Ptolemy's theorem says:

A quadrilateral can be inscribed in a circle if and only if the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

A very interesting fact is that its corollaries are Pythagoras' theorem and the law of cosines.

Is there a generalization of this theorem for conics? And would such theorem have more interesting corollaries?

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    $\begingroup$ A conic is determined by five points, so any quadrilateral is actually inscribed in an infinite family of conics. (See this answer.) And, of course, any pentagon is inscribed in a single conic. A Ptolemy-like result would probably want to provide a condition for which a hexagon is inscribed in some conic. (Alternatively, if we specify a positive $e$, we could ask when a pentagon is inscribed in a conic with eccentricity $e$.) It's not clear what form the condition would take; there are a lot of distances to consider. $\endgroup$ – Blue Oct 11 '16 at 20:41

Too long for a comment ...

A Ptolemy-like Theorem for conic-inscriptable hexagons "can" be obtained from a certain polynomial system.

Let the hexagon have vertices $P_i := (x_i, y_i)$ for $i = 0, ..., 5$ (with index arithmetic done modulo 6), and define $d_{ij} := |\overline{P_i P_j}|$. For convenience, we take $P_0 = (0,0)$ and $P_1 = (d_{01}, 0)$. We can write five equations of the form $$d_{ij}^2 = (x_i - x_j)^2 + (y_i-y_j)^2$$ for the distances between consecutive vertices (only five, because we've already accounted for $d_{01}$), and three for the distances of the "long" diagonals $d_{i,i+3}$. One more equation encodes the condition that the $P_i$ lie on a conic: $$\left| \begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 1 \\ d_{01}^2 & 0 & 0 & d_{01} & 0 & 1 \\ x_2^2 & y_2^2 & x_2 y_2 & x_2 & y_2 & 1 \\ x_3^2 & y_3^2 & x_3 y_3 & x_3 & y_3 & 1 \\ x_4^2 & y_4^2 & x_4 y_4 & x_4 & y_4 & 1 \\ x_5^2 & y_5^2 & x_5 y_5 & x_5 & y_5 & 1 \end{array}\right| = 0$$

That's nine equations in eight unknown $x$- and $y$-values. Theoretically, we can eliminate those unknowns to determine an equation involving only the distances, $d_{ij}$. Unfortunately, a brute force elimination process, using, say, the method of resultants, quickly generates intermediate polynomials with tens of thousands of terms that choke my copy of Mathematica, so I can't tell whether anything resembling a sane relation ultimately emerges. Perhaps a more-clever approach would succeed.

I've also tried exploiting the Ceva-like property of Pascal's Hexagon Theorem, to no avail (so far).


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