Let $ABC$ be a triangle. Let $D,E$ be points on the segment $BC$ such that $BD=DE=EC$. Let $F$ be the mid-point of $AC$. Let $BF$ intersect $AD$ in $P$ and $AE$ in $Q$ respectively. Determine the ratio of the area of the triangle $APQ$ to that of the quadrilateral $PDEQ$.
Problem
Source: RMO 2012
Tags: ratio, geometry, trigonometry, area of a triangle
02.12.2012 13:36
Ans is $\frac{7}{3}$
02.12.2012 13:41
I have not solved the full problem. first use menelaus to find $\frac{AP}{PD}$ and $\frac{AQ}{QE}$. then using the concept of areas in $\triangle ADE$,we will get it.
02.12.2012 13:52
Easily we get $3QE=2AQ,3PD=AP$ so $\frac {APQ}{ADE}=\frac {9}{20}\implies \frac{AQE}{PDEQ}=\frac {9}{11}$
02.12.2012 14:15
subham1729 wrote: Easily we get $3QE=2AQ,3PD=AP$ so $\frac {APQ}{ADE}=\frac {9}{20}\implies \frac{AQE}{PDEQ}=\frac {9}{11}$ Got the same answer
02.12.2012 14:21
I used , mid point theorem and got 9/11 as the answer. which method you all use guys.?
02.12.2012 14:37
subham1729 wrote: Easily we get $3QE=2AQ,3PD=AP$ so $\frac {APQ}{ADE}=\frac {9}{20}\implies \frac{AQE}{PDEQ}=\frac {9}{11}$ Seems to have some connection to the American catastrophe.
02.12.2012 14:39
shivangjindal wrote: I used , mid point theorem and got 9/11 as the answer. which method you all use guys.? Yeah getting 9/11.
02.12.2012 14:54
Sahil wrote: subham1729 wrote: Easily we get $3QE=2AQ,3PD=AP$ so $\frac {APQ}{ADE}=\frac {9}{20}\implies \frac{AQE}{PDEQ}=\frac {9}{11}$ Seems to have some connection to the American catastrophe. So this 2nd December should be also a memorable day
02.12.2012 19:18
22.11.2015 17:41
A solution can also be obtained using coordinate geometry( ) Use oblique coordinates for short computations..
07.10.2017 11:56
Using menelaus twice will also help to solve easily
11.08.2022 15:48
19.10.2022 02:50
lifeismathematics wrote:
07.11.2022 09:13
lifeismathematics wrote:
how in proof you get that $\frac{AC}{EC}$ divide $\frac{BF}{AF}$ is equal to $\frac{AE}{EC}$
14.11.2022 12:20
SolveForChocolate wrote: lifeismathematics wrote:
how in proof you get that $\frac{AC}{EC}$ divide $\frac{BF}{AF}$ is equal to $\frac{AE}{EC}$ where i have written that?
01.12.2022 12:58
lifeismathematics wrote: SolveForChocolate wrote: lifeismathematics wrote:
how in proof you get that $\frac{AC}{EC}$ divide $\frac{BF}{AF}$ is equal to $\frac{AE}{EC}$ where i have written that? so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$
01.12.2022 13:30
Yeah crct
03.12.2022 09:57
lifeismathematics wrote: Yeah crct and? (AC\EC)\(BF\AF)=AE\EC why this is true?
21.09.2024 10:59
Using menelaus , I got 9/11