5. Through the vertex $M$ of some angle, a circle is drawn, intersecting the sides of the angle at points $N$ and $K$, and the bisector of this angle at point $L$. Find the sum of the lengths of segments $M N$ and $M K$, if the area of $M N L K$ is 25, and the angle $L M N$ is $30^{\circ}$.
Answer: $10 \sqrt[4]{3}$.

## Solution.

Let $L P$ and $L Q$ be the perpendiculars to $M N$ and $M K$ respectively.

Point $\mathrm{L}$ lies on the angle bisector and is therefore equidistant from the sides of the angle, which means $L P = L Q$.

Right triangles $N P L$ and $L Q K$ are congruent (by leg and hypotenuse), as well as right triangles $M P L$ and $M Q L$ (similarly, by leg and hypotenuse).

On one hand,

$$
M N + M K = M P + P N + M K = M P + K Q + M K = M P + M Q = 2 M P.
$$

On the other hand,

$$
25 = S_{M N L K} = S_{M P L} + S_{M Q L} = 2 S_{M P L} = M P \cdot P L = M P^2 \cdot \operatorname{tg} 30^{\circ} = \frac{M P^2}{\sqrt{3}} \Rightarrow M P = 5 \sqrt[4]{3}
$$

Thus, we have $\quad M N + M K = 2 M P = 10 \sqrt[4]{3}$.

## Grading criteria are provided in the table:

| Points | Grading Criteria |
| :---: | :--- |
| $\mathbf{7}$ | Complete and justified solution. |
| $\mathbf{6}$ | Justified solution with minor flaws. |
| $\mathbf{5 - 6}$ | The solution contains minor errors, gaps in justification, but is generally correct and can become fully correct after minor corrections or additions. |
| $\mathbf{4}$ | The problem is more solved than not, for example, one of the two (more complex) significant cases is correctly considered. |
| $\mathbf{2 - 3}$ | The problem is not solved, but formulas, diagrams, considerations, or auxiliary statements relevant to the solution of the problem are provided. |
| $\mathbf{1}$ | The problem is not solved, but an attempt at a solution has been made, for example, individual (partial) cases have been considered in the absence of a solution or in the case of an incorrect solution. |
| $\mathbf{0}$ | The solution is absent or does not meet any of the criteria listed above. |

## Ministry of Science and Higher Education of the Russian Federation  Council of Rectors of Universities of Tomsk Region  Open Regional Inter-University Olympiad 2021-2022  MATHEMATICS (10th grade)  Final Stage  Variant 2