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?
 A: 
First note that $\triangle ATB$ is right triangle with $\angle ATB = 90^\circ$.
Drop a perp from $P$ to $BQ$. Then, $AB = PD = 2 \sqrt2$ (as $PQ = 3, QD = 1$).
Next drop a perp from $B$ to $PQ$. Then, $ \displaystyle \triangle QBH \sim \triangle QPD \implies QH = \frac{2}{3}, BH = \frac{4 \sqrt2}{3}$
$ \displaystyle TH = TQ - HQ = \frac43$
$ \displaystyle BT = \sqrt{TH^2 + BH^2} = \frac{4}{\sqrt3}$
Now for a right triangle, inradius is $  \displaystyle \frac{a + b - c}{2} ~, ~ $ where $c$ is hypotenuse and $a$ and $b$ are perpendicular sides.
So inradius of $ \displaystyle \triangle BTH = \frac{2 (\sqrt2 + 1 - \sqrt3)}{3}  $
As $\triangle BTH \sim \triangle BAT, ~ $ inradius of $\triangle BAT$,
$ \displaystyle r = \sqrt{\frac32} \cdot \frac{2 (\sqrt2 + 1 - \sqrt3)}{3} = \frac{2 + \sqrt2 - \sqrt6}{\sqrt3}$
A: 
$\triangle AO_1T: \triangle BO_2T\rightarrow \text{isosceles}$
$\triangle ATB(\text{right})\\\angle O_1AT = x \implies \angle ABT = x\\AO_1T = 180-2x$
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, $\displaystyle A=pr \rightarrow r=\frac{A}{p}\\\displaystyle \therefore r=\frac{\frac{2\sqrt{6}}{3}.\frac{4\sqrt{3}}{3}}{\frac{2\sqrt{6}}{3}+\frac{4\sqrt{3}}{3}+2\sqrt{2}}= 0.5569$
A: 
Note that $\small CF=FB=GE=DA=1$.
From similar triangles, $\small TG=\dfrac13CF=\dfrac13$.
Therefore $\small TE=\dfrac43$.
From Pythagorean theorem to $\small\triangle DCF$, $\small DF=2\sqrt2=AB$.
Therefore $\small AE=\dfrac{2\sqrt2}3$ and $\small BE=\dfrac{4\sqrt2}3$.
Now again from Pythagorean theorem to $\small \triangle ATE$ and $\small\triangle BTE$, we find $\small AT=\dfrac{2\sqrt6}3$ and $\small BT=\dfrac{4\sqrt3}3$.
Using the formula for inradius of a right triangle(1), $$\displaystyle r=\frac12\left(\frac{2\sqrt6}3+\frac{4\sqrt3}3-2\sqrt2\right)=\frac13(\sqrt6+2\sqrt3-3\sqrt2)$$
