# Triangle area on surfaces of constant curvature

I am looking for an elementary derivation of the formula for the area of a geodesic triangle lying in a surface of constant curvature $\kappa$, depending on the angles and side length.

Of course, the formula can easily be derived from the Gauss–Bonnet formula to be $$A = \frac{1}{\kappa}(\alpha + \beta + \gamma - \pi)$$ for $\kappa \neq 0$. However, I would like to have an elementary geometric proof.

Does anybody know a reference?

• For $\kappa > 0$, I believe you can use inclusion-exclusion with 3 geodesics on a sphere. – S. Carnahan May 12 '13 at 12:39
• For $\kappa<0$ Gauss had a similar inclusion-exclusion proof: It is in his collected works (letter to Wolfgang/Farkas Bolyai). – Misha May 12 '13 at 21:31
• For both cases, there are visual proofs, which can be considered elementary: mathoverflow.net/questions/8846/proofs-without-words/… and mathoverflow.net/questions/8846/proofs-without-words/… – Igor Khavkine May 13 '13 at 0:30
• There are some proofs by considering bigons (for spherical geometry) and subdivisions of an ideal triangle (in hyperbolic geometry). You may also consult J. Ratcliffe's book "Foundations of hyperbolic manifolds", some of the first chapters. However, the area of a bigon or ideal triangle is computed by "not very elementary" means there. I agree with Anton, all "elementary proofs" are cheating up to a certain extent. – SashaKolpakov Aug 31 '13 at 23:25

## 2 Answers

M. Berger, Geometrie, vol. V. MR0536874

Edit. Let me sketch a proof for the spherical triangle. Let the sphere have area $$4\pi$$. First you derive the area of digon. It is $$2\alpha$$, where $$\alpha$$ is the angle, by completely elementary reasons. Now consider a triangle. Extend its sides to three full great circles. These three circles make several digons and two equal triangles (the second one is centrally symmetric to the original one). Make a picture showing how these three circles partition the sphere. As the areas of all digons are known the area of a triangle is simply derived by the exclusion-inclusion formula!

Notice: this proof is truly elementary in the sense that it only uses the existence of the area for a digon and triangle, its invariance with respect to rotations, and finite additivity. Euclid COULD give a rigorous proof of this. As rigorous as his investigation of areas of Euclidean triangles.

• Chapter 5 is "Affine-projective relationship" did you really mean this? – Anton Petrunin May 13 '13 at 23:20
• Sorry, I was using Russian edition, where this is called Chapter V. Now I checked the original, and in the original it is VOLUME V. And unfortnately I did not find an English translation:-( – Alexandre Eremenko May 14 '13 at 21:10
• It use the properties of the area which (if you look carefully) already include the original statement inside. – Anton Petrunin May 15 '13 at 4:50

All the "elementary derivations of that type" are cheating (it may look nice but it proves nothing).

The only elementary way to introduce area is adding it as an axiom (which is already kind of cheating). You have to say that there is a additive area-functional on the set of all polygons. Then you probably want to prove that this functional is unique (or include it in the same axiom).

It remains to notice that your functional $A$ satisfies the same properties and nothing left to prove.

• You would need some normalization axiom, in order to distinguish between proportional measures. Do you have a specific one in mind? – Sergei Ivanov May 12 '13 at 22:38
• @Sergei, yes sure, all I wanted to say is that if one knows what is area and curvature then there is nothing to prove. – Anton Petrunin May 13 '13 at 2:17
• Anton: I disagree with what you say. The area of a TRIANGLE is an elementary notion. (The theory of areas of triangles in Euclid is completely rigorous, by all modern standards.) And the formula has a really elementary proof. – Alexandre Eremenko May 14 '13 at 20:47
• @BS, check this question mathoverflow.net/questions/119953/definition-of-area I would be very happy if you know a better answer. – Anton Petrunin May 16 '13 at 18:33
• @Alexandre, Euclid (and Kiselev) did not prove the existence, essentially they add the existence as an axiom, but they did not say that it is an "axiom". This axiom follows from the rest of axioms, but it takes 20 pages at least. Instead of unit square you have to use other normalization (which essentially defines curvature). A rigourous way to introduce area given in "Elementary Geometry From An Advanced Standpoint" by Moise 35 pages Euclidean plane onlyy,and in "Geometry: A Metric Approach with Models" by Millman and Parker 40 pages neutral plane and contains a gap. – Anton Petrunin May 16 '13 at 23:37