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Four points lie on the surface of a sphere. Given the six distances between the points, calculate the radius of the sphere.
This is (allegedly) an advanced high school level problem. However, it is a remembered problem, so it is possibly misremembered (i.e. there might have been some “bice numbers” so that the general solution is unnecessary). But the solution probably does not involve the full machinery of differential and tensor geometry, and it is possible, but unlikely, that it requires knowllege of the Haversine Rule.
First, is there a solution? If the corners are ABCD, the trianghles ABC, ABD, ACD, and BCD are determined. That give syou the Spgerical Excess, and that in turn gives the radius. Alternatively, one has size numbers. The coordinates of the four points would determine R – one point can be placed at the origin, one along one axis, so we have 5 degres of freesom on where the points go, plus the radius, which makes 6, and we have 6 constraints – 6 equations in 6 unknowns.
It seems to be that the key to this must be a relationship between the two diagonals. Until the second diagonal is drawn, everything could be in a plane. But I have not been able to go from there to a high-school level solution.
This is (allegedly) an advanced high school level problem. However, it is a remembered problem, so it is possibly misremembered (i.e. there might have been some “bice numbers” so that the general solution is unnecessary). But the solution probably does not involve the full machinery of differential and tensor geometry, and it is possible, but unlikely, that it requires knowllege of the Haversine Rule.
First, is there a solution? If the corners are ABCD, the trianghles ABC, ABD, ACD, and BCD are determined. That give syou the Spgerical Excess, and that in turn gives the radius. Alternatively, one has size numbers. The coordinates of the four points would determine R – one point can be placed at the origin, one along one axis, so we have 5 degres of freesom on where the points go, plus the radius, which makes 6, and we have 6 constraints – 6 equations in 6 unknowns.
It seems to be that the key to this must be a relationship between the two diagonals. Until the second diagonal is drawn, everything could be in a plane. But I have not been able to go from there to a high-school level solution.