Cyclic Polygons with Rational Sides and Area

Transcription

1 Cyclic Polygons with Rational Sides and Area Ralph H. Buchholz and James A. MacDougall May 2001 Abstract We generalise the notion of Heron triangles to rational-sided, cyclic n-gons with rational area using Brahmagupta s formula for the area of a cyclic quadrilateral and Robbins formulæ for the area of cyclic pentagons and hexagons. We use approximate techniques to explore rational area n-gons for n greater than six. Finally, we produce a method of generating non-eulerian rational area cyclic n-gons for even n and conjecturally classify all rational area cyclic n-gons. Keywords : cyclic, polygon, rational, area 1 Introductory Points Around 200 BC, Archimedes, in a blatant act of anticipatory plagiarism, discovered Hero s formula for the area of a triangle in terms of the lengths of the sides (see section 3). A short while later, in 620 AD, Brahmagupta concocted his less well known formula for the area of a cyclic quadrilateral (see section 4). Then 14 centuries later, Robbins presented the mathematics community with his derivation of the analogous formulæ for the areas of both cyclic pentagons and cyclic hexagons. Motivated by this flurry of activity we considered trying to find cyclic n-gons having rational sides and rational area. In the literature the case of n = 3 is usually called a Heron triangle. In this spirit we denote the case n = 4 by a Brahmagupta quadrilateral, and the cases n = 5, 6 by Robbins pentagon and Robbins hexagon respectively. 2 Notational Lines We will use n-tuples enclosed in square brackets to denote the sides of our polygons, such as the triangle [13, 14, 15], or sometimes in homogeneous form, Revision : January 23,

2 like the square [1 : 1 : 1 : 1]. The area will be denoted by K n (a 1,..., a n ) while for the circumradius we use R n (a 1,..., a n ). We may drop either the subscripts or the functional dependence on the sides (or both) if it is obvious from context. 3 Heron Triangles Recall that Hero s formula for the area, K 3 say, of a triangle in terms of the sides, a, b and c, is simply where s is the semiperimeter. K 3 = s(s a)(s b)(s c) In 1670 Bachet described a method (see [3]) of generating rational solutions to this equation by joining together two appropriately rescaled Pythagorean triangles (ie. right triangles with rational sides and area). Although this does provide a method of generating all Heron triangles, it was Euler who was the first to explicitly parametrize such triangles via [a : b : c] = [(ps + qr)(pr qs) : rs(p 2 + q 2 ) : pq(r 2 + s 2 )] (1) where p, q, r, s are arbitrary rational parameters. Carmichael provided the most economical parametrization [a : b : c] = [n(m 2 + k 2 ) : m(n 2 + k 2 ) : (m + n)(mn k 2 )] (2) requiring only three rational parameters as opposed to Euler s four. Whenever we make a computational search for various cyclic n-gons with rational area it is expedient to seek out ways to reduce the number of cases to consider. One observation, for an integer sided Heron triangle, is that the perimeter is always even. Of course this result is both trivial to prove and largely irrelevant since we have Euler s and Carmichael s parametrizations. Despite this we give a proof, relying only on an area formula, which will be a guide in those cases when we do not have a parametrization to fall back on. First we need a preliminary lemma. Lemma 1 Any Heron triangle with three integer sides has integer area. Proof: Suppose the area K 3 = u/v for some relatively prime integers u, v. Then Hero s formula provides us with 16u 2 = v 2 (a + b + c)( a + b + c)(a b + c)(a + b c). Without loss of generality we can assume that 2 gcd(a, b, c), otherwise we could conclude that v = 1. Next since gcd(u, v) = 1 we see that v 2 16 hence v 4. Restricting to positive v leaves us with just 3 cases. If v = 1 we are done. If v = 2 then the above equation 2

3 reduces to a + b + c 0 (mod 2) which implies that each of the bracketted factors on the right is divisible by 2. Thus 2 divides u a contradiction. The most interesting case is v = 4 which leads to u 2 = (a + b + c)( a + b + c)(a b + c)(a + b c). Since 2 gcd(a, b, c) we must have either 1 or 3 of a, b and c being odd. This implies that there exist 4 integers k 1, k 2, k 3, k 4 such that u 2 = (2k 1 + 1)(2k 2 + 1)(2k 3 + 1)(2k 4 + 1). But u is odd which means that u 2 = 1 (mod 4) so that 2(k 1 + k 2 + k 3 + k 4 ) 0 (mod 4). By re-expressing the k i in terms of a, b, c we get 2(k 1 + k 2 + k 3 + k 4 ) = (a + b + c 1) + ( a + b + c 1)+ + (a b + c 1) + (a + b c 1) = 2(a + b + c 2) 2 (mod 4). which is a contradiction. Now we are ready to prove the following. Theorem 1 The perimeter of an integer sided Heron triangle is even. Proof : Lemma 1 implies that K 3 is an integer so expressing Hero s formula in terms of the perimeter, P say, gives 16K 2 3 = P (P 2a)(P 2b)(P 2c). Clearly, considering both sides modulo 2 leads to P 4 0(mod 2) implying P 0(mod 2). At this stage we find it useful to define the notion of radial decomposability since it provides us with one useful means of determining if an n-gon in general is constructible from smaller rational area m-gons or not. Definition: A cyclic n-gon with rational sides and area is radially decomposable if it can be subdivided into n isosceles Heron triangles each composed of two circumradii and one side of the n-gon. Of course we are secretly interested in the indecomposible cyclic n-gons since they are the fundamental building blocks of all such n-gons. The following well known result is useful in the sequel. 3

4 B A a R 3 b C R 3 R 3 c Figure 1: Radial decomposition of a Heron triangle Theorem 2 Any Heron triangle is radially decomposable. Proof: Let K 3 (a, b, c) denote the area of a triangle with sides a, b and c then the circumradius, R 3, is given by R 3 = abc 4K 3 (a, b, c). Clearly, for a rational-sided triangle the circumradius is rational if and only if the area is rational. Now let K 3 (a, R 3, R 3 ) = A = α m 1 K 3 (b, R 3, R 3 ) = B = β m 2 K 3 (c, R 3, R 3 ) = C = γ m 3 for positive rationals α, β, γ and squarefree integers m 1, m 2 and m 3. Since we are assuming that the triangle (a, b, c) has rational area we observe that α m 1 + β m 2 + γ m 3 Q and it is then a simple matter (yet somewhat surprising) to show that this implies that m 1, m 2 and m 3 are all rational. In the delightful paper (see [6]) by Richard Guy titled Tiling the square with rational triangles he shows how to prove much more in fact an arbitrary sum of quadratic irrationals being rational forces all of them to be rational. 3.1 Special rational area triangles There are in the literature an almost uncountable variety of restricted versions of rational triangles made, and we would like to add one to that fine tradition. 4

