Triangles on a sphere

[davieddy]

If a triangle area A is constructed from arcs of great circles on
a sphere of radius R, then the sum of the interior angles is (pi+e) radians.

What is e/A?

[davieddy]

Slow burner this one. If you don't know the answer I
can promise you it's worth waiting for.
I'll give you a hint in a couple of days.

[mfgoode]

Originally Posted by davieddy

If a triangle area A is constructed from arcs of great circles on
a sphere of radius R, then the sum of the interior angles is (pi+e) radians.

What is e/A?

Kindly explain what the terms stand for.
We (myself) are not mariners nor astronomers to know what 'e' (normally written as E) stands for.
I know, but I dont want to give the game away as its simple as an English pie!

Mally

[fetofs]

Originally Posted by mfgoode

We (myself) are not mariners nor astronomers to know what 'e' (normally written as E) stands for.

I'm a boy and I know! Point is, I don't think anyone who doesn't know what e stands for will come to this forum at all.

[VBCurtis]

I have no idea what E stands for. I'm interested in the puzzle. Or is explaining the meaning of E essentially giving away the solution?
-Curtis

[davieddy]

e stands for "extra" or "excess".
Let i be the sum of the interior angles of a triangle.
Now in a plane this = pi.
I said that on a sphere, i=pi+e

Using a bit of maths, you may infer that e=i-pi.

[S485122]

Not a proof but a hunch.

The area of a sphere of radius R in a 3 dimensional Euclidian space is 4*pi*R[sup]2[/sup].
When the area A of the triangle tends to 0 the sum of the interior angles of the triangle tends to pi and the excess e to 0.
When the triangle's area A is half of that of the sphere the sum of the angles is 3pi and the excess is 2pi. A=2*pi*R[sup]2[/sup].
When the triangle's area T coincides with that of the sphere the sum of its angles is 5pi and the excess is 4pi. A=4*pi*R[sup]2[/sup].

From there I jump to the conclusion that A=e* R[sup]2[/sup] or e/A=1/R[sup]2[/sup]

[Wacky]

Originally Posted by S485122

Not a proof but a hunch

Send that man back to Notre Dame!

As noted previously, the point of interest is "pi+e", "as simple as an English pie!"

But I prefer the American version, they are sweeter.

As me daddy said, "What a waste, I sent you to College and you come back and tell me ''. Any fool knows that 'Pie are round', 'Cornbread are square' "

[davieddy]

Originally Posted by S485122

Not a proof but a hunch.

From there I jump to the conclusion that A=e* R[sup]2[/sup] or e/A=1/R[sup]2[/sup]

Yes.
It seems surprizing at first, that the relationship should
be that simple. That is until you have this hint:

Consider the area of the regions formed by two great circles.

David

[mfgoode]

Originally Posted by fetofs

I'm a boy and I know! Point is, I don't think anyone who doesn't know what e stands for will come to this forum at all.

Well Fetofs, a little knowledge is a dangerous thing!

With reference to the problem it is NOT what (I presume) you mean.

By definition: The spherical excess is the amount by which the sum of the three angles of a spherical triangle ABC exceeds the sum of the three angles of the plane triangle ABC. It is generally denoted by E in spherical trigonometry. No wonder you got flummoxed by the use of 'e'.

Note for Davi: Well known and used symbols are very important and should not be used at random in mathematics !

No wonder you gave up teaching, and it will be a wonder if your students passed their finals at all!

Mally

[davieddy]

Originally Posted by mfgoode

By definition: The spherical excess is the amount by which the sum of the three angles of a spherical triangle ABC exceeds the sum of the three angles of the plane triangle ABC. It is generally denoted by E in spherical trigonometry. No wonder you got flummoxed by the use of 'e'.

Mally

No. You were the one that was flummoxed.
Perhaps you could provide the proof required.
On second thoughts, perhaps you couldn't.

David