Subject: Mathematics — Circles
Question: If $\Delta_{0}$ is the area of $\bigtriangleup$ formed
by joining the points of contact of incircle with the sides of the given
triangle whose area is $\Delta$. Similarly $\Delta_{1},\Delta_{2}$ and
$\Delta_{3}$ are the corresponding area of the $\bigtriangleup$ formed
by joining the points of contact of excircles with the sides. Find the
value of
$$\frac{\Delta_{1}}{\Delta} + \frac{\Delta_{2}}{\Delta} + \frac{\Delta_{3}}{\Delta} - \frac{\Delta_{0}}{\Delta}.$$
Answer: 2
Solution: Let $A_{1},B_{1}$ and $C_{1}$ be the points of contact of
the ex-circle opposite to vertex $A$, with the side $AC,BC$ and $AB$
respectively.
[IMAGE] Then,
$$I_{1}A_{1} = I_{1}B_{1} = I_{1}C_{1} = r_{1}.$$
$$\angle B_{1}I_{1}A_{1} = \pi - \angle B_{1}CA_{1} = \pi - (\pi - C) = C.$$
Similarly,
$$\angle A_{1}I_{1}C_{1} = B,$$
$$\angle C_{1}I_{1}B_{1} = (B + C).$$
Now,
$$\Delta_{1} = \Delta_{A_{1}I_{1}B_{1}} + \Delta_{A_{1}I_{1}C_{1}} - \Delta_{B_{1}I_{1}C_{1}}.$$
$$\Delta_{1} = \frac{1}{2}\left( r_{1}^{2} \right)\left( \sin C + sinB - sinA \right).$$
$$\Delta_{1} = \frac{1}{2}\frac{\Delta^{2}}{(s - a)^{2}}\left( \frac{c}{2R} + \frac{b}{2R} - \frac{a}{2R} \right).$$
$$\Delta_{1} = \frac{1}{2}\frac{\Delta^{2}}{(s - a)^{2}} \cdot \frac{(2s - 2a)}{2R} = \frac{\Delta^{2}}{2R(s - a)} = \frac{r_{1}\Delta}{2R}.$$
Therefore,
$$\frac{\Delta_{1}}{\Delta} = \frac{r_{1}}{2R}.$$
Similarly,
$$\frac{\Delta_{2}}{\Delta} = \frac{r_{2}}{2R},\quad\frac{\Delta_{3}}{\Delta} = \frac{r_{3}}{2R}\quad\text{and}\quad\frac{\Delta_{0}}{\Delta} = \frac{r}{2R}.$$
Thus,
$$\frac{\Delta_{1}}{\Delta} + \frac{\Delta_{2}}{\Delta} + \frac{\Delta_{3}}{\Delta} - \frac{\Delta_{0}}{\Delta} = \frac{1}{2R}\left( r_{1} + r_{2} + r_{3} - r \right).$$
$$\frac{\Delta_{1}}{\Delta} + \frac{\Delta_{2}}{\Delta} + \frac{\Delta_{3}}{\Delta} - \frac{\Delta_{0}}{\Delta} = \frac{1}{2R}(4R) = 2.$$