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Problem 797

AMC 12 late, AIME early

Geometry Difficulty 4.3 Prove it Baltic Way competition problems · Baltic Way · 1992

Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.

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

Solution:

Let $K$ , $L$ , $M$ be the midpoints of the sides $AB$ , $BC$ , $AC$ of a non-obtuse triangle $ABC$ (see Figure 2). Note that the centre $O$ of the circumcircle is inside the triangle $KLM$ (or at one of its vertices if $ABC$ is a right-angled triangle). Therefore $|AK| + |KL| + |LC| > |AO| + |OC|$ and hence $|AB| + |AC| + |BC| > 2(|AO| + |OC|) = 2d$ , where $d$ is the diameter of the circumcircle.

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