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Geometry Difficulty 8.6 Shortlist Prove it Baltic Way

Let $\omega_1$ and $\omega_3$ be two circles, touching externally in a common point $P$ . Let further $\omega_2$ and $\omega_4$ be two circles touching externally in $P$ . Suppose that for $i \in \{1, 2, 3, 4\}$ $\omega_i$ intersect $\omega_{(i \pmod 4)+1}$ again in $A_i$ . Let $\ell_1$ be the common tangent of $\omega_1$ and $\omega_3$ and $\ell_2$ be the common tangent of $\omega_2$ and $\omega_4$ . Show that $A_1, A_2, A_3$ and $A_4$ are con-cyclic if and only if $\ell_1$ and $\ell_2$ are orthogonal.

Solution

Solution. In figure 15, we have
$\angle CNK = \angle ANK = \angle AOK = 2\angle ABK = 2\angle NBK.$
Hence triangle $BNK$ is isosceles, so $NK = NB = NC$ , and therefore $\angle BKC$ is right.
If we let $KC$ intersect $\omega$ at $K'$ , we see that $K', O$ and $B$ are collinear, as $\angle BKK'$ is right.
Let us note that $C$ lies on $OX$ . Indeed,
$\mathrm{Pow}(C, AONKX) = AC \cdot CN = \frac{1}{2} AC \cdot CB = CM \cdot CB = \mathrm{Pow}(C, BOMLX).$
Now,
$\angle LXC = \angle LXO = \angle LBO = \angle LBK' = \angle LKK' = \angle LKC$
Showing that $XLCK$ is cyclic, as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.