# Centroid on the Incircle in Right Triangle

### Solution

We know that

$\displaystyle GI^2=\frac{1}{18}(a^2+b^2+c^2-24Rr+12r^2),$

Since $IG=r,\,$ we derive $a^2+b^2+c^2=6r^2+24Rr,\,$ implying that what we need to prove is

$\displaystyle 648Rr\ge 25(6r^2+24Rr),$

or, equivalently, $48Rr\ge 25\cdot 6r^2,\,$ i.e., $8R\ge 25r,\,$ which is a well-known inequality.

### Acknowledgment

The above problem - by Kadir Altintas and Leo Giugiuc - has been originally posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook group. I am grateful to Leo Giugiuc for keeping me abreast via the CutTheKnotMath facebook page. The solution is by Marian Dinca.

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