Hyperbolic Heron Triangles and Elliptic Curves
aa r X i v : . [ m a t h . N T ] F e b HYPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES
MATILDE LALÍN AND OLIVIER MILA
Abstract.
We define hyperbolic Heron triangles (hyperbolic triangles with “rational”side-lengths and area) and parametrize them in two ways as rational points of certainelliptic curves. We show that there are infinitely many hyperbolic Heron triangleswith one angle α and area A for any (admissible) choice of α and A ; in particular,the congruent number problem has always infinitely many solutions in the hyperbolicsetting. We also explore the question of hyperbolic triangles with a rational medianand a rational area bisector (median splitting the triangle in half). The problem of finding triangles with rational area and side lengths in the Euclideanplane goes at least as far back as ∼
600 A.D with the Indian mathematician Brah-magupta (see [5]). If the triangle is assumed right, this is the classical congruentnumber problem (a number is congruent if it is the area of a right triangle with ra-tional sides). Remarkably, this problem is equivalent to finding (non-torsion) rationalsolutions to the elliptic curve y = x − n x . For non-right triangles, it was shown byGoins and Maddox [5] that Heron triangles are in correspondence with rational pointson the curve y = x ( x − nτ )( x + nτ − ) , where n denotes the area and τ is the tangentof half of an angle.In this paper we investigate the analog problem in the hyperbolic plane. The firstconcept we need to transport is that of rationality. Unlike the Euclidean case wheretrigonometric laws are polynomial in the area, side lengths and sine and cosine of theangles, their hyperbolic counter part involve the hyperbolic sine and cosine of the sidelengths, and the sine and cosine of the area. For instance, in a triangle with side lengths a, b, c and a right angle opposing side a , Pythagoras’ Theorem and the area A are givenby: cosh( a ) = cosh( b ) cosh( c ) and sin( A ) = sinh( b ) sinh( c )cosh( a ) + 1 respectively. It is thus natural to ask that the sine/cosine of the angles be rational, andsimilarly for the hyperbolic functions of the sides, instead of directly asking that thesequantities be rational.For the rationality of the hyperbolic functions on the side lengths, at least two con-ventions have been used in the literature. One such choice was made by Brody andSchettler [1]. They call a triangle rational if the hyperbolic tangent of its side lengths This work is supported by the Swiss National Science Foundation, Project number
P2BEP2_188144 ,by the Natural Sciences and Engineering Research Council of Canada, Discovery Grant ,and by the Fonds de recherche du Québec - Nature et technologies, Projet de recherche en équipe . are rational, and they use this definition to prove a correspondence between rationaltriangles of semi-perimeter tanh − ( σ ) and inradius sinh − ( ρ ) and rational points on thecurve σ ( x y + xy + ρ ( x + xy + y − ρ )( x y + xy ) .A second (stronger) choice is that of Hartshorne and van Luijk [7]. Here they call alength x rational if e x ∈ Q , and then study Pythagorean triples in this context. Thisis the notion we will adopt in this paper.For the area, the Gauss–Bonnet theorem implies that a triangle with angles α, β, γ has area A = π − α − β − γ. The previous discussion together with the above formula for the area suggest a commondefinition of rationality for angles and area: we call an area A and an angle α rationalif the sine and cosine of these quantities are rational (or equivalently if e iA , e iα ∈ Q [ i ] ).Thus a hyperbolic triangle with area A , angles α, β, γ and side lengths a, b, c is a hy-perbolic Heron triangle if e a , e b , e c ∈ Q and e iα , e iβ , e iγ , e iA ∈ Q [ i ] . By the well known circle parametrization, e ix ∈ Q [ i ] if and only if cos( x ) = − t t and sin( x ) = t t for some t ∈ Q . By abuse of terminology we will call t the rational angle (resp. rational area ) of a hyperbolic triangle if its angle (resp. area) is x .Our main result is the following: Theorem 1.
There is a one-to-one correspondence between hyperbolic Heron triangleshaving rational area m and one rational angle u , and rational points satisfying an opencondition (A) (to be described later) on the curve: y = x ( x − n )( x − n ( u + 1)) , where n = m ( m + 1)(2 u − m ( u − . This open condition (A) encodes the fact that the angles are positive and that thearea is < π , and it also excludes two further points on the curve. Computing the rankof the above curve (and proving it is ≥ ) we get: Corollary 2.
For almost all m, u ∈ Q with m, u > and mu < there are infinitelymany hyperbolic Heron triangles with rational area m and one rational angle u . Here “almost” means all values except possibly those lying on finitely many curves { f i ( m, u ) = 0 } of with deg( f i ) ≥ . Setting u = 1 (which corresponds to the case ofone right angle), it is easily verified that the f i ( m, have no rational solution. Thusthe congruent number problem always has infinitely many solutions in the hyperbolicsetting.One can also parametrize hyperbolic Heron triangles using side lengths: Theorem 3.
