Let $r$ be the radius of each of the larger circles.

The sides of the square are equal to $r+1$, the sum of the two radii.

The diagonal of the square is $2r$.

By Pythagoras, $$(r+1)^2+(r+1)^2 = (2r)^2$$Simplifying gives: $$ 2(r+1)^2 = 4 r^2$$ i.e. $$(r+1)^2 = 2r^2$$

so$$r+1 = \sqrt{2}r$$

[$-\sqrt{2}r$ is not possible since $r+1> 0$].

Therefore $(\sqrt{2} - 1)r = 1$.

Hence $r= \frac {1}{\sqrt{2}-1} = \sqrt{2} + 1$.

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*

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