centroid of triangle

[Crazy Gamer]

Triangle ABC has centroid G.Triangles ABG,BCG, AND CAG have centroids G1,G2,G3 respectively.the value of [G1G2G3]/[ABC] CAN BE represented by p/q for positive integers p,q.

find p+q
where[ABCD] denotes the area of ABCD

[KOUSHIK SOM]

Let D, E, F be the midpoints of BC, CA, AB respectively.
Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G1F$
Similarly,$GG_2 = 2G2D$ and$GG_3 = 2G3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$
Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=9+1=10

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