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

## Check the Answer

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 $p+q$=$9+1$=\(10\)

## Other useful links

- https://www.cheenta.com/problem-based-on-lcm-amc-8-2016-problem-20/
- https://www.youtube.com/watch?v=7AlfBAPWEMg