29, 69, 87, 667, \) and \( 2001 \).

4. We also know that \( b - a \mid p(b) - p(a) \). Since \( p(b) - p(a) = 2001 - 1999 = 2 \), it follows that \( b - a \mid 2 \). Therefore, \( b - a \) can be \( -2, -1, 1, \) or \( 2 \).

5. Considering the possible values of \( a \) and \( b \) and the condition \( b - a \mid 2 \), we have the following pairs:
   - \( (a, b) = (1, 3) \)
   - \( (a, b) = (1999, 2001) \)

6. For the pair \( (a, b) = (1, 3) \):
   - Since \( p(1) = p(a) = 1999 \), we have \( p(1) = 1999 \).

7. For the pair \( (a, b) = (1999, 2001) \):
   - We know that \( p(x) - x \) has roots at \( 0, 1999, \) and \( 2001 \). Therefore, \( p(x) - x = x(x - 1999)(x - 2001)r(x) \) where \( r(x) \) is a polynomial with integer coefficients.
   - Evaluating at \( x = 1 \), we get:
     \[
     p(1) = 1 \cdot (1 - 1999) \cdot (1 - 2001) \cdot r(1) + 1 = (-1998)(-2000)r(1) + 1 = 3996000r(1) + 1
     \]
   - Given \( 0 \leq p(1) \leq 10^7 \), we solve for \( r(1) \):
     \[
     0 \leq 3996000r(1) + 1 \leq 10^7
     \]
     \[
     -1 \leq 3996000r(1) \leq 9999999
     \]
     \[
     0 \leq r(1) \leq \frac{9999999}{3996000} \approx 2.5
     \]
     Since \( r(1) \) is an integer, \( r(1) \) can be \( 0, 1, \) or \( 2 \).

8. Therefore, the possible values of \( p(1) \) are:
   - \( r(1) = 0 \): \( p(1) = 1 \)
   - \( r(1) = 1 \): \( p(1) = 3996000 \cdot 1 + 1 = 3996001 \)
   - \( r(1) = 2 \): \( p(1) = 3996000 \cdot 2 + 1 = 7992001 \)

9. Combining all possible values, we have \( p(1) = 1999, 1, 3996001, \) and \( 7992001 \).

The final answer is \( \boxed{1, 1999, 3996001, 7992001} \).

Answer: 1, 1999, 3996001, 7992001


Solution:
2. B.

Since $\left\{\begin{array}{l}x>0, x \neq 1, \\ 2 x^{2}+x-1>0,\end{array}\right.$ therefore, $x>\frac{1}{2}, x \neq 1$.
From $\log _{x}\left(2 x^{2}+x-1\right)>\log _{x} 2-1$, we deduce
$$
\log _{x}\left(2 x^{3}+x^{2}-x\right)>\log _{x} 2 \text {, }
$$

further leading to
$$
\left\{\begin{array} { l }
{ 0 < x < 1, \\
2 x^{3}+x^{2}-x>2 .
\end{array}\right.\right.
$$

Solving, we get $0 < x < 1$ or $x > 1$.
Therefore, the range of $x$ is $x>\frac{1}{2}$, and $x \neq 1$.

Answer: B


At Olympic High School, $\frac{2}{5}$ of the freshmen and $\frac{4}{5}$ of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?
$\textbf{(A)}$ There are five times as many sophomores as freshmen.
$\textbf{(B)}$ There are twice as many sophomores as freshmen.
$\textbf{(C)}$ There are as many freshmen as sophomores.
$\textbf{(D)}$ There are twice as many freshmen as sophomores.
$\textbf{(E)}$ There are five times as many freshmen as sophomores.

Solution:
Let $f$ be the number of freshman and $s$ be the number of sophomores.
$\frac{2}{5}f=\frac{4}{5}s$
$2f = 4s$
$f=2s$
There are twice as many freshmen as sophomores.
$\boxed{\text{D}}$

Answer: D


Let $a, b, c$ be real numbers, and satisfy
$$
a+b+c=15, a^{2}+b^{2}+c^{2}=100 \text {. }
$$

Then the product of the maximum and minimum values of $a$ is $\qquad$

Solution:
$-1 . \frac{25}{3}$.
Since $100-a^{2}=b^{2}+c^{2}$
$$
\geqslant \frac{1}{2}(b+c)^{2}=\frac{1}{2}(15-a)^{2},
$$

Therefore, $3 a^{2}-30 a+25 \leqslant 0$.
Since the maximum and minimum values of $a$ are the two roots of the equation
$$
3 a^{2}-30 a+25=0
$$

and the maximum and minimum values of $a$ can be achieved, by Vieta's formulas, the product of the maximum and minimum values of $a$ is $\frac{25}{3}$.

Answer: \frac{25}{3}


Find the limit $\lim _{x \rightarrow 0}\left(\left(\int_{0}^{x^{2}} \cos t \, dt\right) / x\right)$.

Solution:
Solution. We will use L'Hôpital's rule, as there is an indeterminate form of type «0/0».

$$
\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos x d x}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x^{2}} \cos x d x\right)^{\prime}}{x^{\prime}}=\lim _{x \rightarrow 0} \frac{\cos \left(x^{2}\right) \cdot 2 x}{1}=0
$$

Answer: 0


2 boys and 3 girls went on a trip. Both the boys and the girls carried equally weighted backpacks, and the 5 backpacks together weighed 44 kg. On the way - so that 1 girl could always rest - the boys carried the girls' backpacks, one each on their backs and one together, in their hands, while the girls took the boys' backpacks on their backs, one each. Thus, except for the resting girl, everyone's load increased by the same amount. How much did each boy's and girl's backpack weigh?

Solution:
The weight of the girl's backpack is evenly distributed among the other four hikers. Among them, the girls carry more by the amount that the boys' backpacks are heavier, so each boy's backpack is $1/4$ part heavier than the girls'. Thus, the total weight of the 5 backpacks is 5 and a half times the weight of a girl's backpack, which means a girl's backpack weighs $44: 5.5=8 \text{ kg}$, and a boy's is $1/4$ part more: $10 \text{ kg}$.

Indeed, this way, for 1 boy, $3 \cdot 8: 2=12 \text{ kg}$ is carried, which is $2 \text{ kg}$ more than his own backpack.

Huhn Ágnes (Szeged, Rókusi Ált. Isk. VIII. o. t.)

Answer: 8


Let $N$ denote the number of different ways to distribute $r$ identical items to $n$ people, find $N$.

Solution: Solution: Let the $i$-th person $(1 \leqslant i \leqslant n)$ receive $x_{i}$ items, then $x_{i} \geqslant 0$ and $x_{1}+x_{2}+\cdots+x_{n}=r$, so $N$ is equal to the number of non-negative integer solutions of the indeterminate equation $x_{1}+x_{2}+\cdots+x_{n}=r$, which is $\binom{n+r-1}{r}$.

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


Find all positive integers $k$ such that the indeterminate equation $x^{2}+y^{2}=k x y-1$ has positive integer solutions for $x$ and $y$.

Solution:
Prove that among the positive integer solutions of the given indeterminate equation, the one that minimizes $x+y$ is $\left(x_{0}, y_{0}\right)$.
If $x_{0}=y_{0}$, then $x_{0}^{2} \backslash\left(2 x_{0}^{2}+1\right)$.
Therefore, $x_{0}^{2} \mid 1 \Rightarrow x_{0}=1$.
Thus, $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}=3$.
If $x_{0} \neq y_{0}$, by symmetry, assume $x_{0}>y_{0}$.
By $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}$
$$
\Rightarrow x_{0}^{2}-t y_{0} x_{0}+y_{0}^{2}+1=0 \text {. }
$$

Viewing equation (1) as a quadratic equation in $x_{0}$, by Vieta's formulas, it has two roots $x_{0}$ and $t y_{0}-x_{0}$, and $t y_{0}-x_{0}=\frac{y_{0}^{2}+1}{x_{0}}$.

Since $t y_{0}-x_{0} \in \mathbf{Z}_{+}$, $\left(\frac{y_{0}^{2}+1}{x_{0}}, y_{0}\right)$ is also a positive integer solution of equation (1).
By $x_{0}>y_{0} \geqslant 1$, we have $\frac{y_{0}^{2}+1}{x_{0}}<x_{0}$, i.e.,
$$
\frac{y_{0}^{2}+1}{x_{0}}+y_{0}<x_{0}+y_{0} \text {. }
$$

This contradicts the minimality of $\left(x_{0}, y_{0}\right)$.
In conclusion, $k=\frac{x_{0}^{2}+y_{0}^{2}+1}{x_{0} y_{0}}=3$.

Answer: 3


There are different positive integers written on the board. Their (arithmetic) mean is a decimal number, with the decimal part exactly 0.2016. What is the least possible value of the mean?

(Patrik Bak)

Solution:
Solution. Let $s$ be the sum, $n$ the number and $p$ the integer part of the mean of the numbers on the board. Then we can write

$$
\frac{s}{n}=p+\frac{2016}{10000}=p+\frac{126}{625}
$$

which gives

$$
625(s-p n)=126 n \text {. }
$$

Numbers 126 and 625 are coprime, thus $625 \mid n$. Therefore $n \geqslant 625$.

The numbers on the board are different, that is

$p=\frac{s}{n}-\frac{126}{625} \geqslant \frac{1+2+\cdots+n}{n}-\frac{126}{625}=\frac{n(n+1) / 2}{n}-\frac{126}{625} \geqslant \frac{625+1}{2}-\frac{126}{625}>312$.

The integer $p$ is thus at least 313 and the value of the mean at least 313,2016 .