5 Consider the problem of finding all rational area triangles with two rational sides and one quadratic irrational side. More specifically we wish to find rational area triangles with sides [a, b, c] where a, b Q and c mq for squarefree integers m expressible as the sum of two rational squares. Of course this seemingly arbitrary set of triangles will turn out to be somewhat useful in the sequel. The first result is one which restricts the base segments and the altitude to the irrational side (refer to Figure 2). a h c b c 1 c 2 Figure 2: Special rational area triangle c Theorem 3 Consider any rational area triangle with sides denoted by a, b, c, altitude to side c denoted by h c and the base segments of side c, created by the altitude, denoted by c 1 and c 2. If a, b Q and c mq then h c, c 1, c 2 mq. Proof: If we denote the area of the triangle by then = ah a 2 = bh b 2 = ch c 2 implies that h a, h b Q while h c mq. Pythagoras theorem gives Furthermore, two applications of c 2 1 = a 2 h 2 c c 2 2 = b 2 h 2 c implying that c 2 1 and c 2 2 are both rational. If we let c = c m for rational c then squaring the identity c 1 + c 2 = c leads to c 1 c 2 = c2 m c 2 1 c where the right hand side is clearly rational. Replacing c 2 by c c 1 and collecting the rational parts gives us ( c 2 m c 2 1 c 2 ) 2 m c 1 = 2cm 5

6 hence c 1 mq. Similarly c 2 mq. Following Carmichael [2] we attempt to parametrize all such triangles. So we consider the equations h 2 = a 2 c 2 1 = b 2 c 2 2, c = c 1 + c 2 where a, b Q and h, c 1, c 2, c mq. Now let h = h m, c 1 = c 1 m, c 2 = c 2 m, c = c m for rational h, c1, c 2 and c. Since h 2 = (a c 1 )(a + c 1 ) we can, without loss of generality, introduce a new variable u := a + c 1 where u Q( m). Letting u = u 1 + u 2 m gives us a + c 1 m = u1 + u 2 m a c 1 m = h 2 m u 1 + u 2 m. As long as m Z 2 then we can equate coefficients in this pair of equations to get a = u 1 = h2 mu 1 u 2 1 and c 1 = u 2 = h2 mu 2 mu2 2 u 2 1 mu2 2 which reduce to the single constraint u 2 1 mu 2 2 = h 2 m. Similarly, for the equation h 2 = (b c 2 )(b + c 2 ) we let b + c 2 = v 1 + v 2 m to obtain the constraint v 2 1 mv 2 2 = h 2 m. Notice that if we make the transformations u 1 = mu 1 and v 1 = mv 1 in the above conditions a little rearrangement gives ( )2 ( ) 2 ( )2 ( ) 2 h u2 h v2 m = + = +. u 1 u 1 As mentioned above this is one reason to restrict the squarefree integer m to those expressible as the sum of two rational squares. Since this restriction comes up again in a later setting we make the following observation. Lemma 2 If m = u v 2 0 then the general solution to is given by m = u 2 + v 2 v 1 u = u 0p 2 + 2v 0 p u 0 p v 1 where p Q. v = v 0p 2 2u 0 p v 0 p

7 Proof: Use the chord method. We will consider m to be fixed, and of the correct shape, so that we can calculate a particular u 0 and v 0 satisfying the hypothesis of Lemma 2 (via the Cremona- Rusin algorithm say). So, we can see that there exist rational parameters, g 1, g 2, p and q say, such that and h = g 1 (u 0 p 2 + 2v 0 p u 0 ) u 2 = g 1 (v 0 p 2 2u 0 p v 0 ) u 1 = g 1 (p 2 + 1) h = g 2 (u 0 q 2 + 2v 0 q u 0 ) v 2 = g 2 (v 0 q 2 2u 0 q v 0 ) v 1 = g 2 (q 2 + 1). If we eliminate h between these two pairs of equations we get the conic condition u 0 p 2 + 2v 0 p u 0 = k(u 0 q 2 + 2v 0 q u 0 ) where k = g 2 /g 1 Q. Now completing the square on both sides gives us 1 (u 0 p + v 0 ) 2 u2 0 + v 2 ( 0 1 = k (u 0 q + v 0 ) 2 u2 0 + v 2 ) 0. u 0 u 0 u 0 u 0 Making the affine transformation P := u 0 p + v 0, Q := u 0 q + v 0 rearranging this conic condition leads to and then P 2 kq 2 = (1 k)m. Finally we turn this into homogeneous form by letting P = P/R and Q = Q/R so that P 2 kq 2 = (1 k)mr 2. (3) Using the Hilbert symbol (see [4] pp ) allows us to say that this has a rational solution iff (k, (1 k)m) p = 1 for all prime p including p =. Of course, since m is positive then it is clear that there are real solutions which means that it is sufficient to consider only the finite primes. Notice that the other 3 ways of associating h, u 2, and h, v 2, to their respective right hand sides all lead to a conic similar to (3) the only difference being the value of k. Now if k is a fixed rational which satisfies (k, (1 k)m) p = 1 for all primes p then the equation u 0 p 2 + 2v 0 p u 0 = k(u 0 q 2 + 2v 0 q u 0 ) 7

8 has a particular solution, (p, q) = (p 0, q 0 ) say. This can be used in the chord method to write p and q in terms of the fixed values u 0, v 0, p 0, q 0, k and two free parameters r and s, namely, p = ( u 0p 0 2v 0 )r 2 + 2k(u 0 q 0 + v 0 )rs kp 0 u 0 s 2 u 0 r 2 ku 0 s 2 q = u 0q 0 r 2 2(u 0 p 0 + v 0 )rs + k(u 0 q 0 + 2v 0 )s 2 u 0 r 2 ku 0 s 2. 4 Brahmagupta Quadrilaterals Brahmagupta s formula for the area, K 4 say, of a cyclic quadrilateral in terms of the sides, a, b, c and d, is simply K 4 = (s a)(s b)(s c)(s d) where s is again the semiperimeter. Since we will again be considering radial decomposition we need the circumradius formula (ac + bd)(ad + bc)(ab + cd) R 4 = 4K 4 first derived by Parameśvara around 1430 AD (see [5]). Unlike the Heron triangle case, the following lemma and theorem limiting the perimeter actually form a useful result. Lemma 3 Any Brahmagupta quadrilateral with four integer sides has integer area. Proof: Straightforward analog to the proof of Lemma 1. Theorem 4 The perimeter of an integer sided Brahmagupta quadrilateral is even. Proof : Straightforward analog to the proof of Theorem 1. For n-gons with 4 or more sides another notion of decomposability becomes apparent namely that of diagonal decomposability. Definition: A cyclic n-gon with rational sides and area is diagonally decomposable if it can be subdivided into at least 2 disjoint rational area polygons using a common rational diagonal. Now, initially we had not imagined that there was any relationship between diagonal decomposability and radial decomposability. However, a series of experiments, lemmas and theorems made us grudgingly reverse our view and admit 8

