ABCD are the vertices of a square with A opposite C and side AB = s. The distances of a point P in space from A, B, C, D are a, b, c, d respectively. Show that a^{2} + c^{2} = b^{2} + d^{2}, and that the perpendicular distance k of P from the plane ABCD is given by 8k^{2} = 2(a^{2} + b^{2} + c^{2} + d^{2}) - 4s^{2} - (a^{4} + b^{4} + c^{4} + d^{4} - 2a^{2}c^{2} - 2b^{2}d^{2})/s^{2}.

**Solution**

Let Q be the point of the plane ABCD closest to P. Let O be the center of the square ABCD. Let QO make an angle θ with AC. Then using the cosine rule we have: AQ^{2} = AO^{2} + OQ^{2} - 2AO·OQ cos θ, CQ^{2} = CO^{2} + OQ^{2} + 2CO.OQ cos θ (*). Adding: AQ^{2} + CQ^{2} = 2AO^{2} + 2QO^{2}. But a^{2} = AQ^{2} + k^{2} etc, so a^{2} + c^{2} = 2k^{2} + s^{2} + 2QO^{2}. Similarly, b^{2} + d^{2} = 2k^{2} + s^{2} + 2QO^{2}. We have established that a^{2} + c^{2} = b^{2} + d^{2}.

We have also established that 8k^{2} = 2(a^{2} + b^{2} + c^{2} + d^{2}) - 4s^{2} - 8QO^{2} (**). The angle φ between QO and BD is π/2 - θ. So cos φ = sin θ. Hence, going back to (*), AQ^{2} - CQ^{2} = -4 AO·OQ cos θ, and BQ^{2} - DQ^{2} = ± 4 AO·OQ sin θ. But AQ^{2} - CQ^{2} = a^{2} - c^{2} etc. So (a^{2} - c^{2})^{2} + (b^{2} - d^{2})^{2} = 16 AO^{2}OQ^{2} = 8s^{2}OQ^{2}. Substituting in (**) gives the required result.

© John Scholes

jscholes@kalva.demon.co.uk

14 Dec 1999