ClassRoom 55 54 Shailesh AShirali At Right Angles At Right Angles | |

Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 he ntsurs h hpssoni iue2aepossible. are 2 Figure called in are using shown They example, shapes For the vary. , to unit we squares three where unit notion, of this number of the generalisation allow natural a is A a get we 1. parts, Figure divide equal we two If into edge-to-edge. object squares this unit two joining a by of produced notion the with familiar are a We of generalisation natural a as Polyomino It Prove To How Keywords: Polyominoes, ,perimeter, parity, closedfigure make afewremarks aboutan openproblem. specific theseobjects,and results concerning In we thisarticle, considerandprove two had beenposedandstudiedaboutpolyominoes. “Low Floor High Ceiling” series),questions In aprevious ofthe issueofAtRiA (aspart tagttromino Straight trominoes iue1 ooioaddomino and Monomino 1. Figure Monomino iue2 h w trominoes two The 2. Figure

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hc sashape a is which domino, Domino L- See monomino. 1

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At Right Angles At Right Angles k for 55 54

ClassRoom 55 54 Shailesh AShirali At Right Angles At Right Angles | |

Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 rv It Prove To How he ntsurs h hpssoni iue2aepossible. are 2 Figure called in are using shown They example, shapes For the vary. squares, to unit we squares three where unit notion, of this number of the generalisation allow natural a is polyomino A a get we 1. parts, Figure divide equal we two If into edge-to-edge. object squares this unit two joining a by of produced notion the with familiar domino are a We of generalisation natural a as Polyomino Keywords: Polyominoes, polygon,perimeter, parity, closedfigure make afewremarks aboutan openproblem. specific theseobjects,and results concerning In we thisarticle, considerandprove two had beenposedandstudiedaboutpolyominoes. “Low Floor High Ceiling” series),questions In aprevious ofthe issueofAtRiA (aspart tagttromino Straight trominoes iue1 ooioaddomino and Monomino 1. Figure Monomino iue2 h w trominoes two The 2. Figure . hc sashape a is which domino, Domino L-tromino See monomino. 1

sw nraetenme fui qae,mr ope hpsbcm osbe o xml,w may 4. we Figure example, in For shown possible. one become the shapes as complex such more ‘holes’ squares, with unit shapes of get polyominoes, you number more about the [2].) For increase and [4]. we [1] ormoreequal also As in one See accounts [3]. readable joining in highly by the formed given to refer figure the definition may geometric is “aplane as (This apolyomino edge-to-edge.” squares define may we general, In figures. these of number ou ntonmesascae ihti bet t eiee,adisnme fsds oeta by that 4. Note exceeding sides. not of area number with its polyominoes and all perimeter, ofan its object: area this the with definition, associated numbers two on focus polyomino a An of sides of number and Perimeter For are They possible. are 3 Figure in shown shapes the edge-to-edge, called squares unit four joining by Similarly, nieyut itte.For them. list to you invite oioi ln emti iuefre byjoining formed figure geometric plane a is n-omino n = tetrominoes ra n Area, ,tesae r called are shapes the 5, 4 4 4 4 4 3 3 2 1 Straight h ae ie oteidvda hpsaeshownalongside. are shapes theindividual to given names The . oiois n-omino qaetetromino Square Z-tetromino T-tetromino L-tetromino tetromino Straight L-tromino tromino Straight Domino Monomino n = L ttrsotta hr r 2psil etmne.We pentominoes. possible 12 are there that out turns It pentominoes. ,tesae r called are shapes the 6, Shape n iue4 A 4. Figure qaeuis al it h perimeter the lists 1 Table units. square iue3 h ietetrominoes five The 3. Figure oiowt hole a with 7-omino al 1 Table T hexominoes eiee,P Perimeter, n qa qae det-de nti eto we section this In edge-to-edge. squares equal 10 10 10 8 8 8 8 6 4 syumyges hr r large a are there guess, may you As . Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 Z P n ubro sides of number and ubro ie,k sides, of Number Square 4 8 8 6 4 6 4 4 4 | |

At Right Angles At Right Angles k for 55 54 On examining the data, we see something noteworthy right away: P and k are even numbers in every case. + + + + + +

