Ratio of radii of two circles inscribed in a right isosceles triangle.

Question

There is a right isosceles triangle $\triangle ABC$ with the vertex $B$ facing the hypotenuse.

A circle is inscribed into the triangle with radius $r_1$, then another circle with radius $r_2$ is inscribed in the leftover space close to either $A$ or $C$ but not $B$

What is the ratio $\large{\frac{r_1}{r_2}}$ equal to?

My Attempt:

Let's call the circle with radius $r_1$, $C_1$ and the other circle with radius $r_2$, $C_2$

The smaller circle shall be closer to vertex $A$.

The Length from $A$ to $C_2$s tangents will be called $h_1$, and from these tangents to $C_1$s tangents will be called $h_2$. The length of the legs of the triangle will be called $x$.

If we draw a line from $C_1$ to $A$ we will see $r_1$ and $r_2$ are bases of similar triangles.

This means $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$

If we ignore $C_2$ we can see the triangle is made up of four smaller triangles and a square, since the sum of the area of these shapes will be equal to the area of the triangle: $$\large{2r_1(x-r_1)+r_1^2=\frac{x^2}{2}\\2xr_1-r_1^2=\frac{x^2}{2}\\-r_1^2+2xr_1-\frac{x^2}{2}=0}$$ From the quadratic equation:$$\large{\frac{-2x\mp\sqrt{4x^2-2x^2}}{-2}\\x\mp\frac{x}{\sqrt{2}}}$$ Since we know $r_1$ must be less than $x$ $$r_1=x-\frac{x}{\sqrt{2}}$$

$h_1+h_2$ is exactly half of the hypotenuse, this means $h_1+h_2=\frac{x}{\sqrt{2}}$

From this it follows that $$\large{\frac{h_1+h_2}{r_1}=\frac{\frac{x}{\sqrt{2}}}{x-\frac{x}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}}$$

Since $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$, $\large{\frac{h_1}{r_2}=\frac{1}{\sqrt{2}-1}}$ and $\large{(\sqrt{2}-1)h_1=r_2}$

Because $h_1+h_2=\frac{r_1}{\sqrt{2}-1}$$$\large{h_2=\frac{r_1-r_2}{\sqrt{2}-1}}$$

If we extend a line like so...

We can see $\large{(h_2)^2+(r_1-r_2)^2=(r_1+r_2)^2}$

Answer 1

In $\triangle OPQ$, $ \displaystyle PQ = r_1 - r_2, OQ = r_1 + r_2, \angle POQ = \frac{\pi}{8}$

$\displaystyle \frac{r_1 - r_2}{r_1 + r_2} = \sin \frac{\pi}8$

$ \implies \displaystyle \frac{r_1}{r_2} = \frac{1 + \sin (\pi /8)}{1 - \sin (\pi /8)}$

Answer 2

Look at your third diagram.

Follow the line from vertex A to the center of the smaller circle.

Where that first intersects the smaller circle label as point D.

Let $x$ denote the length of segment AD.

Then

\begin{equation} \frac{r_2}{x+r_2}=\sin\left(\frac{\pi}{8}\right)\tag{1} \end{equation}

and

\begin{equation} \frac{r_1}{x+2r_2+r_1}=\sin\left(\frac{\pi}{8}\right)\tag{2} \end{equation}

Solve equation (1) for $x$ in terms of $r_2$. Then substitute that for $x$ in equation (2).

The resulting equation can be solved for $r_1$ as a fixed constant times $r_2$ from which you can get the desired ratio of $r_1$ to $r_2$.

ALTERNATE APPROACH

This can also be solved using the geometric series:

When $|x|<1$ then

$$ 1+x+x^2+x^3+\cdots =\frac{1}{1-x} $$

Rather than the ratio $\dfrac{r_1}{r_2}$ let us consider its reciprocal $P=\dfrac{r_2}{r_1}<1$.

At present we have two circles, call them $C_1, C_2$. Continue that sequence of circles where $C_3$ bears the same relation to $C_2$ that $C_2$ bears to $C_1$. Keep that going so that we have a sequence of circles in the triangle approaching $A$. Let $D_1=2r_1,D_2=2r_2, D_3=2r_3, \cdots$.

The ratio of the diameters successive diameter will be the same as the ratio of their radii. So each diameter has length $P$ times the length of the diameter of the previous circle.

So the sum of all the circle diameters in the triangle is

$$ S=D_1+D_1P+D_1P^2+D_1P^3+\cdots=\frac{D_1}{1-P}=\frac{2r_1}{1-P} $$

Now we can solve for $S$ another way. It equals the sum of $r_1$ and the distance from the center of $C_1$ to $A$.

$$ S=r_1+r_1\csc\left(\frac{\pi}{8}\right) $$

So we have

\begin{eqnarray} \frac{2r_1}{1-P}&=&r_1+r_1\csc\left(\frac{\pi}{8}\right)\\ 1-P&=&\frac{2}{1+\csc\left(\frac{\pi}{8}\right)}\\ 1-P&=&\frac{2\sin\left(\frac{\pi}{8}\right)}{\sin\left(\frac{\pi}{8}\right)+1}\\ P&=&\frac{1-\sin\left(\frac{\pi}{8}\right)}{1+\sin\left(\frac{\pi}{8}\right)}\\ \frac{r_1}{r_2}&=&\frac{1}{P}=\frac{1+\sin\left(\frac{\pi}{8}\right)}{1-\sin\left(\frac{\pi}{8}\right)}\end{eqnarray}