How to divide a spherical triangle into three equal-area spherical triangles?

Question

The Centroid point (at intersection of medians) divides a planar triangle into three equal-area smaller triangles. In case of spherical triangle, the three geodesics joining the vertex to the midpoint of the opposite side of spherical triangle intersect at a (Centroid-like) common point. But, when geodesic is constructed to join this Centroid-like point to the three vertices, the resultant smaller spherical triangles do not have equal area! A different geometrical method or a formula for implementing the exact division of spherical triangle into three equal-area spherical triangles could not be found in literature.

Answer 1

For any point $P$ on unit sphere, let $P'$ be its antipodal point and $\hat{P}$ the corresponding unit vector. We will use the notation $\mathcal{C}_{P_1P_2\cdots P_n}$ to denote a spherical arc starting from $P_1$, passing through $P_2,\ldots$ and end at $P_n$.

Let $A, B, C$ be any three points on unit sphere, close enough to fit within half of a hemisphere
(i.e. a spherical lune of angle $\frac{\pi}{2}$ ). Let $\Omega_{ABC}$ be the area of spherical $\triangle ABC$. It can be computed using a formula by Oosterom and Strackee

$$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{ \left|\hat{A}\cdot (\hat{B} \times \hat{C})\right|}{1 + \hat{A}\cdot\hat{B} + \hat{B}\cdot\hat{C} + \hat{C}\cdot\hat{A}}$$

In the special case where $A, B$ lies on the equator, symmetric with respect to $x$-axis and $C$ lies on the upper hemisphere, i.e.

$$ \begin{cases} A &= (\cos\alpha,-\sin\alpha,0),\\ B &= (\cos\alpha,+\sin\alpha,0),\\ C &= (x, y, z)\end{cases} \quad\text{ where } \alpha \in (0,\frac{\pi}{2}), z > 0$$ Above formula reduces to $$\tan\left(\frac{\Omega_{ABC}}{2}\right) = \frac{\sin\alpha z}{\cos\alpha + x}$$

This implies the locus of $P$ in upper hemisphere for fixed $\Omega_{ABP} = \Omega_{ABC}$ is the circular arc $\mathcal{C}_{B'CA'}$.

Let $\theta$ be the angle between $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$ at $B'$. The plane holding the locus has the form

$$t ( c + x ) - s z = 0\quad\text{ where }\quad \begin{cases} t &= \tan\left(\frac{\Omega_{ABC}}{2}\right)\\ c &= \cos\alpha\\ s &= \sin\alpha \end{cases}$$ The normal vector of the plane is pointing along the direction $(t, 0, -s)$. This means the tangent vector of $\mathcal{C}_{B'CA'}$ at $B'$ is along the direction $(t, 0, -s ) \times ( -c, -s, 0 ) \propto (s, -c, t)$. Notice the tangent vector of $\mathcal{C}_{B'ABA'}$ at $B'$ is pointing along the direction $(s,-c,0)$, we find

$$\cos\theta = \frac{s^2 + c^2 + 0}{\sqrt{s^2+c^2}\sqrt{s^2+c^2+t^2}} = \cos\left(\frac{\Omega_{ABC}}{2}\right)$$ From this, we can deduce the circular arcs $\mathcal{C}_{B'PA'}$ and $\mathcal{C}_{B'ABA'}$ intersect at an angle $\frac{\Omega_{ABC}}{2}$.

This leads to following construction of the desired "spherical centroid" $X$.

1. Construct the circular arcs $\mathcal{C}_{B'CA'}$ and $\mathcal{C}_{B'ABA'}$,

2. Trisect the angle $\angle AB'C$ - i.e. find a circular arc $\mathcal{B'\tilde{C}A'}$ such that $\tilde{C}$ is lying on same side as $C$ with respect to $AB$ and $\angle AB'\tilde{C} = \frac13 \angle AB'C$.

3. Repeat above procedures for other two sides of $\triangle ABC$ to get circular arcs $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.

4. $X$ will be lying on the common intersection of the three circular arcs $\mathcal{C}_{B'\tilde{C}A'}$, $\mathcal{C}_{C'\tilde{A}B'}$ and $\mathcal{C}_{A'\tilde{B}C'}$.

Answer 2

HINT:

Even if not found in literature we can still attempt a solution, assuming it exists.

CG sph triangle

Let $$ A = A1+ A2 , \, B = B1+ B2 ,\, C = C1 + C2 \tag{1}$$ where A, B, C are vertex angles of spherical triangle.

If the angles subtended at center being identified are $ \gamma1 + \gamma2 + \gamma3 , = 2 \pi , $

$$ A1+ \gamma1 + C2 = B1 + \gamma2 + A2 = C1 + \gamma3 + B2 = ? \tag{2}$$

The Sine and Cosine Rules could perhaps be used ....

Answer 3

As @Narasimham already wrote, equal area means equal angle surplus. I write $D$ for the center you're after. And I use indices $1$ through $3$ to describe the three subtriangles. Then the area equivalence is expressed by

$$A_1+B_1+D_1 = B_2+C_2+D_2 = C_3+A_3+D_3$$

The angles at the corners have to add up to the original angles:

$$A_1+A_3 = A \qquad B_1+B_2 = B \qquad C_2+C_3 = C$$

Furthermore, the outer edge lengths have to remain the same, which you can formulate using the spherical law of cosines:

$$ \cos c = \frac{\cos C+\cos A\cos B}{\sin A\sin B} =\frac{\cos D_1+\cos A_1\cos B_1}{\sin A_1\sin B_1} $$

or after cross multiplication

$$ (\cos C + \cos A\cos B)\sin A_1\sin B_1 = (\cos D_1+\cos A_1\cos B_1)\sin A\sin B \\ (\cos A + \cos B\cos C)\sin B_2\sin C_2 = (\cos D_2+\cos B_2\cos C_2)\sin B\sin C \\ (\cos B + \cos C\cos A)\sin C_3\sin A_3 = (\cos D_3+\cos C_3\cos A_3)\sin C\sin A $$

Since the angles must add up in the center, you also have

$$D_1+D_2+D_3=2\pi$$

so at this point you have nine equations in nine indenterminates, namely all the angles with indices. But the equations involve trigonometric functions.

You could turn them into completely polynomial equations, by not using $A_1$ itself as an indeterminate, but instead $\sin A_1$ and $\cos A_1$, with the added constraint of $\sin^2 A_1+\cos^2 A_1=1$. Likewise for all the other angles. Adding angles would become more complicated; you'd have to use the trigonometric formulas for angle sums. At the end you'd have $18$ indeterminates and sufficiently many equations to describe them all.

At least in theory. I haven't tried this out yet. If I find the time, I'll try to do this for the general case, see if I can get my computer algebra system to eliminate all the variables and give me an explicit condition for e.g. $\cos A_1$. But I doubt this computation will be pretty, and I assume that the result will be too long to be posted here anyway.

If you have a specific instance of this problem, with actual numbers, then the chances are much higher that you'll be able to obtain an exact solution using this approach.