Centroid on the Incircle in Right Triangle

Source

Centroid  on the Incircle in Right Triangle, source

Problem

Centroid  on the Incircle in Right Triangle, problem

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.

 

Related material
Read more...

  • R and r When G on Incircle
  • The Range of r/R in Triangle with Centroid on the Incircle
  • Centroid on the Incircle in Isosceles Triangle
  • Nagel Point on the Incircle in Isosceles Triangle
  • Nagel Point and Centroid on Adjacent Isosceles Triangles
  • |Contact| |Up| |Front page| |Contents| |Geometry| |Store|

    Copyright © 1996-2017 Alexander Bogomolny

     62050674

    Search by google: