Three friends are dividing up a pizza that has not been pre sliced, but only wish to make two straight line cuts, dividing the pie into three even pieces. If the center of the pizza is on the origin of a graph, the radius is six inches, and the pizza is to be divided using only vertical cuts, how far from the origin must the vertical cuts be made?
(In reply to
re: solution by Richard)
I tried two methods both on the upper semicircle
(1) Integation of Sqrt[r^2-x^2] dx
Here I ended up trying to solve
(x Sqrt[r^2 - x^2] + r^2 ArcTan[x/Sqrt[r^2 - x^2]])/2 = (1/6) (pi r^2) /2
which I solved numerically.
(2) Parametization
I set (x, y) as r (Cos[t], Sin[t]) and then calculated the area of the sector (t r^2/2) less the area of the triangle (r^2 Sin[t]Cos[t]/2)
and set it equal to (1/3) (pi r^2) /2. Solving this for t would give me x = r Cos[t].
The equation to be solved looked promising as r (Pi - 3 t + 3 Cos[t] Sin[t]) = 0 or equivalently Sin[2t] = (2 ( 3 t - Pi))/3 but had no luck. So reverting to numerical methods I found t = 1.30..which gives the same approximate value for x.
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Posted by goFish
on 2005-12-30 16:33:13 |
re(2): solution
