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Estonian Math Competitions

Geometry Difficulty 5.0 AIME, harder Prove it Estonia

A triangle with perimeter PP is divided into kk triangular pieces for some k≥2k \ge 2.

a. Show that there exists a piece with perimeter greater than Pk\frac{P}{k}.

b. Show that if the initial triangle is equilateral, then there exists a piece with perimeter at least Pk\frac{P}{\sqrt{k}}.

Solution

a. The sides of the initial triangle are distributed between the pieces. As k≥2k \ge 2, the sides of the pieces must also pass through the interior of the initial triangle. So the sum of the perimeters of the kk pieces is greater than PP. Hence there must exist a piece with perimeter greater than Pk\frac{P}{k}.

b. Let the area of the equilateral triangle be SS. Then there must exist a piece Δ\Delta with area at least Sk\frac{S}{k}. If Δ\Delta were equilateral, it would be similar to the initial triangle with a scale factor of 1k\frac{1}{\sqrt{k}}, so its perimeter would be Pk\frac{P}{\sqrt{k}}. However, among triangles with a fixed area, an equilateral triangle has the smallest perimeter. Therefore the perimeter of Δ\Delta is at least Pk\frac{P}{\sqrt{k}}.

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