CUTCAKE BRIEF ABSTRACT Determine several regular ways to cut a circular cake in thirds. GENERAL INFORMATION FileName: CUTCAKE Full title: Consider how to cut a circular cake in three equal pieces. Three of the ways suggested use only two straight cuts. Last Revision Date: 5 June 1996. Developer: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Consider a circular cake of radius r = 20 cm and constant depth 8 cm. Determine how to cut the cake in three equal pieces. You should offer four approaches and at least three of these approaches should involve two straight cuts - (1) cuts parallel to each other, (2) cuts as secant lines meeting at one point, and (3) cuts meeting at the intersection of a "T" interior to the cake. KEYWORDS Area, geometry, creativity, art, integration, polar coordinates. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Some students may wish to consider the three dimensional geometry, i.e. cut through the cake at non-vertical cuts. We have not considered that and present this problem with the idea that vertical cuts will be used throughout. Prerequisites Integration in rectangular and polar coordinates, although the latter can be circumvented. Time allotment - time management Give this project time for doodling at home and sharing of ideas among students. Thus introduce it in class, permit individual sketching first for about 5-10 minutes, then small group discussions, and then assign for homework - be it in teams or individually Expectations We expect students to use their imagination and to produce mathematically and geometrically interesting cuts which lead to interesting analysis. Future payoffs Students will see an open ended problem and see that they have creative options. Extensions Consider other type regions such as the one below such that regions 1, 2, and 3 are equal. Or how about this one? Again regions 1, 2, and 3 are equal. And here is one in which the three regions are cut from the circle by shifted sine waves! What sine waves could be used? Can one consider three sections such that the OUTER perimeter (a tasty extra thick icing perhaps) is the same for all three sections. Can one consider three sections such that the TOTAL perimeter (a tasty extra thick icing perhaps) is the same for all three sections. Of course dividing the circle into three equal cuts (120 degrees each) through the center gives each participant equal area AND equal perimeter. Are there other "nice" cuts which do this, e.g., can any of our cuts do this? can they be modified to do this? One could also consider the problem of cutting the cake into say n (n > 3) equal pieces and prescribing some geometrical constraints. References and Sources Happy Birthday to you! Abdus Sattar Gazdar, Al-Khwarizmi Research Centre for Promotion of Mathematics Education, 44 North St., Frewville South Australia 5063 AUSTRALIA. The Australian Mathematics Teacher. 1995. 