There is a one-to-one correspondence between hyperbolic Heron triangleshaving two sides of lengths log( v ) and log( w ) with v, w ∈ Q , and rational points satis-fying an open condition (B) (to be described later) on the curve: y = x (cid:0) x − ( v − v − ) (cid:1) (cid:0) x − ( w − w − ) (cid:1) . YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 3
Here the condition (B) describes the fact that the lengths are positive and satisfy thetriangle inequality. It turns out that (perhaps surprisingly) this curve has genericallyrank 0 over Q , suggesting that it is harder to complete two sides to a rational triangle,than it is to find a triangle with a fixed angle and area.In the Euclidean world, a further question one can ask is that of the rationality ofthe medians. This was first asked (and solved) by Euler in the case of one rationalmedian [4]. The problem for Heron triangles having 3 rational medians is problem D21in Guy’s book [6]; it is still open as of today. The two-median problem was solved byBuchholz and Rathbun [2, 3].In the hyperbolic setting we have again two choices in our translation of median: the(hyperbolic) line from one vertex meeting the opposite edge in its midpoint, or the linefrom one vertex separating the triangle into two triangles of equal area. We will callthe first one the median and the second one the area bisector . These two lines are notthe same in general (one can easily be convinced by considering ideal triangles), butthey coincide in an isosceles triangle (for the lines passing through the apex).The first (negative) result along those lines concerns the simple case of equilateraltriangles. Proposition 4.
There are no equilateral hyperbolic Heron triangles. Moreover, equi-lateral triangles with rational sides or rational angles have no rational medians/areabisectors.
With similar methods as those used for Theorem 1 and Theorem 3, we can pa-rametrize hyperbolic triangles with one rational median and Heron triangles with onerational area bisector using elliptic curves. However, these curves are quite complicated,so we will present them in Section 4 and 5 respectively. We will only give one corollaryof this parametrization, in the case of medians:
Theorem 5.
For almost all values u, w ∈ Q there are infinitely many hyperbolic tri-angles having rational side lengths, two of which given by a = 2 log( u ) and b = log( w ) ,and one rational median (intersecting side a ). Here almost all means all but those cut out by a curve in u, w .The paper is organized as follows. Section 1 and Section 2 are dual to each other, andcover the parametrization of Heron triangles in terms of angles (Theorem 1) and sides(Theorem 3) respectively. Section 3 is focused on medians and area bisectors in thesimple case of equilateral triangles. Finally, in Section 4 we give the parametrization ofhyperbolic triangles with rational side lengths and one rational median, while Section 5is devoted to the dual computation of the parametrization of Heron triangles with onerational area bisector.1.
Hyperbolic heron triangles – Angle parametrization
Finding the angle parametrization.
In this section we give the parametriza-tion of hyperbolic Heron triangles in terms of angles and area, and prove Theorem 1and Corollary 2. Note that throughout this paper we only consider the case of non-degenerate bounded triangles (i.e., with no vertex at infinity), so that its angles andarea are always positive and its side lengths finite.
MATILDE LALÍN AND OLIVIER MILA
Let α, β, γ > denote the angles of a hyperbolic triangle. Assume they are rational(as defined in the introduction) i.e., that e iα , e iβ , e iγ ∈ Q [ i ] . Since the area A = π − α − β − γ we get that A is also rational: e iA ∈ Q [ i ] .If we denote by a (resp. b, c ) the side length opposite α (resp. β, γ ), the hyperboliclaw of cosines (for the angles) reads:(1) sin( β ) sin( γ ) cosh( a ) = cos( α ) + cos( β ) cos( γ ) and similar formulas involving cosh( b ) , cosh( c ) . As the angles are rational, it followsthat cosh( a ) , cosh( b ) , cosh( c ) ∈ Q . To get a hyperbolic Heron triangle, it thus onlyremains to find the condition that sinh of a, b, c are also rational.The (inverse of) the hyperbolic law of sines gives sinh( a )sin( α ) = sinh( b )sin( β ) = sinh( c )sin( γ ) . Multiplying by sin( α ) sin( β ) sin( γ ) we get: sinh( a ) sin( β ) sin( γ ) = sinh( b ) sin( α ) sin( γ ) = sinh( c ) sin( α ) sin( β ) . Call this quantity ∆ . Squaring (1) we get sin( β ) sin( γ ) (sinh( a ) + 1) = (cos( α ) + cos( β ) cos( γ )) and thus:(2) ∆ = (cos( α ) + cos( β ) cos( γ )) − sin( β ) sin( γ ) ∈ Q . Using trigonometric identities, we can rewrite this as a symmetric expression in α, β, γ : = cos( − α + β + γ ) + cos( α − β + γ ) + cos( α + β − γ )+ cos( α + β + γ ) + cos(2 α ) + cos(2 β ) + cos(2 γ ) + 1 , and we have that ∆ ∈ Q if and only if all three of sinh( a ) , sinh( b ) , and sinh( c ) arerational. Substituting γ = π − A − α − β and expanding the cosines, we obtain (writing c A = cos( A ) , s A = sin( A ) , etc...):(3) = ( c A − s A ) (cid:2) ( c α c β − s α s β ) − ( c α s β + c β s α ) (cid:3) − c A s A (cid:2) c α s α ( c β − s β ) + c β s β ( c α − s α ) (cid:3) − c A (cid:2) ( c α c β − s α s β ) − ( c α s β + c β s α ) + 2 c α + 2 c β − (cid:3) +4 s A ( c α s α c β + c β s β c α ) + 2 c α + 2 c β − . We wish to express this in terms of rational angles. Setting cos( A ) = − m m , cos( β ) = − u u , cos( α ) = − t t , sin( A ) = m m , sin( β ) = u u , sin( α ) = t t , and w = ( m + 1)( u + 1)( t + 1)∆ , equation (3) rewrites as: w =4 m ( mu − m − u )( mt − t − m ) (cid:2) ( mu − m − u ) t (4) + ( − mu − u + 2) t − mu + m + 2 u (cid:3) . YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 5
Finally, setting n = m ( m + 1)(2 u − m ( u − and applying the change of variables y = − (2 u − m ( u − t (cid:2) u − m ( u − mu − m + u ) m t − m u − m − mu − u + 2)(2 u − m ( u − m t + 2(2 u − m ( u − m − ( mu − m + u ) mtw + (2 u − m ( u − m w (cid:3) x = (2 u − m ( u − t (cid:2) − mu − m + u ) mt + 2( m u + m − mu + 2) mt + 2(2 u − m ( u − m + mw (cid:3) , with inverse t = − m (2 u − m ( u − x − ( m + u ) ( m + 1)) y + ( m + u )(1 − mu )( x − n ) w =2 m (2 u − m ( u − y + ( m + u )(1 − mu )( x − n )) − × (cid:2) x − m + 1)( u + m ) x + m ( m + 1) (2 u − m ( u − m u + 2 u + 2 mu + 2 m u + 4 u + 6 mu + 3 m ) x − m ( m + 1) ( u + m ) ( u + 1)(2 u − m ( u − + 2( m + 1) u ( u + m )( mu − y (cid:3) we get the equation:(5) y = x (cid:0) x − n (cid:1)(cid:0) x − n ( u + 1) (cid:1) . Its discriminant is u ( u + 1) n = 2 u m ( u + 1) ( m + 1) (2 u − m ( u − .We are ready to complete the proof of Theorem 1. Proof of Theorem 1.
It just remains to exhibit the open condition (A), which encodesthe fact that the parameters give rise to an actual hyperbolic triangle. Assume firstthat the inverse of the change of variables is defined; this happens everywhere but onthe line y + ( m + u )(1 − mu )( x − n ) = 0 . We claim that the conditions on m, u, t ,along with their meaning are given by the following table: The first three conditionsCondition Translation m > area A > t > α > u > β > tm + tu + mu − tmu − t − m − u ) > γ > mu < α + β + γ < π . Table 1.
Conditions for Theorem 1follow trivially from the circle parametrization, and so does the fourth one if one writesthe formula for the rational angle corresponding to γ . For the fifth one, assuming thefour others, we know that < A, α, β, γ < π , and we must find a condition encoding A + α + β + γ = π . By construction, any solution to (5) gives rise to a set of parameterssatisfying this last equation modulo π ; thus we have A + α + β + γ ∈ { π, π } and MATILDE LALÍN AND OLIVIER MILA we wish to eliminate the π case. For such a set of parameters, the π case happensexactly when any two elements in A, α, β, γ sum up to > π . Hence we can just pickthe condition A + β < π , which translates as mu < if one writes the formula for therational angles.Now if m, u are given parameters with m, u > and mu < , the conditions fromTable 1 can be simplified to the following open condition on the variable t :(A) < t < − mum + u . Using the above change of variables, we get an open condition in terms of x, y encodingthe desired properties.It remains to treat the case of points where the expression for t is not defined, i.e.,points of the curve (5) lying on the line y + ( m + u )(1 − mu )( x − n ) = 0 . There are 3such points: the torsion point T = ( n, , the point P = (cid:0) ( m + 1)( m + u ) , u ( m + 1) ( m + u )( mu − (cid:1) and the third point − ( P + T ) . Looking back at points on (4) mapping to the line, wefind only one: the point ( w, t ) with t = 2 m ( m + u )(1 − mu )(2 u − m ( u − m u − m u − m u + m + 4 m u + 6 m u + u . Its image is the point P . So whenever this point exists and fulfills the condition (A)we get a pre-image for P , and otherwise we do not. The condition (A) thus needs tobe strengthened to exclude those points. (cid:3) Rank computations.
In this section we compute the rank of the above curve(fixing one parameter u or m ) and prove Corollary 2. We will write E m,u for the curvegiven by (5), and E m (resp. E u ) for the same curve seen over C ( u ) (resp. C ( m ) ). Lemma 1.1.
