X1908–X2000

Bicentric Pairs PU(k) & Orthocorrespondents

Midpoints, crosssums, crossdifferences, trilinear poles, and trilinear products of bicentric pairs PU(k) (X1908–X1982), Vega transforms, and orthocorrespondents (X1992–X2000).
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) ( A R_b ): Euclid I.22 Guide Circle (A) ( B R_a ): Euclid I.22 Guide Circle (B) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) circumcircle: Circumcircle of △ABC incircle: Incircle of △ABC A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) R_b = Radius Point R_b (0.8571, 0.0000) R_a = Radius Point R_a (0.2857, 0.0000) C = Vertex C (0.6122, 0.5999) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_1 = INCENTER (0.5714, 0.2333) X_1908 = MIDPOINT OF PU(8) (0.5880, 0.2495) X_1909 = CROSSSUM OF PU(9) (0.5132, 0.1854) X_1912 = IDEAL POINT OF PU(9) (0.0794, 0.5669) X_1917 = TRILINEAR 6th POWER POINT (0.6636, 0.4181) X_1923 = BICENTRIC SUM OF PU(13) (0.6447, 0.3295) X_1925 = CROSSSUM OF PU(13) (0.4347, 0.1429) X_1938 = IDEAL POINT OF PU(15) (-0.0178, 0.3902) X_1941 = CROSSSUM OF PU(17) (0.5282, 0.3794) X_1960 = BICENTRIC SUM OF PU(25) (1.2500, 1.2247) X_1977 = BICENTRIC DIFFERENCE OF PU(42) (0.8105, 0.1754) X_1979 = CROSSSUM OF PU(41) (1.3801, -0.0730) X_1981 = VEGA TRANSFORM OF X(647) (0.7074, 0.6447) X_1983 = VEGA TRANSFORM OF X(523) (0.9661, 0.8719) X_1984 = VEGA TRANSFORM OF X(1020) (0.4361, -0.0651) X_1986 = HATZIPOLAKIS REFLECTION POINT (0.5352, 0.5340) X_1990 = ORTHIC-AXIS INTERCEPT OF LINE X(4)X(6) (0.6502, 0.7938) A B C X3 X1 X1908 X1909 X1912 X1917 X1923 X1925 X1938 X1941 X1960 X1977 X1979 X1981 X1983 X1984 X1986 X1990 (a, b, c) = (5, 6, 7) | AB = 1

X1908–X2000 (Bicentric Pairs PU(k) & Orthocorrespondents) — 16 representative centers (from 93 in collection) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Group Preamble — Centers from Bicentric Pairs, 1908-1982

For a definition of a bicentric pair (e.g., the 1st and 2nd Brocard points) click Tables at the top of this page. Suppose P and U are a bicentric pair. Many operations on P and U result in triangle centers. Among these are trilinear and barycentric product, bicentric sum, bicentric difference, crosssum, and crossdifference. For definitions of these, click Tables. At the time this section is added to ETC (September 15, 2003), bicentric pairs

P(1),U(1); P(2),U(2); …; P(42),U(42)

are defined in Tables. In this present section, the abbreviation PU(n) means the bicentric pair P(n),U(n).

Group Preamble — Orthocorrespondents, 1992-2006

Suppose P is a point in the plane of triangle ABC. The perpendiculars through P to the lines AP, BP, CP meet the lines BC, CA, AB, respectively, in collinear points. Let L denote their line. The trilinear pole of L is the orthocorrespondent of :g:`P <P*>`. This definition is introduced in Bernard Gibert , Orthocorrespondence and Orthopivotal Cubics, Forum Geometricorum 3 (2003) pages 1-27.

If P is given in barycentrics by P =\(p : q : r\), then the orthocorrespondent of P has barycentrics
\(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where

\(f(a,b,c) = a^{2}qr + (-pS_{A} + qS_{B} + rS_{C})p\),

where \(S_{A}\) = (\(b^{2}\) + \(c^{2}\) - \(a^{2})\)/2, and \(S_{B}\) and \(S_{C}\) are defined cyclically.