Will this be the case for polyominoes with larger numbers of sides? We shall show that the answer is Yes. − − − + + + The proofs we offer are very instructive, as they use the notionof parity. + + + − + + + + Proof that the perimeter P is even. The sides of the unit squares making up a polyomino give riseina (a) (8 90◦) (4 90◦)=360◦ (b) 8 90◦ = 720◦ natural way to two mutually perpendicular directions which we may regard as a pair of coordinate axes. × − × × Let these axes be drawn; the outer boundary of a polyomino is now entirely composed of segments of unit Figure 6. Total turning angle for a polyomino length, each parallel to the x-axis or the y-axis. From this reasoning, we deduce that 90(a b) is a multiple of 360, and therefore that a b is a multiple Now let us take a walk around the outer boundary of the polyomino, advancing in steps of unit length and − − of 4. Hence a b is an even number. Therefore, a + b =(a b)+2b is an even number as well. That is, marking a dot at each lattice point along our path. (We must fix a direction for our tour; let us assume that − − we walk in the counterclockwise direction.) The number of steps we take clearly equals the perimeter P of k is even. So the number of sides of the polyomino is an even number. the polyomino. With each step, our location changes in the following way: either the x-coordinate changes by 1 unit, or the y-coordinate changes by 1 unit; but not both at the same time. Hence if P1 and Unsolved problems P are two successive lattice points along the path, then P P is one of the following: 2 1 − 2 There is just 1 monomino, and just 1 domino; there are 2 trominoes, and 5 tetrominoes. Thesenumbers ( ) (1, 0), ( 1, 0), (0, 1), (0, 1). give rise to an interesting but difficult problem. For any positive integer n, let f n denote the number of − − different n-ominoes. Care is needed in interpreting the word ‘different.’ We regard two shapes as ‘the Let a, b, c, d be the respective number of steps of each of the above types, as we traverse the outer same’ if they are geometrically congruent to each other (‘congruent’ in the usual, Euclidean sense of that boundary. Then: word); and two shapes are ‘different’ if they are not congruent to one another. With this understanding, P = a + b + c + d. we find the following values taken by the function f.

After a full circuit, the total change in the x-coordinate must be 0; hence a = b. In the same way, we must n 1234 5 6 7 have c = d, because the total change in the y-coordinate must be 0. Hence: ··· f(n) 1 1 2 5 12 35 108 ··· P = 2a + 2c = 2(a + c). The 12 possible pentominoes are depicted in Figure 7. We will leave it to you to sketch the 35possible It follows that P is an even number. hexominoes. Or you may refer to [3] for some sketches. Figure 5 illustrates the meanings of the parameters a, b, c, d for the polyomino depicted in Figure 4, the path being described by the arrows.

← ←

↓ ↑ → ← a = 4, c = 4 • ↓ ↑ ↓ ↑ b = 4, d = 4 • ← ↓ ↑

→ → → Figure 5. Traversing the outer boundary of the 7-omino shown earlier

Figure 7. The 12 pentominoes Proof that the number of sides is even. Suppose that the polyomino is a k-sided polygon (or k-gon for Source: https://simple.wikipedia.org/wiki/Pentomino#/media/File:Pentominos.svg. short), with k vertices. (Note that the polygon need not be convex.) Let the vertices be P1, P2,...,Pk. As we traverse the outer boundary in the counterclockwise direction, let the angle through which we need to Now for the problems that these numbers suggest: (a) Given n, is there a simple way of computing the turn at vertex Pi be θi (angles measured in degrees); then θi = 90 for each i. Let a be the number of ( ) ( ) ± value of f n ? (b) Is there a simple formula for f n ? As of now, the answers to both questions appear to be vertices where we make a +90◦ turn, and let b be the number of vertices where we make a 90◦ turn; ( ) − No, and the only known way to compute f n for higher values of n is to use computer-assisted then k = a + b, and the total turning angle is 90(a b)◦. − enumeration, based on clever . Using such means, the sequence of values of f has been obtained Now, for any polygon, the total turning angle as we traverse the outer boundary is necessarily a multiple of to many terms (the first two values are f(1)=1 and f(2)=1): 360◦. In the case of a convex polygon, the total turning angle is exactly 360◦, but for with ± 1, 1, 2, 5, 12, 35, 108, 369, 1285, 4655, 17073, 63600, 238591, 901971, 3426576, 13079255, regions of non-convexity and/or holes, the total turning angle may be a higher multiple of 360◦. Figure 6 50107909, 192622052, 742624232, 2870671950, 11123060678, 43191857688, 168047007728, shows examples: (a) where the total turning angle is 360◦; (b) where the total turning angle is 720◦. 654999700403, 2557227044764, 9999088822075, 39153010938487, 153511100594603, ....

