Let the three points A,B,C be the vertices of a moving spherical triangle on the surface of a sphere. The triangle moves so that while the vertices A,B remain fixed, the angle BCA at the vertex C stays constant. What is the locus of the moving vertex C? Is there a special name for the curve traced out by C? If A,B,C were the vertices of a plane triangle, the corresponding locus would be the arc of a circle. I have made some calculations, which-if they do not contain errors-lead me to believe that, in the spherical case, the locus I am seeking is not the arc of a small circle (on the sphere). But, if so, I do not know what type of curve it is.
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asked Jul 23, 2015 at 18:46
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$\begingroup$ It is quite obvious that "the angle BCA at the vertex C" means "the angle on the spherical surface", since otherwise the problem is trivial, but maybe it is the case to state it clearly. Nice question. $\endgroup$– Jack D'AurizioCommented Jul 23, 2015 at 23:46
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1 Answer
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Let $O$ be the centre of the sphere. Then the spherical $BCA$ angle $\gamma$ is the angle between the planes through $O,A,C$ and $O,B,C$. So we can find our locus in the following way: we take a plane $\pi$ through $O,A$ and find a plane $\tau$ through $O,B$ such that the angle between $\pi$ and $\tau$ equals $\gamma$. By intersecting $\pi\cap\tau$ with the sphere we find every point of our locus. As an alternative, we may exploit the spherical cosine rule: $$ \cos C = \frac{\cos c-\cos a\cos b}{\sin a\sin b}\tag{1}$$ where $c=\widehat{AOB}$ and so on. Notice that both $c$ and $C$ are fixed, so everything boils down to a not-so-trivial constraint between $a$ and $b$.
answered Jul 24, 2015 at 0:36
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$\begingroup$ Thanks alot for your answer. I wonder if there exists a fairly simple formula in spherical polar coordinates which will yield a picture of what this locus actually looks like. $\endgroup$ Commented Jul 24, 2015 at 18:36