Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

An image of a square is projected onto a sphere (radius $R$) as above (the dot is the centre of the sphere, and the red projection is marked out where the line from the centre-dot to a point on the square's perimeter intersects the sphere's surface). The square touches the sphere with one vertex only (this is also the point where the plane the square inhabits is tangent to the sphere). A (red) kite is the projection formed on the sphere.

enter image description here

It's fairly evident, using symmetry, to find that there are 2 pairs of kite's edges (is geodesic the correct term?) which have equal length (blue and red above). Similarly two of the angles ($\beta$) are the same.

It's easy to find $P$ and $Q$ given $R$ and the square's edge-length, but how does one find the angles $\alpha$ and $\beta$ given $P$ and $Q$?

(I assume $\gamma=\frac{\pi}{4}$ because the square touches the sphere at the bottom right vertex).

This originated from a question about the red kite's area (easy to get is you know $\alpha$, $\beta$ and $\gamma$), but whilst I'm interested how to find $P$ and $Q$ now for its own sake, is there a more direct way that only involves $P$ and $Q$ to find the area (this is not the same as a formula that only involves $P$, $Q$ and the area but whose derivation requires $\alpha$, $\beta$ and $\gamma$)?

Apologies that the question is very basic, this is my first experimentation with spherical geometry.

share|cite|improve this question
When you said "project": which projection to the sphere are you using? In other words, where is the big black dot in reference to the sphere? – Willie Wong Dec 21 '12 at 12:45
Duly edited, thanks. – Alyosha Dec 21 '12 at 15:10
What is the actual question here? In one part you ask about angles $\alpha$ and $\beta$, in another you want to compute an area. As the former is written with emphasis, I'd have considered this the core question, but the currently accepted answer only refers to the latter. – MvG Dec 23 '12 at 14:48
Both. I realise it would have been better to make two questions, the latter was more of an afterthought. Both answers helped; I somewhat arbitrarily chose Willie's on account of his links leading to further insights into spherical geometry, even though that wasn't part of the question. – Alyosha Dec 23 '12 at 18:04
up vote 3 down vote accepted

Is there a more direct way that only involves $P$ and $Q$ to find the area (this is not the same as a formula that only involves $P$, $Q$ and the area but whose derivation requires $α$, $β$ and $γ$)?

In short, no.

You are using the gnomonic projection on the sphere. This projection has the great advantage that great circles gets mapped to straight lines and vice versa, which means that as you noted, all four of the curves shown that bounds your "kite" are geodesic segments.

For geodesic polygons on the sphere, a corollary of the Gauss-Bonnet Theorem is that

The angle surplus (the sum of the internal angles minus the total sum for the corresponding polygon in Euclidean space) is related to the area of the polygon via a factor of proportionality that depends only on the radius of the sphere.

See also Girard's theorem.

Now, by the properties of the gnomonic projection (which can be had by symmetry considerations; take four squares that join together at the tangent point) the angles $\beta$ and $\gamma$ are necessarily right angles (measure $\pi/2$). Hence: to find the area of the kite is exactly equivalent to finding the value of the angle $\alpha$.

share|cite|improve this answer
Is the statement in the second quotation provable without using differential geometry? – Alyosha Dec 21 '12 at 21:41
@Alyosha: considering that Girard discovered the statement more than two hundred years before differential geometry was formulated, I would say yes. See – Willie Wong Jan 3 '13 at 12:51

If you had no problem computing $P$ and $Q$, then I guess that computing the length of the diagonal (the one along the axis of symmetry) should be easy as well. Let's call this length $S$. Then you have a triangle and know three edge lengths, so you can apply the spherical law of cosines to compute any angle. In particular, you get

\begin{align*} \cos\beta &= \frac{\cos\frac SR-\cos\frac PR\cdot\cos\frac QR}{\sin\frac PR\cdot\sin\frac QR} \\ \cos\frac\alpha2 &= \frac{\cos\frac PR-\cos\frac QR\cdot\cos\frac SR}{\sin\frac QR\cdot\sin\frac SR} \end{align*}

share|cite|improve this answer

When reading further into the problem, I found this excellent for proofs of spherical trigonometric formulae, especially as it didn't use vectors, which in hindsight overcomplicated things, like here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.