IMFM
Institute of Mathematics, Physics and Mechanics JADRANSKA 19, 1000 LJUBLJANA, SLOVENIA
Preprint series Vol. 48 (2010), 1117
issn 2232-2094
DIOPHANTINE STEINER TRIPLES Bojan Hvala
Ljubljana, March 29, 2010
Diophantine Steiner triples
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BOJAN HVALA
Computer programs for dynamic geometry provide a very effective tool for motivating students. It is my experience that a certain effort in preparing an adequate applet and spending some time on a careful presentation can result in a great change of the atmosphere in the classroom. On the other hand, these programs also open new perspectives in geometry exploration and can provoke new interesting research questions related to the already known topics. The possibility of experimenting and the chance of visual examination of the results adds new dimensions to research and can also be appealing to non-mathematical audiences.
A good example, which can be presented in geometry class as an application of inversion is an applet related to the Steiner Porism. The picture below shows the main idea of the applet.
With three sliders we can dynamically vary the integer values of the radius R of the large circle K, the radius r of the small circle k (the two circles shown bold), and the distance d between the centres. After choosing a point T on K, the applet starts creating the Steiner chain of circles L1, L2,... inscribed into the region between K and k. The first circle Li touches K at T and also touches k. Next circles L2,L3... touch one another successively and they all touch the two circles K and k. If the n-th circle Ln also happens to touch the first circle L1, we say that L1,...,Ln is a closed Steiner chain of length n (cf. [3, 6, 7, 10]). The picture on the right shows a closed Steiner chain of lenght 6, while that on the left shows a Stainer chain, that is not closed. Note that a Steiner chain may also close after several loops around the central circle, but we we won't consider this situation here; for us, a closed chain means that it closes in the first round.
The applet provides an experimental method for finding the integer values of R, r and d for which Steiner chain closes.
In this paper, we find all integer triples (R, r, d) giving rise (as above) to closed Steiner chains.
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1. Steiner identity and Steiner triples
The famous result, called Steiner Porism, states that if the chain is closed for one starting point T, then it is closed for all starting points on K. It is also well known that
the existence of a closed Steiner chain of the length n > 3, is equivalent to the fact that parameters R, r, d and n are connected by the so called Steiner identity (cf. [7]):
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(R - r)2 - d2 = 4Rr tan2(n).	(1)
Let us start our investigation with clarifying some notions. A triple (R, r, d) of real numbers is called an acceptable triple, if 0 < r < R; d > 0, and R > r + d.
Acceptable triples therefore define two circles with different radii such that the larger circle contains the smaller one. For every acceptable triple we can then pick a point T on K and start constructing the Steiner chain.
An acceptable triple (R, r, d) of real numbers is called a Steiner triple, if the Steiner chain in the configuration described above closes. When a Steiner triple (R, r, d) is integer
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valued, we call it Diophantine Steiner triple, abbreviated DS triple. A DS triple is primitive if the greatest common divisor of the members is equal to 1: D(R,r,d) — 1.
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As is done in similar circumstances, we will restrict our search of DS triples to primitive DS triples. We will start with the investigation of the possible lengths of the associated Steiner chains.
2.The length n of the Steiner chain
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Theorem 1 Let (R, r, d) be a DS triple with the Steiner chain of length n. The only possible values for n are n = 3, n = 4 or n = 6.
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Proof: It follows from (1) that if (R, r, d) is a DS triple with the length of the Steiner chain equal to n, then yn — tan2(n) is a rational number. We will see bellow that yn is a root of a certain polynomial pn with integer coefficients. We will get the result investigating the rational roots of pn.
First, recall that einx = (eix)n gives us the formulas for sinnx, cosnx, from where we can obtain the formula for tan nx:
sin nx _	© - © tan2 x + ...
tan nx —	— tan x • Tn\ o
cos nx	1 — ('2J tan2 x + ...
Considering this equality for x — n, we see that yn — tan2(n) is a root of a polynomial
S	pn(y)— (n) - (n) y +(n) y2 - (n) y3 + ....
