## Maclaurin's generation of a hyperbola related to the equilateral triangle

Consider an equilateral triangle ABC and its centroid (middle) D. Draw through D an arbitrary line intersecting sides AB, AC at points X, Y respectively. Find the locus of the intersection point P of lines BY, AX.

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 (x0, y0,z0) is (y0+z0)x+(x0-z0)y+(x0-y0)z=0. At point P corresponding to X(s,t,0) the tangent is t2x-s2y-(s-t)2z=0 and its intersection point with line XD satisfies y+z=0, thus it is a point Z on the tangent of the conic at A.
[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 x2+x-1=0, which has two real roots. Thus the conic, intersecting the line at infinity at two distinct points, is a hyperbola.
[8] Take r1, r2 to be the roots of the previous quadratic and calculate the tangents at the points at infinity (asymptotes) of the curve. Their intersection point O (center of the conic) turns out to be (3,1,1). Thus it is on AK, which is described by y-z=0. Besides, from its coordinates follows that OA/OK=2/3.
[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 dAC from AC is dAC = 2QK. It is easily seen that QK=AK/5.
[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 Chasles-Steiner definition of a conic (see Chasles_Steiner.html ). The two bundles of lines are L(B) and L(C) (L(X) denoting all lines through X). The homographic relation between lines of the two bundles is established by intersecting AB, AC with a line XY passing through the fixed point D. The homographic relation corresponds to line BY of L(B) line CX of L(C).
[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 .