Spieker Conic and generalization of Nagel lineGiven A'B'C' as the median triangle, A'D, B'E & C'F are 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. (On the left, scroll down to see the measurement 'w' = NS/SG at the bottom). NOTE: If a security pop-up menu appears in your browser, please choose RUN/ALLOW to let the applet run properly. It is completely safe & can be trusted. If you have the very latest Java on your PC or Apple Mac, and experience problems with the applet loading, please go here for additional information on Java settings that should resolve the issue. Drag points A, B, C, D or E. Also drag D or E onto the extensions of the sides. What do you notice? Read my article A generalization of the Spieker circle and Nagel line Michael de Villiers, 9 November 2007, Created with GeoGebra |