An equilateral triangle PQR is drawn inside a unit square LQST such that Q corresponds to the common vertex of the square and the triangle.

Determine the maximum possible area of the triangle PQR.

Let the coordinates of the square be

  L(1,0), Q(0,0), S(0,1), and T(1,1).

For maximum area the coordinates of
the triangle are

  P(1,x), Q(0,0), and R(x,1).

Let s be the side length of the triangle

  s^2 = 1^2 + x^2 = (x-1)^2 + (1-x)^2

      or

  x^2 - 4x + 1 = 0

      or

  x = 2 - sqrt(3)

The area of the triangle is

  Area = (1/2)*s^2*sin(60)

       = (1/2)*(4x)*(sqrt(3)/2)

       = 2*sqrt(3) - 3

      ~= 0.4641016  

Posted by Bractals on 2007-09-30 19:00:52