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Geometry Difficulty 6.5 National olympiad Prove it

Let A1A2A3A_{1} A_{2} A_{3} be a triangle, and let ω1\omega_{1} be a circle in its plane passing through A1A_{1} and A2A_{2}. Suppose there exists circles ω2,ω3,…,ω7\omega_{2}, \omega_{3}, \ldots, \omega_{7} such that for k=2,3,…,7k=2,3, \ldots, 7, circle ωk\omega_{k} is externally tangent to ωk−1\omega_{k-1} and passes through AkA_{k} and Ak+1A_{k+1} (indices  mod 3)\bmod 3). Prove that ω7=ω1\omega_{7}=\omega_{1}.

Solution

The idea is to keep track of the subtended arc AiAi+1^\widehat{A_{i} A_{i+1}} of ωi\omega_{i} for each ii. To this end, let β=∡A1A2A3,γ=∡A2A3A1\beta=\measuredangle A_{1} A_{2} A_{3}, \gamma=\measuredangle A_{2} A_{3} A_{1} and α=∡A1A2A3\alpha=\measuredangle A_{1} A_{2} A_{3}. ! Initially, we set θ=∡O1A2A1\theta=\measuredangle O_{1} A_{2} A_{1}. Then we compute
∡O1A2A1=θ∡O2A3A2=−β−θ∡O3A1A3=β−γ+θ∡O4A2A1=(γ−β−α)−θ \begin{aligned} & \measuredangle O_{1} A_{2} A_{1}=\theta \\ & \measuredangle O_{2} A_{3} A_{2}=-\beta-\theta \\ & \measuredangle O_{3} A_{1} A_{3}=\beta-\gamma+\theta \\ & \measuredangle O_{4} A_{2} A_{1}=(\gamma-\beta-\alpha)-\theta \end{aligned}
and repeating the same calculation another round gives
∡O7A2A1=k−(k−θ)=θ \measuredangle O_{7} A_{2} A_{1}=k-(k-\theta)=\theta
with k=γ−β−αk=\gamma-\beta-\alpha. This implies O7=O1O_{7}=O_{1}, so ω7=ω1\omega_{7}=\omega_{1}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.