Somebody asked this question on AskMe.com, and I thought it was interesting enough to copy and ask the enthusiasts here:
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I was stargazing with some friends one night when one of them was amazed to find that three stars formed an equilateral triangle. Another suggested that any three stars would form an equilateral triangle if viewed at from the right direction.

Is this true? Given any three (non-collinear) stars, can you find a direction from which they will appear as an equilateral triangle?

If true, can you specify the direction?
If not, can you find a counterexample?

Is this question better thought of as a vector algebra question about projections onto a plane

or as a 3-d cartesian coordinates question?
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Intuitively, I know the hypothesis is true, but I'm not sure what would be the simplest method to calculate the appropriate direction (vector), given three stars. I would probably start by creating a coordinate system with one star at the origin (0,0,0), the second one at (1,0,0), and the third one at an arbitrary point in the first quadrant (a,b,0). Then it gets interesting. . . .

A tetrahedron is a 4-sided, perfect solid. Its base is an equilateral triangle. Let's say you have a model of a tetrahedral skeleton. No matter how you hold it, you can only see the equilateral triangle if your point of view, the centroid of the base, and the vertex are all collinear. Otherwise, you will see three randomly placed points. It also works in reverse. If you imagine three rays coming from your point of view, that which when a plane intersects them the distances to your eyes are congruent, then any three points in space may be rotated to "become" an equilateral triangle.
Cut a piece of cardboard into a scalene triangle of any sort. If you rotate it the right way, it will look like an equilateral triangle.

It seem that with an equalateral triangle you will see it as an equalateral anywhere along a line thorugh centroid and perpendicular to the plane of the triangle.

Interisting thing as you are at the plane of the three equalateral stars, they will be level with your field of vision, 120 deg. apart.

Going a little further: Start makeing an isosceles triangle by reduceing the apex angle from 60 deg. on down. Now it seem that you will see an equalateral triangle anywhere along two seperate lines. Picture holding your paper triangle and slowly rotate the apex away until equalateral then same by rotating apex toward you.
Would the angle you are on with respect to the plane of the triangle be1.5 x apex angle ?

If the apex angle on this isosceles triangle increases from 60 deg. then things get too wierd for me !!

Help ! Dr.G, maiku or diane

Great question Regards Mike B. (AKA Byter)

That is almost exactly what I said!

Someone on Askme.com finally answered this one. I don't have time right now to see if the answer makes sense, but I'll post it and let the other enthusiasts decide:

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IQGuru gave this response on 6/27/2002:

Hi!

This problem is really a rather simple cartesian coordinate problem (IF you can think in three dimensions) because any three points in space DEFINE a flat plane...

IN THE FLAT PLANE:
(1) Draw the triangle your three stars make upon their flat plane.
(2) Pick any two of the stars and bisect the distance between them (locate the Midpoint between them).
(3) Construct a circle, using said Midpoint as center, and the distance from said Midpoint to your third star as the radius.
(4) Construct the perpedicular bisectors, of all three sides of your triangle; draw and extend each bisector to intersect with the circle twice. Also notice that all three bisectors also intersect at a single point; label that point "P1".

ON THE SPHERE:

(5) Now think of your circle as a sphere. Porjecting everything perpendicular to your original plane, your three extended bisectors become great circles on that sphere.
(6) The projection of "P1" onto the sphere, perpendicular to your plane, (label it "P2") becomes the point in space from which your three stars will appear to be equilateral.

FINAL OBSERVATIONS:

Because you can perpendicularly project your flat-plane triangle onto either half of your sphere (either hemisphere) there are TWO similar points for "P2" (one on each hemisphere) where the three stars will appear equilateral. Additionally, the above exercise can be repeated, starting with any 2 of the 3 stars. Thus, there are SIX locations in space from which your three stars will appear equilateral...

PLEASE RATE THIS ANSWER.
Warm regards, "IQ"


IQGuru gave this follow-up answer on 6/27/2002:

ONE LAST OBSERVATION.
Each of the six locations, once determined, can also (of course) be extended to an infinity of other points (where the stars appear equilateral) by constructing a line from the midpoint of the original triangle, through each located point... "IQ"

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If this is right, then it looks like the Askme.com "experts" beat the Answerpool.com "enthusiasts" on this one. (Of course, I'm both and I didn't answer it, so I guess I'm not one to criticize.)

Still Starstruck

After looking over IQGuru's answer I copied, I can say it's definitely wrong. Intuitively, a triangle should only appear equilateral from two directions (one from each side), not six! Also, IQGuru (who is apparently misnamed) is using the wrong center for the triangle. He's using the circumcenter, or intersection of perpendicular bisectors. In an obtuse triangle, the circumcenter isn't even in the triangle's interior, which should be a clue that it's not going to give an equilateral view of the triangle.

The key to this problem is the centroid of the triangle, which is the intersection of the medians (a median goes from a vertex to the midpoint of the opposite side). I don't have a general answer for this problem yet, but I do have three examples, including a trivial one, and the answers are all rays radiating from the centroids.

