Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A triangle on a sphere is composed of points $A$, $B$ and $C$. The $\alpha$, $\beta$ and $\gamma$ denote the angles at the corresponding points of the triangle:

Spherical triangle

The Girard's theorem states that the surface area of any spherical triangle:

$$ A = R^2 \cdot E $$

where $R$ is the radius of the sphere and $E$ is the excess angle of $(\alpha + \beta + \gamma - \pi)$

I'm wondering how to derive this formula.

Could you please help explain this clearly?

share|improve this question
    
    
TeX tip: Notice the difference between $\alpha$ + $\beta$ + $\gamma$ - $\pi$ and $\alpha + \beta + \gamma - \pi$. In the former, the minus sign looks like a hyphen instead of a minus sign. There's a reason why that's not standard usage. I fixed this in the posting. –  Michael Hardy Feb 16 '12 at 18:06
    
@Nunoxic, no. That question is about how to determine the angles between ABC, ACB and BAC. –  ezpresso Feb 16 '12 at 18:06

1 Answer 1

up vote 5 down vote accepted

Consider the following three parts of the sphere: let $P_A$ be the lune created from the triangle $ABC$ plus the triangle adjacent across the $BC$ line segment, plus the opposite lune (on the opposite side of the sphere), and similarly for parts $P_B$ and $P_C$.

The area of $P_A$ is $4\alpha R^2$: the total area of the sphere is $4\pi R^2$, and the area of $P_A$ is certainly proportional to $\alpha$.

Notice now that $P_A\cup P_B\cup P_C$ is the entire sphere, and that $P$'s intersect at the triangle + the opposite triangle. We thus have:

area($P_A$) + area($P_B$) + area($P_C$) = area of the sphere + 2 area of the triangle +2 area of the opposite triangle.

As the two triangles have the same area, you get your formula.

share|improve this answer
    
You really should explain what they cut the circle to 4 parts actually means. The first part is the made up of the lunes $ABA'C$ and $A'B'AC'$ put together, for example. Otherwise this is a good argument by symmetry. –  anon Feb 16 '12 at 19:53
    
@anon: I meant "they cut the sphere"; thanks for spotting it! –  user8268 Feb 16 '12 at 19:57
    
I've edited it to reflect what I believe you were saying. Additionally, if we regard $A'$, $B\,'$ and $C\,'$ respectively as the points opposite to $A$, $B$ and $C$ on the sphere, we have the decomposition $$\begin{array}{} P_A=ABA'C\;\cup\;A'B\,'AC\,' \\ P_B= BCB\,'A\;\cup\; B\,'C\,'BA \\ P_C= CAC\,'B\;\cup\; C\,'A'CB\,' \end{array}$$ –  anon Feb 16 '12 at 20:18
    
$ P_B = BCB'A \cup B'C'BA' $? –  ezpresso Feb 16 '12 at 21:02
    
Yes, forgot that last accent mark, thanks. –  anon Feb 17 '12 at 16:22

Your Answer

 
discard

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.