How to fairly share a watermelon
HHow to fairly share a watermelon
Timoteo Carletti naXys, Namur Institute for Complex Systems, University of Namur, rempart de la vierge 8, B5000 Namur, Belgium Duccio Fanelli Universit`a degli Studi di Firenze, Dipartimento di Fisica e Astronomia,CSDC and INFN, via G. Sansone 1, 50019 Sesto Fiorentino, Italy
Alessio Guarino Universit´e de La R´eunion – Laboratoire Icare EA 7389, France
Geometry, calculus and in particular integrals, are too often seen by young students as technicaltools with no link to the reality. This fact generates into the students a loss of interest with aconsequent removal of motivation in the study of such topics and more widely in pursuing scientificcurricula. With this note we put to the fore a simple example of practical interest where the aboveconcepts prove central; our aim is thus to motivate students and to reverse the dropout trend byproposing an introduction to the theory starting from practical applications. More precisely, we willshow how using a mixture of geometry, calculus and integrals one can easily share a watermeloninto regular slices with equal volume.
I. INTRODUCTION TO THE MAIN QUESTION
What better than a fresh watermelon to fight the heat wave in these sunny and hot summer days? In Italy it isvery common to buy watermelons weighting as much as 15 Kg for few euros. Sharing the watermelon is hence anopportunity for a party with many friends. A recurring question is then: how can one slice the watermelon intoequal volume parts to have a fair sharing among friends?
This question is particularly interesting if the slices are notcut along the longitudinal direction, namely the longer one. Cuts along the transversal direction are in fact oftenperformed to return slices which can be more straightforwardly manipulated. In the following we will show thatsymmetry helps: for the first cuts up to a certain number of friends, one can indeed proceed by dividing each portioninto two identical parts. To go further, when the number of guests is large, we propose a rule of the thumb whichfollows a simple mathematical analysis: this will be referred as the “2 / b . Themajor axis measures a , with a > b (See Figs. 1 and 2). The total volume of the watermelon is thus V tot = 4 πb a/ n friends: to obtain n slices we need to perform n − < λ a < λ a < · · · < λ n − a < a , along the semi-major axis where to slice, in such a waythat each obtained part shares the same volume. This latter should hence equal 1 /n of the volume of one fourth ofthe watermelon, V = πb a/ V tot / n = 4): Problem : find λ i ∈ (0 , i = 1 , . . . , n −
1, such that 0 < λ < λ < · · · < λ n − < V i = V /n ,where V i is the volume of the slice obtained by respectively cutting at positions λ i a and λ i − a . a r X i v : . [ phy s i c s . e d - ph ] S e p A) Longitudinal view B) Lateral viewC) Top view x a + y b + z b = 1 a = 2 , b = 1 CA B
Watermelon . Different views of a watermelon (panels A, B and C). Bottom right: a (spheroid) ellipsoid with equalmedian and semi-minor axes, b , and semi-major axis, a > b whose cartesian equation is given by x /a + y /b + z /b = 1.The geometrical shape depicted in the bottom right panel corresponds to the choice a = 2 and b = 1 in arbitrary units. Tohelp visualising the geometrical setting, we show in panels A, C, as well as in the bottom right panel, the semi-major axis a and the semi-minor axis b . longitudinalcuttransversalcut b
First few cuts . Cutting along the symmetry axes of the watermelon can easily return 4 equal volume pieces that willconstitute the starting point for successive cuts. The semi-major axis a and the semi-minor axis b are displayed in Figure. II. COMPUTATIONS NEEDED TO SOLVE THE PROBLEM.
Let us consider 0 < λ < V ( λa ) of the portion of spheroid contained between the centralface and the cut at distance λa from this face. Consider a tiny slice, shaped as a circular sector with central (opening)angle of π , i.e. a half-disk, at a generic distance x measured from the central face, with thickness dx (see Fig. 4).Then, its volume is given by dV = π/ × (cid:96) ( x ) dx , where (cid:96) ( x ) = b (cid:112) − x /a is the radius of the sector of the disk.The slice is indeed a portion of a cylinder of height dx and whose base is a sector of a disk with opening π and radius (cid:96) ( x ); the radius depends on the position of the slice, it equals b (the semi-minor and medium axis) for x = 0 and it b) c) < < < < a) a a a aV V V V V = V = V = V angle = ⇡/ FIG. 3:
Where to cut next?
