The point P guaranteed by the Three Tangents Theorem is the C-perspector of the triangles ABC and UVW.
Of course, for ΔABC, C is an inconic, and ΔUVW its contact triangle. For ΔUVW C is a circumconic, and ΔABC its tangential triangle.
Observe that the concepts here are projective. If we look at an embedding, we can discuss the centre of a conic as the pole of the line at infinity.
Earlier, we defined duality with respect to a non-degenerate conic. Here we look at a simpler form of duality. This associates the point (pole) p with the line (polar) pTx = 0. Using exactly the same techniques as before, we have
(1) La Hire's Theorem If P is on the polar of Q, then Q is on the polar of P.
(2) The dual of the conic with matrix M is the conic with matrix M*.
Of course, as before, the dual of a point conic (locus) is a line conic (envelope), and vice versa.
In this approach, the dual of the vertices of the basic triangle, i.e. 1:0:0, 0:1:0, 0:0:1, are the lines x = 0, y = 0, z = 0. These are the sidelines of the triangle! It follows that we have
The dual of a circumconic is an inconic and vice versa.
The Three Tangents Theorem can be generalized as
If the reference triangle ABC is not self-polar with respect to the conic C : fx2+gy2+hz2+2pyz+2qzx+2rxy=0, then ΔABC and its polar triangle are in perspective, with perspector [1/(qr-fp),1/(rp-gq),1/(pq-hr)].
If M is the matrix for C, then the vertices of the polar triangle are the poles of the sidelines, and so correspond to the rows of M*. Then the equation for the line joining A to the pole of BC is (pr-gq)y=(pq-hr)z, and similarly for the lines through B and C. Provided the coefficients do not vanish, we get the stated result. If one of the coefficients vanishes, then the perspector is A,B or C. If two vanish, then any point on one sideline serves as perspector. If all three vanish, then M* and hence M is diagonal, so the triangle is self-polar.
(1) Any circumconic has an equation of the form pyz + qzx + rxy = 0, with pqr ≠ 0.
(2) Any inconic has an equation of the form p2x2 + q2y2 + r2z2 - 2qryz - 2rpzx - 2pqxy = 0, with pqr ≠ 0.
TC2(1) follows since a circumconic must pass through 1:0:0, 0:1:0, 0:0:1. Part (2) follows fromTC1.
In each part the condition pqr ≠ 0 is needed to ensure that the conic is non-degenerate.
We refer to the conic in (1) as the circumconic C(P) defined by the point P p:q:r, and that in (2) as the inconic I(P) defined by the point P p:q:r.
(1)The C(P)-perspector is P.
(2) The I(P)-perspector is the isotomic congugate of P.
If we use duality with respect to a conic C, the polar of a point Q on C is the tangent at Q, and the pole of a tangent to C is its point of contact. Thus, if C is a circumconic for the basic triangle, the tangents at the vertices correspond to the rows of its matrix M. Then the vertices of the tangential triangle correspond to the columns of M* (since these are obtained from two rows of M). It is then trivial to check that the lines joining the appropriate vertices meet at the point p:q:r. For an inconic, we see that the vertices of the contact triangle are given directly by the rows of M*, and we find that the lines concur at 1/p:1/q:1/r.
Since we are looking at an embedding on a plane, the centre of the conic C is most easily obtained as the pole of the line at infinity. Working in barycentric coordinates, the line at infinity has vector (1,1,1) so that we have
(1) If P is p:q:r then C(P) has centre p(q+r-p):q(r+p-q):r)p+q-r).
(1) If P is p:q:r then I(P) has centre q+r:r+p:q+p, the complement of P.
The inconic with perspector P = p:q:r has centre p(q+r):q(r+p):r(p+q).
Note that the centre is given here as the complement of the isotomic conjugate of the perspector. It follows that the perspector can be recovered as the isotomic conjugate of the complement of the centre.
Simply by working through the algebra, we can verify that we have
Theorem TC5 (Yiu)
If the C(P) has centre Q, then C(Q) has centre P.
This offers an easy way to construct the circumconic with perspector P. We can construct the circumconic with centre P, and find its perspector Q. Then the required circumconic it that with centre Q.
If C is an inconic or circumconic for which the C-perspector coincides with the centre, then C is the Steiner inellipse or Steiner circumellipse. These are the ellipses
(1) inellipse x2+y2+z2-2yz-2zx-2xy=0,
(2) circumellipse yz+zx+xy=0.
In each case, the centre/perspector is X(2) = 1:1:1, the centroid.
Suppose that C is a circumconic, and that the perspector is P = p:q:r. Using TC4, this coincides with the centre if and only if there is a constant k with p = kp(q+r-p), and similarly for q,r. Then q+r-p=r+p-q=p+q-r, so p=q=r.
For an inconic, we use TC4' and obtain the same result.
The equations follow at once. We can check that the conics always ellipses by showing that neither curve meets the line at infinity x+y+z=0. Note that the inellipse equation may be rewritten (x+y+z)2-4(yz+zx+xy)=0. Thus the problem reduces to showing that the circumellipse does not meet the line at infinity. This leads to the equations x+y+z=0 and x2+y2+xy = 0. The only real solution is x=y=z=0.
