## Solution.

Let $ABC$ be a right triangle with a right angle at vertex $C$, and let $\beta=\angle ABC = 60^{\circ}$. If we rotate the triangle $ABC$ around vertex $C$ by $30^{\circ}$ in the positive direction, we obtain triangle $A' B' C$ as shown in the following sketch.

![](https://cdn.mathpix.com/cropped/2024_05_30_d455efeee432fadf0574g-04.jpg?height=326&width=622&top_left_y=979&top_left_x=637)

The angle $\beta=60^{\circ}, \angle A'CB=90^{\circ}-30^{\circ}=60^{\circ}$, so triangle $CDB$ is equilateral with side length 4.

Using the Pythagorean theorem, we get $|CA|=\sqrt{|AB|^2-|CB|^2}=\sqrt{8^2-4^2}=4\sqrt{3}$.

Since $\overline{CF}$ is the altitude to the hypotenuse of the right triangle $CA'B'$, we have $|CF|=\frac{|CA'|\cdot|CB'|}{|A'B'|}=\frac{4\sqrt{3}\cdot 4}{8}=2\sqrt{3}$.

Thus, $|BF|=|BC|-|CF|=4-2\sqrt{3}$.

The angle $\angle BEF=30^{\circ}$ (since it is the angle between the line segment $\overline{AB}$ and its rotated counterpart $\overline{A'B'}$), and since $\angle EBF=60^{\circ}$, triangle $FEB$ is half of an equilateral triangle.

Therefore, $|BE|=2\cdot|BF|=8-4\sqrt{3}$ and $|FE|=\frac{\sqrt{3}}{2}\cdot|BE|=4\sqrt{3}-6$.

Now, $P_{FEB}=\frac{1}{2}|FE|\cdot|BF|=\frac{1}{2}(4\sqrt{3}-6)\cdot(4-2\sqrt{3})=14\sqrt{3}-24$.

Furthermore, the area of triangle $CDB$ is $P_{CDB}=\frac{4^2\sqrt{3}}{4}=4\sqrt{3}$.

Finally, the area of the common part of triangles $ABC$ and $A'B'C$ is

$P=P_{CDB}-P_{FEB}=4\sqrt{3}-(14\sqrt{3}-24)=24-10\sqrt{3}$.

Note: A student who rotates the triangle in the opposite direction should obtain the same result and receive full credit.

## SCHOOL COMPETITION IN MATHEMATICS

2nd grade - high school - B variant
January 3, 2024.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.
## Task B-1.7.

In a right-angled triangle $ABC$, the right angle is at vertex $C$, $|AB|=8$, and $\angle ABC=60^{\circ}$. The triangle $ABC$ is rotated around vertex $C$ by $30^{\circ}$ counterclockwise, mapping it to triangle $A'B'C$. What is the area of the common part of triangles $ABC$ and $A'B'C$?