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China Hong Kong Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Hong Kong

Let $I$ and $O$ be the incentre and circumcentre of $\triangle ABC$ , respectively. Assume $\triangle ABC$ is not equilateral (so $I \neq O$ ). Prove that $\angle AIO \leq 90^\circ$ if and only if $2BC \leq AB + CA$ .

Solution

Let $AI$ meet $(ABC)$ again at $D$ . Recall that $DB = DI = DC$ . Applying Ptolemy's theorem to $ABCD$ , we obtain
$AD \times BC = AB \times CD + AC \times BD = DI(AB + AC).$
Note that $\angle AIO \le 90^\circ$ if and only if $AI \ge ID$ . This is equivalent to
$2 \le \frac{AD}{DI} = \frac{AB + AC}{BC}.$
This holds if and only if $2BC \le AB + AC$ as desired.

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