CCUTTING CAKES AND KISSING CIRCLES
ALEXANDER M ¨ULLER-HERMES
Abstract.
To divide a cake into equal sized pieces most people use a knife anda mixture of luck and dexterity. These attempts are often met with varyingsuccess. Through precise geometric constructions performed with the knifereplacing Euclid’s straightedge and without using a compass we find methodsfor solving certain cake-cutting problems exactly. Since it is impossible toexactly bisect a circular cake when its center is not known, our constructionsneed to use multiple cakes. Using three circular cakes we present a simplemethod for bisecting each of them or to find their centers. Moreover, given acake with marked center we present methods to cut it into n pieces of equalsize for n = 3 , Introduction
Dividing a cake among a given number of people is a relevant problem at manybirthday parties. Mathematically, this problem has often been studied in the con-text of “fair division” going back to Hugo Steinhaus [Ste49]. Here, every cake-eatingparticipant should receive a share they themselves consider to be fair. Many proto-cols for fair division have been studied (see for instance [DS61, Str80, Woo80]), butthey do not seem to be practical at birthday parties with more than two partici-pants who might lack the enthusiasm to perform complicated protocols, or who arenot able to accurately compare the sizes of different pieces of cake. For example,the author himself finds it even difficult to cut a cake into two pieces from whichthe larger piece cannot be selected immediately. In this article, we present precisegeometric constructions for how to cut circular cakes into equal sized pieces. Ourconstructions are based on the following assumptions:(1) Cakes are perfect circles.(2) Straight lines can be carved into the cake (or the table) using a knife.It is our firm opinion that geometric considerations of cake-cutting should only usethe knife, which is considered equivalent to a straightedge (i.e. unmarked ruler) inEuclidean geometry. Non-circular cakes are beyond the scope of this article, butwe challenge the reader to find techniques to cut whatever shape of bakery theymight encounter into pieces of equal size.2.
History of cake cutting: The Poncelet-Steiner theorem
With the assumptions stated above, we can cut cakes by geometric constructionsfollowing the Euclidean axioms, but without using the compass. To substitute forthe compass the cakes themselves may be used as preexisting circles. The study ofsuch constructions has a long history, and we will start with the following theoremproved by Jakob Steiner [Ste33] in 1833 after being conjectured by Victor Poncelet.We recommend [D¨or65] for a well-written exposition of its proof.
Theorem 1 (Poncelet-Steiner) . Any construction in Euclidean geometry can beperformed using the straightedge alone provided that a circle with its center is given.
Date : August 27, 2020. a r X i v : . [ m a t h . HO ] A ug ALEXANDER M¨ULLER-HERMES
By the Poncelet-Steiner theorem it is possible to cut a cake with marked centerinto n pieces of equal size whenever the regular n -gon is constructible in Euclideangeometry, i.e. whenever n is a product of distinct Fermat primes and a power of 2.Many cakes encountered in the wild have their center marked by some kind ofdecoration, but cakes with unmarked or inaccurately marked center are also quitecommon. Unfortunately, the conclusion of the Poncelet-Steiner theorem is no longertrue, when the center of the given circle is not known. This observation is due toDavid Hilbert as noted in [Cau12], and an exposition of his argument can be foundin [RT57]. As a consequence, it is impossible to exactly cut a cake in half, whenthe center is not known and only a knife may be used. Luckily, having more cakesavailable saves the day as shown by Detlef Cauer [Cau12] in 1912. It is possibleto construct the center of a circle c using the straightedge alone if either anothercircle intersecting c is given, or two other circles are given in any position. By thePoncelet-Steiner theorem this implies the following theorem. Theorem 2 (Cauer) . Any construction in Euclidean geometry can be performedusing the straightedge alone provided that either two intersecting circles, or threecircles in arbitrary position are given.
Since cakes, unlike circles drawn on a piece of paper, are movable objects, wecan push two cakes together so that they touch in a point. By Cauer’s theoremtwo cakes are then enough to perform general Euclidean constructions using a knifealone. While we provide all ingredients for cutting pairs of touching cakes in thisarticle, we do not go into the details of these methods since they come with adrawback: In the constructions known to the author (see for instance Figure 1 for abisection) auxiliary lines have to be carved into the table rather than into the surfaceof the cake. While we have not excluded this possibility in the assumptions statedabove, it might create conflict in practice and the aspiring cake-cutter might notget invited anymore to parties with cakes to cut. In the following, we will presentconstructions that only use the surface of the cakes but require at least three cakesor a marked center. Specifically, we present such constructions for cutting a cakeinto n pieces of equal size where n = 2 , , Figure 1.
