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 Square in a Sector (Posted on 2007-06-06)
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 No Rating

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 No formula but: | Comment 1 of 5
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 0.44721360105 0.461995198 0.42837299120 0.523372891 0.41768125135 0.585786438 0.41421356150 0.651084740 0.41768125165 0.721508222 0.42837299180 0.800000000 0.44721360 195 0.890725289 0.47596315210 1.000000000 0.51763809225 1.138071187 0.57735027240 1.322780956 0.66402333255 1.587627458 0.79410449270 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 A-Z
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 2007-06-06 16:55:57

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