Pie with square - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: Pie with square (/thread-6497.html) Pie with square - Pekis - 07-03-2016 01:46 AM 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 RE: Pie with square - Paul Dale - 07-03-2016 08:06 AM 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 RE: Pie with square - klesl - 07-04-2016 03:34 AM 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. RE: Pie with square - Paul Dale - 07-04-2016 04:02 AM 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 RE: Pie with square - Pekis - 07-04-2016 04:16 AM 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 RE: Pie with square - SlideRule - 07-14-2016 03:00 PM 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