5. Quadrilateral $ABCD$ is inscribed in a circle, and its diagonals intersect at point $P$. Points $K, L$, and $M$ are the midpoints of sides $AB, BC$, and $CD$ respectively. The radius of the circle circumscribed around triangle $KLP$ is 1. Find the radius of the circle circumscribed around triangle $LMP$.

Problem 1 Answer: 18 passengers.
Solution. Let $y_{k}$ be the number of passengers in the car with number $k, k=1,2,3, \ldots, 10$. According to the problem, $\sum_{k=1}^{10} y_{k}=270$. Additionally, it is stated that $y_{2} \geq y_{1}+2, y_{3} \geq y_{1}+4, \ldots, y_{9} \geq y_{1}+16, \quad y_{10} \geq y_{1}+18$. Adding these inequalities,

$$
y_{2}+y_{3}+\ldots+y_{9}+y_{10} \geq y_{1}+2+y_{1}+4+\ldots+y_{1}+16+y_{1}+18
$$

or

$$
270-y_{1} \geq 9 y_{1}+90
$$

From this, we get $y_{1} \leq 18$. From the conditions $y_{10} \geq y_{1}+18$ and $y_{10} \leq 2 y_{1}$, it follows that $2 y_{1} \geq y_{1}+18$ or $y_{1} \geq 18$. Comparing the obtained inequalities, we conclude that $y_{1}=18$. The set $y_{1}=18, y_{2}=20$, $\ldots, y_{10}=36$ satisfies the conditions of the problem.

Problem 2 Answer: $x=\frac{8 \pi}{5}$.

Solution. Introduce the notations: $u=\sin x, v=\sin 2 x, w=\sin 3 x$. Then the equation becomes

$$
(u+v+w)^{3}=u^{3}+v^{3}+w^{3}
$$

Rewrite the equation as

$$
(u+v+w)^{3}-u^{3}-\left(v^{3}+w^{3}\right)=0
$$

Using the formulas for the difference and sum of cubes, as well as the difference of squares, factorize the left-hand side of the equation

$$
\begin{aligned}
& (u+v+w)^{3}-u^{3}-\left(v^{3}+w^{3}\right)=(v+w)\left((u+v+w)^{2}+u(u+v+w)+u^{2}\right)- \\
& -(v+w)\left(v^{2}-v w+w^{2}\right)=(v+w)\left[\left((u+v+w)^{2}-v^{2}\right)+\left(u^{2}-w^{2}\right)+u(u+w)+v(u+w)\right]= \\
& =(v+w)(u+w)(u+2 v+w+u-w+u+v))=3(v+w)(u+w)(u+v)
\end{aligned}
$$

As a result, the equation takes the form

$$
3(v+w)(u+w)(u+v)=0
$$

Returning to the original variables, we get the equation:

Case $1 . \sin x+\sin 2 x=0$.

$\left[\begin{array}{c}2 x=-x+2 \pi k \\ 2 x=\pi+x+2 \pi m\end{array} \rightarrow\left[\begin{array}{c}x=\frac{2 \pi k}{3}, \\ x=\pi+2 \pi m, m, k \in Z .\end{array}\right.\right.$ The largest solution on the interval $(0 ; 2 \pi)$ is $x=\frac{4 \pi}{3}$

Case $2 . \sin x+\sin 3 x=0$.

$\left[\begin{array}{c}3 x=-x+2 \pi k \\ 3 x=\pi+x+2 \pi m\end{array} \rightarrow\left[\begin{array}{c}x=\frac{\pi k}{2}, \quad \text { The largest solution on the interval }(0 ; 2 \pi) \text { is } x=\frac{3 \pi}{2} \\ x=\frac{\pi}{2}+\pi m, m, k \in Z .\end{array} \quad\right.\right.$.

Case $3 . \sin 3 x+\sin 2 x=0$.

$\left[\begin{array}{c}3 x=-2 x+2 \pi k \\ 3 x=\pi+2 x+2 \pi m\end{array} \rightarrow\left[\begin{array}{c}x=\frac{2 \pi k}{5}, \quad \text { The largest solution on the interval }(0 ; 2 \pi) \text { is } x=\frac{8 \pi}{5} \\ x=\pi+2 \pi m, m, k \in Z .\end{array}\right.\right.$.

