What is a triangle

[Boof-head]

It's not a triangle if one of the sides is the curve of a hyperbola.

This is a comment made by someone (at another site) who appears to know quite a bit of math.

They appear to also believe, that I don't know as much of it as them.

 

However, I believe this person has made a mistake. Does anyone else, and if not, why not?

[sanctus]

Well, in Euclidean space I would say he is right, but it is just a neccessary (not sufficient )condition...on the sphere for example depending on how you define the triangle, you can (actually have to I would say) have the sides being the curve of a hyperbola or parabola.

That is a t least how I see it.

[Boof-head]

What's a "hyperbolic triangle"? Is it Euclidean?

[sanctus]

I never really talked about a hyperbolic triangle. I would define a triangle as the figure you get when connecting 3 points with 3 lines with the lines being geodesics. So in a space with negative curvature (like a saddle point), you would get a kind of hyperbolic triangle. Maybe it is easier to imagine on a space with positive curvature (like a sphere), when you connect the three points on the sphere with geodesics, they are not straight lines...hope that helps

[Boof-head]

Right, if you have three geodesics that intersect on the sphere at 3 points, you have a three-sided figure with sides that are curved because so is the surface.

If you draw a single geodesic it's a great circle, this looks 'straight' when you look along it, if you draw two others (three geodesics) so they intersect as an equilateral - it's an equilateral triangle if you project it the right way onto a plane.

 

Also, you can find an article in wikipedia headed "hyperbolic triangle"; this is the figure with an apex on the unit hyperbola touching the unit circle at (1,0). It has an area equal to the area of the triangle in a quadrant of the unit circle. This is how you derive hyperbolic trig relations, I believe.

[sanctus]

So, it looks like the person has not made a mistake, do you agree?

[C1ay]

See Spherical Trigonometry for an example of a triangle with curved sides. These sides are not hyperbolic though since these arcs would close into a circle if continued. There are triangles in hyperbolic geometry though which are non-Euclidean. It would be my opinion that a 3 sided figure which only has one side that is a hyperbola is not a hyperbolic triangle though.

[Boof-head]

This is what another mathematician reckons:

 

 

Hyperbolic triangle

From Wikipedia, the free encyclopedia

 

 

In mathematics, the term hyperbolic triangle has more than one meaning.

 

Contents

 

* 1 Euclidean geometry

* 2 Hyperbolic geometry

* 3 See also

* 4 References

 

[edit] Euclidean geometry

 

In the foundations of the hyperbolic functions sinh, cosh and tanh, a hyperbolic triangle is a right triangle in the first quadrant of the Cartesian plane

 

Does Cartesian plane mean "Euclidean"?

[CraigD]

What is a triangle[?]

In the most general sense I can think of, a triangle is three vertexes in the context of a metric, such that when the metric is applied to the 3 lines defined by the three vertexes, the triangular inequality holds. That is, the length of no side can exceed the sum of the length of the other two.

 

Beyond that, I don’t think there are any restrictions on what a triangle can be. Different metrics result in various interesting properties in addition to the triangular inequality, such as more exact equalities similar to the triangular inequality, such as the cosine rule, and rules about the sum of interior angles [math]S[/math], such as the Euclidean metric’s [math]S=\pi[/math], the spherical’s [math]S>\pi[/math], and the hyperbolic’s [math]S<\pi[/math].

 

You can have triples of vertexes in spaces with metrics that violate the triangular inequality. These spaces are called “non-triangular”. You could call such triples “triangles”, but to my intuition, they’d be lacking their essential triangle-ness. The study of such spaces, and even more general spaces in which no metric exists, is usually considered a branch of topology

Hyperbolic triangle

From Wikipedia, the free encyclopedia

…

In the foundations of the hyperbolic functions sinh, cosh and tanh, a hyperbolic triangle is a right triangle in the first quadrant of the Cartesian plane

 

Does Cartesian plane mean "Euclidean"?

In the context of the wikipedia quote, yes.

