8,9

In triangle $ABC$, side $AB$ is equal to $\frac{5 \sqrt{2}}{2}$, side $BC$ is equal to $\frac{5 \sqrt{5}}{4}$. Point $M$ lies on side $AB$, point $O$ lies on side $BC$, and lines $MO$ and $AC$ are parallel. Segment $BM$ is 1.5 times longer than segment $AM$. The bisector of angle $BAC$ intersects line $MO$ at point $P$, which lies between points $M$ and $O$, and the radius of the circumcircle of triangle $AMP$ is $\sqrt{2+\sqrt{2}}$. Find side $AC$.
Let $\angle B A C=a, R$ - the radius of the circumcircle of triangle $A M P (R=\sqrt{2+\sqrt{2}})$. Then

$$
A M=2 \cdot \frac{1}{5} A B=\sqrt{2}, \angle M P A=\angle P A C=\angle B A P=\frac{\alpha}{2}
$$

Therefore

$$
M P=A M=\sqrt{2}, M P=2 R \sin \frac{\frac{\alpha}{2}}{}
$$

From this, we find that

$$
\sin \frac{\alpha}{2}=\frac{M P}{2 R}=\frac{\sqrt{2}}{2 \sqrt{2+\sqrt{2}}}
$$

$$
\begin{gathered}
1-\cos a=2 \sin 2 \frac{\alpha}{2}=2\left(\frac{\sqrt{2}}{2 \sqrt{2+\sqrt{2}}}\right) 2=\frac{1}{2+\sqrt{2}} \\
\cos a=1-2 \sin 2 \frac{\alpha}{2}=1-\frac{1}{2+\sqrt{2}}=\frac{1+\sqrt{2}}{\sqrt{2}(\sqrt{2}+1)}=\frac{1}{\sqrt{2}}
\end{gathered}
$$

Therefore, $a=45^{\circ}$. By the cosine theorem

$$
B C 2=A B 2+A C 2-2 A B \cdot A C \cos a
$$

Let $A C=x$. We get the quadratic equation

$$
\frac{125}{16}=\frac{25}{2}+x 2-2 \cdot \frac{5 \sqrt{2}}{2} \cdot x \cdot \frac{1}{\sqrt{2}}
$$

or

$$
x 2-5 x+\frac{75}{16}=0
$$

Its roots are: $x=\frac{5}{4}$ or $x=\frac{15}{4}$. According to the problem condition

$$
M P<O M=\frac{3}{5} A C
$$

Therefore, only the second root fits

## Answer

$\frac{15}{4}$.