# Either of two unit circles passes through the center of the other. How do you prove that the common area is 2/3pi-sqrt 3/2 areal units?

Sep 20, 2016

Proved in the explanation

#### Explanation:

The common area is enclosed by unit circle arcs subtending

$\angle {120}^{o} = \frac{2}{3} \pi$ radian, at the respective centers.

For this arc, sector area of a unit circle

$= \frac{\frac{2}{3} \pi}{2 \pi}$(area of unit circle)=(2/3pi)/(2pi)(pi)=pi/3#

One half of the common area

= this sector area less area of the inner triangle, of sides $\left[1 , \sqrt{3} , 1\right]$

$= \frac{\pi}{3} - \left(\frac{1}{2}\right) \left(\sqrt{3}\right) \left(\frac{1}{2}\right)$

Twice this is the common area =$\frac{2}{3} \pi - \frac{\sqrt{3}}{2}$ areal units.

I welcome a graphical depiction,.from an interested reader.

Sep 20, 2016

see explanation.

#### Explanation:

1) Equilateral triangle area ${A}_{e} = \sqrt{\frac{3}{4}} \times {1}^{2} = \frac{\sqrt{3}}{4}$

2) yellow area ${A}_{y} = \frac{\pi}{6} - \frac{\sqrt{3}}{4} = \frac{2 \pi - 3 \sqrt{3}}{12}$

3) Common area ${A}_{c} = 2 {A}_{e} + 4 {A}_{y}$

${A}_{c} = 2 \times \frac{\sqrt{3}}{4} + 4 \left(\frac{2 \pi - 3 \sqrt{3}}{12}\right)$

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

$= \frac{3 \sqrt{3} + 4 \pi - 6 \sqrt{3}}{6}$

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