Pie with square

07032016, 01:46 AM
Post: #1




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 

07032016, 08:06 AM
(This post was last modified: 07042016 04:32 AM by Paul Dale.)
Post: #2




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: \( \pi (r  \frac d 2)^2  (ed)^2 = \frac {4\pi r^2} 9 \) which is a messy quadratic in d.  Pauli 

07042016, 03:34 AM
Post: #3




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.


07042016, 04:02 AM
Post: #4




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 

07042016, 04:16 AM
(This post was last modified: 07042016 03:42 PM by Pekis.)
Post: #5




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(1x^2),x,0,d')d'*e'), which must be equal to PI/9 (with integ(SQRT(1x^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(1x^2),x,d',1)(PI/9)/2e'*(e'd'), which is naturally equal to PI/9 ! (with integ(SQRT(1x^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 

07142016, 03:00 PM
Post: #6




RE: Pie with square
Sorry for the delay but I finally got around to my confirmation of the solution: thought others might enjoy.
[attachment=3757] ps: this is an adobe pdf of an excel spreadsheet. BEST! SlideRule 

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