9 that they were the same concept, at least for quadrilaterals. The same appears to be true in the 5-gon case at which point we were on the verge of believing it to be generally true. But a surprise was in store for us when we toyed with cyclic hexagons. Sadly, this will have to wait while we first build up to our 4-gon conclusion. 4.1 Decomposable Quadrilaterals First we note that Euler, produced a paremetrization of all radially decomposable Brahmagupta quadrilaterals (see [3]). Here we study their relationship to diagonally decomposable quadrilaterals. Lemma 4 In any rational-sided, cyclic quadrilateral one diagonal is rational if and only if the other diagonal is rational. Proof: Let a, b, c, d denote the sides of the cyclic quadrilateral and u 1, u 2 the two diagonals as shown in Figure 3. Using Hero s formula and Brahmaguptas a b u 2 u 1 d c formula leads to Figure 3: One rational diagonal implies two rational diagonals K 3 (a, b, u 1 ) + K 3 (c, d, u 1 ) = K 4 (a, b, c, d) which when expanded to remove the square roots provides a quartic polynomial in u 1, namely (a 2 cd + abc 2 + abd 2 + b 2 cd abu 2 1 cdu 2 1) 2 = 0. Solving for u 1 gives us u 1 = (ac + bd)(ad + bc). (ab + cd) 9

10 Now either repeat the calculation with the other diagonal u 2 or just permute the sides to obtain (ac + bd)(ab + cd) u 2 =. (ad + bc) Clearly, if u 1 is rational then so is u 2 and vice versa. Next we consider the relationship between the two types of diagonal decomposition. In fact, we readily obtain Lemma 5 Any Brahmagupta quadrilateral with one rational diagonal is diagonally decomposable along either diagonal. Proof: As usual we let a, b, c, d denote the sides of the quadrilateral and u 1, u 2 denote the two diagonals. If we assume that u 1 is rational then Lemma 4 implies that u 2 is rational. Furthermore, the rationality of the area leads to Now letting K 3 (a, b, u 1 ) + K 3 (c, d, u 1 ) = K 3 (a, d, u 2 ) + K 3 (b, c, u 2 ) Q. A = s 1 (s 1 a)(s 1 b)(s 1 u 1 ), where s 1 = (a + b + u 1 )/2 B = s 2 (s 2 c)(s 2 d)(s 2 u 1 ), where s 2 = (c + d + u 1 )/2 C = s 3 (s 3 a)(s 3 d)(s 3 u 2 ), where s 3 = (a + d + u 2 )/2 D = s 4 (s 4 b)(s 4 c)(s 4 u 2 ), where s 4 = (b + c + u 2 )/2 gives us the conditions A, B, A + B Q C, D, C + D Q from which we conclude that A, B, C, D Q. We can now prove the first half of our claim connecting diagonal and radial decomposability. Theorem 5 Any diagonally decomposable quadrilateral is radially decomposable. Proof: We consider four cases depending on whether the circumcentre lies on one of the diagonals, one of the sides, inside or outside the quadrilateral. Case (i). Without loss of generality, suppose that O lies on AC then we have K 3 (ABO) = K 3 (BOC) = 1 2 K 3(ABC) Q K 3 (ADO) = K 3 (DOC) = 1 2 K 3(ADC) Q. 10

11 B O A C D Figure 4: Case (ii). If O lies inside the quadrilateral (as shown in Figure 4) then since ABCD is diagonally decomposable we can say that the area K 3 (ABC) is rational. Now by Theorem 2 we infer that K 3 (ABO) and K 3 (BOC) are rational. Similarly, since K 3 (ACD) is radially decomposable we find that K 3 (AOD) and K 3 (DOC) are rational. Case (iii). If O lies on the side AB say, then the radial decomposability of ABC implies that K 3 (BOC) is rational. Similarly, we get K 3 (AOD) and K 3 (DOC) rational from the radial decomposability of ADC. Case (iv). If O lies outside the quadrilateral then a similar argument to case (ii) will apply. To obtain the converse we require the following technical lemma which was supplied to us by a colleague (Garry Hughes) who based the result on Bretschneider s formula (see [7]). Lemma 6 If two isosceles Heron triangles are joined along their respective repeated sides then the extra diagonal, created by joining the extreme vertices, is rational. Referring to Figure 5 if a, b, R, A, B Q then e Q. A a R b B R e R Figure 5: Joined isosceles triangles 11

12 Proof: Given an arbitrary quadrilateral with sides a, b, c, d and diagonals e and f Bretschneider s formula provides the area, namely, K 4 = 1 4 4e2 f 2 (a 2 + c 2 b 2 d 2 ) 2. Specializing this to the case shown in Figure 5 gives 16(A + B) 2 = 4e 2 R 2 (a 2 b 2 ) 2 whence e = 1 16(A + B)2 + (a 2R 2 b 2 ) 2. Now using Heron s formula on each sub triangle implies that A = a 4 4R2 a 2 Q B = b 4 4R2 b 2 Q. Hence there exist rationals m and n such that 4R 2 a 2 = m 2 and 4R 2 b 2 = n 2 implying that we can write A = am 4 B = bn 4. Substituting this into the expression for e gives us e = 1 (am + bn)2 + (a 2R 2 b 2 ) 2 = 1 (am + bn)2 + (a 2R 2 b 2 )(n 2 m 2 ) = 1 (an + bm) 2 2R ( a 2 B + b 2 ) A = 2 R ab showing that e is rational. Clearly, using Lemma 6 immediately implies that any radially decomposable quadrilateral must have at least one rational diagonal. Combining this with Lemma 5 leads to a proof of the following: Theorem 6 Any radially decomposable quadrilateral is diagonally decomposable. So, the Eulerian quadrilaterals (mentioned earlier) completely describe the radially (and diagonally) decomposable quadrilaterals. 12

13 4.2 Restricted Families of Brahmagupta Quadrilaterals Applying various restrictions to the sides of a quadrilateral lead to simpler cases that are invariably decidable. For example, if the sides of a cyclic quadrilateral are in an arithmetic progression or a geometric progression then the area can never be rational (see [1]). In the case of two equal sides it turns out to be fairly trivial to completely parametrize such isosceles Brahmagupta quadrilaterals. Suppose that the sides satisfy [a, b, c, d] = [a, b, c, c], then the area is given by K 2 4(a, b, c, c) = (a + b) 2 (2c a + b)(2c + a b) or k 2 = 4c 2 (a b) 2 where k = K 4 (a, b, c, c)/(a+b). Now this last equation is a homogeneous quadratic and as such is amenable to the standard chord method of parameterizing all rational points given one such point which is easily found by inspection. When we turn the handle on this process we get ga = 8pr gb = 8pq gc = 4p 2 + q 2 2qr + r 2 gk = 4p 2 q 2 + 2qr r 2 (4) where g is simply the greatest common divisor of the 4 right hand sides and p, q, r are arbitrary integer parameters. This family will turn out to be quite useful since we will use it to generate larger radially indecomposable n-gons. As a final example we consider another three parameter family constrained so that the sides are of the form [a, b, c, d] = [x m, x + m, x n, x + n] for rational x, m, n. In this case, the semiperimeter is s = 2x so that the area is simply given by K 2 4(a, b, c, d) = (x 2 m 2 )(x 2 n 2 ). Without loss of generality there exist rational parameters λ, α and β so that x 2 m 2 = λα 2 x 2 n 2 = λβ 2 making the area, K 4 = λαβ, automatically rational. The two equations x 2 = m 2 + λα 2 and x 2 = n 2 + λβ 2 have general solutions [m : α : x] = [r 2 λs 2 : 2λrs : r 2 + λs 2 ] [n : β : x] = [p 2 λq 2 : 2λpq : p 2 + λq 2 ] 13

