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Four friends want to share a pizza. However, the pizza became badly misshapen en route; though it is still flat, it is now shaped like some arbitrary polygon.

a poorly pepperoni'd polygonal pizza

Prove that it is still possible to cut the pizza into four portions of equal area, using exactly two straight cuts which are perpendicular to each other.

Clarification: The two cuts divide the plane into four regions, and the four "portions" are defined to be the pizza parts contained in these regions. A portion might consist of several disconnected pieces.

(The cuts are allowed to go outside the pizza. The pizza is placed on a cutting board, and then a pizza cutter is rolled in a straight line twice, such that the two lines are perpendicular. The pizza may not be moved before making the second cut. The pepperoni distribution doesn't matter).

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    $\begingroup$ To confirm, if the 2 perpendicular cuts result in 5 pieces where 3 have the same area and the other 2 combined also have the same area, that is a failure? $\endgroup$ – Joel Rondeau Oct 20 '15 at 16:10
  • $\begingroup$ this is going to have to do with angles -_- $\endgroup$ – Spacemonkey Oct 20 '15 at 16:15
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    $\begingroup$ This most definitely reduces to some application of the Intermediate Value Theorem, but I am interested in how one would go about showing the continuity here. $\endgroup$ – Fimpellizieri Oct 20 '15 at 16:21
  • $\begingroup$ @Fimpellizieri I have a feeling the answer is going to be like the table-leg problem we had some time ago, and the solution is going to involve rotating the cuts 90 degrees. $\endgroup$ – 2012rcampion Oct 20 '15 at 16:22
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    $\begingroup$ The pepperoni distribution always matters. $\endgroup$ – user1717828 Oct 21 '15 at 0:55
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By continuity, for any cut angle, there's a unique cut that slices the pizza into two equal-area segments. Moreover, the cut placement is continuous in the angle, since for close angles, the lines are nearly parallel and their intersection point must lie within the convex hull of the pizza, a finite range.

Now, consider a cut at some angle and the perpendicular angle. These must intersect and so divide the pizza into four sections, with any two adjacent sections totaling half the pizza. So, the pairs of opposite sections have the same area. It remains to make the area for each pair equal.

Do this by continuously varying the initial angle over a range of 90 degrees. From the start to the end, the original and perpendicular cut will have switched, so the respective areas will have switched. Since the cut placement is continuous in the angle, the sizes of the sections also vary continuously, and so by the intermediate value theorem, must be equal for some angle.

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  • $\begingroup$ Nice solution! I didn't expect to see an answer by you in the one random puzzle I saw on the side of the page. It was in the "Hot Network Questions," though. $\endgroup$ – mbomb007 Oct 20 '15 at 21:50
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Draw two perpendicular lines, such that line1 & line2 intersect at P & line1 is at some angle theta (say 0) to x-axis.
Number the 4 quadrants Q1,Q2,Q3,Q4.
Move the intersection P along line1 , from very far left to very far right. At some position, the area on the left must equal the area on the right, because at very far left, the area of the 2 left quadrants will be 0, while at very far right, the area of the 2 right quadrants will be 0, and at some intermediate position, the areas will be equal.

Do this movement for P moving along line2 , from very far up to very far down. Again the 2 upper quadrants will be equal to the 2 lower quadrant at some point along line2.

Basically , we have moved P all over the plane , when line1 was at theta, and we have found a position where Q1+Q2=Q3+Q4=A/2 & Q2+Q3=Q4+Q1=A/2. Now Q1=Q3 & Q2=Q4, or opposite quadrants are equal.
We get Q1=A/2-Q2.
If Q1=Q2, we are done. If not Q1 is-less-than Q2 Or Q1 is-more-than Q2.

Now repeat this process, but use theta from 0 to 90 Degrees. The locus of the points (for each theta) with this movement must be continuous.
Starting at theta=0 and going to theta=90, we see that Q1 and Q2 exchange their final values. So, starting at Q1 is-less-than Q2, we get Q1 is-more-than Q2. So, at some theta, Q1=Q2. So all 4 quadrants are equal at this point of intersection, at this angle.

Summary : We search over all intersection points, at all angles, which will give us at least one solution, because of continuity.

enter image description here

Here, start with the horizontal gray line and the vertical rid line on the left. Intersection is P, with 4 quadrants. We see that area is Q1+Q4=0, considering the areas of the pizza in the quadrants.
We move P slowly towards right, to get the center red line and the rightmost left line. We end up with Q2+Q3=0.
So, at some point, we have Q1+Q4=Q2+Q3, shown as the central red line.

Now, move the gray line upwards and similarly get a new P which has Q1+Q4=Q2+Q3. In the example given, the locus of P which has Q1+Q4=Q2+Q3 is the central red line, but in general, it may be some other continuous curve C.
In the curve C, when we move P from top to bottom, we start with Q1+Q2=0 and end with Q3+Q4=0, so at some point, Q1+Q2=Q3+Q4.
If Q1=Q2 here , we are done.
Else we must repeat the process with a new theta.
When we vary theta from 0 to 90, we will see that we move from Q1 is-less-than Q2 to Q1 is-more-than Q2 , so at some intermediate point, Q1=Q2, so all 4 quadrants are equal.

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    $\begingroup$ My eyes are bleeding. My smarts ache, can we get a picture? $\endgroup$ – Spacemonkey Oct 20 '15 at 18:15
  • $\begingroup$ @Spacemonkey , I will try to add a small graphic now, but it is late here . . . maybe tomorrow I will ad a better graphic. $\endgroup$ – Prem Oct 20 '15 at 19:13
  • $\begingroup$ "The locus of the points (for each theta) with this movement must be continuous." This is, for the most part, the difficulty of the problem and is mostly glossed over here. $\endgroup$ – Fimpellizieri Oct 20 '15 at 20:27
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Take a line and roll it along until it divides the pizza into two equal halves. Then take a second line, perpendicular to the first and roll it along until it divides the pizza into two equal halves.

The diagonally opposite quadrants will have equal mass.

How find the focus of the centre of the cross for all orientations of the lines. And imagine transitioning smoothly from the original cross to one at right angles to it (which will be in the same location).

The one at right angles will be heavier where the previous one was lighter and vice versa. So between these there must be one where they're all the same (since each of the transitional states has diagonally opposite quadrants equal).

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  • $\begingroup$ Good point. It's the moment about all the lines that's equal. And you can't eat moment. $\endgroup$ – Dr Xorile Oct 20 '15 at 21:13
  • $\begingroup$ So I fixed it. You just need to use the focus of points where the perpendicular lines each bisect the pizza. $\endgroup$ – Dr Xorile Oct 20 '15 at 21:37
  • $\begingroup$ This is how I thought of it, but it wouldn't be specific enough for a proof needed for math homework (a rigorous proof). I'd throw in references to the Intermediate Value Thm or something, despite the solution being easily visible as you described. $\endgroup$ – mbomb007 Oct 20 '15 at 21:45
  • $\begingroup$ Depends if you want rigour or clarity I guess. I was aiming at the easiest way that someone could see that it was true $\endgroup$ – Dr Xorile Oct 20 '15 at 22:44
  • $\begingroup$ The only bit that might take a bit of thought is the continuity. But I think that's clear from the continuity of the pizza. If the pizza were a fractal it might not work. $\endgroup$ – Dr Xorile Oct 20 '15 at 22:46

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