Given a conic (c), two triangles ABC, A'B'C' are called conjugate with respect to (c) if the vertices of each are the poles of the sides of the other triangle. There are several facts connected with this relation: [1] The relation is well defined and symmetric. [2] The two triangles ABC, A'B'C' are point-perspective with respect to some point O. [3] The two triangles are axis-perspective with respect to some line (e), which is the polar of O with respect to (c).
Considering the question as a problem of the projective plane (see ProjectivePlane.html ) a conic is represented through an invertible symmetric 3x3 matrix M and the corresponding quadratic equation of the form xtMx=0. For a point z, the corresponding polar line p(x)=0 is given then by p(x) = ztMx = 0. The symmetry of this equation is the reason for the pole-polar reciprocity. In fact, the symmetry wtMv = vtMw=0, means that if the polar pv of v passes through w then also the polar pw of w passes through v. Let now {x1, x2, x3} represent the vertices of ABC and {p1, p2, p3} be the corresponding polars with respect to (c). Let {y1, y2, y3} represent the vertices of A'B'C', y1 being the intersection of p2,p3, y2 being the intersection of p3,p1 etc.. y1 being on p2, its polar passes through x2 (B). y1 being also on p3, its polar passes through x3 (C). Thus the polar of y1 (A') coincides with side BC. Analogous argument is valid for the other points y2, y3, showing that the conjugation relation is well defined and symmetric. To show [2] define the three numbers t3 = p1(x2) = x1tMx2 = x2tMx1 = p2(x1) (by the symmetry of matrix M). Analogously t1 = p2(x3), t2 = p3(x1). Any line through the intersection point A' of p2(x)=0, p3(x)=0 is represented through p2(x)-kp3(x)=0. Such a line passing through A(x1) satisfies p2(x1)-kp3(x1)=0 ==> k=t3/t2, hence the line being given by: t2p2-t3p3 = 0. (I) Analogously we obtain the equations of the other lines (BB' and CC') through equations: t3p3-t1p1 = 0 , (II) t1p1-t2p2 = 0. (III). Obviously last equation is the negative sum of the first two, hence the three lines pass through the same point. The last assertion [3] is a consequence of the previous. In fact, B* being on the polar of B' and the polar of B implies that B* is the pole of BB'. Analogously C*, A* are the poles of CC' and AA' correspondingly. Since AA', BB', CC' pass through O, their poles are contained in the polar of O.
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Conjugate triangles ![[0_0]](ConjugateTriangles_files/ConjugateTriangles_0_0_0.png)
![[0_1]](ConjugateTriangles_files/ConjugateTriangles_0_0_1.png)
![[0_2]](ConjugateTriangles_files/ConjugateTriangles_0_0_2.png)