Recreational Mathematics

Transcription

1 Recreational Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 2009 Chapters 1 40 Monday, July 31 Monday 6/22 6/29 7/6 7/13 7/20 7/27 8/3 Wednesday 6/24 7/1 7/8 7/15 7/22 7/29 Friday 6/26 *** 7/10 7/17 7/24 7/31

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3 Contents 1 Representation of a number in different bases Base b-representation of a number A number from its base b-representation Balanced division by an odd number Balanced base b representation of a number Arithmetic in balanced base 3 representations The Josephus problem and its generalization The Josephus problem The generalized Josephus problem J(n, k) The nim game The nim sum The game Nim Some number games The game of battle and other subtraction games Wythoff s game Greatest common divisor gcd(a,b) as an integer combination of a and b The two-jug problem Fibonacci numbers Fibonacci numbers Zeckendorf representation Pythagorean triangles Primitive Pythagorean triples Rational angles

4 iv CONTENTS Some basic properties of primitive Pythagorean triples A Pythagorean triangle with an inscribed square When are x 2 px ± q both factorable? Dissection of a square into Pythagorean triangles Points at integer distances from the sides of a primitive Pythagorean triangle triangles in the square The regular pentagon The golden ratio The diagonal-side ratio of a regular pentagon Construction of a regular pentagon with a given diagonal Construction of 36, 54, and 72 angles Integer triangles with a 60 or 120 angle Integer triangles with a 60 angle Integer triangles with a 120 angle Isosceles triangles equal in perimeter and area Cheney s card trick Three basic principles The pigeonhole principle Arithmetic modulo Permutations of three objects A variation of Cheney s card trick A matrix card trick A 3-card trick Basic geometric constructions Some basic construction principles Geometric mean Harmonic mean A.M G.M. H.M Equal subdivisions of a segment

5 CONTENTS v 17 Angle Bisectors Angle bisector theorem The angle bisectors Steiner-Lehmus Theorem Two problems on median, altitude, and bisector AMM A related problem Construction of a triangle from three given lengths The problem of three angle bisectors The classical triangle centers The centroid The circumcircle and the circumcenter The incenter and the incircle The orthocenter and the Euler line The excenters and the excircles Construction of a triangle from three given points Some examples Wernick s construction problems The nine-point circle The nine-point circle Feuerbach s theorem Triangles with nine-point center on the circumcircle Connelly s construction problems The area of a triangle Heron s formula for the area of a triangle Heron triangles Heron triangles with sides < Heron triangles with consecutive sides Heron triangles with sides in arithmetic progression Heron triangles with integer medians Heron triangles and incircles Enumeration of Heron triangles by inradii Inradii in arithmetic progression

6 vi CONTENTS 23.3 Division of a triangle into two subtriangles with equal incircles More about Heron triangles Heron triangle as lattice triangle Heron triangles with square areas Figurate numbers Triangular numbers Palindromic triangular numbers Pentagonal numbers Palindromic pentagonal numbers The polygonal numbers P k,n Which triangular numbers are squares? Sums of consecutive squares Sum of squares of natural numbers Sums of powers of consecutive integers Counting triangles An equilateral triangle cut up into smaller equilateral triangles Scalene triangles of sidelengths n Number of integer triangles with perimeter n Polygonal triples Double ruling of S Primitive Pythagorean triple associated with a k-gonal triple Triples of triangular numbers k-gonal triples determined by a Pythagorean triple Maxima and minima without calculus Infinite Series Power series The Bernoulli numbers and the tangent function The Euler numbers and the secant function Rearrangements of conditionally convergent series

7 CONTENTS vii 31 Some circle constructions Circle in circular segment Circles tangent to two given tangent circles First solution Second solution Neighbours of circle in a circular segment Neighbors of circle between two tangent circles The Arbelos Archimedes twin circle theorem Incircle of the arbelos Construction of incircle of arbelos Archimedean circles in the arbelos Constructions of the incircle Routh and Ceva theorems Barycentric coordinates Cevian and traces Area and barycentric coordinates Prime and perfect numbers Infinitude of prime numbers Euclid s proof Fermat numbers The prime numbers below Primes in arithmetic progression The prime number spirals The prime number spiral beginning with The prime number spiral beginning with Perfect numbers Charles Twigg on the first 10 perfect numbers Mersenne primes Some division problems AMM E AMM E AMM E AMM E

8 viii CONTENTS 36 Transposable integers k-right-transposable integers k-left-transposable integers Permutations The universal sequence of permutations The position of a permutation in the universal sequence Cycle decompositions The disjoint cycle decomposition of a permutation Dudeney s puzzle Dudeney s Canterbury puzzle Card tricks from permutations Digital trivia 1315

9 Chapter 1 Representation of a number in different bases 1.1 Base b-representation of a number Let b be a fixed positive integer. To write an integer n in base b, keep on dividing the number by b until the quotient is 0, and record the remainders, which are integers between 0 and b 1, from right to left. The resulting sequence, which always begins with a nonzero leftmost digit, is the representation of n in base b. Example 1.1. (a) 123 = [ ] 2 divisor quotient remainder

10 102 Representation of a number in different bases (b) 123 = [11120] 3 divisor quotient remainder A number from its base b-representation Given the base b representation of an integer: n = [a 0 a 1 a k 1 a k ] b, to find the integer in its decimal form, calculate a sequence of numbers n 0, n 1,..., n k as follows: (i) n 0 = a 0, (ii) for i = 1, 2,...,k, n i = a i + b n i 1. Then, n = n k. Example 1.2. (a) [ ] 2 = 123. (b) [23147] 11 = Example 1.3. Ask your friend to think of a number between 1 and 31. Then ask her if the number appears in card (a), (b), (c), (d), or (e). Now add up the numbers from the lower left hand corners of the card which she answers yes. That is the number she has chosen.

