A231739 Integer areas of the Lucas Central triangles of integer-sided triangles.

2775, 11100, 24975, 34125, 44400, 69375, 99900, 135975, 136500, 177600, 224775, 277500, 307125, 335775, 399600, 468975, 543900, 546000, 624375, 710400, 801975, 853125, 899100, 1001775, 1110000, 1223775, 1228500, 1343100, 1467975, 1598400, 1672125, 1734375

Offset: 1

Author

Michel Lagneau, Nov 13 2013

Keywords

• nonn

Comments

Consider a reference triangle ABC and externally inscribe a square on the side BC. Now join the new vertices S_AB and S_AC of this square with the vertex A, marking the points of intersection Q_ABC and Q_ACB. Next, draw lines perpendicular to the side BC through each of Q_ABC and Q_ACB. These points cross the sides AB and AC at Q_AB and Q_AC, respectively, resulting in an inscribed square Q_ABC Q_ACB Q_AB Q_AC. The circumcircle through A, Q_AB, and Q_AC is then known as the Lucas A-circles (Panakis 1973, p. 458; Yiu and Hatzipolakis 2001), and repeating the process for other sides gives the corresponding B- and C-circles.

The Lucas central triangle is the triangle L1 L2 L3 formed by the centers of the Lucas circles of a given reference triangle ABC.

The Lucas central triangle has side lengths

a' = 2*R*(a*b*c+b^2*R+c^2*R)*a/((a*c+2*b*R)*(a*b+2*c*R));

b' = 2*R*(a*b*c+a^2*R+c^2*R)*b/((b*c+2*a*R)*(a*b+2*c*R));

c' = 2*R*(a*b*c+a^2*R+b^2*R)*c/((b*c+2*a*R)*(a*c+2*b*R)).

Its area is given by

S' = a*b*c*R^2*sqrt(u)/v where:

u=3*a^2*b^2*c^2+4*a*b*c*(a^2+b^2+c^2)*R+4*(a^2*b^2+a^2*c^2+b^2*c^2)*R^2 ;

v=(b*c+2*a*R)*(a*c+2*b*R)*(a*b+2*c*R);

R = a*b*c/sqrt((a+b+c)*(b+c-a)*(c+a-b)*(a+b-c)) = a*b*c/(4*S) where R is the circumradius of the reference triangle ABC and S its area.

Properties of this sequence:

The side lengths of the Lucas central triangles are rational numbers, sometimes integers, for example a(n) = 136500, ...

The primitive Lucas central triangles are 2775, 34125, ...

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

The following table gives the first values (S', S, a, b, c, a', b', c') where S' is the area of the Lucas central triangles, S is the area of the initial triangles ABC, a, b, c are the integer sides of ABC and a', b', c' are the sides of the Lucas central triangles.

+--------+---------+------+------+------+----------+---------+---------+

| S' | S | a | b | c | a' | b' | c' |

+--------+---------+------+------+------+----------+---------+---------+

| 2775 | 8214 | 111 | 148 | 185 | 975/14 | 580/7 | 185/2 |

| 11100 | 32856 | 222 | 296 | 370 | 975/7 | 1160/7 | 185 |

| 24975 | 73926 | 333 | 444 | 555 | 2925/14 | 1740/7 | 555/2 |

| 34125 | 115248 | 490 | 490 | 588 | 545/2 | 545/2 | 300 |

| 44400 | 131424 | 444 | 592 | 740 | 1950/7 | 2320/7 | 370 |

| 69375 | 205350 | 555 | 740 | 925 | 4875/14 | 2900/7 | 925/2 |

| 99900 | 295704 | 666 | 888 | 1110 | 2925/7 | 3480/7 | 555 |

| 135975 | 402486 | 777 | 1036 | 1295 | 975/2 | 580 | 1295/2 |

| 136500 | 460992 | 980 | 980 | 1176 | 545 | 545 | 600 |

| 177600 | 525696 | 888 | 1184 | 1480 | 3900/7 | 4640/7 | 740 |

| 224775 | 665334 | 999 | 1332 | 1665 | 8775/14 | 5220/7 | 1665/2 |

| 277500 | 821400 | 1110 | 1480 | 1850 | 48755/7 | 5800/7 | 925 |

| 307125 | 1037232 | 1470 | 1470 | 1764 | 1635/2 | 1635/2 | 900 |

+--------+---------+------+------+------+----------+---------+---------+

Examples

			2775 is in the sequence. We use two ways:
First way: from the initial triangle (111, 148, 185) the formula in the comments gives directly the area of the Lucas central triangle: S' = a*b*c*R^2*sqrt(u)/v where:
R = a*b*c/4S = a*b*c/(4*sqrt(s(s-a)(s-b)(s-c))) = 111*148*185/(4*sqrt(222(222-111)(222-148)(222-185))) = 111*148*185/(4*8214) = 185/2 with S=8214.
sqrt(u) = sqrt(3*a^2*b^2*c^2+4*a*b*c*(a^2+b^2+c^2)*R+4*(a^2*b^2+a^2*c^2+b^2*c^2)*R^2) = sqrt(154007727700225) = 12409985.
v = (b*c+2*a*R)*(a*c+2*b*R)*(a*b+2*c*R) = 116291549487925
And S' = 111*148*185*(185/2)^2*12409985/116291549487925 = 2775.
Second way: by calculation of the sides a', b', c' and by using Heron's formula. With the formulas given in the link, we find
a' = 975/14;
b' = 580/7;
c' = 185/2.
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 = (975/14 + 580/7 + 185/2)/2 = 245/2;
We find S'= 2775.
		

Links

• Peter J. C. Moses, Circles and Triangle Centers Associated with the Lucas Circles, Forum Geom. 5, 97-106, 2005.
• Wolfram MathWorld, Lucas Central Triangle
• Wolfram MathWorld, Lucas Circles
• P. Yiu and A. P. Hatzipolakis, The Lucas Circles of a Triangle, Amer. Math. Monthly 108, 444-446, 2001.

Crossrefs

Cf. A188158.

Programs

Mathematica

nn=2000;lst={};Do[s =(a + b + c)/2;area2=s (s-a)(s-b)(s-c);If[area2>0,R = a*b*c /(4*Sqrt[area2]);u = a*b*c*R^2 * Sqrt[3*a^2*b^2*c^2 + 4*a*b*c*(a^2 + b^2 + c^2)*R + 4*(a^2*b^2 + a^2*c^2 + b^2*c^2)*R^2];v = (b*c + 2*a*R)*(a*c + 2*b*R)*(a*b + 2*c*R);If[IntegerQ[u/v], AppendTo[lst, u/v]]],{a,nn},{b,a},{c,b}];Union[lst]