Let us now investigate two different cases: when n is odd and when n is even. We have:
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2k + 1 \ , / t\k k
P2k+i(y) — (2k + 1) -( 3 )y +... + (-1)ky
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^	P2k(y) — (2k) - (23k)y + ... + (-1)k-1 2kyfc_1
We shall first consider the odd case. For odd n > 5 we have 0 < yn = tan2(n) < tan2() = 1, and therefore yn is not an integer. But yn is a root of the polynomial pn = p2k+i = (—1)kyk + • • • + n with integer coefficients. The only possible rational roots of pn are the divisors of n and therefore integers. Hence, yn = tan2(n) is not rational and therefore for odd values n,n > 5, there is no DS triples with corresponding Steiner chain of length n.
Let us now consider the even case and prove a similar conclusion for even n > 8. To prove this we will need the following two Lemmas.
Lemma 1 For k > 4 we have: tan2(2k) < k.
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Proof: Since the function f (x) = tan2(x) is convex on the interval [0, 8], its graph on this interval lies under the segment with the vertices (0, f (0)) and (f, f (f)). Using
the fact f (f) = 3 — 2\/2, this observation can be written as f (x) < ^-j2^2) x. Applying this inequality for x = ; k > 4, we obtain
^	tan2(2k) < US-*® < ..
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Lemma 2 For every even number n = 2k and for all odd numbers m, 1 < m < 2k — 1 the binomial coefficient is even.
Proof: Since (1 + x)2k = (1 + 2x + x2)k, we have (1 + x)2k = (1 + x2)k(mod 2). Therefore the coefficients by the odd powers of x on the left hand side are even. □
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Using these Lemmas we will prove that, for k > 4, y2k = tan2(n) is not a rational
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number. We know that y2k is a root of the polynomial
£	p2k = ( — W1 + (-> + 2k.
Since all the coefficients are even (Lemma 2), we can divide by 2 and see that y2k is a root of a polynomial with integer coefficients of the type (—1)k-1 kyk-1 + ... + k. The positive rational roots of such polynomials are of the form p, where p and q are divisors of k. The smallest such positive rational number is 1. Lemma 1 states that 0 < y2k < 1, and therefore y2k is not among the rational candidates for the roots. This proves that for k > 4 the number y2k = tan2() is not rational. Therefore there is no DS triples with corresponding Steiner chain of length n for even n > 8. □
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3. Diophantine equations and solutions
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According to Theorem 1 we only have DS triples with Steiner chains of lenghts n = 3, n = 4, and n = 6. Therefore, the relation (1) simplifies into one of the following three equations:
R2 - 14Rr + r2 - d2	=0	n	=3	(2)
R2 - 6Rr + r2 - d2	=0	n	=4	(3)
3R2 - 10Rr + 3r2 - 3d2	=0	n	=6	(4)
Finding all DS triples means finding general solutions of these three quadratic Dio-phantine equations. The equations are of the same type as the equations that recently appeared in Math. Gazette, namely in papers [2, 4, 8, 9]; and our method of solving these equations is quite similar. Furthermore, it turns out that Diophantine Steiner triples are in fact closely connected to the triples which appear in the above-mentioned papers. These new findings will be presented in another, more geometrically orientated paper.
The table below presents all the primitive solutions (depending on parameters s and t) of the equations above and therefore yields a way to get a list of all primitive DS triples.
In all cases: s,t € N and D(s,t) = 1. Some additional limitations regarding s and t are specified for each case separately.
