## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_964e2f56f6f4ff54223dg-11.jpg?height=468&width=1105&top_left_y=306&top_left_x=390)

In triangle $A V D$, $\angle V A D = \frac{\alpha}{2}$ and $\angle A D V = \frac{180^\circ - \alpha}{2} = 90^\circ - \frac{\alpha}{2}$. Therefore, triangle $A V D$ is a right triangle, and $\angle U V X = 90^\circ$. For the same reason, the other angles of quadrilateral $X Y U V$ are right angles, so this quadrilateral is a rectangle.

From triangle $A V D$, we have $|D V| = b \sin \frac{\alpha}{2}$ and $|A V| = b \cos \frac{\alpha}{2}$.

2 points

From triangle $D X C$, we have $|D X| = a \sin \frac{\alpha}{2}$.

1 point

From triangle $A B U$, we have $|A U| = a \cos \frac{\alpha}{2}$.

1 point

Then the lengths of the sides of the rectangle are:

$|V X| = a \sin \frac{\alpha}{2} - b \sin \frac{\alpha}{2} = (a - b) \sin \frac{\alpha}{2}$ and $|V U| = a \cos \frac{\alpha}{2} - b \cos \frac{\alpha}{2} = (a - b) \cos \frac{\alpha}{2}$.

1 point

The area of the rectangle is $P = |V X| \cdot |V U| = (a - b)^2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$.

1 point

Since $\sin \frac{\alpha}{2} = \frac{3}{5}$, it follows that $\cos \frac{\alpha}{2} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$.

The desired difference is calculated as follows:

$48 = (a - b)^2 \cdot \frac{3}{5} \cdot \frac{4}{5} \Rightarrow (a - b)^2 = 100 \Rightarrow a - b = 10 \text{ cm}$.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

28th February 2017.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.
## Task B-2.7.

Let $A B C D$ be a parallelogram with side lengths $|A B|=a \mathrm{~cm}$ and $|B C|=b \mathrm{~cm} (a>b)$ and an acute angle $\alpha$. The area of the quadrilateral formed by the intersection of the angle bisectors of the internal angles of the parallelogram is $48 \mathrm{~cm}^{2}$, and $\sin \frac{\alpha}{2}=\frac{3}{5}$. Calculate the difference $a-b$.