5657 At Right Angles | Vol. 5, No. 1, March 2016 Vol. 5, No. 1, March 2016 | At Right Angles 5657 On examining the data, we see something noteworthy right away: P and k are even numbers in every case. + + + + + +

Will this be the case for polyominoes with larger numbers of sides? We shall show that the answer is Yes. − − − + + + The proofs we offer are very instructive, as they use the notionof parity. + + + − + + + + Proof that the perimeter P is even. The sides of the unit squares making up a polyomino give riseina (a) (8 90◦) (4 90◦)=360◦ (b) 8 90◦ = 720◦ natural way to two mutually perpendicular directions which we may regard as a pair of coordinate axes. × − × × Let these axes be drawn; the outer boundary of a polyomino is now entirely composed of segments of unit Figure 6. Total turning angle for a polyomino length, each parallel to the x-axis or the y-axis. From this reasoning, we deduce that 90(a b) is a multiple of 360, and therefore that a b is a multiple Now let us take a walk around the outer boundary of the polyomino, advancing in steps of unit length and − − of 4. Hence a b is an even number. Therefore, a + b =(a b)+2b is an even number as well. That is, marking a dot at each lattice point along our path. (We must fix a direction for our tour; let us assume that − − we walk in the counterclockwise direction.) The number of steps we take clearly equals the perimeter P of k is even. So the number of sides of the polyomino is an even number. the polyomino. With each step, our location changes in the following way: either the x-coordinate changes by 1 unit, or the y-coordinate changes by 1 unit; but not both at the same time. Hence if P1 and Unsolved problems P are two successive lattice points along the path, then P P is one of the following: 2 1 − 2 There is just 1 monomino, and just 1 domino; there are 2 trominoes, and 5 tetrominoes. Thesenumbers ( ) (1, 0), ( 1, 0), (0, 1), (0, 1). give rise to an interesting but difficult problem. For any positive integer n, let f n denote the number of − − different n-ominoes. Care is needed in interpreting the word ‘different.’ We regard two shapes as ‘the Let a, b, c, d be the respective number of steps of each of the above types, as we traverse the outer same’ if they are geometrically congruent to each other (‘congruent’ in the usual, Euclidean sense of that boundary. Then: word); and two shapes are ‘different’ if they are not congruent to one another. With this understanding, P = a + b + c + d. we find the following values taken by the function f.

After a full circuit, the total change in the x-coordinate must be 0; hence a = b. In the same way, we must n 1234 5 6 7 have c = d, because the total change in the y-coordinate must be 0. Hence: ··· f(n) 1 1 2 5 12 35 108 ··· P = 2a + 2c = 2(a + c). The 12 possible pentominoes are depicted in Figure 7. We will leave it to you to sketch the 35possible It follows that P is an even number. hexominoes. Or you may refer to [3] for some sketches. Figure 5 illustrates the meanings of the parameters a, b, c, d for the polyomino depicted in Figure 4, the path being described by the arrows.

← ←

↓ ↑ → ← a = 4, c = 4 • ↓ ↑ ↓ ↑ b = 4, d = 4 • ← ↓ ↑

→ → → Figure 5. Traversing the outer boundary of the 7-omino shown earlier

Figure 7. The 12 pentominoes Proof that the number of sides is even. Suppose that the polyomino is a k-sided polygon (or k-gon for Source: https://simple.wikipedia.org/wiki/Pentomino#/media/File:Pentominos.svg. short), with k vertices. (Note that the polygon need not be convex.) Let the vertices be P1, P2,...,Pk. As we traverse the outer boundary in the counterclockwise direction, let the angle through which we need to Now for the problems that these numbers suggest: (a) Given n, is there a simple way of computing the turn at vertex Pi be θi (angles measured in degrees); then θi = 90 for each i. Let a be the number of ( ) ( ) ± value of f n ? (b) Is there a simple formula for f n ? As of now, the answers to both questions appear to be vertices where we make a +90◦ turn, and let b be the number of vertices where we make a 90◦ turn; ( ) − No, and the only known way to compute f n for higher values of n is to use computer-assisted then k = a + b, and the total turning angle is 90(a b)◦. − enumeration, based on clever algorithms. Using such means, the sequence of values of f has been obtained Now, for any polygon, the total turning angle as we traverse the outer boundary is necessarily a multiple of to many terms (the first two values are f(1)=1 and f(2)=1): 360◦. In the case of a convex polygon, the total turning angle is exactly 360◦, but for polygons with ± 1, 1, 2, 5, 12, 35, 108, 369, 1285, 4655, 17073, 63600, 238591, 901971, 3426576, 13079255, regions of non-convexity and/or holes, the total turning angle may be a higher multiple of 360◦. Figure 6 50107909, 192622052, 742624232, 2870671950, 11123060678, 43191857688, 168047007728, shows examples: (a) where the total turning angle is 360◦; (b) where the total turning angle is 720◦. 654999700403, 2557227044764, 9999088822075, 39153010938487, 153511100594603, ....

