Turn a sphere twice about different axes. Euler says one turn would have done the job. Finding its axis is the interesting bit.

Contents:

Give it a spin

Keep the centre fixed, pick two axes, and rotate. Euler’s rotation theorem says that whatever final position you get, a single rotation about some diameter gets you there too.

The routes only need to agree at the end. The shortcut needn’t retrace the original motion. That would be asking quite a lot of a shortcut.

First turn: xSecond turn: y
P

Ready. Find the axis in the construction below.

Try two turns, then compare the single turn.

A good place to look

Draw a great circle and keep its starting position as a ghost. After the sphere moves, the two circles cross at .

Call the point that arrives there , and the point that reaches . In short, . The arcs and have equal lengths: moving a rigid sphere doesn’t stretch anything.

Now bisect the angle . Also bisect the arc , then draw the great circle perpendicular to it at its midpoint. Call an intersection of these two bisectors .

Some construction. Why should it help?

Step 1 / 6

Pick a circle

Blue is a great circle before the turns; dashed red is its final position. Mark their crossing A.

Original circleTransported circleBisectors / axis
αAaO
Read the diagram steps
  1. Pick a circleBlue is a great circle before the turns; dashed red is its final position. Mark their crossing A.
  2. Follow Aα arrives at A; A arrives at a. The two marked surface arcs have equal lengths.
  3. Bisect the angleSplit the angle between those arcs at A. The dashed gold circle follows its bisector.
  4. Bisect the arcThe other gold circle bisects Aa at a right angle. At their crossing O, OA = Oa.
  5. Move the triangleThe triangles agree in lengths and orientation. Move αAO: α reaches A, A reaches a, and O returns to O.
  6. There’s your axisJoin O to the centre. Both ends of this diameter return to themselves. Now try the single turn.
Step through the construction, then move the sphere.

Now make it stay

The perpendicular bisector gives . The angle bisector, together with , makes the spherical triangles and congruent: two sides and the included angle agree. So .

Let the sphere move. Triangle becomes , where is wherever lands. All its side lengths match triangle .

There are two possible third vertices with those distances, on opposite sides of the great circle through and . The angle bisector puts and on opposite sides of , so the triangles have the same orientation. An actual turn cannot reflect one. This picks , so .

We’ve found a point that comes back to itself.

There’s your axis

The centre stays put too. The line joining it to is therefore a fixed diameter. Rigidity leaves the rest of the sphere one freedom: turning about that line. There’s your single rotation.

This is Euler’s geometric argument from the original supplement, §§24–26, written in 1775 and published in 1776.

What if the construction collapses?

If nothing moves, any diameter works. If is fixed, use its diameter. When , use the preserved great circle perpendicular to the centre– line: its direction reverses. Choose there, with image . If , done. Otherwise their non-antipodal arc midpoints are fixed; if , a quarter-circle neighbour is fixed. Coincident circles either give a fixed pole or this reversed-circle case. These are geometric completions of the exceptional cases; Euler doesn’t spell them all out there.