11 1.1 Base b-representation of a number (a) (b) (c) (d) (e)

12 104 Representation of a number in different bases Exercise 1. Complete the multiplication table in base Multiply in base 7: [12346] 7 [06] 7 = [12346] 7 [15] 7 = [12346] 7 [24] 7 = [12346] 7 [33] 7 = [12346] 7 [42] 7 = [12346] 7 [51] 7 = 3. Ask your friend to write down a polynomial f(x) with nonnegative integer coefficients. Ask her for the value of f(1). She returns 7. Ask her for the value of f(8). She returns What is the polynomial? 1.2 Balanced division by an odd number Given a positive integer b and a string of b consecutive integers, for every integer a, there are unique integers q and r so that a = bq + r with r in the given string of b consecutive integers. In particular, if b is odd, say b = 2c + 1, we may choose the string of remainders to be c, (c 1),..., 1, 0, 1,..., c 1, c.

13 1.3 Balanced base b representation of a number 105 If we write a = bq + r, c r c, we say that this is the balanced division of a by the odd number b. 1.3 Balanced base b representation of a number Let b be a positive odd number. The balanced base b representation of an integer n is obtained by repeatedly performing balanced divisions by b and recording, from right to left, the remainders, which are integers in the range b 1 2,, 1, 0, 1,..., b 1 2 in place of the negative remainders 1, 2, 3, 4,... 1, 2, 3, 4,.... We shall write Examples (1) 100 = divisor quotient remainder The balanced ternary form of a number expresses it as a sum and/or differences of distinct powers of 3: 100 = Here is a simple application of the balanced ternary expansion of numbers.

14 106 Representation of a number in different bases Example 1.4. Suppose we have a beam balance and a set of seven standard weights 1, 2, 4, 8, 16, 32, 64 units. It is possible to weigh every integer units from 1 to 127 with these seven standard weights. For example, since = , balance is achieved by putting a weight of 100 units on one pan, and standard weights of 4, 32 and 64 units on the other. However, with a set of five standard weights 1, 3, 9, 27, 81, it is possible to weigh every integer units from 1 to 121. For example, since = 11101, a weight of 100 units can be balanced by putting it with a standard weight 9 units on one pan and the standard weights 1, 27, 81 on the other pan. 1.4 Arithmetic in balanced base 3 representations Example =

15 Chapter 2 The Josephus problem and its generalization 2.1 The Josephus problem There are n people forming a circle. In succession, every second person is escorted out from the circle, and the last one is awarded a prize. Who is the prize winner? Examples (1) n = 10: 2 6 * (2) n = 21. After the removal of the 10 even numbered ones and then

16 108 The Josephus problem and its generalization the first, there are the 10 odd numbers 3, 5,..., 19, 21. The survivor is the 5-th of this list, which is 11. Theorem 2.1. Let J(n) be the prize winner in the Josephus problem for n people. Example 2.1. J(2n) =2J(n) 1, J(2n + 1) =2J(n) + 1. J(100) =2J(50) 1 =2(2J(25) 1) 1 = 4J(25) 3 =4(2J(12) + 1) 3 = 8J(12) + 1 =8(2J(6) 1) + 1 = 16J(6) 7 =16(2J(3) 1) 7 = 32J(3) 23 =32(2J(1) + 1) 23 = 64J(1) + 9 =73. There is an almost explicit expression for J(n): if 2 m is the largest power of 2 n, then J(n) = 2(n 2 m ) + 1. Corollary 2.2. The binary expansion of J(n) is obtained by transferring the leftmost digit 1 of the binary expansion of n to the rightmost. J(100) = J([ ] 2 ) = [ ] 2 = = 73. Exercise 1. For what values of n is J(n) = n? 2. For what values of n is J(n) = n 1?

17 2.2 The generalized Josephus problem J(n,k) The generalized Josephus problem J(n, k) The generalized Josephus problem J(n, k) asks for the prize winner J(n, k) when beginning with 1, every k-th member (counting cyclically) is escorted out from the circle originally formed by n people. Example 2.2. J(10, 3): J(10, 3) = 4. * For n = 10, here are the sequences of elimination depending on the values of k. The last column gives the prize winners. k J(10, k) Positions 2 and 6 are no-luck positions for the Josephus problem of 10 people and random choice of k.

18 110 The Josephus problem and its generalization Exercise 1. Make a list of the no-luck positions of the Josephus problem for n = 4, 5,...,9. 2. For n = 7, there is only one no-luck position 1. This means that one other position is most likely to receive a prize? Which one is it?

19 Chapter 3 The nim game 3.1 The nim sum The nim sum of two nonnegative integers is the addition in their base 2 notations without carries. If we write 0 0 = 0, 0 1 = 1 0 = 1, 1 1 = 0, then in terms of the base 2 expansions of a and b, a b = (a 1 a 2 a n ) (b 1 b 2 b n ) = (a 1 b 1 )(a 2 b 2 ) (a n b n ). The nim sum is associative, commutative, and has 0 as identity element. In particular, a a = 0 for every natural number a. Example 3.1. (a) 6 5 = [110] 2 [101] 2 = [011] 2 = 3. (b) = [100010] 2 [111001] 2 = [011011] 2 = 27. Here are the nim sums of numbers 15:

20 112 The nim game The game Nim Given three piles of marbles, with a, b, c marbles respectively, players A and B alternately remove a positive amount of marbles from any pile. The player who makes the last move wins. Theorem 3.1. In the game nim, the player who can balance the nim sum equation has a winning strategy. Therefore, provided that the initial position (a, b, c) does not satisfy a b c = 0, the first player has a winning strategy. For example, suppose the initial position is (12, 7, 9). Since 12 9 = 5, the first player can remove 2 marbles from the second pile to maintain a balance of the nim sum equation = 0, thereby securing a winning position. Remarks. (1) This theorem indeed generalizes to an arbitrary number of piles. (2) The Missère Nim game: Suppose now we change the rule: in the Nim game, the player who makes the last move loses. Here is a winning strategy: Play as for ordinary Nim, until you can move to a position in which all piles have just one marble.

