Equilateral hyperbolic triangle
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,
Let's construct a hyperbolic equilateral triangle.
gives us an equilateral triangle since the length of all
sides is the same. We can prove this affirmation from proving that the
length of a side is the same than the other two. To belong A in the circumference of inversion
that transforms B in
C we have that the length of AB is the same as the length of AC. But
the length of AB is also the
same as the BC since we
have constructed A in the
hyperbolic circumference with center in B and radius BC and, by definition, all the
points in a circumference equidist from the center and the distance is
- Draw any
Euclidean ray, starting at a point M in the boundary line.
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.
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
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
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.
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
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
prove that the triangle is equilateral it is only necessary to see that
the length of each one of the sides it is the same one. The length of
the side AC is the same as
the length of the side AB to
be B and C two points
of the hyperbolic circumference with center A. The length of the
side AB is also the same as
the length of the side BC to
be A and C two
points of the circumference with center B. So, the
length of the three sides is the same.
- 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
with center in the second point and that passes through the
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.