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A231328 Integer areas of the reflection triangles of integer-sided triangles.

%I A231328 #12 Feb 16 2025 08:33:20
%S A231328 18,72,90,162,180,252,288,360,450,540,630,648,720,810,882,990,1008,
%T A231328 1152,1440,1458,1512,1620,1638,1800,1890,2160,2178,2250,2268,2520,
%U A231328 2592,2772,2880,2970,3042,3240,3528,3672,3960,4032,4050,4158,4410,4500,4608,4680,4860
%N A231328 Integer areas of the reflection triangles of integer-sided triangles.
%C A231328 The triangle A'B'C' obtained by reflecting the vertices of a reference triangle ABC about the opposite sides is called the reflection triangle (Grinberg 2003).
%C A231328 The area of the reflection triangle is given by
%C A231328 A' = A*t/(a^2*b^2*c^2) where A is the area of the reference triangle of sides (a, b, c) and
%C A231328 t=-(a^6-b^2*a^4-c^2*a^4-b^4*a^2-c^4*a^2-b^2*c^2*a^2+b^6+c^6-b^2*c^4-b^4*c^2)/(a^2*b^2*c^2).
%C A231328 See the link for the side lengths of the reflection triangles.
%C A231328 Properties of this sequence:
%C A231328 The areas corresponding to the primitive reflection triangles are 18, 90, 180, 252, 540,...
%C A231328 The non-primitive triangles of areas 4*a(n),9*a(n),...,p^2*a(n),... are in the sequence.
%C A231328 It appears that one of the side of the reflection triangles equals the greatest side of the initial triangle (see the table below), and the initial triangles are Pythagorean triangles => a(n) = 3*A009112(n).
%C A231328 The following table gives the first values (A, A', a, b, c, a', b', c') where A' is the area of the reflection 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 reflection triangles.
%C A231328 -------------------------------------------------------------------------
%C A231328 |  A'  |   A |  a |  b |   c |       a'         |       b'         |  c'|
%C A231328 -------------------------------------------------------------------------
%C A231328 |  18  |   6 |  3 |  4 |   5 |  9*sqrt(17)/5    |  4*sqrt(97)/5    |  5 |
%C A231328 |  72  |  24 |  6 |  8 |  10 | 18*sqrt(17)/5    |  8*sqrt(97)/5    | 10 |
%C A231328 |  90  |  30 |  5 | 12 |  13 |  5*sqrt(1321)/13 | 36*sqrt(41)/13   | 13 |
%C A231328 | 162  |  54 |  9 | 12 |  15 | 27*sqrt(17)/5    | 12*sqrt(97)/5    | 15 |
%C A231328 | 180  |  60 |  8 | 15 |  17 |  8*sqrt(2089)/17 | 45*sqrt(89)/17   | 17 |
%C A231328 | 252  |  84 |  7 | 24 |  25 |  7*sqrt(5233)/25 | 72*sqrt(113)/25  | 25 |
%C A231328 | 288  |  96 | 12 | 16 |  20 | 36*sqrt(17)/5    | 16*sqrt(97)/5    | 20 |
%C A231328 | 360  | 120 | 10 | 24 |  26 | 10*sqrt(1321)/13 | 72*sqrt(41)/13   | 26 |
%C A231328 | 450  | 150 | 15 | 20 |  25 |  9*sqrt(17)      |  4*sqrt(97)      | 25 |
%C A231328 | 540  | 180 |  9 | 40 |  41 | 27*sqrt(1609)/41 | 40*sqrt(2329)/41 | 41 |
%C A231328 | 630  | 210 | 12 | 35 |  37 | 36*sqrt(1241)/37 | 35*sqrt(2521)/37 | 37 |
%C A231328 | 648  | 216 | 18 | 24 |  30 | 54*sqrt(17)/5    | 24*sqrt(97)/5    | 30 |
%C A231328 .......................................................................
%D A231328 D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.
%H A231328 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/ReflectionTriangle.html">MathWorld: Reflection Triangle</a>
%e A231328 18 is in the sequence. We use two ways:
%e A231328 First way: with the triangle (3, 4, 5) the formula A' = A*t/(a^2*b^2*c^2) gives directly the result: A'= 18 where the area A = 6 is obtained by Heron's formula A =sqrt(s*(s-a)*(s-b)*(s-c))= sqrt(6*(6-3)*(6-4)*(6-5)) = 6, where s is the semiperimeter.
%e A231328 Second way: by calculation of the sides a', b', c' and by using Heron's formula. We obtain from the formulas given in the link:
%e A231328 a' = 9*sqrt(17)/5;
%e A231328 b' = 4*sqrt(97/5);
%e A231328 c' = 5.
%e A231328 Now, we use Heron's formula with (a',b',c'). We find A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
%e A231328 s1 =(a'+b'+c')/2 = (9*sqrt(17)/5+ 4*sqrt(97/5)+ 5)/2. We find A'= 18.
%t A231328 nn = 300 ; lst = {}; Do[s = (a + b + c)/2 ; If[IntegerQ[s],area2 = s (s-a)(s-b) (s-c); If[area2 > 0 && IntegerQ[Sqrt[area2] + (a^2 + b^2 + c^2)/8], AppendTo[lst, Sqrt[area2] + (a^2 + b^2 + c^2)/8]]],{a,nn},{b,a},{c,b}] ; Union[lst]
%Y A231328 Cf. A188158, A009111, A009112.
%K A231328 nonn
%O A231328 1,1
%A A231328 _Michel Lagneau_, Nov 07 2013