Let $ABC$ be a triangle with incenter $I$ and incircle $\omega$. It is given that there exist points $X$ and $Y$ on the circumference of $\omega$ such that $\angle BXC=\angle BYC=90^\circ$. Suppose further that $X$, $I$, and $Y$ are collinear. If $AB=80$ and $AC=97$, compute the length of $BC$.

Let's think step by step and output the final answer within \boxed{}.
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