Hubbry Logo
Pizza theoremPizza theoremMain
Open search
Pizza theorem
Community hub
Pizza theorem
logo
8 pages, 0 posts
0 subscribers
Be the first to start a discussion here.
Be the first to start a discussion here.
Pizza theorem
Pizza theorem
Pizza theorem
View on Wikipedia
from Wikipedia
Example of application of the theorem with eight sectors: by cutting the pizza along the blue lines and alternately taking one slice each, proceeding clockwise or counterclockwise, two diners eat the same amount (measured in area) of pizza.
Proof without words for 8 sectors by Carter & Wagon (1994a).

In elementary geometry, the pizza theorem states the equality of two areas that arise when one partitions a disk in a certain way.

The theorem is so called because it mimics a traditional pizza slicing technique. It shows that if two people share a pizza sliced into 8 pieces (or any multiple of 4 greater than 8), and take alternating slices, then they will each get an equal amount of pizza, irrespective of the central cutting point.

Statement

[edit]

Let p be an interior point of the disk, and let n be a multiple of 4 that is greater than or equal to 8. Form n sectors of the disk with equal angles by choosing an arbitrary line through p, rotating the line ⁠n/2⁠ − 1 times by an angle of ⁠2π/n⁠ radians, and slicing the disk on each of the resulting ⁠n/2⁠ lines. Number the sectors consecutively in a clockwise or anti-clockwise fashion. Then the pizza theorem states (Upton 1968):

The sum of the areas of the odd-numbered sectors equals the sum of the areas of the even-numbered sectors.

History

[edit]

The pizza theorem was originally proposed as a challenge problem by Upton (1967). The published solution to this problem, by Michael Goldberg, involved direct manipulation of the algebraic expressions for the areas of the sectors. Carter & Wagon (1994a) provide an alternative proof by dissection. They show how to partition the sectors into smaller pieces so that each piece in an odd-numbered sector has a congruent piece in an even-numbered sector, and vice versa. Frederickson (2012) gave a family of dissection proofs for all cases (in which the number of sectors is 8, 12, 16, ...).

Generalizations

[edit]
12 sectors: green area = orange area

The requirement that the number of sectors be a multiple of four is necessary: as Don Coppersmith showed, dividing a disk into four sectors, or a number of sectors that is not divisible by four, does not in general produce equal areas. Mabry & Deiermann (2009) answered a problem of Carter & Wagon (1994b) by providing a more precise version of the theorem that determines which of the two sets of sectors has greater area in the cases that the areas are unequal. Specifically, if the number of sectors is 2 (mod 8) and no slice passes through the center of the disk, then the subset of slices containing the center has smaller area than the other subset, while if the number of sectors is 6 (mod 8) and no slice passes through the center, then the subset of slices containing the center has larger area. An odd number of sectors is not possible with straight-line cuts, and a slice through the center causes the two subsets to be equal regardless of the number of sectors.

Mabry & Deiermann (2009) also observe that, when the pizza is divided evenly, then so is its crust (the crust may be interpreted as either the perimeter of the disk or the area between the boundary of the disk and a smaller circle having the same center, with the cut-point lying in the latter's interior), and since the disks bounded by both circles are partitioned evenly so is their difference. However, when the pizza is divided unevenly, the diner who gets the most pizza area actually gets the least crust.

As Hirschhorn et al. (1999) note, an equal division of the pizza also leads to an equal division of its toppings, as long as each topping is distributed in a disk (not necessarily concentric with the whole pizza) that contains the central point p of the division into sectors.

[edit]

Hirschhorn et al. (1999) show that a pizza sliced in the same way as the pizza theorem, into a number n of sectors with equal angles where n is divisible by four, can also be shared equally among n/4 people. For instance, a pizza divided into 12 sectors can be shared equally by three people as well as by two; however, to accommodate all five of the Hirschhorns, a pizza would need to be divided into 20 sectors.

Cibulka et al. (2010) and Knauer, Micek & Ueckerdt (2011) study the game theory of choosing free slices of pizza in order to guarantee a large share, a problem posed by Dan Brown and Peter Winkler. In the version of the problem they study, a pizza is sliced radially (without the guarantee of equal-angled sectors) and two diners alternately choose pieces of pizza that are adjacent to an already-eaten sector. If the two diners both try to maximize the amount of pizza they eat, the diner who takes the first slice can guarantee a 4/9 share of the total pizza, and there exists a slicing of the pizza such that he cannot take more. The fair division or cake cutting problem considers similar games in which different players have different criteria for how they measure the size of their share; for instance, one diner may prefer to get the most pepperoni while another diner may prefer to get the most cheese.

