The area of infinitely many circles inside a right angle triangle A right angle triangle with sides $s_1$ and $s_2$, and hypotenuse $h=\sqrt{s_1^2+s_2^2}$ is containing infinity many circles as follows; $\omega_1$ is the circle which is tangential to $s_1,s_2,h$ (i.e. the in-circle). $\omega_2$ is the circle which is tangential to $\omega_1,s_1,h$.  $\omega_3$ is the circle which is tangential to $\omega_2,s_1,h$.  $\omega_4$ is the circle which is tangential to $\omega_3,s_1,h$. In general, for $n>1$, $\omega_n$ is the circle which is tangential to $\omega_{n-1},s_1,h$.


Knowing $s_1$ and $s_2$, how can we determine the total area of these
  infinitely many circles?

Any help would be really appreciated. THANKS!
 A: 
Let $r_1$, $r_2$, $r_3$ … be the radii. The area of the triangle is,
$$\frac12 r_1\left(s_1+s_2+\sqrt{s_1^2+s_2^2}\right)=\frac12s_1s_2$$
Then,
$${r_1}=\frac12\left(s_1+s_2-\sqrt{s_1^2+s_2^2}\right)
=\frac{s_1}2(1+t-\sqrt{1+t^2}),\>\>\>\>\>\>t=\frac{s_2}{s_1}\tag 1$$
Also, the radii form a geometric series with ratio $a$ and,
$$AO = r_1+2r_2+2r_3+...= \frac{2r_1}{1-a}-r_1= \sqrt{(s_1-r_1)^2+r_1^2}$$
Solve for the ratio with (1),
$$a = \frac{\sqrt{(1+\sqrt{1+t^2})^2+t^2}-t}{\sqrt{(1+\sqrt{1+t^2})^2+t^2}+t}\tag 2$$
The total area, thus, is
$$A=\pi(r_1^2+r_2^2+r_3^2+...) = \frac{\pi r_1^2}{1-a^2}
=\frac{\pi\left(s_1+s_2-\sqrt{s_1^2+s_2^2}\right)^2}{4(1-a^2)}$$
where $a$ is given by (2).
A: 
$$\frac {x}{r_3}=\frac{x+r_2+r_3}{r_2}=\frac{x+r_1+2r_2+r_3}{r_2}$$
$$\frac{r_2+r_3}{r_2-r_3}=\frac{r_1+r_2}{r_1-r_2}$$
$$\Rightarrow r_1, r_2, r_3 are in GP.$$
Thus, radius of all the circles will form an infinite decreasing GP
$$ r_1+r_1cot\theta=s_1 \Rightarrow r_1=\frac{s_1}{1+cot\theta}$$
$$r_1=\frac{s_1}{1+cot\theta}=\frac{\Delta}{s}$$ 
$$\Rightarrow cot\theta=\frac {s_1+h}{s_2}$$
$$ r_1.cot\theta -r_2. cot\theta=2\sqrt{r_1r_2}$$
$$ \Rightarrow \frac{r_1}{r_2}-2\sqrt{\frac{r_1}{r_2}}tan\theta-1=0$$
$$ \Rightarrow \sqrt{\frac{r_1}{r_2}}= |tan\theta \pm sec\theta|$$
$Sum of area= \pi(r_1^2+(r.r_1)^2.....)$
$$S=\frac{\pi r_1^2}{1-r^2}$$
A: The first circle $\omega_1$ is the incircle of the right triangle. Its radius is $\rho=\frac{s_1+s_2-h}2$.
The centres of the circles all lie on the bisector of the left corner. Denote the radii of two successive circles by $R$ and $r$, with $R\gt r$. The line segment connecting their centres has length $R+r$, and its vertical projection has length $R-r$. Thus, with $\alpha$ denoting the angle between $h$ and $s_1$, we have
$$
\frac{R-r}{R+r}=\sin\frac\alpha2=\sqrt{\frac{1-\cos\alpha}2}\;.
$$
Solving for $\frac rR$ yields
$$
\frac rR=\frac{1-\sin\frac\alpha2}{1+\sin\frac\alpha2}\;.
$$
The radii of the circles form a geometric progression, and the sum of their areas is
\begin{eqnarray}
\sum_{k=0}^\infty\pi\rho^2\left(\frac rR\right)^{2k}
&=&
\pi\rho^2\frac1{1-\left(\frac rR\right)^2}
\\
&=&
\pi\rho^2\frac{\left(1+\sin\frac\alpha2\right)^2}{4\sin\frac\alpha2}
\\
&=&
\pi\rho^2\left(\frac{3-\frac{s_1}h}{4\sqrt{2\left(1-\frac{s_1}h\right)}}+\frac12\right)\;.
\end{eqnarray}
If the right triangle is isosceles, this is
\begin{eqnarray}
\pi h^2\left(\frac{\sqrt2-1}2\right)^2\left(\frac{3-\frac1{\sqrt2}}{4\sqrt{2\left(1-\frac1{\sqrt2}\right)}}+\frac12\right)
&=&
\pi h^2\left(\frac{22-15\sqrt2}{32\sqrt{2-\sqrt2}}+\frac{3-2\sqrt2}8\right)
\\
&=&
\pi h^2\left(\frac38-\frac1{2\sqrt2}+\frac{\sqrt{274-193\sqrt2}}{32}\right)
\\
&\approx&0.1683h^2\;,
\;.
\end{eqnarray}
compared to the triangle area $\frac14h^2$, so in this case roughly two thirds of the triangle is covered.
In the limit $\frac{s_1}h\to0$, the area of the incircle goes to zero, and so does the total area of the circles, since they lie in a small corner right next to the incircle. More interesting is the limit $\frac{s_2}h\to0$, that is, $\frac{s_1}h\to1$. Here, too, the area of the incircle goes to zero, but the circles fill the entire length of the triangle, covering a finite area.
To first order, $s_1=\sqrt{h^2-s_2^2}\approx h-\frac12\frac{s_2^2}h$, and $\rho\approx\frac{s_2}2$, so the total area goes as
$$
\frac{\pi s_2^2}4\cdot\frac{2h}{4s_2}=\frac\pi8hs_2\;,
$$
compared to the area $\frac12hs_2$ of the triangle, so in this case a proportion $\frac\pi4$, a bit less than four fifths, of the area is covered by the circles – not surprisingly, since in this limit they are arranged between approximately parallel lines and thus approximately cover the proportion of area that the incircle of a square covers.
Expressing the proportion of covered area in terms of the angle $\alpha$ and simplifying yields
$$
\frac{(\cos\alpha+\sin\alpha-1)^2}{\sin2\alpha}\left(\frac{3-\cos\alpha}{8\sin\frac\alpha2}+\frac12\right)\pi\;.
$$
Here’s a plot for $\alpha\in[0,\frac\pi2]$:

