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# Centroid Problem: Ratio of the areas of two Triangles

Try this beautiful problem from Geometry based on Centroid. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Centroid.

## Centroid Problem: Ratio of the areas of two Triangles

$\triangle ABC$ has centroid $G$.$\triangle ABG$,$triangle BCG$, and $\triangle CAG$ have centroids $G_1$, $G_2$, $G_3$ respectively. The value of $\frac{[G1G2G3]}{[ABC]}$ can BE represented by $\frac{p}{q}$ for positive integers $p$ and $q$.

Find $p+q$ where$[ABCD]$ denotes the area of ABCD.

• $14$
• $10$
• $7$

### Key Concepts

Geometry

Triangle

centroid

Answer: $10$

Question Papers

Pre College Mathematics

## Try with Hints

$\triangle ABC$ has centroid $G$.$\triangle ABG$,$triangle BCG$, and $\triangle CAG$ have centroids $G1$,$G2$,$G3$ respectively.we have to find out value of $\frac{[G1G2G3]}{[ABC]}$ i.e area of $\frac{[G1G2G3]}{[ABC]}$

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 can you find out $GG_1,GG_2,GG_3$ ?

Can you now finish the problem ……….

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$
Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$

can you finish the problem……..

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