# Area of a triangle $\propto\pi-\alpha-\beta-\gamma$

A hyperbolic geometry is a non-Euclidian geometry with constant negative curvature. It has the property that given a line and a point, many lines can be drawn containing the point that never meet the given line. The picture below (Circle Limit by M.C.Escher) is a conformal map of hyperbolic geometry to Euclidian plane.

How do you show that in a space with hyperbolic geometry, area of a triangle with angles $\alpha$, $\beta$, and $\gamma$ is $\Delta\propto\pi-\alpha-\beta-\gamma$?

A similar result exists for sphere: $\Delta=R^2(\alpha+\beta+\gamma-\pi)$. I could prove this, but I suspect my proof cannot be modified for hyperbolic geometry. My proof for the sphere is fairly trivial, it makes use of symmetry and a Venn diagram.

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See this MathOverflow answer ( mathoverflow.net/questions/8846/proofs-without-words/… ) for a picture-proof of the result, along with an appropriate reference. Elsewhere in the thread ( mathoverflow.net/questions/8846/proofs-without-words/… ) is a series of images that likely captures the gist of your own proof of the spherical result. – Blue Aug 20 '10 at 22:21
Yeah that's essentially my proof. – KalEl Aug 20 '10 at 23:10

I direct you to chapter 1 of Fuchsian Groups by Svetlana Katok. It has the following proof of the theorem (at least where I learned it) with illustrations:

Let us define hyperbolic area to be: $$\mu(A)=\int_{A}\frac{dxdy}{y^2}.$$

It can be verified that for all $T \in$ $\text{PSL}(2,\mathbb{R})$ that $\mu(A)=\mu(T(A)).$ (This fact will be useful to us later).

So let us consider a triangle in this hyperbolic plane $H$. We will use the upper half plane model.

Case 1: One vertex of the triangle belongs to $\mathbb{R}\cup {\infty}.$

Then the angle at that vertex is 0. We can use transformations (recall $T$ from above?) from $\text{PSL}(2,\mathbb{R})$ to change two sides of $A$ into vertical geodesics; hence the base of the triangle is a Euclidean semicircle orthogonal to $\mathbb{R}.$

Then we can see that $$\mu(A)=\int_{A}\frac{dxdy}{y^2}=\int_{a}^{b}dx\int_{\sqrt{1-x^2}}^{\infty}\frac{dy}{y^2}=\int_{a}^{b}\frac{dx}{\sqrt{1-x^2}}.$$ When we substitute $x=cos(\theta),$ this integral becomes: $$\mu(A)=\int_{\pi - \alpha}^{\beta}\frac{-\sin(\theta)d\theta}{\sin\theta}=\pi-\alpha-\beta.$$

Case 2: $A$ has no vertices in $\mathbb{R}\cup{\infty}.$ Let the triangle have vertices $A, B,$ and $C.$ Then let the geodesic connecting $A$ and $B$ intersect the real axis at $D$. Then $\mu(A)=\mu(ACD)-\mu(BCD)$ and both are triangles as in the previous case. (You can work out the rest of the details accordingly - draw a picture)

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1. Observe that defect of a (hyperbolic or spherical) polygon (i.e. the difference between the sum of its angles and the sum of angles of a Euclidean polygon with the same number of vertices) is an additive congruence-invariant function.
2. Up to proportionality, there exists only one additive congruence-invariant function (for Euclidean plane it is sometimes called Bolyai–Gerwien theorem; one may say that it just means that area is well-defined).

Hence the result.

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Thanks but I have almost no idea what you are talking about. Some more references? – KalEl Aug 21 '10 at 7:18
I don't know any good reference. But maybe reading a proof of Pick's formula that follows the same plan (additivity of the function in question + uniqueness of area) will help (unfortunately the plan is not quite explicit in that proof, but still). – Grigory M Aug 21 '10 at 11:01

There is also a differential-geometric proof.

For a surface with constant Gaussian curvature $K$ Gauss-Bonnet formula yields $\text{area}\cdot K+\text{sum of exterior angles}=2\pi$ aka $\text{area}=\text{defect}/K$. (This gives not only proportionality but also the coefficient.)

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This is really a comment which complements earlier responses:

For spheres the "formula" for the area of spherical triangle is known as Girard's Theorem:

http://en.wikipedia.org/wiki/Girard%27s_theorem#Girard.27s_theorem

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Also, to complement the earlier remark about the Bolyai-Gerwien Theorem there is also the Dehn Invariant: en.wikipedia.org/wiki/Dehn_invariant – Joseph Malkevitch Aug 20 '10 at 23:32