If follows that if X =\(x : y : z\) in trilinears, then the orthocorrespondent of X has trilinears
\(g(a,b,c) : g(b,c,a) : g(c,a,b)\), where

\(g(a,b,c) = yz + (-x \cos A + y \cos B + z \cos C)x\).

Pairs (i,j) for which the orthocorrespondent of Xi is Xj include the following:
(1,57), (4,2), (11,651), (13,13), (14,14), (15,62), (16,61), (30,2), (98,287), (100,1332), (101,1331), (103,1815), (105,1814), (106,1797), (107,648), (108,651), (109,1813), (111,895), (112,110), (115,110), (125,648), (132,287), (1560,895), (1566,677), (1785,57)

The orthocorrespondent of every point on the line at infinity is the centroid. Two orthoassociate points (i.e., an inverse pair in the polar circle, such as X112 and X115) share the same orthocorrespondent.

X_1978 — TRILINEAR PRODUCT OF PU(41)

X1978TRILINEAR PRODUCT OF PU(41)

As the trilinear product of Steiner circumellipse antipodes, X1978 lies on conic {{A, B, C,X668,X789}} with center X6376 and perspector X75. (Randy Hutson, July 11, 2019)
X_1981 — VEGA TRANSFORM OF X_647

X1981VEGA TRANSFORM OF X647

lies on the line [\(x:y:z\)] that has x,y,z as coefficients. (The line [\(x:y:z\)] is the trilinear polar of the isogonal conjugate of X.) Thus, the Vega transform of X647 lies on the Euler line.
X_1986 — HATZIPOLAKIS REFLECTION POINT

X1986HATZIPOLAKIS REFLECTION POINT

Let HaHbHc be the orthic triangle of triangle ABC. Let AB be the reflection of A in Hc, and define AC, BC, BA, CA, CB functionally. Then the nine-point circles of the triangles
X_1989 — ISOGONAL CONJUGATE OF X_323

X1989ISOGONAL CONJUGATE OF X323

X1989 plays a major role in the theory of special isocubics, as presented in Chapter 6 of Jean-Pierre Ehrmann and Bernard Gibert,, “Special Isocubics in the Triangle Plane,” downloadable from Bernard Gibert, Cubics in the Triangle Plane.
X_1991 — 2nd VAN LAMOEN PERPENDICULAR BISECTORS POINT

X19912nd VAN LAMOEN PERPENDICULAR BISECTORS POINT

Erect squares inwardly on the sides of triangle ABC. Two edges emanate from A; let P and Q be their endpoints. Let a’ be the perpendicular bisector of PQ, and define b’ and c’ cyclically. Then a’, b’, c’ concur in X1991. See X591 for the 1st Van Lamoen perpendicular bisectors point, constructed from outwardly drawn squares.
X_1992 — ORTHOCORRESPONDENT OF X_2

X1992ORTHOCORRESPONDENT OF X2

Let HaHbHc be the orthic triangle. X1992 is the radical center of the 2nd Lemoine circles of triangles AHbHc, BHcHa, CHaHb. (Randy Hutson, July 31 2018)
X_1994 — ORTHOCORRESPONDENT OF X_5

X1994ORTHOCORRESPONDENT OF X5

Let OAOBOC be the Kosnita triangle. Let A' be the trilinear pole of line OBOC, and define B' and C' cyclically. The lines AA', BB', CC' concur in X1994. (Randy Hutson, March 21, 2019)
X_1995 — ORTHOCORRESPONDENT OF X_6

X1995ORTHOCORRESPONDENT OF X6

Let P be a point in the plane of ABC and not on the circumcircle. Let (OA) be the circle tangent to the circumcircle at A and passing through P. Let A' be the antipode of A in (OA). Let LA be the tangent to (OA) at A'. Define LB and LC cyclically. Let TA = LB∩LC, and define TB and TC cyclically. Triangle TATBTC is homothetic to the tangential triangle. When P = X6, the center of homothety is X1995. (Randy Hutson, June 7, 2019)