Albert1
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very good solution !chisigma said:[sp]Let suppose that the side of the square is 2. In this case, if x is the radius of the 'small circle', for the theorem of Pythagoras it must be...
$\displaystyle (1-x)^{2} + 1 = (1+x)^{2}$
... so that is $\displaystyle x = \frac{1}{4}$...[/sp]
Kind regards
$\chi$ $\sigma$