5657 At Right Angles | Vol. 5, No. 1, March 2016 Vol. 5, No. 1, March 2016 | At Right Angles 5657

58 58

59 59 4. 4. References 3. 2. 1. thepresentmoment. 3 at that [4]) unanswered and to remain [1] (see shown kindcontinue been their has of it However, andothers questions two These 3. 2. 1. thepresentmoment. 3 at that [4]) unanswered and to remain [1] (see shown kindcontinue been their has of it However, andothers questions two These At Right Angles At Right Angles 69720375229712477164533808935312303556800 69720375229712477164533808935312303556800 At Right Angles At Right Angles . isthesmallestpositiveintegerthatexactlydivisiblebyeachnumberfrom1to100? Polyomino. Polyomino. ade,M PloiosadFutFe etnls”C.1 nMri ade’ e ahmtclDvrin rmScientific 1966. Polyomino. from 150-161, Diversions pp. Mathematical Schuster, New Gardner’s and Martin Simon York: in New 13 American. Ch. .” Fault-Free and “Polyominoes M. Gardner, al .W .adCxtr .S M. S. H. Coxeter, and R. W. W. Ball, ade,M PloiosadFutFe etnls”C.1 nMri ade’ e ahmtclDvrin rmScientific 1966. Polyomino. from 150-161, Diversions pp. Mathematical Schuster, New Gardner’s and Martin Simon York: in New 13 American. Ch. Rectangles.” Fault-Free and “Polyominoes M. Gardner, al .W .adCxtr .S M. S. H. Coxeter, and R. W. W. Ball, If anyofyourstudentscomeupwithawell-thought-out response,doshareitwithus. See: https://www.facebook.com/photo.php?fbid=10206841050456928&set= http://mathworld.wolfram.com/Polyomino.html http://mathworld.wolfram.com/Polyomino.html https://en.wikipedia.org/wiki/Polyomino https://en.wikipedia.org/wiki/Polyomino — Adapted fromMoloyDe’s Facebookpage“MathBelieveItorNot”. | |

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Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 Which isthesmallestnumberwhichdoes for movement inIndia. He asaneditor istheauthorofmanymathematicsbooksforhighschoolstudents,andserves Centre inRishi Valley School(AP).He hasbeencloselyinvolved withtheMath Olympiad SHAILESH SHIRALI Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 At RightAngles. He. He [email protected]. [email protected]. Would youbelievethat. Would youbelievethat. In howmanywayscanyouarriveatthisnumber? Our [email protected]. gm.992345247471454&type=3&theater. 3he.NwYr:Dvr p 0-1,1987 109-113, pp. Dover, York: New ed. 13th Essays, and Recreations Mathematical 3he.NwYr:Dvr p 0-1,1987 109-113, pp. Dover, York: New ed. 13th Essays, and Recreations Mathematical isDirector andPrincipal ofSahyadri School(KFI),Pune, andHead andHead oftheCommunity . 72 n < f ( n n ) < not dividethisnumber? 4 . . 65 65 n o l oiieintegers positive all for n.

exploration Keywords: Surd, irrational, telescopingsum,arithmeticprogression, exploration Keywords: Surd, irrational, telescopingsum,arithmeticprogression, ti iet e nisac firtoa ubr digu oa to up adding numbers irrational of instance number! an rational see to nice is It problem. following the in involving lies vignette origin classroom Its a present surds. we note short this In Surds with Exploration An ti iet e nisac firtoa ubr digu oa to up adding numbers irrational of instance number! an rational see to nice is It problem. following the in involving lies vignette origin classroom Its a present surds. we note short this In Surds with Exploration An eosretefloigatrteuulse frtoaiigthe rationalising of step usual the denominators: after following the observe We eosretefloigatrteuulse frtoaiigthe rationalising of step usual the denominators: after following the observe We Problem. Problem. 4 4 ======+ + 4 4 4 4 4 4 16 16 1 1 2 2 − − − − − − 4 4 √ √ − − 2 3 2 2 3 2 2 2 4 4 ipiytefloigexpression: following the Simplify ipiytefloigexpression: following the Simplify √ √ √ √ 12 12 + + = = + + 3 3 3 3 1 1 2 2 + + 2 1 2 1 2 2 + + √ √ √ √ . . 2 2 3 2 3 2 √ √ 3 3 √ √ + + 1 1 12 12 3 3 + + 3 3 2 2 − − 4 4 − − √ √ − − 2 2 2 2 2 2 2 2 √ √ 8 8 √ √ √ √ + + 2 2 3 3 2 2 Nt h eecpccancellation.) telescopic the (Note Nt h eecpccancellation.) telescopic the (Note + + 2 2 + 1 1 + + √ √ 2 2 2 2 2 2 2 √ √ 1 1 √ √ √ √ 8 8 + + 2 2 2 2 − − 2 2 − − 2 2 − − 4 4 + + 2 2 2 2 2 2 √ √ 2 2 1 1 + + 2 2 . 1 1 (1) (1) Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 Vol. 5,No.1,March 2016 Bharat Karmarkar Bharat Karmarkar | |

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At Right Angles At Right Angles At Right Angles At Right Angles 58 58 59 59 Class Room