14 where p, q, r, s are rational. In the case where the greatest common divisors of the two right hand sides are one we can equate the two expressions for x to get which has the solution r 2 + λs 2 = p 2 + λq 2 [r : s : p : q] = [u 2 λv 2 + λw 2 : 2uv : u 2 + λv 2 λw 2 : 2uw]. Now recall that x 2 m 2 λα 2 = 0 and x 2 n 2 λβ 2 = 0 so that we obtain the conditions λ(λ 1)(λ + 1)u 2 v 2 (u 2 λv 2 + λw 2 ) = 0 λ(λ 1)(λ + 1)u 2 w 2 (u 2 + λv 2 λw 2 ) = 0 which imply, through the non-degeneracy of the quadrilateral, that u 0, v 0, w 0, λ 0, λ 1 and λ ±u 2 /(v 2 w 2 ) leaving only the case λ = 1. Thus we get the family x = u 4 + 2u 2 v 2 + 2u 2 w 2 + v 4 2v 2 w 2 + w 4 m = u 4 6u 2 v 2 + 2u 2 w 2 + v 4 2v 2 w 2 + w 4 n = u 4 + 2u 2 v 2 6u 2 w 2 + v 4 2v 2 w 2 + w Indecomposable Quadrilaterals What about the indecomposable ones? We note that there are infinitely many such indecomposable Brahmagupta quadrilaterals by simply considering rectangles with side lengths 2m and 2n for integer m and n such that 0 < m < 2n + 1. The circumradius, R, is simply given by R = m 2 + n 2 and satisfies the inequality hence cannot be rational. n < R < n + 1 For a general Brahmagupta quadrilateral if we allow a rational similarity transformation then it is possible to show that the circumradius must be one of 1, 2, 5, 10, 13, 17,..., m,... where m = u 2 + v 2 for integers u and v in particular, no Brahmagupta quadrilateral can have a circumradius of 3. If we assume that the sides and area of a general Brahmagupta quadrilateral are rational then Parameśvara s formula for the circumradius immediately allows us to conclude that R 4 (a, b, c, d) Q( m) 14

15 A a R b B R R D d R c C Figure 6: A cyclic quadrilateral for some m Q /Q 2. So when we apply Hero s formula to the isosceles triangles in Figure 6 we get A = s(s a)(s R)(s R) = (s R) s(s a) where s = R + a/2. Similarly, for the other three triangles giving us A = a 4R2 a 4 2, B = b 4R2 b 4 2 C = c 4R2 c 4 2, D = d 4R2 d 4 2. Now we infer from A + B + C + D Q and 4R 2 Q that A, B, C, D Q. If we let R := p m say, for rational p and squarefree integer m then since R must satisfy equations of the form 4R 2 α 2 = β 2 for rational α and β we see that m satisfies ( ) 2 ( ) 2 α β m = +. 2p 2p Characterising such integers is a well studied problem and the upshot is that the subset of squarefree ones are given by the following criterion m = p i prime where p i = 2 or p i 1(mod 4) and e i {0, 1}. This proves our earlier result on the sequence of possible circumradii of indecomposable quadrilaterals. Notice p ei i 15

16 that for every m expressible as a sum of two integer squares, u 2 + v 2 say, we can always find a Brahmagupta quadrilateral with a circumradius of m by simply considering the rectangle with sides [2u, 2v, 2u, 2v]. Can we determine all Brahmagupta quadrilaterals of a given circumradius? Of course, we are really interested in the case when R = m. If we attempt to use a method analogous to that of Euler, in the R = 1 case, we find it will lead to a hyperelliptic condition. So to construct a Brahmagupta quadrilateral with radius m we choose 3 free parameters, p, q and r say, so that Lemma 2 gives us a = 2(u 0p 2 + 2v 0 p u 0 ) p b = 2(u 0q 2 + 2v 0 q u 0 ) q c = 2(u 0r 2 + 2v 0 r u 0 ) r If we substitute these into Brahmagupta s area formula we end up with C : K 2 4 = αd 4 + βd 3 + γd 2 + δd + ɛ (5) where α, β, γ, δ, ɛ are functions of p, q, r. Notice that equation 5 is an elliptic curve in general which would seem to allow infinitely many rational pairs (d, K 4 ) for a given choice of p, q, r. However, these do not all lead to quadrilaterals with a circumradius of m. This extra constraint means we should set d = 2(u 0 t 2 + 2v 0 t u 0 )/(t 2 + 1) which leads to the hyperelliptic condition D : y 2 = octic(t) (6) where y = K 4 (t 2 + 1) 2. By Faltings theorem equation 6 has only finitely many rational solutions. This condition contrasts with the analogous situation for Euler in which he was guaranteed that the 4th isosceles triangle would have rational area and rational circumradius. An alternative approach is based on the results of Section 3.1. As above, we fix the circumradius to R = m and use Parameśvara s Theorem and the proof of Lemma 4 to express the diagonals u 1 and u 2 in terms of the circumradius. In particular, recall that the diagonals are given by (ac + bd)(ad + bc) u 1 = (ab + cd) (ac + bd)(ab + cd) u 2 = (ad + bc) while the circumradius is R 4 = (ac + bd)(ad + bc)(ab + cd) 4K 4. 16

17 Combining these leads to so that u 1, u 2 mq. u 1 = 4K 4 m ad + bc u 2 = 4K 4 m ab + cd Now we would like to deduce that the areas of the 4 subtriangles in Figure 3 are rational. Without loss of generality we focus on just one such triangle, [a, b, u 1 ] say. We have a, b Q and u 1, R 4 mq. Recalling that R 3 = abc 4K 3 (a, b, c) means that if we let u 1 = u 1 m then K 3 (a, b, u 1 ) = abu 1 m 4 m = abu 1 4 clearly implying that the area of [a, b, u 1 ] is rational. Similarly, the other three triangles also have rational area. Conversely, if any triangle [a, b, c] has two rational sides a and b say and third side c in mq then the circumradius is also in mq. Hence the problem of determining all Brahmagupta quadrilaterals is reduced to that of finding rational area triangles with sides [a, b, c] such that a, b Q and c mq. This is precisely what was done in Section 3.1. To complete this process all we need to do is glue together two such triangles with supplementary angles subtending the diagonal. These triangles are distinguished by the ordering of the u 0 and v 0 solution to m = u v Robbins Pentagons Since Robbins area formula was the real motivation for this section (and in fact the entire paper) we briefly restate it. Theorem 7 (Robbins) Consider a cyclic pentagon with sides a 1,..., a 5, and area K 5. If σ 1,..., σ 5 are the symmetric polynomials in the squares of the sides, u = 16K 2 5, t 2 = u 4σ 2 + σ 2 1, t 3 = 8σ 3 + σ 1 t 2, t 4 = 64σ 4 + t 2 2 and t 5 = 128σ 5 then the area satisfies the degree 7 condition ut t 2 3t t 3 3t 5 18ut 3 t 4 t 5 27u 2 t 2 5 = 0. 17

