# Two Triangles In A Conic

Here is a proof of the theorem that was suggested by Vladimir Nikolin

Two triangles ABC and DEF intersect at points H = AB \cap FD,\, I = FD \cap BC,\, J = BC \cap DE,\, K = DE \cap CA,\, L = CA \cap EF,\, G = EF \cap AB. It will be shown that ABCDEF lie on a conic if and only if the lines HK,\, IL and JG are concurrent.

The proof is based on the properties of involutions. If ABCDLG are considered as fixed points, the pencil of conics through ABCD intersects the line LG in pairs of points in involution. The theorem will be proved if it can be shown that this involution can be replicated using the intersections of the lines. The three degenerate conics in the pencil are the pairs of lines (BC,\,AD) , (AB,\,CD) and (BD,\,AC) and their intersections with LG give three pairs of the involution. It will then be sufficient to find a projectivity from the line LG to itself that agrees with these three pairs.

The usual way of defining a projectivity on the line would be to find a construction taking a point E on the line to its corresponding point F. This doesn't seem to be easily done so we seek the relation between E and F that is given by the concurrency of the lines. For points E and F on LG , let ED meet BC and CA at J and K respectively and let FD meet BC and AB at I and H respectively. The claim is that, if E and F are distinct and the lines HK,\, IL and JG are concurrent, then E and F are pairs of the involution.

Diagram

The applet allows the points E and F to be slid independently along the line GL. It is instructive to move one of them keeping the other fixed to see how the other points and lines move. Slide them towards a pair that is given by one of the degenerate conics and observe what happens. For instance, if E is on AC and F is on BD, then the lines HK and IL coincide so the concurrency occurs. Let F approach E and note that concurrency always occurs since FD = ED. Note also that there seems to be just one other point F for concurrency for each E. This suggests that, if we express GL in parametric form, the relation between the parameters for E and F should be quadratic in each.

Assuming homogeneous coordinates, let a point be represented by an ordered triple \textbf u\,=(x,y,z). The points E and F can be described in the parametric form \textbf u + s\textbf v and \textbf u + t\textbf v respectively. The points K and J will take the form \textbf k + s\textbf k' and \textbf j + s\textbf j' and the points H and I will take the form \textbf h + t\textbf h' and \textbf i + t\textbf i'. Coefficients for the equations of lines can also be written as triples so the line GJ will depend only on s and be described by \textbf p + s\textbf p' forming a pencil of lines as s varies and similarly the line LI will be described by \textbf q + t\textbf q'. Since the coefficients for the line HK are determinants of order 2 in the coordinates of the points, they will take the form \textbf r + s\textbf r' + t\textbf r'' + st\textbf r''' . The condition that the lines are concurrent is then that the determinant that has these triples as rows is zero. It would be very complicated to calculate this determinant from the given fixed points but it will be a polynomial of the form

a +bs +ct+dst+es^{2}t +fst^{2} +gs^{2}t^{2}

However, F=E satisfies the concurrency condition for all E and therefore

a +(b+c)s+ds^{2}+(e+f)s^{3}+gs^{4}

is zero for all values of s implying that

a=0,\,c=-b,\,d=0,\,f=-e,\,g=0
.

The condition becomes b(s-t)+est(s-t)=0 so t=b/es, an involution.