Every so often I see spherical geometry/trigonometry questions on this site, and this has piqued my curiosity about geometric constructions on the sphere.

Constructions for the equilateral triangle, square and regular pentagon on a sphere are long known, even though for the square and pentagon you can't use the diagonal/side ratio from planar geometry. Instead we can get the square from the diagonals being perpendicular bisectors of each other, and the pentagon yields to the cube/dodecahedron relationship (The face diagonal of a regular dodecahedron are also the edges of inscribed cubes).

But what about the higher Fermat prime sided regular polygons, for which such relationships are not evident? How, if at all, can these polygons be constructed directly on a sphere?

  • $\begingroup$ Are you interested in constructing a regular polygon of a particular size on a sphere, or of any size? When you mention dodecahedron, you will only get a pentagon of a particular size. And what kind of construction are you looking for? Compass and "geodesic"-edge? $\endgroup$
    – robjohn
    Commented Jun 17, 2017 at 23:59
  • $\begingroup$ You can then construct concentric pentagons of any other size. The central angles of the dodecahedral face provides the necessary template. Likewise, a regular $n$-gon of one size gives central angles for corresponding polygons of any size. One size implies all others. $\endgroup$ Commented Jun 18, 2017 at 0:05
  • $\begingroup$ Compass and geodesic edges, yes. Think of the straightedge as a string you pull taut and it makes great circular arcs on the sphere. $\endgroup$ Commented Jun 18, 2017 at 0:17
  • $\begingroup$ I swear there was a recent article in an MAA journal about constructions on the sphere; I wrote to the authors. Can't find it now. What is shown is just the minimum, that we can construct all lengths $x$ on the unit sphere such that $\sin x \in E,$ where $E$ is the smallest field containing the rationals that is closed under taking square roots of positive elements. Going for that to th 17-gon is still a big leap, but it is a proof of existence. $\endgroup$
    – Will Jagy
    Commented Jun 18, 2017 at 0:26
  • 1
    $\begingroup$ until I find that, here is the hyperbolic plane zakuski.utsa.edu/~jagy/papers/Intelligencer_1995.pdf $\endgroup$
    – Will Jagy
    Commented Jun 18, 2017 at 0:39

1 Answer 1


The quickest description is to say that the constructible angles on the sphere are exactly the same as the constructible angles in the plane.

There are a few oddities. It is not possible to trisect a segment on the sphere, as this would involve trisecting an angle; think of the segment as part of the equator, and the north pole as part of a triangle with that.

It is possible to square the circle on the unit sphere. Not impressive, really. A hemisphere is both a square and a circle. There are a countable set of pairs of circular caps and symmetric quadrilaterals of the same area, where both are constructible. One such quadrilateral is the face of an inscribed cube: if we construct the perpendicular bisector of any edge and then draw the small circle K having that edge as diameter, the distance between either endpoint of the chosen edge and the intersections of circle K with the perpendicular bisector equals the circle radius required to match the area of the cube face.

Anyway, i wrote my article about the hyperbolic plane, not the unit sphere, but this material matches up.


April 2016, me to E. Rykken

Dear Prof. Rykken, I have just read your article with Myles Dworkin on constructions on the sphere, and I have some suggestions. A little late in the day, of course. In preparing my Intelligencer article on constructions in the hyperbolic plane, I did not actually write out the analogous statements on the unit sphere. However, it seemed to me very likely that the angles that can be constructed on the sphere are exactly the same as the angles that can be constructed in the plane. As to your final paragraph, I would expect there to be a countable set of pairs of circles and "squares" on the unit sphere, both constructable, with equal area. However, i would expect these both need be constructed at the same time, that there is no construction to begin with one and reach the other. Anyway, here are a copy of my old article and a page with Marvin's later article that incorporates some of it.


  • $\begingroup$ you might like the example I added regarding squaring a circle. $\endgroup$ Commented Oct 7, 2022 at 23:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .