X395

MIDPOINT OF X14 AND X16

X395 is the center of the (equilateral) pedal triangle of X16, as well as the circumcenter of the pedal triangle of X14.
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 14, 16 }: Central Line L(14, 16) { 2, 6 }: Centroid–Symmedian Line { 5, 13 }: Central Line L(5, 13) { 61, 140 }: Central Line L(61, 140) { 3, 398 }: Central Line L(3, 398) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_14 = 2nd ISOGONIC CENTER (0.9134, -0.1143) X_16 = 2nd ISODYNAMIC POINT (1.8453, 2.3239) X_2 = CENTROID (0.5374, 0.2000) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_5 = NINE-POINT CENTER (0.5561, 0.2489) X_13 = 1st ISOGONIC CENTER (FERMAT POINT, TORRICELLI POINT) (0.5879, 0.2819) X_61 = ISOGONAL CONJUGATE OF X(17) (0.5764, 0.2283) X_140 = MIDPOINT OF X(3) AND X(5) (0.5281, 0.1755) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_398 = CROSSPOINT OF X(4) AND X(18) (0.5945, 0.2099) X_395 = MIDPOINT OF X(14) AND X(16) (1.3794, 1.1048) A B C X14 X16 X2 X6 X5 X13 X61 X140 X3 X398 X395 (a, b, c) = (5, 6, 7) | AB = 1

X395 (MIDPOINT OF X14 AND X16) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(\cos(B - C) + 2 \cos(A + \pi /3) : :\)
Search Value (6, 9, 13): \(-1.205389861913\)
▸ Incident Central Lines15 lines

{2, 6}, {3, 398}, {5, 13}, {14, 16}, {15, 549}, {39, 618}, {53, 472}, {61, 140}, {115, 530}, {187, 531}, {202, 495}, {216, 465}, {466, 577}, {532, 624}, {533, 619}

▸ Functional Relations13 relations

Notes

Let F = X14, the 2nd Fermat point. Let Ab be the point on CA and Ac the point on AB such that FAbAc is an equilateral triangle. Let A' be the midpoint of FAbAc, and define B' and C' cyclically. The triangle A'B'C', here named the 8th Fermat-Dao equilateral triangle, is perspective to ABC. The circumcircle of A'B'C' is the pedal circle of X14 and (X16), and the centroid of A'B'C' is X395. The perspector of ABC and A'B'C' is X2, and the centroid of A'B'C' is X395. See X16247, X16267, and X16459. (Based on notes from Dao Thanh Dao, March 15, 2018)

X395 is the pole of the Euler line wrt every conic passing through X14, X16, X18 and X62. (Randy Hutson, January 17, 2020)

You can view 7th and 8th Fermat-Dao equilateral triangles (Fermat-Dao_7-8.gif).