The area of infinitely many circles inside a right angle triangle

Question

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!

Answer 1

$$\frac {x}{r_3}=\frac{x+r_2+r_3}{r_2}=\frac{x+r_1+2r_2+r_3}{r_2}$$

$$\frac{r_2+r_3}{r_2-r_3}=\frac{r_1+r_2}{r_1-r_2}$$

$$\Rightarrow r_1, r_2, r_3 are in GP.$$

Thus, radius of all the circles will form an infinite decreasing GP

$$ r_1+r_1cot\theta=s_1 \Rightarrow r_1=\frac{s_1}{1+cot\theta}$$

$$r_1=\frac{s_1}{1+cot\theta}=\frac{\Delta}{s}$$

$$\Rightarrow cot\theta=\frac {s_1+h}{s_2}$$

$$ r_1.cot\theta -r_2. cot\theta=2\sqrt{r_1r_2}$$

$$ \Rightarrow \frac{r_1}{r_2}-2\sqrt{\frac{r_1}{r_2}}tan\theta-1=0$$

$$ \Rightarrow \sqrt{\frac{r_1}{r_2}}= |tan\theta \pm sec\theta|$$

$Sum of area= \pi(r_1^2+(r.r_1)^2.....)$

$$S=\frac{\pi r_1^2}{1-r^2}$$

Answer 2

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).

Answer 3

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 $\alpha\in[0,\frac\pi2]$:

Answer 4

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}