Pizza Cutter (Posted on 2005-02-04)

	Submitted by David Shin
Rating: 3.6667 (3 votes)
Solution:	(Hide)
You will always end up with half the pizza. Let the four slices you eat be colored black, and the other four slices colored white. Lemma. Suppose two chords AC and BD in a circle of radius R intersect at P making four right angles and giving rise to the four line segments PA, PB, PC, PD with lengths a, b, c, d, respectively. Then a^2+b^2+c^2+d^2=4R^2. Proof. Let D' be the point diametrically opposite D, so that BDD' is a right triangle. The Pythagorean Theorem then yields (2R)^2=(c-a)^2+(b+d)^2. A similar construction in the other direction tells us that (2R)^2=(a+c)^2+(b-d)^2. Expanding and combining yields the result. Now return to the main problem. We assume in all that follows that the circle's radius is 1. Choose polar coordinates with the origin at P and the polar axis lying along one of the cuts; let the first slice counterclockwise from the polar axis be black. Let r(t) denote the distance from P to the circle at angle t with the polar axis. The area of the black region is then (1/2)(INT(r^2(t),t,0,pi/4)+INT(r^2(t),t,pi/2,3pi/4)             +INT(r^2(t),t,pi,5pi/4)+INT(r^2(t),t,3pi/2,7pi/4)). Combining into one integral, and making variable substitutions, we have (1/2)INT(r^2(t)+r^2(t+pi/2)+r^2(t+pi)+r^2(t+3pi/2),t,0,pi/4), which, by the lemma, is simply (1/2)INT(4,t,0,pi/4) = pi/2, as required. ------------------------------ In the comments, Richard also presents a solution where he expresses the area of the black slices as a function of the position and orientation of the cutter, and shows that this function has zero derivative with respect to the orientation. Since an orientation where the cutter cuts through the center of the pizza yields half of the pizza, we can conclude that the function is a constant one.

	Subject	Author	Date
	Answer	K Sengupta	2008-03-03 11:00:26
	re(3): thoughts -- Somewhat messy solution	Richard	2005-02-09 18:21:00
	re(2): thoughts -- Somewhat messy solution	Richard	2005-02-09 05:51:47
	wrong solution	sagar	2005-02-06 08:34:29
	solution	sagar	2005-02-06 06:04:15
	re: thoughts	Richard	2005-02-06 01:14:54
	1/2 much less doubtful	Richard	2005-02-05 07:49:32
	thoughts	Charlie	2005-02-05 02:37:54
	1/2 Doubtful	Richard	2005-02-04 22:32:13
	(Partial) Solution	Eric	2005-02-04 20:59:00
	Conjecture	Jer	2005-02-04 18:19:17