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 6 of 7: Solve for the squared distance from the center to Q

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In plain words

All dependence on the specific choice of A, B, C disappears, revealing a hidden invariant.

|Q|^2 = 3R^2 - 2d^2

Detailed analysis

The terms $-2Q\cdot P$ cancel from both sides of the previous equation, leaving $|Q|^2+d^2 = 3R^2-d^2$ , i.e. $|Q|^2 = 3R^2-2d^2$ — a value depending only on the fixed radius $R$ and the fixed distance $d=|OP|$ , not on the choice of $A,B,C$ .