18 Our first result is the by now obvious: Lemma 7 Any Robbins pentagon with five integer sides has integer area. Proof : Substituting K 5 = r/s into Theorem 7 and clearing denominators gives us 0 = 16r 2 T s 2 T 2 3 T s 8 T 3 3 t 5 18s 8 T 3 T 4 t s 10 r 4 t 2 5 (7) where T 3 := s 2 t 3 = 8σ 3 s 2 + σ 1 (16r 2 4s 2 σ 2 + s 2 σ 2 1) T 4 := s 4 t 4 = 64σ 4 s 4 + (16r 2 4s 2 σ 2 + s 2 σ 2 1) 2. Now observe that there are integers K and L so that T 3 = s 2 K + 16σ 1 r 2 and T 4 = s 2 L r 4 which when substituted into (7) lead to 16r 2 ((16r 2 ) 6 + s 2 L ) + s 2 K = 0. Since gcd(r, s) = 1 this last equation implies that s or s We can assume without loss of generality that 2 gcd(a 1, a 2, a 3, a 4, a 5 ), otherwise we would have 2 2i σ i for i {1, 2, 3, 4, 5} and 2 2i T i for i {3, 4}. But these imply that 2 r forcing s = 1 and we are done. Thus we have 14 positive cases to consider for s. Case (i) If s = 2 then r is odd so we can write r = 2R + 1. Meanwhile equation (7) quickly reveals that σ (mod 2) which implies that σ 1 0 (mod 2) and hence we have either 2 or 4 odd sides. The values of σ i (mod 4) vary as a function of the number of odd sides so that we end up with two subcases. (a) If there are 2 odd sides then recalling the definition of the σ i show that there exist integers S i such that (σ 1, σ 2, σ 3, σ 4, σ 5 ) = (4S 1 + 2, 4S 2 + 1, 2S 3, 2S 4, 2S 5 ). Substituting all this into equation (7) gives us and T 3 = 2 6 S (2S 1 + 1)((4R 2 + 4R + 1) (4S 2 + 1) + (4S S 1 + 1)) = 2 5 (2K + 1) T 4 = 2 11 S ((4R 2 + 4R + 1) (4S 2 + 1) + (4S S 1 + 1)) 2 = 2 8 (8L + 1) 18

19 implying that 0 = 2 28 (2R + 1) 2 (8L + 1) (2K + 1) 2 (8L + 1) (2K + 1) 3 S (2K + 1)(8L + 1)S (2R + 1) 4 S 2 5. Dividing this last equation by 2 28 and reducing the result modulo 4 leads to the following contradiction 0 (4R 2 + 4R + 1)(8L + 1) 3 + (4K 2 + 4K + 1)(8L + 1) 2 (mod 4) 2 (mod 4) completing this subcase. (b) If there are 4 odd sides then there exist integers S i such that (σ 1, σ 2, σ 3, σ 4, σ 5 ) = (4S 1, 2S 2, 4S 3, 4S 4 + 1, 2S 5 ). Substituting all this into equation (7) gives us and implying that T 3 = 2 7 S S 1 ((4R 2 + 4R + 1) 2S 2 + 4S 2 1) = 2 6 K T 4 = 2 10 (4S 4 + 1) ((4R 2 + 4R + 1) 2S 2 + 4S 2 1) 2 = 2 8 (4L + 1) 0 = 2 28 (2R + 1) 2 (4L + 1) K 2 (4L + 1) K 3 S K(4L + 1)S (2R + 1) 4 S 2 5. Dividing this last equation by 2 28 and reducing the result modulo 4 again leads to a contradiction, namely, completing this case. 0 (4R 2 + 4R + 1)(4L + 1) 3 (mod 4) 1 (mod 4) Case (ii) As for Heron triangles the most work is required when s = 4. This time we note that 2 gcd(a 1, a 2, a 3, a 4, a 5 ) implies that either an odd number of the sides are odd or 2 or 4 are odd. (a) The three subcases including an odd number of sides force σ 1 to be odd and since r is also odd we let r = 2R + 1 σ 1 = 2S

20 for integer R and S. Substituting these into the defining equations for T 3 and T 4 gives us and T 3 = 2 7 σ (2S + 1)(4R 2 + 4R + 1 4σ 2 + 4S 2 + 4S + 1) = 2 7 σ (2S + 1)(2(R 2 + R σ 2 + S 2 + S) + 1) = 2 5 (2K + 1) T 4 = 2 14 σ (4R 2 + 4R + 1 4σ 2 + 4S 2 + 4S + 1) 2 = 2 14 σ (2(R 2 + R σ 2 + S 2 + S) + 1) 2 = 2 10 (4L + 1). Substituting these into equation (7) gives 0 = 2 30 (4R 2 + 4R + 1)(4L + 1) (4K 2 + 4K + 1)(4L + 1) (2K + 1) 3 σ (2K + 1)(4L + 1)σ (2R + 1) 4 σ 2 5. Dividing out the factor of 2 30 and reducing the resulting equation modulo 4 leads to 0 2 (mod 4) a contradiction which completes this subcase. (b) If there are 2 odd sides then leads to (σ 1, σ 2, σ 3, σ 4, σ 5 ) = (4S 1 + 2, 4S 2 + 1, 2S 3, 2S 4, 2S 5 ) T 3 = 2 5 (2K + 1) and T 4 = 2 8 (8L + 1) which when substituted into equation (7) leads to 0 (2R + 1) 2 (8L + 1) 3 (mod 4), which is impossible. (c) If there are 4 odd sides then (σ 1, σ 2, σ 3, σ 4, σ 5 ) = (4S 1, 2S 2, 4S 3, 4S 4 + 1, 2S 5 ) leads to T 3 = 2 6 K and T 4 = 2 8 (4L + 1) 20

21 which when substituted into equation (7) gives which is impossible. 0 (2R + 1) 2 (4L + 1) 3 (mod 4), Case (iii) If s = 2 i for i {3,..., 14} then again we readily obtain r 14 0 (mod 2) which contradicts gcd(r, s) = 1 and completes the proof. Just as in the Heron triangle case we observe that this Lemma leads to the following. Theorem 8 The perimeter of an integer sided Robbins pentagon is even. Proof : By Lemma 7 it is clear that the area is integral so if we consider the equation satisfied by the area, K 5, from Theorem 7 modulo 2 we get t 2 3t 2 4 0(mod 2). Substituting for t 3, t 4 and the resulting t 2 gives σ1 14 0(mod 2) or a a a a a 2 5 0(mod 2) or a 1 + a 2 + a 3 + a 4 + a 5 0(mod 2). The major result in this section is that a cyclic pentagon with rational sides and area (ie. a Robbins pentagon) has either 5 rational diagonals or no rational diagonals. As yet, we have been unable to show either that indecomposable Robbins 5-gons do not exist or find an example through a computational search. Theorem 9 Any Robbins pentagon has either zero or five rational diagonals and in the latter case is diagonally decomposable. Proof : If the pentagon has one rational diagonal, u 4 say, then it must decompose into a triangle and a quadrilateral such that K 3 (a, b, u 4 ) + K 4 (c, d, e, u 4 ) Q implies that K 3 (a, b, u 4 ) Q and K 4 (c, d, e, u 4 ) Q. But by Theorem 2 we find that R 3 (a, b, u 4 ) Q implies that R 4 (c, d, e, u 4 ) Q. Now since quadrilateral [c, d, e, u 4 ] is radially decomposable we infer that u 1 and u 2 are rational. Similar arguments applied to other quadrilaterals reveal that all diagonals, and the areas of all sub-triangles and all sub-quadrilaterals are rational. We can in fact prove more, namely the equivalence of diagonal and radial decomposability. 21

