Gossard perspector

In geometry the Gossard perspector (also called the Zeeman–Gossard perspector) is a special point associated with a plane triangle. It is a triangle center and it is designated as X(402) in Clark Kimberling's Encyclopedia of Triangle Centers. The point was named Gossard perspector by John Conway in 1998 in honour of Harry Clinton Gossard who discovered its existence in 1916. Later it was learned that the point had appeared in an article by Christopher Zeeman published during 1899 – 1902. From 2003 onwards the Encyclopedia of Triangle Centers has been referring to this point as Zeeman–Gossard perspector.

Definition

H, H_A, H_B, H_C, H_g are orthocenters, and G, G_A, G_B, G_C, G_g are centroids of triangles ABC, AEF, BFD, CDE, A_gB_gC_g respectively.

Gossard triangle

Let ABC be any triangle. Let the Euler line of triangle ABC meet the sidelines BC, CA and AB of triangle ABC at D, E and F respectively. Let A_gB_gC_g be the triangle formed by the Euler lines of the triangles AEF, BFD and CDE, the vertex A_g being the intersection of the Euler lines of the triangles BFD and CDE, and similarly for the other two vertices. The triangle A_gB_gC_g is called the Gossard triangle of triangle ABC.

Gossard perspector

Let ABC be any triangle and let A_gB_gC_g be its Gossard triangle. Then the lines AA_g, BB_g and CC_g are concurrent. The point of concurrence is called the Gossard perspector of triangle ABC.

Properties

• Let A_gB_gC_g be the Gossard triangle of triangle ABC. The lines B_gC_g, C_gA_g and A_gB_g are respectively parallel to the lines BC, CA and AB.
• Any triangle and its Gossard triangle are congruent.
• Any triangle and its Gossard triangle have the same Euler line.
• The Gossard triangle of triangle ABC is the reflection of triangle ABC in the Gossard perspector.

Trilinear coordinates

The trilinear coordinates of the Gossard perspector of triangle ABC are

( f ( a, b, c ) : f ( b, c, a ) : f ( c, a, b ) )

where

f ( a, b, c ) = p ( a, b, c ) y ( a, b, c ) / a

where

p ( a, b, c ) = 2a⁴ − a²b² − a²c² − ( b² − c² )²

and

y ( a, b, c ) = a⁸ − a⁶ ( b² + c² ) + a⁴ ( 2b² − c² ) ( 2c² − b² ) + ( b² − c² )² [ 3a² ( b² + c² ) − b⁴ − c⁴ − 3b²c² ]

In the figure, DEF is the Euler line of triangle ABC. The line XYZ moves parallel to the line DEF. The triangle A'B'C'  remains congruent to triangle ABC whatever be the position of the line XYZ. The blue 'inverted' triangle is the Gossard triangle of triangle ABC.

Generalisations

The construction yielding the Gossard triangle of a triangle ABC can be generalised to produce triangles A'B'C'  which are congruent to triangle ABC and whose sidelines are parallel to the sidelines of triangle ABC.

Generalisation 1

This result is due to Christopher Zeeman.

Let l be any line parallel to the Euler line of triangle ABC. Let l intersect the sidelines BC, CA, AB of triangle ABC at X, Y, Z respectively. Let A'B'C'  be the triangle formed by the Euler lines of the triangles AYZ, BZX and CXY. Then triangle A'B'C'  is congruent to triangle ABC and its sidelines are parallel to the sidelines of triangle ABC.

Generalisation 2

Paul Yiu's generalisation of Gossard triangle.

This generalisation is due to Paul Yiu.

Let P be any point in the plane of the triangle ABC different from its centroid G.

Let the line PG meet the sidelines BC, CA and AB at X, Y and Z respectively.

Let the centroids of the triangles AYZ, BZX and CXY be G_a, G_b and G_c respectively.

Let P_a be a point such that YP_a is parallel to CP and ZP_a is parallel to BP.

Let P_b be a point such that ZP_b is parallel to AP and XP_b is parallel to CP.

Let P_c be a point such that XP_c is parallel to BP and YP_c is parallel to AP.

Let A'B'C'  be the triangle formed by the lines G_aP_a, G_bP_b and G_cP_c.

Then the triangle A'B'C'  is congruent to triangle ABC and its sides are parallel to the sides of triangle ABC.

When P coincides with the orthocenter H of triangle ABC then the line PG coincides with the Euler line of triangle ABC. The triangle A'B'C'  coincides with the Gossard triangle A_gB_gC_g of triangle ABC.

Generalisation 3

Let ABC be a triangle. Let H and O be two points, and let the line HO meets BC, CA, AB at A₀, B₀, C₀ respectively. Let A_H and A_O be two points such that C₀A_H parallel to BH, B₀A_H parallel to CH and C₀A_O parallel to BO, B₀A_O parallel to CO. Define B_H, B_O, C_H, C_O cyclically. Then the triangle formed by the lines A_HA_O, B_HB_O, C_HC_O and triangle ABC are homothetic and congruent, and the homothetic center lies on the line OH. If OH is any line through the centroid of triangle ABC, this problem is the Yiu's generalization of the Gossard perspector theorem.