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This nice problem is dedicated to Dirk Laurie (1946-2019), who gave it to me about 10 years or so ago. He was an outstanding mathematician, and colleague for almost 40 years, and passed away suddenly in 2019. He and I worked together on the SA Mathematics Olympiad Committee for many years. He was also the South African IMO team leader on several occasions as well as the Chief Coordinator at IMO 2014 in Cape Town. Read a short tribute to him at: "Passing of a remarkable South African mathematician" or this scanned report (in Afrikaans) in the 18 August 2019 Rapport newspaper at: "Briljantste' mens skryf oor 'Onvolkome Kennis'".

**Problem**: Given a cyclic hexagon *ABCDEF* with alternates sides *AB* = *CD* = *EF* = *R* (the radius of the circle)^{1}, prove that the respective midpoints *G*, *H* and *I* of the other set of alternate sides *FA*, *BC* and *DE* form an equilateral triangle.

Dirk Laurie Tribute Problem

**Challenge**

1) Can you *explain why* (**prove that**) the above result is true?

2) Though the problem can probably be most easily attacked using trigonometry, the real challenge is finding a purely geometric solution. Can you do so?

3) If you get stuck, have a look at *Pompe's Hexagon Theorem*. Or alternatively, try using Napoleon Related Variation 2 on this page *Related Variations & Generalizations of Napoleon's Theorem*.

4) Can you generalize the result? Hint: is it necessary that the outer six vertices of the equilateral triangles that form the hexagon are cyclic?

5) Check your exploration in 4 above at: *Some further generalizations*.

Read the following solutions (2020) by Michael Fried (Israel) to *Dirk Laurie's Hexagon Problem*.

**Footnote 1**: In general, a cyclic hexagon with a set of alternate sides equal has 3 pairs of adjacent angles equal and can be viewed as one possible generalization of an *isosceles trapezium*. More information is available at *Alternate sides cyclic-2n-gons*.

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Created 6 April 2020 by Michael de Villiers; updated 24 July 2020.