IMO Hk TST

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Hong Kong

Let the angle bisectors of $\angle A$ , $\angle B$ , $\angle C$ of $\triangle ABC$ intersect the circumcircle of $\triangle ABC$ at $P$ , $Q$ , $R$ respectively. Prove that $AP + BQ + CR > BC + CA + AB$ .

Solution

Let $I$ be the incentre of $\triangle ABC$ . It is well-known that $PB = PI = PC$ , etc. By the triangle inequality, we have
$\begin{aligned} & BI + CI > BC, \\ & CI + AI > CA, \\ & AI + BI > AB, \\ & 2IP = BP + CP > BC, \\ & 2IQ = CQ + AQ > CA, \\ & 2IR = AR + BR > AB. \end{aligned}$
Adding these inequalities and dividing both sides by 2, we obtain the desired inequality.

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