Solve the equation $5x + 2|x| = 3x$

$(|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0)$.

Solution:
# Solution

Transform the equation to the form

$$
2 x+2|x|=0
$$

1) Solve the equation for $x \geq 0$.

For $x \geq 0$, we get

$$
\begin{gathered}
2 x+2 x=0 \\
4 x=0 \\
x=0
\end{gathered}
$$

The obtained value of $x$ satisfies the condition $x \geq 0$. Therefore, $x=0$ is not a solution to the equation.

2) Solve the equation for $x<0$.

For $x<0$, we get

$$
2 x-2 x=0
$$

The obtained equality is true for any value of $x$. Therefore, the solution will be any number from the considered interval, i.e., $x<0$.

Considering both cases, we get the solution to the equation: $x \leq 0$.

## Criteria

Only the first case is considered and correctly solved - 2 points.

Only the second case is considered and correctly solved - 3 points.

Both cases are considered, but the obtained equation is not solved in each of them - 1 point.

The first case is correctly solved, and in the second, the equation is obtained but no conclusion is made about its solution or the conclusion is incorrect - 3 points.

The second case is correctly solved, and in the first, the equation is obtained and solved, but the condition $x \geq 0$ is not checked, and the correct answer is obtained - 5 points.

A typo or arithmetic error that does not affect the solution process and the answer - 6 points.

If the solution is correct, but the answer is given in the form $x<0, x=0$, the task is fully credited - 7 points.

Answer: x\leq0


The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$.

Solution:
Solution. The number $N$ can be represented as

$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2 \cdot 2 \cdot \ldots \cdot 2 \cdot(1 \cdot 2 \cdot \ldots \cdot 2017 \cdot 2018)}{2018!}=(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2^{2018}
\end{aligned}
$$

We obtained the product of odd numbers and a power of two.

Answer: 2018.

Answer: 2018


Solution:
4. 10.

Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$

Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10-a}>0 \text{. }
$$

Contradiction.
Therefore, $10-a-b=0$, which means $a+b=10$.
Thought 2: The geometric meaning of this problem is: $a$ is the x-coordinate of the intersection point $A$ of $y=\lg x$ and $y=10-x$, and $b$ is the x-coordinate of the intersection point $B$ of $y=10^{x}$ and $y=10-x$. By the symmetry of the graphs of two inverse functions, $A$ and $B$ are symmetric about the line $y=x$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=x \\
y=10-x
\end{array}\right.
$$

we get $C(5,5)$. Since $C$ is the midpoint of $A$ and $B$, we have $\frac{a+b}{2}=5$, which means $a+b=10$.

Answer: 10


Given trapezoid $ABCD$. On its lateral side $CD$, a point $M$ is chosen such that $CM / MD=4 / 3$. It turns out that segment $BM$ divides diagonal $AC$ into two segments, the ratio of whose lengths is also $4 / 3$. What values can the ratio $AD / BC$ take? If necessary, round the answer to 0.01 or write the answer as a common fraction.

Solution:
Answer. $7 / 12 \approx 0.58$.

Solution. Let segment $B M$ intersect segment $A C$ at point $N$. If $C N / N A = 4 / 3$, then $C N / N A = 4 / 3 = C M / M D$, from which, by the converse of Thales' theorem (or from the similarity of triangles $C N D$ and $C A D$), lines $M N$ and $A D$ are parallel; but this is impossible, because then lines $B C$ and $M N$ would also be parallel, but they intersect at point $B$. Therefore, $N A / N C = 4 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-23.jpg?height=546&width=1022&top_left_y=1703&top_left_x=517)

Extend segment $B M$ to intersect line $A D$ at point $K$. Triangles $B C N$ and $K A N$ are similar by two angles, from which $K A / B C = N A / N C = 4 / 3$. Triangles $B C M$ and $K D M$ are also similar by two angles, from which $K D / B C = M D / M C = 3 / 4$. It remains to note that

$$
\frac{A D}{B C}=\frac{K A - K D}{B C}=\frac{K A}{B C}-\frac{K D}{B C}=\frac{4}{3}-\frac{3}{4}=\frac{7}{12} \approx 0.58 .
$$

Answer: \frac{7}{12}\approx0.58


If $p$ and $q$ are positive integers, $\max (p, q)$ is the maximum of $p$ and $q$ and $\min (p, q)$ is the minimum of $p$ and $q$. For example, $\max (30,40)=40$ and $\min (30,40)=30$. Also, $\max (30,30)=30$ and $\min (30,30)=30$.

Determine the number of ordered pairs $(x, y)$ that satisfy the equation

$$
\max (60, \min (x, y))=\min (\max (60, x), y)
$$

where $x$ and $y$ are positive integers with $x \leq 100$ and $y \leq 100$.

Solution:
Solution 1

We consider two cases: $y<60$ and $y \geq 60$.

We use the facts that $\min (a, b) \leq a, \min (a, b) \leq b, \max (a, b) \geq a$, and $\max (a, b) \geq b$. In other words, the minimum of two numbers is less than or equal to both numbers and the maximum of two number is greater than or equal to both numbers.

We also use the fact that if $a \leq b$, then $\min (a, b)=a$ and $\max (a, b)=b$. Note that the case of $a=b$ does not affect these equations.

Case 1: $y<60$

If $y<60$, then $\min (x, y) \leq y$ and $y<60$, so $\min (x, y)<60$.

Since $\min (x, y)<60$, then LS $=\max (60, \min (x, y))=60$.

Also, $\max (60, x) \geq 60$, so since $y<60$, then $\mathrm{RS}=\min (\max (60, x), y)=y$.

Therefore, in the case $y<60$, the equation is satisfied whenever $60=y$, which is never true.

Therefore, there are no ordered pairs $(x, y)$ with $y<60$ that satisfy the equation.

Case 2: $y \geq 60$

If $x<y$, then $\min (x, y)=x$ and so LS $=\max (60, x)$.

If $x<y$, then both 60 and $x$ are no larger than $y$, so the largest of 60 and $x$ is no larger than $y$. In other words, $\max (60, x) \leq y$ and so $\mathrm{RS}=\min (\max (60, x), y)=\max (60, x)$.

If $x<y$, the equation becomes $\max (60, x)=\max (60, x)$ and so is true for every ordered pair $(x, y)$.

If $x \geq y$, then $\min (x, y)=y$ and so $\mathrm{LS}=\max (60, y)=y$ since $y \geq 60$.

If $x \geq y$, then $x \geq 60$ so $\max (60, x)=x$ and so $\mathrm{RS}=\min (x, y)=y$.

If $x \geq y$, the equation becomes $y=y$ and so is true for every pair $(x, y)$.

Therefore, the given equation is satisfied by all pairs $(x, y)$ with $y \geq 60$.

Since $1 \leq x \leq 100$, there are 100 possible values for $x$.

Since $60 \leq y \leq 100$, there are 41 possible values for $y$.

Therefore, there are $100 \times 41=4100$ ordered pairs that satisfy the equation.

## Solution 2

In this solution, we use the facts that if $a \leq b$, then $\min (a, b)=a$ and $\max (a, b)=b$.

We also label as $(*)$ the original equation $\max (60, \min (x, y))=\min (\max (60, x), y)$.

We consider the possible arrangements of the integers $60, x$ and $y$.

There are six possible arrangements. We examine the various pieces of the left and right sides of the given equation for each of these orders, moving from the innermost function out on each side:

$$
\begin{array}{|c||c|c||c|c|}
\text { Case } & \min (x, y) & \mathrm{LS}=\max (60, \min (x, y)) & \max (60, x) & \mathrm{RS}=\min (\max (60, x), y) \\
\hline 60 \leq x \leq y & =x & =\max (60, x)=x & =x & =\min (x, y)=x \\
60 \leq y \leq x & =y & =\max (60, y)=y & =x & =\min (x, y)=y \\
x \leq 60 \leq y & =x & =\max (60, x)=60 & =60 & =\min (60, y)=60 \\
x \leq y \leq 60 & =x & =\max (60, x)=60 & =60 & =\min (60, y)=y \\
y \leq 60 \leq x & =y & =\max (60, y)=60 & =x & =\min (x, y)=y \\
y \leq x \leq 60 & =y & =\max (60, y)=60 & =60 & =\min (60, y)=y
\end{array}
$$

If $60 \leq x \leq y$, then $(*)$ is equivalent to $x=x$, so all $(x, y)$ with $60 \leq x \leq y$ satisfy $(*)$. If $60 \leq y \leq x$, then $(*)$ is equivalent to $y=y$, so all $(x, y)$ with $60 \leq y \leq x$ satisfy $(*)$. If $x \leq 60 \leq y$, then $(*)$ is equivalent to $60=60$, so all $(x, y)$ with $x \leq 60 \leq y$ satisfy $(*)$. If $x \leq y \leq 60$, then $(*)$ is equivalent to $60=y$, so only $(x, y)$ with $y=60$ and $x \leq 60$ satisfy
$(*)$. These $(x, y)$ are already accounted for in the third case.

If $y \leq 60 \leq x$, then $(*)$ is equivalent to $60=y$, so only $(x, y)$ with $y=60$ and $x \geq 60$ satisfy $(*)$. These $(x, y)$ are included in the second case.

If $y \leq x \leq 60$, then $(*)$ is equivalent to $60=y$, so only $y=60$ and $60 \leq x \leq 60$, which implies $x=60$. In this case, only the pair $(60,60)$ satisfies $(*)$ and this pair is already included in the first case.

Therefore, we need to count the pairs of positive integers $(x, y)$ with $x \leq 100$ and $y \leq 100$ which satisfy one of $60 \leq x \leq y$ or $60 \leq y \leq x$ or $x \leq 60 \leq y$.

From these three ranges for $x$ and $y$ we see that $y \geq 60$ in all cases and $x$ can either be less than or equal to 60 or can be greater than or equal to 60 and either larger or smaller than $y$.

In other words, these three ranges are equivalent to the condition that $y \geq 60$.

Therefore, $(*)$ is satisfied by all pairs $(x, y)$ with $y \geq 60$.

Since $1 \leq x \leq 100$, there are 100 possible values for $x$.

Since $60 \leq y \leq 100$, there are 41 possible values for $y$.

Therefore, there are $100 \times 41=4100$ ordered pairs that satisfy the equation.

ANSWER: 4100

## Part B

Answer: 4100


In the parking lot, only two-wheeled motorcycles and four-wheeled cars are parked, and the number of motorcycles is more than the number of cars. The two types of vehicles have a total of 84 wheels. After 3 vehicles are driven away, the number of wheels in the parking lot is 3 times the number of vehicles. Originally, there were $\qquad$ motorcycles parked in the parking lot.

Solution:
\begin{tabular}{l}
$\mathbf{3 7}$ \\
\hline 16
\end{tabular}

Answer: 16


In the analysis of bank accounts, it was found that the remaining balance on each of them is more than 10 rubles. It also turned out that there is a group of clients, each of whom has the same amount of money on their account. This amount is a number consisting of only ones. If you add up all the money on the accounts of this group of clients, the resulting sum will also be represented by a number consisting of only ones. Find the smallest number of clients in the group for which this is possible, if the group has more than one person.

Solution:
# Solution

This problem is equivalent to the following.

Find the smallest natural number $m$, for which there exist natural numbers $n$ and $k$, such that $n>k>1$ and $\underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{k} \cdot m$.

Obviously, $m>9$. If $m=\overline{a b}$, where $a \geq 1$, then the equality $\underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{k} \cdot \overline{a b}$ means that $b=1$. However, in this case, regardless of $a$, the second digit from the right in the number $\underbrace{11 \ldots 1}_{k} \cdot \overline{a b}$ is $a+1$, and this cannot be 1. Therefore, $m \geq 100$. Clearly, $n=100$ does not satisfy the condition because $\underbrace{11 \ldots 1 \cdot 100}_{k}=\underbrace{11 \ldots 100}_{k}$.

On the other hand, $n=101$ works, since 101$\cdot$11=1111.

Answer: 101.

Answer: 101


Given 5 distinct positive numbers can be divided into two groups such that the sums of the numbers in each group are equal, how many different ways are there to divide these numbers into such groups?

Solution:
[Solution] (1) If the number of elements in the two groups with equal sums are 1 and 4, respectively, i.e., the largest number among the 5 numbers equals the sum of the other 4 numbers, then there is obviously only 1 way to divide them.
(2) Suppose the number of elements in the two groups with equal sums are 2 and 3, respectively: $\left\{m_{1}, m_{2}\right\}$, $\left\{m_{3}, m_{4}, m_{5}\right\}$, and $m_{1}+m_{2}=m_{3}+m_{4}+m_{5}$. From (1), we know that if there is another valid grouping, it must still be a 2,3 grouping. Assume without loss of generality that it is $m_{1}+m_{5}=m_{2}+m_{3}+m_{4}$, but this leads to $m_{5}=m_{2}$, which is a contradiction.
Therefore, there is only 1 way to group the numbers that satisfies the requirements of the problem.

Answer: 1


Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased his usual speed to arrive exactly at 9:00?

Solution:
Answer: By $30 \%$.

Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be increased by $\frac{520}{3} \div \frac{400}{3}=1.3$ times, i.e., by $30 \%$.

Answer: 30


Find all positive integers $n$ such that $n^4 - n^3 + 3n^2 + 5$ is a perfect square.

Solution:
To find all positive integers \( n \) such that \( n^4 - n^3 + 3n^2 + 5 \) is a perfect square, we will consider two cases: when \( n \) is even and when \( n \) is odd.

