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A231328
Integer areas of the reflection triangles of integer-sided triangles.
0
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
1,1
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.
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| A' | A | a | b | c | a' | b' | c'|
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| 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 |
.......................................................................
REFERENCES
D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.
EXAMPLE
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.
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Nov 07 2013
STATUS
approved