Given that AB=12, CD=1, angle BAC=90∘, and the semicircle is tangent to BC,find the radius of the semicircle.
+9519 Let r be the radius.
Let O be the midpoint of AD, i.e., the center of the semicircle.
Let T be the point where the line BC touches the semicircle.
As BA and BT are both tangents to the circle, meeting at the same point:
BT = BA = 12
Then, using Pythagorean theorem on \(\triangle CTO\), \(CO^2 = CT^2 + OT^2\)
Notice that CO = CD + DO = 1 + r, and OT is one of the radii of the semicircle.
\((1 + r)^2 = CT^2 + r^2\\ CT = \sqrt{2r + 1}\)
Also, CA = CD + DO + OA = 1 + r + r = 1 + 2r
Using Pythagorean theorem on \(\triangle CAB\),
\((1 + 2r)^2 + 12^2 = \left(\sqrt{2r + 1} + 12\right)^2\)
Let \(t = \sqrt{2r + 1}\).
\(t^4 + 144 = (t + 12)^2\\ t^4 - t^2 - 24t = 0\\ t(t^3 - t -24) = 0\)
Factorising further using Factor Theorem, (t - 3) is a factor of (t3 - t - 24).
By long division, t3 - t - 24 = (t - 3)(t2 + 3t + 8).
\(t(t - 3)(t^2 + 3t + 8) = 0\)
Checking the discriminant of (t2 + 3t + 8), there are no real t satisfying t2 + 3t + 8 = 0.
Also, t = 0 would imply r = -1/2, which isn't possible either.
The only possible scenario is t = 3.
Recall the definition of t (defined above)
\(\sqrt{2r + 1} = 3\\ 2r + 1 = 9\\ \boxed{r= 4}\)
P.S. Considering the difficulty of this problem, I am suspecting this is some kind of math contest problems 
