Integer-sided equable shapes Shapes with integer side lengths and with equal area and perimeter. Rectangles ab = (a + b) 1 = 1 a + 1 b Trapezia 6 8 14 1 4 3 0
Triangles 6 10 8 P = = 4 13 1 P = = 30 8 10 17 P = = 36 6 9 0 1 1 P = = 4 9 7 9 0 P = = 60 1 6
A parametrisation of equable trapezia Any trapezium can be split into a triangle and a parallelogram. a a b Furthermore, if the trapezium is integer-sided, then so are both the triangle and the parallelogram. Let the triangle have perimeter P and area. Then the trapezium has Perimeter = P + a Area = a+b b Then (P + a)b = (a + b) a = P > b(p ) ( b) To determine all integer-sided equable trapezia we simply need to consider all integer-sided triangles with P > and check if a = b(p ) is an integer. ( b)
The (3,4,) triangle a = b(p ) ( b) = 6b (6 b). 3 b = 3 a = 3 4 6 6 b = 4 a = 6 3 10 b = a = 1 1 4 3 0 The (,,6) triangle a = b(p ) ( b) = b 1 b. b = 6 a = 8 The (,,8) triangle a = b(p ) ( b) = 3b 1 b 6. b = 8 a = 6 14
What lies between 1 and 1? The (3,4,) triangle is an obvious example of an integer-sided triangle with Area Perimeter a rational number less than 1. As we shall see, it actually has the smallest such ratio. Furthermore, it can also be used to construct two further examples, the (,,8) and (,,6) triangles. Area 6 1 1 Perimeter 1 18 16 Ratio 1 3 3 4 This raises the questions: What ratios of Area Perimeter are possible? Can a general method be found for determining all integer-sided triangles for which the ratio is a rational number less than 1? What new equable trapezia can be formed using these triangles? Much mathematical attention has been given to triangles whose areas are integer multiples of their perimeters. The problem of determining all integersided triangles for which the ratio is a rational number less than 1 does not appear to have received attention. However, this problem does have an interesting solution. We shall discover that there are two infinite families of solution triangles, for which the ratios of Area Perimeter converge to and 3, respectively.
A standard method for considering integer-sided triangles is based upon the incircle. z C z Semi-perimeter s = x + y + z Sides x y a = s x, b = s y, c = s z Area, = xyz(x + y + z) A x y B We are interested in triangles such that Area Perimeter is a rational number less than one and it will prove convenient to consider this as two separate conditions: xyz < 4(x + y + z) (x + y + z)xyz is the square of a rational number. It would considerably simplify our analysis if the perimeter of the triangle were even, since then x, y and z would all be integers. Fortunately this can be proved to be the case, as follows. Suppose that s were an odd integer, then X = x, Y = y and Z = z would all be odd integers.then (X + Y + Z)XYZ would be both the square of a rational number and an integer and is therefore a square number. If X Y Z (modulo 4), then (X + Y + Z)XYZ 3Z 4 3 (modulo 4). Otherwise, without loss of generality, we can suppose X + Y 0 (modulo 4), and then (X + Y + Z)XYZ 3Z 3 (modulo 4). In both cases we have a square congruent to 3 (modulo 4), which is impossible. Therefore x, y and z are positive integers. For the remaining analysis we shall suppose, without loss of generality, that x y z.
Case 1. If yz > 4. Then the inequality xyz < 4(x + y + z) can be used to limit the sizes of x, y and z and can be rewritten as (xz 4)(yz 4) < 4z + 16 yz 4 < 4z + 16 y < 4 z + 4 + 16 z Running through the possibilities for z :- If z = 1, then 4 < y < 4 + 0 and so y =,6,7 or 8. If z =, then < y < + 8 and so y = 3 or 4. If z 3, then z y < 1 (4 + ) and so y = z = 3. 3 The following table shows the values of x determined by these cases and the expression which we require to be a square number. z y Square? 1 x < 4(x + 6) x 3 x(x + 6) 1 6 6x < 4(x + 7) 6 x 13 6x(x + 7) 1 7 7x < 4(x + 8) 7 x 10 7x(x + 8) 1 8 8x < 4(x + 9) x = 8 8x(x + 9) 3 6x < 4(x + ) 3 x 9 6x(x + ) 4 8x < 4(x + 6) 4 x 8x(x + 6) 3 3 9x < 4(x + 6) 3 x 4 9x(x + 6) The various cases can be quickly checked by computer. The only solution is x = y = 3, z =. This gives the (,,6) triangle that has already been noted.
Case. If yz 4. The possibilities are listed in the table. z y (x + y + z)xyz 1 1 x(x + ) 1 x(x + 3) 1 3 3x(x + 4) 1 4 4x(x + ) 4x(x + 4) Three of these possibilities, (y, z) = (1,1), (4,1) and (,), require the solution in positive integers of a Diophantine equation of the form x(x + n) = a, for n =, 4 and. Then a = x + b, where b < n and so b = 1 or. The only solution is n =, b =, x = 4 which gives the (,,8) triangle that has already been noted. Dealing with the two remaining possibilities requires more effort. x(x + 3) = a Then x and a must both be divisible by 3. Furthermore, a is even. Let x = 3X and a = 6A, then X(X + 1) = A. X and X + 1 can have no common factor and so one is a square and the other is twice a square. Therefore X is the smaller of U and V, where U and V are positive integers such that 3x(x + 4) = a U = V ± 1. In this case we can let x = X and a = 6A. Then X(X + ) = 3A, where the greatest common divisor of X and X + is either 1 or. A similar argument to that used above leads to two equations. U = 3V + 1 and U = 3V.
