Question

In a triangle $ABC$, let $G$ denote its centroid and let $M,N$ be points in the interiors of the segments $AB,AC$, respectively, such that $M,G,N$ are collinear. If $r$ denotes the ratio of the area of triangle $AMN$ to the area of $ABC$ then

• A. $r=1/2$
• B. $r>1/2$
• C. $4/9\le r<1/2$
• D. $4/9<r$

Answer 1

The correct option is C $4/9\le r<1/2$

$\underset{AB}{\to }=\underset{b}{\to },\underset{AC}{\to }=\underset{c}{\to }\underset{AM}{\to }=\lambda \left(\underset{b}{\to }\right)\underset{AN}{\to }=\mu \left(\underset{c}{\to }\right)$

Let $G$ divides $MN$ in the ratio $K:1$

So, $\frac{K\mu \left(\underset{c}{\to }\right)+\lambda \left(\underset{b}{\to }\right)}{K+1}=\frac{\underset{b}{\to }+\underset{c}{\to }}{3}$

$\Rightarrow \frac{K\mu }{K+1}=\frac{1}{3}\frac{\lambda }{K+1}=\frac{1}{3}K=\frac{\lambda }{\mu }\frac{1}{\lambda }+\frac{1}{\mu }=\frac{1}{3}A.M\ge G.M\frac{\frac{1}{\lambda }+\frac{1}{\mu }}{2}\ge \sqrt{\frac{1}{\lambda }.\frac{1}{\mu }}\Rightarrow \sqrt{\lambda \mu }\ge \frac{2}{3}$

Now, $\frac{Area△AMN}{Area△ABC}=\frac{\frac{1}{2}\lambda \mu |\left(\underset{b}{\to }\right)\times \left(\underset{c}{\to }\right)|}{\frac{1}{2}|\left(\underset{b}{\to }\right)\times \left(\underset{c}{\to }\right)|}$

$=\lambda \mu$

using $\frac{1}{\lambda }+\frac{1}{\mu }=\frac{1}{3}$

Ratio $=\frac{3{\lambda }^2}{3\lambda -1}$

which has a maximum value $\frac{1}{2}$

So option $C$ is correct