Great circles and spherical triangles

Try

On the unit sphere, join the three positive coordinate-axis points by shorter great-circle arcs. Each angle is a right angle, so their sum is 3π/2. Eight congruent spherical octants partition the sphere, giving this triangle area π/2.

This is a concrete counterexample to transporting Euclidean triangle sums onto a curved surface without changing the metric. A spherical line means a great circle: the intersection of the sphere with a plane through its center. Smaller latitude circles are not spherical geodesics.

Predict

Predict why antipodal points are exceptional. They lie on every great circle whose plane contains their common diameter, so the joining great circle is not unique. Distinct nonantipodal points span one plane through the origin and have a unique joining great circle.

Predict whether measuring a straight chord through the interior of the sphere gives surface distance. It does not. The shortest surface arc has length Rθ on a sphere of radius R, where θ is the angle between the radius vectors.

Experiment

Compare a quarter-circle arc with its chord.

Sphere radius	Central angle	Arc length	Chord length
1	π/2	π/2	√2
2	π/2	π	2√2

Both lengths scale with radius, but they are not equal. The angle calculation uses normalized radius vectors: cos θ=(p·q)/R². For triangles in this lesson, choose nondegenerate shorter arcs bounding the intended spherical region; long arcs and complementary regions require separate conventions.

Understand

For unit vectors p,q, the short-arc distance is arccos(p·q). To see the minimizing mechanism, rotate p to the north pole. Along any surface path, the absolute rate of change of polar angle is bounded by its spherical speed. Therefore path length is at least the required change in polar angle. The meridian arc attains the bound. At antipodes the same lower bound π is attained by every meridian half-circle.

For the cosine law set p=(0,0,1), q=(sin c,0,cos c), and r=(sin b cos A,sin b sin A,cos b). Here b and c are angular side lengths, and A is the angle between their tangent directions at p. Taking q·r gives cos a=cos b cos c+sin b sin c cos A. This is the spherical counterpart of the hyperbolic formula, with the corresponding changes of sign and functions.

For radius R, use a/R,b/R,c/R inside the trigonometric functions. The quantities a,b,c then denote actual lengths rather than angular lengths.

Apply

The octant example has perpendicular arc tangents at each vertex, establishing its three right angles directly. Its area calculation follows by symmetry and the sphere’s area 4π, so it does not need the general spherical-excess theorem.

For a general simple geodesic triangle, Gauss–Bonnet supplies Area=R²(A+B+C−π), with the chosen region and interior angles specified. This is a theorem dependency, not a conclusion inferred from one octant.

Repair “The sphere is just neutral geometry with no parallels.” Antipodal incidence and global ordered-line axioms fail. Every pair of great circles meets, but that single fact does not describe the entire change of geometry.

Continue

State whether coordinates are unit vectors or radius-R vectors, and whether sides are angular measures or lengths. Continue to the elliptic quotient to remove the antipodal incidence duplication while retaining a different global order structure.