A problem involving a tetrahedron Let $ABCD$ a tetrahedron. We know the angle
$$\angle{ACB}=45^\circ$$the sum
$$\overline{AD}+\overline{BC}+\frac{\overline{AC}}{\sqrt2}=90$$and that the volume is $4500.$ We also know (but I don't know if this can be useful for the solution) that $\overline{CD}^2$ is integer: how much is $\overline{CD}^2?$
My problem is that I don't know how to relate the data I have in my possession. In particular how can I use the given sum of those three edges?

 A: This is an expanded version of the comment of Zerox. Let $h$ be the (length of the) height from $D$, the distance between $D$ and the plane of $\Delta ABC$. Then:
$$
\begin{aligned}
30^3 &= 27000
=6\operatorname{Volume}[ABCD]
=h\cdot 2\operatorname{Area}[ABC]
=h\cdot AC\cdot BC\cdot\sin 45^\circ
\\
&=h\cdot \frac{AC}{\sqrt 2}\cdot BC\ .
\\
\text{Then:}
\\
30 &=\left(h\cdot \frac{AC}{\sqrt 2}\cdot BC\right)^{1/3}
\\
&\le\left(AD\cdot \frac{AC}{\sqrt 2}\cdot BC\right)^{1/3}
\\
&\qquad\qquad\qquad\text{with equality iff  $h=AD$, i.e. $DA\perp (ABC)$}
\\
&\le\frac 13\left(AD+ \frac{AC}{\sqrt 2}+ BC\right)
\\
&\qquad\qquad\qquad\text{with equality iff  $AD=\frac{AC}{\sqrt 2}= BC$}
\\
&=\frac 13\cdot 90=30\ .
\end{aligned}
$$
Since the beginning and the end show both the value $30$, we have equalities in between, so we know $DA=30$ is the height from $D$, the segment $DA$ being thus perpendicular on the plane $(ABC)$, and in particular also on $AC$, and $AC/\sqrt 2=30$ and $BC=30$. The triangle $\Delta DAC$ has thus a right angle in $A$, giving
$$
DC^2 = DA^2+AC^2=30^2+(30\sqrt 2)^2=30^2(1+2)=2700\ .
$$
$\square$
