Solution:

This proof uses some neat algebra, and the result that

x² + y² > 2xy with x² + y² = 2xy exactly when x = y.

(You can prove this result by starting with (x – y)² >= 0 and expanding the brackets before rearranging.)

Back to challenge 268…

|OP|² + |OQ|² + |OR|² + |OS|² =

½(|OP|² + |OQ|²) + ½ (|OQ|² + |OR|²) + ½ (|OR|² + |OS|²) + ½ (|OS|² + |OP|²) >=

|OP||OQ| + |OQ||OR| + |OR||OS| + |OS||OP|

Now, ½ |OP||OQ| > ½ |OP||OQ|sin(POQ) = area of triangle OPQ, with ½ |OP||OQ| = ½ |OP||OQ|sin(POQ) = area of triangle OPQ when POQ is a right angle.

Therefore

|OP||OQ| + |OQ||OR| + |OR||OS| + |OS||OP| >=

2 x area OPQ + 2 x area OQR + 2 x area ORS + 2 x area OSR =

2 (area OPQ + area OQR + area ORS + area OSR) = 2 x area PQRS = 2A

So if we know that |OP|² + |OQ|² + |OR|² + |OS|² = 2A, then all the conditions for equality must hold:

|OP| = |OQ| = |OR| = |OS|

and angles POQ, QOR, ROS, SOP must all be right angles.

Therefore, the diagonals PR and QS of the quadrilateral PQRS must be perpendicular and of equal length.

Hence PQRS must be a square.

Challenge 268: A Traditional Task