IIT-JEE 2000Question 1
Let $ABC$ be a triangle with incentre $I$ and inradius $r$. Let $D,E,F$ be the feet of the perpendiculars from $I$ to the sides $BC,CA$ and $AB$ respectively. If $r_1,r_2$ and $r_3$ are the radii of circles inscribed in the quadrilaterals $AFIE$, $BDIF$ and $CEID$ respectively, prove that $$\frac{r_1}{r-r_1}+\frac{r_2}{r-r_2}+\frac{r_3}{r-r_3}=\frac{r_1r_2r_3}{(r-r_1)(r-r_2)(r-r_3)}.$$
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. Each quadrilateral (e.g. $AFIE$) is a kite with two right angles. Its inradius gives $\frac{r_1}{r-r_1}=\cot\frac A2$, and the claim becomes the identity $\sum\cot\frac A2=\prod\cot\frac A2$. Formulas used. (1) For a polygon with an incircle: inradius $=\dfrac{\text{area}}{\text{semi-perimeter}}$(2) $AF=AE=r\cot\frac A2$; and in any triangle $\sum\cot\frac A2=\prod\cot\frac A2$Step 1 - The kite. $AF=AE$ and $IF=IE=r$, with right angles at $F$ and $E$. So $AF+IE=AE+IF$, and the kite has an incircle (its centre is on $AI$).Step 2 - Its inradius. Area $=2\cdot\frac12\,r\cdot AF=r\cdot AF$ and semi-perimeter $=AF+r$. By (1), $$r_1=\frac{r\cdot AF}{r+AF}.$$ Step 3 - The ratio. $r-r_1=\dfrac{r^2}{r+AF}$, so $$\frac{r_1}{r-r_1}=\frac{r\cdot AF}{r+AF}\cdot\frac{r+AF}{r^2}=\frac{AF}{r}=\cot\frac A2\quad\text{by (2)}.$$ In the same way, $\frac{r_2}{r-r_2}=\cot\frac B2$ and $\frac{r_3}{r-r_3}=\cot\frac C2$. Step 4 - Conclude. The required equation is $\sum\cot\frac A2=\prod\cot\frac A2$, which holds in every triangle by (2) (since $\frac A2+\frac B2+\frac C2=\frac\pi2$). Hence proved.