# Circle in the corner

What is the ratio of the two circles?

Solution
My preferred solution is to solve the problem by considering an infinite sequence of circles, that, in their limit, completely cover the diagonal. If we let the first circle have radius $R$, and the next, radius $\alpha R$, then the third sized circles with have radii of $\alpha^2 R$ (by self similarity) and so on. We will get the following identities $$2\sqrt{2}R = 2R + 4\alpha R + 4\alpha^2 R + 4\alpha^3 R + \dots$$ Factorising the right hand side and tidying up a little gives $$2\sqrt{2} -2= 4\alpha(1 + \alpha + \alpha^2 + \dots)$$ $$\sqrt{2} - 1= \frac{2\alpha}{1-\alpha}$$ Rearranging this for $\alpha$ gives $$\alpha = \frac{\sqrt{2}-1}{\sqrt{2}+1} = 3 - 2\sqrt{2}$$ The required ratio is therefore $1 : 3-2\sqrt{2}$