## Infinite Circles

#### Problem

A circle is inscribed inside an equilateral triangle and an infinite set of circles are nested inside such that each circle touches the previous circle and the edges of the triangle act as tangents.

What fraction of the large red circle do the infinite set of smaller circles represent?

#### Solution

We begin by considering the relationship between one circle and the next.

Let the radius of the large circle be $R$, the radius of the small circle be $r$, and the hypotenuse of the small right angled triangle be $x$.

$\therefore sin(30^o) = \dfrac{1}{2} = \dfrac{r}{x} \Rightarrow x = 2r$

In other words, for a $30^o$ right angle triangle the hypotenuse is twice the height. As the hypotenuse of the big right angle triangle is given by $x + r + R$ we get the following.

$$\begin{eqnarray}x + r + R & = & 2R\\\therefore 2r + r + R & = & 2R\\\therefore 3r & = & R\end{eqnarray}$$That is, the radius of each subsequent circle is $\frac{1}{3}$ of the radius of the previous circle.

$\therefore \pi \left(\dfrac{R}{3}\right)^2 + \pi \left(\dfrac{R}{9}\right)^2 + ... = \pi R^2 \left(\dfrac{1}{9} + \dfrac{1}{81} + ...\right)^2$

Using the sum to infinity for a geometric progression: $S_{\infty} = \dfrac{a}{1-r},$ where the first term, $a = \dfrac{1}{9}$, and the common ratio, $r = \dfrac{1}{9},$ we get $\dfrac{1}{9} + \dfrac{1}{81} + ... = \dfrac{ \dfrac{1}{9} }{\dfrac{8}{9}} = \dfrac{1}{8}$.

Hence the infinite set of smaller circles must represent exactly $\dfrac{3}{8}$ the area of the large red circle.

Show that all of the circles occupy approximately 83% of the triangle.