A209432 - OEIS

%I #21 Jun 12 2017 17:33:18

%S 24,96,216,294,300,324,384,600,810,864,1176,1200,1296,1452,1536,1920,

%T 1944,2400,2520,2646,2700,2904,2916,3240,3456,4056,4320,4704,4800,

%U 4950,5184,5400,5808,6144,6300,6936,7260,7290,7350,7500,7680,7776,8064,8100,8214

%N Area A of the triangles such that A and the sides are integers and there exists at least one square inscribed in the triangle whose sides are also integers.

%C Every triangle has three inscribed squares (squares in its interior such that all four of the square's vertices lie on sides of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). However, in the case of a right triangle, two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two distinct inscribed squares. Within a given triangle, a longer common side is associated with a smaller inscribed square. If an inscribed square has sides of length x and the triangle has a side of length a, part of which side coincides with a side of the square, then x, a, and the triangle's area A are related according to x = 2Aa/(a^2+2A).

%C Property of this sequence: the numbers of the form 24*k^2 are in the sequence.

%C Theorem: Consider a triangle whose area A and sides (a, b, c) are integers such that there exists at least one square inscribed in this triangle whose sides x are also integers. Then, if the smallest side a = min {a, b, c} of this triangle is of the form a = 4k, k integer, then x = 3k and A = 24k^2.

%C Proof: Let k be an integer, and let the sides of a triangle be a = 4k, b = 13k, c = 15k. Then s = (a+b+c)/2 = 16k and A = sqrt(s(s-a)(s-b)(s-c) = 24k^2. With x = 2Aa/(a^2+2A), we find x = 3k.

%H Herbert Bailey and Duane Detemple, <a href="http://www.jstor.org/stable/2690699">Squares inscribed in angles and triangles</a>, Mathematics Magazine 71(4), 1998, 278-284.

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

%F A = sqrt(p*(p-a)*(p-b)*(p-c)) with p = (a+b+c)/2 (Heron's formula);

%F Sides of the three squares: x1 = 2*A*a/(a^2+2*A); x2 = 2*A*b/(b^2+2*A); x3 = 2*A*c/(c^2+2*A).

%e 294 is in the sequence because for (a, b, c) = (21, 28, 35) => x1 = 2*21*294/(2*294+21^2) = 12348/1029 = 12 is the integer value of the side of the square inscribed in the triangle (21, 28, 35) whose area equals 294 and whose side coincides with the side [21] of this triangle. But we also have a second square with the side x2 = 2*28*294/(2*294+28^2) = 16464/1372 = 12 whose side coincides with the side [28] of the same triangle.

%p with(numtheory):T:=array(1..1500):k:=0:nn:=500: for a from 1

%p to nn do: for b from a to nn  do: for c from b to nn  do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and (irem(2*a*s,2*s+a^2) = 0 or irem(2*b*s,2*s+b^2) = 0 or irem(2*c*s,2*s+c^2) = 0) then k:=k+1:T[k]:= s: else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

%t nn=500;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0<area2<=nn^2&&IntegerQ[Sqrt[area2]]&&(IntegerQ[2*a* Sqrt[area2]/(2* Sqrt[area2]+a^2)] || IntegerQ[2*b* Sqrt[area2]/(2* Sqrt[area2]+b^2)] || IntegerQ[2*c* Sqrt[area2]/(2* Sqrt[area2]+c^2)]),AppendTo[lst, Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]

%Y Cf. A120062, A188158, A208984.

%K nonn

%O 1,1

%A _Michel Lagneau_, Mar 09 2012