### Hubert Shutrick

Some of the postings on cut-the-knot treat the question whether there is a conic passing through six points two on each side. A simple way to test it is to consider, for each side, the intersection of its extension with line through the two given points on the other two sides that nearest to it. If those three points are collinear, then there is a conic through the six given points by Pascal's criterion applied to one of the hexagons with the points as vertices.

Here I give a more complicated criterion that also relates to the triangle ABC. The method I have used here is also part of a more comprehensive treatment by Bernard Gibert.

**Preliminaries** The criterion will depend on the following facts about involutions on lines.

An involution can be considered as a set of pairs of points on the line that harmonically divide the two fixed points i.e. the two pairs that are the same point counted twice. It is uniquely determined by two of its pairs because the fixed points can be constructed or calculated from them. The fixed points can be imaginary but we only need the case when they are real.

Given two pairs (A,B) and (F,G) in the segment AB, the fixed points can be constructed as follows: find the radical axis of the circles with diameters AB and FG and let O be the point where it intersects the line; let A' and B' on opposite sides of O on the radical axis be such that OA' = OA and OB' = OB; the fixed points are where the circle with diameter A'B' intersect the line.

The conics through four points form a pencil whose members intersect any line that does not pass through one of the points at pairs of points that are in involution. The fixed points are where one of the conics is tangent to the line.

**Lemma**. There is a conic, ellipse in this case, touching BC at D, CA at E and AB at F if and only if AD, BE and CF are concurrent.

This is true because the pencil of conics that are tangent at E and F give an involution on BC such that (B,C) is a pair from the degenerate conic AB.AC and the point D' where EF meets BC is a fixed point because of the degenerate conic EF.EF. The complete quadrilateral BCFE gives that AB harmonically divides DD' if and only if the Cevians are concurrent so that D is a fixed point of the involution.

**Criterion**. Given a pair of points on each of the three segments AB, BC, CA, consider the involution on each side containing the pair and the pair of vertices of the triangle. Construct the fixed point between each pair for these three involutions. There is a conic through the six given points if and only if the Cevians whose feet are the fixed points are concurrent.

**Proof**. If the conic exists, the pencil of conics passing through the pairs on two of the sides includes one that is tangent to the third side at the fixed point of its involution. Therefore, the pencil of conics that pass through one of the pairs and touch this third side gives the involution that includes the other pair, which can then also be replaced by its fixed point. The pencil of conics tangent at these two fixed points give the known involution on the other side so there is conic in that pencil that is tangent at its fixed point. This can only happen if the Cevians are concurrent by the lemma.

**Theorem**. If a triangle has two sets of concurrent Cevians, then their feet lie on a conic.

The proof should be obvious from the following diagram where the triangle has been taken as the frame of reference for homogeneous coordinates with (1,1,1) at one of the intersection points and (s,t,1) representing the other. The points (0,0,1), (1,0,1), (s,0,1), (1,0,0) can be parametrized by x/z, that is 0,1,s,\infty and a simple calculation checks that the fixed points have parameters \pm\surd{s}.

The Z axis can be parametrized with slope y/x.