22 b a u 4 u 5 c u 3 u 1 u 2 e d Figure 7: A cyclic pentagon Theorem 10 Any Robbins pentagon is diagonally decomposable if and only if it is radially decomposable. Proof : DD RD : Any 3 connected sides and the appropriate diagonal can be used to form a diagonally decomposable 4-gon which by Theorem 5 must be radially decomposable. Since this applies to all such sub-4-gons we see that the original 5-gon is radially decomposable. RD DD : A single application of Lemma 6 followed by Theorem 9 forces any radially decomposable pentagon to have 5 rational diagonals and be diagonally decomposable. At this stage we are lacking a formula for the diagonal of a cyclic pentagon. Unsurprisingly, they satisfy a degree 7 polynomial. Suppose we fix on the diagonal u 1 say in Figure 7, then we can equate the circumradii of the triangle and quadrilateral either side of the diagonal to get R 3 (c, d, u 1 ) = R 4 (a, b, e, u 1 ) or or cdu 1 4K 3 (c, d, u 1 ) = (au1 + be)(ae + bu 1 )(ab + eu 1 ) 4K 4 (a, b, e, u 1 ) (au 1 + be)(ae + bu 1 )(ab + eu 1 )K 2 3(c, d, u 1 ) (cdu 1 ) 2 K 2 4(a, b, e, u 1 ) = 0. The other 5 diagonals can be obtained by simply permuting the sides or they can be combined into a single representation by labelling the edges [a 0, a 1, a 2, a 3, a 4 ] 22

23 and the corresponding diagonals as [u 0, u 1, u 2, u 3, u 4 ] with u i opposite a i to get: (a i u i + a i+1 a i+4 )(a i a i+4 + a i+1 u i )(a i a i+1 + a i+4 u i )K 2 3(a i+2, a i+3, u i ) (a i+2 a i+3 u i ) 2 K 2 4(a i, a i+1, a i 1, u i ) = 0 for i = 0, 1, 2, 3, 4 and the subscripts are taken modulo 5. With this in hand a series of Monte Carlo tests were run in Magma to calculate the required field extension for a specific diagonal of a random cyclic pentagon. In each case the other 4 diagonals all required the same extension field of Q suggesting that this may well be generally true. A series of searches was initiated using Robbins formula in an attempt to find an indecomposable Robbins pentagon. The results of the exploration of all such pentagons with perimeter less than 400 revealed no such beast (see Table 1). perimeter sides radius area diagonals 68 [7,7,15,15,24] 25/2 276 [336/25,20,24,117/5,25] 72 [7,15,15,15,20] 25/2 342 [20,24,24,25,117/5] 172 [16,16,25,52,63] 65/ [2016/65,39,63,253/5,65] 176 [16,25,33,39,63] 65/ [39,52,60,60,65] 178 [9,20,20,51,78] 325/ [143/5,504/13,65,1161/25,75] 178 [16,25,25,52,60] 65/ [39,600/13,63,56,836/13] 182 [16,25,33,52,56] 65/ [39,52,323/5,60,312/5] 182 [25,25,33,39,60] 65/ [600/13,52,60,63,65] 184 [16,25,39,52,52] 65/ [39,56,65,312/5,60] 186 [25,33,33,39,56] 65/ [52,3696/65,60,323/5,837/13] 188 [25,33,39,39,52] 65/ [52,60,312/5,65,63] 218 [13,13,40,68,84] 85/ [2184/85,51,84,304/5,85] 220 [9,20,51,65,75] 325/ [143/5,65,406/5,70,78] 220 [20,20,51,51,78] 325/ [504/13,65,25806/325,75,406/5] 224 [9,20,65,65,65] 325/ [143/5,75,78,78,70] 224 [13,36,40,51,84] 85/ [805/17,68,77,75,85] 226 [20,20,51,65,70] 325/ [504/13,65,406/5,75,78] 234 [13,36,40,68,77] 85/ [805/17,68,84,75,408/5] 236 [13,40,40,68,75] 85/ [51,1200/17,84,77,1364/17] 238 [12,12,55,55,104] 325/ [7752/325,65,1232/13,371/5,100] 240 [13,40,51,68,68] 85/ [51,77,85,408/5,75] 240 [36,36,40,51,77] 85/ [5544/85,68,77,416/5,85] 242 [36,40,40,51,75] 85/ [68,1200/17,77,84,1443/17] 246 [36,40,51,51,68] 85/ [68,77,408/5,85,416/5] 256 [22,39,48,62,85] 1105/ [289/5,1305/17,442/5,1127/13,91] 266 [35,35,35,44,117] 125/ [336/5,336/5,75,11753/125,100] 278 [35,35,44,44,120] 125/ [336/5,75,10296/125,100,527/5] 292 [12,55,55,65,105] 325/ [65,1232/13,100,100,1395/13] 292 [45,45,50,50,102] 425/ [1386/17,85,1500/17,104,105] 294 [12,55,55,68,104] 325/ [65,1232/13,507/5,100,2668/25] 306 [45,50,50,76,85] 425/ [85,1500/17,102,105,104] 314 [29,29,60,60,136] 425/ [24128/425,85,1848/17,532/5,125] 318 [55,65,65,65,68] 325/ [100,104,104,105,507/5] 334 [35,35,44,100,120] 125/ [336/5,75,120,100,125] 340 [35,44,44,100,117] 125/ [75,10296/125,120,527/5,3116/25] 346 [35,44,75,75,117] 125/ [75,527/5,120,120,3116/25] 354 [35,44,75,100,100] 125/ [75,527/5,125,120,117] 370 [17,17,87,105,144] 145/ [4896/145,100,144,3237/29,145] 372 [17,24,87,100,144] 145/ [203/5,105,143,116,145] 374 [29,60,60,85,140] 425/ [85,1848/17,125,125,2405/17] 376 [17,24,87,105,143] 145/ [203/5,105,144,116,4200/29] 378 [24,24,87,100,143] 145/ [6864/145,105,143,3483/29,145] 384 [29,60,60,99,136] 425/ [85,1848/17,663/5,125,3531/25] 390 [25,25,59,136,145] 3625/ [1430/29,406/5,4375/29,2529/25,150] 398 [29,60,85,99,125] 425/ [85,125,140,136,136] Table 1: Robbins pentagons Finally, we end this section by revealing our lack of knowledge (and stamina) by presenting the Pentagon Conjectures: Conjecture 1 The denominator of the circumradius takes the form 2 α 3 β. Conjecture 2 The area of a Robbins pentagon is divisible by 6. 23

