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**Construction**: Construct equilateral △*ABC*. 1) Draw a circle with centre *A* and arbitrary radius *AD*. 2) Construct a circle with centre *B* and radius *BD*. Label as *P* the other intersection of the two circles centred at *A* and *B*. 3) 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 *A*. Then ∠*DAF* = ∠*DBE* = ∠*ECF* = 120° and *ADBECF* is a Haag hexagon. The construction and properties of a Haag hexagon is mentioned in the notes of the famous Dutch artist M.C. Escher (see Schattschneider, 1990, p. 90).

**Reference**

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

A Construction of the Haag hexagon (and more)

**Challenges**

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?

**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 above - click on the 'Link to Haag Tiling' button on the bottom right of the sketch.

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.

**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 above, that irrespective of the position of point *D*, ∠*DAF* = 2∠*BAC*, ∠*DBE* = 2∠*ABC* and ∠*ECF* = 2∠*BCA*. Click on the 'Link to General Triangle' button on the bottom right of the sketch to view the case for a general triangle.

**Challenges & Further Investigation**

1) Can you explain why (prove that) the angle relationship above is valid for a general Haag hexagon, and for a Haag polygon?

2) Drag the red vertices until the circumcentre *O* of △*ABC* coincides with *P*. What do you notice about the sides of the hexagon *ADBFCE* in this case? Can you prove your observations?

3) For the hexagon in 2), can you produce a tiling?

**Conjecturing: Haag Octagon & Haag Dodecagon**

Click on the relevant "Link' buttons in the above sketch to navigate to dynamic sketches for a Haag octagon *AEBFCGDH* and a Haag dodecagon *AGBHCIDJEKFL*, respectively created from a parallelogram *ABCD* and a hexagon *ABCDEF* with opposite sides parallel. Drag the points as indicated. What do you notice? Can you make some conjectures?

**Challenge**

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; updated to *WebSketchpad*, 26 January 2022.