Half Inscribed

Let's call the radius of the semicircle as $r$ . The length of $AC$ is $2r+1$ . The length of $BC$ is, by Pythagorean Theorem, $\sqrt{(2r+1)^2+12^2 }=\sqrt{4r^2+4r+145}$ .

Look the image bellow:

The triangle $CMN$ is similar to the triangle $ABC$ . Then we have:

$\frac{\sqrt{4r^2+4r+145}}{r+1}=\frac{12}{r}$

Solving this equation we get:

$x=4$ or $x=-\frac{1}{2}$ or two others complex solutions. The only possible solution for the problem is $\boxed{x=4}$

Let $M$ be the midpoint of $AD$ and the center of the circle; let $T$ be the point where the circle touches $BC$ .

Then $\angle MTC = 90^\circ = \angle BAC$ , proving that $\triangle MTC \sim \triangle BAC$ .

The Pythagorean Theorem gives $MC^2 = MT^2 + CT^2\ \ \therefore\ \ (r+1)^2 = r^2 + CT^2\ \ \therefore\ \ CT = \sqrt{(r+1)^2 - r^2} = \sqrt{2r+1}.$

Similarity gives the ratios $\frac{AC}{TC} = \frac{AB}{TM}\ \ \therefore\ \ \frac{2r+1}{\sqrt{2r+1}} = \frac{12}{r},$ which upon squaring becomes $(2r + 1)r^2 = 12^2.$ This cubic equation is not easy to solve, but when we realize that Brilliant wants an integer value, it becomes obvious that $r$ must be a divisor of 12, such that $12/r$ is odd; this requires that $r = 4$ or $r = 12$ , and it is easy to see that only $\boxed{r = 4}$ works.

First I'd like to thank Mr. @Victor Paes Plinio for the image; it follows by Pythagorean Theorem on the triangle $ABC$ that if we take $x$ for the lenght of the segment $\overline{MC}$ and $R$ for the radius of the circle

$(2R + 1)^2 = x^2 + 24x,$

given that $\overline{AB} = \overline{BM}$ since both $A$ and $M$ are tangent points to the circle. Now, given the second right triangle $MNC$ we have $x^2 = 2R + 1$ . Feeding this information into the first result gives us the following equation for $x$ :

$x^4 - x^2 - 24x = x (x^3 - x - 24) = 0.$

If we want $x \neq 0$ we must solve a third degree polynomial equation. That can be done via Girard's relations, which will show you that the only real solution is $x = 3$ . Feeding it back into our equation relating $x$ to $R$ will result in

$3^2 = 2R + 1 \Rightarrow R = 4.$

I'm borrowed the diagram below from Victor Paes Plinio.

Let $N$ be the center of the circle and $r$ be the radius of the circle, then $r = DN = AN = NM$ and $AB = 12 \implies BM = 12$ .

Let $CM = x$ .

$\triangle{NMC} \sim \triangle{ABC} \implies \dfrac{12}{r} = \dfrac{12 + x}{r + 1} \implies x = \dfrac{12}{r}$

For $\triangle{NMC}$ the Pythagorean theorem $\implies (\dfrac{12}{r})^2 + r^2 = (r + 1)^2 \implies 2r^3 + r^2 - 144 = 0 \implies$ $(r - 4)(2r^2 + 9r + 36) = 0 \implies r = 4, \:\ r = \dfrac{-9}{4} \pm \dfrac{3}{4}\sqrt{23}i$ .

Since $r$ must be real and positive $\implies r = 4$ .

seeing AB = 12 makes us think of a (5,12,13) right triangle. Another is (9,12,15). Seeing if it is 9,12,15 we find the radius is 4. this works with the formula rs=a if ABC is turned into an isosceles triangle with ABC flipped over. Checking the 5, 12, 13 we get the radius is 12/5 which is incorrect and does not fit the shape. Therefore the radius is 4.

(not considering negative and complex solutions we end up with 4)

Solutions that are only one equation are quite elegant !!

Triangle ABC is a 3-4-5 right triangle. Since side AB is 12 units long (3x4), side AC must be 9 units long (3x3), and DC=1, so the diameter of the semicircle is 9 - 1 = 8 = 2r, so r = 4.

Let the center of the semicircle be $O$ , its radius be $r$ and the foot of the perpendicular from $O$ to $BC$ be $E$ . We note that $\triangle ABC$ and $\triangle EOC$ are similar. From $\triangle ABC$ , we have $\tan \theta = \dfrac {2r+1}{12}$ and from $\triangle EOC$ , $\cos \theta = \dfrac r{r+1}$ . Then, we have:

$\begin{aligned} \tan \theta & = \frac {2r+1}{12} \\ \sin \theta & = \frac {2r+1}{12} \cos \theta \\ \sin^2 \theta & = \left(\frac {2r+1}{12} \right) \cos^2 \theta \\ \sin^2 \theta + \cos^ \theta & = \left(\left(\frac {2r+1}{12}\right)^2 + 1 \right) \cos^2 \theta \end{aligned}$

$\begin{aligned} \implies \left(\left(\frac {2r+1}{12}\right)^2 + 1 \right) \color{#3D99F6} \cos^2 \theta & = 1 & \small \color{#3D99F6} \text{Note that }\cos \theta = \frac r{r+1} \\ \left(\left(\frac {2r+1}{12}\right)^2 + 1 \right) \color{#3D99F6} \left(\frac r{r+1}\right)^2 & = 1 \\ 4r^4 + 4r^3 + 145r^2 & = 144r^2+288r+144 \\ 4r^4 + 4r^3 + r^2 - 288r - 144 & = 0 \\ (r-4)(2r+1)(2r^2+9r+36) & = 0 \\ \implies r & = \boxed{4} & \small \color{#3D99F6} \text{Since }r > 0 \end{aligned}$

I don't know how I did this, but I visualised the BC tangent being at 1/3 the circumference, 1/3 of 12 is 4.

I think, there is a mistake/unclarity. It's nowhere said that the semicircle touches A. It's shown, yes, but never trust a picture ;)

Thank Mr. @Victor Paes Plinio for the image.It's easy to find $\frac{CM}{MN}$ = $\frac{CA}{AB}$ . We can get this equation $\frac{\sqrt{2r+1}}{r}$ = $\frac{2r+1}{12}$ . Let's do this replacement: $t=\sqrt{2r+1}$ . Sort out the equation,then we have: $t(t+1)(t-1)=24$

It is easy to see that a solution of this equation is $t=3$ ,which means $x=4$ .

When $t>0, x>1$ , $f(t)=t^3-t$ increase monotonously. So x=4 is the only solution.

Can't it be 7.5 ???