The ranks of the K surfaces E m ( C ( u )) and E u ( C ( m )) fulfill the inequal-ities: ≤ rk( E m ( C ( u ))) , rk( E u ( C ( m ))) ≤ . Moreover, the point P ( m, u ) = (cid:0) ( m + 1)( m + u ) , u ( m + 1) ( m + u )( mu − (cid:1) is a point of infinite order on E m,u , and the torsion group is isomorphic to Z / Z × Z / Z ,with the points of order two given by (0 , , ( n, , and ( n ( u + 1) , .Proof. The lower bound follows from the fact that P is on the curve E m,u and of infiniteorder, which is easily verified by inspection. Alternatively one can see that the heightpairing of the Mordell–Weil group on E u (resp. on E m ) gives h ( P ) = 2 .For the upper bound, we apply Tate’s algorithm [11, IV.9]. For E u ( C ( m )) we seethat the singularities m = 0 , ± i, uu − are all of type I ∗ in the Kodaira classification,and so the number of components of each singular fiber is 5. Now by the Shioda–Tateformula (see [10, Corollary 1.5], or alternatively [9, Corollary 6.7)]) we have ρ ( E u ) = rk( E u ( C ( m ))) + 2 + 4 · (5 −
1) = rk( E u ( C ( m ))) + 18 YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 7 where ρ ( E u ) is the Picard number of E u . Since E u is a K surface, we have ρ ( E u ) ≤ ,and thus rk( E u ( C ( m ))) ≤ .For E m ( C ( u )) , we use the same formula. The singularities at u = 0 and u = ∞ areof type I , the ones at u = ± i of type I and the ones at the roots of mu − u − m oftype I ∗ . Applying the Shioda–Tate formula again, ρ ( E m ) = rk( E m ( C ( u )) + 2 + 2 · (4 −
1) + 2 · (2 −
1) + 2 · (5 −
1) = rk( E m ( C ( u )) + 18 . Thus rk( E m ( C ( u ))) ≤ also in this case.Finally, it is immediate to see that (0 , , ( n, , and ( n ( u +1) , are points of order 2.Recall that the Euler characteristic satisfies χ = 2 for K surfaces. We combine thiswith the bound on the rank and Table (4.5) in [8] to conclude that E u (resp. E m )has torsion isomorphic to Z / Z × Z / Z or Z / Z × Z / Z . Then one can check directlythat the points of order 2 cannot be written as twice a point in E u ( C ( m )) (resp. in E m ( C ( u )) ). (cid:3) If we set u = 1 in the case of E u , the singularities are at m = 0 , ± i, ∞ and theprevious Lemma still applies. In this particular case, we can identify a second point ofinfinite order given by Q ( m ) = (cid:0) m ( m + 1) , i m ( m − (cid:1) . One can prove that the height pairing of the Mordell–Weil group on E gives h ( Q ) = 2 ,and that h P, Q i = 0 , showing that P ( m, and Q ( m ) are independent, and that therank of E ( C ( m )) is exactly 2.We move on to the proof of Corollary 2. Proof of Corollary 2.
We need to show that for each fixed m, u > with mu < , thereare infinitely many rational points on the curve E m,u satisfying the open condition (A).By a theorem of Poincaré and Hurwitz (see [12, Satz 11, p. 78]) the rational points E m,u ( Q ) of E m,u are dense in E m,u ( R ) provided E m,u ( Q ) is infinite and both connectedcomponents of E m,u ( R ) contain a rational point. Hence the corollary will follow bydensity, once these two conditions are proven.Now by Lemma 1.1, E m,u ( Q ) has positive rank, so it is infinite. And since the 3non-trivial torsion points are given by (cid:0) , (cid:1) , (cid:0) n, (cid:1) , (cid:0) n ( u + 1) , (cid:1) and are rational, we conclude that both components have rational points. (cid:3) Hyperbolic heron triangles – Side length parametrization
Finding the side lengths parametrization.
This section is devoted to theparametrization of hyperbolic Heron triangles using side lengths. The arguments arequite similar to those of Section 1, but using the (dual) hyperbolic law of cosines forthe side lengths.Let a, b, c denote the side lengths of a (non-degenerate bounded) hyperbolic triangle,and assume that e a , e b , e c are rational. Let α (resp. β, γ ) be the angles opposing thesides of length a (resp. b, c ). By the law of cosines (for the side lengths)(6) sinh( b ) sinh( c ) cos( α ) = cosh( b ) cosh( c ) − cosh( a ) MATILDE LALÍN AND OLIVIER MILA and hence cos( α ) is rational (and thus also cos( β ) and cos( γ ) for similar reasons).Multiplying the hyperbolic law of sines sin( α )sinh( a ) = sin( β )sinh( b ) = sin( γ )sinh( c ) by sinh( a ) sinh( b ) sinh( c ) , we get sin( α ) sinh( b ) sinh( c ) = sin( β ) sinh( a ) sinh( c ) = sin( γ ) sinh( a ) sinh( b ) . Call this quantity ∆ . We have that ∆ is rational if and only if sin( α ) , sin( β ) , sin( γ ) are rational.As in Section 1, we square (6) to get sinh( b ) sinh( c ) (1 − sin( α ) ) = (cosh( b ) cosh( c ) − cosh( a )) . Hence ∆ = sinh( b ) sinh( c ) − (cosh( b ) cosh( c ) − cosh( a )) ∈ Q . As a side note, observe that if A denotes the area of the triangle, we have sin( A ) = − sin( α ) sin( β ) sin( γ ) + sin( α ) c β c γ + sin( β ) c α c γ + sin( γ ) c α c β where c α = cos( α ) , etc... Since ∆ sin( α ) ∈ Q , and similarly for β, γ , we conclude that sin( A ) = r ∆ for some r ∈ Q . Thus a hyperbolic triangle with rational side lengths hasrational area exactly when ∆ ∈ Q , i.e., when all its angles are rational.Rewriting the equation for ∆ in a symmetric way, we get: ∆ = 1 − cosh( a ) − cosh( b ) − cosh( c ) + 2 cosh( a ) cosh( b ) cosh( c ) . Letting u = e a , v = e b and w = e c (so that u, v, w ∈ Q ), this equation rewrites as: u v w ∆ = ( uv − w )( uw − v )( vw − u )( uvw − . We introduce the following change of variables y = 2 u ( v − w − v w − vw ( u − vw ) , x = ( v − w − uvw − vw ( vw − u ) , with inverse ∆ = ( v − w − v w − y (cid:0) x + ( v − w − (cid:1)(cid:0) v w x + ( v − w − (cid:1) ,u = v w x + ( v − w − vw (cid:0) x + ( v − w − (cid:1) . This leads to the desired equation:(7) y = x (cid:0) x − ( v − v − ) (cid:1) (cid:0) x − ( w − w − ) (cid:1) . Its discriminant is given by ( v − v − ) ( w − w − ) ( v − w − vw − ) ( vw − v − w − ) .We can now proceed towards the proof of Theorem 3. YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 9
Proof of Theorem 3.
It suffices to exhibit the open condition (B) that ensures the pa-rameters give rise to a hyperbolic triangle. First, all the side lengths must be positive,whence u > , v > and w > . Second, the three triangle inequalities must besatisfied, i.e., u < vw, v < uw and w < uv . Assuming v, w are fixed and > , we getthe condition:(B) max (cid:16) vw , wv (cid:17) < u < vw. Under this condition, the above change of variable is always defined (and so is its inversesince x is necessarily positive). Thus the theorem is proven. (cid:3) Rank Computations.
This section is similar to 1.2. Let E v,w denote the curvegiven by (7), and let E v (resp. E w ) denote the same curve seen over C ( w ) (resp. C ( v ) )).Our goal is to give bounds on the rank of E v and E w ; since the equation for E v,w issymmetric in v and w , it is enough to consider the curve E v . Lemma 2.1.
The rank of the K surface E v ( C ( w )) satisfies ≤ rk( E v ( C ( w ))) ≤ . Moreover, the point R ( v, w ) = (cid:0) − vw ( v − v − )( w − w − ) , ivw ( v − v − )( w − w − )( vw − v − w − ) (cid:1) is a point of infinite order on E v,w .In addition, the torsion group is isomorphic to Z / Z × Z / Z , generated by S ( v, w ) = (cid:0) ( v − v − )( w − w − ) , i ( v − v − )( w − w − )( v − − w − )( vw + 1) (cid:1) and S ( v, w ) = (cid:0) ( v − v − ) , (cid:1) . Proof.
The proof of this result is very similar to that of Lemma 1.1. As before, thelower bound for the rank follows from the fact that R is on the curve and of infiniteorder, which is easily verified by inspection. More formally, one can see that the heightpairing of the Mordell–Weil group on E v gives h ( R ) = .We observe that E v has singularities at w = 0 , ∞ , ± of type I in the Kodairaclassification, while the singularities at w = ± v, ± v − are of type I . By the Shioda–Tate formula, we have ρ ( E v ) = rk E v ( C ( w )) + 2 + 4 · (4 −
1) + 4 · (2 −
1) = rk E v ( C ( w )) + 18 . Since E v is a K surface, we have ρ ( E v ) ≤ , and thus rk( E v ( C ( w ))) ≤ .One can directly check that S and S generate a subgroup isomorphic to Z / Z × Z / Z . Combining the information about the rank and the Euler characteristic withTable (4.5) in [8] we conclude that the torsion group is given exactly by Z / Z × Z / Z . (cid:3) Even if Lemmas 1.1 and 2.1 are very similar from the point of view of the arithmeticof the involved K surfaces, they contain a fundamental difference for the geometricproblem. In the case of Lemma 1.1, the point P ( m, u ) is defined over Q ( m, u ) andgenerates a nontrivial solution to the question of the Heron hyperbolic triangle withgiven area and angle. In contrast, the point R ( v, w ) of Lemma 2.1 is certainly not defined over Q ( v, w ) , and we speculate that there is no point of infinite order over Q ( v, w ) . This results in the following: we do not know a priori whether there is aHeron hyperbolic triangle with given sides v and w . This will depend on the choice of v and w , much like the classical congruent number problem depends on the choice ofthe area. 3. Equilateral Triangles
This short section covers the specific case of (non-degenerate bounded) equilateraltriangles.
Proposition 3.1.
There are no equilateral hyperbolic Heron triangles.Proof.