### Case 1: \( n \) is even
1. Let \( n = 2k \).
2. We need to check if \( n^4 - n^3 + 3n^2 + 5 \) can be written as \( m^2 \) for some integer \( m \).
3. Consider the expression \( (n^2 - k + 1)^2 \):
   \[
   (n^2 - k + 1)^2 = n^4 - 2kn^2 + k^2 + 2n^2 - 2k + 1
   \]
4. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   n^4 - n^3 + 3n^2 + 5 < (n^2 - k + 1)^2
   \]
   Simplifying, we get:
   \[
   0 < \frac{7n^2}{4} + n + 4 \quad \text{(TRUE)}
   \]
5. Now consider \( (n^2 - k + 2)^2 \):
   \[
   (n^2 - k + 2)^2 = n^4 - 2kn^2 + k^2 + 4n^2 - 4k + 4
   \]
6. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   (n^2 - k + 2)^2 > n^4 - n^3 + 3n^2 + 5
   \]
   Simplifying, we get:
   \[
   \frac{5n^2}{4} - 2n - 4 > 0 \quad \Rightarrow n \ge 3
   \]
7. Testing \( n = 2 \):
   \[
   2^4 - 2^3 + 3 \cdot 2^2 + 5 = 16 - 8 + 12 + 5 = 25 \quad \text{(which is a perfect square)}
   \]
   Therefore, \( n = 2 \) is a solution.

### Case 2: \( n \) is odd
1. Let \( n = 2k + 1 \).
2. We need to check if \( n^4 - n^3 + 3n^2 + 5 \) can be written as \( m^2 \) for some integer \( m \).
3. Consider the expression \( (n^2 - k)^2 \):
   \[
   (n^2 - k)^2 = n^4 - 2kn^2 + k^2
   \]
4. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   n^4 - n^3 + 3n^2 + 5 > (n^2 - k)^2
   \]
   Simplifying, we get:
   \[
   0 < \frac{3n^2}{2} + n + \frac{9}{2} \quad \text{(TRUE)}
   \]
5. Now consider \( (n^2 - k + 1)^2 \):
   \[
   (n^2 - k + 1)^2 = n^4 - 2kn^2 + k^2 + 2n^2 - 2k + 1
   \]
6. We want to compare this with \( n^4 - n^3 + 3n^2 + 5 \):
   \[
   (n^2 - k + 1)^2 > n^4 - n^3 + 3n^2 + 5
   \]
   Simplifying, we get:
   \[
   5n^2 - 6n - 11 > 0 \quad \Rightarrow n \ge 3
   \]
7. Testing \( n = 1 \):
   \[
   1^4 - 1^3 + 3 \cdot 1^2 + 5 = 1 - 1 + 3 + 5 = 8 \quad \text{(which is not a perfect square)}
   \]
   Therefore, there are no solutions for odd \( n \).

### Conclusion:
The only solution is \( n = 2 \).

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

Answer: n = 2


Find all real numbers $x$ that satisfy the inequality
$$
\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}
$$

Solution:
Solve: First, by
$$
\left\{\begin{array}{l}
3-x \geqslant 0 \\
x+1 \geqslant 0
\end{array}\right.
$$

we get $-1 \leqslant x \leqslant 3$. Since in the interval $-1 \leqslant x \leqslant 3$,
$$
\sqrt{3-x}-\sqrt{x+1}
$$

decreases as $x$ increases, solving the equation
$$
\begin{array}{c}
\sqrt{3-x}-\sqrt{x+1}=\frac{1}{2}, \\
\frac{7}{4}-2 x=\sqrt{x+1}, \\
64 x^{2}-128 x+33=0, \\
x=\frac{8 \pm \sqrt{31}}{8} .
\end{array}
$$

Upon verification, $x=\frac{8-\sqrt{31}}{8}$ is a solution to (1), meaning that when $x=\frac{8-\sqrt{31}}{8}$, the value of $\sqrt{3-x}-\sqrt{x+1}$ is $\frac{1}{2}$.
Therefore, the solution to the inequality is $-1 \leqslant x<\frac{8-\sqrt{31}}{8}$.

Answer: -1\leqslantx<\frac{8-\sqrt{31}}{8}


Given point $P$ is on the line $x+2 y-1=0$, point $A$ is on the line $x+2 y+3=0$, the midpoint of $P Q$ is $M\left(x_{0}, y_{0}\right)$, and $y_{0}>x_{0}+2$, then the range of $\frac{y_{0}}{x_{0}}$ is $\qquad$ .

Solution:
Answer $\left(-\frac{1}{2},-\frac{1}{5}\right)$.
Analysis Note that the two lines are parallel, so the locus of points is a line parallel to the two lines and equidistant from them, with the equation $x+2 y+1=0$, i.e., $M\left(x_{0}, y_{0}\right)$ satisfies $x_{0}+2 y_{0}+1=0$. The points that satisfy the inequality $y_{0}>x_{0}+2$ are in the upper left of the line $y=x+2$. The problem is transformed into finding the range of $\frac{y_{0}}{x_{0}}$ for the point $M\left(x_{0}, y_{0}\right)$ on the ray $x_{0}+2 y_{0}+1=0\left(x_{0}<-\frac{5}{3}\right)$, and the geometric meaning of $\frac{y_{0}}{x_{0}}$ is the slope of the line connecting $M\left(x_{0}, y_{0}\right)$ and the origin, so $k_{O M}=\frac{y_{0}}{x_{0}} \in\left(-\frac{1}{2},-\frac{1}{5}\right)$.

Answer: (-\frac{1}{2},-\frac{1}{5})


Let $a, b \in \mathbb{R}^{+}, i^{2}=-1$, and there exists $z \in \mathbb{C}$, such that
$$
\left\{\begin{array}{l}
z+\bar{z}|z|=a+b i, \\
|z| \leqslant 1 .
\end{array}\right.
$$

Then the maximum value of $a b$ is

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

Let $z=x+y i\left(x, y \in R, i^{2}=-1\right)$, substitute into the given equation and by the definition of equality of complex numbers we get
$$
\left\{\begin{array}{l}
a=x \cdot\left(1+\sqrt{x^{2}+y^{2}}\right), \\
b=y \cdot\left(1-\sqrt{x^{2}+y^{2}}\right) .
\end{array}\right.
$$

From $|z| \leqslant 1$ we get $0<x^{2}+y^{2}<1$, and $0<x, y$. From (1) and (2) we get $2 a b=2 x y\left(1-\left(x^{2}+y^{2}\right)\right) \leqslant\left(x^{2}+y^{2}\right)(1$ $\left.-x^{2}-y^{2}\right) \leqslant\left(\frac{x^{2}+y^{2}+1-x^{2}-y^{2}}{2}\right)^{2}=\frac{1}{4}$. Therefore, $a b \leqslant$ $\frac{1}{8}$. When $x=y$ and $x^{2}+y^{2}=1-x^{2}-y^{2}$, i.e., $x=y=\frac{1}{2}$, $a b=\frac{1}{8}$. Hence, the maximum value of $a b$ is $\frac{1}{8}$.

Answer: \frac{1}{8}


If $a, b, c, d, e, f, g, h, k$ are all 1 or -1, try to find the maximum possible value of
$$
a e k - a f h + b f g - b d k + c d h - c e g
$$

Solution: 7.5. Since each term in the expression $a e k - a f h + b f g - b d k + c d h - c e g$ is either 1 or -1, the value of the expression is even. However, the expression cannot equal 6. If it did, then $a e k$, $b f g$, and $c d h$ would all have to be 1, making their product 1, while $a f h$, $b d k$, and $c e g$ would all have to be -1, making their product -1. But these two products should be equal, which is a contradiction. The value of the expression can be 4, for example, when $a=b=c=e=h=k=1$ and $d=f=g=-1$, the value of the expression is 4. Therefore, the maximum possible value is 4.

Answer: 4


Try to find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$

has positive integer solutions.

Solution:
Solution: Without loss of generality, let $x \leqslant y \leqslant z$. Clearly, $z^{2} \mid\left(x^{3}+y^{3}\right)$, so, $z^{2} \leqslant x^{3}+y^{3}$.
But $x^{3} \leqslant x z^{2}, y^{3} \leqslant y z^{2}$, hence we have
$$
z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y) \text {. }
$$

Thus, we have
$$
\begin{array}{l}
x^{3}+y^{3} \geqslant z^{2} \geqslant\left[n x^{2} y^{2}-(x+y)\right]^{2}, \\
n^{2} x^{4} y^{4}<2 n x^{2} y^{2}(x+y)+x^{3}+y^{3}, \\
n x y<2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}} .
\end{array}
$$

Notice that the left side of inequality (1) increases with $x, y$, while the right side decreases with $x, y$, i.e., the upper bound of $n x y$ does not exceed $2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}$, and the lower bound of $2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}$ is no less than $n x y$. It is easy to see that $x=1$.
Substituting $x=1$ into equation (1) yields
$n y<2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}$.
Equation (2) can only hold when $y \leqslant 3$.
But $z^{2} \mid\left(1+y^{3}\right)$, so it can only be $y=1,2(y=3$ when, $z^{2} \mid 28, z^{2}=1,4$, which contradicts $z \geqslant y$), accordingly, $z$ $=1,3$. At this point, $n=3,1$.

Answer: n=3,1


Cars A and B start from locations $A$ and $B$ simultaneously and travel back and forth between $A$ and $B$ at a constant speed. If after the first meeting, Car A continues to drive for 4 hours to reach $B$, while Car B only drives for 1 hour to reach $A$, then when the two cars meet for the 15th time (meetings at $A$ and $B$ are not counted), they have driven $\qquad$ hours.

Solution:
【Analysis】Let the two cars meet after $t$ hours, the speed of car A is $v_{1}$, and the speed of car B is $v_{2}$. From the problem, we have: $4 v_{1}=t v_{2}$, $(t+4) v_{1}=(t+1) v_{2}$. Solving these equations, we get $t=2$. Therefore, it takes car A 6 hours to complete the journey, and car B 3 hours. Interestingly, when car A completes one trip, car B completes the round trip. Thus, excluding the meetings at points A and B, the two cars meet every 6 hours. They meet 2 hours after starting in opposite directions, and 4 hours after starting in the same direction. The 15th trip is in opposite directions, so $6 \times 14+2=86$ (hours).

【Solution】Let the two cars meet after $t$ hours, with car A's speed being $v_{1}$ and car B's speed being $v_{2}$. Then:
$$
\begin{array}{l}
4 v_{1}=t v_{2}, \\
(t+4) v_{1}=(t+1) v_{2},
\end{array}
$$

Solving these, we get $t=2$.
So it takes car A 6 hours to complete the journey, and car B 3 hours. Excluding the meetings at points A and B, the two cars meet every 6 hours. They meet 2 hours after starting in opposite directions, and 4 hours after starting in the same direction. The 15th trip is in opposite directions, so the two cars meet for the 15th time after:
$$
\begin{array}{l}
6 \times(15-1)+2 \\
=6 \times 14+2 \\
=84+2 \\
=86 \text { (hours) }
\end{array}
$$

Answer: When the two cars meet for the 15th time, they have been traveling for 86 hours.

Answer: 86


A square of paper is cut into two pieces by a single straight cut. Which of the following shapes cannot be the shape of either piece?
A An isosceles triangle
B A right-angled triangle
C A pentagon
D A rectangle
E A square

Solution:
Solution
E

Assuming one of the options cannot be formed, we answer the question by eliminating four of the options. The diagrams below show examples of how a square could be cut to form the first four shapes, with a dotted line representing the position of the cut.
Hence we conclude that the shape which cannot be obtained is a square.

Answer: E


If $a$, $b, c \in \mathbf{R}$, and $a, b, c$ satisfy the system of equations
$$\left\{\begin{array}{l}
a^{2}-b c-8 a+7=0 \\
b^{2}+c^{2}+b c-6 a+6=0
\end{array}\right.$$

Try to find the range of values for $a$.

Solution: Solve: From (1) + (2), we get $b^{2}+c^{2}=-a^{2}+14 a-13$. From (2) - (1), we get $b+c= \pm(a-1)$, so by the corollary, $2(-$ $\left.a^{2}+14 a-13\right) \geqslant[ \pm(a-1)]^{2}$, solving this yields $1 \leqslant a \leqslant 9$, hence the range of values for $a$ is $1 \leqslant a \leqslant 9$.

Answer: 1 \leqslant a \leqslant 9


There are 1994 points on a circle, which are painted in several different colors, and the number of points of each color is different. Now, take one point from each color set to form a polygon with vertices of different colors inside the circle. To maximize the number of such polygons, how many different colors should the 1994 points be painted, and how many points should each color set contain?

