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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.4 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle with $AB \neq AC$ . Its incircle has center $I$ and touches the side $BC$ at point $D$ . Line $AI$ intersects the circumcircle $\mathcal{C}$ of triangle $ABC$ at $M$ and $DM$ intersects again $\mathcal{C}$ at $P$ . Prove that $\widehat{API}=90^{\circ}$ .

Solution

Assume that $AB < AC$ . We have $MB = MC = MI$ . Indeed, from the hypothesis it follows
$\widehat{IBM} = \frac{1}{2}(\widehat{A} + \widehat{B}) = \widehat{BIM},$
hence triangle $BMI$ is isosceles.

Furthermore, since
$\widehat{BPM} = \frac{1}{2} \widehat{A} = \widehat{MBC}$
one has $\triangle DBM \sim \triangle BPM$ , hence $MB^2 = MD \cdot MP$ . It follows $MI^2 = MD \cdot MP$ , so $\triangle MDI \sim \triangle MIP$ , and consequently
$\begin{equation*} \widehat{MPI} = \widehat{MID}. \tag{1} \end{equation*}$
Notice now that $ID \parallel OM$ , implying $\widehat{MID} = \widehat{AMO}$ . Let $N$ be the antipodal point of $M$ in circle $\mathcal{C}$ . Finally, we have
$\text{API = APM - MPI = APM - AMN = 1 2 ( AM - AN ) = 1 2 MN = 90}$

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