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The function is strictly convex

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August 29, 2010
functiongeometryConvex FunctionsalgebraConvexityIMO Shortlist

Problem Statement

Let DD be the interior of the circle CC and let A∈CA \in C. Show that the function f:D→R,f(M)=∣MA∣∣MM′∣f : D \to \mathbb R, f(M)=\frac{|MA|}{|MM'|} where M′=AM∩CM' = AM \cap C, is strictly convex; i.e., f(P)<f(M1)+f(M2)2,∀M1,M2∈D,M1≠M2f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2 where PP is the midpoint of the segment M1M2.M_1M_2.
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