 

What the quote means, however, is just that the hyperbolic functions can be described geometrically in terms of the Euclidean coordinates of a hyperbola, and that only the first quadrant of the Euclidean plane – that is, non-negative coordinates – need be used.

 

The hyperbolic triangle to which it refers, which isn’t essential to defining triangles or hyperbolic functions, is an ordinary one on the Euclidean plane, such as the one sketched in the attached thumbnail.

[Boof-head]

When you draw a unit circle and then a unit hyperbola touching at (1,0), you have the point (1,1) forming the upper right apex of a square on (1,0). a triangle with a hypotenuse from (0,0) to (1,1) has the same area as the three-sided figure with the same base ([0,1]). and it has a hyperbolic curve as the other side, with a hypotenuse which is longer and subtends a smaller angle.

 

This figure with three sides [is] not a hyperbolic triangle, the straight-sided figure in your diagram above this is?

[CraigD]

When you draw a unit circle and then a unit hyperbola touching at (1,0), you have the point (1,1) forming the upper right apex of a square on (1,0). a triangle with a hypotenuse from (0,0) to (1,1) has the same area as the three-sided figure with the same base ([0,1]). and it has a hyperbolic curve as the other side, with a hypotenuse which is longer and subtends a smaller angle.

I don’t quite follow your description, Boof-head. Can you sketch it?

This figure with three sides in not a hyperbolic triangle, the straight-sided figure in your diagram above this is?

I wouldn’t call a triangle on the Euclidean plane with 2 sides straight lines, one side a section of a hyperbola, a hyperbolic triangle, because all 3 sides of a hyperbolic triangle of this kind should be sections of hyperbolas. Because a hyperbola approaches a straight line as its coordinates approach infinity, a hyperbolic triangle could approach such a figure.

 

The figure I sketched is a completely different kind of figure, as described in the “Euclidean Geometry” section of the wikipedia article “hyperbolic triangle” that you quoted.

 

This is the first I’ve read of the hyperbolic triangle defined this way. As the article lacks footnoting, and this section isn’t supported by its only online-readable reference, some skepticism that this meaning exists other than in the imagination of the article’s author, and/or the imaginations of some others and some less common texts, is in order.

 

It’s easy to derive parametric equations for the vertexes of the triangle defined this way, one of which I sketched above, for example:

[math](0,0)[/math]

[math](a,a) [/math]

[math](a +\frac{1}{4a}, a -\frac{1}{4a})[/math]

where the domain is [math]a \ge \frac12[/math]

 

It wouldn’t be much more work to derive a 3-case parametric equations for the points of the sides of this triangle, useful equations if you’re rendering sketches of these triangles with a computer.

 

The meaning of “hyperbolic triangle” I knew before, which I learned in school, and which I believe is most common, is a triangle on the hyperbolic plane, the second meaning given in the wikipedia article. Projected onto the Euclidean plane, this is a triangle in which the lines are sections of a hyperbola. The most commonly mentioned property of these triangles, in my experience, is that the sum of their interior angles is less than [math]\pi[/math].

[Boof-head]

I don’t quite follow your description, Boof-head. Can you sketch it?

Draw a horizontal line; label the left end (0,0), the right end (1,0). Construct a square on this line (assume the line is the lower edge, iow). draw a diagonal from (0,0) to (1,1). Two triangles with an area 1/2 that of the unit square. Now construct a unit hyperbola on the rhs of the square. Draw a line from (0,0) to the hyperbola so it subtends the greatest angle; what is the area of the figure with 3 sides, the rh side of which is a section of the hyperbola? Is it also 1/2?

 

As the article lacks footnoting, and this section isn’t supported by its only online-readable reference, some skepticism that this meaning exists other than in the imagination of the article’s author, and/or the imaginations of some others and some less common texts, is in order.

You should take that issue up with the authors; perhaps they need to learn a bit more math?

 

The angle the line tangent to the hyperbola subtends, wrt to the line (0,0),(1,0), can be subdivided; say you divide it by 2, then again. How small does a subdivided angle need to be, before the angle subtended on the lowest point (at 0,1)of the hyperbola = a right angle? When does a subdivided triangle = a right triangle?

