2
$\begingroup$

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 $B_5 \ne B_1$?

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ Have you check it with geogebra (for sphere, for example)? $\endgroup$ Jul 8, 2017 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 $\endgroup$ Jul 8, 2017 at 12:20

1 Answer 1

4
$\begingroup$

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 $A,B$.)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.