This value can be attained by numbers $1,2, \ldots, 624$ and 751 . We get

$$
\frac{1+2+\cdots+624+751}{625}=\frac{312 \cdot 625+751}{625}=313+\frac{126}{625}=313,2016 .
$$

Answer: 313.2016


A cube with edge length $n$ ($n$ is a positive integer) is painted red on its surface, and then it is cut into $n^{3}$ smaller cubes with edge length 1. It is found that the number of smaller cubes with only one face painted red is exactly 12 times the number of smaller cubes with exactly two faces painted red. Then $n$ equals (

Solution:
26
【Answer】The number of small cubes with only one face painted red is $6(n-2)^{2}$; the number of small cubes with exactly two faces painted red is $\frac{4(n-2) \times 6}{2}=12(n-2)$; so $12(n-2) \times 12=6(n-2)^{2} \Rightarrow n-2=24 \Rightarrow n=26$

Answer: 26


If $x=4$ and $y=2$, which of the following expressions gives the smallest value?
(A) $x+y$
(B) $x y$
(C) $x-y$
(D) $x \div y$
(E) $y \div x$

Solution:
When $x=4$ and $y=2, x+y=4+2=6, x y=4 \times 2=8, x-y=4-2=2, x \div y=4 \div 2=2$, and $y \div x=2 \div 4=\frac{1}{2}$.

Therefore, the expression which gives the smallest value when $x=4$ and $y=2$ is $y \div x$.

ANSWER: (E)

Answer: E


In a convex hexagon $ABCDEF$, $\angle A=\angle B=\angle C=\angle D=\angle E=\angle F$, and $AB+BC=11, FA-CD=3$, find $BC+DE$.

Solution:
From the problem, we know that $A F // C D, A B // E D$, so we can complete the hexagon to form a parallelogram $M C N F$, as shown in Figure 1.4.13. It is clearly a parallelogram, one of its internal angles is $60^{\circ}$, and $\triangle A M B$ and $\triangle D E N$ are both equilateral triangles. Thus,
$$
\begin{aligned}
M C & =A B+B C=11 \\
F A=M F-A M & =C N-A B=C D+D E-A B
\end{aligned}
$$

Therefore, $F A-C D=E D-A B$, and since $F A-C D=3$, then $E D - A B=3$, hence $B C+D E=14$.
Note: This problem can also be solved by completing it into an equilateral triangle or a trapezoid.

Answer: 14


Call a day a [i]perfect[/i] day if the sum of the digits of the month plus sum of the digits of the day equals the sum of digits of the year. For example, February $28$th, $2028$ is a perfect day because $2+2+8=2+0+2+8$. Find the number of perfect days in $2018$.

Solution:
1. First, we need to determine the sum of the digits of the year 2018:
   \[
   2 + 0 + 1 + 8 = 11
   \]
   Therefore, for a day to be perfect, the sum of the digits of the month plus the sum of the digits of the day must equal 11.

2. We will analyze each month to find the days that satisfy this condition.

3. **Month 1 (January):**
   - Sum of the digits of the month: \(1\)
   - We need the sum of the digits of the day to be \(11 - 1 = 10\).
   - Possible days: \(19\) (1+9=10) and \(28\) (2+8=10).
   - Total: 2 days.

4. **Month 2 (February):**
   - Sum of the digits of the month: \(2\)
   - We need the sum of the digits of the day to be \(11 - 2 = 9\).
   - Possible days: \(9\) (0+9=9), \(18\) (1+8=9), and \(27\) (2+7=9).
   - Note: February in 2018 does not have a 29th day.
   - Total: 3 days.

5. **Month 3 (March):**
   - Sum of the digits of the month: \(3\)
   - We need the sum of the digits of the day to be \(11 - 3 = 8\).
   - Possible days: \(8\) (0+8=8), \(17\) (1+7=8), and \(26\) (2+6=8).
   - Total: 3 days.

6. **Month 4 (April):**
   - Sum of the digits of the month: \(4\)
   - We need the sum of the digits of the day to be \(11 - 4 = 7\).
   - Possible days: \(7\) (0+7=7), \(16\) (1+6=7), and \(25\) (2+5=7).
   - Total: 3 days.

7. **Month 5 (May):**
   - Sum of the digits of the month: \(5\)
   - We need the sum of the digits of the day to be \(11 - 5 = 6\).
   - Possible days: \(6\) (0+6=6), \(15\) (1+5=6), and \(24\) (2+4=6).
   - Total: 3 days.

8. **Month 6 (June):**
   - Sum of the digits of the month: \(6\)
   - We need the sum of the digits of the day to be \(11 - 6 = 5\).
   - Possible days: \(5\) (0+5=5), \(14\) (1+4=5), and \(23\) (2+3=5).
   - Total: 3 days.

9. **Month 7 (July):**
   - Sum of the digits of the month: \(7\)
   - We need the sum of the digits of the day to be \(11 - 7 = 4\).
   - Possible days: \(4\) (0+4=4), \(13\) (1+3=4), and \(22\) (2+2=4).
   - Total: 3 days.

10. **Month 8 (August):**
    - Sum of the digits of the month: \(8\)
    - We need the sum of the digits of the day to be \(11 - 8 = 3\).
    - Possible days: \(3\) (0+3=3), \(12\) (1+2=3), \(21\) (2+1=3), and \(30\) (3+0=3).
    - Total: 4 days.

11. **Month 9 (September):**
    - Sum of the digits of the month: \(9\)
    - We need the sum of the digits of the day to be \(11 - 9 = 2\).
    - Possible days: \(2\) (0+2=2), \(11\) (1+1=2), and \(20\) (2+0=2).
    - Total: 3 days.

12. **Month 10 (October):**
    - Sum of the digits of the month: \(1 + 0 = 1\)
    - We need the sum of the digits of the day to be \(11 - 1 = 10\).
    - Possible days: \(19\) (1+9=10) and \(28\) (2+8=10).
    - Total: 2 days.

13. **Month 11 (November):**
    - Sum of the digits of the month: \(1 + 1 = 2\)
    - We need the sum of the digits of the day to be \(11 - 2 = 9\).
    - Possible days: \(9\) (0+9=9), \(18\) (1+8=9), and \(27\) (2+7=9).
    - Total: 3 days.

14. **Month 12 (December):**
    - Sum of the digits of the month: \(1 + 2 = 3\)
    - We need the sum of the digits of the day to be \(11 - 3 = 8\).
    - Possible days: \(8\) (0+8=8), \(17\) (1+7=8), and \(26\) (2+6=8).
    - Total: 3 days.

15. Summing up all the perfect days:
    \[
    2 + 3 + 3 + 3 + 3 + 3 + 3 + 4 + 3 + 2 + 3 + 3 = 36
    \]

The final answer is $\boxed{36}$

Answer: 36


Solution:
5.2. A boy presses a vertical rod against a rough horizontal surface with his thumb. Then he gradually tilts the rod, keeping the component of the force directed along the rod applied to its end unchanged. At an angle of inclination of the rod to the horizontal $\alpha=70^{\circ}$, the rod starts to slide along the surface. Determine the coefficient of friction between the surface and the rod, if in the vertical position the normal reaction force of the plane was 21 times the weight of the rod. Round your answer to the hundredths.

Answer. $\{0,35\}$

Answer: 0.35


Given an arithmetic sequence $a_{1}, a_{2}, \cdots, a_{1000}$, the sum of the first 100 terms is 100, and the sum of the last 100 terms is 1000. Then $a_{1}=$ $\qquad$ .

Solution:
3. 0. 505 .

Let the common difference of the arithmetic sequence be $d$. Then
$$
\begin{array}{l}
100 a_{1}+4950 d=100, \\
100 a_{1}+94950 d=1000 .
\end{array}
$$

Therefore, $d=0.01, a_{1}=1-0.495=0.505$

Answer: 0.505


In triangle $ABC$, points $X$ and $Y$ are taken on sides $AC$ and $BC$ such that $\angle ABX = \angle YAC$, $\angle AYB = \angle BXC$, and $XC = YB$. Find the angles of triangle $ABC$.

Solution:
For the external angles $BXC$ and $AYB$ of triangles $ABX$ and $CAY$, we write the equalities $\angle BXC = \angle ABX + \angle BAX$, $\angle AYB = \angle CAY + \angle YCA$ (see the figure). Since by the condition $\angle BXC = \angle AYB$, $\angle ABX = \angle CAY$, then $\angle BAX = \angle YCA$, which means that triangle $ABC$ is isosceles, $AB = BC$.

For triangles $XBC$ and $YAB$, two sides and the angle not between them are equal: $\angle BXC = \angle AYB$, $XC = YB$, $BC = AB$. Such triangles are either equal or

$\angle XBC + \angle YAB = 180^\circ$ (we will prove this below), but the second case is impossible since $\angle XBC + \angle YAB < \angle ABC + \angle CAB = 180^\circ - \angle ACB < 180^\circ$. Therefore, triangles $XBC$ and $YAB$ are equal, and thus $\angle ABC = \angle BCA$, which means that triangle $ABC$ is equilateral.

Now we will prove the stated fact. Mark a point $B'$ on the ray $XB$ such that $XB' = YA$. Point $B'$ can be either inside or outside the segment $XB$ (see the figure).

Triangles $XB'C$ and $YAB$ are equal by two sides and the angle between them ($XC = YB$, $XB' = YA$, $\angle CXB' = \angle BYA$). Therefore, $CB' = CB$. If $B$ and $B'$ coincide, triangles $XBC$ and $YAB$ are equal. If $B'$ does not coincide with $B$, then in the isosceles triangle $B'CB$, the angles $B'BC$ and $BB'C$ are equal, and thus,

$\angle XBC + \angle XB'C = 180^\circ$, and $\angle XBC + \angle YAB = 180^\circ$.

## Answer

All angles are $60^\circ$.

Answer: 60


Given a fifth-degree polynomial $f(x)$ with the leading coefficient of 1 satisfies: $f(n)=8 n, n=1,2, \cdots, 5$, then the coefficient of the linear term of $f(x)$ is $\qquad$ .

Solution: According to the problem, $f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+8 x$, so its coefficient of the linear term is $5!\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+8=120+60+40+30+24+8=282$

Answer: 282


$ABCD$ is a rectangle with $AB = CD = 2$. A circle centered at $O$ is tangent to $BC$, $CD$, and $AD$ (and hence has radius $1$). Another circle, centered at $P$, is tangent to circle $O$ at point $T$ and is also tangent to $AB$ and $BC$. If line $AT$ is tangent to both circles at $T$, find the radius of circle $P$.

Solution:
1. **Identify the given information and set up the problem:**
   - Rectangle \(ABCD\) with \(AB = CD = 2\).
   - Circle centered at \(O\) is tangent to \(BC\), \(CD\), and \(AD\) with radius \(1\).
   - Another circle centered at \(P\) is tangent to circle \(O\) at point \(T\) and is also tangent to \(AB\) and \(BC\).
   - Line \(AT\) is tangent to both circles at \(T\).

2. **Determine the coordinates and positions:**
   - Place the rectangle \(ABCD\) in the coordinate plane with \(A = (0, 0)\), \(B = (2, 0)\), \(C = (2, 2)\), and \(D = (0, 2)\).
   - The circle centered at \(O\) is tangent to \(BC\), \(CD\), and \(AD\), so its center \(O\) is at \((1, 1)\) with radius \(1\).

3. **Analyze the second circle centered at \(P\):**
   - The circle centered at \(P\) is tangent to circle \(O\) at point \(T\) and is also tangent to \(AB\) and \(BC\).
   - Let the radius of the circle centered at \(P\) be \(r\).

4. **Determine the coordinates of \(P\):**
   - Since the circle is tangent to \(AB\) and \(BC\), the center \(P\) must be at \((2 - r, r)\).

5. **Use the tangency condition at point \(T\):**
   - The distance between the centers \(O\) and \(P\) is \(1 + r\) (sum of the radii).
   - The distance between \(O\) at \((1, 1)\) and \(P\) at \((2 - r, r)\) is given by:
     \[
     \sqrt{(2 - r - 1)^2 + (r - 1)^2} = \sqrt{(1 - r)^2 + (r - 1)^2} = \sqrt{2(1 - r)^2} = \sqrt{2} |1 - r|
     \]
   - Since \(1 - r\) is positive (as \(r < 1\)), we have:
     \[
     \sqrt{2} (1 - r) = 1 + r
     \]

6. **Solve the equation:**
   \[
   \sqrt{2} (1 - r) = 1 + r
   \]
   \[
   \sqrt{2} - \sqrt{2} r = 1 + r
   \]
   \[
   \sqrt{2} - 1 = (1 + \sqrt{2}) r
   \]
   \[
   r = \frac{\sqrt{2} - 1}{1 + \sqrt{2}}
   \]

7. **Rationalize the denominator:**
   \[
   r = \frac{\sqrt{2} - 1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{(\sqrt{2} - 1)(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})}
   \]
   \[
   r = \frac{\sqrt{2} - 2 - 1 + \sqrt{2}}{1 - 2} = \frac{2\sqrt{2} - 3}{-1} = 3 - 2\sqrt{2}
   \]

The final answer is \(\boxed{3 - 2\sqrt{2}}\).

Answer: 3 - 2\sqrt{2}


Solution:
Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$

Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$.

Answer: 1


A steamboat travels from A to B in 3 hours and, with the same engine power, from B to A in \(4 \frac{1}{2}\) hours. How long does a vessel propelled only by the current take for the journey from A to B?

Solution:
}

We can assume the distance-time law of uniform motion \(s=vt\) in both directions of the river. For the trip downstream, this means \(v=v_{D}+v_{S}\), and for the trip upstream, \(v=v_{D}-v_{S}\), where \(v_{D}\) is the boat's own speed and \(v_{S}\) is the river's current speed. Therefore,

\[
\begin{array}{r}
s=\left(v_{D}+v_{S}\right) \cdot(3 \text{ h})=\left(v_{D}-v_{S}\right) \cdot(4.5 \text{ h}) \\
v_{D}=\frac{4.5 \text{ h}+3 \text{ h}}{4.5 \text{ h}-3 \text{ h}} v_{S}=5 v_{S}
\end{array}
\]

and thus \(s=\left(v_{D}+v_{S}\right) \cdot(3 \text{ h})=6 v_{S} \cdot(3 \text{ h})\). For a boat that is only carried by the current, \(s=v_{S} t\); with the above equation, we have

\[
s=6 v_{S} \cdot(3 \text{ h})=v_{S} t
\]

From this, the travel time for a vessel driven only by the current is \(t=18 \text{ h}\).

Answer: 18


Let the set $A=\{1,2, \cdots, n\}$, and let $S_{n}$ denote the sum of the elements in all non-empty proper subsets of $A$, and $B_{n}$ denote the number of subsets of $A$, then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}=$ $\qquad$ .

Solution:
8. $\frac{1}{4}$.

It is easy to know that $B_{n}=2^{n}$.
First, we examine the number of times the number 1 appears in all non-empty proper subsets of $A$. It is easy to see that it appears 1 time in the one-element sets, i.e., $\mathrm{C}_{n-1}^{0}$ times, $\mathrm{C}_{n-1}^{1}$ times in the two-element sets, $\mathrm{C}_{n-1}^{2}$ times in the three-element sets, $\cdots \cdots$, and $\mathrm{C}_{n-1}^{n-2}$ times in the $(n-1)$-element sets. Therefore, the total number of times the number 1 appears in all non-empty proper subsets of $A$ is $\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-2}$. Thus,
$$
S_{n}=(1+2+\cdots+n)\left(\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-2}\right)=\frac{n(n+1)}{2}\left(2^{n-1}-1\right) .
$$

Therefore,
$$
\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}=\frac{n(n+1)\left(2^{n-1}-1\right)}{2 n^{2} \times 2^{n}}=\frac{1}{4} .
$$

Answer: \frac{1}{4}


The sequence of integers $u_{0}, u_{1}, u_{2}, u_{3}, \cdots$ satisfies $u_{0}=1$, and for each positive integer $n$
$$u_{n+1} u_{n-1}=k u_{n}$$

where $k$ is some fixed positive integer. If
$$u_{2000}=2000$$

find all possible values of $k$.

Solution:
［Solution］ Let $u_{1}=u$. By the problem statement,
$$u_{2}=\frac{k u_{1}}{u_{0}}=k u$$

If $u=0$, then $u_{1}=0, u_{2}=0$. If $u_{t-1}=0$, then $k u_{t}=u_{t-1} \cdot$ $u_{t+1}=0$, hence $u_{t}=0$. Therefore, for any positive integer $n, u_{n}=0$. This contradicts $u_{2000}=2000$. Thus, $u \neq 0$.

By the problem statement, after calculation we get
$$\begin{array}{ll}
u_{3}=\frac{k u_{2}}{u_{1}}=k^{2}, & u_{4}=\frac{k u_{3}}{u_{2}}=\frac{k^{2}}{u}, \\
u_{5}=\frac{k u_{4}}{u_{3}}=\frac{k}{u}, & u_{6}=\frac{k u_{5}}{u_{4}}=1, \\
u_{7}=\frac{k u_{6}}{u_{5}}=u . &
\end{array}$$

From this, we can see that the given sequence is a periodic sequence with a period of 6. Therefore, by
$$2000=6 \times 333+2$$

we have
$$u_{2000}=u_{2}=k u,$$

Thus, $k$ should satisfy
$$\left\{\begin{array}{l}
k u=2000, \\
\frac{k}{u} \text { is an integer, }
\end{array}\right.$$

Answer: k u=2000


In that year, there were no rematches, and 75 teams participated in the matches. How many matches were played for the cup?

Solution:
Note: after a loss, the team is eliminated.

## Solution

Since one team is eliminated after each game, a total of 74 matches were played.

## Answer

74 matches.

Answer: 74


Through the midpoint of the hypotenuse of a right triangle, a perpendicular is drawn to it. The segment of this perpendicular, enclosed within the triangle, is equal to c, and the segment enclosed between one leg and the extension of the other is equal to 3c. Find the hypotenuse.

Solution:
Let $M$ be the midpoint of the hypotenuse $AB$ of the right triangle $ABC$, $N$ and $K$ be the points of intersection of the perpendicular to $AB$ passing through point $M$ with the leg $AC$ and the extension of the leg $BC$ respectively. Denote $AM = BM = x$. Due to the obvious similarity of triangles $AMN$ and $KMB$,

$\angle BKM = \angle BAC = \alpha$, so $MN : AM = BM : MK$, or $c / x = x / 4c$, from which $x = 2c$. Therefore, $AB = 2x = 4c$.

## Answer

$4c$.

Answer: 4c


The maximum value of the function $y=3 \sqrt{x-1}+4 \sqrt{5-x}$ is $\square$ $\qquad$ .

Solution:
According to the problem, $y>0$ and $x \in[1,5]$, so by the Cauchy-Schwarz inequality, we have: $\square$
$$y^{2}=(3 \sqrt{x-1}+4 \sqrt{5-x})^{2} \leqslant\left(3^{2}+\right.$$
$\left.4^{2}\right)\left[(\sqrt{x-1})^{2}+(\sqrt{5-x})\right]=100$, equality holds if and only if $\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}$, i.e., $x=\frac{61}{25}$. Therefore, the maximum value of the function $y=3 \sqrt{x-1}+4 \sqrt{5-x}$ is 10.

Answer: 10


Solution:
4. The number 2090 is written as the product of six different integers. Determine the smallest value of the sum of these numbers.

## Solution.

Let's factorize the number 2090 into prime factors:

$$
\begin{aligned}
2090 & =209 \cdot 10 \\
& =11 \cdot 19 \cdot 2 \cdot 5
\end{aligned}
$$

1 POINT

If the number is written as the product of six different integers and it has four prime factors, then two of the factors of this product must be the numbers 1 and -1.