24 Conjecture 3 The denominators of the diagonals of a Robbins pentagon are of the form p 1 (mod 4) p. 6 Robbins Hexagons Again we used Robbins formula as our starting point and began a search for rational area cyclic hexagons. Theorem 11 (Robbins) Consider a cyclic hexagon with sides a 1,..., a 6, and area K 6. If σ 1,..., σ 5 are the symmetric polynomials in the squares of the sides, σ 6 = a 1 a 2 a 3 a 4 a 5 a 6, u = 16K 2 5, t 2 = u 4σ 2 + σ 2 1, t 3 = 8σ 3 + σ 1 t 2 16σ 6, t 4 = t σ σ 1 σ 6 and t 5 = 128σ t 2 σ 6 then the area satisfies either the degree 7 condition ut t 2 3t t 3 3t 5 18ut 3 t 4 t 5 27u 2 t 2 5 = 0. or the same condition with σ 6 replaced by its negative. As in all previous cases so far it is sufficient to consider only even perimeters. Lemma 8 Every integer sided Robbins hexagon has integer area. Proof : A tedious analog to the proof of Lemma 7. Theorem 12 Every integer sided Robbins hexagon has an even perimeter. Proof : Use Lemma 8 and consider Robbins formula modulo 2. We expected to find Eulerian examples, however as was the case for Brahmagupta quadrilaterals, we found radially indecomposable examples as well. Consider the cyclic hexagon formed from the sides [2, 5, 2, 5, 11, 5] as shown in Figure 8. Symmetry allows us to observe that we have two isosceles trapezia joined at the (initially unknown) central diagonal. Since Robbins formula gives us the area of this hexagon, namely 54, we can sum two quadrilateral areas to reveal that the unknown diagonal is rational, in fact equals 41/5. Armed with this we can verify that the circumradii of the two sub-quadrilaterals (and consequently the whole hexagon) are R 4 (2, 5, 2, 41/5) = = R 4 (5, 11, 5, 41/5). In the computational searches for Robbins 6-gons with perimeter less than 400 a total of 424 distinct examples have turned up. Only 3 of these have six distinct sides, while 7 are radially decomposable, the remainder being non-eulerian (see Table 2). 24

25 / Figure 8: A radially indecomposable Robbins hexagon sides area circumradius [7, 7, 15, 15, 15, 15] /2 [16, 16, 25, 25, 33, 63] /2 [16, 25, 25, 25, 33, 60] /2 [16, 16, 25, 25, 52, 52] /2 [16, 25, 25, 33, 33, 56] /2 [16, 25, 25, 33, 39, 52] /2 [25, 25, 33, 33, 39, 39] /2 [10, 19, 26, 40, 47, 52] /8 17 [7, 14, 22, 25, 55, 73] /2 221 [7, 14, 22, 25, 62, 70] /2 221 Table 2: A selection of Robbins hexagons 6.1 Restrictions on central diagonals Suppose we label the vertices of a cyclic hexagon by A 1,..., A 6 and the sides by a 1,..., a 6 as in Figure 9. Let D ij denote the major diagonal from vertex A i to vertex A j, and let d ij denote the minor diagonals. Then we have the following result. Lemma 9 If a Robbins hexagon has 1 rational central diagonal then the circumradius and all diagonals are expressible in the form p m for some rational p and squarefree integer m. Proof : First observe that a rational central diagonal and rational area force the two quadrilaterals, either side of the central diagonal to have rational area. Suppose, without loss of generality, that D 14 Q. Then the quadrilateral A 1 A 2 A 3 A 4 has a circumradius of (a1 a 2 + a 3 D 14 )(a 1 a 3 + a 2 D 14 )(a 1 D 14 + a 2 a 4 ) R = 4K 4 (a 1, a 2, a 3, D 14 ) 25

26 A 1 A 6 a 6 a 1 A 2 a 5 a 2 A 5 a 4 a 3 A 3 A 4 Figure 9: Diagonal constraints and a diagonal of (a1 a 2 + a 3 D 14 )(a 1 a 3 + a 2 D 14 )(a 1 D 14 + a 2 a 4 ) d 13 =. (a 1, a 2, a 3, D 14 ) So the squarefree portions of the radicands of R and d 13 are the same and hence we can write R = p 1 m and d13 = p 2 m. Similarly, the other diagonal of this quadrilateral is expressible as d 24 = p 3 m. Since the quadrilateral A 1 A 4 A 5 A 6 has the same circumradius the same result applies to its two minor diagonals, d 15 = p 4 m and d46 = p m. Finally, we consider the quadrilateral A 1 A 3 A 4 A 5 for which Ptolemy s theorem gives a 4 d 13 + a 3 d 15 = d 35 D 14 so that we can express the minor diagonal as d 35 = a 4p 2 m + a3 p 4 m D 14 = p 6 m for some rational p 6. A similar argument gives d 26 = p 7 m. Theorem 13 If one central diagonal of a Robbins hexagon is rational then all three central diagonals are rational. Proof : Suppose D 14 Q then applying Ptolemy s theorem to the quadrilateral A 1 A 2 A 4 A 5 gives D 14 D 25 = a 1 a 4 + d 24 d

27 A little algebra and the proof of Lemma 9 gives D 25 = a 1a 4 + p 3 m p4 m D 14 = a 1a 4 + p 3 p 4 m D 14 so that D 25 is rational. A similar argument on quadrilateral A 1 A 6 A 3 A 4 leads to D 36 Q. 7 General Cyclic n-gons During the searches for various rational area cyclic n-gons it seemed to be the case that examples were easier to come by when dealing with an even number of sides. The reason for this is probably related to the fact that for such n-gons we have both Euler s generation method as well as the quadrilateral generation method at our disposal whereas for an odd number of sides only Euler s construction works. So far, no rational area 7-gon has turned up, despite a search of all perimeters up to 133. In the case of cyclic polygons with more than six sides we have no explicit formula to work with and so we are forced to use an approximate technique to discover rational area cyclic n-gons. The process works as follows: 1. select the number of sides, n, and a perimeter size, p, to exhaust over, 2. select the first set of n monotonically increasing positive integers with the property that a i < j i a j for all i, j [1..n] and i a i = p, 3. calculate the approximate circumradius using Newtons method applied to maximising the total area of n isosceles triangles, 4. use the approximate circumradius to calculate the approximate area, 5. use continued fractions to determine if the real area looks like an integer. 6. now use a partition number enumeration algorithm to determine the next set of n sides satisfying property 2 and recurse on step 3. Once a putative area has been calculated it is usually the case that one can use a little geometry to prove that this is in fact the true area. Consider the 8-gon shown in Figure 10 which was found to have an area very close to 171. By symmetry we can arrange the sides any way we like without altering the area, so we have a rectangle surrounded by two trapezia. Clearly, the area of the 8-gon is given by K 8 = 2A + B where Brahmagupta s formula provides A = (5 + L)/4 (5 + L)(15 L) and B = 9L. Substituting for A, B and K 8 and squaring to remove the square gives 27

