+129849 We have a pizza [ of some radius - it doesn't really matter ] that we would like to share among three people. It has been decided that two vertical cuts will be made that will divide the pizza into three equal parts......the question is.....where should the cuts be made???
Hint : This problem could be soved with some Calculus, but it's not really necessary...some simple Trig and/or Geometry should also do the trick.....However....I anticipate that you will need a Computer Algebra System [ WolframAlpha will do ] in order to derive the final answer.....Good luck !!!!
+118667 challenge accepted.
I am only looking at the semicircle on the right.
The coloured in brown bit plus the two little sectors must be half the size of the segment on the far right.
\(cos\theta = \frac{h}{r}\\ h=rcos\theta\\ Area\;of\;brown\; triangle=0.5r^2sin(2\theta)\\ Area \;of\;one\;little\;sector=\frac{(0.5\pi-\theta)}{2\pi}\pi r^2=\frac{(0.5\pi-\theta)r^2}{2}\\ \text{add these three areas together and get}\\ \text{half of middle slice is }\\ 0.5r^2sin(2\theta)+2*\frac{(0.5\pi-\theta)r^2}{2}\\ =0.5r^2sin(2\theta)+(0.5\pi-\theta)r^2\\ =r^2[0.5sin(2\theta)+0.5\pi-\theta]\\~\\ \text{area of segment }=0.5\pi r^2-r^2[0.5sin(2\theta)+0.5\pi-\theta]\\ \text{area of segment }=r^2[0.5\pi-0.5sin(2\theta)-0.5\pi+\theta]\\ \text{area of segment }=r^2[\theta-0.5sin(2\theta)]\\~\\ r^2[\theta-0.5sin(2\theta)]=2*r^2[0.5sin(2\theta)+0.5\pi-\theta]\\ \theta-0.5sin(2\theta)=sin(2\theta)+\pi-2\theta\\ 3\theta-1.5sin(2\theta)=\pi\\ \)
https://www.wolframalpha.com/input/?i=3x-1.5*sin(2x)%3Dpi++++++++++radians
\(\theta \approx 1.30266\;radians\)
\(h=rcos\theta\\ h\approx 0.2649r\)
Here is what your Pizza should look like. The measurements are of course approximate.
+129849 Good job, Melody!!!........
I did it a little differently.......I'll show my mehod in a day or two.....!!!!
+129849 Here's my approach.....for simplicity.....I set the radius = 1.......we can adjust for this in the final answer....
Lett DE and BC be perpendicular diameters......now imagine that one of the cuts will be made along chord FH parallel to BC......and connect the center of the circle, at A with F and let point I lie on BC such that IF ll AG
The area bounded by the irregular shape BFGA [ rectangle IFGA and the area between the semi-chord IF and the minor arc BF ] will be (1/4) of (1/3) of the area of the pizza = (1/3)* (1/4) pi * r^2 = pi/ 12
And let angle BAF = θ......and because IF and AG are parallels cut by a transversal, angle FAB = angle GFA
And the area of the sector ABF = (1/2)r^2* θ = (1/2) (1)^2 * θ = θ/2
And the area of right triangle AGF = (1/2) AG * GF = (1/2) * (r)sinθ * (r)cosθ = (1/2)* (1)* sinθ* (1)*cosθ = (1/2)sinθcosθ
And these two area comprise the total area BFGA
So we have that
θ/2 + (1/2)sinθcosθ = pi/12 multiply through by 2
θ + sin θ cos θ = pi / 6
Using Wolframalpha to solve → θ ≈ 0.268133
And the distance from the center to the cut is given by r* sin (0.268133) = r * 0.264932 = 0.264932* r
Just as Melody found !!!!