21 3.2 The game Nim 113 Exercise In each of the following nime games, it is your turn to move. How would you ensure a winning position? (a) 3, 5, 7 marbles. 1 (b) 9, 10, 12 marbles. 2 (c) 1, 8, 9 marbles. 3 (d) 1, 10, 12 marbles. 4 1 Take one marble from the pile with 7. 2 It is possible to remove an appropriate number of marbles from any pile. 3 This is a losing position. 4 Take one from the pile with 12.

22 Chapter 4 Some number games 4.1 The game of battle and other subtraction games Starting with a given positive integer N, two players alternately subtract a positive amount less than a given positive number d < N. The one who gets down to 0 wins. Theorem 4.1. The player who secures a multiple of d has a winning strategy. Two more subtraction games: (1) Subtraction of square numbers. Beginning with a given number N, two players alternately subtract a positive square number. The one who gets 0 wins. (2) Subtraction of primes. Beginning with a given number N, two players alternately subtract a prime number. The one who gets 0 wins. 4.2 Wythoff s game Wythoff s game is a variant of Nim. Given two piles of marbles, a player either removes an arbitrary positive amount of marbles from any one pile, or an equal (positive) amount of marbles from both piles. The player who makes the last move wins. We describe the position of the game by the amounts of marbles in the two piles. If you can make (2, 1), then you will surely win no matter how your opponent moves. Now, to forbid your opponent to get to this position, you should occupy (3, 5).

23 202 Some number games The sequence of winning positions: starting with (a 1,b 1 ) = (1, 2), construct (a k,b k ) by setting a k := min{c : c > a i,b i, i < k}, b k :=a k + k. Here are the 18 smallest winning positions for Wythoff s game: Here is a succinct description of the Wythoff sequence. Theorem 4.2. The winning positions of Wythoff s game are the pairs ( nϕ, nϕ 2 ), where ϕ = is the golden ratio. Appendix: Beatty s Theorem Theorem 4.3 (Beatty). If α and β are positive irrational numbers satisfying = 1, then the sequences α β and α, 2α, 3α,... β, 2β, 3β,... form a partition of the sequence of positive integers. Proof. (1) If an integer q appears in both sequences, then there are integers h and k such that From these, Combining these, we have q < hα < q + 1, q < kβ < q + 1. h q + 1 < 1 α < h q, k q + 1 < 1 β < k q. h + k q + 1 < 1 < h + k, q

24 4.2 Wythoff s game 203 and q < h + k < q + 1, an impossibility. This shows that an integer q can appear in at most one of the sequences. (2) Now suppose an integer q does not appear in any of these sequence. Then there are integers h and k such that From these, Combining these, we have (h 1)α < q < q + 1 < hα, (k 1)β < q < q + 1 < kα. h q + 1 > 1 α > h 1 q k q + 1 > 1 β > k 1. q h + k q + 1 > 1 > h + k 2, q and h + k > q + 1 > q > h + k 2, an impossibility. This shows that every integer q appears in at least one of the sequences. From (1) and (2) we conclude that every integer appears in exactly one of the sequences.

25 Chapter 5 Greatest common divisor 5.1 gcd(a, b) as an integer combination of a and b It is well known that the gcd of two (positive) integers a and b can be calculated efficiently by repeated divisions. Assume a > b. We form two sequences r k and q k as follows. Beginning with r 1 = a and r 0 = b, for k 0, let q k = rk 1 r k, r k+1 = mod(r k 1,r k ) := r k 1 q k r k. These divisions eventually terminate when some r n divides r n 1. In that case, gcd(a,b) = r n. If, along with these divisions, we introduce two more sequences (x k ) and (y k ) with the same rule but specific initial values, namely, x k+1 =x k 1 q k x k, x 1 = 1, x 0 = 0; y k+1 =y k 1 q k y k, y 1 = 0, y 0 = 1. then we obtain gcd(a,b) as an integer combination of a and b: 1 gcd(a,b) = r n = ax n + by n. 1 In each of these steps, r k = ax k + by k.

26 206 Greatest common divisor k r k q k x k y k 1 a b a a a b b b r 1 x 1 y 1. n 1 r n 1 q n 1 x n 1 y n 1 n r n q n x n y n n It can be proved that x n < b and y n < a. Theorem 5.1. Given relatively prime integers a > b, there are unique integers h, k < a such that ak bh = 1. Proof. Clearly, x n and y n are opposite in sign. Take (k,h) = (x n, y n ) or (b + x n,a y n ) according as x n > 0 or < 0. Corollary 5.2. Let p be a prime number. For every integer a not divisible by p, there exists a positive integer b < p such that ab 1 is divisible by p. Exercise 1. Find the gcd of the following pair of numbers by completing the second column of each table. Express the gcd as an integer combination of the given numbers by completing the last two columns. r k q k x k y k Somebody received a check, calling for a certain amount of money in dollars and cents. When he went to cash the check, the teller

