The locus is a hyperbola. The proof could be deduced from the general theorem on Maclaurin's generation of conics (see Maclaurin.html ). Here I give the slightly simpler proof as an exercise in the use of trilinear coordinates (see Trilinears.html ). The trilinears are defined through the projective base {A,B,C,D}, D being the centroid of the triangle.

In these system the four basic points have coordinates {A=(1,0,0), B=(0,1,0), C=(0,0,1), D=(1,1,1)}. Side-lines of the triangles correspond to equations BC(x=0), CA(y=0), AB(z=0). The line at infinity corresponds to equation x+y+z=0 and its intersection with line ax+by+cz=0, is the point at infinity of the later and has coordinates (b-c, c-a, a-b). This will be used below. Now the calculation:

[1] A point X on AB has coordinates (s,t,0).

[2] The equation of line XD is tx-sy+(s-t)z=0. Its intersection point Y with AC is given by (t-s, 0, t).

[3] Lines BY and CX are given correspondingly by equations: tx + (s-t)z = 0, tx - sy = 0.

[4] Their intersection has coordinates x = s(t-s), y=t(t-s), z=st, which by elliminating (s,t) gives the equation:

yz - zx - xy = 0.

[5]This is a conic passing through the vertices of the triangle. Its perspector is the point S(-1,1,1). The tangents at the vertices can be calculated, using general arguments, from their relation to the perspector. They are also easily calculated directly from the equation: At A(1,0,0) the tangent is y+z=0, at B(0,1,0) is x-z=0 and at C(0,0,1) is x-y=0. It follows, by calculating the point at infinity of the first, that it is parallel to BC(x=0). The other two tangents pass obviously through point D, later seen then to be the pole of line BC with respect to the conic.

[6] More general, the matrix of the conic being

its tangent at (x

[7] Substituting into the equation x+y+z=0 of the line at infinity, the parametric expression of the curve given in [4], we find that t/s satisfies the quadratic x

[8] Take r

[9] AK has equation y-z=0 and its intersection with the conic is (1,2,2). Thus Q is the point on AK whose distance d

[10] Since the transformation (x,y,z)-->(x,z,y) leaves the conic invariant, it is symmetric with respect to the median AK.

[11] By considering the complete quadrilateral of the four points {X,Y,B,C} one verifies that the intersection point P' of BC with AP is the pole of line XY with respect to the conic.

[12] The conic generation through the intersection point P, studied here, is an example of the

[13] Points {D,S} are harmonic conjugate to {A,K}.

[14] The simple conic discussed here can be used to generate conics with analogous properties for arbitrary triangles. This is discussed in ConicsMaclaurin2.html .

ConicsMaclaurin2.html

Maclaurin.html

Trilinears.html

Produced with EucliDraw© |