Painting - Heptagon from Ten Equal Lines

This is one of a series of paintings in which Crockett Johnson explored ways of constructing the regular heptagon. The construction is his own, and a drawing for it is attached to the back of the painting. By an arrangement of ten equal line segments, he produced three sides and two angles of a regular heptagon. Two sides and one angle are actually shown in the painting.

Crockett Johnson supposed that four equal isosceles triangles, constructed with six equal line segments, were arranged as shown in his figure to form sides of a rhombus and of a parallelogram within it. Two adjacent sides of the rhombus also served as the long sides of equal triangles oriented in the opposite direction. Finally, a line parallel to one of these sides passed through points of intersection of the sides of triangles.

More specifically, in the drawing triangles BAF, DAR, DKE, and HBE are arranged within rhombus ABED, and around a central parallelogram. Two other equal triangles DES and BAG are also included. AFand EJ intersect at a point C and EK and BH at a point P. The tenth line, UL parallel to BE, passes through points C and P. Crockett Johnson claimed that BCPE represents three sides of a regular heptagon. His argument appears in his papers. The painting shows only the ten equal lines described in the title.

The sections of the rhombus are in black, white, and rose, with a purple background. There is a wooden frame painted purple. This oil painting on masonite is #109 in the series. It is marked on the back: HEPTAGON FROM TEN EQUAL LINES (/) Crockett Johnson 1973. Taped to the back is a sheet of paper with an explanation that is entitled: HEPTAGON FROM TEN EQUAL LINES.

Currently not on view