Kepler's Triangle

Here is another construction for you all. (It must seem like I must do nothing but these, but in my defense, I've been sick as a dog for a long time and I seem to be fit for nothing but mathematics when I'm sick.)

Given points A and B, draw

1. circle AB
2. circle BA intersecting circle AB at points C and D,
3. circle CD intersecting circle AB at point E≠D and circle BA at point F≠D,
4. circle AF,
5. circle BE intersecting circle AF at points G and H,
6. circle CH,
7. circle HE intersecting circle CH at point I;

then triangle CDI is Kepler's triangle. (Yes, that Kepler, the pre-eminent astronomer.) This triangle is the unique right triangle with edges forming a geometric progression: 1:x:x². (Curiously, the Great Pyramid of Giza when it was built—it has now weathered considerably—had proportions matching Kepler's triangle to three decimals.)

I found this by doodling around and ending up with this bizarre construction:

Lines EABF and IDJ sure look parallel, don't they? But that's weird, I was just doodling randomly and these circles have pretty arbitrary radii, so it wouldn't make sense for the intersection points to line up so nicely. So I just had to prove to myself that they were, in fact, parallel. I'll spare you my original trigonometric proof, which involved walking through five triangles using the law of cosines,* and give you a much simpler sketch using our circle-circle intersection formula. Due to Kurt Hofstetter, CH has a radius of √3ϕ. One can use the Pythagorean theorem to derive that circle HE has a radius of √6. The intersection points of circles CH and HE are therefore located at distance ((√3ϕ)²-(√6)²+(√3ϕ)²)/(2√3ϕ)=√3 from C to H. But CD=√3. Therefore CDI is a right angle, and lines AB and ID are, indeed, parallel. But the magic trick is in applying the Pythagorean theorem to triangle CDI to find DI=√3√ϕ—but √ϕ is a pretty funny number, isn't it? It means that CD:DI:CI=1:√ϕ:ϕ—which is indeed 1:x:x²—and therefore triangle CDI is Kepler's triangle. Crazy!

* Fun fact: back in high school, I ignored all the various theorems that my geometry teacher wanted me to use and instead drew tons and tons of triangles, proving whatever construction was requested using trigonometry. This was nice and easy to do—no thinking required—but the proofs routinely ran to multiple pages. About halfway through the school year my teacher stopped bothering to check my proofs. I know this since I started giving her faulty proofs in order to test her. ;)

Construction of a Regular Heptagon

The construction of a regular heptagon was unknown to the Greeks, and was only finally managed by Renaissance geometers.

Given points A and B, draw

1. circle AB,
2. circle BA intersecting circle AB at points C and D,
3. circle CD intersecting circle AB at point E≠D and circle BA at point F≠D,
4. circle EB intersecting circle BA at point G,
5. circle FG intersecting circle AB at point H;

then BH is the side of a regular heptagon, and may be copied around the edge of circle AB to form the other sides.

...just kidding! The regular heptagon is impossible to draw with a straightedge and compasses (or indeed with compasses alone), and this fact was known at least as early as Kepler. In fact, the regular heptagon wasn't even unknown to the ancients: Archimedes managed to construct one with only slightly more sophisticated tools. The one I've constructed here is just an approximation, though a very good one, and is related to a construction by Albrect Dürer.