Higher dimensions

[edit]

Brailov (2021), Brailov (2022), Ehrenborg, Morel & Readdy (2022), and Ehrenborg, Morel & Readdy (2023) extend this result to higher dimensions, i.e. for certain arrangements of hyperplanes, the alternating sum of volumes cut out by the hyperplanes is zero.

Compare with the ham sandwich theorem, a result about slicing n-dimensional objects. The two-dimensional version implies that any pizza, no matter how misshapen, can have its area and its crust length simultaneously bisected by a single carefully chosen straight-line cut. The three-dimensional version implies the existence of a plane cut that equally shares base, tomato and cheese.

See also

[edit]

References

[edit]
[edit]
Revisions and contributorsEdit on WikipediaRead on Wikipedia

Pizza theorem

View on Grokipedia
from Grokipedia
The pizza theorem, also known as the cheese pizza theorem, is a result in stating that if a disk is divided into 2N2N equiangular sectors of equal (measured at PP) using NN straight cuts passing through an arbitrary interior point PP (not necessarily ), then for even N≥4N \geq 4, the total area of the alternating sectors (e.g., the "gray" slices) equals the total area of the remaining sectors (e.g., the "white" slices). This surprising equality holds regardless of the position of PP within the disk, as long as it is interior, and applies to the classic case of N=4N=4 (eight 45-degree slices), where the sums of the areas of every other slice are identical. The theorem was first proposed as an unsolved problem by L. J. Upton in Mathematics Magazine in 1967. Originally posed in the context of fairly dividing a circular pizza among two people by assigning alternate slices, the theorem highlights a counterintuitive property of areas in polar coordinates, where the non-central cutting point distorts individual slice sizes but preserves the balance in alternating sums. The first proof for even NN was provided by Michael Goldberg in Mathematics Magazine in 1968, using geometric arguments involving inscribed and circumscribed circles around the cutting point. Subsequent proofs have employed integration in polar coordinates, symmetry properties of the disk, or , demonstrating that the area contributions from regions inside and outside the circle centered at PP with radius to the disk's center cancel out appropriately. The theorem generalizes to higher even numbers of slices (12, 16, etc.) and has been extended to nn-dimensional balls and other convex bodies bounded by quadratic hypersurfaces, where alternating volume sums vanish under similar symmetric cutting arrangements. For odd N≥3N \geq 3, the areas do not balance equally unless PP coincides with the center, leading to the related "pizza conjecture" resolved in the , which specifies the direction of imbalance based on the position of the center relative to the slices. These results have applications in problems, , and educational demonstrations of area preservation under non-radial partitions.

Core Concepts and Statement

Geometric Setup

The pizza theorem concerns a disk, defined as a closed two-dimensional region in the bounded by a of finite radius. This disk represents the pizza in the theorem's , with the boundary serving as the crust. Within this disk, consider an arbitrary interior point pp, which need not coincide with the disk's center. From pp, draw nn rays emanating outward to intersect the boundary , where n≥8n \geq 8 and nn is a multiple of 4. These rays divide the disk into nn sectors, each subtending an equal of 2πn\frac{2\pi}{n} radians at pp. The sectors are labeled consecutively around the disk, with adjacent sectors designated as odd-numbered (e.g., 1, 3, 5, ...) and even-numbered (e.g., 2, 4, 6, ...). For illustration, take the case n=8n = 8, corresponding to four straight cuts through pp at 45-degree intervals, akin to slicing a pizza off-center. Imagine a circular pizza with pp closer to one edge; the eight resulting wedge-like sectors vary in shape and area due to pp's offset position, yet alternate sectors (odd versus even) form the basis for the theorem's division. This setup highlights how the rays create non-uniform pieces despite equal angles. A prerequisite for understanding sector areas is the formula for the area of a circular sector with fixed radius rr and angle θ\theta in radians: 12r2θ\frac{1}{2} r^2 \theta. In the pizza theorem, however, radii vary along each ray from pp to the boundary, complicating direct application but underscoring the need for more advanced area computations.

