### 6th Putnam 1946

Problem A3

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 a2 + c2 = b2 + d2, and that the perpendicular distance k of P from the plane ABCD is given by 8k2 = 2(a2 + b2 + c2 + d2) - 4s2 - (a4 + b4 + c4 + d4 - 2a2c2 - 2b2d2)/s2.

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: AQ2 = AO2 + OQ2 - 2AO·OQ cos θ, CQ2 = CO2 + OQ2 + 2CO.OQ cos θ (*). Adding: AQ2 + CQ2 = 2AO2 + 2QO2. But a2 = AQ2 + k2 etc, so a2 + c2 = 2k2 + s2 + 2QO2. Similarly, b2 + d2 = 2k2 + s2 + 2QO2. We have established that a2 + c2 = b2 + d2.

We have also established that 8k2 = 2(a2 + b2 + c2 + d2) - 4s2 - 8QO2 (**). The angle φ between QO and BD is π/2 - θ. So cos φ = sin θ. Hence, going back to (*), AQ2 - CQ2 = -4 AO·OQ cos θ, and BQ2 - DQ2 = ± 4 AO·OQ sin θ. But AQ2 - CQ2 = a2 - c2 etc. So (a2 - c2)2 + (b2 - d2)2 = 16 AO2OQ2 = 8s2OQ2. Substituting in (**) gives the required result.