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?
Lets explore a bit and reduce two of the sides: 4 and 5 get cut down to 3.5 and 4.5. Now put the sides in the cyclic order 2, 3.5, 6, 4.5. Note that opposite pairs of sides are equal 2+6 = 8 = 3.5+4.5.
Then we'd have a side order consistent with a tangential quadrilateral. Tangential quadrilaterals have a proper incircle tangent to all four sides.
The largest inradius for a tangential quadrilateral occurs when the quadrilateral is also a cyclic quadrilateral. A cyclic quadrilateral has a proper circumradius containing all four vertices and has the largest area of any quadrilateral with a given set of side lengths.
The combination of tangential and cyclic makes it into a bicentric quadrilateral. So lets have our cut-down quadrilateral be a bicentric quadrilateral.
There are two area formulas of interest:
We have Area = inradius*semiperimeter from being tangential
And also Area = sqrt(product of sides) from being bicentric.
Then inradius*semiperimeter = sqrt(product of sides)
R*(2+6) = sqrt(2*6*3.5*4.5)
R = 3*sqrt(21)/8 = 1.718
This is already larger than Steven's "shortcut approach". And at the same time is from a quadrilateral that had two of its sides cut down and so is actually smaller than the 2,4,5,6 quadrilateral in the problem. So the max radius actually sought is extremely likely to exceed R=1.718
My proposition for solving the actual quadrilateral: arrange the sides in the order 2,4,6,5 and set them to be a cyclic quadrilateral and see how large of a circle can be placed inside.
Bicentric exploration
