Octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle $\Omega$ with center $O$. It is known that $A_1A_2\|A_5A_6$, $A_3A_4\|A_7A_8$ and $A_2A_3\|A_5A_8$. The circle $\omega_{12}$ passes through $A_1$, $A_2$ and touches $A_1A_6$; circle $\omega_{34}$ passes through $A_3$, $A_4$ and touches $A_3A_8$; the circle $\omega_{56}$ passes through $A_5$, $A_6$ and touches $A_5A_2$; the circle $\omega_{78}$ passes through $A_7$, $A_8$ and touches $A_7A_4$. The common external tangent to $\omega_{12}$ and $\omega_{34}$ cross the line passing through ${A_1A_6}\cap{A_3A_8}$ and ${A_5A_2}\cap{A_7A_4}$ at the point $X$. Prove that one of the common tangents to $\omega_{56}$ and $\omega_{78}$ passes through $X$.
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Tags: octagon, circles, radical axis, concurrency, common tangents, geometry
Dis_2020
12.08.2021 18:14
Any ideas? It's clear that $X$ lies on the common radical axis of $\omega_{12}$, $\omega_{78}$ and $\omega_{34}$, $\omega_{56}$, but question is how to prove that tangents from $X$ to $\omega_{12}$ and $\omega_{78}$ coincide. Also it is easy to see that this radical axis passes through ${A_1A_6}\cap{A_3A_8}$ and ${A_5A_2}\cap{A_7A_4}$ and the centers of the circles $\omega_{12}$, $\omega_{34}$, $\omega_{56}$ and $\omega_{78}$ form parallelogram; and we have that points of tangency of the common tangent to $\omega_{34}$, $\omega_{56}$, $A_6$ and $A_7$ are concyclic (so are points of tangency of the common tangent to $\omega_{12}$, $\omega_{78}$, $A_2$ and $A_3$). It is enough to prove that points of tangency of the common tangent to $\omega_{12}$, $\omega_{78}$ and $\omega_{34}$, $\omega_{56}$ are concyclic.
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gvole
05.06.2024 17:50
What a problem. Rename the circles to $\omega_1$, $\omega_2$, and so on. You can find the two isosceles trapezoids whose lateral sides are tangent to $\omega_1$ and $\omega_3$, and the other pair as well. These lateral sides have equal lengths due to the parallel condition, so we conclude that the common tangent segments of $\omega_1$ and $\omega_3$ have equal lengths as for the other pair. Now some more parallelogram action gives that the two points which lie on the desired line (forgot what they're called) have equal powers wrt. $\omega_1$ and $\omega_4$, and likewise for $\omega_2$ and $\omega_3$, so the desired line is the shared radical axis of these two pairs. We also have that $O$ is the midpoint of both $O_1O_3$ and $O_2O_4$, so $O_1O_2O_3O_4$ is a parallelogram. Now since the desired line is the common radical axis you can see that it's sufficient to show that the common tangent segments of $\omega_1$ and $\omega_2$ have the same length as those of $\omega_3$ and $\omega_4$ (just consider the intersection of the common tangents and then the powers/lengths of tangent segments etc.). This is now trivial to show since we can express this condition just using the radii and distances from $O_i$ to the common radical axis. You just abuse the parallelogram condition, equations from PoP, and the equation you get from the other two pairs with equal common tangent segments mentioned above.