I found this problems three years ago. But I never have been a proof. Recently I posted in math.stackexchange.com. I am looking for a solution of the following problems:

A chain of six circles associated with a conic.

Let $A_1, ..., A_6$ and $B_1, ..., B_6$ be 12 points lying on a conic, and suppose that for $i=1, ..., 5$ through $A_i, A_{i+1}, B_{i+1}, B_i$ passes a circle $(O_i)$. Then through $A_6, B_6, A_1, B_1$ as well passes a circle $(O_6)$. Let $P_1, P_4$ be intersection points of $(O_1)$ and $(O_4)$; the same for $P_2, P_5$ and $P_3, P_6$. Show that:

  1. Three lines $O_1O_4$, $O_2O_5$, and $O_3O_6$ have a common point $O$.

  2. Six points $P_1, ..., P_6$ lie on a circle with center in $O$.

Drawing to the problem

My remark: With six points and six lines we get the Pascal theorem. With 12 points and six circles we have this problem

  • $\begingroup$ You can see the problem in $\endgroup$ – Oai Thanh Đào Mar 29 '16 at 14:59
  • 4
    $\begingroup$ Your second question follows from the fact that if $ABCD$, $A_1ABB_1$, $B_1BCC_1$, and $C_1CDD_1$ are cyclic quadrilaterals then $A_1B_1C_1D_1$ is a cyclic quadrilateral if and only if $A_1ADD_1$ is a cyclic quadrilateral. This fact can be proved by a straightforward angle chase. $\endgroup$ – zeb Apr 1 '16 at 2:34
  • $\begingroup$ Have you tried solving it by brute force calculation? You can parameterize the conic, pick $A_1,A_2,A_3,A_4,A_5,A_6,B_1$ for generic values of parameters, and proceed to calculate $B_2,...,B_6$, as well as $O_1,...,O_6$. It might be beyond the usual software's capabilities, but it might not be. $\endgroup$ – Lev Borisov Apr 6 '16 at 22:12
  • $\begingroup$ @LevBorisov I checked it by Geogebra many times, But I can not calculate. Please see: Sequences of Concyclic Points on a Conic and Geogebra A chain of six circles associated with a conic $\endgroup$ – Oai Thanh Đào Apr 7 '16 at 1:33
  • $\begingroup$ And if points do not belong to a conic, this fails? $\endgroup$ – Fedor Petrov Apr 7 '16 at 5:54

Since $С_1$, $C_2$, and your conic $\alpha$ pass through $A_2$ and $B_2$ we get that $A_1B_1$ and $A_3B_3$ are parallel (it calls three conic theorem http://mathworld.wolfram.com/ThreeConicsTheorem.html). Applying it several times to your construction you get that $C_6$ exists.

Intersection of $O_1O_4$, $O_2O_5$ and $O_3O_6$ follows from the Pappus theorem applied to the triple parallel lines $O_1O_2$, $O_3O_4$, and $O_5O_6$ and the triple $O_2O_3$, $O_4O_5$, and $O_6O_1$.

Upd. Regarding the second question: @zeb was almost right. We need lemma with little bit different combinatorics: $(abcd)$, $(aba_1b_1)$, $(abc_1d_1)$, $(cda_1b_1)$, $(cdc_1d_1)$ => $(a_1b_1c_1d_1)$.

Lets show that $P_1$, $P_4$, $P_3$, and $P_6$ lie on a circle. For that we note that the following quadruples circumscribed $(A_2B_2A_5B_5)$, $(A_2B_2P_1P_4)$, $(A_2B_2P_3P_6)$, $(A_5B_5P_1P_4)$, $(A_5B_5P_3P_6)$ and then apply the Lemma.

It is clear that center of this circle is $O$ because it is lie on the corresponded perpendicular bisectors of $P_iP_{i+3}$. Therefore all $P_i$ lie on the fixed circle with center at $O$.

  • $\begingroup$ Is your proof complete ? @akopyan $\endgroup$ – Oai Thanh Đào Apr 7 '16 at 10:12
  • $\begingroup$ Sorry, I do not understand, what do you mean. $\endgroup$ – Arseniy Akopyan Apr 7 '16 at 10:37
  • $\begingroup$ I mean: Do you proof $P_1, P_2, P_3, P_4, P_5, P_6$ lie on a circle? $\endgroup$ – Oai Thanh Đào Apr 7 '16 at 11:32
  • $\begingroup$ I thought it was answered by @zeb. But now I do not think his observation solves the problem. So, I do not know how to prove it $\endgroup$ – Arseniy Akopyan Apr 7 '16 at 11:44
  • $\begingroup$ Following your suggestion: one question one topic. So, I removed the second question. So I think @zeb did not answer the question above. $\endgroup$ – Oai Thanh Đào Apr 7 '16 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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