Panel a): cutting along the longitudinal direction by dividing the angle π into n equal parts( n = 4 in the panel) can be hard. Panel b): cutting along the transversal direction, i.e. parallel to the central face (hatched inthe figure), requires identifying the cuts positions, 0 < λ a < λ a < · · · < λ n − a < a (again n = 4 in the panel). Panel c): theslices obtained by cutting at the positions given by the procedure presented in panel b) should all have the same volume. vanishes at x = a (see panel b) of Fig. 3). In conclusion we get for the volume V ( λa ) V ( λa ) = π (cid:90) λa (cid:96) ( x ) dx = π b (cid:90) λa (cid:18) − x a (cid:19) dx = π b aλ (cid:18) − λ (cid:19) . (1)
Geometry for the computation of the volume of the slice . We ideally slice the large piece of watermelon intoarbitrarily tiny slices of width dx (on the left panel). The latter can be considered as a very tiny sector of disk, with opening π and radius (cid:96) ( x ), the latter depending on the position x of the slice (on the middle panel). Such an arbitrary thin slice hasa volume dV = π/ × (cid:96) ( x ) dx (on the right panel). Indeed it can be considered as an infinitesimally tick cylinder (with hight dx ), whose basis is a sector of disk with central angle π . Given the integer n ≥ λ i ∈ (0 ,
1) such that 0 < λ < λ < · · · < λ n − <
1, then we get for the volumes V i (defined on page 1) V = V ( λ a ) , V i +1 = V ( λ i +1 a ) − V ( λ i a ) ∀ i = 1 , . . . , n − V n = V ( a ) − V ( λ n − a ) . (2)Our problem is thus equivalent to solve the equations1 n = V V = 32 λ (cid:18) − λ (cid:19) , n = V i V = 32 λ i (cid:18) − λ i (cid:19) − λ i − (cid:18) − λ i − (cid:19) ∀ i = 2 , . . . , n − n = V n V = V ( a ) V − λ n − (cid:18) − λ n − (cid:19) . (3)Stated differently, let p ( λ ) = λ (cid:16) − λ (cid:17) , the first equation of (3) is equivalent to looking for a root of p ( λ ) = 1 /n .Then given λ one can plug this value in the next equation for i = 2 and solve for λ the equation p ( λ ) = 1 /n + p ( λ ) =2 /n , where we used that fact that p ( λ ) = 1 /n . We can iterate the above steps and get: ∀ i = 2 , . . . , n − in = p ( λ i ) . The last equation involving V n results into the identity p (1) = 1, i.e. the volume of the whole piece is V .In conclusion given n we have to solve n − p ( λ ) = 32 λ (cid:18) − λ (cid:19) = f n , (4)where f n ∈ { /n, . . . , ( n − /n } . Before proceeding, let us observe that the above equation always admits one andonly one solution in the interval (0 , p ( λ ) vanishes at λ = 0 and its first derivatives vanishesat λ = ±
1. Further, it can be easily proven that it has a maximum at λ = 1 where it reaches the value p (1) = 1 anda minimum at λ = −
1. The polynomial is thus steadily increasing from 0 up to 1. Hence, any constant horizontalline set at f n ∈ (0 ,
1) intersects the polynomial in just one point belonging to the interval (0 ,
1) (see Fig. 5 for agraphical representation of this claim). One can of course solve analytically the above equations, by making use of
FIG. 5:
Existence and uniqueness of the solution p ( λ ) = f n . We show the polynomial p ( λ ) for λ ∈ (0 ,
1) (black curve)and a generic horizontal line set at f n ∈ (0 ,
1) (blue dashed line). the explicit formulae for roots of the third order polynomials. Notice however that a good approximation for the firstslices can be obtained by setting λ app = f n × / . The adequacy of this latter ansatz is readily assessed by noticingthat p ( λ app ) = f n − f n × /
27 is very close to the required value, f n , since the cubic term f n × /
27 can be safelyneglected.Consider as an example n = 2. We need to share every one fourth of watermelon into 2 equal volume parts.According to the above recipe, we perform the cut at λ app a = a × / × / a/
3, namely assuming a 1 / / n = 3 we need to cut at λ a and λ a , for λ and λ solutions of Eq. (3). In analogy with the above, we canapproximate λ and λ by using the / -rule, yielding λ (1) app = / × / λ (2) app = / × /
3. Thus the first slicecan be obtained with a cut at λ (1) app a = a × / × / a/ . λ (2) app a = a × / × / a/