The equation u = v ± 1 A method of solving this equation was known to the Pythagoreans in c00 BC. The details of this method were described by Theon of Smyrna in c100 AD. Theon demonstrated that infinitely many solutions could be generated from the initial solution u = 1, v = 1. The subsequent solutions are given by repeating the calculations The first few solutions are as shown. u = u + v, v = u + v. u 1 3 7 17 41 99 39 77 v 1 1 9 70 169 408 The fractions are known as the convergents to. 1 1, 3, 7, 17 1, 41 9, 99 70, 39 169, 77 408, A similar method can be used to solve other equations of the type u = Av + 1. Various Greek mathematicians appear to have been able to produce solutions to such equations but no trace remains of their methods. One absurdly challenging problem is the so called Cattle problem of Archimedes, c0 BC, which required the solution of the equation u = 4 79 494v + 1.
The Pell equation, u = Av + 1 A general procedure to solve any such equation, for A positive and squarefree, was in use at the time of Bhramagupta, c600 AD. It is unclear (and controversial) as to whether this method, known as the Indian method, was of Greek origin. The problem was rediscovered by Fermat, c160, who challenged mathematicians to give a general solution. The solution by the first President of the Royal Society, Viscount William Brouncker, was accepted by Fermat as being correct. However, Fermat pointed out that Brouncker had not given a general proof, of the type that he could himself obtain, by the method of descente applied in a quite special manner. It should be noted that a full solution required: a proof that a solution was always possible; a method of finding a first solution; a method of generating infinitely many solutions from one solution; a proof that all solutions can be obtained in this manner. In the 18th century, Lagrange completed Euler s work on these equations and gave the first rigorous proof of the method. We only require the solutions of the equations u = 3v and u = 3v + 1. These solutions are given by the convergents to 3. u v = 1 1, 1, 3, 7 4, 19 11, 6 1, 71 41, 97 6,
Can the (3, x + 1, x + ) triangle be used? P = (x + 3) x + 1 x + = x(x + 3) = xp b(p ) is an integer ( b) 3 = b + fg, where f is a factor of b and g is a factor of P. f (b, b) = (b, ) (b, xp). g (P, b) = (P, P b) (P, P b) = (b(b x), P b). If b = 3 f 3 and g (3(x 3), x + 3) 7. Therefore 3 {1,3,9,7,81}. If b = x + 1 f (x + 1, x(x + 3)) 4 and g (x + 1, x + ) 4. Therefore (x + 1) {1,,4,8,16}. If b = x + f (x +, x(x + 3)) 4 and g ((x + ), x + 4) 4. Therefore (x + ) {1,,4,8,16}. Since = x(x + 3), each possibility gives a quadratic equation for x. The only valid solutions are those with x = 3 i.e. the (3,4,) triangle.
Can the (4, x + 1, x + 3) triangle be used? P = (x + 4) x + 1 x + 3 = 3x(x + 4) b(p ) ( b) is an integer 4 = b + fg, where f is a factor of b and g is a factor of P. f (b, b) = (b, ) (b, 3x(x + 4)). g (P, P b) = ((x + 4)(x + 16), P b). If b = 4 f 4 and g ((x + 4)(x + 16), x + 4) 11. Therefore 4 {1,,4,8,16,3,64,7,14,8,6,11,4,448}. If b = x + 1 f (x + 1, 3x(x + 4)) 9 and g ((x + 4)(x + 16), x + 7) 7. Therefore (x + 1) {1,3,9,7,81,43}. If b = x + 3 f (x + 3, 3x(x + 4)) 9 and g ((x + 4)(x + 16), x + ) 11. Therefore (x + 3) {1,3,9,11,33,99}. Since = 3x(x + 4), each possibility gives a quadratic equation for x. The only valid solutions are those with x = i.e. the (3,4,) triangle.
The (3, x + 1, x + ) triangles U = V 1 = 6UV P = 1V 4UV 3U + 1 3U + U and V are the numerators and denominators of the odd terms in the sequence of convergents to. P = U V U 1 3 1 1, 3, 7, 17 1, 41 9, 99 70, 39 169, 77 408, 4UV 3U U 3 3U 1 U = V + 1 = 6UV P = 6U U and V are the numerators and denominators of the even terms. P = V U 1 1, 3, 7, 17 1, 41 9, 99 70, 39 169, 77 408
The (4, x + 1, x + 3) triangles U = 3V + 1 = 1UV P = 8U 6UV 4U 3 4U 1 U and V are the numerators and denominators of the even terms in the sequence of convergents to 3. P = 3V U 3 U 3 4 1 1, 1, 3, 7 4, 19 11, 6 1, 71 41, 97 6, 3UV U + 1 U 1 4 U + 3 U = 3V = 6UV P = 1V U and V are the numerators and denominators of the odd terms. P = U V 3 1 1, 1, 3, 7 4, 19 11, 6 1, 71 41, 97 6,