2 POINTS

Since the number is positive and we want the smallest value of the sum, we need to have an even number of negative factors, and the factors with the largest absolute values should be negative.
$2090=(-19) \cdot(-11) \cdot(-5) \cdot 2 \cdot(-1) \cdot 1$
2 POINTS
The smallest value of the sum is $-19+(-11)+(-5)+2+(-1)+1=-33$.
1 POINT

TOTAL 6 POINTS

Answer: -33


Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution:
The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$.  The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$.  Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die.  Therefore the probability that he is using the fair die is $\frac{1}{17}$, and the probability that he is using the biased die is $\frac{16}{17}$.  The probability of rolling a third six is
\[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\]
Therefore, our desired $p+q$ is $65+102= \boxed{167}$

Answer: 167


Let $a, b, c$ be real numbers. Then, for any real number $x$, the necessary and sufficient condition for the inequality $a \sin x+b \cos x+c>0$ to always hold is
A. $a, b$ are both 0, and $c>0$
B. $\sqrt{a^{2}+b^{2}}=c$
C. $\sqrt{a^{2}+b^{2}}<c$

Solution:
Solve:
$a \sin x + b \cos x + c = \sqrt{a^{2} + b^{2}} \sin (x + \theta) + c > 0$ always holds $\Rightarrow -\sqrt{a^{2} + b^{2}} + c > 0$ $\Rightarrow \sqrt{a^{2} + b^{2}} < c$. Therefore, choose $C$.

Answer: C


Convert $\cos \alpha+\cos 3 \alpha+\cos 5 \alpha$ $+\cos 7 \alpha$ into a product.

Solution:
Let $z=\cos \alpha+i \sin \alpha$. Then
$$
\begin{aligned}
\text { Original expression }= & \frac{1}{2}\left(z+z^{-1}+z^{3}+z^{-3}+z^{8}\right. \\
& \left.+z^{-5}+z^{7}+z^{-7}\right) \\
= & \frac{1}{2}\left(z+z^{-1}\right)\left(z^{2}+z^{-2}\right) \\
& \cdot\left(z^{4}+z^{-4}\right) \\
= & \frac{1}{2} \cdot 2 \cos \alpha \cdot 2 \cos 2 \alpha \cdot 2 \cos 4 \alpha \\
= & 4 \cos \alpha \cos 2 \alpha \cos 4 \alpha .
\end{aligned}
$$

Answer: 4 \cos \alpha \cos 2 \alpha \cos 4 \alpha


Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

Solution:
Label the point of [intersection](https://artofproblemsolving.com/wiki/index.php/Intersection) as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the [law of cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_cosines),

\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}
Since we are looking for the earliest possible intersection, $20$ seconds are needed. Thus, $8 \cdot 20 = \boxed{160}$ meters is the solution.
Alternatively, we can drop an altitude from $C$ and arrive at the same answer.

Answer: 160


Let $ABCD$ be a cyclic quadrilateral, and now four relations are given:
(1) $\sin A=\sin C$;
(2) $\sin A+\sin C=0$;
(3) $\cos B+\cos D=0$;
(4) $\cos B=\cos D$.

The number of relations that always hold is
(A) one;
(B) two;
(C) three;
(D) four.

Solution:
1. (B)

Since $A B C D$ is a cyclic quadrilateral, $C=180^{\circ}-A$, and $A$ and $C$ cannot be $0^{\circ}$ or $180^{\circ}$, therefore equation (1) always holds; equation (2) never holds.

Similarly, since $D=180^{\circ}-B$, equation (3) always holds; equation (4) only holds when $B=D=90^{\circ}$. Therefore, among the four given equations, only (1) and (3) always hold.

Answer: B


Given $3 \sin ^{2} \alpha+2 \sin ^{2} \beta=1,3(\sin \alpha+\cos \alpha)^{2}-2(\sin \beta+\cos \beta)^{2}=1$, then $\cos 2(\alpha+\beta)=$ $\qquad$ .

Solution: (2) $-\frac{1}{3}$

Answer: -\frac{1}{3}


Given $x \geqslant 0, y \geqslant 0, z \geqslant 0, x+y+z=1$, find the maximum value of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2}$.

Solution:
9. Suppose $x \geqslant y \geqslant z$, then
$$\begin{aligned}
& x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2} \\
= & x^{2} \cdot\left[(1-x)^{2}-2 y z\right]+y^{2} z^{2}+x^{2} y^{2} z^{2} \\
= & x^{2}(1-x)^{2}-2 x^{2} y z+y^{2} z^{2}+x^{2} y^{2} z^{2} \\
\leqslant & \frac{1}{16}-\left(x^{2} y z-y^{2} z^{2}\right)-\left(x^{2} y z-x^{2} y^{2} z^{2}\right) \\
\leqslant & \frac{1}{16} .
\end{aligned}$$

When $x=y=\frac{1}{2}, z=0$, the equality holds.

Answer: \frac{1}{16}


Determine the smallest values that the expression $V=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ can take, given that the real numbers $a$, $b$, $c$ satisfy the pair of conditions

$$
\begin{aligned}
a+3 b+c & =6 \\
-a+b-c & =2
\end{aligned}
$$

Solution:
SOLUTION. By adding both equations, we find that $b=2$. Substituting $b$ into one of them yields $c=-a$. Therefore, $V=(a-2)^{2}+(2+a)^{2}+(-2 a)^{2}$. After squaring and adding, we find that $V=6 a^{2}+8 \geqq 8$. Equality occurs precisely when $a=0, b=2$ and $c=0$.

The sought minimum value of the expression $V$ is thus equal to 8.

GUIDING AND SUPPLEMENTARY PROBLEMS:

Answer: 8


Which digit of $.12345$, when changed to $9$, gives the largest number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: When dealing with positive decimals, the leftmost digits affect the change in value more.  Thus, to get the largest number, we change the $1$ to a $9 \rightarrow \boxed{\text{A}}$.

Answer: A


Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:

- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.

How many degrees does angle $B A M$ measure?

Solution:
Answer: 44.

Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does not exceed the angle $PAK$, and these angles are equal only if points $M$ and $P$ coincide. Therefore, $M$ is this point of tangency.

![](https://cdn.mathpix.com/cropped/2024_05_06_c47c8a04f76ef67ab4e5g-27.jpg?height=403&width=311&top_left_y=935&top_left_x=575)

The radius $KM$ of the circle is perpendicular to the tangent $AM$. Also, in the right triangle $AMK$, the leg $MK$ is half the hypotenuse $AK$, so $\angle MAK=30^{\circ}$. Additionally, from the condition, we get that $\angle BAC=180^{\circ}-2 \cdot 53^{\circ}=74^{\circ}$. Therefore,

$$
\angle BAM=\angle BAC-\angle MAK=74^{\circ}-30^{\circ}=44^{\circ}
$$

Answer: 44


Bob, a spherical person, is floating around peacefully when Dave the giant orange fish launches him straight up 23 m/s with his tail. If Bob has density 100 $\text{kg/m}^3$, let $f(r)$ denote how far underwater his centre of mass plunges underwater once he lands, assuming his centre of mass was at water level when he's launched up. Find $\lim_{r\to0} \left(f(r)\right) $. Express your answer is meters and round to the nearest integer. Assume the density of water is 1000 $\text{kg/m}^3$.

Solution:
1. **Initial Setup and Assumptions:**
   - Bob is launched upwards with an initial velocity \( u = 23 \, \text{m/s} \).
   - Bob's density \( \sigma = 100 \, \text{kg/m}^3 \).
   - Water's density \( \rho = 1000 \, \text{kg/m}^3 \).
   - We need to find the limit of \( f(r) \) as \( r \to 0 \), where \( f(r) \) is the depth Bob's center of mass plunges underwater.

2. **Energy Conservation:**
   - Using the work-energy theorem, the initial kinetic energy of Bob is given by:
     \[
     \frac{1}{2} m u^2
     \]
   - The work done by the buoyant force \( W_{\text{buy}} \) and the gravitational potential energy change \( mgh \) must be considered.

3. **Buoyant Force and Work Done:**
   - The buoyant force \( F_{\text{buy}} \) is given by Archimedes' principle:
     \[
     F_{\text{buy}} = \rho V g
     \]
     where \( V \) is the volume of Bob.
   - The work done by the buoyant force as Bob plunges to depth \( h \) is:
     \[
     W_{\text{buy}} = F_{\text{buy}} \cdot h = \rho V g h
     \]

4. **Volume and Mass Relationship:**
   - Bob's volume \( V \) can be expressed in terms of his mass \( m \) and density \( \sigma \):
     \[
     V = \frac{m}{\sigma}
     \]

5. **Energy Conservation Equation:**
   - Equating the initial kinetic energy to the work done by the buoyant force and the gravitational potential energy change:
     \[
     \frac{1}{2} m u^2 = \rho V g h - mgh
     \]
   - Substituting \( V = \frac{m}{\sigma} \):
     \[
     \frac{1}{2} m u^2 = \rho \left( \frac{m}{\sigma} \right) g h - mgh
     \]
   - Simplifying:
     \[
     \frac{1}{2} u^2 = \left( \frac{\rho}{\sigma} - 1 \right) gh
     \]

6. **Solving for \( h \):**
   - Rearranging the equation to solve for \( h \):
     \[
     h = \frac{\frac{1}{2} u^2}{\left( \frac{\rho}{\sigma} - 1 \right) g}
     \]
   - Substituting the given values \( u = 23 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( \sigma = 100 \, \text{kg/m}^3 \), and \( g = 9.81 \, \text{m/s}^2 \):
     \[
     h = \frac{\frac{1}{2} \cdot 23^2}{\left( \frac{1000}{100} - 1 \right) \cdot 9.81}
     \]
     \[
     h = \frac{\frac{1}{2} \cdot 529}{(10 - 1) \cdot 9.81}
     \]
     \[
     h = \frac{264.5}{9 \cdot 9.81}
     \]
     \[
     h = \frac{264.5}{88.29}
     \]
     \[
     h \approx 2.99 \, \text{m}
     \]

The final answer is \( \boxed{3} \, \text{m} \).

