Unless I'm mistaken, you can just mimic this

An interesting geometry problem, in an even more interesting place on the ELMO shortlist... By Pascal's Theorem on $A_1A_2\ldots A_6$, we have $B_1, B_2, A_1A_6 \cap A_3A_4$ are collinear. By the converse of Pascal's Theorem on $B_1B_6B_3B_8B_5B_2$ this implies that the six points $B_i$ $(i\neq4,7)$ lie on a conic. A suitable cyclic permutation of indices, combined with the fact that five points determine a unique conic, solves the problem.

To make Catalyst's argument more explicit: Let $\omega$ denote the circumcircle of cyclic octagon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$. Let $f$ be the quartic that vanishes on the lines $A_1 A_2$, $A_3 A_4$, $A_5 A_6$, and $A_7 A_8$, and let $g$ be the quartic that vanishes on the lines $A_2 A_3$, $A_4 A_5$, $A_6 A_7$, and $A_8 A_1$. Let $P$ be a point on $\omega$ other than one of the $A_i$, and let $t$ be a real number such that $h = f + tg$ vanishes at $P$. Let $\gamma$ be the quartic curve defined by $h = 0$. Then $\gamma$ has 9 points in common with $\omega$ (a conic), namely the eight $A_i$ and $P$. But by Bezout's theorem, a quartic and a conic can have at most $4 \cdot 2 = 8$ points in common, unless the conic is actually a factor of the quartic. The quotient of $\gamma$ and $\omega$ is another conic, say $\omega'$. Note that at each $B_i$, $f = g = 0$, so $h = 0$. Hence, each $B_i$ lies on $\gamma$, which means each $B_i$ lies on $\omega'$.

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.

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 $B_2B_5B_8B_3B_6B_1$,it follows that these points are conconic if and only if the lines $\overline{A_1A_7}, \overline{B_8A_3A_4B_3}, \overline{B_1B_2}$ are collinear. This follows by Pascal's on $A_1A_2A_3A_4A_8A_7$. Repeating this cyclically gives the result.

Let $\lambda=(A_6,A_4;A_2,A_8)_\omega=(A_2,A_8;A_6,A_4)_\omega$. Let $A_2A_3\cap A_4A_5=X$ and $A_2A_3\cap A_5A_8=Y$. Pascal's on $A_1A_2A_3A_4A_5A_8$ gives that $Y$ lies on $B_1B_8$. Now $B_1(B_2,B_4;B_6,B_8)=(B_2,X;A_2,Y)\stackrel{A_5}=(A_6,A_4;A_2,A_8)=\lambda$. Similarly $B_3(B_2,B_4;B_6,B_8)=B_5(B_2,B_4;B_6,B_8)=B_7(B_2,B_4;B_6,B_8)=\lambda$ as well, finishing.

We will do some preliminary stuff before. $\bullet$ Apply Pascal's on $A_1A_2A_3A_4A_5A_6$ to get that $\overline{B_1B_2}$, $\overline{B_3B_8}$, $\overline{A_1A_6}$ concur. $\bullet$ Apply converse of Pascal's on $B_1B_2B_5B_8B_3B_6$ to get its coconic because of the previous point. See that if we shift the indices by $3$ then we get that $B_4B_5B_8B_3B_6B_1$ is coconic and see that $B_1B_2B_4B_6B_7$ is common between them and hence $B_1B_2B_3B_4B_5B_6B_8$ is coconic. Finally shift the index by $1$ now to see that $B_1B_2B_3B_4B_5B_6$ is common between them and all the $B_i$'s are convered between them and hence the whole thing is coconic. Remark: My favourite algebra problem of all time.

Let $\mathcal C$ be the circle. Define $l_i$ to be the linear polynomial which vanishes on $A_iA_{i+1}$. Consider $$f = l_1 l_3 l_5 l_7, \qquad g = l_2 l_4 l_6 l_8.$$ Notice that the $B_i$ vanish on both $f$ and $g$. Now, notice that there exists a point $P$ on $\mathcal C$ 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 $\mathcal C$ it must factor as a conic by some other degree 2 curve. Notably $\mathcal C$ contains none of the $B_i$, so the $B_i$ lie on a conic, as desired.

What an interesting algebra problem. This is just Pascal's spam. Note that by Pascal's Theorem on hexagon $A_1A_2A_3A_4A_5A_6$ we have that points $B_1$ , $B_2$ and $X=\overline{A_1A_6} \cap \overline{A_3A_4}$ are collinear. Now, note that this implies that $B_1B_6 \cap B_8B_5 = A_1$ , $B_6B_3 \cap B_5B_2 = A_6$ and $B_3B_8 \cap B_1B_2 = X$ are collinear, so hexagon $B_1B_6B_3B_8B_5B_2$ is coconic. Similarly one can show that $B_3B_4B_5B_6B_8B_1$ is coconic and $B_6B_7B_8B_1B_3B_4$ is coconic, which since a conic is determined uniquely by 5 points implies that all points $B_i$ for $1\le i \le 8$ lie on a conic $\mathcal{C}$ as desired.