Let $a,b,c$ denote the lengths of $BC, CA, AB$, respectively. Let $M$ be the midpoint of $BC$, $D$ be the foot of the altitude from $A$ to $BC$, $K$ the foot of the altitude from $M$ to the line parallel to $BC$ through $A$, and $H$ the orthocentre. It is obvious that $M, O, K$ are collinear. Moreover, by Euler, $O, G, H$ are collinear with $OH = 3 \cdot OG = 3$. Hence $AK = OH = MD = 3$. Because $ABC$ is acute and $\angle{C}=45$, we have that $CD > CM$ \[ 3 = MD = CD - CM = AC \sin \angle{C} - \frac{1}{2} AB = \frac{b \sqrt{2} - a}{2}. \]We find another equation in terms of $b$ and $a$. Note that $3 = OH = OA^2 - AH^2$. As the circumradius, $OA$ is easily calculable with the law of sines, \[ OA = \frac{c}{2 \sin \angle{C}} = \frac{c \sqrt{2}}{2} .\]On the other hand, triangles $AGH$ and $AMD$ are similar so $\frac{AG}{AM} = \frac{AH}{AD}$. However, it is well known that $\frac{AG}{AM} = \frac{2}{3}$ by homothety. Combined with the fact that $AH = \frac{b\sqrt{2}}{2},$ we obtain \[9 = OH^2 = \frac{c^2}{2} - \frac{2b^2}{9} . \]Using the law of cosines, we can substitute \[ c^2 = a^2 + b^2 - 2ab \sin{C} = a^2 + b^2 - ab \sqrt{2} .\]Thus we have the system of equations \[ \begin{cases} b\sqrt{2} - a = 6 \\ \frac{a^2 + b^2 - ab \sqrt{2}}{2} - \frac{2b^2}{9} = 9 \end{cases} \]Substitute $b = \frac{a + 6}{\sqrt{2}}$ to obtain \[ \frac{a^2 + \left( \frac{a + 6}{\sqrt{2}} \right)^2 + a \left( \frac{a+6}{\sqrt{2}} \right) \sqrt{2} } {2} - \frac{2 \left( \frac{a + 6}{\sqrt{2}} \right)^2}{2} .\]Simplifying gives \[ 5a^2 - 48 a - 144 = (a-12)(5a+12) = 0. \]Clearly $a$ is positive so $a=BC=12$.