27 5.2 The two-jug problem 207 made a mistake and paid him the amount which was written as cents, in dollars, and vice versa. Later, after spending $ 3.50, he suddenly realized that he had twice the amount of the money the check called for. What was the amount on the check? 5.2 The two-jug problem There are two jugs S and L, of capacity a and b units. We assume a and b relatively prime, with a < b. Suppose there are x units of water in one of the jugs. How can one run tap water to measure exactly y units for a given arbitrary y? It is enough to consider x,y < a. For each integer k, denote by k a the unique integer satisfying k = a q + k a, 0 k a < a. Proposition 5.3. For every integer x, the sequence (x + kb) a, k = 0, 1, 2...,a 1, is a permutation of 0, 1,..., a 1; similarly if b is replaced by b. Let x be an nonnegative integer < a. (1) Suppose the smaller jug S contains exactly x units of water. Fill L and from it fill S. Now, L has b (a x) = x a + b units. By emptying L into S we are finally left with (x + b) a units in L, which we may transfer into S. We call this operation p. (2) Now, suppose there are x units in the larger jug L. Repeatedly use S to fill L. When the latter is full, say, after t times, the smaller jug S contains ta (b x) = x b + ta = (x b) a units of water, which we may transfer back into L. We call this operation q. Note that p and q are inverses; they undo one another. Repeated applications of p yield the sequence (x+kb) a. On the other hand, repeated applications of q yield (x kb) a. By the proposition above, within a steps, any desired (integer units < a) of water can be measured. Example 5.1. Suppose a = 11, b = 19, and x = 3. How can we measure 7 units? (1) The equation 7 = (3+19k) 11 = (3+8k) 11 ; 8k+3 7 (mod 11); 8k 4 (mod 11); 2k 1 (mod 11). Clearly, k = 6. (2) One the other hand, the equation 7 = (3 19k) 11 has solution k = 11 6 = 5. A sequence of 5 operations q will do the job.

28 Chapter 6 Fibonacci numbers 6.1 Fibonacci numbers The Fibonacci numbers F n are defined recursively by F n+1 = F n + F n 1, F 0 = 0, F 1 = 1. The first few Fibonacci numbers are n F n Here are some basic facts about Fibonacci numbers. 1. gcd(f m, F n ) = F gcd(m,n). 2. F n+1 F n 1 F 2 n = ( 1) n. 3. F 1 + F F n = F n+2 1. Example 6.1. The conversion from miles into kilometers can be neatly expressed by the Fibonacci numbers. miles kilometers

29 210 Fibonacci numbers How far does this go? Taking 1 meter as inches, what is the largest n for which F n miles can be approximated by F n+1 kilometers, correct to the nearest whole number? Zeckendorf representation Every positive integer n can be expressed uniquely as a sum of distinct Fibonacci numbers n = F k1 + F k2 + + F kr, with k 1 >> k 2 >> >> k r >> 0, where by k >> h we mean k h + 2. For example, 123 = , 500 = , 600 = , 700 = , 800 = , 900 = , 1000 = Two players compete in the following game: There is a pile containing n chips. The first player removes any number of chips except that he cannot take the whole pile. From then on, the players take turns, each removing one or more chips but not more than twice as many chips as the preceding player has taken. The player who removes the last chip wins. Here is a winning strategy for the first player. Let µ(n) be the smallest Fibonacci number summand in the Zeckendorf representation of n. There is a winning move if and only if n is not a Fibonacci number, and the winning move is to remove the whole pile if permissible, otherwise exactly µ(n) chips. 1 Answer: n = 11, 89 miles 144 km; but 144 miles km. The next Fibonacci number is 233. However, on the reverse side, 233 km can be taken as 144 miles.

30 Chapter 7 Pythagorean triangles 7.1 Primitive Pythagorean triples A Pythagorean triangle is one whose sidelengths are integers. An easy way to construct Pythagorean triangles is to take two distinct positive integers m > n and form (a,b,c) = (2mn, m 2 n 2, m 2 + n 2 ). Then, a 2 + b 2 = c 2. We call such a triple (a,b,c) a Pythagorean triple. The Pythagorean triangle has perimeter p = 2m(m + n) and area A = mn(m 2 n 2 ). B m 2 + n 2 2mn A m 2 n 2 C A Pythagorean triple (a,b,c) is primitive if a, b, c do not have a common divisor (greater than 1). Every primitive Pythagorean triple is constructed by choosing m, n to be relatively prime and of opposite parity.

31 302 Pythagorean triangles Rational angles The (acute) angles of a primitive Pythagorean triangle are called rational angles, since their trigonometric ratios are all rational. A sin cos tan 2mn m 2 n 2 m 2 +n 2 m 2 +n 2 B m2 n 2 m 2 +n 2 2mn m 2 n 2 2mn m 2 n 2 m 2 +n 2 2mn More basic than these is the fact that tan A 2 and tan B 2 are rational: tan A 2 = n m, tan B 2 = m n m + n. This is easily seen from the following diagram showning the incircle of the right triangle, which has r = (m n)n. B m(m n) (m + n)n I r r (m + n)n (m n)n A m(m n) (m n)n C Some basic properties of primitive Pythagorean triples 1. Exactly one leg is even. 2. Exactly one leg is divisible by Exactly one side is divisible by The area is divisible by 6.

