Heron Triangles With Sides in Arithmetic Progression - DocServer

Heron Triangles With Sides in Arithmetic Progression J. A. MacDougall School of Mathematical and Physical Sciences, University of Newcastle,NSW, Australia 2308 February 2, 2005 In a recent article [2] in this journal, Charles Fleenor demonstrates the existence of Heron triangles having sides whose lengths are consecutive integers. He presents a list of examples of such triangles and observes a number of interesting relationships that appear to hold among their side lengths. The purpose of this article is to show how to derive all possible triangles satisfying these conditions (there are an infinite family of them) and to explain why the relationships he observes are indeed true. The more general problem of Heron triangles with sides having lengths in any arithmetic progression is also discussed and a complete solution is found, but using a different method. In addition, it is shown that there are no cyclic quadrilaterals having integer area and sides of consecutive integer lengths.

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Consecutive integer sides

A triangle with sides of lengths a, b, c is a Heron triangle if a, b and c are all integers and the area is also an integer. It is most convenient to calculate the area using Heron’s formula: p A = s(s − a)(s − b)(s − c) where s is the semiperimeter, (a + b + c)/2. Although this formula is called Heron’s formula there is speculation, if not solid evidence, that this formula was actually discovered by Archimedes several hundred years before the time of Heron (see [2], p.124). Let the sides of the triangle be b − 1, b and b + 1 . Then s = 3b/2, and substitution of these into the formula yields p b 3(b2 − 4) A= . 4

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If b were odd, then 3(b2 − 4) would be odd and then A would not be an integer. Consequently b is even and if we set b = 2x we then have p A = x 3(x2 − 1). So we have a Heron triangle if and only if 3(x2 − 1) is a square. Then we must have x2 − 1 = 3y 2 , i.e. x2 − 3y 2 = 1 which is one of the well-known Diophantine equations of the form x2 − dy 2 = 1, called Pell equations. The solutions to the Pell equations can be found with the help of continued fractions. Specifically, if d is any squarefree positive integer, then find the convergents (successive truncations) hn /kn of the continued fraction expansion of d. A standard theorem (for example, see [4], chap.7) says that all solutions to the Pell equation must be of the form√(x, y) = (hn , kn ) for some n. In our case, d = 3 and the continued fraction for 3 is [1; 1, 2] = 1 +

1 1+

1 2+

1+

. 1

1

2+ 1

..

.

The convergents are 2 5 7 19 26 71 97 265 , , , , , , , ,··· 1 3 4 11 15 41 56 153 and we find that the pairs (2, 1), (7, 4), (26, 15), (97, 56), ... corresponding to every second convergent are solutions to our equation. These yield the values of b and the corresponding triangles listed in Fleenor’s article, the first few of which are shown in the table below. How do we know that there will be an infinite number of solutions amongst these convergents? Well, there is a simple way to generate all solutions once you have found the first one. Again, a standard theorem in number theory ([4], chap. 7) points the way. Suppose (x1 , y1 ) is the smallest solution to x2 − dy 2 = 1. Then all positive solutions are of the form (xn , yn ) where ³ √ √ ´n xn + yn d = x1 + y1 d For many small values of d the smallest solution can easily √ be found by inspection, without having to compute the convergents of d. In our case where d = 3, it is easy to spot the smallest solution which is (2, 1). Then all values of x satisfying the equation √ are found by taking the rational component of the irrational expression (2 + 3)n . Computation (with the help of Derive) gives the solutions x = 2, 7, 26, 97, 362, 1351, etc. and the corresponding sides b = 2x for the triangles as listed in the table.

