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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<area2 && IntegerQ[t], AppendTo[lst, t]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

CROSSREFS

Cf. A188158.

Sequence in context: A234564 A234557 A245857 * A072596 A309310 A330702

Adjacent sequences:  A227876 A227877 A227878 * A227880 A227881 A227882

KEYWORD

nonn

AUTHOR

Michel Lagneau, Oct 25 2013

STATUS

approved

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Last modified June 7 01:26 EDT 2020. Contains 334836 sequences. (Running on oeis4.)