1
$\begingroup$

In triangle $ABC$,points $E$ and $D$ are on side $AC$ and point $F$ is on side $BC$ such that AE=ED=DC and $BF:FC$ =2:3. $AF$ intersects $BD$ and $BE$ at points $P$ and $Q$, respectively. Find the ratio of the area $EDPQ$ to the area of $ABC$

taken from the 2017 IMC held in India

I assumed ABD was equilateral so that the area of ABE would be equal to BED but didn’t gain anything from that

$\endgroup$
  • $\begingroup$ Are you familiar with using mass points? $\endgroup$ – Anirudh Jul 6 at 18:43
1
$\begingroup$

Let $k\in DC$ such that $FK||BD$.

Thus, $DK:KC=2:3.$

Let $DC=5x$.

Thus, $DK=2x$ and $AD=10x$, which gives $$\frac{AP}{PF}=\frac{AD}{DK}=\frac{10x}{2x}=5.$$ Similarly, let $M\in EC$ such that $FM||BE.$

Thus, $EM:MC=2:3,$ which gives $EM=4x$ and $$\frac{AQ}{QF}=\frac{AE}{EM}=\frac{5x}{4x}=\frac{5}{4}.$$ From here we obtain: $$AQ:QP:PF=10:5:3.$$ Now, $$S_{\Delta BPQ}=\frac{5}{18}S_{\Delta ABF}=\frac{5}{18}\cdot\frac{2}{5}S_{\Delta ABC}=\frac{1}{9}S_{\Delta ABC}=\frac{1}{3}S_{\Delta BED},$$ which gives $$S_{EDPQ}=\frac{2}{3}S_{\Delta BED}=\frac{2}{9}S_{\Delta ABC}$$ and we are done!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.