Subject: Mathematics — Circles
Question: Let $\bigtriangleup ABC$ be equilateral on side $BA$
produced, we choose a point $P$ such that $A$ lies between $P$ and $B$.
We now denote "$a$" as the length of a side of $\bigtriangleup ABC$;
$r_{1}$ as the radius of incircle of $\bigtriangleup PAC$; and $r_{2}$
as the radius of the excircle of $\bigtriangleup PBC$ with respect to
side $BC$. Determine the sum $\left( r_{1} + r_{2} \right)$ as a
function of "$a$" alone.
Answer: $\frac{a\sqrt{3}}{2}$
Solution: Looking the figure, we see that
$\angle T_{1}O_{1}R = 60^{\circ}$ and it is the supplement of
$\angle T_{1}AR = 120^{\circ}$ \[as an exterior angle for
$\bigtriangleup ABC$\].
Hence,
$$\angle AO_{1}R = 30^{\circ}.$$
[IMAGE] Similarly, we obtain $\angle BO_{2}S = 30^{\circ}$.
Since tangents drawn to a circle from external points are equal, we
have
$$T_{1}T_{2} = T_{1}A + AB + BT_{2} = RA + SB + AB.$$
$$T_{1}T_{2} = r_{1}\tan 30^{\circ} + a + r_{2}\tan 30^{\circ} = \frac{r_{1} + r_{2}}{\sqrt{3}} + a,$$
and
$$T'_{1}T'_{2} = T'_{1}C + CT'_{2} = CR + CS = (a - RA) + (a - SB).$$
$$T'_{1}T'_{2} = 2a - \left( \frac{r_{1} + r_{2}}{\sqrt{3}} \right).$$
Since common external tangents to two circles are equal,
$$T_{1}T_{2} = T'_{1}T'_{2}.$$
Hence,
$$\frac{r_{1} + r_{2}}{\sqrt{3}} + a = 2a - \frac{r_{1} + r_{2}}{\sqrt{3}}.$$
Hence we find that
$$r_{1} + r_{2} = \frac{a\sqrt{3}}{2}.$$