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?
I've been meaning to give one of these a try, even though it's probably not my strong suit...so here goes...
Using general triangle formulae: Area = .5*h*b
In this case, b = BC = BD+1, and h is measured from A to BC.
Because BA=AD=1, BD = 2cos(ABD) and h = sin(ABD).
Therefore Area = .5 * sin(ABD) * (2cos(ABD) + 1)
This can be simplified through product formulas to
Area = .5sin(2ABD) + .5sin(ABD)
which goes is maximized when
0 = cos(2ABD) + .5cos(ABD)
Therefore ABD is about 53.624 degrees and the Area is about .880
There, now I can read Bractal's solution and see if I got it right.
My attempt
