A regular hexagon area H has its vertices on the perimeter of a convex polygon of area A. Prove that 2A ≤ 3H. When do we have equality?
Solution
Solution by Bazavan Eduard
We use the basic result that there is at least one support line through any point on the boundary of a convex set. The support line has the property that all points of the convex set lie on one side of the line (or on the line itself). Since each vertex of the hexagon lies on the boundary of the polygon, there is a support line through each vertex of the hexagon.
Let the hexagon be ABCDEF. Take points P, Q outside the hexagon so that ABP, BCQ are equilateral triangles. Let the support lines through A, B, C, D, E, F be LA, LB, LC, LD, LE, LF respectively. Then these lines bound a hexagon H' which contains the convex polygon, so it is sufficient to show that area H' ≤ 3H/2. Let LA and LB meet at X and let LB and LC meet at Y. The area of H' is H plus the area of the triangle ABX and the area of the five corresponding triangles on the other sides. We claim that area ABX + area BCY ≤ H/6. It follows that the area of all six triangles is ≤ H/2 and so the required result follows.
Extend XY to meet AP at R and CQ at S. Since the hexagon area H is regular, AP is parallel to CQ and hence angle ARB = angle QSB. Also angle ABR = angle QBS and AB = BQ. So triangles ABR and QBS are congruent and hence have equal area. So area ABX + area BCY <= area ABR + area BCS = area QBS + area BCS = area BCQ = H/6, which establishes the claim. If we have equality then area ARX = 0, so LA must coincide with the line AB or the line FA. Without loss of generality, it coincides with FA. Then LB must coincide with the line BC, so H' is an equilateral triangle whose sides contain three sides of the hexagon. It is easy to see that this does indeed give equality.
© John Scholes
jscholes@kalva.demon.co.uk
5 Sep 2002