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A228435 Integer areas of the mixtilinear triangles of integer-sided triangles.

%I A228435 #21 Feb 16 2025 08:33:20
%S A228435 6,14,24,54,56,84,96,112,124,126,130,150,154,156,210,216,224,266,294,
%T A228435 336,342,344,350,366,384,436,448,456,486,496,504,520,531,546,570,600,
%U A228435 616,624,630,660,686,726,756,814,840,864,896,924,1008,1014,1064,1116,1134
%N A228435 Integer areas of the mixtilinear triangles of integer-sided triangles.
%C A228435 The mixtilinear triangle is the triangle connecting the centers of the mixtilinear incircles.
%C A228435 A circle that in internally tangent to two sides of a triangle and to the circumcircle is called a mixtilinear incircle. There are three mixtilinear incircles, one corresponding to each angle of the triangle.
%C A228435 The area of the mixtilinear triangle is given by
%C A228435 A' = A*(a^3 - a^2*b - a*b^2 + b^3 - a^2*c + 6*a*b*c - b^2*c - a*c^2 - b*c^2 + c^3)/(a + b + c)^3
%C A228435 where A is the area of the reference triangle of sides (a, b, c).
%C A228435 See the link for the side lengths of the mixtilinear triangles.
%C A228435 Properties of this sequence:
%C A228435 The primitive mixtilinear triangles are 6, 14, 84, 112, 124, 130, ...
%C A228435 The nonprimitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
%C A228435 It appears that if the triangles are isosceles, one of the sides of the mixtilinear triangles is an integer, for example, for the triangles of areas a(n) = 112, 342, 448, 1008, 1368, 1792, 2800, 3078, 4032, 5472, ... corresponding to a triangle with an integer side of values respectively 8, 24, 16, 24, 48, 32, 40, 72, 48, 96, ...
%C A228435 The following table gives the first values (A', A, a, b, c, a', b', c') where A' is the area of the mixtilinear triangles, A is the area of the initial triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the mixtilinear triangles.
%C A228435 -------------------------------------------------------------------------------
%C A228435 A' |   A| a| b|  c|        a'         |        b'        |          c'
%C A228435 -------------------------------------------------------------------------------
%C A228435 6  |  54| 9|12| 15|  9*sqrt(5)/4      | 4*sqrt(13)/3     |  5*sqrt(37)/12
%C A228435 14 | 126|15|28| 41|135*sqrt(1717)/196 |28*sqrt(757)/27   | 41*sqrt(16045)/5292
%C A228435 24 | 216|18|24| 30|  9*sqrt(5)/2      | 8*sqrt(13)/3     |  5*sqrt(37)/6
%C A228435 56 | 504|30|56| 82|135*sqrt(1717)/98  |56*sqrt(757)/27   | 41*sqrt(16045)/2646
%C A228435 84 | 756|39|42| 45|104*sqrt(445)/147  | 7*sqrt(37)/3     | 45*sqrt(197)/49
%C A228435 96 | 864|36|48| 60|  9*sqrt(5)        |16*sqrt(13)/3     |  5*sqrt(37)/3
%C A228435 112| 972|45|45| 72| 20*sqrt(2)        |20*sqrt(2)        |        8
%C A228435 124|1116|39|62| 85|351*sqrt(8605)/961 |31*sqrt(853)/27   |680*sqrt(79957)/25947
%C A228435 126|1134|45|84|123|405*sqrt(1717)/196 |28*sqrt(757)/9    | 41*sqrt(16045)/1764
%C A228435 ...............................................................................
%H A228435 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MixtilinearTriangle.html">Mixtilinear triangles</a>
%e A228435 6 is in the sequence. We use two ways:
%e A228435 First way: from the initial triangle (9, 12, 15) the formula given in the comments gives directly the area of the mixtilinear triangle: A' = 54*(9^3 - 9^2*12 - 9*12^2 + 12^3 - 9^2*15 + 6*9*12*15 - 12^2*15 - 9*15^2 - 12*15^2 + 15^3)/(9 + 12 + 15)^3 = 6, where the area of the initial triangle A = 54 is obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(18*(18-9)*(18-12)*(18-15)) = sqrt(2916) = 54, where s=18 is the semiperimeter.
%e A228435 Second way: by calculation of the sides a', b', c' and by use of Heron's formula. With the formulas given in the link, we find
%e A228435 a’ = 9*sqrt(5)/4;
%e A228435 b’ = 4*sqrt(13)/3;
%e A228435 c’ = 5*sqrt(37)/12.
%e A228435 Now, we use Heron's formula with (a',b',c'). We find A' = sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
%e A228435 s1 = (a' + b' + c')/2 = (9*sqrt(5)/4 + 4*sqrt(13)/3 + 5*sqrt(37)/12)/2.
%e A228435 We find A' = 6.
%t A228435 nn = 500; lst = {}; Do[s = (a + b + c)/2; t = a^3 - a^2*b - a*b^2 + b^3 - a^2*c + 6*a*b*c - b^2*c - a*c^2 - b*c^2 + c^3; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 && IntegerQ[Sqrt[area2] * t/(a + b + c)^3], AppendTo[lst, Sqrt[area2] * t/(a + b + c)^3]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y A228435 Cf. A188158.
%K A228435 nonn
%O A228435 1,1
%A A228435 _Michel Lagneau_, Nov 10 2013