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Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.
1

%I #12 Apr 25 2013 09:13:09

%S 108,384,432,768,972,1536,1728,2700,3072,3456,3888,5292,6144,6912,

%T 8748,9600,10800,12288,13068,13824,15552,17280,18252,18816,18900,

%U 19200,21168,24300,24576,27000,27648,31104,31212,34560,34992,37632

%N Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

%C Let a triangle with the angles (A, B, C) and the sides opposite the angles labeled (a, b, c). The length of the perpendicular bisectors is given by (x, y, z) where:

%C x is the perpendicular bisector passing through the midpoint of the segment BC = a;

%C y is the perpendicular bisector passing through the midpoint of the segment AC = b;

%C z is the perpendicular bisector passing through the midpoint of the segment AB = c.

%C We obtain the relations:

%C x = (a/2)*tg B if x intersects AB or (a/2)* tg C if x intersects AC;

%C y = (b/2)* tg A if y intersects AB or (b/2)* tg C if y intersects BC;

%C z = (c/2)*tg A if z intersects AC or (c/2) *tg B if z intersects BC.

%C The area A of the triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.

%C Finally, we obtain:

%C x = (a/2) * min {tg B, tg C }; y = (b/2) * min {tg A, tg C }; z = (c/2) * min {tg A, tg B } with tg A = 4*A/(b^2+c^2-a^2) ; tg B = 4*A/(c^2+a^2-b^2) ; tg C = 4*A/(a^2+b^2-c^2).

%C Properties of this sequence:

%C The numbers of the form 108*n^2, 384*n^2, 768*n^2, 17280*n^2, 18900*n^2 are in the sequence because the area of the primitive triangles (15, 15, 18), (24, 32, 40), (40, 40, 64), (120, 288, 312), (150, 255, 315) are 108, 384, 768 , 17280 and 18900 respectively.

%C There exists three class of numbers included into a(n) :

%C Case (i) : a subset of isosceles triangles;

%C Case (ii) : a subset of right triangles;

%C Case (iii) : other (neither isosceles nor right triangle).

%H Ray Chandler, <a href="/A182171/b182171.txt">Table of n, a(n) for n = 1..65</a>

%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/Triangle.html">MathWorld: Triangle</a>

%F x = Min{2*a*A/(c^2+a^2-b^2) , 2*a*A/(a^2+b^2-c^2)};

%F y = Min{2*b*A/(a^2+b^2-c^2) , 2*b*A/(b^2+c^2-a^2)};

%F z = Min{2*c*A/(c^2+a^2-b^2) , 2*c*A/(b^2+c^2-a^2)}.

%e Primitive solutions follow:

%e Area, ( a, b, c), ( x, y, z), Case

%e 108, (15, 15, 18), (10, 10, 12), Isosceles,

%e 384, (24, 32, 40), (16, 12, 15), Right,

%e 768, (40, 40, 64), (15, 15, 24), Isosceles,

%e 17280, (120, 288, 312), (144, 60, 65), Right,

%e 18900, (150, 255, 315), (100, 68, 84), Other,

%e 27000, (255, 255, 450), (68, 68, 120), Isosceles,

%e 34560, (312, 312, 576), (65, 65, 120), Isosceles,

%e 53760, (272, 400, 448), (255, 150, 168), Other,

%e 54000, (240, 450, 510), (225, 120, 136), Right,

%e 91476, (429, 462, 495), (364, 308, 330), Other,

%e 95256, (252, 819, 945), (168, 104, 120), Other,

%e 96768, (336, 720, 960), (126, 105, 140), Other.

%p zz:=evalf(1/10^6):k:=0:nn:=350:

%p for a from 15 to nn do:

%p for b from a to nn do:

%p for c from b to nn do:

%p p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):

%p u:= a^2+b^2-c^2:v:= b^2+c^2-a^2 :w:=c^2+a^2-b^2:

%p if s>0 then s1:=sqrt(s):else fi:

%p if u>0 then u0:=u:else u0:=zz:fi:

%p if v>0 then v0:=v:else vo:=zz:fi:

%p if w>0 then w0:=w:else w0:=zz:fi:

%p a0:= evalf(2*a*s1/w0):a1:=evalf(2*a*s1/u0): b0:= evalf(2*b*s1/u0):b1:=evalf(2*b*s1/v0): c0:= evalf(2*c*s1/w0):c1:=evalf(2*c*s1/v0):

%p if a0<a1 then x:= a0:else x:=a1:fi:

%p if b0<b1 then y:=b0:else y:=b1:fi:

%p if c0<c1 then z:=c0:else z:=c1:fi:

%p if s1=floor(s1) and x=floor(x) and y=floor(y) and z=floor(z) then print(s1):else fi:

%p od:

%p od:

%p od:

%Y Cf. A188158, A210643, A181924, A181925, A181928.

%K nonn

%O 1,1

%A _Michel Lagneau_, Apr 16 2012

%E More terms from _Ray Chandler_, Apr 24 2013