9, we obtain a slice 14% smaller (in volume) of the desired one. Theerror is getting worse because the approximated cuts follow a linear profile, λ app = / × f n , while the curvature ofthe watermelon is responsible for the cubic term in p ( λ ).For a generic n ≥ λ ( i ) app a = a × / × i/n for i = 1 , . . . , n −
2. More precisely let us consider the ratio of the i -th approximated volume V appi , that is the volumeobtained with a cut at λ ( i ) app a (instead of λ i a ), computed from Eq. (3) evaluated at λ ( i ) app a , with V the initial volumeof the watermelon piece (one fourth of the full volume): V appi V = p ( λ ( i ) app ) − p ( λ ( i − app ) = in − i n − i − n + 427 ( i − n = 1 n −
427 3 i + 3 i + 1 n . One can immediately realise that the relative error grows with i and is largest for the last cut, i = n − n as compared to their exact homologues (horizontal dashed lines). Onecan see that for all n the first few approximated volumes are quite accurate, while the disagreement increases forfurther cuts at fixed n . V appi V n = n = n = n = FIG. 6:
Approximated volumes . Each column represents the relative approximated volumes V appi /V for a choice of n . Thehorizontal dashed lines represent the correct equal share for the corresponding n , i.e. 1 /n . III. THE EXPERIMENT
To challenge our findings and the underlying hypotheses, (i.e. the watermelon can be correctly approximated bya spheroid and the approximated cuts provide equal volumes), we performed the experiment and cut one fourth ofa watermelon into n (supposedly equal volume) portions. In line with the spirit of the note, we decided to use onlytools available in our kitchen [9, 10] so as to make our experiment easily reproducible by everyone. The experimenthas been filmed and a movie is available at [11].For the experiment we used a small watermelon weighting approximately 4 Kg. Of course the results are independentfrom the watermelon size. The first step is to measure the watermelon volume. To accomplish this task we make useof the Archimedes principle [12, 13]: a body immersed in water, will displace a volume of water equal to the volumeof the body. We cut the watermelon into two equal parts (longitudinal cut as shown in Fig. 2), we insert each part ina sufficiently large bowl and we fill it with water until a reference mark is reached. We then remove the watermelon:the volume of water in the bowl is hence lower (for the sake of completeness we refer the reader to the discussion atthe end of this section). To measure this volume we use a measuring cup to fill again the bowl. Once the water levelhas reached again the initial reference mark, the added volume of water is equal to the volume of the watermelon (SeeFig. 7). In our case we got ∼ . a) b) c)d) Fill the bowl with water Remove theWatermelonFill again the bowl with water e) f)The volume of the watermelon is the volume of added water FIG. 7:
Measure a volume by using the Archimedes principles . Panel a): insert in the bowl the watermelon whosevolume is to be measured. Panel b): fill the bowl with water up to a reference mark. Panel c): remove the watermelon. Panelsd)-e): to measure the drop in volume, fill again the bowl with water up to the reference mark. Panel f): the volume of thewatermelon equals the added volume of water.
We are now ready to prepare the n slices. To this end we cut the two half of watermelon into two parts each,yielding 4 quarters. The first quarter is sliced into n = 2 equal volume portions, from the second n = 3, from thethird n = 4 and finally n = 5 slices from the last one. In each case the positions of the cuts have been determined byusing the approximation λ ( i ) app a = a × / × i/n for i = 1 , . . . , n − / Slice number n = 2 n = 3 n = 4 n = 51 490 300 230 1702 510 290 220 1703 ∗
330 230 1604 ∗ ∗
240 1405 ∗ ∗ ∗
Volumes of the slices . We report the volumes of the watermelon slices (expressed in cm ) obtained in theexperiment. Each column corresponds to a choice of n , while the rows refer to the slice number. The symbol ∗ means that theslice is not allowed for the choice of n . During the experiment we could also obtain an accessory result: a measure of the density of the watermelon, ρ ,namely the quantity of matter contained in the (watermelon) volume. To do this we use a kitchen scale to weight thewatermelon. This first half of watermelon weights ∼ .
