A right triangle and a point inside that divides it into three equal areas $ABC$ is a right-angled triangle ($\measuredangle ACB=90^\circ$). Point $O$ is inside the triangle such that $S_{ABO}=S_{BOC}=S_{AOC}$. If $AO^2+BO^2=k^2,k>0$, find $CO$.

The most intuitive thing is to note that $AO^2+BO^2=k^2$ is part of the cosine rule for triangle $AOB$ and the side $AB:$ $$AB^2=c^2=AO^2+BO^2-2AO.BO\cos\measuredangle AOB\\ =k^2-2AO.BO\cos\measuredangle AOB$$ From here if we can tell what $2AO.BO\cos\measuredangle AOB$ is in terms of $k$, we have found the hypotenuse of the triangle (with the given parameter). I wasn't able to figure out how this can done.
Something else that came into my mind: does the equality of the areas mean that $O$ is the centroid of the triangle? If so, can some solve the problem without using that fact?

Suppose $G$ is the centroid of triangle $ABC$, and produce $CG$ to meet $AB$ at its midpoint $M$: the area of triangle $CMB$ is $1/2$ the area of triangle $ABC$, because base $MB$ is $1/2$ of base $AB$ and the altitude is the same. For the same reason the area of triangle $CGB$ is $2/3$ the area of triangle $CMB$, because $CM=2/3\,CB$. Hence:
$$
Area_{CGB}={2\over3}Area_{CMB}={2\over3}\cdot{1\over2}Area_{ABC}
={1\over3}Area_{ABC}
$$
and the same reasoning can be repeated for triangles $AGB$ and $CGA$. It follows that point $O$ in your problem is the centroid of $ABC$ and its distance from every vertex is $2/3$ of the corresponding median.
By Pythagoras theorem we then have:
$$
\left({3\over2}OB\right)^2=\left({1\over2}AC\right)^2+BC^2\\
\left({3\over2}OA\right)^2=\left({1\over2}BC\right)^2+AC^2\\
$$
Adding these equalities we thus get:
$$
{9\over4}(OA^2+OB^2)={5\over4}(AC^2+BC^2)
$$
and finally:
$$
k^2=OA^2+OB^2=
{5\over9}(AC^2+BC^2)={5\over9}AB^2
={5\over9}\left({3}OC\right)^2={5}OC^2,
$$
where I also used that hypotenuse $AB$ is twice the corresponding median and thus $3OC$.
