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 take AB and AD as unit vectors and x=BAD, then the area ABD = (1/2) Sin(x).

ABD is isosceles so angle ADC=PI-(PI-x)/2=y and the area ADC= (1/2) Sin(y)= -(1/2)Cos(x/2)

Differentiating we get a maximum when

(1/2)Cos(x)-(1/4)Sin(x/2)=0.

The required angle is of course given by (PI-x)/2

Posted by goFish on 2005-09-19 21:47:10