Division of Spherical triangle into two Equal area spherical triangles
 Consider the spherical triangle illustration below:
 In the spherical triangle formed by the vertices ABC, the sides BC, CA and AB are arcs of the great circle,
whose angular lengths are A, B and C respectively.
 The angle subtended between the arcs at A, B and C vertices are a, b and c respectively.
 It is now desired to find the point D on BC such that, the (solid angle) area of the triangle ABD and ACD
are equal.
 If S is the area of the spherical triangle ABC then, S=a+b+c-
 The equation for area of spherical triangle ABD is: x+y+b-=S/2=(a+b+c-
a  c b 
a  c b 
 y
x
2
2
 a  c b  
 a  c b  
(1)
 cos y  cos 
 cos x  sin 
 sin x
2
2




cos y  cos b cos x
 cos y  cos C sin b sin x  cos b cos x (2)
 Again in triangle ABD , cos C 
sin b sin x
 From Eq. (1) and Eq. (2), we get,
 a  c b  
 a  c b  
cos 
 cos x  sin 
 sin x  cos C sin b sin x  cos b cos x
2
2




 a  c b 
 x y 
  cos b

 tan x 
 a  c b  
sin bCosC  sin 

2


cos 

2
 The value of x can be got from the above formula, to find the position D that divides the triangle ABC into
two equal area triangles.
 Similarly, it is possible to divide the triangle ABC into two equal area smaller triangles with B or C as the
common vertex point
 Reference: I. Todhunter, Spherical Trignometry, 1886
Division of Spherical triangle into three Equal area spherical triangles
 Consider the spherical triangle illustration below:
 In the spherical triangle formed by the vertices ABC, the sides BC, CA and AB are arcs of the great circle,
whose angular lengths are A, B and C respectively. The angle subtended between the arcs at A, B and C
vertices are a, b and c respectively.
 It is now desired to find the point D within ABC such that the (solid angle) area of the triangle ABD, BCD
and CAD are equal.
 The unknown variables (x, y, z, u, v, w) are as described in the above figure, and it is now sought to solve for
x, y and z values.
 Area of the spherical triangle ABC is S=a+b+c-. Then, area of the 3 smaller spherical triangles are s=S/3.
 Consider the triangle BCD . Its area can be stated by the expression,
s  y  (c  z )  u    u     c  s    y  z 
 cos u  cos    c  s  cos  y  z   sin   c  s  sin  y  z 
 cos u   cos  c  s  cos y cos z  sin y sin z   sin  c  s  sin y cos z  cos y sin z 
 Again in triangle BCD,
cos u  cos A sin y sin  c  z   cos y cos  c  z 
 cos u  cos A sin y  sin c cos z  cos c sin z   cos y  cos c cos z  sin c sin z 
(1)
(2)
 Equating the right hand side terms of Eq. (1) and Eq. (2) and dividing by cos y cos z , we get,
 cos  c  s 1  tan y tan z   sin  c  s  tan y  tan z   cos A tan y  sin c  cos c tan z    cos c  sin c tan z 


 tan y sin  c  s   cos A sin c   cos A cos c  cos  c  s   tan z  cos  c  s   cos c   sin  c  s   sin c  tan z
 tan y 
cos  c  s   cos c   sin  c  s   sin c  tan z
sin  c  s   cos A sin c   cos A cos c  cos  c  s   tan z
or
tan y 
t1  t2 tan z
t3  t4 tan z
where t1  cos  c  s   cos c, t 2  sin  c  s   sin c, t3  sin  c  s   cos A sin c, & t4  cos A cos c  cos  c  s 
(3)
 Eq. (3) has a homographic form of relation between the unknowns tan y and tan z, where the terms t1, t2, t3 &
t4 can be directly calculated from the starting spherical triangle.
 Similar homographic expressions relating tan z with tan x, and tan x with tan y can be found by considering
the triangles CAD and ABD respectively, as
tan z 
t5  t6 tan x
t7  t8 tan x
where t5  cos  a  s   cos a, t6  sin  a  s   sin a,
t7  sin  a  s   cos B sin a, & t8  cos B cos a  cos  a  s 
and
t  t tan y
tan x  9 10
where t9  cos  b  s   cos b, t10  sin  b  s   sin b,
t11  t12 tan y
t11  sin  b  s   cos C sin b, & t12  cos C cos b  cos  b  s 
(4)
(5)
 The three unknowns of tanx, tany and tanz can be solved from the cyclic homographic relations (3)-(5) as
follows:
 Substituting Eq. (4) in Eq. (3) we get,
 t  t tan x 
t1  t2  5 6

t7  t8 tan x  t1t7  t2 t5   t1t8  t2t6  tan x t13  t14 tan x

tan y 


 t5  t6 tan x  t3t7  t4 t5   t3t8  t4 t6  tan x t15  t16 tan x
t3  t 4 
(6)

 t7  t8 tan x 
where t13  t1t7  t2 t5 , t14  t1t8  t 2t6 , t15  t3t7  t4t5 & t16  t3t8  t4 t6
 And, again substituting Eq. (5) into Eq. (6), we get,
 t  t tan x 
t13  t14  9 10

t11  t12 tan x  t13t11  t14 t9   t13t12  t14 t10  tan x t17  t18 tan x

tan y 


 t9  t10 tan x  t15t11  t16 t9   t15t12  t16t10  tan x t19  t20 tan x
t15  t16 

 t11  t12 tan x 
where t17  t13t11  t14t9 , t18  t13t12  t14t10 , t19  t15t11  t16t9 & t20  t15t12  t16t10
 t20 tan 2 y   t19  t18  tan y  t17  0
 The above quadratic equation of tan y can be solved as:
tan y 
  t19  t18  
 t19  t18 
2t20
2
 4t20t17
 For all arbitrary possible spherical triangles (checked for several random instances of spherical triangle ABC
), the determinant term of the quadratic equation (term with square root) turns out to be identically zero. Hence,
tan y is always found to exhibit a unique solution given by,
  t19  t18 
tan y 
2t20
 Eq. (4) and Eq. (5) can then be used to solve for tan z and tan x. Thus, the centre point of D which divides the
spherical triangle ABC into three equal area spherical triangles is determined.