The largest solution among $\frac{4 \pi}{3}, \frac{3 \pi}{2}, \frac{8 \pi}{5}$ is $x=\frac{8 \pi}{5}$.

Problem 3 Answer: all 100 digits after the decimal point are nines.

Solution. Consider $(7+4 \sqrt{3})^{2023}$. Use the binomial formula

$$
(7+4 \sqrt{3})^{2023}=\sum_{k=0}^{2023} C_{2023}^{k} 7^{2023-k} \cdot(4 \sqrt{3})^{k}=a+b \sqrt{3}
$$

where $a=\sum_{k=0}^{1011} C_{2023}^{2 k} 7^{2023-2 k} \cdot(4 \sqrt{3})^{2 k} \in N, b=\sum_{k=0}^{1011} C_{2023}^{2 k+1} 7^{2022-2 k} \cdot(4 \sqrt{3})^{2 k} \in N$.

Now consider $(7-4 \sqrt{3})^{2023}:$

$$
(7-4 \sqrt{3})^{2023}=\sum_{k=0}^{2023} C_{2023}^{k} 7^{2023-k} \cdot(-4 \sqrt{3})^{k}=a-b \sqrt{3}
$$

where $a=\sum_{k=0}^{1011} C_{2023}^{2 k} 7^{2023-2 k} \cdot(4 \sqrt{3})^{2 k} \in N, b=\sum_{k=0}^{1011} C_{2023}^{2 k+1} 7^{2022-2 k} \cdot(4 \sqrt{3})^{2 k} \in N$.

Adding the obtained equalities, we get

$$
(7+4 \sqrt{3})^{2023}+(7-4 \sqrt{3})^{2023}=a+b \sqrt{3}+a-b \sqrt{3}=2 a \in N
$$

From this, we have

$$
\begin{aligned}
& (7+4 \sqrt{3})^{2023}=2 a-(7-4 \sqrt{3})^{2023}=2 a-\frac{(7-4 \sqrt{3})^{2023} \cdot(7+4 \sqrt{3})^{2023}}{(7+4 \sqrt{3})^{2023}}=2 a-\frac{1}{(7+4 \sqrt{3})^{2023}}= \\
& =2 a-1+1-\frac{1}{(7+4 \sqrt{3})^{2023}}, \quad 2 a-1 \in N .
\end{aligned}
$$

Notice that $\frac{1}{(7+4 \sqrt{3})^{2023}}0
$$

Then, considering the non-negativity of $x$, we find $x=\frac{\sqrt{2 \pm \sqrt{3}}}{4}$. It remains to check the condition $x \leq \frac{1}{2}$ :

$$
\frac{\sqrt{2 \pm \sqrt{3}}}{4} \vee \frac{1}{2} \leftrightarrow \sqrt{2 \pm \sqrt{3}} \vee 2 \leftrightarrow 2 \pm \sqrt{3} \vee 4 \leftrightarrow \pm \sqrt{3} \vee 2 \rightarrow \pm \sqrt{3}<2
$$

Therefore, both roots $x=\frac{\sqrt{2+\sqrt{3}}}{2}$ and $x=\frac{\sqrt{2-\sqrt{3}}}{2}$ satisfy the condition $x \leq \frac{1}{2}$.

Problem 5 Answer: $R=1$.

Solution. Consider triangles $A B P$ and $D C P$. They are similar by two equal angles ( $\angle B A C=\angle B D C, . \angle A B D=\angle A D C$ as angles inscribed in a circle and subtending the same arc). Therefore, the sides of these triangles are proportional:

$$
\frac{A B}{D C}=\frac{A P}{D P}=\frac{B P}{P C}
$$

Now consider triangles $A K P$ and $D C P$. By the condition $A K=\frac{A B}{2}, D M=\frac{D C}{2}$, so $\frac{A K}{D M}=\frac{A B}{D C}$. As a result, we have $\frac{A K}{D M}=\frac{A P}{D B}$. Therefore, triangle $A P K$ is similar to triangle $D P M$ by an equal angle and proportional sides. Then the angle $A P K$ is equal to the angle $D P M$ as angles in similar triangles, lying opposite proportional sides.

By the problem's condition, $K L$ and