Triangle ABC has a point D on side BC such that BA=AD=DC=1. What is angle ABD when the area of the triangle is maximized, and what is the maximum area?

If we call angle DAC=x, then ACD=x, ABD=2x, and BAC=2π-3x. If we set A at (0,0), then C is at (2.cos(x),0), D at (cos(x), sin(x)) and B at (cos(2π-3x), sin(2π-3x))= (-cos(3x), sin(3x)).

The area of the triangle is ½*2cos(x)*sin(3x)= cos(x)*(3sin(x)-4sin(x)&exp3;). To seek the maximum, we find the derivative and equate to 0; after some algebra, and writing s for sin(x)² we find 16s²-18s+3=0.

Working backwards, we get x=arcsin(√((9+√33)/16))= 26.8124 degrees, so ABD is twice that; 53.6248 degrees.