Solution:
Let 1994 points be colored with $n$ colors, and the number of points of each color, in ascending order, is $m_{1}1$. In fact, if $m_{1}=1$, because $m_{1} m_{2} \cdots m_{n}m_{k+1} m_{k}$.
Therefore, when $m^{\prime}{ }_{n+1}, m^{\prime} \star$ replace $m_{\star+1}, m_{\star}$, the value of $M$ increases, leading to a contradiction:
(2) $m_{i+1}-m_{i}=2$ holds for exactly one $i$. For this:
(i) $m_{i+1}-m_{i}=2$ holds for at most one $i$. If otherwise, there exist positive integers $j, k$ such that $1 \leqslant j < k \leq n$ and $m_{j+1} - m_{j} = m_{k+1} - m_{k} = 2$. Then, $m_{j+1} = m_{j} + 2$ and $m_{k+1} = m_{k} + 2$. If we replace $m_{k+1}, m_{j}$ with $m^{\prime}{ }_{k+1}, m^{\prime} j_{j}$, the value of $M$ increases, leading to a contradiction.
(ii) If $m_{i+1}-m_{i}=1$ for all $i=1,2, \cdots, n-1$, then $\sum_{i=1}^{n} m_{i} = n m_{1} + \frac{1}{2} n(n-1) = 1994 = 2 \times 997 \Rightarrow n(2 m_{1} - 1 + n) = 4 \times 997$. Since $n$ and $2 m_{1} - 1 + n$ are one odd and one even, and $2 m_{1} - 1 + n > n$, and 997 is a prime number, the only solution is $n=4$, $2 m_{1} - 1 + n = 997$, giving $m_{1} = 497 = 495 + 22$. If we take $m^{\prime} = m_{k+2} - 2$, then $m_{k} < 2$ and $m_{k+2} > 4$. Hence, $2 m^{\prime} > m_{k+2}$. Replacing $m_{k+2}$ with 2 and $m^{\prime}$, the value of $M$ increases, leading to a contradiction. Therefore, $m_{1} = 2$.
(II) From (I), we can assume the number of points of each color is 2, 3, ..., $k-1$, $k+1$, $k+2$, ..., $n$, $n+1$, $n+2$ ($k \leqslant n-1$).
We have $\frac{1}{2}(n+1)(n+4) - k = 1994$.
This gives $\left\{\begin{array}{l}n^{2} + 5n - 3984 > 0, \\ n^{2} + 3n - 3982 \leqslant 0 .\end{array}\right.$

Solving, we get $61 \leqslant n < 62$.
Taking $n=61$, we have $k = \frac{1}{2}(n+1)(n+4) - 1994 = 31 \times 65 - 1994 = 21$. Therefore, 1994 points can be colored with 61 colors, with the number of points of each color being $2, 3, \cdots, 19, 20, 22, 23, \cdots, 61, 62, 63$. In this case, the resulting polygon is a 61-sided polygon, and the number of such polygons is the maximum. (Xu Jingling, Jiugu Middle School, Susong County, Anhui Province)

Answer: 61


Find the smallest natural number $n$ with the following properties: (a) In decimal representation it ends with 6. (b) If we move this digit to the front of the number, we get a number 4 times larger.

Solution: 1. From the conditions of the problem we have $n=10 x+6$ and $4 n=$ $6 \cdot 10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40 n=6 \cdot 10^{m+1}+n-6 \Rightarrow n=2\left(10^{m+1}-1\right) / 13$. Hence we must find the smallest $m$ such that this fraction is an integer. By inspection, this happens for $m=6$, and for this $m$ we obtain $n=153846$, which indeed satisfies the conditions of the problem.

Answer: 153846


Let $x_{1}, \ldots, x_{100}$ be positive reals satisfying $x_{i}+x_{i+1}+x_{i+2} \leqslant 1$ for all $i \leqslant 100$ (with the convention $x_{101}=x_{1}$ and $x_{102}=x_{2}$). Determine the maximum value that

$$
S=\sum_{i=1}^{100} x_{i} x_{i+2}
$$

can take.

Solution:
The idea here is to look at two consecutive terms of this sum and try to connect them using the hypothesis:

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

using the inequality $x(1-x) \leqslant 1 / 4$ for all $x \in[0,1]$, which is the case for $x_{i+1}+x_{i+2}$. It remains to break down the sum into packets of four:

$$
S=\sum_{k=1}^{50}\left(x_{2 k-1} x_{2 k+1}+x_{2 k} x_{2 k+2}\right) \leqslant \sum_{k=1}^{50} \frac{1}{4}=\frac{25}{2}
$$

It remains to show that this value can be achieved. For this, we look at the case of equality of the inequalities we have used. Notably, for the inequalities used to be equalities, it is necessary that $x_{2 k}+x_{2 k+1}=\frac{1}{2}$. On the other hand, at least one of the two terms $x_{2 k-1}$ and $x_{2 k+2}$ must be $1 / 2$ to satisfy $x_{i}+x_{i+1}+x_{i+2}=1$. We then arrive at the construction $x_{2 k}=0$ and $x_{2 k+1}=1 / 2$ for all $k$, which we verify allows us to reach the maximum value.

## 5 LTE and Zsigmondy (Pierre-Marie)

## Course Elements

You can find the proofs of these course elements at the following address: https : //maths-olympiques.fr/wp-content/uploads/2017/09/arith_lte.pdf.

Theorem 1 (Lifting The Exponent Lemma, or LTE).

Let $p$ be a prime number, $x$ and $y$ integers coprime with $p$.

A) Case $p \neq 2$:

(i) If $p \mid x-y$ then $v_{p}\left(x^{n}-y^{n}\right)=v_{p}(x-y)+v_{p}(n)$

(ii) If $p \mid x+y$ and $n$ is odd then $v_{p}\left(x^{n}+y^{n}\right)=v_{p}(x+y)+v_{p}(n)$

B) Case $p=2$:

(i) If $2 \mid x-y$ and $n$ is even then $v_{2}\left(x^{n}-y^{n}\right)=v_{2}(x-y)+v_{2}(x+y)+v_{2}(n)-1$.

(ii) If $2 \mid x-y$ and $n$ is odd then $v_{2}\left(x^{n}-y^{n}\right)=v_{2}(x-y)$.

(iii) If $4 \mid x-y$ then $v_{2}\left(x^{n}-y^{n}\right)=v_{2}(x-y)+v_{2}(n)$.

Theorem 2 (Zsigmondy's Theorem).

Let $x>y>0$ be coprime integers. Let $u_{k}=x^{k}-y^{k}$ and $v_{k}=x^{k}+y^{k}$ for $k \geqslant 1$. We call a primitive prime factor of a term $s_{n}$ a prime number $p$ that divides $s_{n}$ but no previous term of the sequence $s$.

A) Except for the cases

- $n=2$ and $x+y$ is a power of 2,

- $n=6, x=2, y=1$,

every $u_{n}$ has a primitive prime factor.

B) Except for the case $n=3, x=2, y=1$, every $v_{n}$ has a primitive prime factor.

## Exercises

Answer: \frac{25}{2}


Find the smallest positive integer $n$ with the property that in the set $\{70, 71, 72,... 70 + n\}$ you can choose two different numbers whose product is the square of an integer.

Solution:
To find the smallest positive integer \( n \) such that in the set \(\{70, 71, 72, \ldots, 70 + n\}\), you can choose two different numbers whose product is the square of an integer, we need to identify pairs of numbers whose product is a perfect square.

1. **Prime Factorization of Numbers in the Set:**
   - \(70 = 2 \cdot 5 \cdot 7\)
   - \(71\) is a prime number.
   - \(72 = 2^3 \cdot 3^2\)
   - \(73\) is a prime number.
   - \(74 = 2 \cdot 37\)
   - \(75 = 3 \cdot 5^2\)
   - \(76 = 2^2 \cdot 19\)
   - \(77 = 7 \cdot 11\)
   - \(78 = 2 \cdot 3 \cdot 13\)
   - \(79\) is a prime number.
   - \(80 = 2^4 \cdot 5\)
   - \(81 = 3^4\)
   - \(82 = 2 \cdot 41\)
   - \(83\) is a prime number.
   - \(84 = 2^2 \cdot 3 \cdot 7\)
   - \(85 = 5 \cdot 17\)
   - \(86 = 2 \cdot 43\)
   - \(87 = 3 \cdot 29\)
   - \(88 = 2^3 \cdot 11\)
   - \(89\) is a prime number.
   - \(90 = 2 \cdot 3^2 \cdot 5\)
   - \(91 = 7 \cdot 13\)
   - \(92 = 2^2 \cdot 23\)
   - \(93 = 3 \cdot 31\)
   - \(94 = 2 \cdot 47\)
   - \(95 = 5 \cdot 19\)
   - \(96 = 2^5 \cdot 3\)
   - \(97\) is a prime number.
   - \(98 = 2 \cdot 7^2\)

2. **Identifying Pairs with Perfect Square Products:**
   - We need to find two numbers \(a\) and \(b\) such that \(a \cdot b\) is a perfect square.
   - The product \(a \cdot b\) is a perfect square if and only if the exponents of all prime factors in the product are even.

3. **Finding the Smallest \( n \):**
   - We need to find the smallest \( n \) such that there exists a pair \((a, b)\) in \(\{70, 71, 72, \ldots, 70 + n\}\) with \(a \cdot b\) being a perfect square.
   - From the prime factorizations, we see that \(72 = 2^3 \cdot 3^2\) and \(98 = 2 \cdot 7^2\).
   - The product \(72 \cdot 98 = (2^3 \cdot 3^2) \cdot (2 \cdot 7^2) = 2^4 \cdot 3^2 \cdot 7^2\).
   - The exponents of all prime factors in \(72 \cdot 98\) are even: \(2^4 \cdot 3^2 \cdot 7^2\), which is a perfect square.

4. **Conclusion:**
   - The smallest \( n \) such that the set \(\{70, 71, 72, \ldots, 70 + n\}\) contains two numbers whose product is a perfect square is \(98 - 70 = 28\).

The final answer is \(\boxed{28}\).

Answer: 28


The base area of the rectangular prism is 4, and the length of the diagonal is 4. Then the maximum value of the lateral surface area of the rectangular prism is $\qquad$

Solution:
$3.16 \sqrt{2}$.
As shown in Figure 10, let the length of the rectangular prism be $x$, the width be $y$, and the height be $z$. According to the problem, we have
$$
\left\{\begin{array}{l}
x y=4 \\
x^{2}+y^{2}+z^{2}=16 .
\end{array}\right.
$$

Thus, $z^{2}=16-\left(x^{2}+y^{2}\right) \leqslant 16-2 x y=8$.
This gives $0<z \leqslant 2 \sqrt{2}$.
And when $x=y=2$, $z=2 \sqrt{2}$. At this point,
$$
\begin{aligned}
S_{\text {worm }} & =2(x+y) z=2 z \sqrt{x^{2}+y^{2}+2 x y} \\
& =2 z \sqrt{\left(16-z^{2}\right)+8}=2 z \sqrt{24-z^{2}} \\
& =2 \sqrt{z^{2}\left(24-z^{2}\right)}=2 \sqrt{144-\left(z^{2}-12\right)^{2}} .
\end{aligned}
$$

At this time, $S_{m}$ reaches its maximum value, $S_{m}=2(2+2) \times 2 \sqrt{2}=16 \sqrt{2}$.

Answer: 16 \sqrt{2}


A stenotypist has a long manuscript to transcribe. For the first half, she types 15 pages per day, and for the second half, she types 25 pages per day.

What is her average daily output overall?

Solution:
If the manuscript has $2 M$ pages, she needs $\frac{M}{15}$ days for the first half, $\frac{M}{25}$ days for the second half, and in total for $2 M$ pages:

$$
\frac{M}{15}+\frac{M}{25}=\frac{8}{75} M \text { days }
$$

Thus, she writes on average $\frac{2 M}{\frac{8}{75} M}=\frac{75}{4}=18.75$ pages per day.

Answer: 18.75


Given the parabola $y=ax^{2}(a>0)$ and the line $y=bx+c(b \neq 0)$ have two common points, their x-coordinates are $x_{1}, x_{2}$, and the line $y=bx+c$ intersects the x-axis at the point $\left(x_{3}, 0\right)$. Then the relationship satisfied by $x_{1}, x_{2}, x_{3}$ is ( ).
(A) $x_{1}+x_{2}=x_{3}$
(B) $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$
(C) $x_{3}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}$
(D) $x_{1} x_{2}+x_{2} x_{3}=x_{3} x_{1}$

Solution:
4. B.

Solving the system of equations $\left\{\begin{array}{l}y=a x^{2}, \\ y=b x+c\end{array}\right.$ yields $a x^{2}-b x-c=0$.
Therefore, $x_{1}+x_{2}=\frac{b}{a}, x_{1} x_{2}=-\frac{c}{a}$.
Substituting $y=0$ into $y=b x+c$, we get $x_{3}=-\frac{c}{b}$.
Thus, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{b}{c}$.
Hence, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$.

Answer: B


If his next roll is between $1$ and $5$ inclusive, Bowser will shoot his ``Zero Flame" that sets a player's coin and star counts to zero. Fortunately, Theo has a double dice block, which lets him roll two fair $10$-sided dice labeled $1$-$10$ and take the sum of the rolls as his "roll". If he uses his double dice block, what is the probability he escapes the Bowser zone without losing his coins and stars?

Solution:
To solve this problem, we need to determine the probability that the sum of two rolls of a fair 10-sided die is not between 1 and 5 inclusive.

1. **Calculate the total number of possible outcomes:**
   Each die has 10 faces, so when rolling two dice, the total number of possible outcomes is:
   \[
   10 \times 10 = 100
   \]

2. **Identify the outcomes where the sum is between 1 and 5 inclusive:**
   We need to find the pairs \((a, b)\) such that \(1 \leq a + b \leq 5\). Let's list these pairs:
   - For \(a + b = 2\): \((1, 1)\)
   - For \(a + b = 3\): \((1, 2), (2, 1)\)
   - For \(a + b = 4\): \((1, 3), (2, 2), (3, 1)\)
   - For \(a + b = 5\): \((1, 4), (2, 3), (3, 2), (4, 1)\)

   Summarizing, the pairs are:
   \[
   (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)
   \]
   There are 10 such pairs.

3. **Calculate the probability of the sum being between 1 and 5 inclusive:**
   The number of favorable outcomes is 10, and the total number of possible outcomes is 100. Therefore, the probability is:
   \[
   P(\text{sum between 1 and 5}) = \frac{10}{100} = \frac{1}{10}
   \]

4. **Calculate the probability of escaping the Bowser zone:**
   The probability of escaping the Bowser zone is the complement of the probability of landing in the Bowser zone:
   \[
   P(\text{escape}) = 1 - P(\text{sum between 1 and 5}) = 1 - \frac{1}{10} = \frac{9}{10}
   \]

The final answer is \(\boxed{\frac{9}{10}}\).

