Circle ω is inscribed in unit square ABCD, and points I and E lie on ω such that C, I, and E are collinear. Find, with proof, the greatest possible area for triangle AIE.
I couldn't solve it, but I can get you started on this problem
After thinking of this for an eternity, I only came up with that the largest triangle you can make into a circle is an inscribed equilateral triangle with points XYZ.
If you have that inscribed equilateral triangle XYZ that is inscribed in a circle
to fit into the restrictions of the problem.
you will have to deform the equilateral triangle to do so.
You have to have vertice X as close to A (vertice of square) as possible this makes A your new point on of the triangle instead of X. Then you have to make sure vertices Y and Z are collinear with C (vertice of square)
Y and Z have to be as far as apart as possible while it has to be as close to edge CD of square as possible while remaining in circle w.
That was with my imagination, I am not sure if this is entirely correct.
I have not answered the question that you have asked. (I don't know how)
Like CalculatorUser I have just played with it.
Obviously the answer has to be between 0 and 0.5 but I know you want an exact answer with a prrof.
I graphed it using Geogebra. On my computer the line CE can move to give different points of contact for E and I
To the accuracy Geogebra can give the maximum area is 0.25.