What is the value of the inradius of triangle $PHQ$?

Question

For refrence: The right triangle $ABC$, right at $B$, the height $BH$ is drawn. Let $P$ and $Q$ be the intersections of triangles $AHB$ and $BHC$, $PQ$ intersects at $E$ the $BH$, where $\frac{BE}{EH} = 5\sqrt2$ and the inradius of the triangle $ABC$ measures $10$. Calculate the inradius of triangle $PHQ$.(Answer: $1$)

My progress..It is a complicated drawing ..not to scale...

I found it a difficult question and didn't get much out of it.... Maybe there is some relationship between the inradius but I don't know

I I know this(by property): $ r_1+r_2+R = BH\\ r_1+r_2+10 = BH$

Also: $\triangle MPH \sim \triangle NQH \implies\\ \frac{r_1}{r_2} = \frac{PH}{QH}$

By Geogebra $\triangle PQH$ is right?? $\implies$ T.Poncelet: $PH+HQ = PQ+2r$

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Answer 1

As $\angle PHQ = 90^\circ$ and $HE$ is its angle bisector, the incenter of $\triangle PHQ$ must lie on the segment $HE$. We mark it as $I$.

Next we note that $PM = MH = r_1$ and $QN = NH = r_2$. That leads to $PH = r_1 \sqrt2, QH = r_2 \sqrt2$.

As $\triangle AHB \sim \triangle BHC$ and they are right triangles, their inradii must be in the ratio of their hypotenuse.

So, $ \displaystyle \frac{r_1}{r_2} = \frac{c}{a} ~ $. But in right triangle $\triangle PHQ$, we also have $ \displaystyle \frac{PH}{HQ} = \frac{r_1}{r_2}$.

So, $\triangle PHQ \sim \triangle ABC \sim \triangle AHB \sim \triangle BHC \implies \angle HPQ = \angle A$.

If $r$ is the inradius of $\triangle PHQ$, $HI = r \sqrt2$

Using similarity of triangles,

$ \displaystyle \frac{AH}{r_1} = \frac{PH}{r} \implies AH = \frac{\sqrt2 r_1^2}{r} $. Similarly, $\displaystyle CH = \frac{\sqrt2 r_2^2}{r}$

Adding, $ \displaystyle AC = \frac{2 (r_1^2 + r_2^2)}{r \sqrt2} = \frac{PQ^2}{r \sqrt2} \implies \frac{AC}{PQ} = \frac{PQ}{r \sqrt2}$

As $\dfrac{AC}{PQ} = \dfrac{10}{r}, PQ = 10 \sqrt2$

$PH + HQ - PQ = 2r \implies \sqrt2 (r_1 + r_2) - 10 \sqrt2 = 2 r$. Using $r_1 + r_2 = BH - 10$, we get $ ~ BH - 20 = \sqrt2 r$.

Given $PI$ is angle bisector of $\angle HPE$ and $\angle BPI = 90^\circ$ so it follows that pencil $P (HE, IB)$ is harmonic. So we have,

$\dfrac{HI}{IE} = \dfrac{HB}{BE}$

$\dfrac{HI}{HE-HI} = 1 + \dfrac{1}{5 \sqrt2}$

$ (10 \sqrt2 + 1) HI = (5 \sqrt2 + 1) HE = BH$

$(20 + \sqrt2) r = BH = 20 + \sqrt2 r$

$ \implies r = 1$

Answer 2

$S$ the point of touching incircle ABH with BH. $T$ is the point of intersection of straight lines $PS$ and $QN$. Triangles $PES$ and $PQT$ are right and similar. $PS=r_1$, $PT=r_1+r_2$, $QT=r_2-r_1$. From similarity:

$$ES=PS\cdot QT/PT=r_1\frac{r_2-r_1}{r_1+r_2}$$

$$EH=ES+SH=r_1\frac{r_2-r_1}{r_1+r_2}+r_1=\frac{2r_1r_2}{r_1+r_2}$$

From similarity of triangles $AHB$, $CHB$ and $ABC$ and Pythagoras theorem follows $$R^2=r_1^2+r_2^2\Rightarrow 2r_1r_2=(r_1+r_2)^2-R^2\Rightarrow EH=r_1+r_2-\frac{R^2}{r_1+r_2}$$

$$BH=BE+EH=5\sqrt{2}\cdot EH+EH=EH(5\sqrt{2}+1)$$

Also $BH=r_1+r_2+R$, then

$$r_1+r_2+R=\left(r_1+r_2-\frac{R^2}{r_1+r_2}\right)(5\sqrt{2}+1)$$

Let $r_1+r_2=xR$, then

$$x+1=\left(x-\frac{1}{x}\right)(5\sqrt{2}+1)\Rightarrow x=(x-1)(5\sqrt{2}+1)\Rightarrow x=1+\frac{\sqrt{2}}{10}$$

The incircle radius of tectangular triangle $PHQ$ is $$\frac{PH+HQ-PQ}{2}=\frac{r_1\sqrt{2}+r_2\sqrt{2}-\sqrt{2r_1^2+2r_2^2}}{2}=R\sqrt{2}\ \frac{x-1}{2}=\frac{R}{10}=1$$

The answer will transform to 2 if one changes $5\sqrt{2}$ to $\frac{5}{\sqrt{2}}$.