Geometry of a PizzaDate: 5/31/96 at 2:41:58 From: Anonymous Subject: Geometry with circles I have a pizza. The radius is 10 inches long. The pizza was cut into 16 equal slices. When 1 slice was left, my brother and I both wanted it, so we agreed to cut it in half, but I like the crust more than he does, so we decided to cut it in a way perpendicular to the "usual way." In other words, the two pieces would not be symmetrical. One piece would contain all topping, and the other would contain some topping and some crust. Finally, the question: How far up the radius from the center of the circle will I need to cut so we will both have an equal area of pizza? Note: the area of the crust and topping make no difference. Just think of it as a regular circle. My first strategy was to simply find where to cut a triangle in the same way above, but I had no luck. At this point I can think of no other plan. Could you help me out? Thanks! Date: 5/31/96 at 3:30:28 From: Doctor Pete Subject: Re: Geometry with circles This is a pretty ugly problem (the answer is a little messy). I'm not sure what level of math you are familiar with, but the solution I've got requires trigonometry. So, take a piece (1/16th) of the pizza. Then its area is Pi*r^2/16 = 100*Pi/16 = 25*Pi/4. Half of this, 25*Pi/8, is the area of the slice you want. So if you were to cut the slice perpendicularly at its axis of symmetry ("the usual way") at some point, x inches, away from the center, you'd get a triangle (the topping part), and a funny-shaped trapezoid-like thingy (the crust part). So the area of the triangle must be 25*Pi/8. Here's a picture (it's a bit wider than it actually is): . /|\ / |A\ / | \ / | \ / |x \ / | \ / |_ \ /_______|_|_____\ b where angle A = Pi/16 (this is easy to see). But tan(A) = b/x, so b = x tan(A). It follows that the area of this triangle is b*x = x^2* tan(A) = x^2*tan(Pi/16). But this is equal to 25*Pi/8, so solving for x: x = sqrt(25*Pi/(8*tan(Pi/16))) = 5/2 * sqrt(Pi/(2*tan(Pi/16))) = about 7.025 . Now, what's so nasty about this answer? Well, what is tan(Pi/16)? Most people would be content to leave the solution at that, but using some half-angle identities, we have: tan(Pi/16) = sqrt((1-cos(Pi/8))/(1+cos(Pi/8))) = sqrt((1-sqrt((1+cos(Pi/4))/2))/(1+sqrt((1+cos(Pi/ 4))/2))) sqrt(1-sqrt((1+sqrt(2)/2)/2)) = ----------------------------- sqrt(1+sqrt((1+sqrt(2)/2)/2)) = sqrt(7+4*sqrt(2)-(4+2*sqrt(2))*sqrt(2+sqrt(2))). -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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