Circles with common tangent

Question

I am trying to solve this simple geometry problem but I am always tangled in so many equations it makes my head spin. I tried solving it via similar triangles but i cant seem to eliminate all the unwanted variables. Please help.

I have to prove $ r_1\times r_3=(r_2)^2$

Thank you

Answer 1

We have \begin{align} \triangle AOB:\quad |AO| &= \frac{r_1}{\sin\phi} ,\\ |AP|&=|AO|+|OP|=\frac{r_1}{\sin\phi}+r_1+r_2 ,\\ \triangle APC:\quad |AP| &= \frac{r_2}{\sin\phi} , \end{align}
which gives \begin{align} r_2&= \frac{r_1(1+\sin\phi)}{1-\sin\phi} . \end{align}

Similarly,

\begin{align} |AQ| &= |AP|+|PQ|=\frac{r_1}{\sin\phi}+r_1+2r_2+r_3 \\ &= \frac{r_1}{\sin\phi}+r_1+ \frac{2r_1(1+\sin\phi)}{1-\sin\phi} +r_3 ,\\ \triangle AQD:\quad |AQ| &= \frac{r_3}{\sin\phi} ,\\ \end{align}

hence

\begin{align} r_3&= \frac{r_1(1+\sin\phi)^2}{(1-\sin\phi)^2} ,\\ \end{align}

and

\begin{align} r_1r_3&=r_2^2 \end{align}

follows.

Answer 2

Due to the way they are constructed, the triangles $ABO,ACP,ADQ$ are similar, the figures $ABCPO$, $ACDQP$ are homothetic and $\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}.$

Answer 3

We have $h'$s of contacting circles from common apex A, the series of similar triangles:

$$ h_2-h_1=2r_1, h_3-h_2=2r_2, h_4-h_3=2r_3 \tag1$$

$$ \sin \alpha = \frac {(h_2-h_1)/2}{(h_2+h_1)/2}= \frac {h_2-h_1}{h_2+h_1} =\frac {h_3-h_2}{h_3+h_2}\tag2$$

Apply componendo/dividendo to each pair

$$\frac{1+\sin \alpha}{1-\sin \alpha}= \frac{h_2}{h_1} = \frac{h_3}{h_2} =\frac{h_4}{h_3} = = k \tag3$$

which gives successive geometric means that are also called powers of circle corresponding to extreme $h$ values product of each pair:

$$ r_1=\sqrt{h_1h_2},\, r_2=\sqrt{h_2h_3},\, r_3=\sqrt{h_3h_4} \, \tag4$$

$$ \frac{r_2^2 }{r_1 r_3}=\frac{h_2 h_3}{\sqrt{h_1h_2h_3h_4}}=\sqrt{\frac{h_2 h_3}{h_1 h_4}}=\sqrt{k/k}=1 \tag5$$