Geometric inequality with areas

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Source: Czech and Slovak Olympiad 2015, National Round, Problem 5

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Radar

#1 h Apr 1, 2015, 1:33 PM • 2 Y

Y by Adventure10, Mango247

In given triangle $\triangle ABC$ , difference between sizes of each pair of sides is at least $d>0$ . Let $G$ and $I$ be the centroid and incenter of $\triangle ABC$ and $r$ be its inradius. Show that $[AIG]+[BIG]+[CIG]\ge\frac{2}{3}dr,$ where $[XYZ]$ is (nonnegative) area of triangle $\triangle XYZ$ .

This post has been edited 1 time. Last edited by Radar, Apr 1, 2015, 1:40 PM

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#2 h Apr 1, 2015, 3:13 PM • 2 Y

Y by Adventure10, Mango247

This problem is trivial with barycentric coordinates.

Let $\triangle{ABC}$ be our reference triangle, so that $A = (1, 0, 0)$ and $B = (0, 1, 0)$ and $C = (0, 0, 1).$ Also let $a = BC, b = CA,$ and $c = AB.$ It is well-known that $G = (1 : 1 : 1)$ and $I = (a : b : c)$ so we find that $[AIG] = \frac{[ABC]}{3(a + b + c)}\begin{vmatrix} 1 & 0 & 0 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = \frac{[ABC]|b - c|}{3(a + b + c)}.$ Finding similar equations for $[BIG]$ and $[CIG]$ we find that $[AIG] + [BIG] + [CIG] = \frac{[ABC](|b - c| + |c - a| + |a - b|)}{3(a + b + c)} = \frac{r(|b - c| + |c - a| + |a - b|)}{6}.$ Now assume WLOG that $a \le b \le c.$ Since $b - a \ge d$ and $c - b \ge d$ we have that $c - a \ge 2d$ which implies that $\frac{r(|b - c| + |c - a| + |a - b|)}{6} \ge \frac{2dr}{3}$ as desired.

This post has been edited 1 time. Last edited by Wolstenholme, Apr 1, 2015, 3:14 PM

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#3 h Oct 9, 2018, 2:12 PM • 2 Y

Y by Adventure10, Mango247

Equality iff $a-b=b-c$ where sides are named such that $a\ge b\ge c$

This post has been edited 1 time. Last edited by WolfusA, Oct 9, 2018, 2:12 PM

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