Let P and Q be points on a unit circle with center O such that angle POQ is x degrees ( x ≤ 270 ).
Let the bisector of angle POQ intersect the circle at point A.
Let KLMN be a square with vertices K and L on line segments OP and OQ and vertices M and N on the arc PAQ.
Give the area of the square in terms of x.
If x = 45, write the area of the square as (a+b√c)/d where a, b, c, and d are integers.
Put the center of the circle at the origin with A at (1,0). The square is now symmetric to the x-axis so we only need consider the first quadrant corners.
They are (sqrt(1-y^2) , y) and (y/tan(x/2) , y) and the area is 4y^2 [x is the angle as stated in the problem, not an x coordinate]
The larger x-coordinate is the sum of the smaller and a side of the square so we have the equation:
sqrt(1-y^2) = 2y + y/tan(x/2)
area = 4y^2 = 4 /(1/tan^2(x/2) + 4/tan(x/2) + 5)
=4sin^2(x/2) / (cos^2(x/2) + 4sin(x/2)cos(x/2) + 5sin^2(x/2))
=(2-2cos(x))/(3-2cos(x)+2sin(x)) [many steps omitted]
This is the best I could simplify to.
Substituting x=45 gives (2-sqrt(2))/3
Posted by Jer
on 2007-06-07 12:06:23