ISL 2002 G2

by Wolstenholme, Aug 3, 2014, 5:07 AM

Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]

Proof:

It is clear that $ F $ is the Fermat point of $ \triangle{ABC} $ so constructing points $ P, Q $ such that triangles $ \triangle{ACP} $ and $ \triangle{ABQ} $ are equilateral and have interiors disjoint from the interior of $ \triangle{ABC} $, we have that $ F = BP \cap CQ $.

Now it is clear that quadrilateral $ AFCP $ is cyclic so the area of $ \triangle{AFC} $ is maximal when $ F $ is the midpoint of minor arc $ \widehat{AC} $ of the circumcircle of quadrilateral $ AFCP $. In this case its area is one-third that of $ \triangle{ACP} $ so $ \frac{PD}{FD} \geq 3 \Longrightarrow \frac{FD}{FP} \leq \frac{1}{4} $. Similarly $ \frac{FE}{FQ} \leq \frac{1}{4} $.

Since $ F, E, Q $ are collinear and $ F, D, P $ are collinear, this implies that $ 4DE \leq PQ $ so it suffices to show that $ PQ \leq AB + AC $.

But $ PQ \leq AP + AQ = AB + AC $ by the Triangle Inequality so we are done.

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