0
$\begingroup$

I have 3 points on a unit sphere identified by their XYZ coordinates. They form a spherical triangle. If I'm not mistaken, perpendicular bisectors of a spherical triangle intersect in a single point, just like on a plane. What is the easiest way to calculate its coordinates?

Is this point collinear with the circumcenter of the planar triangle and the center of the sphere?

$\endgroup$
  • $\begingroup$ Not in one point but in two points. $\endgroup$ – Moishe Kohan Jun 15 '14 at 16:04
  • $\begingroup$ Well, yes, but one of the points is on the other side of the sphere, and I'm not interested in it. $\endgroup$ – Alexey Jun 16 '14 at 6:54
0
$\begingroup$

After some thinking, I realized that of course the intersections of perpendicular bisectors of a spherical triangle are collinear with the circumcenter of a planar triangle with the same vertices and the center of the sphere.

Let's examine vertices A and B of a triangle, with D the midpoint of the planar edge and E its spherical counterpart. If O is the center of the sphere, then OAB is an isosceles triangle, and O,D,E are collinear. Therefore both planar and spherical perpendicular bisectors line in a single plane perpendicular to OAB that passes through OE.

If we repeat this with vertices B and C we will see that both planar and spherical perpendicular bisector intersections must lie on the same line, since two planes cross along a line. And since both planes contain O, they are all collinear.

This means that my approach of finding the circumcenter of the planar triangle and normalizing the vector to obtain the point I need is perfectly valid.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.