32 7.1 Primitive Pythagorean triples 303 Theorem 7.1 (Fermat). The area of a Pythagorean triangle cannot be a square (number). Appendix: Primitive Pythagorean triples < 1000 m, n a, b, c m, n a, b, c m, n a, b, c m, n a, b, c 2, 1 3, 4, 5 3, 2 5, 12, 13 4, 1 15, 8, 17 4, 3 7, 24, 25 5, 2 21, 20, 29 5, 4 9, 40, 41 6, 1 35, 12, 37 6, 5 11, 60, 61 7, 2 45, 28, 53 7, 4 33, 56, 65 7, 6 13, 84, 85 8, 1 63, 16, 65 8, 3 55, 48, 73 8, 5 39, 80, 89 8, 7 15, 112, 113 9, 2 77, 36, 85 9, 4 65, 72, 97 9, 8 17, 144, , 1 99, 20, , 3 91, 60, , 7 51, 140, , 9 19, 180, , 2 117, 44, , 4 105, 88, , 6 85, 132, , 8 57, 176, , 10 21, 220, , 1 143, 24, , 5 119, 120, , 7 95, 168, , 11 23, 264, , 2 165, 52, , 4 153, 104, , 6 133, 156, , 8 105, 208, , 10 69, 260, , 12 25, 312, , 1 195, 28, , 3 187, 84, , 5 171, 140, , 9 115, 252, , 11 75, 308, , 13 27, 364, , 2 221, 60, , 4 209, 120, , 8 161, 240, , 14 29, 420, , 1 255, 32, , 3 247, 96, , 5 231, 160, , 7 207, 224, , 9 175, 288, , , 352, , 13 87, 416, , 15 31, 480, , 2 285, 68, , 4 273, 136, , 6 253, 204, , 8 225, 272, , , 340, , , 408, , 14 93, 476, , 16 33, 544, , 1 323, 36, , 5 299, 180, , 7 275, 252, , , 396, , , 468, , 17 35, 612, , 2 357, 76, , 4 345, 152, , 6 325, 228, , 8 297, 304, , , 380, , , 456, , , 532, , , 608, , 18 37, 684, , 1 399, 40, , 3 391, 120, , 7 351, 280, , 9 319, 360, , , 440, , , 520, , , 680, , 19 39, 760, , 2 437, 84, , 4 425, 168, , 8 377, 336, , , 420, , , 672, , 20 41, 840, , 1 483, 44, , 3 475, 132, , 5 459, 220, , 7 435, 308, , 9 403, 396, , , 572, , , 660, , , 748, , , 836, , 21 43, 924, , 2 525, 92, , 4 513, 184, , 6 493, 276, , 8 465, 368, , , 460, , , 552, , , 644, , , 736, , , 828, , , 920, , 1 575, 48, , 5 551, 240, , 7 527, 336, , , 528, , , 624, , , 816, , , 912, , 2 621, 100, , 4 609, 200, , 6 589, 300, , 8 561, 400, , , 600, , , 700, , , 800, , , 900, , 1 675, 52, , 3 667, 156, , 5 651, 260, , 7 627, 364, , 9 595, 468, , , 572, , , 780, , , 884, , 2 725, 108, , 4 713, 216, , 8 665, 432, , , 540, , , 756, , , 864, , 1 783, 56, , 3 775, 168, , 5 759, 280, , 9 703, 504, , , 616, , , 728, , 2 837, 116, , 4 825, 232, , 6 805, 348, , 8 777, 464, , , 580, , , 696, , 1 899, 60, , 7 851, 420, , 2 957, 124, , 4 945, 248, , 6 925, 372, 997

33 304 Pythagorean triangles 7.2 A Pythagorean triangle with an inscribed square How many matches of equal lengths are required to make up the following configuration? Suppose the shape of the right triangle is given by a primitive Pythagorean triple (a,b,c). The length of a side of the square must be a common multiple of a and b. The least possible value is the product ab. There is one such configuration consisting of (i) two Pythagorean triangles obtained by magnifying (a, b, c) a and b times, (ii) a square of side ab. The total number of matches is (a + b)(a + b + c) + 2ab = (a + b + c)c + 4ab. The smallest one is realized by taking (a,b,c) = (3, 4, 5). It requires 108 matches. How many matches are required in the next smallest configuration?

34 7.3 When are x 2 px ± q both factorable? When are x 2 px ± q both factorable? For integers p and q, the quadratic polynomials x 2 px+q and x 2 px q both factorable (over integers) if and only if p 2 4q and p 2 +4q are both squares. Thus, p and q are respectively the hypotenuse and area of a Pythagorean triangle. p 2 + 4q = (a + b) 2, p 2 4q = (b c) 2. b a p b p b a q b a a q a b c = p q x 2 px + q x 2 + px q (x 2)(x 3) (x 1)(x + 6) (x 3)(x 10) (x 2)(x + 15) (x 5)(x 12) (x 3)(x + 20) The roots of x 2 px+q are s a and s b, while those of x 2 +px q are s c and s. Here, s is the semiperimeter of the Pythagorean triangle. 7.4 Dissection of a square into Pythagorean triangles Here is the smallest square which can be dissected into three Pythagorean triangles and one with integer sides and integer area

35 306 Pythagorean triangles 7.5 Points at integer distances from the sides of a primitive Pythagorean triangle Let (a,b,c) be a primitive Pythagorean triangle, with vertices (a, 0), (0,b), and (0, 0). The hypotenuse is the line bx + ay = ab. The distance of an interior point (x,y) to the hypotenuse is 1 (ab bx ay). We c seek interior points which are at integer distances from the hypotenuse. With the parameters (6,1) we have the Pythagorean triangle (35,12,37). Here the five points (29,1), (23,2), (17,3), (11,4), (5,5) are at distances 1, 2, 3, 4, 5 from the hypotenuse Another example: with parameters (5,2) we have the triangle (21,20,29). Here we have the interior points (8,11), (16,2), (3,13), (11,4), (6,6), (1,8), (4,1) at distances 1, 2, 3, 4, 6, 8, 11 from the hypotenuse. The arrangement is not as regular as the previous example Exercise 1. A man has a square field, 60 feet by 60 feet, with other property adjoining the highway. He put up a straight fence in the line of 3

36 7.5 Points at integer distances from the sides of a primitive Pythagorean triangle 307 trees, at A, P, Q. If the distance between P and Q is 91 feet, and that from P to C is an exact number of feet, what is this distance? A 60 B 60 P? 91 Q D C 2. Give an example of a primitive Pythagorean triangle in which the hypotenuse is a square. 3. Give an example of a primitive Pythagorean triangle in which the even leg is a square. 4. Give an example of a primitive Pythagorean triangle in which the odd leg is a square. triangle to be a square? 5. Find the smallest Pythagorean triangle whose perimeter is a square (number). 6. Find the shortest perimeter common to two different primitive Pythagorean triangles. 7. The number of primitive Pythagorean triangle with a fixed inradius is always a power of Show that there are an infinite number of Pythagorean triangles whose hypotenuse is an integer of the form For each natural number n, how many Pythagorean triangles are there such that the area is n times the perimeter? How many of these are primitive? 10. Find the least number of toothpicks (of equal size) needed to form

