# Minimum Path Via Circle

James Tanton provides another imaginative problem on Twitter.

“I am at point A and want to walk to point B via some point, any point, P on the circle. What point P should I choose so that my journey A → P → B is as short as possible?”

Hint: I got ideas for a solution from two of my posts, “Square Root Minimum” and “Maximum Product”.

# Barrier Minimal Path Problem

This is a nifty little problem from the Quantum math magazine.

“Two ants stand at opposite corners of a 1-meter square. A barrier was placed between them in the form of half a 1-meter square attached along the diagonal of the first square, as shown in the picture. One ant wants to walk to the other. How long is the shortest path?”

See the Barrier Minimal Path Problem for solutions.

# Pairwise Products

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

Problem
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

# Maximum Product

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

Problem
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.