Let $\triangle ABC$ have $AB=6$, $BC=7$, and $CA=8$, and denote by $\omega$ its circumcircle. Let $N$ be a point on $\omega$ such that $AN$ is a diameter of $\omega$. Furthermore, let the tangent to $\omega$ at $A$ intersect $BC$ at $T$, and let the second intersection point of $NT$ with $\omega$ be $X$. The length of $\overline{AX}$ can be written in the form $\frac{m}{\sqrt{n}}$ for positive integers $m$ and $n$, where $n$ is not divisible by the square of any prime. Find $100m+n$.

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