Tag Archives: Colin Hughes

Pairwise Products

This 2005 four-star problem from Colin Hughes at Maths Challenge is also a bit challenging.

For any set of real numbers, R = {x, y, z}, let sum of pairwise products,
________________S = xy + xz + yz.
Given that x + y + z = 1, prove that S ≤ 1/3.”

Again, I took a different approach from Maths Challenge, whose solution began with an unexplained premise.

See the Pairwise Products

Circular Rendezvous Mystery

Here is yet another surprising result from Colin Hughes at Maths Challenge.

It can be shown that a unique circle passes through three given points. In triangle ABC three points A’, B’, and C’ lie on the edges opposite A, B, and C respectively. Given that the circle AB’C’ intersects circle BA’C’ inside the triangle at point P, prove that circle CA’B’ will be concurrent with P.”

I have to admit it took me a while to arrive at the final version of my proof. My original approach had some complicated expressions using various angles, and then I realized I had not used one of my assumptions. Once I did, all the complications faded away and the result became clear.

See Circular Rendezvous Mystery.

Consecutive Product Square

This problem from Colin Hughes at Maths Challenge is a most surprising result that takes a bit of tinkering to solve.

We can see that 3 x 4 x 5 x 6 = 360 = 19² – 1. Prove that the product of four consecutive integers is always one less than a perfect square.”

The result is so mysterious at first that you begin to understand why the ancient Pythagoreans had a mystical relationship with mathematics.

See the Consecutive Product Square.

(Update 11/12/2020) Generalization and Visual Proof
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Maximum Product

This 2007 four-star problem from Colin Hughes at Maths Challenge is definitely a bit challenging.

For any positive integer, k, let Sk = {x1, x2, … , xn} be the set of [non-negative] real numbers for which x1 + x2 + … + xn = k and P = x1 x2 … xn is maximised. For example, when k = 10, the set {2, 3, 5} would give P = 30 and the set {2.2, 2.4, 2.5, 2.9} would give P = 38.25. In fact, S10 = {2.5, 2.5, 2.5, 2.5}, for which P = 39.0625.

Prove that P is maximised when all the elements of S are equal in value and rational.”

I took a different approach from Maths Challenge, but for me, it did not rely on remembering a somewhat obscure formula. (I don’t remember formulas well at my age—only procedures, processes, or proofs, which is ironic, since at a younger age it was just the opposite.) It is also clear from the Maths Challenge solution that the numbers were assumed to be non-negative.

See Maximum Product.