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Problem 2002

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.3 Prove it IMO Hk TST · Hong Kong

Let $O$ be the centre of the excircle, touching the side $BC$ of $\triangle ABC$ . $M$ is the midpoint of $AC$ and $P$ is the intersection of lines $MO$ and $BC$ . Show that if $\angle BAC = 2\angle ACB$ , then $AB = BP$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $AO$ meet $BC$ at $D$ . Since
$\angle DAC = \angle BAD = \frac{1}{2} \angle BAC = \angle ACD,$
we have $DA = DC$ . Consider $\triangle OCA$ and $\triangle OCD$ . By considering their areas, we find that
$\frac{AO}{OD} = \frac{[OCA]}{[OCD]} = \frac{AC \times d(O, AC)}{CD \times d(O, CD)} = \frac{AC}{CD}.$

The last equality holds since $d(O, AC) = d(O, CD)$ is the A-exradius. Now, applying Menelaus' theorem to $\triangle ACD$ , we obtain
$\frac{DP}{PC} \times \frac{CM}{MA} \times \frac{AO}{OD} = 1.$
Using $CM = MA$ and $\frac{AO}{OD} = \frac{AC}{CD} = \frac{AC}{AD}$ , this becomes
$\frac{PD}{PC} = \frac{AD}{AC}.$

By the angle bisector theorem, we know that $AP$ bisects $\angle DAC$ . Therefore, we get
$\angle BAP = \angle BAD + \angle DAP = \angle ACP + \angle PAC = \angle APB.$
Thus, $AB = BP$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.