Answer: \frac{9}{10}


Solve the equation

$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x}+\ldots+\frac{1}{x}=3
$$

where $x$ is a positive integer.

Solution:
## Solution.

Multiplying both sides of the equation by $x$, we have

$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$

The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1, a_{n}=1$,

$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{1-x+1}{-1}+1=x-1 . \text { Since } S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n
$$

then $\frac{x-1+1}{2} \cdot(x-1)=3 x, x^{2}=7 x, x=7(x \neq 0)$.

Answer: $x=7$.

Answer: 7


In how many ways can the number $6 k(k \in \mathbb{N})$ be written as the sum of three natural numbers? Records that differ only in the order of the addends are considered the same.

Solution:
Solution. Let $6 k=x_{1}+x_{2}+x_{3}$, where $x_{1}, x_{2}, x_{3}$ are natural numbers such that $x_{1} \leq x_{2} \leq x_{3}$. If $x_{1}=2 l-1$, where $1 \leq l \leq k$, then $x_{2}$ can be equal to one of the numbers

$$
2 l-1,2 l, 2 l+1, \ldots,\left[\frac{6 k-2 l+1}{2}\right]=3 k-l
$$

and for each of these $3 k-3 l+2$ possibilities, the number $x_{3}$ is equal to $6 k-x_{1}-x_{2}$. If $x_{1}=2 l$, where $1 \leq l \leq k$, then $x_{2}$ is equal to one of the numbers

$$
2 l, 2 l+1, \ldots, 3 k-l
$$

and for each of these possibilities, the number $x_{3}$ is uniquely determined. Therefore, the desired number is equal to

$$
\sum_{l=1}^{k}(3 k-3 l+2)+\sum_{l=1}^{k}(3 k-3 l+1)=\sum_{l=1}^{k}(6 k-6 l+3)=k(6 k+3)-6 \frac{k(k+1)}{2}=3 k^{2}
$$

Answer: 3k^2


For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$?

Solution:
Solution 1
We know by [De Moivre's Theorem](https://artofproblemsolving.com/wiki/index.php/De_Moivre%27s_Theorem) that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number) $t$ and all [integers](https://artofproblemsolving.com/wiki/index.php/Integer) $n$.  So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the [trigonometric identities](https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities) $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ hold for all real $u$.  If our original equation holds for all $t$, it must certainly hold for $t = \frac{\pi}2 - u$.  Thus, the question is equivalent to asking for how many [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$.
$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$.  We know that two [complex numbers](https://artofproblemsolving.com/wiki/index.php/Complex_number) are equal if and only if both their [real part](https://artofproblemsolving.com/wiki/index.php/Real_part) and [imaginary part](https://artofproblemsolving.com/wiki/index.php/Imaginary_part) are equal.  Thus, we need to find all $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$.
$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$.  So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$.  Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work.  There are $\boxed{250}$ of them in the given range.

Solution 2
This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$. Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$. Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$. Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$. Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$.

Solution 3
We can rewrite $\sin(t)$ as $\cos\left(\frac{\pi}{2}-t\right)$ and $\cos(t)$ as $\sin\left(\frac{\pi}{2}-t\right)$. This means that $\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}$. This theorem also tells us that $e^{\frac{\pi i}{2}}=i$, so $\sin t + i\cos t = \frac{i}{e^{it}}$. By the same line of reasoning, we have $\sin nt + i\cos nt = \frac{i}{e^{int}}$.
For the statement in the question to be true, we must have $\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}$. The left hand side simplifies to $\frac{i^n}{e^{int}}$. We cancel the denominators and find that the only thing that needs to be true is that $i^n=i$. This is true if $n\equiv1\pmod{4}$, and there are $\boxed{250}$ such numbers between $1$ and $1000$. Solution by Zeroman

Solution 4
We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is $cis(n(90^{\circ}-t))$, and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$. Canceling gets $n \equiv 1 \mod 4$, and thus there are $\boxed{250}$ values of n.
-AlexLikeMath

Solution 5(CHEAP)
Let $t=0$. Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$. Thus, the answer is $\boxed{250}$.

Solution 6
We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we have a total of $1000 \cdot\frac{1}{4} = \boxed{250}$ integers $n$ that work.

Answer: 250


Solution:
\[
\begin{aligned}
\sum_{k=1}^{n} k^{3} & =\sum_{j=1}^{3}\binom{n+1}{j+1} \Delta^{j} O^{3} \\
& =\binom{n+1}{2} \cdot \Delta O^{3}+\binom{n+1}{3} \Delta^{2} O^{3}+\binom{n+1}{4} \Delta^{3} O^{3} \\
& =\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4} \\
& =\frac{n^{2}(n+1)^{2}}{4} .
\end{aligned}
\]

Answer: \frac{n^{2}(n+1)^{2}}{4}


For a given $n \in N$, find the number of all different natural number triples whose sum is $6 n$.

Solution:
[Solution] Let $(x, y, z)$ be a triplet of natural numbers such that $x+y+z=6n$ and $x \leqslant y \leqslant z$. When $k=1,2, \cdots, n$, all triplets $(x, y, z)$ for which $x=2k-1$ are:
$$
\begin{array}{l}
(2k-1, 2k-1, 6n-4k+2), (2k-1, 2k, 6n-4k+1), \\
\cdots, (2k-1, 3n-k, 3n-k+1),
\end{array}
$$

All triplets $(x, y, z)$ for which $x=2k$ are:
$$
\begin{array}{l}
(2k, 2k, 6n-4k), (2k, 2k+1, 6n-4k-1), \\
\cdots, (2k, 3n-k, 3n-k).
\end{array}
$$
The number of triplets in (1) and (2) is
$$
\begin{aligned}
S_{k} & =[(3n-k)-(2k-2)]+[(3n-k)-(2k-1)] \\
& =6(n-k)+3 .
\end{aligned}
$$

Summing up, the total number of all triplets is
$$
S=\sum_{k=1}^{n} S_{k}=\sum_{k=1}^{n}[6(n-k)+3]=6n^2-3n(n+1)+3n=3n^2 \text{. }
$$

Answer: 3n^2


In the right-angled triangle $\mathrm{ABC}$, the angle at vertex $B$ is $30^{\circ}$. The center of the square constructed outward on the hypotenuse $\mathrm{ABC}$ is $D$. What is the measure of the angle $A D B$?

Solution:
Let's denote the sought angle $A D B$ by $\delta$, then

$$
B A D \varangle=180^{\circ}-\left(75^{\circ}+\delta\right)=105^{\circ}-\delta
$$

We write the sine theorem in the triangle $A B D$:

$$
\begin{aligned}
\frac{\sin \delta}{\sin \left(105^{\circ}-\delta\right)} & =\frac{\sqrt{3}}{\sqrt{2}}, \quad \text { hence } \\
\sin \delta & =\frac{\sqrt{3}}{\sqrt{2}}\left(\sin 105^{\circ} \cos \delta-\cos 105^{\circ} \sin \delta\right)
\end{aligned}
$$

Using the angle sum formulas, we calculate the sine and cosine of $105^{\circ}$:

$$
\begin{aligned}
& \cos 105^{\circ}=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}=\frac{1}{2} \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\frac{1-\sqrt{3}}{2}\right) \\
& \sin 105^{\circ}=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}=\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}+\frac{1}{2} \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}+1}{2}\right)
\end{aligned}
$$

Substitute the values obtained into equation (1):

$$
\begin{aligned}
& \sin \delta=\frac{\sqrt{3}}{\sqrt{2}}\left[\frac{\sqrt{2}}{2} \frac{\sqrt{3}+1}{2} \cos \delta-\frac{\sqrt{2}}{2} \frac{1-\sqrt{3}}{2} \sin \delta\right] \\
& \sin \delta=\frac{\sqrt{3}(\sqrt{3}+1)}{4} \cos \delta-\frac{\sqrt{3}(1-\sqrt{3})}{4} \sin \delta
\end{aligned}
$$

Divide by $\cos \delta \neq 0$:

$$
\operatorname{tg} \delta=\frac{3+\sqrt{3}}{4}-\frac{\sqrt{3}-3}{4} \operatorname{tg} \delta
$$

Thus,

$$
\begin{gathered}
\operatorname{tg} \delta\left(1+\frac{\sqrt{3}-3}{4}\right)=\frac{3+\sqrt{3}}{4} \quad \text { and } \\
\operatorname{tg} \delta=\frac{3+\sqrt{3}}{1+\sqrt{3}}=\frac{(3+\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{3+\sqrt{3}-3 \sqrt{3}-3}{1-3}=\sqrt{3}
\end{gathered}
$$

Therefore, $\delta=A D B \varangle=60^{\circ}$.

II. solution. In the quadrilateral $A B D C$, the angles at the opposite vertices $A$ and $D$ are right angles, so their sum is $180^{\circ}$. This means that the quadrilateral is cyclic, and a circle can be inscribed around it.

![](https://cdn.mathpix.com/cropped/2024_05_02_7f8170a68b4295f75174g-1.jpg?height=441&width=442&top_left_y=1899&top_left_x=836)

$A C B \varangle=A D B \varangle$ are inscribed angles subtending the same arc, hence they are equal, which implies that $A D B \varangle=60^{\circ}$.

Answer: 60


Let the set $A=\{1,2, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies $17 \mid(a+b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) The number of a set of binary subsets of $A$, which are pairwise disjoint and have property $P$, is how many?

Solution:
Analysis According to the problem, we classify the set using the remainder of division by 17.
Solution (1) Divide $1,2, \cdots, 366$ into 17 categories based on the remainder when divided by 17: $[0],[1], \cdots,[16]$ (where [i] represents the category of numbers that leave a remainder of $i$ when divided by 17). Since $366=17 \times 21+9$, the categories $[1], [2], \cdots, [9]$ each contain 22 numbers; the categories $[10],[11], \cdots,[16]$ and $[0]$ each contain 21 numbers.
(1) When $a, b \in[0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$;
(2) When $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{21}^{1}=462$;
(3) When $a \in[8], b \in[9]$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{22}^{1}=484$.
Therefore, the number of subsets of $A$ with property $P$ is $210+462 \times 7+484=3928$.
(2) To ensure that the binary subsets do not intersect, when $a, b \in[0]$, 10 subsets can be paired; when $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, 21 subsets can be paired for each; when $a \in[8], b \in[9]$, 22 subsets can be paired.
Thus, the number of pairwise disjoint subsets with property $P$ is $10+21 \times 7+22=179$.

Answer: 179


Calculate the definite integral:

$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x
$$

Solution:
## Solution

$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x=\int_{1 / 8}^{1} \frac{15}{(x+3) \sqrt{(x+3) x}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{x+3}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{1 \cdot x-(x+3) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{-3}{x^{2}} d x= \\
& =-\frac{3}{2 x \sqrt{(x+3) x}} d x \Rightarrow \frac{d x}{x \sqrt{(x+3) x}}=-\frac{2 d t}{3} \\
& x=\frac{1}{8} \Rightarrow t=\sqrt{\frac{\frac{1}{8}+3}{\frac{1}{8}}}=\sqrt{1+24}=5 \\
& x=1 \Rightarrow t=\sqrt{\frac{1+3}{1}}=\sqrt{4}=2
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{1 / 8}^{1} \frac{15}{\frac{x+3}{x} \cdot x \sqrt{(x+3) x}} d x=\int_{5}^{2} \frac{15}{t^{2}} \cdot\left(-\frac{2 d t}{3}\right)=\int_{5}^{2} \frac{-10 d t}{t^{2}}= \\
& =\left.\frac{10}{t}\right|_{5} ^{2}=\frac{10}{2}-\frac{10}{5}=5-2=3
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-15$ »

Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last modified: 21:56, 18 June 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 11-16

## Material from PlusPi

Answer: 3


Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution:
Examine the first term in the expression we want to evaluate, $\frac{\sin 2x}{\sin 2y}$, separately from the second term, $\frac{\cos 2x}{\cos 2y}$.

The First Term
Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$, we have:
$\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}$

The Second Term
Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2.
Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$, we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$.
Equation 1 then becomes:
$\sin x = 3\sin y$.
Equation 2 then becomes:
$\cos x = \frac{1}{2} \cos y$
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$
Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:
$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$
Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$.
So, $\sin^2 y = \frac{3}{35}$.
Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$), we can solve for $\sin^2 x$:
$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$
Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$, we can solve for $\frac{\cos 2x}{\cos 2y}$.
$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$
$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$
Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$.
Plugging in the numbers we got back into the original equation :
We get $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$.
So, the answer is $49+58=\boxed{107}$.

Answer: 107


A triangle has side lengths: 3, 4, 5. Calculate the radius of the inscribed circle (the circle inside the triangle and tangent to all three sides of the triangle).

Solution:
Let's call $A, B, C$ the vertices of the triangle, $a=BC, b=CA, c=AB$ the lengths of the three sides, $I$ and $r$ the center and the radius of the inscribed circle. The heights from $I$ to the triangles $IAB, IBC$, and $ICA$ are all equal to $r$, so the areas of these three triangles are: area $(IAB)=\frac{rc}{2}$, area $(IBC)=\frac{ra}{2}$, area $(ICA)=\frac{rb}{2}$, from which we deduce: $\operatorname{area}(ABC)=\frac{r(a+b+c)}{2}$. The triangle with sides $3,4,5$ is a right triangle, its area is: $\frac{3 \times 4}{2}=6=\frac{r(3+4+5)}{2}$. It follows that $r=1$.

