Pie with square
07-03-2016, 01:46 AM
Post: #1
 Pekis Member Posts: 124 Joined: Aug 2014
Pie with square
Hello,

An old aunt invited 8 of her nephews to come and eat a fruit tart with her. Previously professor of mathematics, she decided to cut the tart in 9 equal but unusual parts: If you were one of her nephews, could you check if the parts were really equal, computing the values of d (yellow strip's width) and e (orange square's width), in % of the pie radius, before eating it ?

Have fun 07-03-2016, 08:06 AM (This post was last modified: 07-04-2016 04:32 AM by Paul Dale.)
Post: #2 Paul Dale Senior Member Posts: 1,766 Joined: Dec 2013
RE: Pie with square
The diagram is labelled badly Anyway, e is easy:

$$e^2 = \frac {\pi r^2} 9$$
$$\therefore e = \frac { r \sqrt \pi } 3$$

The conversion to a percentage is left up to the reader.

I started working on d using $$2 \int_0 ^{\frac d 2} \sqrt {r^2 - x^2} dx$$ but got lost in the LaTex expression morass far too quickly.

An afterthought: it might be better working on the red bits for d: $$\pi (r - \frac d 2)^2 - (e-d)^2 = \frac {4\pi r^2} 9$$ which is a messy quadratic in d.

- Pauli
07-04-2016, 03:34 AM
Post: #3
 klesl Member Posts: 106 Joined: Mar 2016
RE: Pie with square
If I was a one of her nephews I would cut and weight 3 different pieces- orange, red and yellow. This the same method how to find area of peaks in GC.
07-04-2016, 04:02 AM
Post: #4 Paul Dale Senior Member Posts: 1,766 Joined: Dec 2013
RE: Pie with square
If we're looking for interesting solutions, a quickly constructed hatchet planimeter would be able to measure the comparative areas of the pieces and if they are equal, the are the same volume. Assuming a cylindrical tart (which is kind of implied).

- Pauli
07-04-2016, 04:16 AM (This post was last modified: 07-04-2016 03:42 PM by Pekis.)
Post: #5
 Pekis Member Posts: 124 Joined: Aug 2014
RE: Pie with square
Well, for curious people, here is the solution:

1) I will consider that the circle has a radius equal to 1, in order to express the d and e values directly in % of the radius of a circle of any size
2) The surface of the circle is then PI*1^2=PI, which implies that each part must have a surface equal to PI/9
3) All the parts with the same color have the same form
4) I will work with e'=e/2 and d'=d/2 to simplify calculations

Surface of the orange square:
e'^2*4, which must be equal to PI/9
=> e'=SQRT(PI)/6=0.295409 approx.
=> e=2*e'=SQRT(PI)/3
e=0.590818 approx.

Surface of the top yellow strip:
2*(integ(SQRT(1-x^2),x,0,d')-d'*e'), which must be equal to PI/9

(with integ(SQRT(1-x^2),x,0,d') to integrate the right part, -d'*e' to discard it's orange part, and *2 to add the left part)

=> d'=0.251508 approx. (using a solver)
=> d=2*d'
d=0.503016 approx.

Since 5 parts (4 yellow and 1 orange) of the surface of the pie are now known as 5*(PI/9), the 4 remaining red parts which have the same form share 4*(PI/9) => each has it's surface equal to PI/9:

Check Surface of the left top red part:
integ(SQRT(1-x^2),x,d',1)-(PI/9)/2-e'*(e'-d'), which is naturally equal to PI/9 !

(with integ(SQRT(1-x^2),x,d',1) to integrate starting from the right, -(PI/9)/2 to discard the half of the left yellow strip, and -e'*(e'-d') to discard it's small orange part)

Thanks for reading 07-14-2016, 03:00 PM
Post: #6 SlideRule Senior Member Posts: 1,325 Joined: Dec 2013
RE: Pie with square
Sorry for the delay but I finally got around to my confirmation of the solution: thought others might enjoy.
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