		R	r	d	
n = 3	(a)	3t2 + 4s2 + 7st	st	|3t2 - 4s21	t is odd, s is not divisible by 3
	(b)	12t2 + s2 + 7st	st	112t2 - s2|	s is odd, s is not divisible by 3
	(c)	6t2 + 2s2 + 7st	st	16t2 - 2s2|	t, s are odd, s is not divisible by 3
n = 4		t2 + 2s2 + 3st	st	|t2 - 2s21	t is odd
n = 6	(a1)	3 (4s2 + t2 + 5st)	st	1 |4s2 - t2|	t is odd, one of s,t is congruent 1(mod 3) and the other congruent 2(mod 3)
	(a2)	4s2 + t2 + 5st	3st	|4s2 -12|	t is odd, either s = t ^ 0(mod 3) or one of s, t = 0(mod 3) and s = t(mod 3)
	(b1)	3 (2t2 + 2s2 + 5st)	st	2 |t2 - s2|	t, s are odd, s = t = 0(mod 3)
	(b2)	2t2 + 2s2 + 5st	3st	2 |t2 - s2|	t, s are odd, s = t(mod 3)
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In the proofs of the results above, the following Lemma will be frequently used: Lemma 3 Let p be a prime number and x, y, u positive integers, satisfying the relation:
xy = pu2.
If D(x, y) = 1, then one of the factors x, y is of the form ps2 and the other one is of the form t2; s,t € N.
Proof: Since the prime number p divides xy and D(x,y) = 1, it must divide one of the factors, for instance x = pxi. Therefore we have yxi = u2. Write u as the product of prime numbers and the result follows immediately, since D(x1,y) = 1. □
Going back to prooving that the above table presents the primitive solutions to Dio-phantine Equations (2),(3) and (4), we have to consider the three cases separately. Since the techniques of these three considerations are quite similar, we will present here just the proofs of the cases n = 3 and n = 6. The easiest case n = 4 is omitted.
Proof of the case n = 3: A straightforward check shows us that R, r, d of the cases (a), (b), (c) above satisfy equation (2). Let us prove that the triple (R, r, d) of the case (a) is primitive. Suppose p is a prime, dividing R, r and d. Note that t and d are odd, so p = 2. Note also that s and d are not divisible by 3, so p = 3. Since p divides all three elements of the set {R + d, R — d, r} = {6t2 + 7st, 8s2 + 7st, st}, we have p|6t2 and p|8s2. Since p = 2 and p = 3, p divides both t and s, which contradicts D(s,t) = 1. This contradiction proves that (R, r, d) is primitive. In cases (b) and (c) the proof that (R, r, d) is primitive is similar.
The presented triples are therefore primitive and satisfy equation (2). Let us prove
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that all the primitive solutions of (2) are listed above. First we write (2) in the form
(R - 7r + d)(R - 7r - d) = 48r2	(5)
and prove that D((R - 7r + d), (R - 7r - d)) is equal either to 2 or to 4. Let p € {2,3} be a prime, dividing both numbers. Then p divides the sum and the difference of those numbers and therefore p divides d and R - 7r. It follows from (5) that p divides r as well, and therefore also R, contradicting D(R, r, d) = 1. In the case of p = 3 the proof is similar. Therefore we have:
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D((R - 7r + d), (R - 7r - d)) = 2"
Let us check that the only possible values of m are 1 and 2. The possibility m > 3 would, together with (5) imply that r, d, R - 7d and R are even, again contradicting the fact that the triple is primitive. In the case m = 0, one of the factors in the left hand side of equation (5) would be divisible by 16 and the other one would be odd. Thus, the difference is odd. While substracting the two numbers one would get 2d, which is a contradiction. So, the only possibilities left are m = 1, 2.
In case m = 2 we have (R - 7r + d) = 4x1, (R - 7r - d) = 4x2 for some x1, y1 € Z and D(x1,y1) = 1. Relation (5) can now be rewritten as x1y1 = 3r2. It follows from Lemma 3 that one of the factors is equal to 3t2 and the other to s2. Note that D(s,t) = 1 and r = st. We get two possibilities, both of them leading to the solution (c).
