Suppose $O$ is circumcenter of triangle $ABC$. Suppose $\frac{S(OAB)+S(OAC)}2=S(OBC)$. Prove that the distance of $O$ (circumcenter) from the radical axis of the circumcircle and the 9-point circle is \[\frac {a^2}{\sqrt{9R^2-(a^2+b^2+c^2)}}\]
I suppose the notation $S(OAB)$ is the same as $S_{\triangle OAB}$ or $[OAB]$
I haven't gave a thought in it, but I believe it can be solved by using the formula $OH^2=9R^2-(a^2+b^2+c^2)$.
$S_{ABC}=3S_{BOC}(1)$let $AO$ intersect $BC$ at $A'$ from $(1)$ we can get$\frac {AO}{OA'}=2$ so Euler line is parallel to $BC$.
in triangle $AOH$ use pythagores theorem and $OH^2=9R^2-(a^2+b^2+c^2)$ formula .
with Casi theorem we can calculate the distancse
the statement is wrong, if $\triangle ABC$ is acute, then the distance should be $\frac {a^2}{4\sqrt{9R^2-(a^2+b^2+c^2)}}$. and I believe that the result will be different when $\triangle ABC$ is obtuse.