Triangle circle maximization problem

The Endeavour 2024-10-16

Let a, b, and c be the sides of a triangle. Let r be the radius of an inscribed circle and R the radius of a circumscribed circle. Finally, let p be the perimeter. Then the previous post said that

2prR = abc.

We could rewrite this as

2rR = abc / (a + b + c)

The right hand side is maximized when a = b = c. To prove this, maximize abc subject to the constraint a + b + c = p using Lagrange multipliers. This says

[bc, ac, ab] = λ[1, 1, 1]

and so ab = bc = ac, and from there we conclude a = b = c. This means among triangles with any given perimeter, the product of the inner and outer radii is maximized for an equilateral triangle.

The inner radius for an equilateral triangle is (√3 / 6)a and the outer radius is a/√3, so the maximum product is a²/6.

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