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# Understand the problem

Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$. If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively,
prove that$$OG \leq \sqrt{ R ( R - 2r ) } .$$

Balkan MO 1996

Geometry
##### Difficulty Level
Easy
Let $I$ be the incentre. Euler’s theorem says that $OI^2=R(R-2r)$. Hence the result actually proves that $OG\le OI$.

Do you really need a hint? Try it first!

The distance $OG$ is easily computable from standard formulae. For example, one can use complex numbers by assuming that $O=0$ and $|A|=|B|=|C|=R$.
The centroid is given by $\frac{A+B+C}{3}$. Hence $OG^2=\frac{|A+B+C|^2}{9}=\frac{(A+B+C)(\overline{A}+\overline{B}+\overline{C})}{9}=\frac{3R^2+A\overline{B}+A\overline{C}+B\overline{A}+B\overline{C}+C\overline{A}+C\overline{B}}{9}=R^2-\frac{(2R^2-A\overline{B}-B\overline{A})+(2R^2-A\overline{C}-C\overline{A})+(2R^2-B\overline{C}-C\overline{B})}{9}=R^2-\frac{|A-B|^2+|B-C|^2+|C-A|^2}{9}=R^2-\frac{a^2+b^2+c^2}{9}$.
Show that $R(R-2r)=R^2-\frac{abc}{a+b+c}$.
Combining all the hints, the problem reduces to proving that $(a^2+b^2+c^2)(a+b+c)\ge 9abc$. This follows from     $\frac{a^2+b^2+c^2}{3}\ge 3(abc)^{\frac{2}{3}}$ and $\frac{a+b+c}{3}\ge 3(abc)^{\frac{1}{3}}$.