89 Kg for a volume of ∼ .
95 l; for the second half, ∼ .
73 Kgand a volume of ∼ .
81 l. The density results thus (we take the average of the two measures) ρ ∼
12 (1 .
89 Kg / .
95 l + 1 .
73 Kg / .
81 l) = 0 .
97 Kg / l = 0 .
97 g / cm . Observe that the density is smaller than 1 g / cm which can be assumed to be the density of the water at theexperimental conditions. Wa can conclude that the watermelon is less dense than water and thus it (almost) floats.Indeed the watermelon is composed for a large fraction by water and the weight of the few fibres (heavier than water)is compensated by the air (way lighter than water) trapped into the watermelon pulp. Let us observe that in [14]the density of the watermelon has been estimated to 0 .
94 g / cm , assuming a spherical watermelon and measuring itsbuoyancy. (1) app = 10 . ⇥ ⇥
12 = 3 . ( i ) app = 9 . ⇥ ⇥ i (1) app ⇠ . (2) app ⇠ . n = 2 n = 3 n = 4 ( i ) app = 10 . ⇥ ⇥ i (1) app ⇠ . (2) app ⇠ . (3) app ⇠ . ( i ) app = 9 . ⇥ ⇥ i (1) app ⇠ . n = 5 (2) app ⇠ . (3) app ⇠ . (4) app ⇠ . FIG. 8:
The approximated cuts realised in the experiment . We report the approximated cuts realised in the experimentusing the formula λ ( i ) app a = a × / × i/n , where a is the semi-major axis of the spheroid (the watermelon) for several values of n : n = 2 (top left panel), n = 3 (top right panel), n = 4 (bottom left panel) and n = 5 (bottom right panel). Let us conclude this section with an observation about the application of the Archimedes principles. As we haveseen the watermelon is slightly lighter than water and thus it can not be completely immersed; to measure the volumeof the watermelon part we are interested in, we have thus to force it into the water.
IV. CONCLUSION
We have provided a straightforward recipe to (transversally) cut a (spheroid) watermelon into an arbitrary numberof pieces with equal volume. Interestingly enough, after the first few cuts dictated by the symmetry, i.e. divide by 2along the semi-major and minor axes, a “new rule” of the thumb emerges where the ratio 1 / / × /n where n identifies the number of (equal volume) slices that one wants to recover.Let us conclude that the same result, and thus an identical rule, holds true for a spherical watermelon. In thelatter cases one could always invoke the spherical symmetry and thus cut slices with an angle of 2 π/n (sort of applewedges); however for large n it can be difficult correctly identify such thin angles, while cutting vertically is normallyeasier.The theory has been complemented with an experiment (a movie describing the experiment is also available [11]),realised by using tools available in our kitchen and thus hopefully reproducible in the students houses.As a final comment, notice that the relative distribution of (watermelon) skin has not been considered in thecomputation. The obtained equal volume slices display an uneven distribution of edible and non edible material.Accounting for this (i.e. producing equal volume slices of edible pulp) is in principle possible (as a straightforwardgeneralisation of the method proposed) at the price of a more cumbersome computation which goes beyond the scopeof this pedagogical note. [1] W. Wollman and F. Lawrenz, Identifying potential “dropouts” from college physics classes , Journal of Research in ScienceTeaching, R´enovation de l’enseignement des sciences physiques et formation des enseignants. Regards didactique ,(2013), De Boeck Sup´erieur (Louvain La Neuve, Belgium) ISBN : 9782804175443, DOI : 10.3917/dbu.boivin.2013.01 [InFrench][6] R. Taconis, M.G.M. Ferguson-Hessler and H. Broekkamp,
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