Five, (20 points) For the parabola $y^{2}=2 p x(p>0)$ with focus $F$, does there exist an inscribed isosceles right triangle such that one of its legs passes through $F$? If it exists, how many are there? If not, explain the reason.

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As shown in Figure 4, let \( A\left(x_{A}, y_{A}\right) \), \( B\left(x_{B}, y_{B}\right) \), \( C\left(x_{C}, y_{C}\right) \), and the inclination angle of \( AC \) be \( \theta \). Clearly, \( \theta \neq 0 \), \( \frac{\pi}{2} \). Suppose \( \theta \in \left(0, \frac{\pi}{2}\right) \). By the focal chord length formula, we have
\[
\begin{array}{l}
|A C|=\frac{2 p}{\sin ^{2} \theta} . \\
\because\left\{\begin{array}{l}
y_{A}-y_{C}=|A C| \sin \theta=\frac{2 p}{\sin \theta}, \\
y_{A}+y_{C}=2 p \frac{x_{A}-x_{C}}{y_{A}-y_{C}}=2 p \cot \theta,
\end{array}\right.
\end{array}
\]
\[
\left\{\begin{array}{l}
y_{C}-y_{B}=|C B| \cos \theta=|A C| \cos \theta=\frac{2 p \cos \theta}{\sin ^{2} \theta}, \\
y_{C}+y_{B}=2 p \frac{x_{C}-x_{B}}{y_{C}-y_{B}}=-2 p \tan \theta .
\end{array}\right.
\]

From (1), we get \( y_{C}=p \frac{\cos \theta-1}{\sin \theta} \),
From (2), we get \( y_{C}=p \frac{\cos ^{2} \theta-\sin ^{3} \theta}{\sin ^{2} \theta \cos \theta} \),
\(\therefore \cos\) squared minus \(\sin ^{3} \theta\),
which means \(\cos \theta+\sin \theta=\tan \theta\),
or equivalently, \(\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)=\tan \theta\).
The problem reduces to whether (3) has a solution in \(\left(0, \frac{\pi}{2}\right)\). From the graphs of trigonometric functions, (3) has exactly one intersection point in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), indicating that there exists an isosceles right triangle that satisfies the given conditions. By symmetry, when \(\theta \in \left(\frac{\pi}{2}, \pi\right)\), there is also one intersection point.

In summary, there exist exactly two isosceles right triangles that satisfy the given conditions.