3. In $\triangle A B C$, $A B=A C=5, B C=6$, the orthocenter $H$ of $\triangle A B C$ satisfies $\overrightarrow{A H}=m \overrightarrow{A B}+n \overrightarrow{B C}$. Then $m+n=$ $\qquad$
3. $\frac{21}{32}$.

In $\triangle A B C$, let the altitude from $A$ to $B C$ be $A D$. Draw $H E / / B C$, intersecting $A B$ at point $E$. Then
$$
\begin{array}{l}
A D=4, \\
H D=B D \tan \angle D B H=B D \cot \angle A C B \\
=3 \times \frac{3}{4}=\frac{9}{4} .
\end{array}
$$

Thus, $m=\frac{A E}{A B}=\frac{A H}{A D}=\frac{7}{16}$,
$$
n=\frac{E H}{B C}=\frac{E H}{2 B D}=\frac{7}{32} \text {. }
$$

Therefore, $m+n=\frac{21}{32}$.