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From: Archimedes Plutonium on 27 Jan 2010 03:05

Enrico, I think I found an easy way of doing this problem, and of
course
in Euclidean.

I think I can even do a ascii art to illustrate.

Here we have a equilateral triangle in Euclidean.
/ \
/ \
------

Now the question I want to know is at what juncture does a
concave outwards sided triangle (Elliptic geometry) disallow
the reverse concavity of concave inward sided triangle due to
the impossibility of the vertices to match?

So rather than seeking an analytical solution I love to always
go for models of experiments to arrive at the answer.

So in the equilateral triangle above the question is what type of
a arc of those three sides will allow for the reverse concavity to
preserve the vertices.

It is my suspicion that a 36 by 36 degree latitude longitude on the
globe
is the upper bound.

So with the given triangle I can place that center of the circle to
give
me various arcs of the sides of the triangle.

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
From: Enrico on 27 Jan 2010 20:49
On Jan 27, 1:05 am, Archimedes Plutonium
<plutonium.archime...(a)gmail.com> wrote:
> Enrico, I think I found an easy way of doing this problem, and of
> course
> in Euclidean.
>
> I think I can even do a ascii art to illustrate.
>
> Here we have a equilateral triangle in Euclidean.
>    / \
>   /   \
>  ------
>
> Now the question I want to know is at what juncture does a
> concave outwards sided triangle (Elliptic geometry) disallow
> the reverse concavity of concave inward sided triangle due to
> the impossibility of the vertices to match?
>
> So rather than seeking an analytical solution I love to always
> go for models of experiments to arrive at the answer.
>
> So in the equilateral triangle above the question is what type of
> a arc of those three sides will allow for the reverse concavity to
> preserve the vertices.
>
> It is my suspicion that a 36 by 36 degree latitude longitude on the
> globe
> is the upper bound.
>
> So with the given triangle I can place that center of the circle to
> give
> me various arcs of the sides of the triangle.
>
> Archimedes Plutoniumwww.iw.net/~a_plutonium
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies

===========================================

This is the way I see it:

Given an equilateral triangle with vertices A, B, and C;
Viewed with A uppermost, B on the lower left. and
C on the lower right:

A
B C

The maximally hyperbolic case for triangle ABC occurs
when three identical circles are brought together touching
pairwise at points A, B, and C.

Call these 3 circles X, Y, and Z.

Circle X must pass thru points A and B
Circle Y must pass thru points B and C
Circle Z must pass thru points C and A

Like this:

X A Z
B C
Y

If this whole thing is drawn on a sphere S, the maximum inward
concavity possible will depend on the size of triangle ABC relative
to the size of the sphere S. I guess that's why you specified
lattitude and longitude lines a couple of posts ago.

To change the concavity toward convex, move the centers of
circles X, Y, and Z equally away from the center of ABC,
while retaining the constraint that:

Circle X must pass thru points A and B
Circle Y must pass thru points B and C
Circle Z must pass thru points C and A

X, Y, and Z will increase in size until they are all great circles on
S.
At this point, AB, BC, and CA will be straight lines (on the surface
of
the sphere), and triangle ABC will have sides of zero concavity.

If you continue moving the ceners of circles Z, Y, and Z in the same
directions as before, the great circles will begin to shrink and the
arcs
AB, BC, and CA will become increasingly convex. The limiting case
seems to be the point at which the 3 arcs AB, BC, and CA merge into
a circle and the vertices of the triangle become indiscernable.


For your question: (Assuming a spherical surface)

"So in the equilateral triangle above the question is what type of
a arc of those three sides will allow for the reverse concavity to
preserve the vertices. "

The answer is:
Concave: The smallest circles X, Y, and Z such that

Circle X must pass thru points A and B
Circle Y must pass thru points B and C
Circle Z must pass thru points C and A

Convex: The dual of the concave situation above:
A circle centered on the triangle ABC and passing through points A, B,
and C.

Disclaimer - I don't think this works if the surface area of the
equilateral
triangle is greater than or equal to 1/8 of the surface area of the
sphere.

The task of converting all this into numbers is not one I care to have
to
learn how to do before producing a result. I don't have the time or
interest.



Enrico

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