# Geometry Problem from BDMO Past Papers.

If the difference of areas of outer and inner circles of an equilateral hexagon $ABCDEF$ is $\pi$, what is the area of the hexagon?

• So what have you tried? Jan 10 '18 at 20:41
• Let $R$ be the radius of the outer circle and $r$ of the inner circle. How are they linked? Express the area between the circles and the area of the hexagon in terms of $R$ and $r$.
– user491874
Jan 10 '18 at 20:43

Let $R$ be a radius of the biggest circle.
Thus, $\frac{R\sqrt3}{2}$ is a radius of the small circle, which gives $$\pi R^2-\pi\cdot\frac{3}{4}R^2=\pi$$ or $$R=2.$$ Id est, $$6\cdot\frac{\sqrt3\cdot2^2}{4}=6\sqrt3$$ it's needed area.
• If $AB=AO=BO=R$ and $OD$ be an altitude of $\Delta ABO$ then $OD=\sqrt{R^2-\frac{R^2}{4}}=\frac{R\sqrt3}{2}.$ Jan 11 '18 at 10:11
Take every gon of hexagon of length r. Then the radius of outer circle is $r$ and that of inner circle is ${\sqrt 3\over 2}r$. Therefore the area difference is $\pi r^2-{3\over 4}\pi r^2=\pi\to r=2$ then the area of hexagon is ${{3\sqrt 3}\over{2}}r^2=6\sqrt 3$.