37 308 Pythagorean triangles Challenge: Two pairs of primitive Pythagorean triples with almost equal perimeters In a class in Number Theory the professor gave four students the assignment of finding a fairly large primitive Pythagorean triangle using the well known formula for the legs: a = 2mn, b = m 2 n 2, c = m 2 + n 2, where m and n are relatively integers, not both odd. The four students produced four entirely different primitive triangles, but on comparing them it was found that two of them had the same perimeter, while the other two also had the same perimeter, this perimeter differing from the first one by 2. This interested the class greatly, and much time was spent in an effort to find other such sets, only to discover that there were only four such sets with perimeters less than 500,000. Can you find at least one such set? perimeter m + n 2m m n a b c

38 7.5 Points at integer distances from the sides of a primitive Pythagorean triangle 309 Challenge: Cross number puzzle on primitive Pythagorean triples B, 3D, 9B 29B, 7A, 21D 12B, 11U, 20U 2D, 6D, 5B 19U, 15D, 7D 22D, 18B, 15U 27A, 2D, 26D 20A, 8D, 8A 16A, 31A, 33A 5D, 3A, 25B 30D, 14A, 9A 16B, 24D, 23B 28D, 35A, 3U 30U, 9U, 13D 22A, 32U, 32D 4U, 21A, 21D 19U, 17D, 10A 32U, 34A, 33A The answers are distinct 2- and 3-digit decimal numbers, none beginning with zero. Each of the above sets of answers is a primitive Pythagorean triple, in increasing size, so that the third member is the hypotenuse. A = across, B = back, D = down, U = up. For example, 1B has its tens and units digits in the squares labelled 2 and 1 respectively; 11U is a 3-digit number with its tens and units digits in squares 16 and 11 respectively. 1 R. K. Guy, Problem 1153, Crux Math., 12 (1986) 139.

39 Chapter triangles in the square

40 triangles in the square

41 Chapter 9 The regular pentagon 9.1 The golden ratio The construction of the regular pentagon is based on the division of a segment in the golden ratio. 1 Given a segment AB, to divide it in the golden ratio is to construct a point P on it so that the area of the square on AP is the same as that of the rectangle with sides PB and AB, i.e., AP 2 = AB PB. A construction of P is shown in the second diagram. M A P B A P B Suppose PB has unit length. The length ϕ of AP satisfies ϕ 2 = ϕ In Euclid s Elements, this is called division into the extreme and mean ratio.

42 314 The regular pentagon This equation can be rearranged as ( ϕ 1 ) 2 = Since ϕ > 1, we have ϕ = 1 2 ( ). Note that AP AB = ϕ ϕ + 1 = 1 ϕ = 2 = This explains the construction above The diagonal-side ratio of a regular pentagon Consider a regular pentagon ACBDE. It is clear that the five diagonals all have equal lengths. Note that (1) ACB = 108, (2) triangle CAB is isosceles, and (3) CAB = CBA = ( ) 2 = 36. In fact, each diagonal makes a 36 angle with one side, and a 72 angle with another. C A P B E It follows that (4) triangle PBC is isosceles with PBC = PCB = 36, (5) BPC = = 108, and (6) triangles CAB and PBC are similar. Note that triangle ACP is also isosceles since (7) ACP = APC = 72. This means that AP = AC. D

43 9.3 Construction of a regular pentagon with a given diagonal 315 Now, from the similarity of CAB and PBC, we have AB : AC = BC : PB. In other words AB AP = AP PB, or AP 2 = AB PB. This means that P divides AB in the golden ratio. 9.3 Construction of a regular pentagon with a given diagonal To construct a regular pentagon ACBDE with a given segment AB as a diagonal. (1) Divide AB in the golden ratio at P. (2) Construct the circles A(P) and P(B), and let C be an intersection of these two circles. (3) Construct the circles A(AB) and B(C) to intersect at a point D on the same side of BC as A. (4) Construct the circles A(P) and D(P) to intersect at E. Then ACBDE is a regular pentagon with AB as a diameter. 9.4 Construction of 36, 54, and 72 angles Each of the following constructions begins with the division of a segment AB in the golden ratio at P. C A P B

44 316 The regular pentagon C C A 36 P B 54 A 72 P B D Example 9.1. (Odom s construction of the golden ratio) Let D and E be the midpoints of the sides AB and AC of an equilateral triangle ABC. If the line DE intersects the circumcircle of ABC at F, then E divides DF in the golden ratio. A D M E F O B C Example 9.2. Given a segment AB, erect a square on it, and an adjacent one with base BC. If D is the vertex above A, construct the bisector of angle ADC to intersect AB at P. Then P divides AB in the golden ratio. D A P B C

45 9.4 Construction of 36, 54, and 72 angles 317 Example 9.3. In the following diagram, the three circles are congruent if and only if the center of the middle circle divides the radius OA in the golden ratio. 2(R x) R 2x R O x R x K R x A Proof. Let R be the radius of the large circle. If OK = x, then 2x R R = R x R + x (2x R) + R R = (R x) + (R + x) R + x x R = Thus, x(r + x) = R 2, or x 2 = R(R x). This means that K divides OA in the golden ratio. R R + x.