The red line indicates the final cut bisecting the larger cake.
UTTING CAKES AND KISSING CIRCLES 3 Bisecting an odd number of cakes
Our method for bisecting cakes is based on an elementary construction usingkissing circles. We say that a pair of circles c , c is kissing in a point S if thefollowing conditions are satisfied:(1) The intersection satisfies c ∩ c = { S } .(2) The circles c and c lie on opposite sides of their common tangent in S .See Figure 2 for an illustration. Figure 2.
Kissing circles.Given a pair of kissing circles and a point P on one of the circles we can constructa point P (cid:48) on the other circle by passing P through the kissing point: Definition 3 (Passing through the kissing point) . Consider circles c and c kissingin a point S . For a point P on c we construct a point P (cid:48) on c as follows:(1) If P = S , then we set P (cid:48) = S .(2) If P (cid:54) = S , then we set P (cid:48) to be the intersection point different from S of theline P S with c .We will say that P (cid:48) is obtained by passing P through the kissing point S . Suppose now that we are given an odd number of cakes. To bisect one of themwe first push them together such that they form a closed chain of kissing circles asin Figure 3, i.e. circles c , . . . , c n +1 such that the pair ( c i , c i +1 ) is kissing in thepoint S i and the pair ( c n +1 , c ) is kissing in the point S n +1 . Starting with a point P on c , we construct a point P on c by passing P through the kissing point S . Repeating this successively for the other circles leads to points P , . . . , P n +1 on c , . . . , c n +1 as in Figure 3. Finally, we construct a point Q on c by passing P n +1 through the kissing point S n +1 . We claim that the chord QP is a diameterand hence bisects the first cake.To make the previous statement precise we formulate the following theorem alsocovering the case of an even number of cakes. Theorem 4.
Consider circles c , . . . , c n such that each pair ( c i , c i +1 ) is kissing in apoint S i and the pair ( c n , c ) is kissing in a point S n . Starting with a point P on c we construct points P , . . . , P n such that each point P i +1 is obtained by passing thepoint P i through the kissing point S i . Let Q be the point on c obtained by passing P n through the kissing point S n . Then, we have the following two statements:(1) If n is even, then Q = P .(2) If n is odd, then the chord QP is a diameter of c . ALEXANDER M¨ULLER-HERMES
Figure 3.
Construction of a diameter (red) using three and fivecircular cakes. Dotted lines should only be carved slightly into thesurface of the cakes.While the construction outlined above works for every starting point P it isrecommendable to use P = S for bisecting cakes in practice. By doing so, oneneeds to carve one auxiliary line less, and extending the final cut also bisects thesecond cake c . In the next section, we will present a proof of Theorem 4. Forthis we will need to understand the operation of “passing through a kissing point”which will require a bit of elementary geometry of circles.4. Why does it work?
Our proof will only use Thales’ theorem and its converse. Given a point R ona circle and a chord c as in Figure 4, the inscribed angle α is the angle formed bythe two lines connecting the ends of the chord c with R . The following theorem iswell-known: Theorem 5 (Thales) . Every inscribed angle over a diameter is a right angle. More-over, whenever an inscribed angle over a chord is a right angle, then that chord isa diameter.
Figure 4.
The inscribed angle α over a chord c .Recall that two points on a circle are called antipodal if the chord connectingthem is a diameter, i.e. passes through the center. Using Thales’ theorem we canunderstand how antipodal points are passed through a kissing point. Lemma 6 (Passing antipodal points through a kissing point) . Consider circles c and c kissing in a point S and let P and Q be antipodal points on c . By passing P UTTING CAKES AND KISSING CIRCLES 5 and Q through the kissing point S we obtain points P (cid:48) and Q (cid:48) on c . The followingstatements are true:(1) The points P (cid:48) and Q (cid:48) are antipodal as well.(2) The diameters d = P Q and d = P (cid:48) Q (cid:48) are parallel.(3) The direction of the rays −−→ P Q and −−−→ P (cid:48) Q (cid:48) is reversed.See Figure 5 for an illustration. Figure 5.
