Consider the following shape:
Three identical circles are tanged inside a circle.
Which is greater:
- The area of the shaded regions.
- Twice the area of the unshaded regions.
My attempt: I tried to solve it approximately. Let $R=2$ be the radius of the big circle, and its area is $4\pi$. The diameter of the smaller circles is about $R$ and their radius is $R/2=1$, the area of each small circle is $\pi$. So, the area of the shaded region is $3\pi$. The area of the unshaded region is about $4\pi - 3\pi =\pi$. Twice of it is about $2\pi$ which is smaller than $3\pi$, the area of the shaded region.
However, the correct answer is 2!


Let $r$ be a radius of the little circle and $R$ be a radius of the big circle.
Thus, $$\frac{2}{3}\cdot\frac{2r\sqrt3}{2}+r=R $$ and $$r=(2\sqrt3-3)R.$$ Now, we can see that $$3(2\sqrt3-3)^2<2\left(1-3(2\sqrt3-3)^2\right).$$