1970 IMO Problems/Problem 1

Problem

( Proposed by Poland ) Let $\displaystyle M$ be a point on the side $\displaystyle AB$ of $\displaystyle \triangle ABC$ . Let $\displaystyle r_1, r_2$ , and $\displaystyle r$ be the inscribed circles of triangles $\displaystyle AMC, BMC$ , and $\displaystyle ABC$ . Let $\displaystyle q_1, q_2$ , and $\displaystyle q$ be the radii of the exscribed circles of the same triangles that lie in the angle $\displaystyle ACB$ . Prove that

$\displaystyle \frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$ .

Solution

We use the conventional triangle notations.

Let $\displaystyle I$ be the incenter of $\displaystyle ABC$ , and let $\displaystyle I_{c}$ be its excenter to side $\displaystyle c$ . We observe that

$r \left[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) \right] = c$ ,

and likewise,

$\begin{matrix} c & = & \displaystyle q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\ & = & \displaystyle q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}$