0

Well, I think the title already explains my question. Given a sphere and an ordered sequence of inner angles ($\alpha$, $\beta$, $\gamma$, $\delta$) how many spherical quadrangles do there exist that have that sequence as angles and the added property that three of the edges need to have the same size and the fourth edge needs to have a different size.?

I was told that this on the sphere there might be several quadrangles with quite different appearances, but I can't find any reference explaining this in more detail or bounding the number of possible quadrangles.

You can assume that the angles satisfy the conditions necessary to be the angles of a spherical quadrangle.

nvcleemp
  • 297

2 Answers2

2

There is no bound on the sphere either. Fix an angle $\alpha > 0.$ For any $\beta > 0,$ Construct the triangle with angles $$ \frac{\pi}{2} + \alpha, \; \; \frac{\pi}{2} - \beta, \; \; \alpha + \beta. $$ The angle sum is $\pi + 2 \alpha,$ so the area is $2 \alpha.$ Now, find the midpoint of the edge opposite the angle $\frac{\pi}{2} + \alpha,$ and place a copy of the same triangle there rotated 180 degrees ($\pi$) around that point, so that the result is a quadrangle. Because $$ \frac{\pi}{2} - \beta + \alpha + \beta = \frac{\pi}{2} + \alpha, $$ the resulting quadrangle has all four angles equal to $ \frac{\pi}{2} + \alpha.$

However, by the Law of Sines on the sphere, the ratio of (the sines of) the edge $x$ opposite the angle $\alpha + \beta$ and the edge $y$ opposite the angle $ \frac{\pi}{2} - \beta$ is $$ \frac{\sin x}{\sin y} \; \; = \; \; \frac{\sin(\alpha + \beta)}{\sin{(\frac{\pi}{2} - \beta})} = \frac{\sin(\alpha + \beta)}{\cos \beta} = \sin \alpha + \cos \alpha \tan \beta. $$ That is, as we vary $\beta,$ we get uncountably many quadrangles with the same vertex angles that are not congruent.

Will Jagy
  • 139,541
0

The answer for the plane is wrong. You can move any vertex and the two sides incident at it while keeping the opposite vertex and the two sides incident at it fixed, then extend or clip all sides to meet in two more vertices. The result isn't a scaled version of the original quadrangle. This is easiest to imagine for a rectangle or a parallelogram, where you can simply extend two parallel sides.

The same is true for at least some sequences of angles on the sphere. For instance, take two meridians and the equator, which form two right angles, and connect two points at equal latitude on the meridians by a great circle. Now if you move the meridians apart symmetrically, they still form right angles with the equator. The angle they form with the great circle changes, but you can compensate for that by connecting them at a different latitude with a different great circle.

joriki
  • 238,052
  • circles of latitude are not geodesics – Will Jagy Feb 23 '12 at 20:32
  • @Will: Sorry, what nonsense; somehow it had slipped my mind that the sides should be geodesics. I deleted that. – joriki Feb 23 '12 at 20:36
  • 1
    ... and replaced it with a hopefully more sensible example. – joriki Feb 23 '12 at 20:42
  • Thanks, I updated my question to contain a more correct assessment of the plane case.

    I was wondering whether you could limit the number of quadrangles if you add the property that three of the edges need to be equal and one different from the other three.

     – nvcleemp Feb 24 '12 at 09:35
  • @nvcleemp: I think "up to scaling of opposite edges" understates the freedom in constructing quadrangles in the plane. That was just one example that I gave that's easy to visualize; the process that I described is more general than that, and I don't see how it could be described as "scaling" -- certainly not only of opposite edges. I think the statement that there's "a unique quadrangle which has those angles" is simply wrong. On your other question: Do you mean in the plane? In that case, if three of the edge have to be equal, that leaves only one quadrangle up to similarity. – joriki Feb 24 '12 at 11:07
  • No, I mean the case on the sphere. I have a sequence of four angles and the condition that three edges need to be the same and one different. I was hoping that just the sequence of angles would be sufficient to limit the number of possible quadrangles, but - as you demonstrated - this is not the case, so the other conditions will also be needed and I hope that this at least limits the number of quadrangles a bit. – nvcleemp Feb 24 '12 at 13:06
  • @nvcleemp: Yes, it does. The excess of the angle sum over $2\pi$ is the area of the quadrangle (as you can show by splitting it into two triangles, in each of which the excess over $\pi$ is the area), so the area of the quadrangle is fixed. If you fix three of the angles, including the two between the three equal sides, each length of the equal sides will yield one quadrangle, and only one of these will have the right area and thus the right fourth angle. – joriki Feb 24 '12 at 13:20
  • @nvcleemp: You haven't changed the sentence about the unique quadrangle in the plane -- do you disagree with my argument about that? – joriki Feb 24 '12 at 13:21
  • @joriki: no, I just haven't had the time to fix that ;-) – nvcleemp Feb 24 '12 at 16:46