HKMO2015-16Q14

by luimichael, Mar 8, 2017, 1:14 PM

Given ABCD is a parallelogram, E is the mid-point of AD and $DF:FC=1:2$. AF and BF are straight lines intersecting EC at G and H respectively.
To find the ratio of areas between ABCD and $\Delta FGH$.

Solution:
Key step: Show that $\frac {KH}{HC}=\frac 1 3 $ and $\frac {KG}{GE}= \frac 2 3 $.
Then $\Delta FKH = \frac 14 \Delta FKC =\frac 14 (\frac 2 3 )^2 \Delta CDE =\frac 14 (\frac 2 3 )^2 \frac 1 4 [ABCD]=\frac 1{36}[ABCD]$ .
Let $GK=\lambda$ and $HK=\mu$. Then $EK=\frac 3 2 \mu + \mu = \frac 5 2 \mu$ and $KC=\lambda + 3 \lambda = 4 \lambda$.
But $\frac {EK}{KC} = \frac 1 2 \implies \frac {5/2 \mu} {4 \lambda}= \frac 1 2 \implies \frac {\mu}{\lambda} = \frac 4 5$
Therefore, $\Delta FGK= \frac 4 5 \Delta FKH = \frac 4 5 \times \frac 1 {36}[ABCD]$.
Hence, $\Delta FGH = \left ( \frac 4 5 \times \frac 1{36}+\frac 1{36} \right ) [ABCD]=\frac 1 {20}[ABCD]$.
Answer: The required ratio is 20:1.
Alternative Solution:
$\frac {FG}{AG}=\frac{FK}{AE}=\frac{FK}{DE}=\frac 2 3$,

$\frac {FH}{HB}=\frac{FK}{BC}=\frac{FK}{2DE}=\frac12* \frac 2 3 = \frac 1 3$.

$\frac{[ABCD]}{[FGH]}=\frac{[ABCD]}{[AFB]}*\frac{[AFB]}{[FGH]}=\frac 2 1 * \frac{AF*FB}{FG*FH}=2*\frac{AF}{GF}*\frac{FB}{FH}=2*\frac 5 2 *4=20$
Attachments:
This post has been edited 3 times. Last edited by luimichael, Mar 10, 2017, 3:58 AM
Reason: more detail

Comment

0 Comments

a