May 23rd, 2020 at 8:08:05 AM
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Assume the radius is 1

Let M be the midpoint of AC

AOM and COM are side-side-side congruent, so angle AOM = angle COM;

AOM + COM = AM = 180, so AOM and COM are both right angles

Let x be the measure of angle AOM (and COM) in degrees (I couldn't figure out how to enter lowercase theta in Photoshop Elements)

AM = sin x; OM = cos x

The area of the "wedge" of the circle bounded by AO and CO is PI * 2x / 360 = PI x / 180

The area of triangle AOC is 1/2 (2 sin x) (cos x) = 1/2 sin 2x

The area of the region bounded by the circle and AC is PI x / 180 - 1/2 sin 2x

Since this is a piece of the cake, it must be 1/4 of the total area, or PI / 4:

PI x / 180 - 1/2 sin 2x = PI / 4

PI (x - 45) / 90 = sin 2x

Using Excel approximation, I get x = 66.17322941 degrees

Let h be the height of triangle DAE measured from A

tan x = h / (DE / 2)

h = (DE tan x) / 2

The area of DAE = 1/2 DE h = 1/4 (DE)^2 tan x

This is also a piece of the cake, so its area is also PI / 4:

1/4 (DE)^2 tan x = PI / 4

(DE)^2 = PI cot x

DE = about 1.177863

Since the radius is 1, this is the ratio of the bar of the "A" cut to the radius

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