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A230361 Integer areas of the tangential triangles corresponding to the integer-sided triangles with integer areas.

%I A230361 #21 Feb 16 2025 08:33:20
%S A230361 23660,26250,53235,94640,105000,147875,212940,222530,236250,378560,
%T A230361 390390,420000,479115,591500,656250,788970,851760,945000,1286250,
%U A230361 1330875,1561560,1680000,1916460,2126250,2608515,2625000,3176250,3407040,3513510
%N A230361 Integer areas of the tangential triangles corresponding to the integer-sided triangles with integer areas.
%C A230361 The tangential triangle is the triangle (A', B', C') of side lengths (a', b', c') formed by the lines tangent to the circumcircle of a given triangle (A, B, C) of side lengths (a, b, c) at its vertices.
%C A230361 The area of the tangential triangle is given by Area = (1/2)*Area(A,B,C)*|sec A * sec B * sec C|
%C A230361 The side lengths of the tangential triangle are:
%C A230361 a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2|
%C A230361 b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2|
%C A230361 c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2|
%C A230361 In the general case, these sides lengths are rational numbers (see the examples in the table below). But it is possible to find integer sides, for example the triangle (210, 210, 252) generates a tangential triangle (625, 625, 350).
%C A230361 It is possible to find the same tangential triangle with two distinct triangles of side lengths (a1,b1,c1) and (a2,b2,c2); for example, the triangles (105, 105, 175) and (140, 140, 224) generate the same tangential triangle (625/2, 625/2, 175).
%C A230361 The following table gives the first values (S', S, a, b, c, a', b', c') where S' is the area of the tangential triangle (A', B', C'), S is the area of the initial triangle (A, B, C), a, b, c the integer sides of the triangle (A, B, C) and a', b', c' are the integer sides of the tangential triangle (A', B', C').
%C A230361 **************************************************************
%C A230361    S'  *   S   *  a  *  b  *  c  *   a'    *    b'   *  c'
%C A230361 **************************************************************
%C A230361 23660  * 10584 *  84 * 273 * 315 *  182    *  2197/6  * 1625/6
%C A230361 26250  *  5292 * 105 * 105 * 126 *  625/2  *   625/2  *  175
%C A230361 26250  *  9408 * 140 * 140 * 224 *  625/2  *   625/2  *  175
%C A230361 26250  * 24192 * 168 * 360 * 480 *  175    *   625/2  *  625/2
%C A230361 53235  *  8064 * 104 * 160 * 168 * 2197/4  *  1625/4  *  273
%C A230361 94640  * 42336 * 168 * 546 * 630 *  364    *  2197/3  * 1625/3
%C A230361 105000 * 21168 * 210 * 210 * 252 *  625    *   625    *  350
%C A230361 105000 * 37632 * 280 * 280 * 448 *  625    *   625    *  350
%C A230361 147875 * 40320 * 200 * 416 * 504 * 8125/12 * 10985/12 *  455
%C A230361 212940 * 32256 * 208 * 320 * 336 * 2197/2  *  1625/2  *  546
%C A230361 .........................................
%D A230361 Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.
%D A230361 Kimberling, C. Triangle Centers and Central Triangles. Congr. Numer. 129, 1-295, 1998.
%H A230361 C. Kimberling, <a href="http://www.jstor.org/stable/2690608">Central Points and Central Lines in the Plane of a Triangle</a>, Math. Mag. 67, 163-187, 1994.
%H A230361 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/TangentialTriangle.html">MathWorld: Tangential Triangle</a>
%e A230361 26250 is in the sequence because the triangle of integer sides (a, b, c)= (105, 105, 126) generates the tangential triangle (a', b', c') = (625/2, 625/2, 175) where:
%e A230361 a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2| = 625/2
%e A230361 b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2| = 625/2
%e A230361 c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2| = 175
%e A230361 The area of this triangle is given by two possible ways:
%e A230361 First way:
%e A230361 Heron's formula gives Area = sqrt(s'*(s'-a')*(s'-b')*(s'-c')) = sqrt(400*(400-625/2)*(400-625/2)*(400-175)) = 26250 with the semiperimeter s' = (625/2 + 625/2 + 175)/2 = 400.
%e A230361 Second way:
%e A230361 Area of the triangle (a, b, c) = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(168*(168-105)*(168-105)*(168-126)) = 5292 with the semiperimeter s = (105 + 105 + 126)/2 = 168.
%e A230361 Then, we use the formula Area = (1/2)* Area(A,B,C) * |sec A * sec B * sec C| = 2646*5/3*5/3*25/7 = 26250 where:
%e A230361 sec A = 1/cos A = 2*b*c/(b^2+c^2-a^2)= 5/3;
%e A230361 sec B = 1/cos B = 2*a*c/(c^2+a^2-b^2)= 5/3;
%e A230361 sec C = 1/cos C = 2*a*b/(a^2+b^2-c^2)= 25/7.
%t A230361 nn=1500; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); aa=Abs[((b^2+c^2-a^2)*(c^2+a^2-b^2)*(a^2+b^2-c^2))];If[0 < area2 && aa>0&& IntegerQ[Sqrt[area2]* (4*a^2*b^2*c^2)/aa], AppendTo[lst, Sqrt[area2]* (4*a^2*b^2*c^2)/aa]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y A230361 Cf. A188158.
%K A230361 nonn
%O A230361 1,1
%A A230361 _Michel Lagneau_, Oct 16 2013