A Geometry Theorem Looking for a Geometric Proof

[Update: Dan Lawson has proved the theorem without trigonometry. Thanks, Dan!]

I spent a good chunk of last week reading about David Johnson Leisk (1906–1975), who is better known by his nom-de-plum Crockett Johnson. Johnson is most well known as the author of Harold and the Purple Crayon, a children’s book from 1955, and its sequels. Johnson was also the author of the 1940s comic Barnaby.Harold_and_the_Purple_Crayon_(book)

Later in his life Johnson became interested in mathematics. He was particularly interested in geometry, and most specifically in the problems of antiquity (squaring the circle, trisecting the angle, doubling the cube, and constructing regular polygons). He turned many geometric theorems into works of art. Eighty of his paintings are now at the Smithsonian.

Johnson even created new mathematics. I would like to discuss one of his contributions here. (See his article “A Construction for a Regular Heptagon,” The Mathematical Gazette Vol. 59, No. 407 (Mar., 1975), pp. 17-21.)

The heptagon is notewothy because it is the regular polygon with the fewest number of sides that cannot be constructed with compass and straightedge alone. In his article, Johnson gives a way to construct the heptagon using a marked straightedge (this is called a neusis construction). Johnson did not give the first neusis construction of a heptagon—François Viète gave the first such construction in 1593. (Also, Archimedes gave an unorthodox neusis-like construction).

However, Johnson’s proof used trigonometry (including the law of cosines and several trigonometric identities). My question to you is: Is there a purely geometric proof of his result? I played around with it for a little while and couldn’t find one, and I couldn’t find a geometric proof in the literature.

The construction

The key to Johnson’s construction is producing 3:3:1 triangle; that is, a triangle in which the angles are in a 3:3:1 ratio (they would be 3\pi/7, 3\pi/7, and \pi/7). The three vertices of the triangle are three vertices of a regular heptagon. If we construct the circumcircle, then it is easy to construct the four remaining vertices with a compass and straightedge.

Here’s his construction of a 3:3:1 triangle using a marked straightedge—that is, an otherwise ordinary straightedge, but possessing a mark one unit from the end (or equivalently, two marks one unit apart). We begin by drawing a line segment AB of length one (see below, left). Construct a unit line segment AC perpendicular to AB. Also, construct the perpendicular bisector to AB; call it l. Then construct a circle with center B and radius BC. Now we perform the neusis construction with the marked straightedge: Construct a line AD so that D is on l and D is one unit from the circle. (That is, the end of the straightedge is at D, the mark is on the circle, and the edge passes through A.) Then \triangle ADB is the 3:3:1 triangle.


A geometric proof?

Johnson’s proof that \triangle ABD is a 3:3:1 triangle used trigonometry (see his article for details). Is there a geometric proof?

Boiled down to its essence, here’s the question: Suppose \triangle ABD is isosceles and E is on AD. Moreover, suppose ED=1,  AB=1, and BE=\sqrt{2}, prove that 3\angle ADB= \angle DAB=\angle ABD.


Here is a fact that may help. Johnson discovered this fact—ironically—when he was sitting in a café in Syracuse on Sicily (the birth place of Archimedes) playing with the wine and food menus and some tooth picks. If we have an isosceles triangle ABD with the point on BD and on AD  such that DE=EF=BF=AB, then triangle ABD is a 3:3:1 triangle. (This is easy to prove. Suppose \angle ADB=\theta. Then, because \triangle DEF is isosceles, \angle DFE=\theta. So the exterior angle of  \triangle DEF, \angle BEF, is 2\theta. Because  \triangle BEF is isosceles, \angle EBF=2\theta. Lastly, observe that  \triangle ABD and  \triangle ABF are similar, so \angle ABF=\theta. It follows that \angle ABD=\angle BAD=3\theta.)


Thus, a possible route to proving the theorem is to find a point in the original diagram so that DE=EF=BF=AB=1.

If you find a proof, I’d love to hear about it!

Here are two of the paintings Johnson made from his work with the heptagon.

Johnson’s “Heptagon from Its Seven Sides” (1973)
Johnson’s “Heptagon from Its Seven Sides” (1973)


  1. gaurish says:

    lovely post. Archimedes was also able to trisect an angle! see: http://wp.me/p4LQy6-e0

  2. jessemckeown says:

    More Archimedean connections: according to “Triumf der Mathematik” (H. Dörrie), the same argument given here that the wine-menu triangle is 3:3:1 was also used in Archimedes’ own neusis trisection method.

    1. Archimedes’s construction is very unorthodox—and very different from this one. His neusis-like construction is very unusual. He (or supposedly he—some scholars are skeptical that it is his work) performs a construction in which he draws a line through a point so that the two triangles it forms have equal area. Not your typical compass and straightedge move! Once he’s accomplished this, he is able to construct a pi/7 angle. So his goal isn’t to explicitly construct the 3:3:1 triangle, but he does get one along the way. Here’s an article that talks about his construction http://www.jstor.org/stable/41133724

      1. jessemckeown says:

        That is quite interesting, but I wasn’t remembering a heptagon construction at all; rather I was refering to the same thing gaurish mentioned (though I didn’t notice that he did untill quite later).

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