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Let $c$ be an arbitrary conic section. Choose $6$ distinct points on $c$. Draw $6$ lines $t_1,t_2,...t_6$ through these points that are tangents to $c$. Denote points $T_1=t_1\cap t_2,T_2=t_2\cap t_3\cdots T_6=t_6\cap t_1$. Then lines $\leftrightarrow T_1T_4$,$\leftrightarrow T_2 T_5$,$\leftrightarrow T_3,T_6$ meet at 1 point.

I would like to get a reasonable explanation on how this can work or a rigorous proof. I couldn't find a starting point for this. What I am familiar with is the proof of the classic Brianchon's theorem for hexagon circumscribed around a circle, using radix axes, it's easy to prove, but I am struggling with proving this for general conic section. Thanks for any tips and hints.

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  • $\begingroup$ There are two topics important to constructing a proof polarity and duality. (On a conic polar and dual are somewhat 'interchangeable') A few things are interesting about the polar/dual of Brianchon's theorem. $\endgroup$ – Patrick Abraham May 3 '18 at 14:33
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If you have demonstrated that the theorem holds for an arbitrary hexagon circumscribed around a circle, including a non-convex one (i.e. with points numered in different order), then you can build the general proof on that.

Any non-degenerate real conic section is equivalent to any other under a projective transformation. A projective transformation preserves incidence, which implies that tangents will remain tangents. Thus there always exists a projective transformation to turn the general case of a conic into the special case of the circle without loss of generality.

For ellipses, you could show the same using affine transformations only. But for parabolas or hyperbolas, you need projective transformations to turn them into circles.

As Patrick Abraham suggested, another way to demonstrate this is using the point-line duality of projective geometry. Take the theorem statement and exchange the terms “point” with “line”. Exchange “point on line” with “line through point”. Exchange “point where two lines intersect” with “line connecting two points”. Exchange “point on conic” with “tangent to conic”. Exchange “collinear” with “concurrent”. Similar for other related formulations. In the end the theorem about six tangents to a conic leading to three concurrent lines becomes a theorem about six points on a conic leading to three collinear points: Pascal's theorem.

Both suggested approaches make use of projective geometry. That's a very natural setup to use when working with arbitrary conic sections, so if you are not familiar with it, I suggest you familiarize yourself since doing so will likely be easier and more generally useful than finding a non-projective proof for this specific scenario. At least in my personal opinion, which might be biased.

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