## Task A-2.4. (4 points)

Three circles with a radius of $1 \mathrm{~cm}$ have exactly one common point, and their centers are the vertices of an equilateral triangle. Determine the area of the set of all points that belong to two of these circles.

![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-10.jpg?height=346&width=357&top_left_y=235&top_left_x=1458)
## Solution.

Let $A B C$ be the triangle formed by the centers of the given circles, and $S$ their common point. This is also the center of the equilateral triangle $A B C$.

We have $|S A|=|S B|=|S C|=1$.

Let $T$ be the other intersection point of the circles of radius 1 with centers $A$ and $B$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-10.jpg?height=536&width=571&top_left_y=977&top_left_x=731)

Since $A B$ is the perpendicular bisector of segment $\overline{S T}$, we have $\varangle S A T=2 \varangle S A B=60^{\circ}$, so we conclude that triangle $A S T$ is equilateral with side length 1.

(1 point)

The desired area is equal to three times the area of the intersection of two circles, or six times the area of the circular segment between chord $\overline{S T}$ and the circle centered at point $A$.

(1 point)

The area of this segment is the difference between the area of the circular sector $A S T$ and the area of the equilateral triangle $A S T$, i.e.,

$$
P(\text { segment })=P(\text { sector })-P(\text { triangle })=\frac{1}{6} \cdot 1^{2} \cdot \pi-\frac{1}{4} \cdot 1^{2} \cdot \sqrt{3}=\frac{\pi}{6}-\frac{\sqrt{3}}{4}
$$

The desired area is $\pi-\frac{3 \sqrt{3}}{2} \mathrm{~cm}^{2}$.

Note. A student who does not show that triangle $A S T$ is equilateral but uses this fact and obtains the correct result should receive 3 points.