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Let O the circumcenter of triangle ABC and G its centroid. Extend OG until a point P such that OG/GP=1/2. We'll prove that P is the orthocenter H.
Draw the median AA' where A' is the midpoint of BC. Triangles OGA and PGA are similar, since GP=2GO, AG=2A'G and angle OGA'=angle PGA. Then angle OA'G =angle PGA and OA' is parallel AP. But OA' is perpendicular perp BC so AP is perpendicular to BC, that is, AP is a height of the triangle.
Repeating the same argument for the other medians proves that P lies on the three heights and therefore it must be the orthocenter H.
The ratio is $OG/GH=1/2$ since we constructed it that way.
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