Here is a picture of a triangle in position-space. The vertices are the arrows (not just their base points) labelled A, B, C. It is the (relative) orientation of the arrows A, B that determines the 'trajectory' of the line AB - by which we mean the path traced by the reference-point (or arrow-base) across the sphere. This path is always an arc of a circle (not necessarily a great circle, of course).
It appears likely from the picture that the trajectories meet in a point (call it X - near the bottom right of the figure). Can you think of any reason why this should be the case? (Note that the lines in position-space AB, BC, and CA do not meet when produced, even though the trajectories do: the arrow-directions will not agree at X.)
Angles
Consider the angle between the lines AB and AC. To move away from A in the direction of B, the teapot/circumsphere has to be rotated, clockwise, about a certain point of the sphere (call it p). To move away from A in the direction of C necessitates rotation (clockwise) about a different point, q. We define the angle between AB and AC to be the spherical distance between p and q.
Let's have a look at the sum of the angles in the triangle ABC.
To move from B towards A we need to rotate clockwise about p', the antipode of p. To move from B towards C we rotate about r. To move from C towards A we rotate about q' (antipode of q); to move from C towards B we rotate about r'. So the sum of the angles is the sum of the spherical distances
pq + p'r + q'r' = pq + qr + rp'.
This is the distance from p to its antipode p' going via q and r, which will be greater than the shortest distance from p to p', which is π (in radian measure) or 180°, unless q and r happen to lie on a spherical line (or great circle) going from p to p'.
Thus the sum of the angles of the triangle will almost always be greater than 180°. Is it at all likely that p, q and r lie in a straight line? We will get to the bottom of that in the next section.
Combining rotations
So far, we have that rotation about p (through a certain angle) gets us from position A to position B, rotation about r gets us from B to C, and rotation about q gets us from A to C. So we have two ways of getting from A to C: directly, in one rotation about q; or indirectly, via B, rotating first about p, then about r. Remarkably, knowing the locations of the centres p, q and r is enough to tell us the angles of the rotations, as follows.
Rotation about the point p, through an angle θ, say, is equivalent to two reflections in lines through p, separated by an angle θ/2. Similarly, rotation about r through angle ψ is equivalent to reflection in two lines through r, separated by angle ψ/2.
There is nothing to stop us choosing as one of the lines, in both cases, the line pr.
We see that reflection in l1, followed by reflection in pr (taking a to a' and then a'') amounts to rotation about p through angle θ; and that reflection in pr then l2 amounts to rotation about r through angle ψ.
Therefore rotation about p followed by rotation about r is equivalent to reflection in l1 followed by reflection in l2 (since the reflections in pr cancel out).
But the lines l1 and l2 have to meet somewhere (these are spherical lines, remember?), in a point s, say, so the two reflections are equivalent to a rotation about s.
But the rotations about p and then r get us from position A to B and then C, which can be done in one step by rotation about q. So a rotation about s is equivalent to a rotation about q: therefore, either s and q must be the same point, or else antipodal points. In the latter case, we can replace s by its antipode, by producing the lines l1 and l2 in the opposite direction.
So the angles of rotation can be found by drawing lines between the centres p, q and r to form the spherical triangle pqr. The required angles of rotation will then be just twice the angles of the triangle pqr.
As a corollary, we note that if p, q and r are in a straight line, the triangle will collapse. Two of its angles will be zero, the third will be 180°. Doubling up the angles, we find that the rotations will be null in each case: therefore A, B and C were all the same point. If ABC is any non-trivial triangle in position-space, its angles will add to more than 180°.
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