Answer: 3


In a square $A B C D$, the diagonal $A C$ has a length of $10.0 \, \text{cm}$.

a) Construct such a square! Describe and justify your construction!

b) A rectangle $E F G H$ is said to be inscribed in the square $A B C D$ if, with appropriate labeling, $E$ lies on $A B$, $F$ lies on $B C$, $G$ lies on $C D$, and $H$ lies on $D A$. Additionally, $E F \| A C$.

Determine the perimeter of each such rectangle $E F G H$!

Solution:
a) Since in every square the diagonals are perpendicular to each other, equal in length, and bisect each other, the vertices $B$ and $D$ lie first on the perpendicular bisector of $A C$ and second on the circle with radius $\frac{A C}{2}$ centered at the midpoint of $A C$.

Therefore, a square $A B C D$ meets the conditions of the problem only if it can be obtained through the following construction:

1. We draw a segment $A C$ of length $A C=10.0 \, \text{cm}$ and construct its perpendicular bisector.
2. We draw a circle around the midpoint $M$ of the segment $A C$ with radius $\frac{1}{2} A C$.
3. If the circle intersects the perpendicular bisector at two points, let these points be $B$ and $D$. $A B C D$ is the desired square.

Proof: By construction, $B D \perp A C$. Furthermore, by construction, $A M = C M = B M = D M$. Consequently, by the congruence theorem (side-angle-side), the triangles $A M B, B M C, C M D$, and $D M A$ are congruent to each other. Therefore, $A B = B C = C D = D A$.

Since these triangles are also right-angled and isosceles with the vertex at $M$, their base angles are each $45^{\circ}$. Since each of the angles $\angle A B C, \angle B C D, \angle C D A, \angle D A B$ is equal to the sum of two of these base angles, each of them is $90^{\circ}$. Therefore, $A B C D$ is a square, and it has the prescribed diagonal length.

b) Let a rectangle $E F G H$ be inscribed in the square $A B C D$ as specified in the problem statement. Let $F G$ intersect the diagonal $A C$ at $P$ and $H E$ intersect the diagonal $A C$ at $Q$. Since the diagonal of a square bisects the interior angles as a line of symmetry, $\angle C A B = 45^{\circ}$.

Because $E F \parallel A C$, it follows that $\angle C A B = \angle F E B = 45^{\circ}$ as corresponding angles on intersecting parallel lines, and because $\angle F E H = 90^{\circ}$, it follows that $\angle H E A = 45^{\circ}$. Thus, the triangle $A E Q$ is isosceles due to the equal base angles, and it follows that $A Q = Q E$.

Similarly, $A Q = H Q$, so $E H = 2 A Q$. Correspondingly, $F G = 2 C P$. Since $E H = F G$, it follows that $A Q = C P$. Finally, because $E F \parallel Q P$ and $E Q \parallel F P$, it also follows that $E F = Q P$.

For the perimeter $u$ of the rectangle $E F G H$, we have:

$$
u = 2(E F + E H) = 2(Q P + A Q + C P) = 2 A C
$$

The perimeter of any such rectangle $E F G H$ is thus $20.0 \, \text{cm}$.

Solutions from the 3rd Round 1977 taken from [5]

### 4.20 18th Olympiad 1978

### 4.20.1 1st Round 1978, Class 7

Answer: 20\,


A project needs 100 days to complete. Initially, 10 people worked for 30 days and completed $\frac{1}{5}$ of the entire project. Then, 10 more people were added to complete the project. How many days earlier can the task be completed?

Fill in the blank: $\qquad$ days.

Solution:
【Analysis】First, consider this project as a unit "1". According to the work efficiency $=$ work volume $\div$ work time, use the work volume completed by 10 people in 30 days divided by $10 \times 30$ to find out what fraction of the project each worker completes per day; then find out what fraction of the project 10 more people complete per day, and then according to the work time $=$ work volume $\div$ work efficiency, use the remaining work volume divided by the work volume completed per day by 10 more people to find out how many days the remaining project needs; finally, subtract the actual required time from 100 to find out how many days ahead of schedule the task can be completed.

【Solution】Solution: $100-30-\left(1-\frac{1}{5}\right) \div\left[\frac{1}{5} \div(10 \times 30) \times(10+10)\right]$
$$
\begin{array}{l}
=70-\frac{4}{5} \div \frac{1}{75} \\
=70-60 \\
=10 \text { (days) }
\end{array}
$$

Answer: The task can be completed 10 days ahead of schedule.
Therefore, the answer is: 10.

Answer: 10


In September, a trio of friends met at the restaurant twice. The first time, each of them paid separately, so the waiter added the date to each bill and asked each of them for 168 KÄ. Four days later, they had lunch there again and ordered exactly the same as before. This time, however, one of them paid for everyone together. The waiter thus added the date to the bill only once and asked for 486 KÄ. The friends did not think that although the prices on the menu had not changed, their lunch was cheaper than before, and they uncovered the waiter's fraud that day. What was the date?

Hint. Determine what their total expenditure would have been if they had each paid separately the second time as well.

Solution:
If on the day when friends uncovered the waiter's fraud, each of them paid separately, the waiter would have said to each of them $168+4=172$ (CZK). In total, they would have paid $3 \cdot 172=516$ (CZK). By paying for everyone at once, the price was reduced by an amount corresponding to two dates. According to the problem, this price is 486 CZK. Therefore, the amount corresponding to two dates is $516 - 486=30$ (CZK). The number in the date is $30: 2=15$; the friends uncovered the fraud on the 15th of September.

Note. We can verify our result with a final summary: The friends first visited the restaurant on September 11th. Each of them was supposed to pay 157 CZK according to the menu, but they paid $157+11=$ $=168$ (CZK). The second time they met in the restaurant was on September 15th. One person was supposed to pay $3 \cdot 157=471$ (CZK) for everyone according to the menu, but the waiter charged $471+15=486$ (CZK).

Answer: 15


A row of 11 numbers is written such that the sum of any three consecutive numbers is 18. Additionally, the sum of all the numbers is 64. Find the central number.

Solution:
Answer: 8.

Solution. Number the numbers from left to right from 1 to 11.

Notice that the sum of the five central numbers (from the fourth to the eighth) is 64 (the sum of all numbers) $-2 \cdot 18$ (the sum of the numbers in the first and last triplets) $=28$.

Then the sixth (central) number is 18 (the sum of the fourth, fifth, and sixth numbers) +18 (the sum of the sixth, seventh, and eighth numbers) -28 (the sum of the central five numbers) $=8$.

Answer: 8


Solution:
a) The expression in the square brackets can be written as:

$$
\begin{aligned}
& x^{2}+(a+b) x+a b=(x+a)(x+b) \\
& \quad x(x-a)-b(x-a)=(x-a)(x-b)
\end{aligned}
$$

The triple product of these two expressions is

$$
3\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)=3 x^{4}+3 a^{2} x^{2}-3 b^{2} x^{2}+3 a^{2} b^{2}
$$

and thus the dividend in the curly brackets is

$$
3 a^{2} x^{2}+3 b^{2} x^{2}-3 a^{2} b^{2}+3 x^{4}-3 a^{2} x^{2}-3 b^{2} x^{2}+3 a^{2} b^{2}=3 x^{4}
$$

Therefore, the desired quotient is

$$
3 x^{4}: x^{2}=3 x^{2}
$$

b) Since $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$ and $-(b-a)=a-b$, we have

$$
\begin{gathered}
\frac{1}{a-b}-\frac{3 a b}{a^{3}-b^{3}}-\frac{b-a}{a^{3}+b+b^{2}}=\frac{a^{2}+a b+b^{2}-3 a b+(a-b)^{2}}{(a-b)\left(a^{2}+a b+b^{2}\right)}= \\
=\frac{2 a^{2}-4 a b+2 b^{2}}{(a-b)\left(a^{2}+a b+b^{2}\right)}=\frac{2(a-b)^{2}}{(a-b)\left(a^{3}+a b+b^{2}\right)}=\frac{2(a-b)}{a^{2}+a b+b^{2}} \\
c) \quad \frac{a}{(a-2 b)(a-c)}+\frac{2 b}{(2 b-c)(2 b-a)}+\frac{c}{(c-a)(c-2 b)}= \\
=\frac{a}{(a-2 b)(a-c)}-\frac{2 b}{(a-2 b)(2 b-c)}+\frac{c}{(2 b-c)(a-c)}
\end{gathered}
$$

It can be shown that for any two terms of this sum, they are equal to the third term with the opposite sign, provided that $a \neq c, a \neq 2 b, c \neq 2 b$. For example, the first two terms:

$$
\begin{gathered}
\frac{a}{(a-2 b)(a-c)}-\frac{2 b}{(a-2 b)(2 b-c)}=\frac{a(2 b-c)-2 b(a-c)}{(a-2 b)(2 b-c)(a-c)}= \\
=\frac{2 a b-a c-2 a b+2 b c}{(a-2 b)(2 b-c)(a-c)}=-\frac{c(a-2 b)}{(a-2 b)(2 b-c)(a-c)}=-\frac{c}{(2 b-c)(a-c)}
\end{gathered}
$$

and thus our expression is identically equal to 0.

Denes Redly (Pannonhalma g. II. o. t.)

Answer: 0


Four princesses each thought of a two-digit number, and Ivan thought of a four-digit number. After they wrote their numbers in a row in some order, the result was 132040530321. Find Ivan's number.

Solution:
Solution. Let's go through the options. Option 1320 is not suitable because the remaining part of the long number is divided into fragments of two adjacent digits: 40, 53, 03, 21, and the fragment 03 is impossible, as it is not a two-digit number. Option 3204 is impossible due to the invalid fragment 05 (or the fragment 1 from a single digit). Option 2040 results in the impossible fragment 03. Option 0405 is not an option - it is a three-digit number. Option 4053 results in the impossible fragment 03. Option 0530 is impossible. Option 5303 is the only possible one, as option 3032 leads to a fragment of one digit 1.

Answer. 5303.

## CONDITION

Answer: 5303


The density function of a random variable $X$ is given by

$$
p(x)=\frac{c}{1+x^{2}}
$$

Find the value of the parameter $c$.

Solution:
Solution. The density function must satisfy condition $(2.3 .6)$, i.e., the equality

$$
\int_{-\infty}^{+\infty} \frac{c}{1+x^{2}} d x=c \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=1
$$

must hold, from which

$$
c=1 / \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}
$$

The indefinite integral is a standard one:

$$
\int \frac{d x}{1+x^{2}}=\operatorname{arctg} x
$$

Let's compute the improper integral:

$$
\begin{aligned}
& \int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}=\int_{-\infty}^{0} \frac{d x}{1+x^{2}}+\int_{0}^{+\infty} \frac{d x}{1+x^{2}}=\left.\operatorname{arctg} x\right|_{-\infty} ^{0}+\left.\operatorname{arctg} x\right|_{0} ^{+\infty}= \\
& =\operatorname{arctg} 0-\operatorname{arctg}(-\infty)+\operatorname{arctg}(+\infty)-\operatorname{arctg} 0=0-\left(-\frac{\pi}{2}\right)+\frac{\pi}{2}-0=\pi
\end{aligned}
$$

Therefore, $c=1 / \pi$; the density function is

$$
p(x)=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}}, \quad p(x)=\frac{1}{\pi\left(1+x^{2}\right)}
$$

Answer: \frac{1}{\pi}


Find the smallest positive integer that is twice a perfect square and three times a perfect cube.

Solution: Answer: 648. Let $n$ be such a number. If $n$ is divisible by 2 and 3 exactly $e_{2}$ and $e_{3}$ times, then $e_{2}$ is odd and a multiple of three, and $e_{3}$ is even and one more than a multiple of three. The smallest possible exponents are $n_{2}=3$ and $n_{3}=4$. The answer is then $2^{3} \cdot 3^{4}=648$.

Answer: 648


Find the sum of all integers $x$ satisfying $1 + 8x \le 358 - 2x \le 6x + 94$.

Solution:
1. We start with the compound inequality:
   \[
   1 + 8x \le 358 - 2x \le 6x + 94
   \]
   This can be split into two separate inequalities:
   \[
   1 + 8x \le 358 - 2x \quad \text{and} \quad 358 - 2x \le 6x + 94
   \]

2. Solve the first inequality:
   \[
   1 + 8x \le 358 - 2x
   \]
   Add \(2x\) to both sides:
   \[
   1 + 8x + 2x \le 358
   \]
   Simplify:
   \[
   10x + 1 \le 358
   \]
   Subtract 1 from both sides:
   \[
   10x \le 357
   \]
   Divide by 10:
   \[
   x \le 35.7
   \]

3. Solve the second inequality:
   \[
   358 - 2x \le 6x + 94
   \]
   Add \(2x\) to both sides:
   \[
   358 \le 6x + 2x + 94
   \]
   Simplify:
   \[
   358 \le 8x + 94
   \]
   Subtract 94 from both sides:
   \[
   264 \le 8x
   \]
   Divide by 8:
   \[
   33 \le x
   \]

4. Combine the results from the two inequalities:
   \[
   33 \le x \le 35.7
   \]
   The integer solutions within this range are \(33, 34,\) and \(35\).

5. Calculate the sum of these integers:
   \[
   33 + 34 + 35 = 102
   \]

The final answer is \(\boxed{102}\)

Answer: 102


A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{8}{3}$ times. Determine how many liters of liquid the student poured out the first time, if the tank's volume is 1000 liters.

Solution:
# Solution.

1) Let the syrup content in the original solution be $p \%$ and let $\mathcal{X}$ liters of the solution be poured out the first time.
2) Then after pouring out the liquid, there are $(1000-x)$ liters of solution left, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100}$ water.
3) After adding $X$ liters of water, there are 1000 liters of solution in the tank, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100}+x$ liters of water.
4) At the end of all the pouring, there are 1000 liters of solution in the tank with a syrup content of $\frac{3 p}{8} \%$, which is $1000 \cdot \frac{\frac{3 p}{8}}{100}=\frac{30 p}{8}$ liters of syrup and $1000-\frac{30 p}{8}$ liters of water.
5) Then before the last addition of $2 x$ liters of water, there were $(1000-2 x)$ liters of solution in the tank, and in it $\frac{30 p}{8}$ liters of syrup and $1000-\frac{30 p}{8}-2 x$ water. This is the same liquid as in point 3) of the solution, so the ratio of syrup to liquid in it is the same.

We can set up the equation:

$$
\frac{(1000-x) \cdot \frac{p}{100}}{1000}=\frac{\frac{30 p}{8}}{1000-2 x} \Leftrightarrow 2 x^{2}-3000 x+\frac{5}{8} \cdot 1000^{2}=0 \Leftrightarrow
$$

$\Leftrightarrow x \in\{1250 ; 250\}, 1250$ liters does not satisfy the condition of the problem.

Answer: 250.

## Solution variant № 2

1. If a two-digit natural number is decreased by 27, the result is a two-digit number with the same digits but in reverse order. In the answer, indicate the median of the numerical sequence formed from all such numbers.

## Solution.

Let $\overline{x y}=10 x+y$ be the original two-digit number, and $\overline{y x}=10 y+x$ be the number with the digits in reverse order. We get the equation $10 x+y=10 y+x+27$. From the equation, it is clear that the two-digit number is greater than 27. Let's start the investigation with the tens digit equal to 3.

| $\boldsymbol{x}$ | equation | $\boldsymbol{y}$ | number |
| :--- | :--- | :---: | :--- |
| 3 | $30+y=10 y+3+27$ | $y=0$ | 30 does not fit the condition |
| 4 | $40+y=10 y+4+27$ | $y=1$ | 41 |
| 5 | $50+y=10 y+5+27$ | $y=2$ | 52 |
| 6 | $60+y=10 y+6+27$ | $y=3$ | 63 |
| 7 | $70+y=10 y+7+27$ | $y=4$ | 74 |
| 8 | $80+y=10 y+8+27$ | $y=5$ | 85 |
| 9 | $90+y=10 y+9+27$ | $y=6$ | 96 |

These can be the numbers $41,52,63,74,85,96$. The median of the sequence is 68.5.

Answer: 68.5.

Answer: 250


Given that a square has three vertices on the parabola $y=x^{2}$, find the minimum value of the area of such a square.

Solution:
Question 173, Solution: Let the square be $ABCD$, and $A\left(a, a^{2}\right), B\left(b, b^{2}\right), D\left(d, d^{2}\right)$ lie on the parabola $y=x^{2}$. Suppose the slope of line $AB$ is $k (k>0)$, then $b^{2}-a^{2}=k(b-a)$, hence $b=k-a$. Similarly, $d=-\frac{1}{k}-a$. Also,
$$
\begin{array}{l}
|AB|=\sqrt{1+k^{2}}(b-a)=\sqrt{1+k^{2}}(k-2a) \\
|AD|=\sqrt{1+\frac{1}{k^{2}}}(a-d)=\sqrt{1+\frac{1}{k^{2}}}\left(2a+\frac{1}{k}\right)=\sqrt{1+k^{2}} \cdot \frac{2ak+1}{k^{2}}
\end{array}
$$

Noting that $|AB|=|AD|$, we have
$$
\begin{array}{l}
\sqrt{1+k^{2}}(k-2a)=\sqrt{1+k^{2}} \cdot \frac{2ak+1}{k^{2}} \\
\Rightarrow (k-2a)k^{2}=2ak+1 \\
\Rightarrow a=\frac{k^{3}-1}{2(k^{2}+k)}
\end{array}
$$

Thus, $|AB|=\sqrt{1+k^{2}}\left(k-\frac{k^{3}-1}{k^{2}+k}\right)=\frac{(k^{2}+1)\sqrt{1+k^{2}}}{k(k+1)} \geq \frac{2k \frac{k+1}{\sqrt{2}}}{k(k+1)}=\sqrt{2}$, with equality holding if and only if $k=1$. Therefore, the area of the square is no less than 2, and when $A(0,0), B(1,1), D(-1,1)$, the area of square $ABCD$ is exactly 2.

Answer: 2


Let $\left\{a_{n}\right\}$ be a positive geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and
$$
2^{10} S_{30}+S_{10}=\left(2^{10}+1\right) S_{20} \text {. }
$$

Then the common ratio of the sequence $\left\{a_{n}\right\}$ is ( ).
(A) $\frac{1}{8}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 1

Solution:
5. C.

From the given information,
$$
\begin{array}{l}
S_{20}-S_{10}=2^{10}\left(S_{30}-S_{20}\right) \\
\Rightarrow \frac{S_{30}-S_{20}}{S_{20}-S_{10}}=2^{-10}=q^{10} .
\end{array}
$$

Since $a_{n}>0$, we have $q=\frac{1}{2}$.

Answer: C


The positive integers are colored with black and white such that:
- There exists a bijection from the black numbers to the white numbers,
- The sum of three black numbers is a black number, and
- The sum of three white numbers is a white number.

Find the number of possible colorings that satisfies the above conditions.

Solution:
1. **Initial Assumptions and Definitions:**
   - We are given that there is a bijection between black and white numbers.
   - The sum of three black numbers is black.
   - The sum of three white numbers is white.