28 5 9 5 A 5 L B L 5 A Figure 10: An integer area cyclic 8-gon us 4(171 9L) 2 (5 + L) 3 (15 L) = 0. This polynomial in L has precisely one rational root, namely L = 13, which we subsequently use to calculate the circumradii of the trapezia, R A, and rectangle, R B, respectively. First note that the area of the trapezium is simply A = = 27 so that we can use Parameśvara and Pythagoras to obtain ( )(5 R A = )( ) 4 27 = 5 10 and R B = 2 2 = In particular, the two circumradii are identical proving that the 8-gon with sides [5, 5, 5, 9, 5, 5, 5, 9] is a genuine rational area cyclic polygon. 7.1 Euler s parametrization In about 1781 Euler used a relatively simple yet elegant method to generate all radially decomposable cyclic n-gons (see [3] p. 221). Since the circumradius is rational it is sufficient to generate all such n-gons on the unit circle and then scale them up by an arbitrary rational parameter. Euler chooses n 1 rational parameters p 1,..., p n 1 which are used to generate n 1 angles, labelled θ 1,..., θ n 1 via the equations sin(θ i ) = 2p i p 2 i + 1, and cos(θ i) = p2 i 1 p 2 i

29 with the angles satisfying the properties that He also defines an extra angle θ n by θ i π n 1 2, and θ i < π. n 1 θ n = π which implicitly satisfies the latter two conditions. The θ i are in fact the halfangles of the angle subtended by each side, of a corresponding n-gon, at the centre of the circumcircle (see Figure 11). Now since sin(a ± B) and cos(a ± B) i=1 i=1 θ i A n a n A 1 2θ n 2θ 1 a 1 2θ 2 a 2 A 2 Figure 11: A rationally decomposable cyclic n-gon are rationally expressible in terms of sin(a), sin(b), cos(a), and cos(b) Euler finds that all of sin(2θ i ), cos(2θ i ) for i = 1,..., n are rational. Hence he can conclude that the side lengths and wedge areas of the corresponding n-gon, given by a i = 2 2 cos(2θ i ) = 2 sin(θ i ) A i = 1 2 sin(2θ i) are all rational. Hence the n-gon has rational area and rational sides. 7.2 Examples of cyclic 6-gons, 8-gons, 10-gons and 12- gons A simple technique that we found early on allowed us to construct a limited number of rational area cyclic n-gons by tiling an isosceles quadrilateral on the 29

30 a b a a c a R b c c b a c a a b a Figure 12: Cyclic rational area 6, 8, 10 and 12-gons built on a square 4 sides of a square. In this way we obtain a radially indecomposable 6-gon, 8- gon, 10-gon and 12-gon respectively (see figure 12). The basic idea is to equate the circumradius of the inner square and the outer isosceles quadrilateral. The square has circumradius R = c/ (2) while the quadrilateral has R = a a 2 + bc (2a (b c))(2a + (b c)). Equating these two expressions gives us a quadratic in b, namely, c 2 b 2 + 2c(a 2 c 2 )b + (2a 4 4a 2 c 2 + c 4 ) = 0 which has a discriminant of 4a 2 c 2 (2c 2 a 2 ). The only way that b can be rational is for the discriminant to be a rational square, D 2 say. It is easy to show that the implied equation a 2 + D 2 = 2c 2 has a general solution given by a = t( r 2 2rs + s 2 ) D = t(r 2 2rs s 2 ) c = t(r 2 + s 2 ). This general parametrization leads to a rational expression for b in terms of the free parameters r, s, t, namely, b = (c2 a 2 ) ± a 2c 2 a 2 c = t r4 + 4r 3 s 6r 2 s 2 4rs 3 + s 4 (r 2 + s 2. ) 30

31 Thus we get the complete rational parametrization of such n-gons as a = t( r 2 2rs + s 2 )(r 2 + s 2 ) b = t(r 4 + 4r 3 s 6r 2 s 2 4rs 3 + s 4 ) c = t(r 2 + s 2 ) 2. When we enumerate the values of r and s the first non-trivial example we obtain occurs when r = 1, s = 2 and leads to the examples shown in Table 3. n perimeter sides area [5,5,17,25,25,25] [5,5,5,5,17,17,25,25] [5,5,5,5,5,5,17,17,17,25] [5,5,5,5,5,5,5,5,17,17,17,17] 877 Table 3: Specially constructed 6,8,10,12-gons A more general search for rational area cyclic 8-gons revealed those in Table 4. perimeter sides area 34 [2,2,5,5,5,5,5,5] [5,5,5,5,5,5,9,9] [1,1,6,6,10,10,10,10] [5,5,5,5,8,8,13,13] [4,4,10,10,10,10,10,10] [5,5,5,5,11,11,17,17] [1,1,1,15,15,15,15,15] [5,5,5,5,5,19,19,19] [7,7,10,10,10,10,14,14] [3,3,11,11,15,15,15,15] [9,9,9,10,10,10,10,23] [5,5,5,5,14,14,21,21] 570 Table 4: Cyclic 8-gons with rational area 7.3 Isosceles quadrilateral construction Whenever we attempt to build an n-gon with an even number of sides then we have an extra construction available that provides us with non-eulerian examples. Basically this works by simply glueing together rescaled versions of indecomposable Brahmagupta quadrilaterals with the same circumradius. This neatly generalises the method shown in the previous section 7.2. As a preliminary step along the way we now consider the following problem. 31

32 Find two isosceles quadrilaterals with a common circumradius and a common side. Since this results in an important construction we outline the approach. First use the parametrization (4) to fix a particular isosceles Brahmagupta quadrilateral and calculate its circumradius. Next use one of the non-repeated sides as the common side and the fixed circumradius to solve Parameśvara s formula for remaining two sides by assuming the discriminant is a perfect square. Finally, select one of the infinitely many solutions which satisfies a side inequality so that it fits between the existing quadrilateral and the circle. We work out a specific example. Suppose that [a 1, b 1, c 1, c 1 ] = [9, 1, 5, 5] is our starting isosceles quadrilateral so that the area and circumradius are K 4 = 15 and R 4 = Now we wish to find an isosceles Brahmagupta quadrilateral with sides [a 2, b 2, c 2, c 2 ], a circumradius of 5 34/6 and b 2 = b 1 = 1. Thus we want to solve = c 2 c a 2 (2c2 a 2 + 1)(2c 2 + a 2 1) for a 2 and c 2. Squaring this and writing it as a polynomial in a 2 leads to a (18c )a 2 + (18c c ) = 0 which has solutions for rational a 2 if and only if the discriminant is a rational square. The discriminant is given by D := c 2 2( 3 2 c ) so that there exists a rational E satisfying E 2 = 3 2 c Now let c 2 = 5x/3z and E = 5y/z to transform this equation into primitive form x 2 + y 2 = 34z 2 which has the particular solution [x, y, z] = [3, 5, 1] (obtained from the starting quadrilateral) and hence the general solution x z = 3P P Q 3Q 2 P 2 + Q 2, y z = 5P 2 6P Q 5Q 2 P 2 + Q 2. We want 0 < c 2 < 1 so that 9/2 < P/Q < 1/3. For example, the choice (P, Q) = (1, 4) leads to the solution [a 2, b 2, c 2, c 2 ] = [103/17 3, 1, 25/51, 25/51] with area K 4 = / Of course we can now simply repeat the process using 103/17 3 as the common side and so on until we have as many sides as we like. 32