Given that AB=12, CD=1, angle BAC=90∘, and the semicircle is tangent to BC,find the radius of the semicircle.

Guest Jul 7, 2020

#1**+2 **

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 (t^{3} - t - 24).

By long division, t^{3} - t - 24 = (t - 3)(t^{2} + 3t + 8).

\(t(t - 3)(t^2 + 3t + 8) = 0\)

Checking the discriminant of (t^{2} + 3t + 8), there are no real t satisfying t^{2} + 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

MaxWong Jul 7, 2020