What's the measure of the radius of the circle inscribed in the triangle ATB? For reference: Calculate the radius of the circle inscribed
in the triangle ATB

My progress:  I put the trigonometric solution...would a geometric solution be possible?


$\triangle AO_1T: \triangle BO_2T\rightarrow \text{isosceles}$
$\triangle ATB(\text{right})\\\angle O_1AT = x \implies \angle ABT = x\\AO_1T = 180-2x$
By law of cosines,
$AT^2=2+2\cos(2x)=2(1+\cos(2x))=4\cos^2(x)$
$BT^2=4+4-8\cos(2x)=8(1-\cos(2x))=16\sin^2(x)$
From triangle $\triangle ABT$,
$\displaystyle \tan^2(x)=\frac{AB^2}{BT^2}=\frac{\cos^2(x)}{4\sin^2(x)} \implies \tan^4(x)=\frac{1}{4} \implies \tan(x)=\frac{1}{\sqrt{2}}$
Therefore $\cos(x)=\frac{\sqrt{6}}{3}$ and $\sin(x)=\frac{\sqrt{3}}{3}$.
$AT^2=4\cdot\dfrac{6}{9}=\dfrac{8}{3} \rightarrow AT=\dfrac{2\sqrt{6}}{3}\\BT^2=16\sin^2(x)=16\cdot\dfrac{1}{3} \rightarrow BT=\dfrac{4\sqrt{3}}{3}\\AB^2=\dfrac{8}{3}+\dfrac{16}{3}=8 \rightarrow AB=2\sqrt{2}$
But, $\displaystyle A=pr \rightarrow r=\frac{A}{p}\\\displaystyle \therefore r=\frac{\frac{2\sqrt{6}}{3}.\frac{4\sqrt{3}}{3}}{\frac{2\sqrt{6}}{3}+\frac{4\sqrt{3}}{3}+2\sqrt{2}}= 0.5569$
