A certain triangle ABC has D on AB and E on AC such that AD=DB=BC=CE and AE=ED. What is the measure of angle BAC?
(In reply to
solution by Charlie)
Going to http://www.1728.com/cubic2.htm to see how to solve cubic equations, I found the value for x (one side of the small isosceles triangle ADE), to be
((37+9*sqrt(17))^(1/3) + (379*sqrt(17))^(1/3)  2) / 3
The angle is still arccos(1/(2*x)) or
arcsec(2*((37+9*sqrt(17))^(1/3) + (379*sqrt(17))^(1/3)  2) / 3)

Posted by Charlie
on 20051024 15:18:40 