A triangle puzzle

This morning I was continuing my meander through the old “Carnival of Maths” posts and in issue 17 I came across this post from the blog “Let’s play math“. I’m not fully through the post yet, bit I want to discuss a little puzzle I found there.

Unlike some of the puzzles I’ve posted recently, this one only took a few moments for me to see the answer, but I wanted to share my, albeit brief, thought process:

The puzzle:

“If you keep two sides of a triangle the same but let the other side change, you could make an infinite number of triangles. Of all those possible triangles, which one has the largest area?”

My thought process went: There are no side lengths, does it matter? No, Area = Absin(c). The sine curve popped into my head and hence, the angle must be pi/2.

I loved the simplicity of the puzzle and want to try it on my year 11s,12s and 13s. As it is school holidays, I tried it on my social media networks instead. Will Davies (@notonlyahatrack) came back with this: “The one where the angle between them is a right angle as sin 90 is,a maximum”. I think this shows a similar train of thought to me. (Do correct me if I’m wrong Will!)

My brother (@andy_cav25) is a primary teacher, and as such probably doesn’t have trigonometry on the mind as often as us high school maths teachers. He very quickly came up with the correct answer, but using a different method. He thought of the typical definition of base multiplied by perpendicular height, imagined a computer modelling this visualisation and could see that the right angle would give the largest height. I can see this too. Imagine one side as the fixed base, the other free to move, but fixed at one end to the base. The locus of the free and would be a circle, the largest perpendicular height is clearly when the radius is perpendicular, and as such when the triangle is right angled.

Two lovely, simple, solutions to a lovely simple puzzle.