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 A227879 Integer areas of incentral triangles of integer-sided triangles. 1
 70, 280, 360, 480, 630, 1120, 1312, 1440, 1750, 1768, 1920, 2132, 2520, 3240, 3430, 4320, 4480, 5248, 5670, 5760, 7000, 7038, 7072, 7680, 7800, 8470, 8528, 9000, 9240, 10080, 11808, 11830, 12000, 12960, 13720, 13950, 14744, 15750, 15912, 17280, 17640, 17920 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The incentral triangle IJK is the Cevian triangle of a triangle ABC with respect to its incenter. It is therefore also the triangle whose vertices are determined by the intersections of the reference triangle's angle bisectors with the respective opposite sides. The area is given by: A' = 2*A*a*b*c/((a+b)*(b+c)*(c+a)) where A is the area of the original triangle. The side lengths are: a' = a*b*c*sqrt(3+2*(-cos A + cos B + cos C))/((a+b)*(a+c)) b' = a*b*c*sqrt(3+2*(cos A - cos B + cos C))/((b+c)*(b+a)) c' = a*b*c*sqrt(3+2*(cos A + cos B - cos C))/((c+a)*(c+b)) Properties of this sequence: The areas of the original triangles are integers. The primitive triangles with areas a(n) are 70, 360, 480, 630, 1312, ... The nonprimitive triangles with areas 4*a(n) are in the sequence. It appears that if the original triangle is isosceles, a side of the corresponding incenter triangle is integer. The following table gives the first values (A', A, a, b, c, t1, t2, t3) where A' is the area of the incentral triangles, A is the area of the reference triangles ABC, a, b, c the integer sides of the original triangles ABC and t1, t2, t3 are the sides of the incentral triangles. ------------------------------------------------------------------------ A'| A | a | b | c | t1 | t2 | t3 ------------------------------------------------------------------------ 70 | 294| 21| 28| 35|3*sqrt(65)/2 |4*sqrt(85)/3 |7*sqrt(145)/6 280 |1176| 42| 56| 70|3*sqrt(65) |8*sqrt(85)/2 |7*sqrt(145)/3 360 |1452| 55| 55| 66|3*sqrt(89) |3*sqrt(89) | 30 480 |2028| 65| 65|104|4*sqrt(61) |4*sqrt(61) | 40 630 |2646| 63| 84|105|9*sqrt(65)/2 |4*sqrt(85) |7*sqrt(145)/2 1120|4704| 84| 112|140|6*sqrt(65) |16*sqrt(85)/3 |14*sqrt(145)/3 1312|8820| 63| 280|287|36*sqrt(2501)/35|40*sqrt(7585)/63|28*sqrt(9061)/45 1440|5808|110| 110|132|6*sqrt(89) |6*sqrt(89) | 60 1750|7350|105| 140|175|15*sqrt(65)/2 |20*sqrt(85)/3 |35*sqrt(145)/6 1768|8670| 85| 204|221|50*sqrt(13)/3 |12*sqrt(689)/5 |34*sqrt(949)/15 1920|8112|130| 130|208|8*sqrt(61) |8*sqrt(61) | 80 ....................................................... REFERENCES C. Kimberling, Triangle Centers and Central Triangles. Congr. Numer. 129, 1-295, 1998. LINKS Table of n, a(n) for n=1..42. Wolfram MathWorld, Incentral Triangles EXAMPLE 70 is in the sequence because the formula A' = 2*A*a*b*c/((a+b)*(b+c)*(c+a)) gives with the initial triangle (21,28,35): A'= 2*294*21*28*35/((21+28)*(28+35)*(35+21))= 70, with the area A = 294 obtained by Heron's formula A =sqrt(s*(s-a)*(s-b)*(s-c))= sqrt((42*(42-21)*(42-28)*(42-35)) = 294, where s=42 is the semiperimeter. MATHEMATICA nn=1000; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s(s-a)(s-b)(s-c); t= 2*Sqrt[area2]*a*b*c/((a+b)*(b+c)*(c+a)); If[0

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