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n 1 2 3 4 5 6 7 8 9 10

a 3 13 51 193 723 2701 10083 37633 140451 524173

b 4 14 52 194 724 2702 10084 37634 140452 524174

c 5 15 53 195 725 2703 10085 37635 140453 524175

Area 6 84 1170 16296 226974 3161340 44031786 613283664 8541939510 118973869476

Can we find a closed formula for these values? Expanding by the binomial theorem, we find n µ ¶ ³ √ ´n X n n−k √ k 2+ 3 = 2 3 k 0 The terms corresponding to even powers k will give the rational part; hence xn =

n/2 µ ¶ X n k n−2k 3 2 k

k=0

and b = 2xn is the corresponding side of a Heron triangle. This proves that the only Heron triangles with consecutive integer sides are the infinite family beginning with the ones listed in the table below. Fleenor observes that the ratio of corresponding sides in √ successive triangles in the table appears to be approaching the limit of 2 + 3. This can be explained by expressing √ the xn in a different form. The even terms of the binomial expansion of (2 + 3)n can also be given by the expression √ √ (2 + 3)n + (2 − 3)n xn = . 2 √ Notice that (2 − 3)n = (.267...)n goes rapidly to 0 as n increases, so that à √ !n 2+ 3 xn ≈ . 2 Then clearly the ratio xn+1 /xn is approximately 2 +

2

√

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Extension to other arithmetic progressions

The observation that b−1, b and b+1 are three terms in an arithmetic progression suggests that we ask whether Heron triangles exist with sides whose lengths form some other arithmetic progression. Certainly, if we multiply all sides of one of 3

the triangles described above by some integer k, then they will form an A.P. and the area will be multiplied by k 2 , hence an integer. Triangles (15, 20, 25) or (26, 28, 30) would be examples. But are there triangles that don’t arise this way? We call a triangle primitive if its sides have no common factor, and we will look for primitive triangles. We begin as before: let the sides be b − d, b, b + d where 1 ≤ d ≤ b. Then s = 3b/2 and by Heron’s formula, the area is p b 3(b2 − 4d2 ) . A= 4 Reasoning as before, b must be even and so we let b = 2x. Then the expression simplifies to p A = x 3(x2 − d2 ) Again, the triangle is Heron if and only if 3(x2 − d2 ) is a square, so now we are led to the equation: d2 + 3y 2 = x2 , a homogeneous quadratic Diophantine equation. Not all such equations are solvable, but when they are, it is possible to find all solutions using the chord method (for example, see [5]). The solution is always given by a set of parametric equations in 2 variables. In our case all primitive solutions are given by d = |m2 − 3n2 |/g y = 2mn/g x = (m2 + 3n2 )/g where g = gcd(m2 − 3n2 , 2mn, m2 + 3n2 ). Since the equation is homogeneous, any multiple of a primitive solution is also a solution and we can thus assume m and n are relatively prime. The table below displays the primitive solutions and the corresponding triangles generated from them for various choices of m and n. All primitive triangles with b < 150 are listed in order of increasing b.

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m 1 2 1 1 3 3 3 5 2 5 1 7 4 1 6 7 11 3 5 1 11

n 1 1 2 3 5 4 7 3 3 2 5 3 3 4 7 5 1 8 4 7 3

d 1 1 11 13 11 13 23 1 23 13 37 11 11 47 37 13 59 61 23 73 47

a 3 13 15 15 17 25 29 51 39 61 39 65 75 51 85 111 65 73 123 75 101

b 4 14 26 28 28 38 52 52 62 74 76 76 86 98 122 124 124 134 146 148 148

c 5 15 37 41 39 51 75 53 85 87 113 87 97 145 159 137 183 195 169 221 195

The solutions generated by the equations above should, of course, include all those derived in section 1 where d = 1. The reader will notice from the table above that all the values of d seem to be congruent (mod 12) to 1 or -1. This is in fact true and the proof is left for the reader.

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Dividing a Heron triangle into two right-angled triangles

Suppose that in any of the triangles considered in section 1, we erect an altitude on the side of even length (the middle of the three sides). If the triangle is acute, this divides it into two right-angled triangles and Fleenor observes that in all of his examples both new triangles are Heron and the lengths of their bases differ by 4. This indeed happens for all the triangles considered in Section 1 and it is not hard to see why. If the sides are consecutive integers, we let the altitude have length h and suppose it divides the base b into lengths u and v as shown in Fig. ??. Then by Pythagoras’ theorem, using b = 2x, we have h2 + u2

= 4x2 − 4x + 1

h2 + v 2

= 4x2 + 4x + 1.

