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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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re: No formula but: | Comment 2 of 5 |
(In reply to No formula but: by Charlie)

From the table it looks like r for the 45-degree case is 1/sqrt(3). The sine and cosine of 45 degrees is 1/sqrt(2), making the square of the sine of half that angle equal to (1-1/sqrt(2))/2.

Thus A = 4 r^2 (sin(45/2))^2 = (4-4/sqrt(2))/6.  If not for the first division symbol not being multiplication, it would fit the format requested.  To get into the format desired, that's (4 - 4*sqrt(2)/2)/6, or (4-2*sqrt(2))/6 = (2-sqrt(2))/3, so that a=2, b=-1, c=2 and d=6.

  Posted by Charlie on 2007-06-07 10:15:38

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