aa r X i v : . [ m a t h . N T ] M a r The Rational Distance Problem for EquilateralTriangles
Roy Barbara §-1
Abstract
Let (P) denote the problem of existence of a point in the plane of a given triangle T,that is at rational distance from all the vertices of T. Answer to (P) is positive if T has arational side and the square of all sides are rational (see [1]). In [2], a complete solutionto (P) is given for all isosceles triangles with one rational side. In this article, we providea complete solution to (P) for all equilateral triangles.
In all what follows , θ denotes an arbitrary positive real number and T = [ θ ] denotesthe equilateral triangle with side-length θ . For convenience, we say that θ is "good" (or"suitable") if answer to (P) is positive for the triangle T = [ θ ] . Clearly, the property " θ is good" is invariant by any rational re-scaling of θ .It turns out that the good θ must have algebraic degree , , or , and they form asubclass of the positive bi-quadric numbers, that is, the positive roots of equations of theform x + ux + v = 0 , u, v ∈ Q . The general form of such numbers is p α ± √ β, α, β ∈ Q , β ≥ , α ± √ β ≥ that includes positive numbers of the form α, √ α, α ± √ β √ α ± √ β, α, β ∈ Q , α, β ≥ . Notations and conventions : ( x, y ) and ( x, y, z ) denote the g.c.d. (cid:0) xp (cid:1) denotes legen-dre’s symbol. A triangle with side-lengths a, b, c is denoted by T = [ a, b, c ] . A triangleis non-degenerated if it has positive area. A radical is non-degenerated if it is irrational.§-2 The resultsTheorem If θ is good, then, θ is bi-quadric. More precisely, θ = α ± √ β for some α, β ∈ Q , β ≥ , and α positive. Theorem Suppose θ / ∈ Q and θ ∈ Q . Then, θ is good ⇔ θ has the form θ = λ √ p ...p r where λ ∈ Q , λ > , r ≥ , p , ..., p r aredistinct odd primes, p i is either or of the form k + 1 . Theorem Suppose θ = α ± √ β, α, β ∈ Q , α, β > , √ β / ∈ Q . Then, θ is good ⇔ up to a rational re-scaling of θ, θ is described as follows: θ = ( a + b + c ) ± √ where [ a, b, c ] is a non-degenerated primitive integral triangle with area ∆ such that √ / ∈ Q . Remark ∆ is given by Hero’s formula, ∆ = p s ( s − a )( s − b )( s − c ) , s = ( a + b + c ) .Equivalently, √ p a + b + c )( − a + b + c )( a − b + c )( a + b − c ) , and the condition √ / ∈ Q means that this latter radical is non-degenerated. Proofs of theorems 0 and 1
Proof of theorem 0:
Suppose θ good. Let M be a point in the plane of triangle T = [ θ ] ,whose distances from the vertices of T are all rational. the following fundamental relationis well-known (see[3]): a + b + c + θ ) = ( a + b + c + θ ) ( (cid:5) ) Expanding ( (cid:5) ) yields a relation as θ − uθ + v = 0 , where u, v ∈ Q and u = a + b + c > . Solving for θ yields θ = α ± √ β, with α, β ∈ Q and α = u > . Lemma : let q > be a square-free integer. Then we have:The equation x + 3 y = qz has a solution in integers x, y, z, with z = 0 if and only ifany prime factor of q is either or of the form k + 1 Proof : Suppose first that q has only prime factors as or k + 1 . Since the quadraticform x + 3 y , x, y ∈ Z , represents and every prime p = 6 k + 1 , and since the set { x +3 y , x, y ∈ Z } is cosed by multiplication, we conclude that the equation x + 3 y = q.z has a solution in integers x, y, z with z = 1 . Conversely, suppose that x + 3 y = q.z has a solution in integers x, y, z, z = 0 . Picksuch a solution with | z | minimum. Clearly, ( x, y ) = 1 . I claim that q is odd and has