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We explore hyperbolic geometry a little more deeply.

Since the SAS theorem is true in hyperbolic geometry, if we are given two sides and the included in a hyperbolic we need to be able to compute the other sides and . Similarly, ASA and AAS should allow us to calculate the remaining pieces of the triangle. Here’s how:

Theorem (hyperbolic laws of sines and cosines) Given 4ABC in the hyperbolic plane with sides a, b, and c, we have the following:

cosh(a) = cosh(b) cosh(c) − cos(A) sinh(b) sinh(c) cos(A) = cosh(a) sin(B) sin(C) − cos(B) cos(C) sin(A) sin(B) sin(C) = = sinh(a) sinh(b) sinh(c)

The first equation works for SAS , the second for ASA, and the third for AAS. (Note: we might have to go on a mini-tangent about hyperbolic trig functions at this !) For instance, we can look at those isosceles right triangles with legs of length one from the last notes. So a = b = 1 while C = 90◦. The remaining side can easily be found from the first equation above, and gives a result of cosh2(1). The two base angles can be found to have measures of about 32.95◦. Note that the total angle measure of that triangle is about 156◦, nowhere near the 180◦ of a Euclidean triangle. In fact, all hyperbolic triangles have total angle measure less then 180◦. In fact, there is something important about this.

Theorem (AAA for hyperbolic triangles) Given three angles whose sum is less than 180◦, all hyperbolic triangles that have these angles are congruent.

This is, of course, because we can use the second law of cosines to solve for the sides of the triangle. Hyperbolic cosine is a one-to-one function (for positive inputs) and thus we get unique answers.

Theorem (defect of a ) Given triangle 4ABC in the hyperbolic plane, its defect is defined to be π −A−B −C and equals the hyperbolid of the triangle.

So, just as in , the angles determine a triangle and its area is easily found. Now, of course, the angles can only get so small. By moving the vertices toward the x-axis, the closer we get the smaller we make the angles. Dropping the angles all the way to zero by (illegally) putting all the vertices on the x-axis, we find that the triangle has area π. This is despite the fact that the sides of such a triangle are growing infinitely long!

1 A very curious thing happens for regular polygons. Since their angles are smaller than those of a regular polygon in , and can be made arbitrarily small by lengthening the sides as needed, we can fit as many around a point as we want. For instance, we can create an octagon with eight right angles and tile the plane so that four of these fit around every vertex. The easiest way to see these tilings is with an app, and these almost always use a round version of the hyperbolic plane called the Poincare disk model. I can recommend using the Wolfram Demonstrations Project, where there are a number of hyperbolic explorers.

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