Monday, March 2, 2009

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

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

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.

Posted by Antonio Gutierrez at 4:53 PM
Labels: , , , ,

4 comments:

  1. c .t . e. oJanuary 8, 2010 at 8:42 AM

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

    ReplyDelete
    Replies
    1. GregDecember 13, 2020 at 7:36 AM

      Beautiful solution.

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

      Delete
  2. Peter TranOctober 20, 2011 at 5:20 PM

    nice solution

    ReplyDelete
  3. GregDecember 13, 2020 at 7:40 AM

    A 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.

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

    Then divide the sums by p+q+r.

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