Formal Statement

The pizza theorem asserts that a disk divided into nn sectors of equal angular measure 2πn\frac{2\pi}{n} by nn rays emanating from an interior point PP (not necessarily ) satisfies the equality of alternating sector areas when n≥8n \geq 8 and nn is a multiple of 4. Specifically, if the sectors are labeled consecutively as S1,S2,…,SnS_1, S_2, \dots, S_n in angular order around PP, then the sum of the areas of the odd-numbered sectors equals the sum of the areas of the even-numbered sectors. Let AiA_i denote the area of sector SiS_i for i=1,2,…,ni = 1, 2, \dots, n. The theorem states: ∑k=1n/2A2k−1=∑k=1n/2A2k.\sum_{k=1}^{n/2} A_{2k-1} = \sum_{k=1}^{n/2} A_{2k}. This holds irrespective of PP's position within the disk, as long as PP is interior, ensuring that alternating slices can be fairly divided between two parties with each receiving half the total area. The condition that nn must be a multiple of 4 and at least 8 is essential; for smaller or non-qualifying even nn, the equality fails in general. For n=4n=4, dividing the disk into four equal-angle sectors from an off-center PP results in unequal alternating sums, with opposite sectors dominating based on PP's proximity to the boundary. Similarly, for n=6n=6, the odd and even sector sums differ unless PP coincides with the center. A direct is that each alternating sum equals half the disk's total area, since their equality plus the fixed total area πr2\pi r^2 (for radius rr) implies ∑k=1n/2A2k−1=∑k=1n/2A2k=πr22\sum_{k=1}^{n/2} A_{2k-1} = \sum_{k=1}^{n/2} A_{2k} = \frac{\pi r^2}{2}.

Proofs and Mathematical Details

Dissection Proof

The dissection proof offers a visual and intuitive geometric approach to verifying the pizza theorem, demonstrating area equality by rearranging the sectors into two congruent polygons without coordinate calculations. Developed by Carter and (1994), this method partitions the sectors into smaller pieces that can be rotated and translated to match exactly between the two groups, highlighting the theorem's reliance on rather than the position of the cutting point P. For the case of eight sectors, the pizza is divided by four radial cuts through P, each separated by 45 degrees, yielding eight equal-angular sectors numbered consecutively from 1 to 8. The odd-numbered sectors (1, 3, 5, 7) form one , while the even-numbered sectors (2, 4, 6, 8) form the other. To prove equality, pair opposite sectors within each group: 1 with 5 and 3 with 7 for odds; 2 with 6 and 4 with 8 for evens. Each pair is then dissected into congruent triangular and trapezoidal pieces—typically two or three per sector—using lines from the circle's to P and along the cuts. These pieces are rearranged by rotating the entire configuration 90 degrees around the pizza's geometric or reflecting across diameters, mapping pieces from odd sectors (e.g., labeled A and B) precisely onto congruent pieces in even sectors (e.g., a and b). This step-by-step transformation reveals that the odd group assembles into a single identical in shape and size to the one formed by the even group, confirming their areas are equal. Visually, the before-dissection diagram depicts the original off-center sectors shaded alternately to distinguish groups, emphasizing P's . The after-dissection shows the rearranged pieces tiled into two matching polygons placed adjacent or overlaid, with boundaries aligning perfectly to underscore congruence and area parity. The core insight of this proof lies in how the equal-angular cuts facilitate a tiling that neutralizes the distortions from P's off-center location, as the 90-degree ensures pairings cancel out any imbalances across the circle. This extends to higher even numbers of sectors, such as 12 or 16, by subdividing each 90-degree quadrant into equal subsectors (three for 12 total sectors of degrees each, four for 16 total of 22.5 degrees each) while preserving the four primary 90-degree rotations. Pairings follow the same opposite-sector principle, with adjusted dissections into finer pieces that reassemble via similar rotations and translations into congruent polygons for the alternating groups, maintaining area equality through the underlying .

Algebraic Proof

To provide an algebraic proof of the Pizza theorem, consider a disk of radius RR centered at the origin in the plane. Let the point PP through which the cuts pass be located at coordinates (a,b)(a, b) with a2+b2<R2a^2 + b^2 < R^2. The cuts consist of nn lines passing through PP, equally spaced at angular intervals of π/n\pi/n, dividing the disk into 2n2n sectors where nn is even (ensuring 2n2n is a multiple of 4). The boundary rays from PP are thus at angles θk=k⋅(π/n)\theta_k = k \cdot (\pi/n) for k=0,1,…,2n−1k = 0, 1, \dots, 2n-1. Switch to polar coordinates centered at PP, so a point in the disk is given by (r,θ)(r, \theta) with 0≤r≤r(θ)0 \leq r \leq r(\theta), where r(θ)r(\theta) is the radial distance from PP to the disk boundary in direction θ\theta. Solving the disk equation x2+y2=R2x^2 + y^2 = R^2 along the ray yields the quadratic r2+2r(acos⁡θ+bsin⁡θ)+(a2+b2−R2)=0r^2 + 2r (a \cos \theta + b \sin \theta) + (a^2 + b^2 - R^2) = 0. The relevant (positive) root is r(θ)=−(acos⁡θ+bsin⁡θ)+(acos⁡θ+bsin⁡θ)2+R2−(a2+b2).r(\theta) = -(a \cos \theta + b \sin \theta) + \sqrt{(a \cos \theta + b \sin \theta)^2 + R^2 - (a^2 + b^2)}.
Add your contribution
Related Hubs
User Avatar
No comments yet.