Given a quadrilateral with side lengths 2, 4, 5, and 6 units, what is the radius of the largest circle which can be drawn completely inside of it?

If we consider the isosceles triangle with sides (2+4), 5, 6, something approaching this configuration might be the optimal shape of the quadrilateral. The question is: would any outward buckling of the 2+4 side, to make the figure a true quadrilateral, also make for a smaller incircle? I suspect so!

Thus, the solution would be an infinitesimal departure from this triangle, and the solution would asymptotically approach the triangle solution.

In this case, with the base=5, and the two base angles=theta,

cos(theta) = 2.5/6

The incenter is at the intersection of the base angle bisector and the altitude.

The radius of the incircle is

(5/2) tan(theta/2) = (5/2) sqrt [ (1-cos(theta)) / (1+cos(theta)) ]

= (5/2) sqrt(7/17) = 1.604

*Edited on ***September 19, 2021, 2:54 pm**