Subject: Mathematics — Circles
Question: In a $\bigtriangleup ABC$, $r_{A},r_{B},r_{C}$ are the
radii of the circles which touch the incircle and the sides emanating
from the vertices $A,B,C$ respectively. Then find the value of
$\sqrt{r_{A}r_{B}} + \sqrt{r_{B}r_{C}} + \sqrt{r_{C}r_{A}}$ in terms
of $r.$
Answer: $r$
Solution: Let the circle of radius $r_{A}$ touch the sides $AB$ and
$AC$ at $D'$ and $E'$, whereas the incircle touches these sides at $D$
and $E$. Let $O$ and $O'$ be centres of the inscribed circle and that of
the circle with radius $r_{A}$. $O$ and $O'$ lie on the bisector of
angle $A$.
[IMAGE] Also,
$$AO' = O'D'cosec\frac{A}{2} = r_{A}cosec\frac{A}{2},$$
and
$$AO = ODcosec\frac{A}{2} = rcosec\frac{A}{2}.$$
Hence,
$$OO' = r + r_{A} = AO - AO'.$$
$$OO' = rcosec\frac{A}{2} - r_{A}cosec\frac{A}{2}.$$
Therefore,
$$\frac{r_{A}}{r} = \frac{cosec\frac{A}{2} - 1}{cosec\frac{A}{2} + 1} \Rightarrow r_{A} = r\tan^{2}\left( \frac{\pi - A}{4} \right).$$
Similarly,
$$r_{B} = r\tan^{2}\left( \frac{\pi - B}{4} \right),\quad r_{C} = r\tan^{2}\left( \frac{\pi - C}{4} \right).$$
Let
$$X = \frac{\pi - A}{4},\quad Y = \frac{\pi - B}{4},\quad Z = \frac{\pi - C}{4}.$$
Hence,
$$\sqrt{r_{A}r_{B}} + \sqrt{r_{B}r_{C}} + \sqrt{r_{C}r_{A}} = r\left( \tan X\tan Y + tanY\tan Z + tanZ\tan X \right).$$
Since
$$X + Y + Z = \frac{\pi}{2}$$
we have
$$\tan X\tan Y + tanY\tan Z + tanZ\tan X = 1.$$
Therefore,
$$\sqrt{r_{A}r_{B}} + \sqrt{r_{B}r_{C}} + \sqrt{r_{C}r_{A}} = r.$$