10. (20 points) Given a moving point $P$ on the parabola $y^{2}=4 x$, the focus $F(1,0)$, and $O$ as the origin. Find the maximum value of the inradius $r$ of $\triangle O P F$.

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10. From the parabolic equation, we know the point $F(1,0)$.
Thus, we can set $l_{O P}: y=k_{1} x, l_{F P}: y=k_{2}(x-1)$.
Solving these equations simultaneously, we get the point $P\left(\frac{k_{2}}{k_{2}-k_{1}}, \frac{k_{1} k_{2}}{k_{2}-k_{1}}\right)$.
Since point $P$ lies on the parabola, we have
$k_{1}^{2} k_{2}=4\left(k_{2}-k_{1}\right)$.
Let the center of the incircle of $\triangle O P F$ be $\left(x_{0}, y_{0}\right)$.
By symmetry, assume point $P$ is above the $x$-axis. Thus, $y_{0}>0$ is the radius $r$ of the incircle of $\triangle O P F$, and $y_{0}<x_{0}<1$.
Using the distance from a point to a line formula, we have
$$
\frac{\left|k_{1} x_{0}-y_{0}\right|}{\sqrt{1+k_{1}^{2}}}=\frac{\left|k_{2} x_{0}-y_{0}-k_{2}\right|}{\sqrt{1+k_{2}^{2}}}=y_{0} \text {. }
$$

Solving, we get $k_{1}=\frac{2 x_{0} y_{0}}{x_{0}^{2}-y_{0}^{2}}, k_{2}=\frac{2\left(x_{0}-1\right) y_{0}}{\left(x_{0}-1\right)^{2}-y_{0}^{2}}$.
Substituting into equation (1), we get
$$
\begin{array}{l}
\left(x_{0}-1\right) \frac{x_{0}^{2} y_{0}^{2}}{x_{0}^{2}-y_{0}^{2}}=x_{0}\left(x_{0}-1\right)+y_{0}^{2} \\
\Rightarrow\left(y_{0}^{2}-x_{0}\right)\left(y_{0}^{2}+\left(x_{0}^{3}-x_{0}^{2}\right)\right)=0 \\
\Rightarrow y_{0}^{2}=x_{0} \text { or } y_{0}^{2}=x_{0}^{2}-x_{0}^{3} .
\end{array}
$$

But $y_{0}^{2}=x_{0}$ contradicts $y_{0}<x_{0}<1$.
Thus, $y_{0}^{2}=x_{0}^{2}-x_{0}^{3}=x_{0}^{2}\left(1-x_{0}\right)$
$\leqslant 4\left(\frac{\left(1-x_{0}\right)+\frac{1}{2} x_{0}+\frac{1}{2} x_{0}}{3}\right)^{3}=\frac{4}{27}$.
Therefore, $r=y_{0} \leqslant \frac{2 \sqrt{3}}{9}$.
When the center is $\left(\frac{2}{3}, \frac{2 \sqrt{3}}{9}\right)$, i.e., $P\left(\frac{4}{3}, \frac{4 \sqrt{3}}{3}\right)$, $r$ reaches its maximum value.