OFFSET
1,1
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 |
+--------+---------+------+------+------+----------+---------+---------+
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.
EXAMPLE
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.
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Nov 13 2013
STATUS
approved