2. **Case Analysis:**
   - We start by assuming \(1\) is black. By induction, we will show that all odd numbers must be black.
   - Suppose \(1\) is black. Then, \(1 + 1 + k = k + 2\) must be black for all odd \(k\). This implies that all odd numbers are black.

3. **Considering Even Numbers:**
   - Suppose there exists a black even number. By the well-ordering principle, let \(m\) be the smallest black even number.
   - By induction, \(1 + 1 + k = k + 2\) must be black for all even \(k \ge m\). This implies that all even numbers greater than or equal to \(m\) are black.
   - Since \(m\) is the smallest black even number, all even numbers less than \(m\) must be white.

4. **Contradiction with Bijection:**
   - There are infinitely many odd numbers, hence infinitely many black numbers.
   - If there are only finitely many white numbers (as there are only finitely many even numbers less than \(m\)), a bijection between black and white numbers cannot exist.
   - Therefore, the assumption that there exists a black even number is false. Hence, all even numbers must be white.

5. **Alternative Coloring:**
   - If we assume \(1\) is white instead, by similar reasoning, all odd numbers must be white, and all even numbers must be black.

6. **Conclusion:**
   - There are exactly two possible colorings that satisfy the given conditions:
     1. All odd numbers are black, and all even numbers are white.
     2. All odd numbers are white, and all even numbers are black.

The final answer is \(\boxed{2}\)

Answer: 2


In tetrahedron $ABCD$, the dihedral angle between faces $ABC$ and $BCD$ is $30^{\circ}$, the area of $\triangle ABC$ is $120$, the area of $\triangle BCD$ is $80$, and $BC=10$. Find the volume of the tetrahedron.

Solution: 8. Draw $D P \perp$ plane $A B C$ at $P$, and draw $D H \perp B C$ at $H$. Connect $P H$, then $P H \perp B C, \angle D H P=30^{\circ}$. Given $S_{\triangle B C D}=$ $80, B C=10$, we get $D P=\frac{1}{2} D H=8$, thus $V_{A B C D}=\frac{1}{3} D P \cdot S_{\triangle A B C}=320$.

Answer: 320


On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation.
If $b \neq 0$ and $c \neq 0$ these two graphs intersect:
$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis}$

Solution:
Replacing $x$ with $-x,$ we have the graph of $y = a(-x)^2 + b(-x) + c = ax^2 - bx+c.$
We can plug in simple values of $a, b,$ and $c$ for convenient drawing. For example, we can graph $x^2+2x+1$ and $x^2-2x+1.$ We see that the parabolas intersect at $(0, 1),$ which is on the y-axis.
The answer is $\boxed{\textbf{(E)}}.$

Answer: \textbf{(E)}


Given the cryptarithm: ÃÂÃÂÃÂÃÂ + ÃÂÃÂÃÂÃÂ = ÃÂÃÂÃÂÃÂÃÂ¬. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter ÃÂ.

Solution:
Answer: 8

Solution: The rebus can be rewritten as ОСЕНЬ $=($ ЖА + ЛО $) \cdot 101$. First, this means that the last digit of ЖА + ЛО is Ь. Second, if ЖА + ЛО $<100$, the result will be a four-digit number. Let $Ж А+Л О=1 Х Ь$, where $X$ is some digit.

Then ОСЕНЬ $=1 Х Ь 00+1 Х Ь$. If $\mathrm{b}<9$, then the second and fourth digits of this number should match, but they do not. Therefore, $\mathrm{b}=9$. Consequently, $\mathrm{A}+\mathrm{O}=9$. But О is 1, so А is 8.

There are many examples, for instance, $7861+6178=14039$.

7. (4 points) Alice and Bob each have three equal segments. First, Alice breaks one of the segments into two unequal parts. Then, Bob breaks another of the original segments into any two parts. As a result, there are five segments, from which ten sets of three segments can be chosen. Alice wins if at least 4 of these ten sets form a triangle. Otherwise, Bob wins. Who will win if both players play optimally?

Answer: Bob wins.

Solution: Let the original segments have a length of 1, and Alice breaks one of the segments into parts $x$ and $1-x$, with $x<1-x$ for definiteness. Then Bob needs to break another segment into parts $1-y$ and $y$ such that $y$ is very small, specifically, the inequalities $y<x$, $y<(1-x)-x$, and $y<(1-y)-(1-x)$, or $2 y<x$, are satisfied. Clearly, he can choose such a $y$.

Using the triangle inequality: to form a triangle, it is necessary and sufficient that the largest side of the triangle is less than the sum of the other two. With such a choice of $y$, it is not greater than the difference between any two other segments, and therefore cannot form a triangle with any of them. There are exactly 6 sets of three segments that include the segment $y$. Additionally, the remaining segment of length 1 and the segments Alice obtained also do not form a triangle. This is the seventh set of segments that does not form a triangle. Thus, Alice is left with only 3 sets of segments that can form a triangle, and she loses.

Answer: 8


Solution:
1. Let \( S = \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots \). Given the sequence \( (a_n) \) defined by \( a_1 = a_2 = 1 \) and \( a_n = a_{n-1} + a_{n-2} \) for \( n > 2 \), we recognize that \( (a_n) \) is the Fibonacci sequence.

2. We can express \( S \) as:
   \[
   S = \frac{1}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \ldots
   \]

3. To simplify, consider \( 2S \):
   \[
   2S = 1 + \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots
   \]

4. Subtract \( S \) from \( 2S \):
   \[
   2S - S = S = 1 + \left( \frac{1}{2} - \frac{1}{2} \right) + \left( \frac{2}{4} - \frac{1}{4} \right) + \left( \frac{3}{8} - \frac{2}{8} \right) + \ldots
   \]
   Simplifying the terms inside the parentheses:
   \[
   S = 1 + 0 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots
   \]

5. Notice that the remaining series is a geometric series with the first term \( \frac{1}{4} \) and common ratio \( \frac{1}{2} \):
   \[
   \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots = \sum_{k=2}^{\infty} \frac{1}{2^k}
   \]

6. The sum of an infinite geometric series \( \sum_{k=0}^{\infty} ar^k \) is given by \( \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = \frac{1}{4} \) and \( r = \frac{1}{2} \):
   \[
   \sum_{k=2}^{\infty} \frac{1}{2^k} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \cdot 2 = \frac{1}{2}
   \]

7. Therefore, we have:
   \[
   S = 1 + \frac{1}{2}
   \]

8. Solving for \( S \):
   \[
   S = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}
   \]

9. However, this contradicts our earlier steps. Let's re-evaluate the subtraction step:
   \[
   S = 1 + \frac{S}{2}
   \]
   Solving for \( S \):
   \[
   S - \frac{S}{2} = 1 \implies \frac{S}{2} = 1 \implies S = 2
   \]

The final answer is \(\boxed{2}\)

Answer: 2


A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=5 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its image.

Solution:
Answer: -10 Dptr $u \approx 7.1$ cm

Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain that the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{5}{10}=5 \mathrm{~cm} . \quad$ The power of the lens: $D=\frac{1}{d}-\frac{1}{f}=\frac{1}{0.1}-\frac{1}{0.05}=-10$ Dptr. The distance between the source and its image: $s=\sqrt{(x-y)^{2}+(d-f)^{2}}=\sqrt{50} \approx 7.1 \mathrm{~cm}$.

Answer: -10


In the rectangular coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the inequality system $\left\{\begin{array}{l}y \leqslant 3 x, \\ y \geqslant \frac{1}{3} x, \\ x+y \leqslant 100\end{array}\right.$ is .

Solution:
Solution: Fill in 2551. Reason: The geometric meaning of the system of inequalities $\left\{\begin{array}{l}y \leqslant 3 x, \\ y \geqslant \frac{1}{3} x, \quad \text { is: a triangular region } Q \text { formed by the three lines } \\ x+y \leqslant 100\end{array}\right.$ $y=3 x, y=\frac{1}{3} x$ and $x+y=100$. Clearly, region $Q$ is entirely within the first quadrant. To find the number of integer points within region $Q$, we can first consider the number of integer points within the region $\mathbf{R}$ formed by the $x$-axis, $y$-axis, and the line $x+y=100$, and then, based on symmetry, subtract twice the number of integer points within the region $G$ formed by the lines $y=\frac{1}{3} x, x+y=100$, and the $x$-axis, to find the required number of integer points.

Thus, the number of integer points (including the boundaries) within the region $\mathbf{R}$ formed by the $x$-axis, $y$-axis, and the line $x+y=100$ is $1+2+3+\cdots+100+101=\frac{(1+101) \times 101}{2}=5151$.

The number of integer points (excluding the integer points on the boundary $y=\frac{1}{3} x$) within the region $G$ formed by the lines $y=\frac{1}{3} x, x+y=100$, and the $x$-axis is $100+96+92+\cdots+8+4=1300$.

By symmetry, the region formed by the lines $y=3 x, x+y=100$, and the $y$-axis (excluding the integer points on the boundary $y=3 x$) also contains 1300 integer points.
Therefore, the number of integer points that satisfy the conditions is $5151-2 \cdot 1300=2551$.

Answer: 2551


Let $a>0>b>c, a+b+c=1, M=\frac{b+c}{a}, N=\frac{a+c}{b}, P=\frac{a+b}{c}$. Then the size relationship between $M, N, P$ is ( ).
(A) $M>N>P$
(B) $N>P>M$
(C) $P>M>N$
(D) $M>P>N$

Solution:
2. (D).

Since $M=\frac{1}{a}-1, N=\frac{1}{b}-1$,
$$
P=\frac{1}{c}-1, N-P=\frac{c-b}{b c}<0, M>P$. Therefore, $M>P>N$.

Answer: D


The vertices of a rhombus are located on the sides of a parallelogram, and the sides of the rhombus are parallel to the diagonals of the parallelogram. Find the ratio of the areas of the rhombus and the parallelogram, if the ratio of the diagonals of the parallelogram is $k$.

