# Variant №1

№1. The number of male students in the university is $35 \%$ more than the number of female students. All students are distributed between two buildings, with $\frac{2}{5}$ of all male students in the first building and $\frac{4}{7}$ of all female students in the second building. How many students are there in total in the university, if it is known that fewer than 2500 students are in the first building, and more than 2500 students are in the second building?
# Solution:

Let $n$ be the number of girls studying at the university, then the number of boys is $1.35 n = \frac{27}{20} n$, from which we can conclude that $n$ is a multiple of 20, i.e., $n = 20 m$.

Thus, the number of girls is $20 m$, and the number of boys is $27 m$.

In the first building, $\frac{2}{5}$ of all boys study, i.e., $\frac{2}{5} \cdot 27 m = \frac{54}{5} m$ students, and in the second building, $\frac{4}{7}$ of all girls study, i.e., $\frac{4}{7} \cdot 20 m = \frac{80}{7} m$ students. For the condition of integrality to be satisfied, the number $m$ must be a multiple of 35, i.e., $m = 35 k$.

Then in the first building, $\frac{2}{5} \cdot 27 m = \frac{54}{5} m = \frac{54}{5} \cdot 35 k = 378 k$ students, $\frac{3}{7} \cdot 20 m = \frac{60}{7} m = \frac{60}{7} \cdot 35 k = 300 k$ students, and a total of $678 k$ people, and in the second building, $\frac{3}{5} \cdot 27 m = \frac{81}{5} m = \frac{81}{5} \cdot 35 k = 567 k$ students, $\frac{4}{7} \cdot 20 m = \frac{80}{7} m = \frac{80}{7} \cdot 35 k = 400 k$ students, and a total of $967 k$ people.

By the condition of the problem $\left\{\begin{array}{l}678 k < 2500 \\ 967 k > 2500\end{array} \Leftrightarrow\left\{\begin{array}{l}k < \frac{2500}{678} \\ k > \frac{2500}{967}\end{array}\right.\right.$, hence, $k = 3$, and

the total number of students is $678 k + 967 k = 1645 k = 4935$.

Answer: 4935.

## Criteria

| Points | Criteria for awarding points |
| :---: | :---: |
| 10 | Complete solution. The correct answer is justified. |
| 5 | The problem model is correctly formulated and there are some advances in the solution. |
| 0 | The solution does not meet any of the above criteria. |

#2. In triangle $\mathrm{ABC}$, angle C is a right angle. Line MK is perpendicular to the median CM and intersects AC at point K. It is known that CK: KA = 5: 3. Point O is the center of the circle circumscribed around triangle MAK. Find the area of triangle OBC if $\mathrm{BC}=4$.

## Solution.

By the property of the median of a right triangle, $\mathrm{CM}=\mathrm{BM}=\mathrm{AM}=x$. Let $CK=k$, KA $=n$. Triangles $\mathrm{ABC}$ and CKM are right and similar by the acute angle $\angle M A C = \angle M C A$ (MCA is isosceles). Therefore, $x: k = (k+n): 2x$ or $x = \sqrt{\frac{k(k+n)}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_58e9d88e43ec055816cfg-02.jpg?height=522&width=968&top_left_y=1755&top_left_x=567)

By the Law of Sines for triangle MAK, we get: $2 R = \frac{M K}{\sin \angle M A C} = \frac{M K}{\sin \angle M C A} = C K$, i.e., the radii of the circles circumscribed around triangles MAK and CMK with centers at points O and N, respectively, are equal. Since MK is a common chord of equal circles, $N O$ is the perpendicular bisector of it and $N O = \sqrt{4 R^{2} - M K^{2}} = \sqrt{k^{2} - C K^{2} + C M^{2}} = C M$.

By the condition of the problem, $\mathrm{CM} \perp \mathrm{MK} \Rightarrow \mathrm{CM} \| \mathrm{ON} \Rightarrow C M O N$ is a parallelogram, i.e., $\mathrm{OM} = \mathrm{CN} = k / 2, \mathrm{C} N \| \mathrm{OM} \Rightarrow \mathrm{OM} \perp B C$. Therefore, OH is the height of triangle OBC and
$\mathrm{OH} = \mathrm{OM} + \mathrm{MH} = \mathrm{OM} + \frac{A C}{2} = R + \frac{k+n}{2} = \frac{k}{2} + \frac{k+n}{2} = k + \frac{n}{2}$.

By the condition $k: n = 5: 3 \Rightarrow 0 \mathrm{H} = 1.3 k$. From the original triangle, by the Pythagorean Theorem $(\mathrm{BC}=4, \mathrm{AC}=1.6 k, A B=2 x=4 k \sqrt{0.2}): 16 + 2.26 k^{2} = 3.2 k^{2} \Rightarrow k^{2} = 25 \Rightarrow k = 5$.

$$
S_{B O C} = \frac{1}{2} \cdot 4 \cdot 1.3 \cdot 5 = 13
$$

Answer: 13.

## Criteria

| Points | Criteria for awarding points |
| :---: | :---: |
| 10 | Complete solution. The correct answer is justified. |
| 5 | It is proved that CMON is a parallelogram. |
| 0 | The solution does not meet any of the above criteria. |

#3. For what values of the parameter $a$ does there