A: 
Let
$s_1=|BC|$,
$s_2=|AC|$,
$h=|AB|=\sqrt{s_1^2+s_2^2}$,
and the inradius of $\triangle ABC$
\begin{align}
r_0&=\tfrac12\,(s_1+s_2-h)
.
\end{align} 
Construct isosceles 
$\triangle A_0BC_0:\ |BA_0|=|BC_0|$,
with the same incircle.
Let $D_0=\tfrac12\,(A_0+C_0)$.
\begin{align} 
\sin\tfrac\beta2&=
\frac{r_0}{\sqrt{r_0^2+(s_1-r_0)^2}}
=\frac{s_1+s_2-h}{2\,\sqrt{s_1^2+s_2^2-s_2\,h}}
.
\end{align} 
\begin{align} 
|BD_0|&=
r_0\,\Big(1+\frac1{\sin\tfrac\beta2}\Big)
\\
&=
\tfrac12\,(s_1+s_2-h)+\sqrt{s_1^2+s_2^2-s_2\,h}
,\\
|BD_1|&=|BD_0|-2\,r_0
,\\
q&
=\frac{|BD_1|}{|BD_0|}
=1-\frac{2\,r_0}{|BD_0|}
=\frac{1-\sin\tfrac\beta2}{1+\sin\tfrac\beta2}
\\
&=
\frac{d-r_0}{d+r_0}
,\quad
\text{where }\quad d=\sqrt{s_1^2+s_2^2-s_2\,h}
.
\end{align}
\begin{align} 
r_k&=r_0\,q^k
,\\
S_\infty
&
=\pi\,\sum_{k=0}^\infty r_k^2
=\pi\,r_0^2\,\sum_{k=0}^\infty q^{2\,k}
=\frac{\pi\,r_0^2}{1-q^2}
=\frac{\pi\,r_0\,(d+r_0)^2}{4\,d}
.
\end{align}