Solution:
Prove that the center of a rhombus coincides with the center of a parallelogram and consider similar triangles.

## Solution

Let the vertices $M, N, K$, and $L$ of the rhombus $MNKL$ be located on the sides $AB, BC, CD$, and $AD$ of the parallelogram $ABCD$, respectively, and the sides $MN$ and $KN$ of the rhombus are parallel to the diagonals $AC$ and $BD$ of the parallelogram, with $\frac{AC}{BD}=k$. If $\alpha$ is the angle between the diagonals of the parallelogram, then

$$
S_{\mathrm{ABCD}}=\frac{1}{2} AC \cdot BD \sin \alpha, \quad S_{\mathrm{KLMN}}=MN \cdot KN \sin \alpha=MN^2 \sin \alpha
$$

Therefore,

$$
\frac{S_{KLMN}}{S_{ABCD}}=\frac{2 MN^2}{AC \cdot BD}
$$

Note that the center of the rhombus coincides with the center $O$ of the parallelogram. Since $ON$ is the angle bisector of triangle $BOC$, we have

$$
\frac{BN}{CN}=\frac{OB}{OC}=\frac{BD}{AC}
$$

Thus,

$$
\frac{BN}{BC}=\frac{BD}{BD+AC}=\frac{1}{1+k}
$$

From the similarity of triangles $BMN$ and $BAC$, we find that

$$
MN=AC \cdot \frac{BN}{BC}=\frac{AC}{1+k}
$$

Therefore,

$$
\frac{S_{KLMN}}{S_{ABCD}}=\frac{2 MN^2}{AC \cdot BD}=\frac{\frac{2 AC^2}{(1+k)^2}}{AC \cdot BD}=2 \cdot \frac{AC}{BD} \cdot \frac{1}{(1+k)^2}=\frac{2 k}{(1+k)^2}
$$

## Answer

$\frac{2 k}{(1+k)^2}$

Answer: \frac{2k}{(1+k)^2}


Let $n$ be a positive integer, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$ .

Solution: Since $a, b>0$, we have $a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$, thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=$ $\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$, equality holds when $a=b=1$. Therefore, the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is 1.

Answer: 1


A set of five-volume encyclopedias is arranged in ascending order on a bookshelf, i.e., from left to right, from Volume 1 to Volume 5. Now, we want to rearrange them in descending order, i.e., from left to right, from Volume 5 to Volume 1, but each time only the positions of two adjacent volumes can be swapped. What is the minimum number of such swaps needed to achieve the goal?

Solution:
[Solution] By sequentially swapping the 1st volume with the 2nd, 3rd, 4th, and 5th volumes, the 1st volume can be moved to the far right position. Then, by sequentially swapping the 2nd volume with the 3rd, 4th, and 5th volumes, the arrangement of the 5 volumes becomes $(3,4,5,2,1)$. Next, perform the following swaps: $\{3,4\}$, $\{3,5\}$, $\{4,5\}$, resulting in the arrangement $(5,4,3,2,1)$, which involves a total of 10 swaps.

On the other hand, to change the arrangement of the 5 volumes from $(1,2,3,4,5)$ to $(5,4,3,2,1)$, the left-right order of any two volumes $A$ and $B$ must be reversed, which can only be achieved by swapping them when they are adjacent. Therefore, any two volumes must be swapped at least once, totaling at least 10 swaps. In conclusion, it is necessary to perform at least 10 swaps to achieve the required arrangement.

Answer: 10


Points $M, N, P$ are the feet of the altitudes dropped from the vertices of triangle $A B C$ with angles $37.5^{0}, 67.5^{0}, 75^{0}$ to its sides. Find the ratio of the areas of triangles $M N P$ and $A B C$.

Solution:
Answer: $(2 \sqrt{3}+3 \sqrt{2}-\sqrt{6}-4): 8$.

## Final round of the "Rosatom" Olympiad, 10th grade, CIS, February 2020

#

Answer: (2\sqrt{3}+3\sqrt{2}-\sqrt{6}-4):8


Angle $A$ at the vertex of isosceles triangle $A B C$ is $100^{\circ}$. On ray $A B$, segment $A M$ is laid off, equal to the base $B C$. Find angle $B C M$.

Solution:
On the ray $B M$, we lay off the segment $B N$ equal to $A M$. Let $A C = A B = a, B N = A M = B C = b$. The angles at the base $C N$ of the isosceles triangle $C B N$ are each $20^{\circ}$. Then

$$
B M = A M - A B = b - a, \quad M N = B N - B M = b - (b - a) = a, \quad C N = 2 B C \cos 20^{\circ} = 2 b \cos 20^{\circ},
$$

$$
\frac{B M}{M N} = \frac{b - a}{a} = \frac{b}{a} - 1 = \frac{\sin 100^{\circ}}{\sin 40^{\circ}} - 1, \quad \frac{C B}{C N} = \frac{b}{2 b \cos 20^{\circ}} = \frac{1}{2 \cos 20^{\circ}}.
$$

We will prove that $\frac{B M}{M N} = \frac{C B}{C N}$. From this it will follow that $C M$ is the angle bisector of triangle $B C N$, and then

$$
\angle B C M = \frac{1}{2} \angle B C N = \frac{1}{2} \cdot 20^{\circ} = 10^{\circ}
$$

Indeed,

$$
\begin{gathered}
\frac{B M}{M N} = \frac{C B}{C N} \Leftrightarrow \frac{\sin 100^{\circ}}{\sin 40^{\circ}} - 1 = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \\
\Leftrightarrow \frac{\sin 100^{\circ} - \sin 40^{\circ}}{\sin 40^{\circ}} = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \frac{2 \sin 30^{\circ} \cos 70^{\circ}}{2 \sin 20^{\circ} \cos 20^{\circ}} = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \\
\Leftrightarrow \frac{\cos 70^{\circ}}{\sin 20^{\circ}} = 1 \Leftrightarrow \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1
\end{gathered}
$$

Which is what we needed to prove.

## Answer

$10^{\circ}$.

It is known that for some internal point $K$ of the median $B M$ of triangle $A B C$, the angles $B A K$ and $B C K$ are equal. Prove that triangle $A B C$ is isosceles.

## Solution

It is sufficient to prove that $B K \perp A C$. Suppose this is not the case. Let $\angle B M C < 90^{\circ}$, and let $T$ be the point symmetric to vertex $C$ with respect to the line $B M$. Then triangle $A M T$ is isosceles, and $M B$ is the bisector of its external angle at the vertex, so $A T \parallel B K$ and $\angle B T K = \angle B C K = \angle B A K$.

From points $A$ and $T$, lying on the same side of the line $B K$, the base $B K$ of the trapezoid $A T B K$ is seen at the same angle, meaning that this trapezoid is inscribed, and therefore, isosceles. The perpendicular bisector of its base $A T$ passes through the midpoint of the base $B K$, which is impossible, since the perpendicular bisector of the base $A T$ of the isosceles triangle $A M T$ passes through the point $M$, distinct from $K$.

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Answer: 10


Solve the following equation:

$$
\left(\frac{3}{4}\right)^{\lg x}+\left(\frac{4}{3}\right)^{\lg x}=\frac{25}{12}
$$

Solution:
Solution: Let $\left(\frac{3}{4}\right)^{\lg x}=y$, then our equation is

$$
y+\frac{1}{y}=\frac{25}{12}
$$

which means

$$
12 y^{2}-25 y+12=0
$$

from which

$$
y_{1}=\frac{4}{3} \quad \text { and } \quad y_{2}=\frac{3}{4} \text {. }
$$

Therefore,

$$
\left(\frac{3}{4}\right)^{\lg x_{1}}=\frac{4}{3}, \quad \lg x_{1}=-1, \quad x_{1}=10^{-1}=\frac{1}{10}
$$

or

$$
\begin{aligned}
\left(\frac{3}{4}\right)^{\lg x_{2}}= & \frac{3}{4}, \quad \lg x_{2}=1 \quad x_{2}=10 . \\
& \text { Endre Hammer (Hajdúnánás, Kőrösi Csoma Sándor g. IV. o. t.) }
\end{aligned}
$$

Answer: x_1=\frac{1}{10},\,x_2=10


On the side $AB$ of an acute-angled triangle $ABC$, a point $M$ is marked. A point $D$ is chosen inside the triangle. Circles $\omega_{A}$ and $\omega_{B}$ are circumscribed around triangles $AMD$ and $BMD$ respectively. The side $AC$ intersects the circle $\omega_{A}$ again at point $P$, and the side $BC$ intersects the circle $\omega_{B}$ again at point $Q$. The ray $PD$ intersects the circle $\omega_{B}$ again at point $R$, and the ray $QD$ intersects the circle $\omega_{A}$ again at point $S$. Find the ratio of the areas of triangles $ACR$ and $BCS$.

Solution:
# Answer: 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-16.jpg?height=508&width=736&top_left_y=1025&top_left_x=680)

Solution. Notice that

$$
\angle P A M=180^{\circ}-\angle P D M=\angle R D M=\angle R B M
$$

from which it follows that $B R \| A C$. Therefore, the distances from points $B$ and $R$ to the line $A C$ are the same. Thus, triangles $A B C$ and $A C R$ have the same base $A C$ and the same heights dropped to $A C$. Therefore, the areas of these triangles are equal. Additionally,

$$
\angle Q B M=180^{\circ}-\angle Q D M=\angle M D S=180^{\circ}-\angle M A S
$$

from which it follows that $A S \| B C$. Therefore, the distances from points $A$ and $S$ to the line $B C$ are the same. Thus, triangles $A B C$ and $B C S$ have the same base $B C$ and the same heights dropped to $B C$. Therefore, the areas of these triangles are also equal. As a result, we get $S_{A C R}=S_{A B C}=S_{B C S}$.

Answer: 1