51(2): 16-18. POSSIBLE SOLUTION(S) We solve the problem for general circle of radius R and ignore the depth. (1) Cuts parallel to each other. Consider the above circle of radius r centered at the origin O = (0, 0) where the two parallel cuts are AB and CD. We note the center cut ABCD is exaggerated in size, but we do so for effect only. If q (theta) is the angle BOP then we shall determine the area of the center region by doubling the top half above the x-axis. Further, we consider symmetry and the upper region of the circle above line AB is presumed to be the same size as the lower region of the circle below CD. Thinking in a rectangular coordinate system we note that the line y = tan(q) x from O to B is significant as are the coordinates of the P = (r cos(q), r sin(q)) - here q is the angle theta in the diagram. Thus the area of middle section ABCD is two times the sum of the areas of the four congruent triangles BOP, EOB, AOE, AOQ and the two pieces BPM and AQN. In our analysis we use t for the angle q (theta) BOP = 1/2 r Cos[t] r Sin[t]; BPM = Integrate[Sqrt[r^2 - x^2],{x,r Cos[t],r}] 2 Pi r ----- - (r (r ArcTan[ 4 2 2 Cot[t] Csc[t] Sqrt[r Sin[t] ] ------------------------------] + r 2 2 Cos[t] Sqrt[r Sin[t] ])) / 2 MidPiece = 2 (4 BOP + 2 BPM) 2 2 (2 r Cos[t] Sin[t] + 2 Pi r 2 (----- - (r (r ArcTan[ 4 2 2 Cot[t] Csc[t] Sqrt[r Sin[t] ] ------------------------------] + r 2 2 Cos[t] Sqrt[r Sin[t] ])) / 2)) We simplify a bit and determine where this MidPiece area is 1/3 Pi r^2. MidPiece = Expand[MidPiece]/. Sqrt[r^2 Sin[t]^2]->r Sin[t] 2 2 2 Pi r - 2 r ArcTan[Cot[t]] + 2 r Cos[t] Sin[t] eq1 = MidPiece == 1/3 Pi r^2 2 2 2 Pi r - 2 r ArcTan[Cot[t]] + 2 r Cos[t] Sin[t] == 2 Pi r ----- 3 eq2 = Simplify[MidPiece/r^2 == 1/3 Pi r^2/r^2] Pi Pi - 2 ArcTan[Cot[t]] + Sin[2 t] == -- 3 sol = FindRoot[eq2,{t,1.0}] {t -> 0.268133} Thus the angle which we must use for q (theta) is 0.268 radians. tsol = t/.sol[[1]] 0.268133 We confirm the area of the MidPiece IS 1/3 Pi r^2 = 1.0472 r^2. MidPiece/.sol[[1]]//N 2 1.0472 r Pi r^2/3//N 2 1.0472 r And we check the top piece to see if it is one third the area of the circle, or 1/3 Pi r^2 = 1.0472 r^2. It is! TopThird = Integrate[Sqrt[r^2 - x^2] - r Sin[tsol], {x,- r Cos[tsol],r Cos[tsol]}] 2 2 -0.510931 r + 0.255465 r Sqrt[r ] + 2 2 3.63968 Sqrt[r ] r ArcTan[----------------] r TopThird/.{Sqrt[r^2]->r} 2 1.0472 r Hence we should make two horizontal cuts at heights TopCut and BottomCut respectively. TopCut = r Cos[tsol] 0.964267 r BottomCut = - r Cos[tsol] -0.964267 r These two cuts will give us a top, a middle, and a bottom piece each of 1/3 the total area of the cake. Of course we could have used polar integration to obtain the area of the sector of the circle BOM (or even the formula for the area of a sector of a circle) and then proceeded to add in the triangular areas. This leads to the same result. (2) Cuts as secant lines meeting at one point. We consider the polar function r = 2 R Cos[q] and require that the polar region BAC be 1/3 Pi R^2. We integrate from angle -a to angle a (where a = q/2). cut[a_] = Integrate[1/2 (2 R Cos[t])^2,{t,-a,a}] 2 R (2 a + Sin[2 a]) We determine the value of a to be 0.268 radians which will give 1/3 the area of the circle. eq3 = cut[a]/R^2 == 1/3 Pi R^2/R^2 Pi 2 a + Sin[2 a] == -- 3 sol = FindRoot[eq3,{a,1}] {a -> 0.268133} tsol = a/.sol[[1]] 0.268133 We check that the "wedge" (BAC) has area 1/3 area of the circle (1/3 Pi R^2 or 1.0472 R^2) - and it does! cut[tsol] 2 1.0472 R Finally, we check that the top piece above the wedge has area 1/3 the area of the circle