Let's go back to Section 1.3 (page 15), see what was used to prove side-side-side, and decide if it is still true for the sphere. We used Axioms 6-8, which talked about isometries, and a lemma that stated that two distinct circles intersect in zero, one or two points.
Must distinct circles on a sphere intersect in zero, one or two points? We probably can't prove this lemma, but after some experimentation it seems reasonable to assume so and move on.
Axiom 6 (page 13) says that if P and Q are points, we can find an isometry that takes P to Q. Is this true on the sphere? Yes -- there's at least one great circle through P and Q, and we can rotate the sphere with that great circle as its equator until P goes to where Q used to be.
Axiom 7 says that if we have a point P and points Q and R with |PQ| = |PR|, then we can find an isometry that holds P in one place and moves Q to R. If we think of P as one of the poles of our sphere then we can spin the sphere around P until Q is moved to R. So Axiom 7 is true for the sphere.
Axiom 8 says that given any line l there exists an isometry that fixes l but no other points. Given a great circle on the sphere, we can reflect the sphere through the plane that intersects the sphere at that great circle, so Axiom 8 is true for the sphere as well.
Everything we used to prove side-side-side on the plane seems to be true for the sphere, so the side-side-side theorem should be true for the sphere. If we were going to publish this result we would have to go back through the entire proof and make sure the lemma about intersecting circles is true on the sphere, but we don't have time for that in this class.
It seems likely that a similar discussion will convince us that side-angle-side and angle-side-angle are true on the sphere. It turns out that angle-angle-side probably isn't true on the sphere, and the textbook doesn't include angle-angle-side for the plane in Section 1.3. (Coincidence? I don't think so!)
Draw or rubber band a large triangle on your sphere. Do you think its angles might add up to 180°?
If your measurements and triangle constructions were accurate, you'll have discovered that the small triangle had an angle sum close to 180° and the large one didn't. In fact, any triangle with one vertex on the north pole and two vertices on the equator will have an angle sum of 270°!
Theorem 10.1.1: Suppose triangle ABC is a triangle on a sphere of radius ρ. Then the area
of triangle ABC is given by:
|△ABC| = ρ2 (m ∠A + m ∠B + m ∠C - π),
where the angle measures are in radians.
Discussion of Theorem: This theorem almost says that the area of a spherical triangle is proportional to the sum of the angles in the triangle. If the radius of the triangle has length 1, the area of the triangle equals π less than the sum of the angles in the triangle!
Discussion of Proof: This theorem is surprisingly easy to prove once you know the area of a lune of the sphere.
Definition: A lune or luna is the part of the surface of a sphere bounded by two great circles. Wikipedia has some excellent illustrations of lunes, which you might describe as shaped like sections of orange peel.
The surface area of a sphere is 4πρ2. If the great circles bounding a lune meet at angle θ, the lune covers θ/(2π) of the surface area of the sphere. So the lune must have area (θ/(2π)) * (4πρ2) = 2θρ2.
The sides of a spherical triangle determine three great circles which meet in pairs at the vertices of the triangle. The proof of the theorem is based on simply adding up the areas of six lunes formed by these three pairs of great circles.
This web page has an interractive image of a triangle ABC on the surface of a sphere.
Proof: Given a triangle ABC on the sphere, let α be the lune between the great circles AB and AC, let β be the lune between great circles AB and BC, and let γ be the lune between AC and BC.
As we can see from Figure 10.1, the great circles AB, BC and AC cut the "front" of the sphere into seven regions. One of these regions is the triangle we're looking at. Three of them are lunes bounded by the vertical angles of the angles of the triangle -- these three lunes overlap at the "back" of the sphere in a triangle congruent to ABC. The remaining three regions are parts of lunes α, β and γ; in particular, they're the parts of those lunes that are outside triangle ABC. Take a moment to look at Figure 10.1 and understand this paragraph -- this is the key to the proof.
The entire sphere is covered by the six lunes we just described. The triangles on the front and back are covered by three lunes, and the points on the rest of the sphere are either covered by one lune or lie on a one of the great circles AB, AC or BC.
If we add the areas of the six lunes, we'll get the area of the entire
sphere plus some extra because the lunes overlap on the two triangles.
The amount extra we get is equal to four times the area of triangle
ABC -- there are two extra layers of lune on each triangle. Using the
facts that the area of a lune with angle theta is
2θρ2 and that vertical angles have the same
measure, we get the equation:
4πρ2 + 4 |△ABC| = 2 (2 m∠A ρ2) + 2 (2 m∠B ρ2) + 2 (2 m∠C ρ2).
Now we do some algebra:
4 |△ABC| = 4 m∠A ρ2 + 4 m∠B ρ2 + 4 m∠C ρ2 - 4πρ2.
|△ABC| = m∠A ρ2 + m∠B ρ2 + m∠C ρ2 - πρ2.
|△ABC| = ρ2 ( m∠A + m∠B + m∠C - π).//
We conclude that the larger a spherical triangle is, the larger the sum of its angles will be. Can you imagine a spherical triangle that covers nearly an entire hemisphere? Can you imagine an even larger spherical triangle?
Alas, at least two things go wrong:
However, this cloud has a silver lining! It turns out that if all three angles of a pair of spherical triangles are congruent, the two triangles must be congruent. On the sphere, angle-angle-angle can be used to prove congruence.