Circle radius problem Right isosceles triangle ABC inscribed in circle k(O,R). Circle k1(O1,R1) is tangential to AB, AC and circle k. Find R1 in terms of  R.


Join $CO$ and let $Q$ be the intersection with $AO_1$. As $AO_1$ is the bisector of $\angle BAC$ you can find $OQ=R/(1+\sqrt2)$. Then by Pythagoras theorem you can compute $OM=\sqrt{R^2+2RR_1}$ so that $AM=R+\sqrt{R^2+2RR_1}$. You can now exploit the similitude between $AOQ$ and $AMO_1$ to get $AO:AM=OQ:MO_1$, so that finally $R_1=2R(2-\sqrt2)$.