The diameters l and l (red) are parallel. Proof.
It is easy to verify the lemma if either P or Q coincides with the kissingpoint S , and in the following we assume that this is not the case.Note that the two inscribed angles at S over the chords d and d are equal sincethey are opposite. By the converse of Thales’ theorem we conclude that d is adiameter whenever d is a diameter. Figure 6.
Proof of the kissing circle lemma.To prove that d and d are parallel we consider the segment through the kissingpoint S and the centers Z and Z of the circles c and c respectively, and thesegment through the kissing point S and the points Q and Q (cid:48) . Hereby, we obtainthe pair of isoceles triangles colored red in Figure 6. Note that the angles α and β in the figure are equal as they are opposite angles at the point S . Therefore, thetwo angles γ and δ in the figure are equal. As these are the angles between thesegment Z Z and the diameters d and d respectively, we conclude that d and d are parallel. Clearly, the direction of the rays −−→ P Q and −−−→ P (cid:48) Q (cid:48) is reversed sincethey are parallel and the segments P P (cid:48) and QQ (cid:48) intersect in S . (cid:3) ALEXANDER M¨ULLER-HERMES
Now, we can provide an easy characterization of the “passing through the kissingpoint” operation. Consider two circles c and c kissing in a point S . For a point P on c we denote by P (cid:48) the point on c obtained by passing P through the kissingpoint S . Furthermore, let r = −−→ Z P denote the ray emerging from the center Z of c going through P drawn red in Figure 7. By Lemma 6 the point P (cid:48) lies onthe diameter of c that is parallel to the ray r and such that the ray r = −−−→ Z P (cid:48) through P (cid:48) emerging from the center Z of c points in the opposite direction of r . See Figure 7 for an illustration. Figure 7.
Passing P through the kissing point.We summarize this observation in the following proposition. Proposition 7 (Passing through a kissing point revisited) . Consider circles c and c kissing in a point S with centers Z and Z respectively. Passing a point P on c through the kissing point S yields the point P (cid:48) on c for which the ray −−−→ Z P (cid:48) isparallel to the ray −−→ Z P and points in the opposite direction. Now, we can easily prove Theorem 4. Consider a chain of circles c , . . . , c n wherethe pairs ( c i , c i +1 ) are kissing in points S i , and let Z i denote the center of circle c i .Starting with a point P on c , Theorem 4 considers a sequence of points P , . . . , P n such that the point P i +1 is obtained by passing the point P i through the kissingpoint S i . By Proposition 7 each point P i +1 is characterized by the ray −−−−−−→ Z i +1 P i +1 being parallel to the ray −−→ Z i P i and pointing in opposite direction. Therefore, we canvisualize the entire construction as in Figure 8 with rays pointing to each of theconstructed points and reversing direction in each step. Finally, when the chainof circles is closed and the construction returns to the initial circle c we eitherend up at the starting point P when the number of circles is even, or we end upantipodal to P when the number of circles is odd. In the latter case, we have founda diameter of the circle c .5. Cutting cakes into more pieces
So far, we have discussed how to find a diameter of a cake when there are atleast three cakes available. Repeating the construction after rotating the cake, givesanother diameter intersecting the first one in the center of the cake. When cuttingthree or more cakes we may therefore assume that the center is marked on each ofthem. By the Poncelet-Steiner theorem mentioned in Section 2 it is then possibleto cut each of the cakes into 3, 4 or 6 pieces of equal size since the correspondingregular n -gons are constructible in Euclidean geometry. In the following, we willshow how to actually do this in practice. UTTING CAKES AND KISSING CIRCLES 7
Figure 8.
Constructing a sequence of points.Our constructions are based on two tricks developed by Jacob Steiner in [Ste33, § § P Q through points P and Q when the midpoint Z of thesegment P Q is given. The second trick, closely related to the first, constructs themidpoint Z of a segment P Q when a parallel line to the segment is given. Bothstatements follow directly from Ceva’s theorem [Cox69, p.220] and we leave theirproof to the reader.
