Haag Hexagon and its generalization to a Haag Polygon

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Construction: Construct equilateral △ABC. Draw a circle with centre A and arbitrary radius AD. Construct a circle with centre B and radius BD. Label as P the other intersection of the two circles centred at A and B. Draw a circle with centre C and radius CP. Label as E and F respectively the other intersections of circle C with circles B and C. Then ∠DAF = ∠DBE = ∠ECF = 120° and ADBECF is a Haag hexagon. The construction and properties of Haag hexagon is mentioned in the notes of the famous Dutch artist M.C. Escher (see Schattschneider, 1990, p. 90). The sketch below is dynamic: drag D, B or C to change it.

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Construction of the Haag hexagon

1) Can you explain why (prove) ∠DAF = ∠DBE = ∠ECF = 120° from the above construction?
2) Can you explain why (prove) the main diagonals AE, BF and DC of the Haag hexagon ADBECF are concurrent?

Schattschneider, D. (1990). M.C. Escher: Visions of Symmetry. New York: W.H. Freeman & Co.

Haag Tiling
One can create a tiling with the Haag hexagon, by starting with a basic tiling of the plane with equilateral triangles, and then covering the plane by rotating the Haag hexagon by 120° and/or 240° around each vertex of the equilateral triangular grid as shown in the dynamic sketch below.

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Haag Tiling

It is not clear whether Escher actually used the Haag tiling to create some of his art, but it is strongly suggestive that some of his work such as the ‘running men’ shown below displays the same rotational symmetry as the Haag tiling.

Running Men

Haag Polygon
One way of generalizing the Haag hexagon is by applying the Haag ‘circle’ construction mentioned above to a general triangle as well as to other polygons such as quadrilaterals, pentagons, hexagons, etc. to create a ‘Haag polygon’. Though the general Haag hexagon created from a general triangle does not tile and its main diagonals are not concurrent, it still has the Haag property as shown in the dynamic sketch below, that irrespective of the position of point D, ∠DAF = 2∠BAC, ∠DBE = 2∠ABC and ∠ECF = 2∠BCA.

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General Haag Hexagon

1) Can you explain why (prove that) the angle relationship above is valid for a general Haag hexagon, and Haag polygon?
2) Which type of quadrilateral would produce a Haag octagon with adjacent sides collinear; i.e. in other words a degenerate quadrilateral? (If necessary, you could use the 1st sketch below to investigate this question).

Below are dynamic sketches for a Haag octagon and a Haag dodecagon, respectively created from a parallelogram ABCD and hexagon ABCDEF with opposite sides parallel. Drag the points as indicated. What do you notice? Can you make some conjectures?

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Haag Octagon created from parallelogram ABCD

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Haag Dodecagon formed from Hexagon ABCDEF with opposite sides parallel

State your conjectures related to the above two sketches and explain why (prove that) they are true.

Read my paper An Investigation of Some Properties of the General Haag Polygon (pdf, 553 KB, Mathematics in School, 43(3), May 2014, pp. 15-18).

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Created by Michael de Villiers, 27 October 2013.