noprime factor k − . For the purpose of contradiction, we consider two cases: case 1 : q is even. Set q = 2 w, w odd. From x + 3 y = 2 wz , we see that x ≡ y ( mod. . As ( x, y ) = 1 , x and y must be odd, so x + 3 y ≡ mod. . Now, / wz yields wz even. But w is odd, hence z is even, so wz ≡ mod. . We get a contradiction. case 2 : q = p.w for some prime p = 6 k − . x + 3 y = pwz yields x + 3 y ≡ mod.p ) .As ( x, y ) = 1 , p cannot divide y . Hence for some t ∈ Z , yt ≡ mod.p ) . Therefore, x t +3 y t ≡ x t + 3 ≡ mod.p ) , so − ≡ ( xt ) ( mod.p ) . Hence (cid:0) − p (cid:1) = +1 contradicting p = 6 k − . Lemma : Let θ = λ √ q, λ ∈ Q , λ > , q > square-free integer. We have: θ is good ⇔ There are a, b, e, r, s ∈ Q , e = 0 , such that a + 3 b = q (1)( a + e ) + 3( b + e ) = qr (2)( a − e ) + 3( b + e ) = qs (3) Proof : By re-scaling, we take θ = 2 √ q . Let T = ABC = [ θ ] . Choose a x − y axis to getthe coordinates A (0 , √ q ) , B ( −√ q, , C ( √ q, . • Suppose first that θ is good: There is a point M = M ( x, y ) in the plane of T such that M A, M B, M C ∈ Q . Clearly, M = A, B, C . Set w = M Aq , r = M Bwq , s = M Cwq . Then, w, r, s ∈ Q − { } . The Pythagoras relations are:
M A = x + ( y − √ q ) = w q (1 ′ ) M B = ( x + √ q ) + y = w q r (2 ′ ) M C = ( x − √ q ) + y = w q s (3 ′ ) Subtracting (2’) and (3’) yields x = w q ( r − s ) . √ q, that is,2 = α √ q, α ∈ Q (4) Then (2’) gives y ∈ Q , and then (1’) gives y √ q ∈ Q , hence, y = γ √ q, γ ∈ Q .For convenience, we put γ = β + 1 , obtaining y = ( β + 1) √ q, β ∈ Q (5) Due to (4) and (5), equations (1’), (2’), (3’) become after dividing by q : α + 3 β = qw ( α + 1) + 3( β + 1) = qw r ( α − + 3( β + 1) = qw s Set a = αw , b = βw , e = 1 w . Dividing by w , we get precisely relations (1), (2), (3). • Conversely suppose that relations (1), (2), (3) hold with some a, b, e, r, s ∈ Q , e = 0 .Define point M = M ( x, y ) in the plane of T by x = ae √ q, y = ( be + 1) p q We may write:
M A = x + ( y − p q ) = q a e + 3 q b e = qe ( a + 3 b ) = qe .q = ( qe ) M B = (cid:0) ( a + ee ) √ q (cid:1) + (cid:0) ( b + ee ) p q (cid:1) = qe (cid:0) ( a + e ) + 3( b + e ) (cid:1) = qe .qr = ( qre ) M C = (cid:0) ( a − ee ) √ q (cid:1) + (cid:0) ( b + ee ) p q (cid:1) = qe (cid:0) ( a − e ) + 3( b + e ) (cid:1) = qe .qs = ( qse ) Therefore,
M A, M B, M C are all rational.
Proof of theorem 1
Let θ such that θ / ∈ Q and θ ∈ Q : θ can be written as θ = λ √ q, λ ∈ Q , λ > , q > square-free integer. • Suppose first that q is even or has a prime factor k − . By lemma 1, a + 3 b = q, a, b ∈ Q , is impossible. Hence , relation (1) in lemma 2 fails, so θ is not good. • Suppose now that q has only prime factors as or k + 1 . We show that θ is good usingthe characterization of lemma 2:By lemma 1, for some a, b ∈ Q , we have a + 3 b = q . Set e = − q b = − ( a + 3 b )4 b , r = a − b b , s = a + b b . We have ( a + e ) + 3( b + e ) = ( a + 3 b ) + 4 e + 2 e ( a + 3 b ) = q + q b − q b ( a + 3 b )= q b (cid:0) b + q − b ( a + 3 b ) (cid:1) = q b (4 b + a + 3 b − ab − b )= q b ( a + b − ab ) = q ( a − b ) b = q.r and ( a − b ) + 3( b + e ) = ( a + 3 b ) + 4 e − e ( a − b ) = q + q b + q b ( a − b )= q b (cid:0) b + q + 2 b ( a − b ) (cid:1) = q b (4 b + a + 3 b + 2 ab − b )= q b ( a + b + 2 ab ) = q ( a + b ) b = q.s Proof of theorem 2Lemma : Let x, y, z, t be positive real numbers such that x + y + z + t ) = ( x + y + z + t ) ( ⊚ ) Then, any three of x, y, z, t satisfy the triangle inequality.
