Let A1A2A3A4A5A6A7A8 be a cyclic octagon. Let Bi by the intersection of AiAi+1 and Ai+3Ai+4. (Take A9=A1, A10=A2, etc.) Prove that B1,B2,…,B8 lie on a conic. David Yang.
Problem
Source: ELMO Shortlist 2012, A10
Tags: geometry, conics, circumcircle, projective geometry, algebra proposed, algebra
04.07.2012 07:10
Unless I'm mistaken, you can just mimic this
04.07.2012 13:41
An interesting geometry problem, in an even more interesting place on the ELMO shortlist... By Pascal's Theorem on A1A2…A6, we have B1,B2,A1A6∩A3A4 are collinear. By the converse of Pascal's Theorem on B1B6B3B8B5B2 this implies that the six points Bi (i≠4,7) lie on a conic. A suitable cyclic permutation of indices, combined with the fact that five points determine a unique conic, solves the problem.
04.07.2012 18:45
To make Catalyst's argument more explicit: Let ω denote the circumcircle of cyclic octagon A1A2A3A4A5A6A7A8. Let f be the quartic that vanishes on the lines A1A2, A3A4, A5A6, and A7A8, and let g be the quartic that vanishes on the lines A2A3, A4A5, A6A7, and A8A1. Let P be a point on ω other than one of the Ai, and let t be a real number such that h=f+tg vanishes at P. Let γ be the quartic curve defined by h=0. Then γ has 9 points in common with ω (a conic), namely the eight Ai and P. But by Bezout's theorem, a quartic and a conic can have at most 4⋅2=8 points in common, unless the conic is actually a factor of the quartic. The quotient of γ and ω is another conic, say ω′. Note that at each Bi, f=g=0, so h=0. Hence, each Bi lies on γ, which means each Bi lies on ω′.
05.07.2012 07:01
On a sidenote: This problem leads to a sort of approach for some geometry problems, where you use this problem and the converse of pascal (and some projective geometry sometimes) to prove certain things are conics, and then you can pascal those conics.
26.11.2023 07:13
I think this is just Gauss's Octagrammum Mysticum (which afaik predates ELMO). Also, why is this in the A shortlist? By Converse Pascal on B2B5B8B3B6B1,it follows that these points are conconic if and only if the lines ¯A1A7,¯B8A3A4B3,¯B1B2 are collinear. This follows by Pascal's on A1A2A3A4A8A7. Repeating this cyclically gives the result.
21.09.2024 05:30
Let λ=(A6,A4;A2,A8)ω=(A2,A8;A6,A4)ω. Let A2A3∩A4A5=X and A2A3∩A5A8=Y. Pascal's on A1A2A3A4A5A8 gives that Y lies on B1B8. Now B1(B2,B4;B6,B8)=(B2,X;A2,Y)A5=(A6,A4;A2,A8)=λ. Similarly B3(B2,B4;B6,B8)=B5(B2,B4;B6,B8)=B7(B2,B4;B6,B8)=λ as well, finishing.
06.11.2024 22:17
We will do some preliminary stuff before. ∙ Apply Pascal's on A1A2A3A4A5A6 to get that ¯B1B2, ¯B3B8, ¯A1A6 concur. ∙ Apply converse of Pascal's on B1B2B5B8B3B6 to get its coconic because of the previous point. See that if we shift the indices by 3 then we get that B4B5B8B3B6B1 is coconic and see that B1B2B4B6B7 is common between them and hence B1B2B3B4B5B6B8 is coconic. Finally shift the index by 1 now to see that B1B2B3B4B5B6 is common between them and all the Bi's are convered between them and hence the whole thing is coconic. Remark: My favourite algebra problem of all time.
06.11.2024 22:39
Let C be the circle. Define li to be the linear polynomial which vanishes on AiAi+1. Consider f=l1l3l5l7,g=l2l4l6l8.Notice that the Bi vanish on both f and g. Now, notice that there exists a point P on C for which f(P) is nonzero., as otherwise by Bezout's theorem f factors as a conic and some other degree 2 curve, impossible. Consider the nonzero polynomial f(P)g−g(P)f.Clearly this is degree at most 4. However by Bezouts, since vanishes on 9 points of C it must factor as a conic by some other degree 2 curve. Notably C contains none of the Bi, so the Bi lie on a conic, as desired.
10.12.2024 11:05
What an interesting algebra problem. This is just Pascal's spam. Note that by Pascal's Theorem on hexagon A1A2A3A4A5A6 we have that points B1 , B2 and X=¯A1A6∩¯A3A4 are collinear. Now, note that this implies that B1B6∩B8B5=A1 , B6B3∩B5B2=A6 and B3B8∩B1B2=X are collinear, so hexagon B1B6B3B8B5B2 is coconic. Similarly one can show that B3B4B5B6B8B1 is coconic and B6B7B8B1B3B4 is coconic, which since a conic is determined uniquely by 5 points implies that all points Bi for 1≤i≤8 lie on a conic C as desired.