ConicsMaclaurin2

1. Maclaurin's generation of conics

Consider a triangle ABC and a point D not lying on its side-lines. Draw through D an arbitrary line intersecting sides AB, AC at points X, Y respectively. The locus of the intersection point P of lines BY, AX is a conic with the following properties:
[1] The conic passes through the vertices of the triangle.
[2] Its perspector is the harmonic conjugate S of D with respect to {A,K}, K being the intersection of AD with BC.
[3] The tangents to the conic at {B,C} pass through D. Line BC is the polar of D with respect to the conic.
[4] The tangents to the conic at A and Q, Q being the intersection of AD with the conic, meet at a point on BC.
[5] The tangent to the conic at P and line XY meet at a point on the tangent to the conic at A.

The results can be proved directly. They are also consequences of the discussion of the specific example, consisting of the equilateral and its centroid D. For the general triangle and arbitrary location of D one can define the projectivity mapping the vertices of the equilateral to the vertices of the triangle and the centroid of the equilateral to D. All the properties stated above correspond to properties shown in this particular example (see ConicsMaclaurin.html ) and transfer to ABC through the projectivity.

Remark-1 Note that fixing point S and taking as D its harmonic associates {S_A, S_B, S_C} (see TriangleConics.html ) creates the same conic, since in all these cases the perspector is S.
Remark-2 There is an inverse procedure to this conic construction in which we start from the conic and an arbitrary triangle ABC inscribed in this conic. Then for every point P on the conic we draw lines {PB, PC} and define their intersection points {Y,X} with lines {AC, AB} correspondingly. All lines XY pass then through a fixed point D, which is the harmonic conjugate D of the perspector S of the conic with respect to to {A,K}. Point D is also the polar of line BC with respect to the conic. A consequence of this is discussed in ConicsMaclaurin3.html .
Remark-3 This way to construct the conic, through intersections of line-pairs of the line bundles L(B), L(C) (L(X) denoting the set of lines passing through X) could be considered as a canonical way to generate the conic through the Chasles-Steiner recipe (therefore the word canonical). The homography between the two line-bundles needed for such a generation is defined through point D, by mapping CX to BY. Using line coordinates for AB, AC the relation of Y(y) to X(x) is described by a homography i.e. a relation of the form y = (ax+b)/(cx+d).

2. The case of parallels

With the previous settings and the case where point [D] is at infinity, the sides of triangle ABC are intersected by parallel lines (passing through [D]) intersecting sides AB, AC at points X, Y respectively. The locus of the intersection point P of lines BY, AX is again a conic with the following properties:
[1] The conic passes through the vertices of the triangle.
[2] Its perspector is the middle S of AK, K being the intersection of the parallel to [D] from A with BC.
[3] The tangents to the conic at {B,C} are parallel to [D] (pass through [D]). Hence the conic is either degenerate or central (no parabolas).
[4] Line BC is the polar of [D] with respect to the conic. This means that BC is a diameter of the conic, the middle M of BC being the center of the conic. Hence the conic passes also through the symmetric of A with respect to M.
[5] The tangents to the conic at A and Q, Q being the intersection of AD with the conic, meet at a point on BC. In fact BC is the conjugate diameter to the diameter of the conic which is parallel to [D]. Thus, AK meets the conic at Q which is symmetric to A with respect to A.
[6] The tangent to the conic at P and line XY meet at a point on the tangent to the conic at A.
[7] All the conics constructed in this way coincide with five-point-conics (conics passing through five points) the four points being {A,B,C,A'} and the fifth being an arbitrary point Q on the parallel to BC from A'.

Nothing to prove here, since the results are special cases of the previous theorem. A special case of such a conic is the one for which the direction [D] coincides with the antiparallel direction to BC. This is studied in AntiparallelHyperbola.html .

3. Application to quadrangles

Consider the set S of all quadrangles ABCD with given angles at their vertices {A,B,C,D}. There is, up to similarity, exactly one quadrangle in S, such that the diagonal AC bisects the other diagonal BD.

In fact, fix the side BC of the quadrangle and consider the lines BA, CD in the directions of the given angles at B and C correspondingly. The various possible quadrangles having the given angles result by intersecting lines {BA,BD} with lines parallel to a given direction (e). The intersections of the diagonals of these quadrangles move along a conic (c), constructed as in the previous paragraph. The middles of diagonals AC move along a line (f) passing through the middle M of BC, which is also the center of the conic (c). The desired quadrangles will be such that their intersection of diagonals coinicides with one of the two intersection points {H₁, H₂} of the the conic (c) with line (f). Since M is the middle of H₁H₂, only one of the quadrangles will satisfy the condition on angles, whereas the other will have complementary angles to the given.

1. Maclaurin's generation of conics

2. The case of parallels

3. Application to quadrangles

See Also