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Is there a general way to find a area of a super triangle?

I know the definition of a super triangle is the following: In the disk model of hyperbolic plane, the area bounded by three points M, P, and Q is called a super triangle if at least one of these three points belongs to the boundary of the disk.

I also know that we can have multiple scenarios here:

  1. M is on the boundary

  2. M, P are on the boundary

  3. M, P, Q are on the boundary.

Is there a general way to find the area for such a triangle when given all three points? For example, $M = 1$, $P = i$, $Q = 1 + \sqrt{2}$

mathwizz123
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  • "the disk model" is ambiguous; do you mean PoincarĂ©? Beltrami-Klein? something else? (Typically, such a model has radius $1$, so $Q=1+\sqrt{2}$ wouldn't be a point in it.) ... Area is equal to the angular defect of the triangle; that is, $\pi-(\text{angle sum})$. The angle at a point on the disk boundary has measure $0$, so case $(3)$ is easy. Calculating angles at arbitrary points can be a bit complicated. ... Please provide some context about your level of familiarity with the tools and techniques involved here. – Blue Dec 08 '21 at 05:42

1 Answers1

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The area of a hyperbolic triangle (assuming hyperbolic space with curvature $-1$) is equal to $\pi$ minus the sum of its interior angles. If one or more vertices of the triangle are on the boundary of hyperbolic space (usually referred to as ideal points), the interior angle at these vertices is 0. The formula still holds in this case.

Assuming the three points you provide are in the Poincaré half-plane model, the interior angle at $P$ is $\pi/4$ while the angles at $M$ and $Q$ are $0$, so the area of this triangle is $3\pi/4$.

Magma
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