Problem 2

We consider a fixed point

P

in the interior of a fixed sphere. We construct three segments

PA, PB, PC

, perpendicular two by two, with the vertices

A, B, C

on the sphere. We consider the vertex

Q

which is opposite to

P

in the parallelepiped (with right angles) with

PA, PB, PC

as edges. Find the locus of the point

Q

when

A, B, C

take all the positions compatible with our problem.

Step 5 of 7: Substitute back into the Pythagorean identity

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In plain words

Both sides now describe the same squared length two different ways, so they can be set equal and simplified.

|Q|^2-2Q\cdot P+d^2 = 3R^2 - 2(Q\cdot P+2d^2)+3d^2

Detailed analysis

The left side is $|\vec{PQ}|^2=|Q-P|^2$ ; the right side is the sum $3R^2-2(A+B+C)\cdot P+3d^2$ from the previous step with $A+B+C=Q+2P$ substituted in, giving one equation purely in $Q\cdot P$ , $|Q|^2$ , $R$ , $d$ .