Proof : Since x, y, z, t play symmetric roles, it suffices to show that x, y, z satisfy thetriangle inequality. Write ( ⊚ ) as t − ( x + y + z ) t + ( x + y + z − x y − y z − z x ) = 0 The discriminant △ of this trinomial in t must be non-negative. But, △ = 6( x y + y z + z x ) − x + y + z ) that factors as △ = 3( x + y + z )( − x + y + z )( x − y + z )( x + y − z ) .Hence, ( − x + y + z )( x − y + z )( x + y − z ) ≥ . The reader can easily check (using con-traposition) that x, y, z must satisfy the triangle inequality.
Lemma : Let T = ABC = (cid:2) θ (cid:3) . Let a, b, c be positive real numbers satisfying a + b + c + θ ) = ( a + b + c + θ ) Then, there is a point M in the plane of T such that M A = a, M B = b, and M C = c . Proof : By lemma 3, a, b, and θ satisfy the triangle inequality. In particular, a + b ≥ θ . Itfolows that the circle C ( A, a ) intersects the circle C ( B, b ) at two points M and M ( M = M if a + b = θ ) . Set c = M C and c = M C . By the fundamental relation ( (cid:5) ) we have a + b + c + θ ) = ( a + b + c + θ ) and a + b + c + θ ) = ( a + b + c + θ ) .Therefore, c and c are the roots of the trinomial in TT − ( a + b + θ ) T + ( a + b + θ − a b − b θ − θ a ) = 0 Since by hypothesis c is also a root of this trinomial, we must have c = c or c = c .Hence c = c or c = c . Therefore, a, b and c are the distances from either point M or M to the vertices A, B and C of T . Proof of theorem 2 :Let θ > such that θ = α ± √ β, α, β ∈ Q , α, β > , √ β / ∈ Q . • Suppose first that θ is good: let P be a point in the plane of T = ABC = [ θ ] such that P A = a, P B = b, P C = c are all rational. We have a + b + c + θ ) = ( a + b + c + θ ) ( (cid:5) ) By lemma 3, a, b, and c satisfy the triangle inequality. Relation ( (cid:5) ) yields θ − U θ + V = 0 with U = a + b + c and V = a + b + c − a b − b c − c a ( U, V ∈ Q ) .Solving for θ , we get θ = ( a + b + c ) ± p a + b + c )( − a + b + c )( a − b + c )( a + b − c ) ( ⋆ ) θ has algebraic degree , then, the radical in ( ⋆ ) is non-degenerated. In particular,the triangle [ a, b, c ] is non degenerated. Select a sufficiently large positive integer N suchthat N a, N b, N c are all integers and set D = ( N a, N b, N c ) . If we multiply relation ( ⋆ ) by N D , this results in replacing in ( ⋆ ) θ by ND .θ and a, b, c by the integers
N aD , N bD , N cD respectively. As an outcome, we obtain essentially the same relation ( ⋆ ) where θ has beenre-scaled by the rational ND , and where the new symbols a, b, c represent relatively prime positive integers, satisfying the triangle inequality. • Conversely, suppose that for some positive rational λ, θ = λ.θ is described precisely asin theorem 2. Eliminating the radical △ √ p a b + b c + c a ) − a + b + c ) in the relation θ = ( a + b + c ) ± △ √ leads to a + b + c + θ ) = ( a + b + c + θ ) By lemma 4 there is a point M in the plane of T = [ θ ] that is at distances a, b, c fromthe vertices of T . Since a, b, c are integers, then, θ is good. Therefore, θ = λ − θ is alsogood.We end this article with a few exercises:1. Check which are "good" among the radicals: √ , √ , √ , √ , √ , √ .2. Show that the positive real number θ = p
25 + 12 √ is "good".3. Suppose that θ = α + √ β, α, β ∈ Q , α, β > , √ β / ∈ Q , and α < β . Show that θ is not good.4. Produce solution-points to problem (P) for the triangle T = [ √ .5. Let θ = α + β √ q > , α, β ∈ Q , β = 0 , q > square-free integer. Show that θ is not good.6. Suppose that θ = α ± √ β > , α, β ∈ Q , α, β > , √ β / ∈ Q . Write the fraction α in lowest terms as α = mn ( m, n positive integers) and suppose that mn has theform mn = 4 l (8 k + 7) , k, l non-negative integers. Then, prove that θ is not good. References [1] T.G. Berry,
Points at rational distance from the vertices of a triangle , Acta Arith-metica, LXII.4(1992).[2] Roy Barbara and Antoine Karam,
The Rational Distance Problem for Isosceles Tri-angles with one Rational Side , Communications in Mathematics and Applications,Vol.4, Number 2, pp. 169-179 (2013).[3] Wikipedia or Wolfram Mathworld,
Equilateral Triangle .[4] Roy Barbara, the rational distance problem for polygonsthe rational distance problem for polygons