A Solution for the Mathematically Inclined

The answer to my post last week is yes. Here’s the quick (intuitive) proof:

First, note that math_a and math_b is the surface of two spheres in three dimensions, and their intersection is either a circle or a point (we know they are not the same sphere since a and b are distinct and that they must intersect because we know p exists).  If it is a point, then p is that point (and since we have not used gamma we have proved the hypothesis).

If the intersection is a circle, it intersects the plane containing a, b, and c at two points, but only one is inside the triangular prism, and that must be the point p. We know it cannot be a point not on the plane, since if it was, two points would satisfy math_c, and not satisfy the uniqueness condition.

I personally prefer algebraic proofs though, so let’s crank some algebra!

First, without loss of generality, assume:
proof_1
Where:
proof_2
So, the equations become:
proof_3

Then:
proof_4
And
proof_9
And
proof_10

It is clear that the triangular prism does not provide any constraints on p3, so the uniqueness condition enforces that p3=0.  Thus, resolving for p1 provides:
proof_11
Since this point must lie within the triangle a b c, p1 must have the same sign as c, so:
proof_12
Therefore:
proof_13

Category(s): Pseudorandom
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