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.
(In reply to
No formula but: by Charlie)
From the table it looks like r for the 45degree 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 (11/sqrt(2))/2.
Thus A = 4 r^2 (sin(45/2))^2 = (44/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 (42*sqrt(2))/6 = (2sqrt(2))/3, so that a=2, b=1, c=2 and d=6.

Posted by Charlie
on 20070607 10:15:38 