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?

First, agree the sides are 2, 4, 5, 6, in that order (and not one of the two other possible orders) and that it is convex.

Second, agree that by specifying a single interior angle, we fix the shape. So, the problem is now: what is the angle that yields the largest incircle and what is its radius?

Third, agree that if we consider the fixed shape missing one of its sides as an "open triangle", then that open triangle has a unique incircle (a circle tangent to each of the three sides) which is a candidate for being the largest incircle of the fixed quadrilateral. Such candidates are the only candidate circles; a circle tangent to three sides.

Fourth, to continue to be a candidate, that incircle cannot cross the 4th side when the 4th side is put back in place.

Fifth, to continue to be a candidate, that incircle must be the largest of the (up to 4) candidates for that fixed shape as we remove each of the 4 sides one by one.

Finally, as we vary the angle through its full legal range, the largest candidate found is the answer.

The calculations needed are: the quadrilateral shape as a function of the one angle; the incenter when only 3 sides are present, which is the intersection of the two angle bisectors; the radius of the incircle, which is the distance between the tangent point (the intersection of the line of one side with the circumference) and the incenter, and finally, disqualification of incircles crossing the fourth side when the distance between the center and that line is less than the radius.

All of these calculations seen doable, just quite a bit of work. As the the angle slowly stepped in a computer code, a maximum radius is found.

(I believe the connected collection of incenters for one fixed quadrilateral is called its medial strips or straight skeleton)

*Edited on ***September 19, 2021, 7:02 am**