#### Problem

Three points, A, B, and C, are chosen at random such that OABC forms a quadrilateral. The midpoints of each edge, P, Q, R, and S, are joined.

Prove that the quadrilateral PQRS will always be a parallelogram.

#### Solution

We shall prove this result by consideration of vectors.

Let OA = 2a, OB = 2b, and OC = 2c.

So AB = AO + OB = -2a + 2b and CB = CO + OB = -2c + 2b.

The position vectors of P, Q, R, and S respectively will be:

OP = (1/2)OA = a

OS = (1/2)OC = c

OQ = OA + (1/2)AB = 2a + -a + b = a + b

OR = OC + (1/2)CB = 2c + -c + b = b + c

Using these we can obtain vectors for each side of the quadrilateral PQRS:

PQ = PO + OQ = -a + a + b = b

SR = SO + OR = -c + b + c = b

PS = PO + OS = -a + c

QR = QO + OR = -(a + b) + b + c = -a + c

As PQ is parallel with and equal to SR, similarly with PS and QR, we prove that the quadrilateral PQRS is a parallelogram.

Was it necessary to consider PS and QR to prove the result?

Problem ID: 250 (02 Dec 2005)     Difficulty: 3 Star

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