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.
If a smaller circle is drawn through P, centered on O as well, and we call its radius r, then, taking one half of the overall angle x, the half height of the square is r*sin(x/2). Angle MOA (or angle NOA) is arcsin(r*sin(x/2)). By looking at the geometry further, relying on the full width of the square being twice the half height:
cos arcsin(r*sin(x/2))  r*cos(x/2) = 2*r*sin(x/2)
The area is then 4*(r^2)*(sin(x/2))^2
So the first step is to solve the given equation for r, and then substitute x and r into the latter formula.
The values come out:
x Area r (the intermediate needed)
15 0.042974480 0.79410449
30 0.118146030 0.66402333
45 0.195262146 0.57735027
60 0.267949192 0.51763809
75 0.335815630 0.47596315
90 0.400000000 0.44721360
105 0.461995198 0.42837299
120 0.523372891 0.41768125
135 0.585786438 0.41421356
150 0.651084740 0.41768125
165 0.721508222 0.42837299
180 0.800000000 0.44721360
195 0.890725289 0.47596315
210 1.000000000 0.51763809
225 1.138071187 0.57735027
240 1.322780956 0.66402333
255 1.587627458 0.79410449
270 2.000000000 1.00000000
These area results agree with experimentation with Geometer's Sketchpad, and the last one is indeed just the area of the inscribed square.
DEFDBL AZ
pi = ATN(1) * 4
FOR x = 15 TO 270 STEP 15
sxo2 = SIN(x * pi / 180 / 2)
cxo2 = COS(x * pi / 180 / 2)
r = .5
DO
smang = ATN(r * sxo2 / SQR(1  r * r * sxo2 * sxo2))
rhs = COS(smang)
pr = r
r = (rhs / (2 * sxo2 + cxo2) + r) / 2
LOOP UNTIL ABS((pr  r) / r) < .0000000000001#
a = 4 * r * r * sxo2 * sxo2
PRINT USING "#### #.######### #.########"; x; a; r
NEXT
Edited on June 7, 2007, 9:51 am

Posted by Charlie
on 20070606 16:55:57 