(1) A circumconic with perspector P is a parabola if and only if P is on the Steiner inellipse.
(2) An inconic with perspector P is a parabola if and only if P is on the Steiner circumellipse.
We need only check that the centre lies at infinity. This follows by TC3 and TC4(1) for circumconics, and TC3, TC4' for inconics.
Recall that a conic C divides projective plane into three regions:
(1) the points of C, points through which there is one tangent
(2) the interior of C, points through which there are no tangents to C,
(3) the exterior of C, points through which there are two tangents to C.
The perspector of an inconic or circumconic C is in the interior of C
If we apply the Three Points Theorem, we can assume that C is a circle, and that ΔUVW is an equilateral triangle. Then the perspector is the centre, an interior point. Since perspectors and interior are projective concepts, the result follows.
|Brocard inellipse||X(37)||X(6)||Brocard points|
Hyperbola H through A,B,C,X(2),X(7)
As X(2) is on H, the conic has perspector at infinity, so is always a hyperbola.
The centre is X(1146), the persector has barycentric function (b-c)(b+c-a), so is X(522).
Note : the perspector is best found from the equation, using the given points on H.
Circumconic with perspector X(7)
Centre is (b2+c2-3a2+2ab+2ac-2bc)/(b+c-a), not in ETC. This is associated with the incircle.
This is a special case of the more general Theorem TC10 below. This in also includes the case n = 3 of the famous
For n ≥ 3, suppose that there is an n-gon inscribed in a conic C and touching conic I. Then, for any point P on C, there is a unique n-gon with P as a vertex inscribed in C and touching I.
Note that this theorem is only true if we work over the complex numbers. Otherwise, if P is chosen inside I, then we have no polygons touching I with vertex P as there are no tangents from P to I.
For n = 3, we have a triangle inscribed in C and touching I. To simplify the algebra, we shall take the triangle as the reference triangle. Then equations for C and I follow from their perspectors. Theorem TC9 is the case where the perspectors agree.
Suppose that C is the circumconic with perspector p':q':r' and that I is the inconic with perspector p":q":r".
For F = f:g:h on C and in the exterior of I, then there is a unique triangle inscribed in C, touching I with vertex F. The side opposite F touches I at the point F* = p"(g/q'+h/r'):q"(h/r'+f/p'):r"(f/p'+g/q').
It is easy to see that there can be at most one such triangle with vertex at F. The sides of the triangle must touch I, so must be tangents. There are at most two tangents to I through F. Of course, even if two tangents exist, there is no guarantee that the second intersections with C give a side which touches I. This is the main thrust of our
If we take the case where the perspectors agree, the result is rather simpler. Indeed, we can identify the points U and K in terms of the coordinates of F.
Suppose that C and I are the circumconic and inconic with perspector P = p:q:r.
For F = f:g:h on C, the triangle with vertex F inscribed in C and touching I has other vertices
U = pg/q:qh/r:rf/p, K = ph/r:qf/p:rg/q (and F* is the "barycentric sum" of U and K).
Proof of Theorem TC10
For those points of C outside I, we have a triangle. The family of these is the poristic family defined by C and I. If C is entirely outside I, the circumcircle and incircle for example, then the poristic family has one member at each point of C.
Suppose that P is a point in the interior of a conic D.
(A) Let F denote the family of triangles with circumconic D and perspector P.
(1) For U on D, there is a unique member of F containing U.
(2) The members of F have a common inconic, namely that with perspector P.
(3) Any two members of F are triply in perspective, with perspectors on the polar of P with respect to D.
(B) In (A), we may interchange the words circumconic and inconic, and the results remain valid.
(C) If S, T have circumconic C, inconic I with common perspector P, then the perspectors of S and T, and of the contact triangles S', T' are identical.
Here the proof is much shorter than the statement! The results are all defined in projective terms. By the Two Point Theorem, we may assume that D is a circle (in our embedding), and P its centre. By symmetry, a triangle inscribed or circumscribed in D with the centre as perspector must be equilateral. It follows that there is exactly one for each choice of U. The family of equilateral triangles with common circumcircle have common incircle, and vice versa. The remarks about perspective follow from the remark that, if we label two such triangles in the same sense, then the lines joining corresponding vertices are parallel, i.e. meet on the line at infinity, the polar of the centre.
As we have observed, the projective aspects of these families are preserved by projective transformations, so may be calculated from an equilateral triangle and its circumcircle and incircle. Suppose we scale our figure has a circumcircle of radius 2, so the incircle has radius 1. Suppose that the equilateral triangle has vertex U, and that V is the second intersection of UP and the circumcircle, and K,L the intersections of UP with the incircle. If K is the intersection further from U, then it is easy to see that K is the contact point of the incircle and the side opposite U. Then we have |UL|=|LP|=|PK|=|KV|(=1). Then the cross-ratios (U,V,K,P)=3, (U,P,K,L)=-3. Further, as the contact triangle is also equilateral, the side opposite K cuts UP in M with |PM|=|ML|(=1/2). Thus (K,L,M,P)=3. Since cross-ratios are invariant, these values hold for any pair of circumconic and inconic with perspector P.