Trisecting a Circle with Parallel CutsDate: 06/12/2002 at 09:08:18 From: Anonymous Subject: Geometry Hi, I wonder if it's possible to draw two parallel lines in a circle so that you get three pieces that have the same area. Grateful for a quick answer. Date: 06/12/2002 at 09:43:38 From: Doctor Ian Subject: Re: Geometry Hi, Take a look at our FAQ on segments of circles: http://mathforum.org/dr.math/faq/faq.circle.segment.html Pay particular attention to case 13, in which you have the radius of the circle (r) and the angle that subtends the arc of the segment (theta). The area of the segment is theta - sin(theta) r^2 ------------------ 2 We would like for the area of the segment to be 1/3 of the area of the entire circle, since then we could cut another segment just like it off the other side of the circle, and end up with three equal-sized pieces. So let's set the area of the segment equal to 1/3 the area of the entire circle: theta - sin(theta) pi r^2 r^2 ------------------ = ------ 2 3 We can cancel r^2 from both sides to get theta - sin(theta) pi ------------------ = -- 2 3 and multiply both sides by 2 to get 2 pi theta - sin(theta) = ---- = 2.09 3 Note that 2pi/3 radians is 120 degrees, so theta is going to be larger than 120 degrees. How much larger? Could it be 150 degrees? 5pi/6 - sin(5pi/6) = 2.62 - 0.5 = 2.12 which is pretty close. You could try angles a little smaller than 150 degrees if you want to get more precision. (This kind of equation needs to be solved in the same way as a square root, namely by guessing the answer and checking it.) But when you find the angle theta to whatever precision makes you happy, you can cut the pizza, by drawing two diagonals across the pizza, * * * * A B * C * B' A' * * * * such that one of the angles between AA' and BB' is theta. Then two parallel cuts, from A to B and from B' to A', will cut the pizza into thirds. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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