Every triangle has a square which is the maximum size square which can be inscribed in the triangle. For most triangles there is only one way to do so. For an equilateral triangle there are three ways to inscribe the square - one for each side.

The equilateral is not the only triangle with that property; there is one other triangle whose maximum inscribed square can be placed on all three sides. Determine the dimensions of that triangle! (Assume the shortest edge is 1 unit.)

Before realizing, as broll did, that the triangle must be obtuse I derived the following formula for the length of the square along side a of triangle a,b,c:
Sa(a,b,c)=2aA/(a²+2A)
Where A is the triangle area, easily found via Heron's Formula.

The above formula, it turns out, only works if the side is bounded by two acute angles.  So it's still good for the long side of an obtuse triangle.

When I realized the triangle must be obtuse I also realized it must be isosceles.  So I decided a=b=1.
This leads to a new formula for the square on a side bounded by an obtuse angle:
S'a(1,1,c)=2A/(1+2A+√(1-4A²))

As quick graph of Sc(1,1,c)=S'a(1,1,c) yields the solution c=1.5513875 with S=.44861248

So the dimensions of this triangle are 1,1,1.5513875 and for good measure the angles are about 101.736º and 39.132º.

I may try to find a closed form solution...

The solution is one of the roots of a quartic polynomial one of whose roots is 2.  This reduces to the cubic 2x^3-2x^2-3x+2 but Wolfram just gives a closed form solution involving imaginary numbers.  It does give extra digits though:
1.5513875245483203922619525102646238

Edited on February 27, 2016, 3:39 pm

Posted by Jer on 2016-02-27 12:46:15