All of these triangles have the origin (0,0,0) as their centroids, they all lie in the x-y plane, and they all have symmetry about the line y = x (which makes them isosceles). Some of the coordinates are a bit ugly, so I'll define some useful constants here to save myself some typing:

u  ≡  ½ √(2 − √3)

v  ≡  ½ √(2 + √3)


I've specified the coordinates of each triangle in cartesian as well as polar coordinates. I included polar coordinates because they give the distances from the centroid (the origin) to the vertices, which tells you at a glance how "distorted" the triangles are. (In an equilateral triangle, these distances are all the same.) I also included the parametric formulae for the viewing vectors.

The first case is trivial: an equilateral triangle.

Cartesian:  (v, -u),  (-u, v),  (-½ √2, -½ √2)

Polar:  (1, 345°),  (1, 105°),  (1, 225°)

Viewing vector:  x = 0t,  y = 0t,  z = 1t


The second case is a slightly skinnier triangle:

Cartesian:  (1, 0),  (0, 1),  (-1, -1)

Polar:  (1, 0°),  (1, 90°),  (√2, 225°)

Viewing vector:  x = 1t,  y = 1t,  z = 1t


The third case is even skinnier:

Cartesian:  (v, u),  (u, v),  (-½ √6, -½ √6)

Polar:  (1, 15°),  (1, 75°),  (√3, 225°)

Viewing vector:  x = 2t,  y = 2t,  z = 1t


Note that these viewing vectors are only valid from the front side or positive z side, where t > 0. Also, it's assumed the viewer is far enough away that projecting the triangle onto the viewing plane (the plane perpendicular to the viewing vector) causes minimal distortion (a fairly reasonable assumption for star-gazing).

In the trivial case, the triangle is viewed straight-on, and the viewing vector is just the z-axis. As the triangle gets skinnier, notice that the viewing vector's x and y parameters increase, tilting the viewing plane more and more to make the projected triangle equilateral again.

Anyhow, this question is still open, since the Askme.com expert blew it. I could still use some help on generalizing these results.

This message has been edited. Last edited by: silenteuphony,

Now that I think about it, there are more than two viewing vectors (I counted four for my isosceles triangles, and presumably there could be six for a scalene triangle), but I still stand by my solution. All of the viewing vectors for a given triangle are just symmetrical variations of one vector, and the centroid, not the circumcenter, is the source of that vector. Feel free to check my solutions, if you want to do some fun geometry.

General Formula for Isosceles Triangles

I found the general formula for isosceles triangles. Given any isosceles triangle with a centroid at the origin and these polar coordinates:

(a, 225°) ,  (1, 45° - θ ) ,  (1, 45° + θ ) ,   where 1 ≤ a < 2

the equilateral viewing vector is

x  =  y  =  t √[2 (a² - 1) ÷ (4 - a²)]  ,   z  =  t

You don't even need to know what θ is, you can calculate the viewing vector just knowing the distances of the vertices from the centroid (the origin, in this case). Of course, you can also use the formula for smaller or larger triangles; a is just the ratio between the centroid-vertex distances. In these examples, I made the smaller distance 1 for convenience. Also, the centroid doesn't have to be at the origin. Just add the x, y, and z coordinates of the centroid to the x, y, and z formulas for the vector, respectively.

This message has been edited. Last edited by: silenteuphony,

By the way, this solution works for any "skinny" isosceles triangle (where the unique angle is less than or equal to the duplicated angle). The ratio a is always less than 2, no matter how skinny your triangle is. To get the four viewing directions, you can actually view along the equilateral viewing vector from either side of the triangle, and you can get a symmetrical variation of this vector by using the opposite (negative) x and y coordinates. Or to restate the solution in the most general terms:

Given any isosceles triangle with a centroid at the origin and these polar coordinates:

(a, 225°) ,  (1, 45° - θ ) ,  (1, 45° + θ ) ,   where a ≥ 1

the equilateral viewing vector is

x  =  y  =  ± t √[2 (a² - 1) ÷ (4 - a²)]  ,   z  =  t

Again, you only need to know the ratio a to find the solution, you don't need to know q. Also, the viewing vector works from either side of the triangle. This works for any "skinny" isosceles triangle in space, since given any such triangle, we can then choose our coordinate system to satisfy our initial conditions (centroid at the origin with symmetry about the line y = x, and the smallest centroid-vertex distance = 1).

Of course, we still need the generalization for scalene triangles. I haven't heard from any other enthusiasts for a while, so feel free to join in, or I'm stuck answering my own question (if I can). In case this helps, I can tell you how I figured out the isosceles solution. I started with the given triangle:

(a, 225°) ,  (1, 45° - θ ) ,  (1, 45° + θ )

I then drew a segment from the centroid of the original triangle (the origin) to the centroid of the projected equilateral triangle which shares a base with the original triangle. I then used the common base, the fact that the centroid is always one third the distance from any side to the opposite vertex, plus lots of similar triangles, to find the coordinates of the new centroid in terms of a. I then divided these coordinates by the z coordinate, and multiplied everything by t, to get a simple parametric formula. Of course, a scalene solution isn't quite as simple, since the projected triangle doesn't share a base with the original; it only has one common vertex.

Still, you ought to be able to do something similar starting with a triangle with a centroid at the origin and the following polar coordinates:

(1, 0°) ,  (a, θ) ,  (b, φ) ,   where 1 ≤ a ≤ b

Again, you shouldn't need the angles (θ and φ) to find the solution, just a and b.

This message has been edited. Last edited by: silenteuphony,