In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $3$, and $AB = 1$, then find $\frac{AC}{BC}.$
This is impossible
Leg AC would be > 6 while AB (the hypotenuse) would only = 1
+1439 
