I ran into this idea of "Pythagorean Polygons" on a problem from Project Euler, and I thought of an interesting question.

A "Pythagorean Polygon" is defined as a polygon that is cyclic and has its longest side be the diameter of a circle. It also always has integer sides.

Now, is there a way to construct a such polygon with any number of sides, where the number of sides is greater than $3$? Obviously $3$ is true, and $4$ you can do by reflecting a right triangle so that we have an isosceles trapezoid, but I'm not sure what else I can do from here.

  • $\begingroup$ One approach would be to start with a right triangle in a circle, say a 3-4-5. Now show that you can find a point on the arc outside the 3 that makes a rational triangle. Scale up to make the fractions integers. If the proof in the second sentence (which I don't have, else this would be an answer) doesn't depend on the right angle of the 3-4-5 you are done as you can always add a side. $\endgroup$ – Ross Millikan May 1 '15 at 4:41
  • $\begingroup$ Yeah, I see what you're saying, but I'm having trouble drawing every line of the proof together. A proof would be nice to see because I'm having trouble connecting all the pieces of logic together. $\endgroup$ – user231271 May 1 '15 at 5:49
  • $\begingroup$ Since you can always scale things up, you might as well ask for a polygon with rational edge lengths inscribed into half the unit circle. Most generally speaking you then want edge lenths $a_i\in\mathbb Q$ such that $$\sum_{i=1}^{n-1}2\sin^{-1}(a_i/2)=\pi$$ but I have no idea how to tackle this, so it might be better to look into specific polygons first. $\endgroup$ – MvG May 1 '15 at 9:40

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