Problematic Triangle

Geometry Level 4

In $\bigtriangleup ABC$ $D$ is a point on $AB$ such that $\frac {BD}{DA}$ = $\frac {2}{3}$ and $E$ is a point on $BC$ such that $\frac {BE}{EC}$ = $\frac {1}{3}$ . $DC$ and $AE$ intersect at $O$ . Then what is the ratio of $Area$ $of$ $\bigtriangleup EOC$ to $Area$ $of$ $\bigtriangleup ABC$ . The ratio will be of the form $\frac {m}{n}$ enter the answer as $m + n$ .

The answer is 5.

3 solutions

Tasmeem Reza
Jan 31, 2015

Connect $B$ with $O$ .

$\frac{\triangle AEC}{\triangle ABC} = \frac{EC}{BC} = \frac{3}{1+3} = \frac{3}{4}$ $\Rightarrow \triangle AEC = \frac{3}{4} \times \triangle ABC$

Again, $\frac{\triangle EOC}{\triangle BOC} = \frac{EC}{BC} = \frac{3}{1+3}=\frac{3}{4}$ $\Rightarrow \triangle BOC = \frac{4}{3} \times \triangle EOC$

Now, $\frac{\triangle AOC}{\triangle BOC} = \frac{AD}{BD} = \frac{3}{2}$ $\Rightarrow \frac{\triangle AOC}{ \frac{4}{3} \times \triangle EOC} = \frac{3}{2}$ $\Rightarrow \frac{\triangle AOC}{\triangle EOC} = \frac{3}{2} \times \frac{4}{3} = 2$

$\Rightarrow \frac{\triangle AOC}{\triangle EOC} + 1 = \frac{\triangle AEC}{\triangle EOC} = 2+1$

$\Rightarrow \frac{\frac{3}{4} \times \triangle ABC}{\triangle EOC} = 3$

$\therefore \frac{\triangle EOC}{\triangle ABC} = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$

Thus the answer is $1+4=\boxed{5}$ .

Karan Shekhawat
Oct 13, 2014

Simple use of Mass Point concept. See cyclic-Squares videos on you-tube channel for learning this concept.

Can you add more details about how to solve this problem?

Calvin Lin Staff - 6 years, 4 months ago

Niranjan Khanderia
Jul 15, 2018

Problematic Triangle

The answer is 5.

3 solutions

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