The given triangle, with sides $3,4,5$, being a right triangle, we can also reason as follows. Let $D, E, F$ be the projections of $I$ onto the sides $BC, CA, AB$. These are the points of contact of the inscribed circle with the sides of the triangle. If $A$ is the vertex of the right angle, $AEIF$ is a square, and $r=IE=IF=AE=AF$. Now, $AE$ and $AF$ are easy to calculate: if we set $u=AE=AF, v=BF=BD$, $w=CD=CE, u+v=c, v+w=a, w+u=b$, then by adding: $2(u+v+w)=a+b+c$ or finally $u=\frac{-a+b+c}{2}, v=\frac{a-b+c}{2}, w=\frac{a+b-c}{2}$. We deduce: $r=u=\frac{-5+4+3}{2}=1$.

The two methods are very different, but the results are the same. Indeed, $\frac{r(a+b+c)}{2}=\left(\frac{-a+b+c}{2}\right)\left(\frac{a+b+c}{2}\right)=\frac{-a^{2}+(b+c)^{2}}{4}=\frac{bc}{2}=$ area $(ABC)$ because the square of the hypotenuse $a^{2}=b^{2}+c^{2}$.

Let's call $M$ and $N$ the midpoints of $[AB]$ and $[CD]$. Suppose first that $(AD)$ and $(BC)$ intersect at a point $S$, and consider the line $(SM)$, which intersects $(CD)$ at $P$. Using Thales' theorem and the fact that $MA=MB$, we have: $\frac{PD}{PS}=\frac{MA}{MS}=\frac{MB}{MS}=\frac{PC}{PS}$, which implies: $PC=PD$. Therefore, $P$ is the midpoint $N$ of $[CD]$, and the line $(SM)$ is none other than $\Delta: \Delta, (AD)$ and $(BC)$ are indeed concurrent at $S$.

Now suppose that $(AD)$ and $(BC)$ are parallel, and draw the line parallel to $(AD)$ and $(BC)$ passing through $M$. Let $P$ be the intersection of this parallel with $(CD)$. $DPM A$ and $CPMB$ are parallelograms, so $DP=MA=MB=CP$. Once again, $P$ is the midpoint of $[CD]$, so $(MP)$, parallel to $(AD)$ and $(BC)$, is none other than $(MN)=\Delta$.

Answer: 1


A non-empty set $S$ satisfies:
(1) $S \subseteq\{1,2, \cdots, 2 n+1\}, n \in \mathbf{N}_{+}$;
(2) If $a \in S$, then $(2 n+2-a) \in S$.

Then, the number of non-empty sets $S$ that satisfy (1) and (2) is $\qquad$

Solution:
5. $2^{n+1}-1$.

Pair the natural numbers $1,2, \cdots, 2 n+1$ into $n+1$ pairs:
$$
\{1,2 n+1\},\{2,2 n\}, \cdots,\{n, n+2\},\{n+1\} \text {. }
$$
The elements of $S$ are chosen from the above $n+1$ pairs, and there are
$$
C_{n+1}^{1}+C_{n+1}^{2}+\cdots+C_{n+1}^{n+1}=2^{n+1}-1
$$

ways to choose.

Answer: 2^{n+1}-1


In $\triangle ABC$, point $D$ is on $BC$, $\frac{BD}{DC}=\frac{1}{3}$, $E, G$ are on $AB, AD$ respectively, $\frac{AE}{EB}=\frac{2}{3}$, $\frac{AG}{GD}=\frac{1}{2}$, $EG$ intersects $AC$ at point $F$, find $\frac{AF}{FC}$.

Solution: 1. Extend $C B, F E$ to intersect at $H$. For $\triangle A D B$ and the transversal $G E H$, we have $\frac{A G}{G D} \cdot \frac{D H}{H B} \cdot \frac{B E}{E A}=\frac{1}{2} \cdot \frac{D H}{H B} \cdot \frac{3}{2}=1$, so $\frac{D H}{H B}=\frac{4}{3}$, which means $\frac{C H}{H D}=\frac{7}{4}$. For $\triangle A C D$ and the transversal $F G H$, we have $\frac{A F}{F C} \cdot \frac{C H}{H D} \cdot \frac{D G}{G A}=\frac{A F}{F C} \cdot \frac{7}{4} \cdot \frac{2}{1}=1$, solving for $\frac{A F}{F C}=\frac{2}{7}$.

Answer: \frac{2}{7}


Write in ascending order the numbers: $\sqrt{121}$, $\sqrt[3]{729}$, and $\sqrt[4]{38416}$.

Solution:
Factoring the numbers and extracting the roots we have:

$$
\begin{aligned}
& \sqrt{121}=\sqrt{11^{2}}=11 \\
& \sqrt[3]{729}=\sqrt[3]{9^{3}}=9 \\
& \sqrt[4]{38416}=\sqrt[4]{2^{4} \times 7^{4}}=2 \times 7=14
\end{aligned}
$$

Therefore, in ascending order we have: $\sqrt[3]{729}, \sqrt{121}, \sqrt[4]{38416}$.

Answer: \sqrt[3]{729},\sqrt{121},\sqrt[4]{38416}


Carla wrote on the blackboard the integers from 1 to 21. Diana wants to erase some of them so that the product of the remaining numbers is a perfect square.

a) Show that Diana must necessarily erase the numbers 11, 13, 17, and 19 to achieve her goal.

b) What is the minimum number of numbers that Diana must erase to achieve her goal?

Solution:
Solution

a) If Diana decides not to erase the number 11, then the product of the remaining numbers will be of the form $P=11 \times A$. Since 11 is the only multiple of 11 among the numbers written by Carla, $A$ is the product of numbers not divisible by 11, and thus $A$ is not a multiple of 11. Therefore, $P$ would be a multiple of 11 but not a multiple of $11^{2}$, so $P$ would not be a perfect square. Similarly, we can see that Diana must also erase the numbers 13, 17, and 19.

b) If Diana erases only the numbers 11, 13, 17, and 19, the product of the remaining 17 numbers would be

$$
1 \times 2 \times 3 \times \cdots \times 10 \times 12 \times 14 \times 15 \times 16 \times 18 \times 20 \times 21=\left(2^{9} \times 3^{4} \times 5^{2} \times 7\right)^{2} \times 21
$$

This number is not a perfect square, so she needs to erase at least one more number. From the factorization above, it is easy to see that if Diana also erases the number 21, she will achieve her goal. Therefore, the smallest number of numbers that Diana must erase is 5.

Answer: 5


Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be 1, and the endpoints of the line segment $M N$ be $M$ on the ray $A A_{1}$, and point $N$ on the ray $B C$, and $M N$ intersects the edge $C_{1} D_{1}$ at point $L$. Then the minimum value of $M N$ is $\qquad$

Solution:
7.3.

Let $A M=x, B N=y$. Then
$$
M N=\sqrt{x^{2}+y^{2}+1} \text {. }
$$

Draw $L L_{1} \perp D C$ at point $L_{1}$. Thus,
$$
\begin{array}{l}
\frac{1}{x-1}=\frac{A A_{1}}{M A_{1}}=\frac{L M}{L N}=\frac{N C}{C B}=\frac{y-1}{1} \\
\Rightarrow x y=x+y \\
\Rightarrow x^{2} y^{2} \geqslant 4 x y \\
\Rightarrow x y \geqslant 4 .
\end{array}
$$

Therefore, $M N=\sqrt{x^{2}+y^{2}+1} \geqslant \sqrt{2 x y+1} \geqslant 3$,
with equality holding if and only if $x=y=2$.

Answer: 3


The number of the boys taking part of this year competition is 55% of the number of all participants. The ratio of the number of juniors boys to the number of senior boys is equal to the ratio of the number of juniors to the number of seniors. Find the ratio of the number of junior boys to the number of junior girls.

Solution:
1. Let \( b \) be the number of boys and \( g \) be the number of girls participating in the competition. According to the problem, the number of boys is 55% of the total number of participants. Therefore, we have:
   \[
   b = \frac{55}{100}(b + g)
   \]
   Simplifying this equation:
   \[
   b = 0.55(b + g)
   \]
   \[
   b = 0.55b + 0.55g
   \]
   \[
   b - 0.55b = 0.55g
   \]
   \[
   0.45b = 0.55g
   \]
   \[
   \frac{b}{g} = \frac{0.55}{0.45} = \frac{11}{9}
   \]

2. Let \( jb \) be the number of junior boys, \( sb \) be the number of senior boys, \( jg \) be the number of junior girls, and \( sg \) be the number of senior girls. According to the problem, the ratio of the number of junior boys to senior boys is equal to the ratio of the number of juniors to seniors. Therefore, we have:
   \[
   \frac{jb}{sb} = \frac{j}{s}
   \]
   This implies:
   \[
   jb = \frac{j}{s} sb
   \]

3. We need to find the ratio of the number of junior boys to the number of junior girls. From the previous step, we know:
   \[
   \frac{jb}{jg} = \frac{b}{g} = \frac{11}{9}
   \]

The final answer is \( \boxed{ \frac{11}{9} } \).

Answer: \frac{11}{9}


Entrepreneurs Vasiliy Petrovich and Pyotr Gennadiyevich opened a clothing factory called "ViP". Vasiliy Petrovich invested 200 thousand rubles, and Pyotr Gennadiyevich invested 350 thousand rubles. The factory proved to be successful, and after a year, Anastasia Alekseevna approached them with an offer to buy part of the shares. They agreed, and after the deal, each of them owned one-third of the company's shares. Anastasia Alekseevna paid 1,100,000 rubles for her share. Determine which of the entrepreneurs is entitled to a larger portion of this money? In your answer, write the amount he will receive.

Solution:
Answer: 1,000,000. Anastasia Alekseevna paid 1,100,000 for her share, which is one-third, so all the shares of the factory are worth 3,300,000. Let the shares of Vasiliy Petrovich be worth X rubles, then the shares of Petr Gennadiyevich are worth $3,300,000 - X$ rubles. It should be that $200 / 350 = X / (3,300,000 - X)$. From this we get: $350X = 200 * 3,300,000 - 200X$ $550X = 660,000,000$

$X = 1,200,000$ rubles - the value of Vasiliy Petrovich's shares. After the deal, he was left with a third of the shares worth 1,100,000, which means he sold Anastasia Alekseevna shares worth 100,000. The remaining shares she bought for 1,000,000 from Petr Gennadiyevich.

Answer: 1,000,000


Solution:
Solution: The original expression is transformed into
$$
y=\sqrt{(x-6)^{2}+3^{2}}-\sqrt{(x-1)^{2}+2^{2}},
$$

where the fixed points are $A(6,3), B(1,2)$, and point $A$ is higher than point $B$.
$$
\begin{aligned}
\because & |A B|=: \sqrt{(6-1)^{2}+(3-1)^{2}}=\sqrt{26}, \\
& \left|x_{2}-x_{1}\right|=5,
\end{aligned}
$$
$\therefore$ The range is $(-5, \sqrt{26}]$.

Answer: (-5, \sqrt{26}]


Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$.

Solution:
Answer: 12045

Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest odd divisors of the numbers from $n+1$ to $2n$ are $n$ different odd numbers that do not exceed $2n$. Thus, these numbers are $1, 3, 5, \ldots, 2n-1$. If we add the number 220 to the set of numbers, the desired sum will be $1+3+5+\ldots+219-A(220)=\frac{1+219}{2} \cdot 110-55=110^2-55=12045$.

## B-2

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(113)+A(114)+\ldots+A(222)+A(223)$.

Answer: 12537

## B-3

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(115)+A(116)+\ldots+A(226)+A(227)$.

Answer: 12939

## B-4

Let $A(n)$ denote the largest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9$, $A(64)=1$. Find the sum $A(117)+A(118)+\ldots+A(230)+A(231)$.

Answer: 13427

Answer: 12045


Let $ABC$ be a triangle. Find the point $D$ on its side $AC$ and the point $E$ on its side $AB$ such that the area of triangle $ADE$ equals to the area of the quadrilateral $DEBC$, and the segment $DE$ has minimum possible length.

Solution:
1. **Given**: We have a triangle \(ABC\). We need to find points \(D\) on side \(AC\) and \(E\) on side \(AB\) such that the area of triangle \(ADE\) equals the area of quadrilateral \(DEBC\), and the segment \(DE\) has the minimum possible length.

2. **Area Relationship**:
   - The area of triangle \(ABC\) is given by:
     \[
     \text{Area}(ABC) = \frac{1}{2} AB \cdot AC \cdot \sin A
     \]
   - The area of triangle \(ADE\) is given by:
     \[
     \text{Area}(ADE) = \frac{1}{2} AD \cdot AE \cdot \sin A
     \]
   - Since the area of triangle \(ADE\) equals the area of quadrilateral \(DEBC\), we have:
     \[
     \text{Area}(ADE) = \frac{1}{2} \text{Area}(ABC)
     \]
   - Therefore:
     \[
     \frac{1}{2} AD \cdot AE \cdot \sin A = \frac{1}{2} \left( \frac{1}{2} AB \cdot AC \cdot \sin A \right)
     \]
   - Simplifying, we get:
     \[
     AD \cdot AE \cdot \sin A = \frac{1}{2} AB \cdot AC \cdot \sin A
     \]
   - Dividing both sides by \(\sin A\), we obtain:
     \[
     AD \cdot AE = \frac{1}{2} AB \cdot AC
     \]

3. **Minimizing \(DE\)**:
   - To minimize the length of \(DE\), we need to consider the geometric mean. The minimum length of \(DE\) occurs when \(AD = AE\).
   - Setting \(AD = AE\), we have:
     \[
     AD^2 = \frac{1}{2} AB \cdot AC
     \]
   - Solving for \(AD\), we get:
     \[
     AD = \sqrt{\frac{1}{2} AB \cdot AC}
     \]

4. **Conclusion**:
   - The points \(D\) and \(E\) are such that \(AD = AE = \sqrt{\frac{1}{2} AB \cdot AC}\).
   - This ensures that the area of triangle \(ADE\) is half the area of triangle \(ABC\) and the segment \(DE\) is minimized.

The final answer is \( \boxed{ AD = AE = \sqrt{\frac{1}{2} AB \cdot AC} } \).

Answer: AD = AE = \sqrt{\frac{1}{2} AB \cdot AC}


Label each vertex of a cube with one of the integers $1 \sim 8$, with each integer used only once, and the sum of the four numbers on each face being equal. Then there are ( ) different arrangements (if two arrangements can be made to coincide by rotation, they are considered the same arrangement).
(A) 1
(B) 3
(C) 6
(D) 12
(E) 24

Solution:
18. C.

Notice that, the sum of the numbers on each face is equal, and the total sum of the numbers is
$$
1+2+3+4+5+6+7+8=36 \text {. }
$$

Therefore, the sum of the four numbers on each face is 18, and the sum of the numbers at the two vertices of non-coplanar parallel edges (opposite edges) is equal.

Consider the extreme case: If 1 and 8 are not on the same edge, then 1 and 8 are on opposite edges.

Let the other two vertices on the opposite edges be filled with $x, y(x, y \in \{2,3, \cdots, 7\})$.
Thus, $1+x=8+y$
$\Rightarrow 7=8-1=x-y \leqslant 7-2=5$.
Contradiction.
Therefore, 1 and 8 must be on the same edge.
Thus, the sum of the numbers at the two vertices on the other edge is also 9.
Hence, the pairs of numbers on the edges are $1-8$, $2-7,3-6,4-5$.

Since the sum of the four numbers on each face is 36, $(8,2,3,5)$ is on one face, and $(1,7,6,4)$ is on another face, as shown in Figure 6.

Answer: C


Find how many natural numbers $k \in\{1,2,3, \ldots, 2022\}$ have the property that, if 2022 real numbers are written on a circle such that the sum of any $k$ consecutive numbers is equal to 2022, then all 2022 numbers are equal.

Solution:
Solution. Let $k \in \{1,2,3, \ldots, 2022\}$ be a number with the property that, given the real numbers $x_{1}, x_{2}, \ldots, x_{2022}$ written in a circle, the sum of any $k$ consecutive numbers is equal to 2022. For any $n \in \mathbb{N}$, we define $x_{n}=x_{r}$, where $r \in \{1,2,3, \ldots, 2022\}$, such that $n \equiv r \pmod{2022}$.

(A) For any $i \in \mathbb{N}$, we have

$$
x_{i} + x_{i+1} + \ldots + x_{i+k-1} = x_{i+1} + x_{i+2} + \ldots + x_{i+k} = 2022
$$

from which it follows that $x_{i} = x_{i+k}$

2 points

(B) If $(2022, k) = d \geq 2$, noting $x = \frac{2022 \cdot d}{k}$ and writing on the circle, in order, the numbers

$$
x, \underbrace{0,0, \ldots, 0}_{d \text{ times } - 1}, x, \underbrace{0,0, \ldots, 0, \ldots x}_{\text{d times } - 1}, \underbrace{0,0, \ldots, 0}_{\text{d times } - 1}
$$

we observe that the sum of any $k$ consecutive numbers is equal to 2022, without the numbers being equal.

Therefore, the numbers $k$ for which $(2022, k) \neq 1$ are not solutions.

(C) We show that all numbers $k \in \{1,2,3, \ldots, 2022\}$ for which $(2022, k) = 1$ are solutions. Since $(k, 2022) = 1$, there exist $u, v \in \mathbb{N}$ such that $k \cdot u = 2022 v + 1$. Then, for any $m \in \mathbb{N}$, we have:

$$
x_{m} = x_{m + k \cdot u} = x_{m + 2022 v + 1} = x_{m+1}
$$

thus all 2022 numbers are equal. 2 points

(D) In total, there are $\varphi(2022) = 672$ solutions. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_cc89f2f54d84ee4a810cg-5.jpg?height=271&width=268&top_left_y=130&top_left_x=858)

## First selection barrier for OBMJ Iaşi, April 19, 2022  Solutions and grading criteria

Answer: 672


a) Let $n \in \mathbb{N}^{*}$. Calculate $\int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x$.

b) Calculate $\lim _{n \rightarrow \infty} \int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} x}{x^{2}+x+1} d x$.

Solution:
Solution. a) $\int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x=\int_{\frac{1}{n}}^{n} \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x=\left.\frac{2}{\sqrt{3}} \operatorname{arctg} \frac{2 x+1}{\sqrt{3}}\right|_{\frac{1}{n}} ^{n}=\frac{2}{\sqrt{3}}\left(\operatorname{arctg} \frac{2 n+1}{\sqrt{3}}-\operatorname{arctg} \frac{\frac{2}{\sqrt{3}}+1}{\sqrt{3}}\right)$

b) Let $I_{n}=\int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} x}{x^{2}+x+1} d x$. Since $\operatorname{arctg} x+\operatorname{arctg} \frac{1}{x}=\frac{\pi}{2}, \quad \forall x>0$, we have:

![](https://cdn.mathpix.com/cropped/2024_06_07_eb436281f7e263db6a33g-2.jpg?height=214&width=1607&top_left_y=1687&top_left_x=194)

With the change of variable $t=\frac{1}{x}$ we get $\int_{\frac{1}{n}}^{n} \frac{\operatorname{arctg} \frac{1}{x}}{1+\frac{1}{x}+\frac{1}{x^{2}}} \cdot \frac{-1}{x^{2}} d x=\int_{n}^{\frac{1}{n}} \frac{\operatorname{arctg} t}{t^{2}+t+1} d t=-I_{n}$, thus

$I_{n}=\frac{\pi}{2} \int_{\frac{1}{n}}^{n} \frac{1}{x^{2}+x+1} d x-I_{n}$. Considering the result from part a), we get $I_{n}=\frac{\pi}{2 \sqrt{3}}\left(\operatorname{arctg} \frac{2 n+1}{\sqrt{3}}-\operatorname{arctg} \frac{\frac{2}{n}+1}{\sqrt{3}}\right)$.

$\lim _{n \rightarrow \infty} I_{n}=\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\frac{\pi^{2}}{6 \sqrt{3}}$.

Scoring.

| a) Calculation of the integral | $2 \mathrm{p}$ |
| :--- | :--- |
| b) Calculation of $I_{n}$ | $4 \mathrm{p}$ |
| Calculation of $\lim _{n \rightarrow \infty} I_{n}$ | $1 \mathrm{p}$ |

Answer: \frac{\pi^{2}}{6\sqrt{3}}


Let three distinct complex numbers $z_{1}, z_{2}, z_{3}$ satisfy the equation $4 z_{1}^{2}+5 z_{2}^{2}+5 z_{3}^{2}=4 z_{1} z_{2}+6 z_{2} z_{3}+4 z_{3} z_{1}$. Denote the lengths of the sides of the triangle formed by $z_{1}, z_{2}, z_{3}$ on the complex plane, in ascending order, as $a, b, c$, then $a: b: c=$ $\qquad$.

Solution:
$$
\begin{array}{l}
4 z_{1}^{2}+5 z_{2}^{2}+5 z_{3}^{2}=4 z_{1} z_{2}+6 z_{2} z_{3}+4 z_{3} z_{1} \\
\Rightarrow 4\left(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}\right)+\left(z_{2}-z_{3}\right)^{2}=0 \\
\Rightarrow 2\left(z_{1}-z_{2}\right)^{2}+3\left(z_{2}-z_{3}\right)^{2}+2\left(z_{3}-z_{1}\right)^{2}=0
\end{array}
$$

Let $u=z_{1}-z_{2}, v=z_{2}-z_{3}, w=z_{3}-z_{1}$,
$$
\begin{array}{l}
\text { then }\left\{\begin{array}{l}
u+v+w=0, \\
2 u^{2}+3 v^{2}+2 w^{2}=0
\end{array} \Rightarrow 2 u^{2}+3 v^{2}+2(u+v)^{2}=4 u^{2}+5 v^{2}+4 u v=0\right. \\
\Rightarrow 4\left(\frac{u}{v}\right)^{2}+4 \cdot \frac{u}{v}+5=0 \Rightarrow \frac{|u|}{|v|}=\left|\frac{u}{v}\right|=\sqrt{\frac{u}{v} \cdot \overline{\left(\frac{u}{v}\right)}}=\frac{\sqrt{5}}{2} \text {, similarly we get } \\
2 u^{2}+3(u+w)^{2}+2 w^{2}=5 u^{2}+5 w^{2}+6 u w=0 \Rightarrow 5\left(\frac{u}{w}\right)^{2}+6 \cdot \frac{u}{w}+5=0 \Rightarrow \frac{|u|}{|w|}=1 \text {. } \\
\end{array}
$$

Therefore, $|u|:|v|:|w|=\sqrt{5}: 2: \sqrt{5} \Rightarrow a: b: c=2: \sqrt{5}: \sqrt{5}$.

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


How many positive integers $m$ are there for which the straight line passing through points $A(-m, 0)$ and $B(0,2)$ and also passes through the point $P(7, k)$, where $k$ is a positive integer?

Solution: Let the slope of the variable straight line be $a$. Then its equation is: $y=a x+2$ It passes through $A(-m, 0)$ and $P(7, k):\left\{\begin{aligned}-a m+2 & =0 \cdots(1) \\ 7 a+2 & =k \cdots(2)\end{aligned}\right.$ $7(1)+m(2): 14+2 m=k m \Rightarrow m(k-2)=14$ $m=1, k=16$ or $m=2, k=9$ or $m=7, k=4$ or $m=14, k=3$ Number of positive integral values of $m$ is 4 .

Answer: 4


If the value of the fraction $\frac{|x|-2}{3 x-2}$ is negative, then the range of $x$ is ( ).
A. $\frac{2}{3}<x<2$.
B. $x>\frac{2}{3}$ or $x<-2$.
C. $-2<x<2$ and $x \neq \frac{2}{3}$.
D. $\frac{2}{3}<x<2$ or $x<-2$.

Solution: Answer: D

Answer: D


Solution:
## Solution

Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:

$\left|\begin{array}{ccc}x-1 & y-5 & z-(-7) \\ -3-1 & 6-5 & 3-(-7) \\ -2-1 & 7-5 & 3-(-7)\end{array}\right|=0$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-5 & z+7 \\
-4 & 1 & 10 \\
-3 & 2 & 10
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
1 & 10 \\
2 & 10
\end{array}\right|-(y-5) \cdot\left|\begin{array}{cc}
-4 & 10 \\
-3 & 10
\end{array}\right|+(z+7) \cdot\left|\begin{array}{ll}
-4 & 1 \\
-3 & 2
\end{array}\right|=0 \\
& (x-1) \cdot(-10)-(y-5) \cdot(-10)+(z+7) \cdot(-5)=0 \\
& -10 x+10+10 y-50-5 z-35=0 \\
& -10 x+10 y-5 z-75=0 \\
& 2 x-2 y+z+15=0
\end{aligned}
$$

The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:

$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$

Find:

$$
d=\frac{|2 \cdot 1-2 \cdot(-1)+2+15|}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}=\frac{|2+2+2+15|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=\frac{21}{3}=7
$$

## Problem Kuznetsov Analytic Geometry 8-31

Answer: 7


If the quadratic equation with real coefficients $a x^{2}+b x+c=0$ has 2 imaginary roots $x_{1}, x_{2}$, and $x_{1}^{3} \in \mathbb{R}$, then $\frac{a c}{b^{2}}=$

Solution:
Answer: 1
Explanation: Note that $x_{2}=\overline{x_{1}}$, from $x_{1}^{3} \in R \Rightarrow x_{1}^{3}=\overline{x_{1}^{3}}=\left(\overline{x_{1}}\right)^{3}=x_{2}^{3} \Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)=0$
$$
\Rightarrow x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=0 \Rightarrow\left(x_{1}+x_{2}\right)^{2}=x_{1} x_{2} \Rightarrow\left(-\frac{b}{a}\right)^{2}=\frac{c}{a} \Rightarrow b^{2}=a c \Rightarrow \frac{a c}{b^{2}}=1
$$

Answer: 1


Calculate the area of a rectangular trapezoid if its acute angle is $60^{\circ}$, the smaller base is $a$, and the larger lateral side is $b$.

Solution:
Solution.

The height of the trapezoid is $\frac{b \sqrt{3}}{2}$, and the larger base is $a+\frac{b}{2}$.

Therefore, its area $S=\frac{1}{2}\left(a+a+\frac{b}{2}\right) \frac{b \sqrt{3}}{2}=\frac{(4 a+b) b \sqrt{3}}{8}$.

Answer: $(4 a+b) b \sqrt{3} / 8$.

Answer: \frac{(4a+b)b\sqrt{3}}{8}


What is the probability that two cards randomly selected (without replacement) from a standard 52-card deck are neither of the same value nor the same suit?

Solution:
Answer: $\frac{12}{17}$
Solution: After choosing a first card, the second needs to be in one of the other three suits and of a different value. Hence, the answer is $\frac{3 \cdot 12}{52-1}=\frac{12}{17}$.

Answer: \frac{12}{17}


Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$

Solution:
6. Let $(a, b)=d, a=d x, b=d y$, then $(x, y)=1$. Substituting into the condition, we get
$$1+9 x y+9(x+y)=7 d x y$$

Thus, $\square$
$$9<7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y} \leqslant 9+9 \times 2+1=28 \text {. }$$

From this, we know that $d=2,3$ or 4. Substituting these values into the indeterminate equation for $x, y$, we can find that $(a, b)=(4,4),(4,38)$ or $(38,4)$.

