A181925 -id:A181925 - cp's OEIS frontend

A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

108, 384, 432, 768, 972, 1536, 1728, 2700, 3072, 3456, 3888, 5292, 6144, 6912, 8748, 9600, 10800, 12288, 13068, 13824, 15552, 17280, 18252, 18816, 18900, 19200, 21168, 24300, 24576, 27000, 27648, 31104, 31212, 34560, 34992, 37632

Offset: 1

Michel Lagneau, Apr 16 2012

nonn

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:
x is the perpendicular bisector passing through the midpoint of the segment BC = a;
y is the perpendicular bisector passing through the midpoint of the segment AC = b;
z is the perpendicular bisector passing through the midpoint of the segment AB = c.
We obtain the relations:
x = (a/2)*tg B if x intersects AB or (a/2)* tg C if x intersects AC;
y = (b/2)* tg A if y intersects AB or (b/2)* tg C if y intersects BC;
z = (c/2)*tg A if z intersects AC or (c/2) *tg B if z intersects BC.
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.
Finally, we obtain:
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).
Properties of this sequence:
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.
There exists three class of numbers included into a(n) :
Case (i) : a subset of isosceles triangles;
Case (ii) : a subset of right triangles;
Case (iii) : other (neither isosceles nor right triangle).

			Primitive solutions follow:
Area,  ( a,   b,   c),  ( x,   y,   z), Case
  108,  (15,  15,  18),  (10,  10,  12), Isosceles,
  384,  (24,  32,  40),  (16,  12,  15), Right,
  768,  (40,  40,  64),  (15,  15,  24), Isosceles,
17280, (120, 288, 312), (144,  60,  65), Right,
18900, (150, 255, 315), (100,  68,  84), Other,
27000, (255, 255, 450),  (68,  68, 120), Isosceles,
34560, (312, 312, 576),  (65,  65, 120), Isosceles,
53760, (272, 400, 448), (255, 150, 168), Other,
54000, (240, 450, 510), (225, 120, 136), Right,
91476, (429, 462, 495), (364, 308, 330), Other,
95256, (252, 819, 945), (168, 104, 120), Other,
96768, (336, 720, 960), (126, 105, 140), Other.

Ray Chandler, Table of n, a(n) for n = 1..65
Eric W. Weisstein, MathWorld: Triangle

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

zz:=evalf(1/10^6):k:=0:nn:=350:
for a from 15 to nn do:
   for b from a to nn do:
       for c from b to nn do:
             p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):
             u:= a^2+b^2-c^2:v:= b^2+c^2-a^2 :w:=c^2+a^2-b^2:
             if s>0 then s1:=sqrt(s):else fi:
             if u>0 then u0:=u:else u0:=zz:fi:
             if v>0 then v0:=v:else vo:=zz:fi:
             if w>0 then w0:=w:else w0:=zz:fi:
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):
             if a0

x = Min{2*a*A/(c^2+a^2-b^2) , 2*a*A/(a^2+b^2-c^2)};
y = Min{2*b*A/(a^2+b^2-c^2)  , 2*b*A/(b^2+c^2-a^2)};
z = Min{2*c*A/(c^2+a^2-b^2)  , 2*c*A/(b^2+c^2-a^2)}.

More terms from Ray Chandler, Apr 24 2013

cp's OEIS Frontend

A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

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