Romanian Mathematical Olympiad

Denote $M$ the midpoint of the edge $[BC]$. Since triangle $VBC$ is isosceles with $VB = VC$, the points $V$, $H$, $I$ and $M$ are collinear.

From $AC \perp OB$ and $AC \perp OV$ follows $AC \perp (BOV)$, hence $VB \perp AC$, which, together with $VB \perp CH$ leads to $VB \perp (ACH)$, whence $AH \perp VB$. In the same way $AH \perp VC$, therefore $AH \perp (VBC)$, so $AH \perp VM$.

Denote $I'$ the projection of $O$ onto the plane $(VBC)$; then $I'$ belongs to $VM$ and $OI' \parallel AH$. This leads to

$$\frac{OI'}{AH} = \frac{MO}{MA} = \frac{1}{3},$$

hence $AH = 3OI'$. This shows that $OI$ equals the distance from $O$ to $(VBC)$, therefore $I = I'$.

From $OI \perp (VBC)$ and $I$ = incenter of the triangle $VBC$ follows that $O$ has equal distances to the lines $VB$ and $BC$.

Denote $J$ the projection of $O$ onto the line $VB$. Then $OJ = OM$, hence the right triangles $OJB$ and $OMB$ are congruent.

This leads to $m(\angle (VB, (ABC))) = m(\angle VBO) = m(\angle MBO) = 30^\circ$.