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Square in a Sector (Posted on 2007-06-06) Difficulty: 3 of 5
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.

See The Solution Submitted by Bractals    
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Solution | Comment 3 of 5 |

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
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