OFFSET
1,1
COMMENTS
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.
The area of the tangential triangle is given by Area = (1/2)*Area(A,B,C)*|sec A * sec B * sec C|
The side lengths of the tangential triangle are:
a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2|
b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2|
c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2|
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).
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).
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').
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S' * S * a * b * c * a' * b' * c'
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23660 * 10584 * 84 * 273 * 315 * 182 * 2197/6 * 1625/6
26250 * 5292 * 105 * 105 * 126 * 625/2 * 625/2 * 175
26250 * 9408 * 140 * 140 * 224 * 625/2 * 625/2 * 175
26250 * 24192 * 168 * 360 * 480 * 175 * 625/2 * 625/2
53235 * 8064 * 104 * 160 * 168 * 2197/4 * 1625/4 * 273
94640 * 42336 * 168 * 546 * 630 * 364 * 2197/3 * 1625/3
105000 * 21168 * 210 * 210 * 252 * 625 * 625 * 350
105000 * 37632 * 280 * 280 * 448 * 625 * 625 * 350
147875 * 40320 * 200 * 416 * 504 * 8125/12 * 10985/12 * 455
212940 * 32256 * 208 * 320 * 336 * 2197/2 * 1625/2 * 546
.........................................
REFERENCES
Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.
Kimberling, C. Triangle Centers and Central Triangles. Congr. Numer. 129, 1-295, 1998.
LINKS
C. Kimberling, Central Points and Central Lines in the Plane of a Triangle, Math. Mag. 67, 163-187, 1994.
Eric W. Weisstein, MathWorld: Tangential Triangle
EXAMPLE
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:
a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2| = 625/2
b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2| = 625/2
c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2| = 175
The area of this triangle is given by two possible ways:
First way:
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.
Second way:
Area of the triangle (a, b, c) = sqrt(sqrt(s*(s-a)*(s-b)*(s-c))=sqrt(168*(168-105)*(168-105)*(168-126))=5292 with the semiperimeter s = (105 + 1052 + 126)/2 = 168.
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:
sec A = 1/cos A = 2*b*c/(b^2+c^2-a^2)= 5/3;
sec B = 1/cos B = 2*a*c/(c^2+a^2-b^2)= 5/3;
sec C = 1/cos C = 2*a*b/(a^2+b^2-c^2)= 25/7.
MATHEMATICA
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 16 2013
STATUS
approved