Home > Analysis, Geometry > Every constant width set has a circumscribed regular hexagon

## Every constant width set has a circumscribed regular hexagon

Every plane set ${K}$ 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 ${\theta \in [0,\pi]}$. Consider the tangent lines of ${K}$ for the directions ${\theta}$ and ${\theta+\pi/3}$. Since these lines form a parallelogram with the same distances between its pairs of parallel lines, we obtain a rhombus ${ABCD}$ which has all its sides tangent to ${K}$ and has an angle (let’s say ${A}$) of measure ${\pi/3}$. Consider now the tangents at ${K}$ parallel to the diagonal ${BD}$. Since ${K}$ is strictly inside ${ABCD}$, these tangents intersect the rhombus by two segments ${XY}$ and ${ZT}$.

Now, the hexagon ${XYDTZB}$ has all angles equal to ${2\pi/3}$, but may not be regular. Nevertheless, if ${XY=TZ}$ the hexagon is regular. It is a standard continuity argument to prove that we may, in fact, choose such a direction ${\theta}$ for which the correspondinv ${XY}$ and ${TZ}$ have the same length. Consider ${\theta}$ varying on ${[0,\pi]}$. If at the beginning we have (for example) ${XY then at the end the configuration is exactly opposite and we have ${XY>TZ}$. Since the variation is continuous, there must be a ${\theta}$ for which ${XY=TZ}$, and we are done.

References: H.G. Eggleston – Convexity