Theorem 8 (Two tricks by Steiner) . Consider points P and Q .(1) Let Z be the point bisecting the segment P Q and consider a point R not onthe line P Q , and a point A on the segment P R . We construct a point B asthe intersection of QA and ZR . Let C denote the intersection of the line P B with the line QR . Then, the line AC is parallel to the line P Q .(2) Let points A and C be given such that the lines AC and P Q are parallel.We construct a point R as the intersection of P A and QC and a point B as the intersection of P C and QA . Then, the line RB intersects P Q in thepoint Z bisecting the segment P Q
See Figure 9 for illustration.
Figure 9.
The red lines are parallel and Z bisects the segmentbetween P and Q . ALEXANDER M¨ULLER-HERMES
We will now show how to cut a cake with marked center Z into 4 pieces of equalsize. Let P and Q denote a pair of antipodal points on the rim of the cake, i.e. suchthat the chord P Q is a diameter. To cut the cake into four pieces of equal size weneed to construct a diameter perpendicular to
P Q . Note that the center Z bisectsthe diameter P Q . By choosing a point R on the rim of the cake and a point A onthe segment P R , we can use the construction from the first case of Theorem 8 tofind a point C such that AC is parallel to the P Q as in Figure 10. Intersecting theline AC with the circle yields points A (cid:48) and C (cid:48) such that the quadrilateral QC (cid:48) A (cid:48) P is an isoceles trapezoid. Finally, the intersection point B (cid:48) of the diagonals P C (cid:48) and QA (cid:48) lies on a diameter perpendicular to P Q and by cutting along the lines
P Q and ZB (cid:48) we divide the cake into 4 pieces of equal size. Figure 10.
The red lines cut the cake into four pieces of equal size.Next, we present a construction for cutting a cake into 3 pieces of equal size.Again, we assume that a cake with marked center is given and using the previousconstruction we find four points
P, Q, P (cid:48) , Q (cid:48) on its rim such that the quadrilateral QP (cid:48) P Q (cid:48) is a square as in Figure 11.
Figure 11.
The red lines cut the cake into three pieces of equal size.
UTTING CAKES AND KISSING CIRCLES 9
We first note that the segments
P Q (cid:48) and P (cid:48) Q are parallel. After choosing a point R on the rim of the cake, we apply the construction from the second case of Theorem8 to find a point X bisecting the segment P Q (cid:48) . Repeating this construction forthe parallel segments QQ (cid:48) and P P (cid:48) (and another point M on the rim of the cake)we find the point X bisecting the segment QQ (cid:48) . The line X X intersects the rimin the points W and Z such that P (cid:48) W Z is an equilateral triangle. Finally, we cutfrom each point P (cid:48) , W and Z to the center of the cake and obtain 3 pieces of equalsize. Finally, we note that the previous construction also allows to cut the cake into6 pieces of equal size by extending the red segments in Figure 11. Acknowledgements
We thank Chris Perry and Emilie Elkiær for insightful comments and interest-ing discussions about cake-cutting that improved this article. We acknowledgesfinancial support from the European Union’s Horizon 2020 research and innovationprogramme under the Marie Sk(cid:32)lodowska-Curie Action TIPTOP (grant no. 843414).
References [Cau12] Detlef Cauer. ¨Uber die Konstruktion des Mittelpunktes eines Kreises mit dem Linealallein.
Mathematische Annalen , 73(1):90–94, 1912.[Cox69] Harold Scott Macdonald Coxeter. Introduction to geometry. 1969.[D¨or65] Heinrich D¨orrie.
100 Great Problems of Elementary Mathematics: Their History andSolution . Dover, 1965.[DS61] Lester E. Dubins and Edwin H. Spanier. How to cut a cake fairly.
The American Math-ematical Monthly , 68(1P1):1–17, 1961.[RT57] Hans Rademacher and Otto Toeplitz.
Enjoyment of Mathematics: Selections from Math-ematics for the Amateur . Princeton University Press, 1957.[Ste33] Jacob Steiner.
Die geometrischen Konstructionen ausgef¨uhrt mittelst der geraden Linieund eines festen Kreises , volume 60. D¨ummler, 1833.[Ste49] Hugo Steinhaus. Sur la division pragmatique.
Econometrica: Journal of the EconometricSociety , pages 315–319, 1949.[Str80] Walter Stromquist. How to cut a cake fairly.
The American Mathematical Monthly ,87(8):640–644, 1980.[Woo80] Douglas R. Woodall. Dividing a cake fairly.
Journal of Mathematical Analysis and Ap-plications , 78(1):233–247, 1980.
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