46 318 The regular pentagon Exercise Justify the following construction of regular pentagon: Let OA and OY be two perpendicular radii of a circle, center O. (1) Mark the midpoint M of OY and bisect angle OMA to intersect OA at P. (2) Construct the perpendicular to OA at P to intersect the circle at B and E. Y B C M O P A D Then A, B, E are three adjacent vertices of a regular pentagon inscribed in the circle. The remaining two vertices can be easily constructed. E

47 Chapter 10 Integer triangles with a 60 or 120 angle 10.1 Integer triangles with a 60 angle If triangle ABC has C = 60, then c 2 = a 2 ab + b 2. (10.1) Integer triangles with a 60 angle therefore correspond to rational points in the first quadrant on the curve x 2 xy + y 2 = 1. (10.2) Note that the curve contains the point P = ( 1, 1). By passing a line of rational slope t through P to intersect the curve again, we obtain a parametrization of the rational points. Now, such a line has equation y = 1 + t(x + 1). Solving this simultaneously with (10.2) we obtain (x,y) = ( 1, 1) = P, and (x,y) = ( 2t 1 t 2 t + 1, t(2 t) t 2 t + 1 which is in the first quadrant if 1 < t 2. By symmetry, we may 2 simply take 1 < t 1 to avoid repetition. Putting t = q for relatively 2 p prime integers p, q, and clearing denominators, we obtain a =p(2q p), b =q(2p q), c =p 2 pq + q 2, ),

48 402 Integer triangles with a 60 or 120 angle with p < q p. Dividing by gcd(a,b) = gcd(p + q, 3), we obtain the 2 primitive integer triangles with a 60 angle: p q (a, b, c) 1 1 (1, 1, 1) 3 2 (3, 8, 7) 4 3 (8, 15, 13) 5 3 (5, 21, 19) 5 4 (5, 8, 7) 6 5 (24, 35, 31) 7 4 (7, 40, 37) 7 5 (7, 15, 13) 7 6 (35, 48, 43) 8 5 (16, 55, 49) 8 7 (16, 21, 19) 9 5 (9, 65, 61) 9 7 (45, 77, 67) 9 8 (63, 80, 73) 10 7 (40, 91, 79) 10 9 (80, 99, 91) Exercise A standard calculus exercise asks to cut equal squares of dimension x from the four corners of a rectangle of length a and breadth b so that the box obtained by folding along the creases has a greatest capacity. a x The answer to this problem is given by b x = a + b a 2 ab + b 2. 6 How should one choose relatively prime integers a and b so that the resulting x is an integer? 1 For example, when a = 5, b = 8, x = 1. Another example is a = 16, b = 21 with x = 3. 1 Answer: a, b, c, with gcd(p + q, 6) = 3.

49 10.2 Integer triangles with a 120 angle Integer triangles with a 120 angle If triangle ABC has C = 120, then c 2 = a 2 + ab + b 2. (10.3) Integer triangles with a 120 angle therefore correspond to rational points in the first quadrant on the curve x 2 + xy + y 2 = 1. (10.4) Note that the curve contains the point Q = ( 1, 0). By passing a line of rational slope t through P to intersect the curve again, we obtain a parametrization of the rational points. Now, such a line has equation y = t(x+1). Solving this simultaneously with (10.2) we obtain (x,y) = ( 1, 0) = Q, and ( ) 1 t 2 Q(t) = t 2 + t + 1, t(2 + t), t 2 + t + 1 which is in the first quadrant if 0 < t < 1. It is easy to check that Q(t) and Q ( 1 t 1+2t) are symmetric about the line y = x. To avoid repetition we may restrict to 0 < t < Putting t = q for relatively prime integers p, q satisfying q < 3 1p, p 2 and clearing denominators, we obtain a =p 2 q 2, b =q(2p + q), c =p 2 + pq + q 2, with 0 < q < p. Dividing by gcd(a,b) = gcd(p q, 3), we obtain the primitive integer triangles with a 120 : p q (a, b, c) 3 1 (8, 7, 13) 4 1 (5, 3, 7) 5 1 (24, 11, 31) 6 1 (35, 13, 43) 7 1 (16, 5, 19) 7 2 (45, 32, 67) 8 1 (63, 17, 73) 9 1 (80, 19, 91) 9 2 (77, 40, 103) 10 1 (33, 7, 37) 10 3 (91, 69, 139)

50 404 Integer triangles with a 60 or 120 angle Exercise (1) Show that a number c is a sum of two consecutive squares if and only if 2c 1 is a square. (2) Suppose an integer triangle contains a 120 angle with its two arms differing by 1. Show that the length of the longest side is a sum of two consecutive squares.

51 Chapter 11 Isosceles triangles equal in perimeter and area Isaac Newton 1 has given an elegant solution to the so-called Roberval problem: To find two (rational) isosceles triangles which shall be equal in area and perimeter. A q q y D y B CE F 2p 2x Let the triangles be ABC, DEF ; their bases AB = 2p, DE = 2x; their slant sides AC = q, DF = y. Then p+q = x+y and p q 2 p 2 = x y 2 x 2, that is p 2 q 2 p 4 = x 2 y 2 x 4. Divide by the equals p+q and x+y and there will be p 2 q p 3 = x 2 y y 3. In place of y write p + q x and there follows p 2 q p 3 = px 2 + qx 2 2x 3 and so q = p3 + px 2 2x 3. p 2 x 2 Divide through by p x and there comes q = p2 +px+2x 2 p+x. 1 D.T.Whiteside, The Mathematical Papers of Issac Newton, vol. IV, pp , Cambridge University Press, 1971.

