Given ΔABC and A'B'C' as its median triangle, A'D, B'E & C'F as 3 concurrent cevians at N, and L, J and K are the respective midpoints of A'N, B'N and C'N, and X, Y and Z are the midpoints of the sides of A'B'C' as shown. If S is constructed as the common midpoint of XL, YJ and ZK, and the intersections of AN, BN and CN with the sides of the median triangle are P, Q and R as shown, and their respective reflections through S are P', Q' and R', and a conic is drawn through any 5 of P, Q, R, P', Q' and R', then the conic cuts through the 6th point, and is inscribed in the median triangle. The conic is also inscribed in the triangle obtained from the median triangle through a half-turn around S. In addition, the center of the conic (S), the centroid of ABC (G) and N (Ceva point) are collinear, and NS = 3 SG.
Spieker Conic and generalization of Nagel line (July 2004)
1) Drag points A, B, C, D or E. Also drag D or E onto the extensions of the sides. What do you notice? Can you drag until the conic becomes a hyperbola or parabola?
2) View and interact with the 'dual' of this result, which is a generalization of the famous nine-point circle (and Euler line) to a nine-point conic (and associated Euler line) at: Nine Point Conic and generalization of Euler Line.
3) For a proof, read my 2006 article A generalization of the Spieker circle and Nagel line in the journal Pythagoras.
4) Here is also an unedited proof communicated to me by John Silvester, King's College, London in 2007: Silvester Spieker Proof.
Michael de Villiers, first created online 9 November 2007, updated 13 June 2017 with WebSketchpad; 2 August 2021.