Suppose only that the triangle has rational (Cartesian) vertices and rational side lengths a,b,c. Via rigid isometry, transport it so that its vertices become [0,0], [0,a], [x,y]. By elementary trigonometry, with O == angle at origin, D == triangle area, y = b cos O, x = D / 2a . Now via the cosine formula in terms of sides, O is rational; also via the determinant formula for area in terms of vertices, D is rational. So both x,y, are rational, and it follows that every rational triangle is congruent to one with an axial edge. Furthermore as Gene pointed out earlier, both of cos O = (a^2+b^2-c^2) / 2 a b, sin O = 2 D / a b are rational; therefore the translation of a vertex to the origin, the rotation taking the side to the axis, and finally the congruence, have rational projective matrices. Fred Lunnon On 10/5/11, Eugene Salamin <gene_salamin@yahoo.com> wrote:
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun@mailman.xmission.com Sent: Tuesday, October 4, 2011 9:24 AM Subject: Re: [math-fun] Scalenicity?
... But in the meantime I've also noticed that my pet {12, 17, 25}-sided triangle has another curious property: its coordinates may be taken as [0,0], [0,12], [15,20] or [0,0], [0,25], [36,77]/5 or [0,0], [0,17], [180,385]/17 --- that is, scaled up by 85, any chosen edge may lie along the y-axis, and the coordinates remain integer. ... What triangles possess multiple nontrivial poses rational on the square lattice? ...
Fred Lunnon _______________________________________________
Let us call a triangle a Lunnon triangle if (1) its sides are rational, and (2) if any vertex is placed at the origin and a second vertex on the x-axis, then the third vertex has rational coordinates. This means that the altitude from the third vertex is rational, and its foot on the x-axis has rational x-coordinate. Equivalently, the sines and cosines of the angles of the triangle must be rational. From the law of cosines, the cosines of the angles of a rational triangle are automatically rational. If the sines are rational, then it follows from the law of sines that (sin A, sin B, sin C) is a rational triangle, and therefore the cosines are also rational, and thus the triangle is Lunnon, as is any rational scaling of it.
The construction of Lunnon triangles is reduced to the problem of finding triples (A,B,C) of angles that sum to pi and have rational sines. Now I claim that the same solution set is found by taking pairs (A,B) of angles having rational sines and cosines. For then, using the sum of angles formula,
sin C = sin A cos B + cos A sin B,
is rational. On the other hand no solutions are lost, for if sin C is rational as well as sin A and sin B, then we have a Lunnon triangle, and so cos A and cos B must be rational. If sin A and cos A are rational, (sin A, cos A, 1) is a rational Pythagorean triple, so A is an acute angle of some primitive Pythagorean triangle, and we know how to determine all of these.
Clearly all Pythagorean triangles are Lunnon. For a nontrivial example, start with a (3,4,5) triangle and let A = B = asin(3/5) < pi/4. Then C > pi/2 and C = asin(24/25). Thus (3/5,3/5,24/25) ~ (5,5,8) is an isosceles Lunnon triangle. If we let A = B = asin(4/5) > pi/4, we get C < pi/2 and also C = asin(24/25). So another isosceles Lunnon triangle is (4/5,4/5,24/25) ~ (5,5,6).
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun