Area of triangle on sphere

For three points ${A_1, A_2, A_3}$ on the unit sphere in ${\mathbb{R}^3}$, suppose that they do not lie on a great circle, then there is a unique great circle on the sphere joining ${A_1}$ and ${A_2}$, and so on. Obviously there are two segments of the great circle joining each pair of points, we can always declare ${A_1A_2}$ to be the minor segment (i.e. length ${<\pi}$) of the great circle joining ${A_1}$ and ${A_2}$. We can also assume ${A_1, A_2, A_3}$ to be positively oriented, i.e. ${\det (A_1, A_2, A_3)>0}$. Then we can uniquely define the spherical triangle ${\triangle A_1A_2A_3}$ to be the region on the sphere bounded by the segments ${A_1A_2, A_2A_3}$ and ${A_3A_1}$ (in this orientation), such that it always lies on the “left hand side” of this oriented closed curve. We are interested to find the area of ${\triangle A_1A_2A_3}$. After discussing with Ken, Michael, John Ma and Wongting, we have obtained the following results. ☺

Theorem 1 (Ken-Michael-Wongting’s theorem, 2011 (Area formula for triangle))

With the notations as above, suppose ${a_i}$ is the interior angle of ${\triangle A_1 A_2 A_3}$ at the vertex ${A_i}$, then

$\displaystyle \textrm{Area}(\triangle{A_1A_2A_3})= a_1+a_2+a_3-\pi.$

Proof: This can be easily proved by the Gauss-Bonnet theorem, but here I will give a more elementary proof. Suppose ${A'=-A}$ is the antipodal point of ${A}$ for any point ${A}$, it is easy to see that ${\triangle ABC}$ and ${\triangle A'B'C'}$ are congruent (they are mapped to each other by the (orientation reversing) map ${X\mapsto -X}$, which clearly preserves distances), and hence have the same area.

Now consider the region consisting of ${\triangle A_1 A_2 A_3}$ and ${\triangle A_1'A_2 A_3}$, as the arcs ${A_2A_1}$ and ${A_3 A_1}$ spans an angle of ${a_1}$ at ${A_1}$, we clearly have

$\displaystyle \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1' A_2 A_3) = \frac{a_1}{2\pi}\times (\text{\textrm{Area}(whole sphere)})= 2a_1.$

So we have the system

$\displaystyle \begin{array}{rcl} \begin{cases} \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1' A_2 A_3) = 2a_1\\ \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_3 A_2' A_1) = 2a_2\\ \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1 A_3'A_2) = 2a_3. \end{cases} \end{array}$

But by the previous remark, ${\textrm{Area}(\triangle A_1' A_2 A_3) =\textrm{Area}(\triangle A_1 A_2' A_3') }$, so the first equation of the system becomes ${\textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1 A_2' A_3') = 2a_1}$, and noting that a hemisphere is divided into the four triangles ${\triangle A_1 A_2 A_3, \triangle A_1 A_2' A_3', \triangle A_3 A_2' A_1}$ and ${\triangle A_1 A_3' A_2'}$, so we can add the forth equation into the system:

$\displaystyle \begin{array}{rcl} \begin{cases} \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1 A_2' A_3') = 2a_1\\ \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_3 A_2' A_1) = 2a_2\\ \textrm{Area} (\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1 A_3'A_2) = 2a_3.\\ \textrm{Area}(\triangle A_1 A_2 A_3)+\textrm{Area}(\triangle A_1 A_2' A_3')+\textrm{Area}(\triangle A_3 A_2' A_1)+\textrm{Area}(\triangle A_1 A_3' A_2)=2\pi. \end{cases} \end{array}$

Summing the first three equations together and subtracting the last one, we can get the result. $\Box$

The above result is very neat, but given three points ${A_1, A_2, A_3}$, it needs some work to find the three angles ${a_i}$. (They can be found by, in principle, the cosine law on sphere. ) I would like to have a more direct formula using ${A_i}$. So far the best (goodest?) result I can obtain is the following

Theorem 2 (Ken-Michael-Wongting’s theorem, 2011 (Area formula for triangle))
With the same notations as before,

$\displaystyle \tan(\textrm{Area}({\triangle A_1A_2A_3})) = \frac {\det(A_1,A_2,A_3)(w_{1}w_{2}+w_{2}w_{3}+w_{1}w_{3}) - \det(A_1,A_2,A_3)^3} {w_{1}w_{2}w_{3}-\det(A_1,A_2,A_3)^2(w_{1}+w_{2}+w_{3})}$

where ${w_{1}=\langle A_1 \times A_2, A_1\times A_3\rangle, w_{2}=\langle A_2 \times A_1, A_2\times A_3\rangle, w_{3}=\langle A_3 \times A_1, A_3\times A_2\rangle}.$

Proof: By Theorem 1,

$\displaystyle \tan(\textrm{Area}({\triangle A_1A_2A_3}))= \tan (a_1+a_2+a_3)= \frac{\tan a_1+ \tan a_2+ \tan a_3- \tan a_1\tan a_2\tan a_3}{1- \tan a_1\tan a_2-\tan a_2\tan a_3-\tan a_1\tan a_3}. \ \ \ \ \ (1)$

The plane ${OA_1A_2}$ has normal vector ${ n_{A_1A_2}=\frac{A_1\times A_2}{|A_1\times A_2|}}$, and it is easy to see that the angle ${a_1}$ is the angle between ${n_{A_1A_2}}$ and ${-n_{A_3A_1}=\frac{A_1\times A_3}{|A_1\times A_3|}}$. Thus we have (see Lemma 2 of here)

$\displaystyle \sin a _1= \left|\frac{A_1\times A_2}{|A_1\times A_2|}\times \frac{A_1\times A_3}{|A_1\times A_3|}\right| =\frac{\left|\det(A_1,A_2,A_3)A_1\right|}{|A_1\times A_2||A_1\times A_3|} =\frac{\det(A_1,A_2,A_3)}{|A_1\times A_2||A_1\times A_3|},$

and

$\displaystyle \cos a_1=\frac{\langle A_1\times A_2, A_1\times A_3\rangle}{|A_1\times A_2||A_1\times A_3|}.$

Therefore

$\displaystyle \tan a_1 = \frac{ \det(A_1,A_2,A_3)}{\langle A_1\times A_2, A_1\times A_3\rangle}, \text{ and similarly, } \tan a_2 = \frac{ \det(A_1,A_2,A_3)}{\langle A_2\times A_1, A_2\times A_3\rangle},\tan a_3 = \frac{ \det(A_1,A_2,A_3)}{\langle A_3\times A_1, A_3\times A_2\rangle}.$

Plugging these into (1), we can get the result. $\Box$

Question: The formula looks quite complicated, can we simplify this result?

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2 Responses to Area of triangle on sphere

1. Junyan Xu says:

You seems arbitrary in assigning names to theorems …

2. KKK says:

Well, I name those theorems according to their contibution to the (re-)discovery of the results. Of course this doesn’t necessarily mean that they are the first ones to discover them, but in this case it is just due to my ignorance about the literature.