Let α denote the angle of an equilateral hyperbolic triangle. The Heron condition(2) from Section 1 in this case is: ∆ = 2 cos( α ) + 3 cos( α ) − α ) − α ) + 1) . Setting u = ∆ cos( α )+1 , this equation rewrites as u = 2 cos( α ) − . Thus the solutions tothe original equation are parametrized by cos( α ) = u +12 and ∆ = u +3 u , for u ∈ Q .Now squaring the equation for cos( α ) , and setting v = 2 sin( α ) , we get v = − u − u + 3 . Making the change of variables u = x − x +1 , v = y ( x +1) (with inverse x = u − u , y = v ( u − ),we get the Weierstrass form y = x ( x + x + 1) . This has rank 0, and the only nontrivial torsion point is (0 , which does not give anactual triangle since v = 0 . (cid:3) We proceed to prove the second part of Proposition 4:
Proposition 3.2.
If an equilateral hyperbolic triangle has either rational side lengthsor rational angles, then it has no rational median/area bisector.Proof.
First observe that for equilateral triangles, mediators, bisectors, medians, andarea bisectors all coincide, so we are free to use any property of these we want. Consideran equilateral hyperbolic triangle of side lengths a and angles α . Let m denote the lengthof the median, and consider the half triangle defined by one median. This triangle hasangles α, α , π and sides a, a , m .First, assume the length a is rational, i.e., that e a ∈ Q . By Pythagoras’ theorem, cosh( m ) cosh( a ) = cosh( a ) , so that cosh( m ) ∈ Q if and only if p = cosh( a ) ∈ Q . Let t = sinh( m ) . Squaring Pythagoras’ formula, we get the following equation for t : (1 + t ) p = (2 p − i.e. s = 4 p − p + 1 , writing s = pt . Changing variables s = 9 − x x , p = y x , y = 4 p ( − s + 8 p − , x = − s + 8 p − . we get the following elliptic curve: y = x + 10 x + 9 x. YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 11
The curve has rank 0, and the torsion is given by E ( Q ) tors = h ( − , , ( − , i ∼ = Z / Z × Z / Z . Since we are looking for solutions with p = 0 , we need y = 0 . The only torsion pointsto consider are therefore ( − , ± and (3 , ± . However, we also need t = 0 , leadingto s = 0 and x = ± . Therefore, there are no solutions.Now assuming the angle α to be rational (i.e. e iα ∈ Q [ i ] ), the situation is entirelysimilar. From the law of cosines (for the angles) we get: sin( α ) cosh( m ) = cos( α ) ,whence cosh( m ) ∈ Q if and only if p = sin( α ) ∈ Q . We set t = sinh( m ) and square theequation to get the same equation as before: (1 + t ) p = (2 p − . Thus also in this case, the median cannot be rational. (cid:3) Rational medians
The goal of this section is to study hyperbolic triangles with one rational median, inthe same spirit as Euler’s problem [4]. Consider a (non-degenerate bounded) hyperbolictriangle with sides a, b, c having opposite angles α, β, γ (by abuse of notation we will let a, b, c also denote the length of the sides). Let m denote the median at angle α , cuttingside a into two equal parts. Denote by θ the angle at the intersection of m and a , onthe side of β ; the one on the side of γ is π − θ . Applying the cosine theorem in the twotriangles, we get: cosh( b ) = cosh( m ) cosh( a/ − sinh( m ) sinh( a/
2) cos( π − θ )cosh( c ) = cosh( m ) cosh( a/ − sinh( m ) sinh( a/
2) cos( θ ) and thus m ) cosh( a/
2) = cosh( b ) + cosh( c ) . Let us now assume that a, b, c are rational side lengths, i.e., that e a , e b , e c ∈ Q . Inorder for cosh( m ) to be rational, it is thus necessary and sufficient that cosh( a/ berational. Since e a ∈ Q , this is equivalent to e a/ ∈ Q . For sinh( m ) ∈ Q we get thefollowing condition from the above equation: (cosh( b ) + cosh( c )) − a/ = 4 sinh ( m ) cosh ( a/
2) = square . Setting u = e a/ , v = e b , w = e c , we need to solve ( v w + w + vw + v ) u − v w ( u + 1) = t for t, u, v, w ∈ Q . Now applying the change of variables x = 2 w (cid:2) u wv + u ( w + 1) v − u w − u w + ut − w (cid:3) y = 2 uw (cid:2) u w v + 3 wu ( w + 1) v + ( − u w + u w − u w + 2 uwt + u − w ) v + u ( w + 1)( uw + t ) (cid:3) with inverse v = − xuw (cid:2) uw ( w + 1)( u + 1) + u ( w + 1) x − y (cid:3) t = − x uw (cid:2) − x + 8 w ( u + 1) (2 w ( u + 1) + ( w + 1) u ) x − uw ( w + 1)( u + 1) y + 32 u w ( w + 1) ( u + 1) (cid:3) , we get the equation y = x + ( u w + 2(4 u + 7 u + 4) w + u ) x (8) + 8( u + 1) w ( u w + 2( u + 1) w + u ) x + 16 u w ( u + 1) ( w + 1) . If, similarly as previously, we let E u,w denote the elliptic curve given by (8) seen over Q , where u, w ∈ Q are parameters, we get: Theorem 6.
