Tangential Essentials

Label the diagram as seen in the figure. Denote the radius of the semicircle by $R$ , the length of $BC$ by $x$ and the semiperimeter of $\triangle ABC$ by $s$ .

$AB$ is a diameter, hence $\angle ACB = 90{}^\circ$ , thus, $\triangle ABC$ is a right triangle.

By Pythagoras’s theorem on $\triangle ABC$ , $AC=\sqrt{{{26}^{2}}-{{x}^{2}}}$ Moreover, for the inradius of $\triangle ABC$ it holds $r=s-AB=\frac{x+\sqrt{{{26}^{2}}-{{x}^{2}}}-26}{2}\Rightarrow 2r=x+\sqrt{{{26}^{2}}-{{x}^{2}}}-26 \ \ \ \ \ (1)$ $M$ is the midpoint of the cord $AC$ . By the Triangle Midsegment Theorem, $OM=\dfrac{x}{2} \ \ \ \ \ (2)$ At the same time, $OM=OD-DM=R-2r\overset{\left( 1 \right)}{\mathop{=}}\,13-\left( x+\sqrt{{{26}^{2}}-{{x}^{2}}}-26 \right)=39-x-\sqrt{{{26}^{2}}-{{x}^{2}}} \ \ \ \ \ (3)$ Combining $\left( 2 \right)$ and $\left( 3 \right)$ we get

$\begin{aligned} \frac{x}{2}=39-x-\sqrt{{{26}^{2}}-{{x}^{2}}} & \Leftrightarrow \sqrt{{{26}^{2}}-{{x}^{2}}}=39-\frac{3x}{2} \\ & \overset{x<26}{\mathop{\Leftrightarrow }}\,{{26}^{2}}-{{x}^{2}}={{\left( 39-\frac{3x}{2} \right)}^{2}} \\ & \Leftrightarrow \frac{1}{4}{{x}^{2}}-9x+65=0 \\ & \overset{x<26}{\mathop{\Leftrightarrow }}\,x=10 \\ \end{aligned}$ Now, we can calculate the area of $\triangle ABC$ :

$\left[ ABC \right]=\frac{1}{2}BC\cdot AC=\frac{1}{2}\times 10\times \sqrt{{{26}^{2}}-{{10}^{2}}}=\boxed{120}$

Let the triangle be $ABC$ ; $O$ , $P$ , and $Q$ be the centers of the semicircle and the two circles: and $D$ , $E$ , and $F$ , be the tangent points of the two circles as shown. Let $\angle A = \theta$ . Then

$\begin{aligned} AD + DB & = AB \\ r\cot \frac \theta 2 + r \cot \left(45^\circ - \frac \theta 2 \right) & = 26 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + r\cdot \frac {1+t}{1-t} & = 26 \\ \frac {(1+t^2)r}{t(1-t)} & = 26 \\ \implies r & = \frac {26t(1-t)}{1+t^2} \end{aligned}$

And

$\begin{aligned} EF + FO & = EO \\ 2r + 13 \sin \theta & = 13 \\ \implies r & = \frac {13}2 (1- \sin \theta) = \frac {13}2 \left(1 - \frac {2t}{1+t^2} \right) = \frac {13(1-t)^2}{2(1+t^2)} \end{aligned}$

Therefore we have:

$\begin{aligned} \frac {26t(1-t)}{1+t^2} & = \frac {13(1-t)^2}{2(1+t^2)} \\ 4t & = 1 -t \\ \implies t & = \frac 15 \end{aligned}$

The area of $\triangle ABC$ ,

$\begin{aligned} [ABC] & = \frac {26^2}2 \sin \theta \cos \theta = 13 \cdot 26\cdot \frac {2t}{1+t^2} \cdot \frac {1-t^2}{1+t^2} = 13 \cdot 26 \cdot \frac 5{13} \cdot \frac {12}{13} = \boxed{120} \end{aligned}$

Let the diameter be represented by AB = 26 units. Then, 2r+a/2=26/2, a²+b²=26² and (b+a-26)/2=r. A =ab/2 =120 unit²