<s>[INST] 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?
 [/INST] As @DavidQuinn noted, point $O$ is the centroid of the right triangle.
If we place $C$ at $(0,0)$ and $B$ at $(a, 0)$ and $A$ at $(0,b)$, then
$O = \dfrac{1}{3} (a, b )  $
Therefore
$OA^2 + OB^2 + OC^2 = \dfrac{1}{9} \bigg( a^2 + b^2 + 4 a^2 + b^2 + a^2 + 4 b^2 \bigg) = \dfrac{1}{9} ( 6 ) (a^2 + b^2 ) = \dfrac{2}{3} (a^2 + b^2) $
It follows that
$OC^2 = \dfrac{2}{3}(a^2 + b^2) - k^2$
Now, as noted by @LiKwokKeung below, we have that
$OC = \dfrac{1}{3} c $
from which $ a^2 + b^2 = c^2 = 9 OC^2$, hence
$ OC^2 = \dfrac{2}{3} (9 OC^2) - k^2 $
And finally,
$ OC^2 = \dfrac{k^2}{5} $
i.e.
$ OC = \dfrac{k}{\sqrt{5}} $
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