52 406 Isosceles triangles equal in perimeter and area Since this solution gives q in terms of p and x, it does not construct an isosceles triangle equal in perimeter and area to a given one. The last expression for q, however, can be construed as an equation in x: 2x 2 (q p)x p(q p) = 0. Since q > p, it is clear that this quadratic equation has exactly one positive root given by x = 1 ( (q p) + ) (q p)(q + 7p). (11.1) 4 Note that this positive root is p if and only if q = 2p, i.e., the given triangle is equilateral. On the other hand, since (q p)(q + 7p) < q + 3p, we have y = p + q x > p + q 1 4 ((q p) + (q + 3p)) = 1 (p + q) > 0. 2 We conclude, therefore, that to every non-equilateral, isosceles triangle, there corresponds a unique, non-congruent isosceles triangle equal in perimeter and area. We shall construct pairs of such isosceles triangles with integer sides. Suppose, then, that in (11.1) above, p and q are integers. The base 2x of the second isosceles triangle is an integer if and only if (q p)(q + 7p) is a square integer. Writing q p = dh 2 and q + 7p = dk 2 for integers d, h, k, with gcd(h,k) = 1, we have Note that p = d 8 (k2 h 2 ), q = d 8 (k2 + 7h 2 ). gcd(2p,q) = d 8 gcd(2(k2 h 2 ), k 2 + 7h 2 ) = d 8 gcd(k2 + 7h 2, 16h 2 ) = d 8 gcd(k2 + 7h 2, 16) { d, if k and h are both odd, = d, if k and h are of different parity. 8

53 407 For the second isosceles triangle, we have, from (11.1) and the relation y = p + q x, Note also that x = d 4 h(h + k), y = d 4 (k2 hk + 2h 2 ). gcd(2x,y) = d 4 gcd(2h(h + k), k2 hk + 2h 2 ) = d 4 gcd(2(h + k), k2 hk + 2h 2 ) { 1, k odd and h even, = 2, k even and h odd, or k h 0 (mod 4), 4, k h 2 (mod 4). To obtain a pair of isosceles triangles of integer sides without common divisors, we choose, for given relatively prime integers h < k, { 1, k h 2 (mod 4), d = 2, k h 0 (mod 4), 8, k h 1 (mod 2). Consider two isosceles triangles (h,k) := ( d 8 (k2 + 7h 2 ), d 8 (k2 + 7h 2 ), d 4 (k2 h 2 )), and (h,k) := ( d 4 (k2 hk + 2h 2 ), d 4 (k2 hk + 2h 2 ), d h(h + k)). 2 Each of these two isosceles triangles has perimeter d 2 (k2 + 3h 2 ) and area d2 16 h(k2 h 2 ) k 2 + 3h 2. Note that the second triangle (h,k) is the same as ( k h, k+3h ), g g where g = gcd(k h,k + 3h). Example. With h+k 10, we obtain the following pairs of isosceles triangles equal in perimeter and area. (h, k) (h, k ) (h, k) (h, k ) perimeter area (1, 2) (1, 5) (11, 11, 6) (8, 8, 12) (1, 4) (3, 7) (23, 23, 30) (28, 28, 20) (2, 3) (1, 9) (37, 37, 10) (22, 22, 40) (1, 6) (5, 9) (43, 43, 70) (64, 64, 28) (2, 5) (3, 11) (53, 53, 42) (46, 46, 56) (3, 4) (1, 13) (79, 79, 14) (44, 44, 84) (1, 7) (3, 5) (7, 7, 12) (11, 11, 4) (1, 8) (7, 11) (71, 71, 126) (116, 116, 36) (2, 7) (5, 13) (77, 77, 90) (86, 86, 72) (4, 5) (1, 17) (137, 137, 18) (74, 74, 144)

54 408 Isosceles triangles equal in perimeter and area It is interesting to note that none of these beginning examples have integer area for the triangles. Newton actually went on to find triangles subject to the same conditions so that their perpendiculars also are rational, ([2; pp ]). To obtain pairs of such triangles with integer sides (without common divisor), it is enough, by the above analysis, to choose h and k so that k 2 + 3h 2 is the square or an integer. It is well known that k and h must be of the form k = m2 3n 2, h = 2mn g g for relatively prime integers m and n, and g = gcd(m 2 3n 2, 2mn). With m+n 10, we obtain the following pairs of isosceles triangles with integer sides and integer areas, equal in perimeter and area. (m, n) (h, k) (h, k ) (h, k) (h, k ) perimeter area (4, 1) (8, 13) (5, 37) (617, 617, 210) (386, 386, 672) (6, 1) (4, 11) (7, 23) (233, 233, 210) (218, 218, 240) (8, 1) (16, 61) (45, 109) (5513, 5513, 6930) (6514, 6514, 4928) (7, 2) (28, 37) (9, 121) (6857, 6857, 1170) (3802, 3802, 7280) (9, 1) (3, 13) (5, 11) (29, 29, 40) (37, 37, 24)

55 Chapter 12 Cheney s card trick You are the magician s assistant. What he will do is to ask a spectator to give you any 5 cards from a deck of 52. You show him 4 of the cards, and in no time, he will tell everybody what the 5th card is. This of course depends on you, and you have to do things properly by making use of the following three basic principles Three basic principles The pigeonhole principle Among 5 cards at least 2 must be of the same suit. So you and the magician agree that the secret card has the same suit as the first card Arithmetic modulo 13 The distance of two points on a 13-hour clock is no more than 6. We decide which of the two cards to be shown as the first, and which to be kept secret. For calculations, we treat A, J, Q, and K are respectively 1, 11, 12, and 13 respectively. Now you can determine the distance between these two cards. From one of these, going clockwise, you get to the other by traveling this distance on the 13-hour clock. Keep the latter as the secret card.

56 410 Cheney s card trick hours distance clockwise 2 and to 7 3 and to 3 2 and J 4 J to 2 A and to A The following diagram shows that the distance between 3 and 10 is 6, not 7. Q K A J Permutations of three objects There are 6 arrangements of three objects The remaining three cards can be ordered as small, medium, and large. 1 Now rearrange them properly to tell the magician what number he should add (clockwise) to the first card to get the number on the secret card. Let s agree on this: arrangement distance sml 1 slm 2 msl 3 mls 4 lsm 5 lms 6 If you, the assistant, want to tell the magician that he should add 4 to the number (clockwise) on the first card, deal the medium as the second card, the large as the third, and the small as the fourth card. 1 First by numerical order; for cards with the same number, order by suits: < < <.