Triangle circle maximization problem
by John from John D. Cook on (#6RH86)
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] = [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 a^2/6.
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