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<TZ} 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

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  1. October 7, 2013 at 12:33 am

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