Let $(C_1)$, $(C_2)$ be two conics on the same Ellipsoid, (or Hyperboloid, or Paraboloid). Let $A_1, A_2, A_3, A_4$ be four arbitrary points lie on $(C_1)$; $B_1$ be arbitrary point on $(C_2)$. The Plane through $A_i, B_i, A_{i+1}$ meets $(C_2)$ again at $B_{i+1}$ for $i=1, 2, 3, 4$.

Question: $B_5 \equiv B_1$ or $B_5 \ne B_1$?

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    $\begingroup$ Have you check it with geogebra (for sphere, for example)? $\endgroup$ – Fedor Petrov Jul 8 '17 at 12:02
  • $\begingroup$ Yes, I have been checked this result in two days by geogebra. Some case $B_5 \equiv B_1$ some case $B_5 \ne B_1$. But I conjecture that $B_5 \equiv B_1$. I think in case $B_5 \ne B_1$ because restrict of geogebra software dear Dr. @FedorPetrov $\endgroup$ – Cố Gắng Lên Jul 8 '17 at 12:20

Yes, this is true. Applying projective transform you may think that your quadric is a sphere. Then your conics are circles, and you have to prove that $A_1,B_1,A_4,B_4$ are concyclic (or complanar, that is the same). Making inversion in $A_1$ you get a planar problem: $B_1B_2A_2$ and $A_2A_3A_4$ are lines, $B_2A_2A_3B_3$, $B_3A_3B_4A_4$ and $B_1B_2B_3B_4$ are circles, and we need to show that $B_1B_4A_4$ is a line. This is simple angle-chasing: $$\angle (B_1B_4,B_4A_4)=\angle (B_1B_4,B_4B_3)+\angle (B_4B_3,B_4A_4)=\angle (B_1B_2,B_2B_3)+\angle (A_3B_3,A_3A_4)=\angle (B_2A_2,B_2B_3)+\angle (A_3B_3,A_3A_2)=0$$ as desired (I use oriented angles, which are defined modulo $\pi$ and enjoy the property that $\angle (AX,XB)$ is fixed when $X$ runs over a circle passing through $A,B$.)


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