A hyperbolic triangle with rational sides a = 2 log( u ) , b = log( w ) hasa rational median (intersecting side a ) if and only if it corresponds (using the abovechange of variables) to a rational point on the elliptic curve E u,w . Let E u denote the curve E u,w seen over C ( u ) . As previously, we have the followinglemma, a weaker analogue to Lemmas 1.1 and 2.1. Lemma 4.1.
The rank of the K surface E u satisfies ≤ rk( E u ( C ( w ))) ≤ . Moreover the point P ( u, w ) = (cid:0) , u ( u + 1) w ( w + 1) (cid:1) is a point of infinite order on the curve.Proof. The proof is entirely similar to that of Lemmas 1.1 and 2.1. The discriminantof E u,w is given by: u w ( u + 1) ( uw + u − u + u + 1) w )( uw + u − u − u + 1) w ) × ( uw + u + 2( u + u + 1) w )( uw + u + 2( u − u + 1) w ) . Looking at the Kodaira classification, we observe that E u has singularities of type I at w = 0 , ∞ , and of type I for all the 8 others. Thus the Shioda–Tate formula gives ρ ( E u ) = rk( E u ( C ( w ))) + 2 + 2 · (8 −
1) + 8 · (1 −
1) = rk( E u ( C ( w ))) + 16 . Since E u is a K surface, we have ρ ( E u ) ≤ , and thus rk( E u ( C ( w ))) ≤ .The lower bound now follows from the face that P ( u, w ) is of infinite order, whichcan be verified by direct computation. (cid:3) Area bisectors
This section is similar to Section 4, but focussing on hyperbolic triangles with rationalarea bisectors, instead of medians.Consider a hyperbolic triangle with sides a, b, c having opposite angles α, β, γ . Let m denote the area bisector at angle α , cutting α into α and α − α . Denote by θ theangle at the intersection of m and a , on the side of α , and (assume) on the side of β .Thus we have two triangles: one with angles α , β, θ and one with α − α , γ, π − θ .By the law of cosines (for the angles) we have sin( α ) sin( β ) cosh( c ) = cos( θ ) + cos( α ) cos( β ) . YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 13
Combining this with the definition of area bisector π − α − θ − β ) = A i.e. θ = π − A − α − β, we get(9) sin( α ) sin( β ) cosh( c ) = − cos (cid:18) A α + β (cid:19) + cos( α ) cos( β ) . Using trigonometric formulas, this rewrites as sin( α ) − = 1 + 1tan( α ) = 2 + sin( β ) sinh( c ) − (cid:0) A (cid:1) + 2(1 − cosh( c )) sin( β ) sin (cid:0) A + β (cid:1)(cid:0) cos( β ) − cos (cid:0) A + β (cid:1)(cid:1) . Now using the law of cosines again: sin( α ) sin( β ) cosh( c ) = cos( γ ) + cos( α ) cos( β ) and setting w = (cid:0) cos( β ) − cos (cid:0) A + β (cid:1)(cid:1) (sin( α )) − sin( α ) , we get the equation w = s α + c β s α + ( c α c β c A − s α s β c A − s α c β s A − c α s β s A − c α c β ) − c β s α (cid:18) cos (cid:18) A (cid:19) c β − sin (cid:18) A (cid:19) s β (cid:19) − s α ( − ( c α c β c A − s α s β c A − s α c β s A − c α s β s A ) + c α c β ) × (cid:18) sin (cid:18) A (cid:19) c β + cos (cid:18) A (cid:19) s β (cid:19) , where, as usual, s α = sin( α ) , etc...Assume now that our triangle has rational angles as well as rational half-area . Weapply a similar change of variables as in Section 1, namely: cos( A ) = − n n , cos( β ) = − u u , cos( α ) = − t t , sin( A ) = n n , sin( β ) = u u , sin( α ) = t t . Setting w = w ( n +1) ( t +1)( u +1)4 n , and clearing squares, we obtain: w =4( n + u ) ( nu − t + 4( n + u )( nu − n u + 3 n u − n − nu − u + 1) t + ( n u + 2 n u + 8 n u + 11 n u + n − n u − n u − n u − n u + 11 n + 64 n u + 86 n u + 24 nu + u − n − nu − u + 1) t − n + u )( nu − n u + 3 n u − n − nu − u + 1) t + 4( n + u ) ( nu − . We make the following final change of variables: y = 4( nu − n + u ) t (cid:2) n u + 3 n u − n − nu − u + 1)( nu − n + u ) t + ( n u + 2 n u + 8 n u + 11 n u + n − n u − n u − n u − n u + 11 n + 64 n u + 86 n u + 24 nu + u − n − nu − u + 1) t + 8( nu − ( n + u ) − n u + 3 n u − n − nu − u + 1)( nu − n + u ) t + 4( nu − n + u ) w − (2 n u + 3 n u − n − nu − u + 1) tw (cid:3) ,x = 1 t (cid:2) nu − ( n + u ) − n u + 3 n u − n − nu − u + 1)( nu − n + u ) t + ( n u − n u − n u + 2 n u + n + 4 n u − n u − n u − n u + 2 n + 24 n u + 38 n u + 12 nu − n − nu − u ) t + 4( nu − n + u ) w (cid:3) , with inverse t = − (cid:2) nu − x ( n + u ) (cid:3) · (cid:2) (2 n u + 3 n u − n − nu − u + 1)( nu − n − u ) ( n + 1) − (2 n u + 3 n u − n − nu − u + 1) x − y (cid:3) − ,w = − nu − n + u ) (cid:2) (2 n u + 3 n u − n − nu − u + 1) ( nu − n − u ) ( n + 1) − n u + 3 n u − n − nu − u + 1)( nu − n − u ) ( n + 1) y + (2 n u − n u − n u − n u + 2 n + 20 n u + 34 n u + 12 nu − u − n − nu − u − n + 1) x − x + y (cid:3) × (cid:2) (2 n u + 3 n u − n − nu − u + 1) ( nu − n − u ) ( n + 1) − n u + 3 n u − n − nu − u + 1) ( nu − n − u ) ( n + 1) x − n u + 3 n u − n − nu − u + 1)( nu − n − u ) ( n + 1) y + (2 n u + 3 n u − n − nu − u + 1) x + 2(2 n u + 3 n u − n − nu − u + 1) yx + y (cid:3) − , and we get y =( x − ( n + 1) ( nu − u − n ) )( x − ( n + 1)( n u − n u − u − n u (10) + 16 nu − n u + 32 n u − u + 16 n u − nu + n − n − x − ( n + 1) ( nu − u − n ) (3 n u − u + 2 n u − nu − n + 1) ) . Let E n,u denote this elliptic curve (10), where n, u are parameters. The data itencodes is the following. By assumption, A , α, β, γ are all rational. Moreover, as inSection 1, it follows easily from the law of cosines that cosh( a ) , cosh( b ) and cosh( c ) arealso rational. Now by construction, a rational solution to (10) corresponds to a trianglewith sin( α ) rational. In addition, (9) implies that cos( α ) ∈ Q , and thus α is rational.Since the area of the small triangle with angles α , β, θ is rational (it is A ), it followsthat θ is also a rational angle.What is not encoded by the curve E n,u is the rationality of sinh( m ) . It is easy tosee that this is actually equivalent to the original triangle being Heron: since all theangles and areas under consideration are rational, we have sinh( m ) ∈ Q if and only ifthe small triangle (with angles α , β, θ ) is Heron (as explained in Section 1). Since c is YPERBOLIC HERON TRIANGLES AND ELLIPTIC CURVES 15 a side of this triangle, this happens exactly when sinh( c ) ∈ Q , which, in turn, is true ifand only if the original triangle is Heron.Therefore, we have proven: Theorem 7.
A hyperbolic Heron triangle has one rational area bisector if and only ifit corresponds (using the above change of variables) to a rational point of E n,u . This curve is more complicated than the ones of Section 1 and 2. Yet we have thefollowing lemma, analog to Lemma 4.1. As before, we let E n denote the curve E n,u seenover C ( u ) . Lemma 5.1.
The rank of the surface E n satisfies ≤ rk( E n ( C ( u ))) ≤ . Moreover, the point Q ( n, u ) = (cid:16) , ( n + 1) ( nu − u − n ) (3 n u − u + 2 n u − nu − n + 1) (cid:17) is a point of infinite order on the surface E n , and its torsion is either Z / Z , Z / Z , Z / Z × Z / Z , or Z / Z × Z / Z .Proof. It is not hard to see that E n is a K surface. Its discriminant is ( n + 1) ( u + n ) ( nu − ( u + 1) (( u − n − u ) × (cid:16) ( n + 18 n + 1) u + 16 n ( n − u + 2( n − n + 17) u − n ( n − u + n + 18 n + 1 (cid:17) We have singularities at u = − n , u , and the roots of nu − u − n of type I , ± i oftype I , and the roots of the last factor of type I . By the Shioda–Tate formula, ρ ( E n ) = rk( E n ( C ( u ))) + 2 + 4 · (4 −
1) + 2 · (2 −
1) + 4 · (1 −
1) = rk( E n ( C ( u ))) + 16 . Thus the rank is ≤ .The lower bound now follows from the fact that Q ( n, u ) is of infinite order, whichcan be verified by direct computation.Finally, it is immediate to see that (( n + 1) ( nu − u − n ) , is a point of or-der 2. We can conclude that the torsion is either Z / Z , Z / Z , Z / Z × Z / Z , or Z / Z × Z / Z . (cid:3) References [1] N. Brody and J. Schettler. Rational hyperbolic triangles and a quartic model ofelliptic curves.
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Université de Montréal, Pavillon André-Aisenstadt, Département de mathématiqueset de statistique, CP 6128, succ. Centre-ville, Montréal, Québec, H3C 3J7, Canada
Email address : [email protected], Web: dms.umontreal.ca/~mlalin
Centre de recherches mathématiques, Université de Montréal, Pavillon André-Aisenstadt, 2920 Chemin de la tour, Montréal, Québec, H3T 1J4, Canada