# Point within a spherical triangle given areas

Consider a spherical triangle like this:

where $A_1, A_2, A_3,$ and $P$ are points on the sphere and $t_1, t_2, t_3$ are the proportion of the area of the large triangle contained within the small triangle: that is, if $\Omega$ is the area of $A_1 A_2 A_3$, the area of $A_1 A_2 P$ is $t_3 \Omega$.

Given $A_1 A_2 A_3$ and $t_1, t_2, t_3$, what is $P$? Is there an analytic expression for $P$, or can it only be found numerically? Is there an analytic expression if $A_1 A_2 A_3$ is an equilateral spherical triangle (i.e. $A_1, A_2,$ and $A_3$ are equidistant)?

(If this was a planar triangle, this would reduce to simple barycentric coordinates in the plane. In fact, that's where I stole the image from...)

For any point $$X$$ on the unit sphere, let $$\vec{X}$$ be corresponding unit vector.

We will assume

• The spherical triangle $$A_1A_2A_3$$ is small in the sense that it falls within a hemisphere.
• Vertices are ordered in counterclockwise manner. i.e. $$\Delta \stackrel{def}{=} \vec{A}_1 \cdot ( \vec{A}_2 \times \vec{A}_3 ) > 0$$.
• The desired point $$P$$ lies inside $$A_1A_2A_3$$. i.e. $$t_1,t_2,t_3 \ge 0, t_1 + t_2 + t_3 = 1$$.

Express $$P$$ as a "linear combination" of $$A_1,A_2,A_3$$. i.e.

$$\vec{P} = u_1 \vec{A}_1 + u_2 \vec{A}_2 + u_3\vec{A}_3$$ for parameter $$u_1,u_2,u_3$$ to be determined.

Let $$(i,j,k)$$ be any cyclic permutation of $$(1,2,3)$$. Define

• $$v_k = \frac{u_k}{1 + u_1+u_2+u_3} \iff u_k = \frac{v_k}{1 - v_1 - v_2 - v_3}$$
• $$\alpha_k = \vec{A}_i \cdot \vec{A}_j$$ and $$\beta_k = \frac{\alpha_i + \alpha_j}{1 + \alpha_k}$$
• $$\tau = \tan\left(\frac{\Omega}{2}\right)$$, $$\tau_k = \tan\left(\frac{t_k\Omega}{2}\right)$$ and $$\lambda_k = \frac{\tau_k}{\tau}$$.

Apply the formula of solid angle in OP's own answer to triangle $$A_1A_2A_3$$, we obtain $$\tau = \frac{\Delta}{1 + \alpha_1 + \alpha_2 + \alpha_3} = \frac{\Delta}{(1+\alpha_k)(1+\beta_k)}$$

Apply same formula to triangle $$PA_iA_j$$, we obtain

\begin{align} \tau_k &= \frac{\vec{A}_i \cdot (\vec{A}_j \times \vec{P})}{ 1+ \vec{A}_i \cdot \vec{A}_j + \vec{A}_i\cdot\vec{P} + \vec{A}_j\cdot\vec{P}}\\ &= \frac{u_k \Delta}{ 1 + \alpha_k + (u_i + u_j \alpha_k + u_k \alpha_j) + (u_i \alpha_k + u_j + u_k \alpha_i)}\\ &= \frac{u_k \Delta}{ (1 + \alpha_k)(1 + u_i + u_j) + (\alpha_j + \alpha_k)u_k}\\ &= \frac{u_k \Delta}{ (1 + \alpha_k)(1 + u_i + u_j + \beta_k u_k)}\\ &= \frac{v_k \Delta}{ (1 + \alpha_k)(1 - (1 -\beta_k) v_k)}\\ \end{align} This leads to

$$v_k = \frac{1}{\frac{\Delta}{\tau_k(1+\alpha_k)} + 1 - \beta_k} = \frac{1}{\frac{(1+\beta_k)\tau}{\tau_k} + (1-\beta_k)} = \frac{\lambda_k}{(1+\beta_k) + (1-\beta_k)\lambda_k}$$

