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

18, 72, 90, 162, 180, 252, 288, 360, 450, 540, 630, 648, 720, 810, 882, 990, 1008, 1152, 1440, 1458, 1512, 1620, 1638, 1800, 1890, 2160, 2178, 2250, 2268, 2520, 2592, 2772, 2880, 2970, 3042, 3240, 3528, 3672, 3960, 4032, 4050, 4158, 4410, 4500, 4608, 4680, 4860

Offset: 1

Author

Michel Lagneau, Nov 07 2013

Keywords

• nonn

Comments

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).

The area of the reflection triangle is given by

A' = A*t/(a^2*b^2*c^2) where A is the area of the reference triangle of sides (a, b, c) and

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).

See the link for the side lengths of the reflection triangles.

Properties of this sequence:

The areas corresponding to the primitive reflection triangles are 18, 90, 180, 252, 540,...

The non-primitive triangles of areas 4*a(n),9*a(n),...,p^2*a(n),... are in the sequence.

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).

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.

-------------------------------------------------------------------------

| A' | A | a | b | c | a' | b' | c'|

-------------------------------------------------------------------------

| 18 | 6 | 3 | 4 | 5 | 9*sqrt(17)/5 | 4*sqrt(97)/5 | 5 |

| 72 | 24 | 6 | 8 | 10 | 18*sqrt(17)/5 | 8*sqrt(97)/5 | 10 |

| 90 | 30 | 5 | 12 | 13 | 5*sqrt(1321)/13 | 36*sqrt(41)/13 | 13 |

| 162 | 54 | 9 | 12 | 15 | 27*sqrt(17)/5 | 12*sqrt(97)/5 | 15 |

| 180 | 60 | 8 | 15 | 17 | 8*sqrt(2089)/17 | 45*sqrt(89)/17 | 17 |

| 252 | 84 | 7 | 24 | 25 | 7*sqrt(5233)/25 | 72*sqrt(113)/25 | 25 |

| 288 | 96 | 12 | 16 | 20 | 36*sqrt(17)/5 | 16*sqrt(97)/5 | 20 |

| 360 | 120 | 10 | 24 | 26 | 10*sqrt(1321)/13 | 72*sqrt(41)/13 | 26 |

| 450 | 150 | 15 | 20 | 25 | 9*sqrt(17) | 4*sqrt(97) | 25 |

| 540 | 180 | 9 | 40 | 41 | 27*sqrt(1609)/41 | 40*sqrt(2329)/41 | 41 |

| 630 | 210 | 12 | 35 | 37 | 36*sqrt(1241)/37 | 35*sqrt(2521)/37 | 37 |

| 648 | 216 | 18 | 24 | 30 | 54*sqrt(17)/5 | 24*sqrt(97)/5 | 30 |

.......................................................................

Examples

			18 is in the sequence. We use two ways:
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.
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:
a' = 9*sqrt(17)/5;
b' = 4*sqrt(97/5);
c' = 5.
Now, we use Heron's formula with (a',b',c'). We find A'=sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
s1 =(a'+b'+c')/2 = (9*sqrt(17)/5+ 4*sqrt(97/5)+ 5)/2. We find A'= 18.
		

References

• D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.

Links

• Eric W. Weisstein, MathWorld: Reflection Triangle

Crossrefs

Cf. A188158, A009111, A009112.

Programs

Mathematica

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]