[C1ay]

A three sided figure in which one of the sides is a hyperbolic curve is a three sided figure in which one of the sides is a hyperbolic curve, not a hyperbolic triangle. I agree with rpenner....

 

BTW, you can calculate the area of such a figure using calculus, you just can't do it with the formula used for triangles.

[CraigD]

Draw a horizontal line; label the left end (0,0), the right end (1,0). Construct a square on this line (assume the line is the lower edge, iow). draw a diagonal from (0,0) to (1,1). Two triangles with an area 1/2 that of the unit square. Now construct a unit hyperbola on the rhs of the square. Draw a line from (0,0) to the hyperbola so it subtends the greatest angle; what is the area of the figure with 3 sides, the rh side of which is a section of the hyperbola? Is it also 1/2?

So you mean the figure sketched in the attached thumbnail, it’s area shaded red?

 

One such figure has an area of 1/2, but in general, they don’t.

 

This is graphically obvious if you consider the figure with x only slightly greater than 1, such as with upper right vertex [math]\left( \frac{11}{10}, \sqrt{\left(\frac{11}{10}\right)^2 -1} \right) \dot= (1.1, 0.4583)[/math]. Its area, even if you don’t exclude the small part you should subtract from a right triangle with the same vertexes, can’t be greater than about 0.252.

 

The one of these figures that subtends the greatest angle would have it’s upper right vertex at a point of infinity, and have an infinite area.

 

As C1ay mentions, you can derive the area of this figure using fairly simple calculus. It’s area [math]A[/math] is the area of a right triangle with upper right vertex on the hyperbola, minus the area under the parabola, so given that the unit hyperbola is defined by

 

[math]y = \sqrt{x^2-1}[/math]

 

So the area of the figure is

 

[math]A = \frac{x \sqrt{x^2-1}}{2} - \int_1^x \sqrt{a^2-1} \, da = [/math]

 

[math]\frac{x \sqrt{x^2-1}}{2} - \frac{ x \sqrt{x^2-1} -\ln(x + \sqrt{x^2-1})}{2} = [/math]

 

[math]\frac{\ln(x + \sqrt{x^2-1})}{2} [/math]

 

So [math]A = 0.5[/math] only for [math]x = \frac{e^2 +1}{2e} \dot= 1.543[/math].

 

When [math]x = \infty[/math], [math]A = \infty[/math].

[Boof-head]

The one of these figures that subtends the greatest angle would have it’s upper right vertex at a point of infinity, and have an infinite area.

"The greatest angle is subtended by the line y = x, the asymptote."

It’s area A is the area of a right triangle with upper right vertex on the hyperbola, minus the area under the parabola,

As your math shows, this figure has an area = 0.5 when the endpoint of the line from (0,0) to the hyperbola is (x, 1). the y coordinate must = 1.

[CraigD]

As your math shows, this figure has an area = 0.5 when the endpoint of the line from (0,0) to the hyperbola is (x, 1). the y coordinate must = 1.

No, for [math]y=1[/math], [math]x= \sqrt{2}[/math], and the area [math]A= \frac{\ln(\sqrt{2}+1)}{2} \dot= 0.4407[/math]. That is, when the figure has vertexes [math](0,0)[/math], [math](1,0)[/math], and [math]( \sqrt{2},1)[/math]

 

For [math]x = \frac{e^2 +1}{2e} \dot= 1.543[/math], [math]y= \sqrt{ \left(\frac{e^2 +1}{2e} \right)^2 -1} \dot= 1.175[/math], [math]A = \frac12[/math].

 

The only solution I see that has neat, rational solutions for [math]x[/math], [math]y[/math], and [math]A[/math] is [math]x=1[/math], [math]y=0[/math], [math]A=0[/math], not much of a figure, as it’s just the line segment (0,0)-(1,0).

[Boof-head]

Now I'm confused; you said the figure had an area A = 0.5, when x = 1.543.

What's the value of y, when x = 1.543? I was sure it's 1.