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

Median of a planar triangle divides it into two equal-area smaller triangles. Whereas, in a spherical triangle, the geodesic joining the corner to the midpoint of the opposite side does not divide the spherical triangle into two equal-area parts! I have searched several textbooks on 'Spherical Trignometry' including the classic book by I. Todhunter. But, a geometrical method or a formula for implementing the exact division of spherical triangle into two equal-area spherical triangles seems to be unavailable.

• The area of a spherical triangle is the spherical excess: en.wikipedia.org/wiki/… - that should tell you how to divine one angle to make the two parts equal. – Ethan Bolker Apr 5 '16 at 14:45

## 2 Answers Consider the spherical triangle $\mathscr A$$\mathscr B$$ \mathscr C$depicted in the figure, with angular length of its sides a, b, c and angle subtended at the corners A, B, C. Then, the solid angle (or its surface area in case of sphere of unit radius) covered is $E=A+B+C-π$

Let the point $\mathscr D$ on $\mathscr B$$\mathscr C be such that the geodesic \mathscr A$$ \mathscr D$ divides $\mathscr A$$\mathscr B$$ \mathscr C$ into two smaller spherical triangles of equal area. Now, we shall solve for the angle x marked in the figure.

Area of spherical triangle $\mathscr A$$\mathscr B$$ \mathscr D$ is: $x+y+b-π=E/2=(A+B+C-π)/2$

$$y=\frac{A+C-B+π}{2}-x$$ $$cos(y)=cos(\frac{A+C-B+π}{2})cos(x)+sin(\frac{A+C-B+π}{2})sin(x)$$

In $\mathscr A$$\mathscr B$$ \mathscr D$, the spherical trigonometry relation between c and included angles is,

$$cos(y)=cos(c)sin(B)sin(x)-cos(B)cos(x)$$

Eliminating cos(y) in the above two equations, the relation for calculating angle x for the given triangle $\mathscr A$$\mathscr B$$ \mathscr C$ can be got,

$$tan(x)=\frac{cos(B)-sin(\frac{A+C-B}{2})}{sin(B)cos(c)-cos(\frac{A+C-B}{2})}$$

In a similar fashion, $\mathscr A$$\mathscr B$$ \mathscr C$ could be divided into two equal area spherical triangles in two more ways using suitable points (say $\mathscr E$ on $\mathscr C$$\mathscr A and \mathscr F on \mathscr A$$ \mathscr B$). The three geodesic arcs $\mathscr A$$\mathscr D, \mathscr B$$ \mathscr E$ and $\mathscr C$$\mathscr F meet at a common point (say, \mathscr H) within spherical triangle \mathscr A$$ \mathscr B$$\mathscr C. However, the three sub-triangles \mathscr A$$ \mathscr B$$\mathscr H, \mathscr B$$ \mathscr C$$\mathscr H and \mathscr C$$ \mathscr A$$\mathscr H do not generally exhibit equal area. Gauss Bonnet theorem for polygons on sphere (radius a) surface has ( K = 1/a^2 )$$ KA = \Sigma \alpha_i - \pi $$for a spherical triangle and, similarly for a smaller triangle of half such area$$ KA/2 = \Sigma \beta_i - \pi $$taking difference$$ KA/2 = \Sigma(\alpha_i - \beta_i)$\$

So at each vertex this much extra rotation at external angle of produced geodesic side of spherical polygon defines a slimmer triangle. There are a number of ways this can be implemented, there is no unique way. The procedure is general, this way we can bifurcate an icosahedron area also, for example.