Proposed Problem

See complete Problem 262 at:

gogeometry.com/problem/p262_regular_hexagon_inscribed_circle_distance.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, March 2, 2009

### Problem 262: Regular Hexagon inscribed in a circle, sum of distances

Labels:
distance,
equilateral,
hexagon,
regular polygon,
triangle

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draw DKH perpendicular to e ( K on f, H on circle )

ReplyDelete=> GDK, KHF equilat (ang GDK=GKD=60,GDK=GFH arc GBAH)

join H to B => BHK = 60 => AHFG trapezoid => HF = b = KF

GK + KF = f = d + b ( at the same way, e = a + c )

Beautiful solution.

DeleteJust one typo : BHK = 60 => "BHFG regular trapezoid" instead of "AHFG trapezoid".

nice solution

ReplyDeleteA more laborious proof involves applying Ptolemy to 6 quadrilaterals to express e and f as functions of a, b, c and d and p, q and r where p = side of the hexagon, q = chord intercepting the arc made by 2 sides and r that made by 3 sides.

ReplyDeleteFor example : quadrilaterals GCEB, GDEA, GDEB, GCFA, GCFB and GDFA.

Then divide the sums by p+q+r.