## Task 28/66

A sector of a circle with a central angle of $\frac{\pi}{3}$ is divided by a perpendicular to the angle bisector such that the perimeter of one part is equal to the perimeter of the other part.

Which of the two parts has the larger area?
First, the position of the perpendicular $X Y$ to the angle bisector

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-137.jpg?height=345&width=368&top_left_y=313&top_left_x=293)
$S M$ is determined. There are three possibilities for it (figure): Either $X Y$ intersects the circular arc, or $X Y$ coincides with $A B$, or $X Y$ intersects the sides of the central angle.

In the case of the coincidence of $X Y$ and $A B$, the perimeter of the equilateral triangle $S A B$ with $A B = B S = S A = r$ would have to be equal to the perimeter of the circular segment $A M B$. The calculation shows that the perimeter of the triangle is larger than the perimeter of the segment. This also implies that $X Y$ cannot intersect the circular arc; by moving the perpendicular towards $M$, the smaller perimeter would be further reduced, while the larger one would be increased.

Therefore, $X Y$ intersects the sides $S A$ and $S B$, forming an equilateral triangle $S X Y$ with sides $S X = X Y = Y S = x$.

For the perimeters of the two parts to be equal, it must hold that $3 x = 2(r - x) + x + \frac{\pi}{3} r$. Solving this equation for $x$ yields $x = \frac{r}{12}(6 + \pi)$. The areas are

$$
A_{1} = \frac{1}{4} x^{2} \sqrt{3} = \frac{1}{576} r^{2}(6 + \pi)^{2} \sqrt{3} \quad \text{and} \quad A_{2} = \frac{1}{6} \pi r^{2} - A_{1} = \frac{1}{576} r^{2}\left[96 \pi - (6 + \pi)^{2}\right] \sqrt{3}
$$

The calculation shows that $A_{1} < A_{2}$.