# Six conelliptic points

Can you prove the following proposition:

Proposition. Given an arbitrary triangle $$\triangle ABC$$. Let $$D,E,F$$ be the points on the sides $$AB$$,$$BC$$ and $$AC$$ respectively , such that $$\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}=k$$, where $$k$$ is an arbitrary ratio. Let $$G,H,I$$ be the points on the line segments $$EF$$,$$DF$$ and $$DE$$ respectively such that $$CG \perp EF$$ , $$AH \perp DF$$ and $$BI \perp DE$$. Now let $$M$$ be the point on the extension of the segment $$EF$$ beyond $$E$$ such that $$EM=AH$$. Similarly, define the points $$N,J,K,O,P$$ so that the point $$N$$ lies on the extended segment $$DE$$ and $$EN=AH$$ , the point $$J$$ lies on the extended segment $$DF$$ and $$FJ=BI$$ , the point $$K$$ lies on the extended segment $$EF$$ and $$FK=BI$$ , the point $$O$$ lies on the extended segment $$DE$$ and $$DO=CG$$ and the point $$P$$ lies on the extended segment $$DF$$ and $$DP=CG$$ . I claim that the points $$J,K,M,N,O,P$$ lie on an ellipse.

The GeoGebra applet that demonstrates this proposition can be found here.

• Maybe worth noting that you also get an(other) ellipse if you extend all segments in opposite directions. May 28 at 9:46
• It is also worth asking what does one get for $k=1$ or $k\to\infty$. It is certain circumellipse (seemingly the same in both cases) but not the Steiner ellipse. Seems to be tangent to the external angle bisectors. Must have some name... May 28 at 10:12
• Using help from a colleague I clarified that limiting case. Let $A'B'C'$ be the triangle made from exterior angle bisectors of $ABC$. Then $ABC$ is the orthic triangle for $A'B'C'$ and the limiting ellipse in question is the orthic inconic of $A'B'C'$. May 28 at 18:56

Note that there exists a triangle $$\tilde{F}\tilde{D}\tilde{E}$$ with side lengths $$\tilde{F}\tilde{D}=\sqrt{FD}$$, $$\tilde{E}\tilde{D}=\sqrt{ED}$$, $$\tilde{F}\tilde{E}=\sqrt{FE}$$. Make an affine transform $$\Phi$$ which maps $$F,E,D$$ to $$\tilde{F}$$, $$\tilde{E}$$, $$\tilde{D}$$ respectively. I claim that $$\Phi$$-images of $$J,K,M,N,O,P$$ are concylic. For proving this it suffices to check that each quadrilateral $$KJMP$$, $$POJN$$, $$NMOK$$ maps to a cyclic quadrilateral: their circumcircles either all coincide (that we need), or have concurrent radical axes, but these axes are the sides of $$\triangle \tilde{F}\tilde{D}\tilde{E}$$.

The map $$\Phi$$ divides all lengths on line $$ED$$ by $$\sqrt{ED}$$ and all lengths on line $$EF$$ by $$\sqrt{EF}$$. Thus to prove that $$NMOK$$ is cyclic, we should check $$EN\cdot EO/ED=EM\cdot EK/EF. \quad (\diamondsuit)$$ Since $$EN=EM$$, $$(\diamondsuit)$$ reads as $$EO/ED=EK/EF$$, or $$OD/ED=KF/EF$$, or $$OD\cdot EF=ED\cdot KF$$, or $${\rm Area}(CEF)={\rm Area}(DEB)$$. This follows from $$\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$$.

This is more a comment than a solution, using a routine method—the $$p,q$$-method—to reduce the problem by means of simple calculations to that of determining whether the determinant of a $$6\times 6$$ matrix vanishes.

We can assume that the triangle vertices are $$(0,0),(1,0),(p,q)$$—then the intermediary points are $$D=(\lambda,0)$$, $$E=(1-\lambda+\lambda p,\lambda q)$$ and $$F=(\lambda p,\lambda q)$$ with $$\lambda=\frac 1 k$$.

It is then routine to compute the unit normals to $$DE,EF,FD$$ and so $$|IB|,|CG|,|AH|$$ in terms of $$\lambda,p,q$$. This gives the coordinates of your six points $$J,K,M,N,O,P$$ and one can then use the fact that they lie on a conic if and only if the determinant of the $$6\times 6$$ matrix with rows $$[x^2\ xy\ y^2\ x\ y\ 1]$$ as $$(x,y)$$ ranges over the coordinates of the points vanishes.

Up to this point, the calculations can be done by hand but that of the determinant is probably intractable and can be carried out using Mathematica.

An advantage of this approach is that it is easy to add parameters to explore possible generalisations or converses. Suggestions: replace $$\lambda$$ by three distict parameters to determine under which conditions on them the result is still valid; replace the equality conditions $$|FJ|=|BI|$$, etc., by proportionality ones $$|FG|=\rho|BI|$$ ; replace the perpendicular distance by the lengths of lines which meet the sides of $$DEF$$ at a fixed angle.

• Yes but could you explain why this determinantal condition is always satisfied? May 28 at 9:44
• Well we learned as freshmen that 5 points determine a conic and 6 points $(x_i,y_i)$ lie on one iff the 6 by 6 matrix with rows $[x_I^2\ x_iy _i\ y_i^2\ x_i\ y_i\ 1]$ has vanishing determinant. May 28 at 11:03