ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 92
  • Posts: 4020
May 23rd, 2020 at 8:08:05 AM permalink


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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