HYACINTHOS 26907

[Antreas P. Hatzipolakis]:

Let ABC be a triangle:

Consider the sequence of triangles: .... M_2, M_1, M0, M1, M2.....

M0 = ABC
M1 = the medial triangle of M0

M2 = the medial triangle of M1

....

Mn = the medial triangle of Mn-1

 

and 

M_-1 = the antimedial triangle of M0
M_-2 = the antimedial triangle of M_1

The triangles share the same centroid G. Let Kn be the symmedian point of Mn.

Which is the center of the circumhyperbola (A, B, C, G, Kn) ?

For n = 0, it is X(1084), for n = 1 it is X(15449). Which is it for n = -1 (antimedial triangle)?

 

Is it lying on Steiner inellipse for all n's ?

 

 [César Lozada]:

 

 

Let A(n) be the A-vertex of the triangle M(n).For n≠0 A(n) has barycentrics coordinates:

 

A(n) = λ(n) : 1 : 1  

 

where

λ(n) = f(n)/g(n)

 

and f(n), g(n) are recursive functions satisfying:

 

f(n) = 0, for n=1

f(n) = 2* f(n-1)+2*(-1)^n ),  for n>1,

f(n) = (f(n+1)+2*(-1)^n)/2, otherwise

                                                                                                                                          

g(n) = 1, for n=1

g(n) = 2*g(n+1)-(-1)^n, for n>1

g(n) = (g(n+1)-(-1)^n)/2, otherwise

 

Here is a small sequence of values of ( n, λ(n) ) :

(-7, -85/43), (-6, -43/21), (-5, -21/11), (-4, -11/5), (-3, -5/3), (-2, -3), (-1, -1), (1, 0), (2, 2), (3, 2/3), (4, 6/5), (5, 10/11), (6, 22/21), (7, 42/43)

 

Successive results are expressed in barycentric coordinates and apply for n≠0:

 

The symmedian point of M(n) is:

K(n) = a^2* λ(n) +b^2+c^2  :  a^2+ b^2*λ(n)+c^2  :  a^2 +b^2+c^2*λ(n)

and the circum-conic through A, B, C, G, K(n) has barycentric equation:

c(n) = ∑ [ (b^2-c^2)*(a^2*λ(n) +b^2+c^2)*y*z

with center on the Steiner inellipse given by

O(n) = ( (b^2-c^2)*(a^2*λ(n)+b^2+c^2) )^2 : :

and perspector on the line at infinity :

P(n) = (b^2-c^2)*(a^2*λ(n)+b^2+c^2) : :

 

Examples:

 

·         For n = -2:

 

P(-2) = X(3566)

 

O(-2) = X(114)X(14772) ∩ X(115)X(2971)

= ((b^2-c^2)*(3*a^2-b^2-c^2))^2 : :

= on Steiner inellipse and these lines: {32, 1992}, {114, 14772}, {115, 2971}, {800, 11672}, {1084, 6132}, {3037, 9043}

= [ 2.5152939083390140, 0.2743000722929381, 2.2898595510866460 ]

 

·         For n = -1:

 

P(-1) = X(525)

 

O(-1) = complement of X(648)

= ( (b^2-c^2)*(-a^2+b^2+c^2))^2 : :

= X(648)+3*X(1494), 2*X(648)-3*X(3163), 2*X(1494)+X(3163)

= on Steiner inellipse and these lines: {2, 648}, {3, 67}, {4, 9530}, {32, 14376}, {69, 248}, {74, 3184}, {97, 15108}, {115, 127}, {122, 125}, {131, 12358}, {141, 216}, {233, 14767}, {235, 8798}, {253, 393}, {264, 1972}, {340, 401}, {343, 6509}, {440, 4370}, {441, 524}, {594, 4605}, {661, 10933}, {754, 15013}, {868, 8754}, {1084, 6388}, {1086, 2968}, {1146, 8287}, {1216, 10600}, {1632, 2794}, {2453, 10749}, {2632, 7068}, {3150, 7668}, {3548, 7888}, {3620, 10979}, {5095, 15000}, {5099, 14672}, {5596, 20027}, {7687, 10745}, {7813, 14961}, {7821, 11585}, {7826, 10316}, {7873, 12605}, {8541, 14003}

= midpoint of X(i) and X(j) for these {i,j}: {2, 1494}, {69, 287}, {253, 6330}, {264, 1972}, {340, 401}

= reflection of X(i) in X(j) for these (i,j): (3163, 2), (3284, 441), (20028, 141)

= complement of X(648)

= antipode of X(3163) in the Steiner inellipse

= trilinear pole of the line {1650, 5489}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (69, 6389, 577), (122, 125, 13611), (125, 2972, 122), (127, 339, 115)

= [ 4.4256372518807090, 2.2627590796099000, 0.0315371566941210 ]

 

·         For n = 1:

 

P(1) = X(826)

 

O(1) = X(15449)

 

·         For n = 2:

 

P(2) = X(7927)

 

O(2) = X(32)X(14381) ∩ X(2482)X(7819)

= ((b^2-c^2)*(2*a^2+b^2+c^2))^2 : :

= on Steiner inellipse and these lines: {32, 14381}, {2482, 7819}, {3124, 15449}, {5421, 11672}, {6660, 11063}

= [ 1.7570517229278650, 3.4757746372785740, 0.4234120147478900 ]

 

César Lozada