and by symmetry the bottom piece also does. TopThird = Integrate[1/2 (2 R Cos[t])^2,{t,tsol, Pi/2}]//N 2 1.0472 R (3) Cuts meeting at the intersection of a "T" interior to the cake. Here we see from our diagram that the angle q (theta) is the variable to consider. We make the two line cuts AB and PC. We seek theta such that pieces BPC, APC, and APBD each have area 1/3 Pi R^2. Clearly theta will be less than Pi/2. BPC[t_] = (Pi - t)/(2 Pi) Pi R^2 + 1/2 R Cos[t] R Sin[t] 2 2 R (Pi - t) R Cos[t] Sin[t] ----------- + ---------------- 2 2 eq4 = Expand[BPC[t]/R^2] == 1/3 Pi R^2/R^2 Pi t Cos[t] Sin[t] Pi -- - - + ------------- == -- 2 2 2 3 sol = FindRoot[eq4,{t,.7}] {t -> 1.30266} tsol = t/.sol[[1]] 1.30266 And we check to see if the area of region BPC is 1/3 Pi R^2 = 1.047 R^2. It is! BPC[tsol]//N 2 1.0472 R And now for the top area above the T. We see it too is 1/3 Pi R^2. TopArea = Integrate[Sqrt[R^2 - x^2] - R Cos[tsol], {x, - R Sin[tsol], R Sin[tsol]}]//N 2 2 -0.510931 R + 0.255465 R Sqrt[R ] + 2 2 3.63968 Sqrt[R ] R ArcTan[----------------] R TopArea/.{Sqrt[R^2]->R} 2 1.0472 R Solution of tangential circles mentioned in the Extensions section. We could not resist! We have three concentric circles r1(t) = 2 R1 sin(t), r2(t) = 2 R2 sin(t), and r3(t) = 2 R sin(t) We want the regions 1, 2, and 3 to each be 1/3 Pi (2 R)^2. and since there are two variables, R1 and R2, we shall get two conditions in the two unknowns. Area1[R1_] = Integrate[(2 R1 Sin[t])^2,{t,0, Pi}] 2 2 Pi R1 Area2[R1_,R2_] = Integrate[(2 R2 Sin[t])^2,{t,0, Pi}] - Area1[R1] 2 2 -2 Pi R1 + 2 Pi R2 Area3[R2_] = Pi (2 R) ^2 - Area1[R1] - Area2[R1,R2] 2 2 4 Pi R - 2 Pi R2 And now for the solution of the equations. sol = Solve[{Area3[R2] == 1/3 Pi (2 R)^2, Area1[R1] == 1/3 Pi (2 R)^2}, {R1, R2}]//N {{R1 -> -0.816497 R, R2 -> -1.1547 R}, {R1 -> -0.816497 R, R2 -> 1.1547 R}, {R1 -> 0.816497 R, R2 -> -1.1547 R}, {R1 -> 0.816497 R, R2 -> 1.1547 R}} Hence we have the inner radii R1sol and R2sol. R1sol = R1/.sol[[4]] 0.816497 R R2sol = R2/.sol[[4]] 1.1547 R We examine the ratio and see it looks familiar!!! R2sol/R1sol 1.41421 So we get the exact radii and ratios to confirm our suspicion of the ratio being the square root of 2. This suggests that if we take the extensions to more circles the ratios would have interesting ratios. Try it and see!!! solA = Solve[{Area3[R2] == 1/3 Pi (2 R)^2, Area1[R1] == 1/3 Pi (2 R)^2}, {R1, R2}] General::spell1: Possible spelling error: new symbol name "solA" is similar to existing symbol "sol". 2 -2 R {{R1 -> -(Sqrt[-] R), R2 -> -------}, 3 Sqrt[3] 2 2 R {R1 -> -(Sqrt[-] R), R2 -> -------}, 3 Sqrt[3] 2 -2 R {R1 -> Sqrt[-] R, R2 -> -------}, 3 Sqrt[3] 2 2 R {R1 -> Sqrt[-] R, R2 -> -------}} 3 Sqrt[3] R2/R1/.solA[[4]] Sqrt[2] And we check to see the three areas are indeed 1/3 Pi (2R) R^2 = 4.18879 R^2. They are!!! Area1[R1sol]//N 2 4.18879 R Area2[R1sol,R2sol]//N 2 4.18879 R Area3[R2sol]//N 2 4.18879 R We plot a sample solution for visual effect. PolarPlot[{R/R Sin[t], R1sol/R Sin[t], R2sol/R Sin[t]}, {t,0, 2 Pi}] ISSUES IN SOLUTION Students may create some "interesting" geometrically shapes, but the need to use their mathematics to verify that this is in fact a cutting in thirds of the circle is imperative.