Every constant width set has a circumscribed regular hexagon
Every plane set of constant width has a circumscribed hexagon (in the sense that every side of the hexagon meets the boundary of the set).
This rather simple property leads to a very quick proof of the Blaschke Lebesgue Theorem.
Proof: Pick a direction . Consider the tangent lines of for the directions and . Since these lines form a parallelogram with the same distances between its pairs of parallel lines, we obtain a rhombus which has all its sides tangent to and has an angle (let’s say ) of measure . Consider now the tangents at parallel to the diagonal . Since is strictly inside , these tangents intersect the rhombus by two segments and .
Now, the hexagon has all angles equal to , but may not be regular. Nevertheless, if the hexagon is regular. It is a standard continuity argument to prove that we may, in fact, choose such a direction for which the correspondinv and have the same length. Consider varying on . If at the beginning we have (for example) then at the end the configuration is exactly opposite and we have . Since the variation is continuous, there must be a for which , and we are done.
References: H.G. Eggleston – Convexity

October 7, 2013 at 12:33 amBlaschkeLebesgue Theorem  Beni Bogoşel's blog