Subtracting the first from the second yields 5

(v − u)(v + u) = 8x. Since v + u = 2x, we get v − u = 4 as observed. Then the altitude of the triangle is p h = (2x − 1)2 − (x − 2)2 p = 3x2 − 3 A = x which is an integer. Thus both are Heron triangles and their bases are x − 2 and x + 2, i.e., they differ by 4.

2x-1

2x+1 h

u

v 2x

Figure 1: Decomposing a Heron triangle into two right triangles. For the triangles of Section 2, a calculation like that above shows that the altitude divides the base into segment of length x − 2d and x + 2d. Then the altitude turns out to be p 3x2 − 3d2 which is A/x, an integer just as before. So each Heron triangle with sides in arithmetic progression b − d, b, b + d can be decomposed into two rightangled Heron triangles whose bases differ by 4d. A similar calculation shows that erecting an altitude on either of the other sides never produces two Heron triangles. It can, in fact, be shown that any Heron triangle whatsoever has at least one even side and that this decomposition into right triangles holds. The triangles derived in Sections 1 and 2 may not be acute, in which case the altitude to the even side falls outside the base. In this case, however, (in fact, precisely when b < 4d) a similar calculation to that above shows that the Heron triangle can be regarded as a difference of two Heron right-angled triangles. 6

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Generalization to cyclic quadrilaterals

In this section, we look briefly at the question of whether the results above can be extended to cyclic quadrilaterals. While Heron’s formula for the area of a circle is well-known, it is not commonly known that there is a similar formula for the area of a cyclic quadrilateral. If a,b,c and d are the sides and s is the semi-perimeter, then p A = (s − a)(s − b)(s − c)(s − d) It can be seen that Heron’s formula follows as the special case where one side of the quadrilateral shrinks to 0. According to Eves ([2], p.183) this formula first appears in the writing of the Hindu mathematician Brahmagupta in the 7th century A.D. Whether it was discovered by him or whether it came from hitherto unknown earlier Persian or Greek sources is unknown. Brahmagupta is certainly respected for other impressive theorems on cyclic quadrilaterals. Let a = b − 1, c = b + 1 and d = b + 2. Then s = 2b + 1, and substitution into the formula gets us the area p A = (b + 2)(b + 1)b(b − 1) (1) and the area will be an integer if and only if (b + 2)(b + 1)b(b − 1) is the square of an integer. But (b + 2)(b + 1)b(b − 1) = (b2 + b − 1)2 − 1 and squares cannot differ by 1, so the expression can never be an integer square. Thus no such quadrilaterals exist. As we did with triangles, we may attempt to generalise and ask whether there exist cyclic quadrilaterals having integer area and integer sides in arithmetic progression. This leads to a quartic equation corresponding to equation (1) which is more difficult to deal with. Ralph Buchholz and I have shown in [1] that there are no non-trivial solutions to this equation (and therefore, no cyclic quadrilaterals with sides in AP), but the proof uses properties of elliptic curves and is beyond the scope of this article. There ought to be an elementary proof of this fact - it amounts to finding four numbers in AP whose product is a square.

References [1] Buchholz, R. H. & MacDougall, J. A., Heron Quadrilaterals with Sides in Arithmetic Progression, Bull. Aus. Math. Soc., 1999, 263-269. [2] Eves, Howard, An Introduction to the History of Mathematics, 5th ed., Saunders, Philadelphia, 1983 [3] Fleenor, Charles R., Heronian Triangles with Consecutive Integer Sides, J. Recr. Math., 28(2), 1987, 113-115 7

[4] Niven, I., Zuckerman, H.S., Montgomery,H.L., An Introduction to the Theory of Numbers, 5th ed., Wiley, N.Y., 1991 [5] Wright, H.N., First Course in the Theory of Numbers, Wiley, NY, 1962

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