Equilateral hyperbolic triangle

Similar triangles do not exist in the Hyperbolic Geometry. That is because exists a relation between angles and distances given by the angle of parallelism formula. The non-existence of similar hyperbolic triangles implies the existence of a unique hyperbolic equilateral triangle fixed the angle.

As it is known, in Hyperbolic Geometry, the sum of the angles of a triangle is always less than two right angles. So, each angle of an equilateral triangle will not have exactly 60º. It can be seen that for any value among 0 and 60º, both non included, we have an equilateral triangle, which is unique, except isometries.

Let's construct a hyperbolic equilateral triangle.

- Draw any
Euclidean ray, starting at a point M in the boundary line.

- Consider two different points, B and C on the ray plotted in the step (1). These will be two of the three points of the triangle.
- Determine
the
inversion with center in the boundary line that exchanges B and C. The
inversive circumference will cut the ray between B and C.
To find this inversion it is only necessary to find this point, in
fact, we will find the radius of the
circumference from this point. We use that a point is inverse
of another if two conditions are fulfilled: the two points have to be
in the same (Euclidean) line that goes through the center of the
inversion
and the product of the distance from each point to the center has to
be the square of the radius of the circumference. The first
affirmation assures us that the origin of the inversion
we are searching is the intersection point between the ray line and the
boundary line. If we impose that the points B
and C are inverse we have
that the radius is √MB·MC. So, we
plot the circumference with center M
and radius the former
length.
It is necessary that the third point, A,
of the equilateral
ttriangle be on this circumference, since then we can affirm that the
hyperbolic segment AB is
transformed in AC since A is fixed and B and C are inverse.

- Plot the hyperbolic circumference with center B and radius BC.

- Consider one of the two intersection points between this hyperbolic circumference and the circumference of inversion. This point will be the third vertex of the hyperbolic equilateral triangle which we wanted to construct.

To check out empirically that it is really equilateral we can use the length function. If we measure each one of the segments we will see that the three segments have the same length. We can also measure the angles and we will see that all three also have the same amplitude.

We can also construct the equilateral triangle in another way. We will use that the points of a circumference have the property of being all at the same distance from the center. We can follow these steps:

- Mark two of the vertices of the equilateral triangle, A, B.
- Plot the hyperbolic circumference with center in the first point and that passes through the second.
- Plot the hyperbolic circumference with center in the second point and that passes through the first.
- Consider
the intersection points between both circumferences.
These two points are at the same distance from
each one. It is because each one of these points belong to
the two circumferences.

- Plot the hyperbolic triangle that has vertices in the two fixed points and one of the two intersection points of the former step. This triangle is equilateral.

Dragging the fixed vertices we will obtain all the other equilateral triangles. The same as before, we can check out empirically that the triangle obtained is really equilateral.

Triangles

Hyperbolic geometry