Answer: (a, b)=(4,4),(4,38),(38,4)


Solve the inequality $1-27^{x} \leqslant 6 \cdot 9^{3 x}$.

Solution:
## Solution.

Let's write the powers from the inequality with base 3

$$
1-3^{3 x} \leqslant 6 \cdot 3^{6 x}
$$

If we introduce the substitution $t=3^{3 x}$, we get the quadratic inequality $6 t^{2}+t-1 \geqslant 0 . \quad 1$ point

Solving this, we get $t \leqslant-\frac{1}{2}$ or $t \geqslant \frac{1}{3}$.

However, since $t>0$, it must be $t \geqslant \frac{1}{3}$, so we have $3^{3 x} \geqslant 3^{-1}$

and finally $x \geqslant-\frac{1}{3}$.

Answer: x\geqslant-\frac{1}{3}


DeterminaÈi toate funcÈiile $f: R \rightarrow(0, \infty)$, primitivabile, ce verificÄ relaÈia: $F(x)+\ln (f(x))=\ln \left(1+\frac{x}{\sqrt{1+x^{2}}}\right)$, $\forall x \in R$, unde $F: R \rightarrow R$ este o primitivÄ a lui $f$ Èi $F(0)=0$.

Solution:
## Subiectul IV. (20 puncte )

$$
\begin{aligned}
& F(x)=\ln \frac{x+\sqrt{x^{2}+1}}{f(x) \cdot \sqrt{x^{2}+1}} \Leftrightarrow e^{F(x)} \cdot f(x)=1+\frac{x}{\sqrt{x^{2}+1}} \Leftrightarrow\left(e^{F(x)}\right)^{\prime}=1+\frac{x}{\sqrt{x^{2}+1}} \\
& \text { Deci } e^{F(x)}=x+\sqrt{x^{2}+1}+C, F(0)=0 \Rightarrow C=0 \Rightarrow F(x)=\ln \left(x+\sqrt{x^{2}+1}\right) \Rightarrow f(x)=\frac{1}{\sqrt{x^{2}+1}}
\end{aligned}
$$

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


During the physical education class, the entire class lined up by height (all children have different heights). Dima noticed that the number of people taller than him is four times the number of people shorter than him. And LÑnya noticed that the number of people taller than him is three times less than the number of people shorter than him. How many people are there in the class, if it is known that there are no more than 30 people?

Solution:
Answer: 21.

Solution. Let $x$ be the number of people who are shorter than Dima. Then, the total number of students in the class is $x$ (people who are shorter than Dima) $+4 x$ (people who are taller than Dima) +1 (Dima) $=5 x+1$ (total number of people in the class).

Let $y$ be the number of people who are taller than Lёnya. Then, the total number of students in the class is $y$ (people who are taller than Lёnya) $+3 y$ (people who are shorter than Lёnya) +1 (Lёnya) $=4 y+1$ (total number of people in the class).

Then, if we subtract 1 from the total number of children in the class, the resulting number will be divisible by both 4 and 5. That is, it will be divisible by 20. In the required range, there is only one such number - 20, so there are 21 people in the class.

Answer: 21


By how much should ${ }^{4} \log 8$ be multiplied to obtain ${ }^{32} \log 8$?

Solution:
Given that ${ }^{4} \log 8=\frac{3}{2}$ and ${ }^{32} \log 8=\frac{3}{5}$, in the context of the problem,

$$
\frac{3}{2} \cdot x=\frac{3}{5} \text { and thus } x=\frac{2}{5}
$$

(József Kiss, Pápa.)

The problem was also solved by: Eckhart F., Fekete M., Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Krisz A., Munk L., Pichler S., Sárközy E., Schuster Gy., Szécsi I., Tandlich E., Tóth B., Winkler J.

Answer: \frac{2}{5}


A convex quadrilateral's diagonals form a $45^{\circ}$ angle. Drop a perpendicular from each vertex of the quadrilateral to the line of the diagonal connecting the two adjacent vertices. How does the area of the quadrilateral formed by the feet of the perpendiculars compare to the area of the original quadrilateral?

Solution:
Solution. Let the vertices of the quadrilateral be $A, B, C, D$, the feet of the perpendiculars be $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$, and the intersection of the diagonals be $K$. In the right triangle $A A^{\prime} K$,

$$
\cos 45^{\circ}=\frac{K A^{\prime}}{K A}, \quad \text { from which } \quad K A^{\prime}=\frac{\sqrt{2}}{2} K A
$$

Similarly,

$$
K B^{\prime}=\frac{\sqrt{2}}{2} K B, \quad K C^{\prime}=\frac{\sqrt{2}}{2} K C
$$

and

$$
K D^{\prime}=\frac{\sqrt{2}}{2} K D
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_346873209cca15f32ed6g-1.jpg?height=688&width=553&top_left_y=667&top_left_x=775)

$\triangle D^{\prime} K A^{\prime} \sim \triangle D K A$, since the ratio of two sides and the included angle are the same. The similarity ratio is $\frac{\sqrt{2}}{2}$, and the ratio of the areas of the two triangles is the square of this, which is $\frac{1}{2}$.

Similarly, $\triangle A^{\prime} K B^{\prime} \sim \triangle A K B$, $\triangle B^{\prime} K C^{\prime} \sim \triangle B K C$, and $\triangle C^{\prime} K D^{\prime} \sim \triangle C K D$, with the similarity ratio being $\frac{\sqrt{2}}{2}$ and the area ratio being $\frac{1}{2}$ in each case.

Since the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the sum of the areas of the four small triangles, and the area of $A B C D$ is the sum of the areas of the four large triangles, the ratio of the areas of the quadrilaterals is also $\frac{1}{2}$.

Answer: \frac{1}{2}


Represent in the form of an irreducible fraction

$$
6 \frac{3}{2015} \times 8 \frac{11}{2016}-11 \frac{2012}{2015} \times 3 \frac{2005}{2016}-12 \times \frac{3}{2015}
$$

Solution:
Answer: $11 / 112$.

Solution. First solution. Let $a=3 / 2015, b=11 / 2016$. Then

$$
\begin{aligned}
& 6 \frac{3}{2015} \times 8 \frac{11}{2016}-11 \frac{2012}{2015} \times 3 \frac{2005}{2016}-12 \times \frac{3}{2015}=(6+a)(8+b)-(12-a)(4-b)-12 a= \\
& =48+8 a+6 b+a b-48+4 a+12 b-a b-12 a=18 b=18 \cdot \frac{11}{2016}=\frac{18 \cdot 11}{18 \cdot 112}=\frac{11}{112}
\end{aligned}
$$

Second solution. After replacing the mixed fractions with common fractions, we get that the original expression equals

$$
\begin{gathered}
\frac{12093}{2015} \cdot \frac{16139}{2016}-\frac{24177}{2015} \cdot \frac{8053}{2016}-\frac{36}{2015}=\frac{195168927-194697381-72576}{2015 \cdot 2016}= \\
=\frac{398970}{2015 \cdot 2016}=\frac{2015 \cdot 18 \cdot 11}{2015 \cdot 18 \cdot 112}=\frac{11}{112}
\end{gathered}
$$

Answer: \frac{11}{112}


In $\triangle A B C$, if $(\sin A+\sin B)(\cos A+\cos B)=2 \sin C$, then
A. $\triangle A B C$ is an isosceles triangle but not necessarily a right triangle
B. $\triangle A B C$ is a right triangle but not necessarily an isosceles triangle
C. $\triangle A B C$ is neither an isosceles triangle nor a right triangle
D. $\triangle A B C$ is both an isosceles triangle and a right triangle

Solution:
1. A Prompt: Left side $=\sin A \cos A+\sin A \cos B+\cos A \sin B+\sin B \cos B$
$$
\begin{array}{l}
=\frac{1}{2}(\sin 2 A+\sin 2 B)+\sin (A+B) \\
=\sin (A+B) \cos (A-B)+\sin (A+B)
\end{array}
$$

Right side $=2 \sin \left[180^{\circ}-(A+B)\right]=2 \sin (A+B)$
Thus, $\sin (A+B) \cos (A-B)+\sin (A+B)=2 \sin (A+B)$
which simplifies to $\sin (A+B)[\cos (A-B)-1]=0$
Since $\sin (A+B)>0$
it follows that $\cos (A-B)=1$, so $A=B$
Taking $A=B=30^{\circ}, C=120^{\circ}$ and substituting into the condition, we find that it satisfies the condition.
Therefore, $\triangle A B C$ is an isosceles triangle, but not necessarily a right triangle.

Answer: A


The radius of the circumscribed circle of a triangle is 2, and the lengths of all altitudes are integers. Find the sides of the triangle.

Solution:
3.2. Answer: $2 \sqrt{3}, 2 \sqrt{3}, 2 \sqrt{3}$.

Firstly, note that the height of the triangle must be less than the diameter of the circumscribed circle, so all heights have a length of 1, 2, or 3.

From the triangle inequality $a < \frac{1}{2} + \frac{1}{3}$, which contradicts the inequality we have derived. Therefore, among the heights, there will be pairs of equal heights, meaning the triangle is isosceles. Consider the height $h$ at the vertex of the isosceles triangle. By the Pythagorean theorem, the base will be $2 \sqrt{R^{2}-(R-h)^{2}} = 2 \sqrt{4 h - h^{2}}$. Depending on $h$, there are three possible cases:

$h=1$. It is not difficult to verify that this is a triangle with angles $120^{\circ}, 30^{\circ}, 30^{\circ}$ and a side length of 2. The height dropped to the side is $\sqrt{3}$, which is not an integer.

$h=2$. This is a triangle with angles $90^{\circ}, 45^{\circ}, 45^{\circ}$ and a side length of $2 \sqrt{2}$, the height to the side is also $2 \sqrt{2}$, which is not an integer.

$h=3$. This is an equilateral triangle with a side length of $2 \sqrt{3}$, all its heights are 3, which fits the condition of the problem.

Answer: 2\sqrt{3},2\sqrt{3},2\sqrt{3}


If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$

Solution:
4. 2 .

Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\Rightarrow 3^{n} \equiv 1(\bmod 8)
$$
$\Rightarrow n$ must be even.
Let $n=2+2 k(k \in \mathbf{N})$. Then
$$
2016+3^{n}=9\left(224+3^{2 k}\right) \text {. }
$$

So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square.
Let $224+\left(3^{k}\right)^{2}=t^{2}\left(t \in \mathbf{Z}_{+}\right)$. Then
$$
\left(t+3^{k}\right)\left(t-3^{k}\right)=224=2^{5} \times 7 \text {, }
$$

and $t+3^{k}>t-3^{k}$, both are even.
So $\left(t+3^{k}, t-3^{k}\right)$
$$
=\left(2^{4} \times 7,2\right),\left(2^{3} \times 7,2^{2}\right),\left(2^{2} \times 7,2^{3}\right),\left(2^{4}, 2 \times 7\right) \text {. }
$$

Solving gives $k=0$, i.e., $n=2$.

Answer: 2


The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$.

Solution:
Solution. Let

$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$

Replace the element $x_{1}$ with the sum of the others. Then

$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$

Reasoning similarly for the other elements, we get that

$$
2 x_{k}=S, \quad k=1 \ldots n \text {. }
$$

Thus, all elements of the set are equal to each other. Since the sum does not change when one addend is replaced, this addend must be equal to what it is replaced with, i.e.,

$$
x_{1}=x_{2}+x_{3}+\cdots+x_{n}
$$

Considering the equality of the elements, we get $x_{1}=(n-1) x_{1}$, hence, $x_{1}=0$. Therefore, the product of all numbers in the set $M$ is 0.

## Answer: 0.

Answer: 0


Nikola had one three-digit and one two-digit number. Each of these numbers was positive and composed of mutually different digits. The difference between Nikola's numbers was 976.

What was their sum?

Solution:
The largest three-digit number with distinct digits is 987. If this were one of Niko's numbers, the other would have to be 11 (to make the difference 976). This is a two-digit number, but it is not composed of distinct digits, so it could not have been Niko's number.

The next smaller three-digit number with distinct digits is 986. If this were one of Niko's numbers, the other would have to be 10. This is a two-digit number with distinct digits, so it could be a possible Niko's number.

No other pair is feasible, as 10 is the smallest two-digit number. Niko's numbers were 986 and 10, so their sum was 996.

Notes. One can also consider from the smallest two-digit number: If one of Niko's numbers was 10, the other would have to be 986 (to make the difference 976). This is a three-digit number with distinct digits, so it could be a possible Niko's number. For other two-digit numbers with distinct digits (12, 13, 14, 15, ...), the second number would either contain two identical digits (988, 989, 990, 991, ...) or would not be a three-digit number (for two-digit numbers greater than 23). Therefore, 10 and 986 were Niko's numbers.

It is not difficult to identify the suitable pair of numbers. Part of the problem is also the analysis of other possibilities, i.e., the justification that there are no more such pairs. A solution without a proper analysis cannot be rated at the highest level.

#

Answer: 996


The sum of the (decimal) digits of a natural number $n$ equals $100$, and the sum of digits of $44n$ equals $800$. Determine the sum of digits of $3n$.

