X3035

COMPLEMENT OF X11

X3035 is the complement of X11.
< A B C >: Reference Triangle △ABC < M_a M_b M_c >: Medial Triangle △M_a M_b M_c [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 11 }: Central Line L(2, 11) { 3, 119 }: Central Line L(3, 119) { 10, 140 }: Central Line L(10, 140) { 12, 404 }: Central Line L(12, 404) { 36, 529 }: Central Line L(36, 529) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) M_a = Midpoint M_a of BC (0.8061, 0.2999) M_b = Midpoint M_b of CA (0.3061, 0.2999) M_c = Midpoint M_c of AB (0.5000, 0.0000) X_2 = CENTROID (0.5374, 0.2000) X_11 = FEUERBACH POINT (0.7347, 0.0667) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_119 = FEUERBACH ANTIPODE (0.3776, 0.4312) X_10 = SPIEKER CENTER (0.5204, 0.1833) X_140 = MIDPOINT OF X(3) AND X(5) (0.5281, 0.1755) X_12 = {X(1),X(5)}-HARMONIC CONJUGATE OF X(11) (0.5641, 0.2407) X_404 = {X(2),X(3)}-HARMONIC CONJUGATE OF X(21) (0.5538, 0.2429) X_36 = INVERSE-IN-CIRCUMCIRCLE OF INCENTER (1.3333, 1.6330) X_529 = DIRECTION OF VECTOR AX + BX + CX, where X = X(12) (0.8589, 0.6909) X_3035 = COMPLEMENT OF X(11) (0.4388, 0.2666) A B C Ma Mb Mc X2 X11 X3 X119 X10 X140 X12 X404 X36 X529 X3035 (a, b, c) = (5, 6, 7) | AB = 1

X3035 (COMPLEMENT OF X11) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

  • Derived Triangle: Medial Triangle △MaMbMc — vertices Ma, Mb, Mc (Formed by the midpoints of sides BC, CA, AB)

  • Central Lines: {2, 11}

  • Related Centers: X2 (CENTROID), X11 (FEUERBACH POINT)

Trilinears: \(bc[(a - b + c)(a - c)^{2} + (a + b - c)(a - b)^{2}] : :\)
Search Value (6, 9, 13): \(2.622485628110\)
▸ Linear Combinations1 combination

X3035 lies on the Spieker circle.

▸ Incident Central Lines29 lines

{1, 1145}, {2, 11}, {3, 119}, {8, 1317}, {9, 1768}, {10, 140}, {12, 404}, {36, 529}, {40, 1537}, {80, 1698}, {104, 631}, {153, 2551}, {230, 1575}, {405, 2932}, {468, 1861}, {474, 498}, {516, 1638}, {549, 993}, {620, 2787}, {632, 1484}, {676, 2804}, {899, 1818}, {908, 1155}, {960, 2800}, {1125, 1387}, {1532, 2077}, {2801, 3041}, {2826, 3039}, {2827, 3038}

▸ Functional Relations10 relations

Notes

Let P = X100 and G=X2; let GA be the centroid of the triangle BCP. Define GB and GC cyclically. Then G, GA, GB, GC are the vertices of a quadrilateral that is homothetic to the cyclic quadrilateral having vertices A, B, C, P. The center of homothety is X3035. Moreover, X3035 is the centroid of both quadrilaterals and is the Feuerbach point of the medial triangle. (Randy Hutson, 9/23/2011)