## Three Circles

In the figure below, 3 equal circles have been drawn so that each one passes through the centers of the other two. Is the area of overlap of the three triangles more or less than a quarter of the area of a circle?
Source: Julia Robinson Mathematics Festival

SOLUTION

The figure above shows the overlap of the three triangles in close-up. Let $x$ denote the area between a circle and a chord, for example between a circle and chord $\overline{AC}$ and let $y$ denote the area of triangle $ABC$. Then,
Area of overlap = $3x+y$

Area of triangle $ABC$
Triangle $ABC$ is equilateral because its three sides are radii of congruent circles. Let $r$ be the radius. $\overline{BD}$ is an altitude and equals $\frac{r\sqrt 3}{2}$ because triangle $BDC$ is a $30^\circ-60^\circ-90^\circ$ triangle with a hypotenuse of length $r$.
Area of $\triangle ABC=\frac{1}{2}\left (AC\right )\left (BD\right )$
$=\frac{1}{2}\left (r\right )\left (\frac{r\sqrt 3}{2}\right )$
$=\frac{r^2\sqrt 3}{4}$

Calculate $x$
$x$ = area of sector $ABC-$ area of $\triangle ABC$
$=\frac{60}{360}\pi r^2-\frac{r^2\sqrt 3}{4}$

$=\frac{\pi r^2}{6}-\frac{r^2\sqrt 3}{4}$

Area of overlap
Area of overlap = $3x+y$
$=3\left (\frac{\pi r^2}{6}-\frac{r^2\sqrt 3}{4}\right )+\frac{r^2\sqrt 3}{4}$

$=\frac{\pi r^2}{2}-\frac{3r^2\sqrt 3}{4}+\frac{r^2\sqrt 3}{4}$

$=\frac{\pi r^2}{2}-\frac{r^2\sqrt 3}{2}$

$=\left (\frac{\pi-\sqrt 3}{2}\right )r^2$

Compare areas
A quarter of the area of a circle = $\left (\frac{\pi}{4}\right )r^2$.
$\frac{\pi-\sqrt 3}{2}=.7048$
$\frac{\pi}{4}=.7854$
Since $.7048<.7854$ we conclude that the area of overlap is less than a quarter of the area of a circle.

Answer: Area of overlap is less than a quarter of the area of a circle.