Solution:
1. Let \( n \) be a natural number such that the sum of its digits is 100. We can represent \( n \) in its decimal form as:
   \[
   n = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0
   \]
   where \( a_i \) are the digits of \( n \). Therefore, the sum of the digits of \( n \) is:
   \[
   a_k + a_{k-1} + \ldots + a_1 + a_0 = 100
   \]

2. Given that the sum of the digits of \( 44n \) equals 800, we need to analyze the representation of \( 44n \). We can write:
   \[
   44n = 44 \cdot (10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0)
   \]
   Expanding this, we get:
   \[
   44n = 10^{k+1} (4a_k) + 10^k (4a_k + 4a_{k-1}) + 10^{k-1} (4a_{k-1} + 4a_{k-2}) + \ldots + 10 (4a_1 + 4a_0) + 4a_0
   \]

3. The sum of the digits in this representation is:
   \[
   4a_k + 4a_k + 4a_{k-1} + 4a_{k-1} + \ldots + 4a_0 + 4a_0 = 800
   \]
   Simplifying, we get:
   \[
   8(a_k + a_{k-1} + \ldots + a_1 + a_0) = 800
   \]
   Since \( a_k + a_{k-1} + \ldots + a_1 + a_0 = 100 \), we have:
   \[
   8 \cdot 100 = 800
   \]

4. To ensure there are no carries in the sum of the digits of \( 44n \), we must have \( 4a_i + 4a_{i-1} < 10 \) for all \( i \). This implies:
   \[
   a_i + a_{i+1} \leq 2
   \]
   for all \( i \).

5. Now, we need to determine the sum of the digits of \( 3n \). Since \( a_i + a_{i+1} \leq 2 \), multiplying by 3 will not cause any carries. Therefore, the sum of the digits of \( 3n \) is:
   \[
   3 \cdot (a_k + a_{k-1} + \ldots + a_1 + a_0) = 3 \cdot 100 = 300
   \]

The final answer is \(\boxed{300}\).

Answer: 300


find all functions from the reals to themselves such that for every real $x,y$.
$$f(y-f(x))=f(x)-2x+f(f(y))$$

Solution:
Given the functional equation for all real numbers \( x \) and \( y \):
\[ f(y - f(x)) = f(x) - 2x + f(f(y)) \tag{1} \]

1. **Substitute \( x = 0 \) and \( y = f(0) \) into (1):**
   \[ f(f(0) - f(0)) = f(0) - 2 \cdot 0 + f(f(f(0))) \]
   \[ f(0) = f(0) + f(f(f(0))) \]
   \[ \Rightarrow f(f(f(0))) = 0 \]
   This implies there exists some \( a \) such that \( f(a) = 0 \).

2. **Substitute \( x = a \) into (1):**
   \[ f(y - f(a)) = f(a) - 2a + f(f(y)) \]
   Since \( f(a) = 0 \):
   \[ f(y) = -2a + f(f(y)) \tag{2} \]

3. **Replace \( f(f(y)) \) in (1) using (2):**
   \[ f(y - f(x)) = f(x) - 2x + f(f(y)) \]
   \[ f(y - f(x)) = f(x) - 2x + (f(y) + 2a) \]
   \[ f(y - f(x)) = f(x) - 2x + f(y) + 2a \tag{3} \]

4. **Substitute \( y = f(x) \) into (3):**
   \[ f(f(x) - f(x)) = f(x) - 2x + f(f(x)) + 2a \]
   \[ f(0) = f(x) - 2x + (f(x) + 2a) \]
   \[ f(0) = 2f(x) - 2x + 2a \]
   \[ 2f(x) = 2x + f(0) - 2a \]
   \[ f(x) = x + \frac{f(0) - 2a}{2} \]
   Let \( c = \frac{f(0) - 2a}{2} \), then:
   \[ f(x) = x + c \tag{4} \]

5. **Substitute \( f(x) = x + c \) back into (1):**
   \[ f(y - (x + c)) = (x + c) - 2x + f(f(y)) \]
   \[ f(y - x - c) = x + c - 2x + f(y + c) \]
   \[ f(y - x - c) = -x + c + f(y + c) \]
   Since \( f(y) = y + c \):
   \[ f(y - x - c) = -x + c + (y + c) \]
   \[ y - x - c + c = -x + c + y + c \]
   \[ y - x = -x + 2c + y \]
   \[ 0 = 2c \]
   \[ c = 0 \]

Thus, the function \( f(x) = x \) satisfies the given functional equation.

The final answer is \( \boxed{ f(x) = x } \)

Answer: f(x) = x


In a triangular pyramid, two of the three faces are isosceles right triangles, and the other is an equilateral triangle with a side length of 1. Then, the number of such triangular pyramids is $\qquad$.

Solution:
4. 3 cases concerning the tetrahedron $ABCD$ are as follows:
(1) $AB=AC=BC=AD=1, CD=BD=\sqrt{2}$;
(2) $BC=BD=CD=1, AB=AC=AD=\frac{\sqrt{2}}{2}$;
(3) $AB=BC=AC=AD=BD=1, CD=\sqrt{2}$.

Answer: 3


Given the condition $p: \sqrt{1+\sin 2 \alpha}=\frac{4}{3}$ and the condition $q:|\sin \alpha+\cos \alpha|=\frac{4}{3}$. Then $p$ is ( ) of $q$.
(A) a sufficient but not necessary condition
(B) a necessary but not sufficient condition
(C) a necessary and sufficient condition
(D) neither a sufficient nor a necessary condition

Solution:
$-1 . \mathrm{C}$.
$$
\begin{array}{l}
\text { Since } \sqrt{1+\sin 2 \alpha}=\sqrt{(\sin \alpha+\cos \alpha)^{2}} \\
=|\sin \alpha+\cos \alpha|,
\end{array}
$$

Therefore, $p$ is a necessary and sufficient condition for $q$.

Answer: C


Find the mass of the plate $D$ with surface density $\mu=x / y^{5}$, bounded by the curves

$$
\frac{x^{2}}{16}+y^{2}=1, \quad \frac{x^{2}}{16}+y^{2}=3, \quad y=\frac{x}{4}, \quad x=0 \quad\left(y \geq \frac{x}{4}, x \geq 0\right)
$$

Solution:
Solution.

1. The mass of the plate $D$ with surface density $\mu=x / y^{5}$ is determined by the formula

$$
m=\iint_{D} \frac{x}{y^{5}} d x d y
$$

2. We calculate the obtained double integral:

a) define the region $D$ by inequalities in Cartesian coordinates

$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq x / 4, \quad x \geq 0
\end{array}\right\}
$$

Since the region $D$ is bounded by ellipses and lines passing through the origin, it is easier to solve the problem in generalized polar coordinates

$$
\left\{\begin{array}{l}
x=4 \varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$

In this case, $(\varrho, \varphi) \in D^{\prime}$, and the sought mass is determined by the formula

$$
m=\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi
$$

To find the region $D^{\prime}$, we replace $x$ with $4 \varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequalities defining the region $D$:

$$
\left\{\begin{array}{l}
1 \leq \frac{16 \varrho^{2} \cos ^{2} \varphi}{16}+\varrho^{2} \sin ^{2} \varphi \leq 3 \\
\varrho \sin \varphi \geq 4 \varrho \cos \varphi / 4, \quad \varrho \cos \varphi \geq 0
\end{array}\right.
$$

Solving these inequalities with respect to $\varrho$ and $\varphi$, we get

$$
D^{\prime}=\left\{\begin{array}{lc}
& (\varrho, \varphi): \begin{array}{c}
1 \leq \varrho \leq \sqrt{3} \\
\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2}
\end{array}
\end{array}\right\}
$$

b) we transition from the double integral to the repeated integral:

$$
m=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi=\int_{\pi / 4}^{\pi / 2} d \varphi \int_{1}^{\sqrt{3}} 16 \varrho^{-3} \cdot \frac{\cos \varphi}{\sin ^{5} \varphi} d \varrho
$$

c) integrating sequentially, we get

$$
m=16 \int_{\pi / 4}^{\pi / 2} \frac{d \sin \varphi}{\sin ^{5} \varphi} \int_{1}^{\sqrt{3}} \varrho^{-3} d \varrho=\left.\left.16\left(-\frac{1}{4 \sin ^{4} \varphi}\right)\right|_{\pi / 4} ^{\pi / 2}\left(-\frac{1}{2 \varrho^{2}}\right)\right|_{1} ^{\sqrt{3}}=4
$$

Answer. $m=4$ units of mass.

Conditions of the Problem. Find the mass of the plate $D$ with surface density $\mu$, where $D$ is bounded by the given lines.

1. $\mu=2 x+y^{2}, \quad x=4, y=0, y=\sqrt{x}$.
2. $\mu=x^{2}+y, \quad x=1, y=0, y=2 \sqrt{x}$.
3. $\mu=x^{2}+2 y, \quad x=0, y=4, y=x^{2}(x \geq 0)$.
4. $\mu=x+y^{2}, \quad x=0, y=1, y=x^{2} / 4(x \geq 0)$.
5. $\quad \mu=\frac{x-y}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=9$

$$
(x \geq 0, y \leq 0)
$$

6. $\quad \mu=\frac{2 y-x}{x^{2}+y^{2}}$

$x=0, \quad y=0, \quad x^{2}+y^{2}=3, \quad x^{2}+y^{2}=5$ $(x \leq 0, y \geq 0)$.
7. $\quad \mu=\frac{y-x}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=16$

$$
(x \leq 0, \quad y \geq 0)
$$

8. $\quad \mu(x, y)=y, \quad y=0, \quad y=x \sqrt{3}, \quad x^{2}+\frac{y^{2}}{4}=1, \quad x^{2}+\frac{y^{2}}{4}=9$

$$
(y \geq 0, \quad y \leq x \sqrt{3})
$$

9. $\quad \mu(x, y)=\frac{y}{x^{2}}, \quad y=0, \quad y=x, \quad \frac{x^{2}}{4}+y^{2}=1, \quad \frac{x^{2}}{4}+y^{2}=4$ $(y \geq 0, \quad y \leq x)$.
10. $\mu(x, y)=\frac{x}{y^{2}}, \quad x=0, \quad y=x, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=25$

$(x \geq 0, y \geq x)$.

Answers. 1. $m=448 / 15$. 2. $m=11 / 7$. 3. $m=448 / 15$. 4. $m=11 / 7$. 5. $m=2$. 6. $m=3(\sqrt{5}-\sqrt{3})$. 7. $m=4$. 8. $m=52 / 3$. 9. $m=(\sqrt{2}-1) / 2$. 10. $m=18(\sqrt{2}-1)$.

## 12.9. Triple Integral

in Cartesian coordinates

Problem Statement. Calculate the triple integral

$$
\iiint_{\Omega} f(x, y, z) d x d y d z
$$

where the region $\Omega$ is bounded by certain surfaces.

Plan of Solution.

1. Define the region $\Omega$ by a system of inequalities, for example,

$$
\left\{\begin{aligned}
a & \leq x \leq b \\
y_{1}(x) & \leq y \leq y_{2}(x) \\
z_{1}(x, y) & \leq z \leq z_{2}(x, y)
\end{aligned}\right.
$$

2. Transition from the triple integral to the repeated integral:

$$
\iiint_{\Omega} f(x, y, z) d x d y d z=\int_{a}^{b} d x \int_{y_{1}(x)}^{y_{2}(x)} d y \int_{z_{1}(x, y)}^{z_{2}(x, y)} f(x, y, z) d z
$$

3. Using the properties of the definite integral, integrate sequentially first with respect to $z$ (considering $x$ and $y$ as constants), then with respect to $y$ (considering $x$ as a constant), and finally with respect to $x$.

Write the answer.

Answer: 4


In the Cartesian coordinate system, a circle with center at $(1,0)$ and radius $r$ intersects the parabola $y^{2}=x$ at four points $A, B, C, D$. If the intersection point $F$ of $A C$ and $B D$ is exactly the focus of the parabola, then $r=$ $\qquad$

Solution:
5. $\frac{\sqrt{15}}{4}$.

Combining the equations of the circle and the parabola, we get
$$
\left\{\begin{array}{l}
y^{2}=x, \\
(x-1)^{2}+y^{2}=r^{2} .
\end{array}\right.
$$

Eliminating $y$ yields $x^{2}-x+1-r^{2}=0$.
From $\Delta=1-4\left(1-r^{2}\right)>0$, we solve to get $r>\frac{\sqrt{3}}{2}$.
According to the problem, quadrilateral $A B C D$ is an isosceles trapezoid, and the intersection point of $A C$ and $B D$ is the focus $F\left(\frac{1}{4}, 0\right)$ of the parabola.
Thus, points $A$, $F$, and $C$ are collinear.
Let $A\left(x_{1}, \sqrt{x_{1}}\right), C\left(x_{2},-\sqrt{x_{2}}\right)\left(0\frac{\sqrt{3}}{2}$, which meets the requirements of the problem.

Answer: \frac{\sqrt{15}}{4}


The perpendicular line restored at vertex $C$ of parallelogram $ABCD$ to line $CD$ intersects at point $F$ the perpendicular line dropped from vertex $A$ to diagonal $BD$, and the perpendicular line restored from point $B$ to line $AB$ intersects at point $E$ the perpendicular bisector of segment $AC$. In what ratio does side $BC$ divide segment $EF$?

Solution:
Let point $K$ be symmetric to $A$ with respect to $B$. Then $E$ is the circumcenter of triangle $A C K$. On the other hand, since $B K C D$ is a parallelogram, $A F \perp C K$, so $F$ is the orthocenter of triangle $A C K$ (see the figure). Therefore, the median $CB$ divides $E F$ in the ratio $1: 2$ (see problem $\underline{55595}$).

![](https://cdn.mathpix.com/cropped/2024_05_06_43a9f723b796fae20c45g-15.jpg?height=1126&width=1192&top_left_y=843&top_left_x=427)

Answer

$1: 2$.

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Answer: 1:2