In the special case $$A_1A_2A_3$$ is equilateral with side length $$\theta$$, we have

$$\alpha_1 = \alpha_2 = \alpha_3 = \cos\theta \implies \beta_1 = \beta_2 = \beta_3 = \beta \stackrel{def}{=}\frac{2\cos\theta}{1 + \cos\theta}$$

The formula for $$P$$ simplifies to

$$\vec{P} = \frac{v_1 \vec{A}_1 + v_2 \vec{A}_2 + v_3 \vec{A}_3}{1-v_1-v_2-v_3}$$

where $$v_k = \frac{\lambda_k}{(1+\beta) + (1-\beta)\lambda_k},\quad \lambda_k = \frac{\tan\left(\frac{t_k \Omega}{2}\right)}{\tan\left(\frac{\Omega}{2}\right)}$$ In particular, when $$\theta = \frac{\pi}{2}$$, $$A_1A_2A_3$$ is one-eight of the unit sphere. $$\vec{A}_1,\vec{A}_2, \vec{A}_3$$ are orthonormal and $$\beta = 0$$. The coefficients $$v_k$$ become something very simple. $$v_k = \frac{1}{\cot\left(\frac{t_k\pi}{4}\right) + 1}$$

Here's an answer that is correct but probably not as simplified as it could be; feel free to improve on it.

Start with the formula for solid angle of a spherical triangle given on wikipedia, and assume all vectors are of unit length, then for a triangle defined by vertices $\vec a\vec b\vec c$:

$$\tan \left( \frac{\Omega_{abc}}{2} \right) = \frac{\vec a\cdot \vec b\times \vec c}{1 + \left(\vec a \cdot \vec b\right) + \left(\vec a \cdot \vec c\right) + \left(\vec b \cdot \vec c\right)}$$

Let $\vec x$ be the point inside the spherical triangle. The area of the triangle $\vec a \vec b \vec x$ can be grouped and rearranged as so:

$$x \cdot \left(\cot \frac{\Omega_{abx}}{2}\vec a \times \vec b - \vec a - \vec b\right) = 1 + \vec a \cdot \vec b$$

Define $\vec L_{abx} = \cot (\Omega_{abx}/2) \vec a \times \vec b - \vec a - \vec b$ and $h_{abx} = 1 + \vec a \cdot \vec b$.

The other triangles $\vec b\vec c\vec x$ and $\vec c \vec a \vec x$ are similar, resulting in $\vec L_{bcx}, \vec L_{cax}, h_{bcx}$, and $h_{cax}$.(Be careful with the order of points here: they all have to be in counterclockwise order, or else your signs get confused.) These quantities are related to $\vec x$ by the linear system

$$\begin{bmatrix} \vec L_{abx} & \vec L_{bcx} & \vec L_{cax} \end{bmatrix} \vec x = \begin{bmatrix} h_{abx} \\ h_{bcx} \\ h_{cax} \end{bmatrix}$$

where $\begin{bmatrix} \vec L_{abx} & \vec L_{bcx} & \vec L_{cax} \end{bmatrix}$ is the matrix where the first column is the vector $\vec L_{abx}$, etc. Define $\vec h = [h_{abx}, h_{bcx}, h_{cax}]$. Cramer's rule can be used to get the components of $\vec x = [x_1, x_2, x_3]$ :

$$x_1 = \frac{\begin{vmatrix}\vec h & \vec L_{bcx} & \vec L_{cax}\end{vmatrix}}{\begin{vmatrix}\vec L_{abx} & \vec L_{bcx} & \vec L_{cax}\end{vmatrix}}, \, etc.$$

This is an explicit, analytic expression, but is complicated. My attempts to simplify this form have only resulted in frustration; maybe someone with access to a better computer algebra system can whittle it down?