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Weekly Problem 3 - 2011

Stage: 3 and 4 Short Challenge Level: Challenge Level:1

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.

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