# Rotations and the parallel postulate.

If we take a full rotation to be $360^\circ$, then it seems that we can prove the following Starting from the red point, we walk clockwise along the triangle. At each vertex, we must turn through the green angles marked to proceed down the adjacent sides of the triangle. When we return to the red point, we will have turned through one full rotation. This means that the sum of the exterior angles is given as $360^\circ$, implying the interior angles of the triangle sums of $180^\circ$.

The fact that the angles of a triangle sum to $180^\circ$ is well known to be equivalent to the parallel postulate and this made me wonder whether if the fact that a full rotation being $360^\circ$ is also equivalent to the parallel postulate?

I avoided stating the question using "exterior angles of a triangle sums to $360^\circ$" and instead used the more ambiguous term "rotations" to emphasize the fact that rotations seem to be more general. We can for example show that the interior angles of a heptagram sum to $180^\circ$ by noting that three full rotations are made while "walking" thge heptagram. This should generalize to arbitrary closed polygons and seems stronger than the fact that the exterior angles sum to $180^\circ$.

In summary, I would be interested in knowing the connections that this technique has to the parallel postulate as well as if this technique is a "rigorous" way of finding the internal angles of more complex shapes such as the heptagram.

• The parallel postulate is only false in non-Euclidean geometries. I'm not sure if it even can fail to the extent that these external angles sum to a full $360^\circ$. Angles are always defined in terms of (roughly speaking) "local rotations" so the statement is fine. – anon Jan 31 '12 at 15:27
• en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem#Triangles – Will Jagy Jan 31 '12 at 19:49

Notice that, in this case, the sum of the exterior angles is $270^\circ$, not $360^\circ$.
However, in answer to your question about the sum of the interior angles of a polygon, since an external and the corresponding interior angle sum to $180^\circ$, the sum of the exterior angles and the interior angles is $180^\circ\times$ the number of sides. Since, as you have noted, in the Euclidean plane, the sum of the exterior angles is $360^\circ$, we get that the sum of the interior angles of a polygon with $n$ sides is $(n-2)180^\circ$.
• No, locally, a complete rotation has $360^\circ$. However, the parallel postulate is not a local postulate. A triangle whose external angles sum to $270^\circ$ cannot be local. Very small spherical triangles have exterior angles that are just a bit below $360^\circ$. In fact, the sum of the exterior angles of a spherical triangle falls short of $2\pi$ radians by the area of the triangle in steradians. – robjohn Jan 31 '12 at 18:21