In case m = 1, one of the numbers (R - 7r + d), (R - 7r - d) is divisible by 2 and the other (because of (5)) divisible by 8. We have two possibilities: (R - 7r + d) = 8x1 and (R - 7r - d) = 2y1 or (R - 7r + d) = 2x1 and (R - 7r - d) = 8y1. In both cases we have D(x1,y1) = 1 and the relation (5) can be rewritten as x1y1 = 3r2. Applying Lemma 3 again and examining all four possibilities, we conclude that they all lead to solutions (a) and (b).
Remarks in the table about t or s being odd or not divisible by 3 are necessary to ensure that the triple is indeed primitive. □
Note that all three relations of the case n = 3 could be covered by one single relation R = k(6t2 + 2s2 + 7st), r = kst, d = k|6t2 - 2s2|,	(6)
where k — 1 if both s, t are odd, and k — 1 otherwise. In the case when s is even, replace s with 2s in (6) to obtain case (a). Similarly we obtain case (b) when t is even, and case (c) when both s, t are odd.
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Proof of the case n=6: The relation (4), namely: 10Rr — 3(R2 + r2 - d2) implies that one of two radii R and r is divisible by 3.
First consider the case R — 3Ri. The relation (4) can be rewritten in the form:
(9Ri - 5r + 3d)(9Ri - 5r - 3d) — 16r2.
(7)
Again we consider D(9R1 - 5r + 3d, 9R1 - 5r - 3d). With arguments similar to those in the proof of the case n — 3 we see that
D(9Ri - 5r + 3d, 9Ri - 5r - 3d) — 2m
where m — 1 or m — 2.
The case m — 2 gives 9Ri - 5r + 3d — 4xi, 9Ri - 5r - 3d — 4yi, D(xi, yi) — 1 and xiyi — r2, which together with Lemma 3 leads to solution (b1).
In the case m — 1 one of the factors on the left hand side of (7) is of the form 2xi, and the other of the form 8yi where D(xi,yi) — 1. The relation (7) becomes xiyi — r2 with the solution xi — t2, yi — s2, D(s,t) — 1. This leads to the solution (a1).
If one of the numbers s, t were divisible by 3, since R is integer valued it would follow that the other number would also be divisible by 3, contradicting D(s, t) — 1. The requirement about one or two parameters being odd is there to ensure that not all the members of the triple are even. The requirement about congruences (mod 3) ensure that R is integer valued. Since s and t are not divisible by 3, t2 = s2 = 1(mod 3), so d is certainly integer valued. Now consider R. In solution (a1) we have 4s2 +12 + 5st = 5 + 5st(mod 3). The condition s = t(mod 3) ensures that 5 + 5st = 0(mod 3), implying that R is integer valued. We consider solution (b1) in a similar way.
To conclude the proof, we have to consider the case, when instead of R, the radius r is divisible by 3, thus r — 3ri. In this case, instead of (7) we have the relation:
(R - 5ri + d)(R - 5ri - d) — 16r2.
(8)
Using standard techniques we prove that D((R - 5ri + d), (R - 5ri - d)) — 2m for m — 1 or m — 2. The case m — 2 leads us to the solution (b2), while m — 1 leads to the solution (a2). □
Note again that the solutions (a1), (a2), (b1), and (b2) could be presented by a single relation:
R — k(2t2 + 2s2 + 5st), r — 3kst, d — 2k|t2 - s2|,
where k € {|, |, i, 1}. When one of t, s is even, put 2t or 2s instead of t or s in the formula and select k to factor out 2.
The next table brings us a list of DS triples (R, r, d) for the values R < 100.
n — 3
(14,1,1) (15,1,4) (20,1,11) (33, 2,13) (52, 3, 23) (77, 3, 52) (85, 6,11) (91, 5, 44) (95, 4, 61)
(63, 2, 47) (72, 5, 13)
n—4
(6,1,1) (63, 10, 17)
(15, 2, 7) (66, 5, 49)
(20, 3, 7) (72, 7, 47)
(28, 3,17) (77,12, 23)
(35, 6,1) (88,15, 7)
(42, 5, 23) (91, 6, 71)
(45,4, 31) (99,14, 41)
n—6
(3,1,0) (45, 7, 32) (85,12, 63)
(9, 2, 5) (45,14,11) (88, 21,45)
(10, 3, 3) (52, 9, 35) (91, 30, 9)
(18, 5, 7) (55, 18, 7) (99, 8, 85)
(28, 9, 5) (60,11, 39)
(35, 9,16) (63, 20,13)
(42, 5, 33) (77,15, 48)
4. Some ideas for possible further investigations

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A look at the list of DS triples shows that there exist values R allowing triples in all three categories: n = 3, n = 4 and n = 6. Such values are, for instance
R = 63,77,91,130,143,187,195, 209, 221,... A closer inspection of this sequence could be interesting.
• Similarly, we could ask whether there exist different primitive DS triples inside the same category sharing the same first term R. The answer is positive. For instance, such triples for category n = 4 are (2145, 32, 2047) and (2145, 368,17). Similarly, (5005, 348, 887) and (5005, 323,1588) belong to category n = 3 and (5005,1548,1273) and (5005, 732,3657) belong to category n = 6. The description of all such triples could also be of interest.
We noticed that the mapping
f-r, i\ f5R + r + d R + 5r +
(R, r, d) ^ ^-2-,-2-, R + r — dj
which has integer solutions
'5R + r + d R + 5r + d
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maps primitive DS triples of the case n = 3 into DS triples of the case n = 6. Direct calculation shows that if the left-hand triple satisfies identity (2), then the right-hand triple satisfies (4). Some deeper reasons for this correspondence will be explained in our geometry paper mentioned above.
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• We could, in a similar way, investigate Diophantine Poncelet triples, based on Pon-celet Porism ([1, 5, 11]) instead of on Steiner Porism. In Poncelet's case n = 3, the relation (2) is replaced by the Euler triangle formula
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R2 — 2Rr — d2 = 0,
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R = k(s +t)2, r = 2kst, d = k ■ |s2 — t21,
where t, s € N, D(s, t) = 1 and k = 2 when both s and t are odd, and k = 1 otherwise.
For n > 4 it seems that essentially different methods would be needed.
References
[1]	H. J. M. Bos, C. Kers, F. Oort, D. W. Raven, Poncelet's closure theorem, its history, its modern formulation, a comparison of its modern proof with those by Poncelet and Jacobi, and some mathematical remarks inspired by these early proofs. Expositiones Mathematicae 5 (1987), pp. 289-364.
[2]	B. Burn, Triangles with a 60° angle and sides of integer length, Math. Gazette, 87 (March 2003) pp. 148-153.
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[3]	H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, Math. Assoc. Amer., Washington DC (1967).
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[4]	J. Gilder, Integer-Sided Triangles with an Angle of 60°, Math. Gazette, 66 (December 1982) pp. 261-266.
[5]	S. M. Kerawala, Poncelet Porism in Two Circles. Bull. Calcutta Math. Soc. 39 (1947), pp. 85-105.
[6]	C. S. Ogilvy, Excursions in geometry, Dover Publications, New York (1990).
[7]	D. Pedoe, Geometry, a comprehensive course, Dover Publications, New York (1988).
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[8]	E. Read, On integer-sided triangles containing angles of 120° or 60°, Math. Gazette, 90 (July 2006) pp.299-305.
[9]	K. Selkirk, Integer-sided triangles with an angle of 120°, Math. Gazette, 67 (December 1983) pp. 251-255.
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[10] E. W. Weisstein, Steiner Chain. From MathWorld-A Wolfram Web Resource. http://mathworld.wolfram.com/SteinerChain.html
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[11] E. W. Weisstein, Poncelet's Porism. From MathWorld-A Wolfram Web Resource. http://mathworld.wolfram.com/PonceletsPorism.html
BOJAN HVALA
Department of Mathematics and Computer Science, University of Maribor,
FNM, Koroska cesta 160, 2000 Maribor, Slovenia e-mail: bojan.hvala@uni-mb.si