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A regular hexagon is split into two regions by a straight line so that the ratio of the perimeters of these regions is 2:1. Find the maximum ratio of the areas of the two regions.

This problem is inspired by one I saw on Brilliant.org today (credit goes to Digvijay Singh). It went as follows,

An equilateral triangle is split into two regions with equal perimeters. Find the maximum ratio of the areas of the two regions.

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Side length is of no relevance here, hence we can assume it is unit without loss of generality.

The region to max/minimise is a quadrilateral, composed of a red triangle with base $x$ and altitude ${\sqrt3\over2}$, and a blue triangle with base $1-x$ and altitude ${\sqrt3\over2}(1+x)$ (see picture). The area of the region is therefore $$ A={\sqrt3\over4}(1+x-x^2), $$ which attains its maximum for $x=1/2$ and its minimum for $x=0$ or $x=1$.

From the above it is easy to answer the given question: details are left to the reader.

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