# Centroid of a spherical triangle.

I'm trying to find the centroid of a spherical triangle over a unit sphere centered at the origin. I have already seen that I can obtain the centroid of the planar triangle (which are formed by the three position vectors of the triangle vertices) and then project the vector to the sphere surface. But what I'm looking is actually to find the centroid by using spherical trigonometry. Any help is appreciated!

• "..project the vector to the sphere surface " : how do you define the centroid of a spherical triangle ? – G Cab Nov 25 '16 at 23:18
• In agreement with @GCab, I’d like to know what the definition of centroid is to be. I’m moderately handy with spherical trigonometry, but would need the definition. – Lubin Jan 5 '17 at 23:45
• Thank you for your replies. What I meant by centroid is the intersection of the three medians or analogous to them in spherical trigonometry. This because I'm trying to find the centroid/center of mass of a spherical triangle over a sphere. Any help is appreciated – Juan Jan 7 '17 at 20:31

This is a really good question. Firstly google spherical trigonometry & look at the wiki page. Make sure you are familiar with the cosine rule, sine rule & supplemental cosine law. It maybe an idea to look at the first ref (Todhunter) In chapter 7 he answers two similar problems to yours. He does the circumcenter and incenter ... these are good examples for you to sharpen your teeth on.Figures 1 & 2

right in the first figure we have drawn the medians (concurring.) This is slightly cheating , it remians to be seen that they really do concur. In the next figure we have drawn half of the triangle with one median drawn. Two sides & the angle between them are known (the side b, the angle C & the side a/2). We have named the other side x & the angle at vertex A A_1 & the final angle A_+. You need to calculate the sin & cos of all three of these quantities

\ \begin{eqnarray*} \cos x&=&\frac{...}{...} &\sin x=&\frac{...}{...}\\ \cos A_1&=&\frac{...}{...} & \sin A_1=&\frac{...}{...}\\ \cos A_+&=&\frac{\sin c\cos B -\sin b\cos C}{\sqrt{\sin^2 b +2\sin b\sin c \cos A +\sin^2c}} & \sin A_+=&\frac{...}{...}\\ \end{eqnarray*}

I have left some blanks for you to fill in.

Figure 3 In figure 3 we have drawn the triangle formed by two medians & half a side ("one sixth of the original triangle"). We know two angles and one side of this triangle, the angle ${A_+}$, the side $a/2$ & the angle ${C_1}$. (${C_1}$ is the same formula we have for ${A_1}$ with a,b & c & A,B & C cyclically permuted.) We need to calculate the cos and sin of the other 3 unknowns

\begin{eqnarray*} \cos x_1&=&\frac{...}{...} &\sin x_1=&\frac{...}{...}\\ \cos z_2&=&\frac{...}{...} & \sin z_2=&\frac{...}{...}\\ \cos B_*&=&\frac{\sin^2b+\sin b(\sin a\cos C + \sin c\cos A) -\sin a\sin c\cos B}{\sqrt{(\sin^2a +2\sin a\sin b \cos C +\sin^2 a)(\sin^2b +2\sin b\sin c \cos A +\sin^2c)}} & \sin B_*=&\frac{...}{...}\\ \end{eqnarray*} Again we have left lots of blanks for you to fill in. (The calculation of $\sin B_*$ is sublime !)

Now note that $\cos B_*$ is invariant in the interchange of a & c and A & C (together.) The lengths $x_1$ or $z_2$ could have been calculated using any other pair of medians & exactly the same formulea would have been derived; this shows that the three medians concur & therefore the centroid of a spherical triangle exists.

It remains only to give the trilinear coordinates of the point. [Figure 4]
See figure 4, we need the perpendicular distance $t_a$ to each side. \begin{eqnarray*} \sin t_a&=&\frac{\sin b\sin C}{\sqrt{(3+2(\cos a+\cos b+\cos c))}} \\ \end{eqnarray*} $t_b$ & $t_c$ can be obtained similarly. We can rescale these by any symmetric function in $a,b$ & c ... & thus the sin of the trilinear coordinates for the centroid of a spherical triangle is \begin{eqnarray*} \sin b\sin C : \sin c\sin A :\sin a\sin B. \\ \end{eqnarray*}

I am slightly saddened that my original comment (I made a couple of days ago has been removed. I do agree that it was not an answer but I had no other means to post my comment.) I am very pleased that I have caused at least 8 other people to revisit this problem :-)