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
prat007
02.12.2012 13:36
Ans is $\frac{7}{3}$
KOSNITA
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.
subham1729
02.12.2012 13:52
Easily we get $3QE=2AQ,3PD=AP$ so $\frac {APQ}{ADE}=\frac {9}{20}\implies \frac{AQE}{PDEQ}=\frac {9}{11}$
ninepointcircle
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
shivangjindal
02.12.2012 14:21
I used , mid point theorem and got 9/11 as the answer. which method you all use guys.?
Sahil
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.
norhh007
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.
subham1729
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
christerson
02.12.2012 19:18
From mass points using only the cevians $AD$ and $BF$, $\frac{AP}{PD} = \frac{3}{1}$. Using mass points with only cevians $AE$ and $BF$, $\frac{AQ}{QE} = \frac{3}{2}$.
So, as a result of the area of a triangle (using $A = \frac{1}{2}ab\sin{C}$), the ratio $\frac{[APQ]}{[ADE]} = \left(\frac{3}{4}\right)\left(\frac{3}{5}\right) = \frac{9}{20}$.
Then $\frac{[PDEQ]}{[ADE]} = \frac{11}{20}$, so $\frac{[AQE]}{[PDEQ]} = \frac{9}{11}$.
aayush-srivastava
22.11.2015 17:41
A solution can also be obtained using coordinate geometry( ) Use oblique coordinates for short computations..
Raj2000
07.10.2017 11:56
Using menelaus twice will also help to solve easily
lifeismathematics
11.08.2022 15:48
lotus ratio lemmaIn a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$, then $\frac{CP}{CF}=\frac{CE}{AE}(1+\frac{AF}{FB})$ and $\frac{BP}{PE}=\frac{BF}{AF}(1+\frac{AE}{EC})$.
proof of lemmaso construct $EX \parallel FC$. then we have $\triangle {BPF}\sim \triangle {BEX}$ and $\triangle{AXE} \sim \triangle {AFC}$. then we have $\frac{BP}=\frac{BF}{BX}$ and $\frac{AX}{AF}=\frac{AE}{AC}$. then also we have $\frac{BP}{PE}=\frac{BF}{FX}$. so we have $FX=\frac{PE\cdot BF}{BP}$. also,$\frac{AX}{AF}=\frac{AE}{AC}$. so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$. similarly we can get the other ratio $\blacksquare$solving the problemJoin $PE$
in $\triangle{CAB}$
applying lotu's ratio lemma we get $\frac{AQ}{QE}=\frac{AF}{AC}(1+\frac{CE}{EB}) \implies \frac{AQ}{QE}=\frac{3}{2}$.
set $[BPD]=A$, then $[PDE]=A.$
In $\triangle{EAB}$
again applying lotu's ratio lemma we get
$\frac{BP}{BQ}=\frac{BD}{DE}(1+\frac{EQ}{AQ})\implies \frac{BQ}{BQ}=\frac{5}{3}$.
now $[EPB]=\frac{5}{3}\cdot [EQP]$ , so $[EQP]=\frac{6A}{5}$, now similarly $[APQ]=\frac{3}{2}\cdot [PEQ]$, so we have $[APQ]=\frac{9A}{5}$ and hence we get $\frac{[APQ]}{[PDEQ]}=\boxed{\frac{9}{11}}$.
InvincibleZ
19.10.2022 02:50
lifeismathematics wrote:
lotus ratio lemmaIn a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$, then $\frac{CP}{CF}=\frac{CE}{AE}(1+\frac{AF}{FB})$ and $\frac{BP}{PE}=\frac{BF}{AF}(1+\frac{AE}{EC})$.
proof of lemmaso construct $EX \parallel FC$. then we have $\triangle {BPF}\sim \triangle {BEX}$ and $\triangle{AXE} \sim \triangle {AFC}$. then we have $\frac{BP}=\frac{BF}{BX}$ and $\frac{AX}{AF}=\frac{AE}{AC}$. then also we have $\frac{BP}{PE}=\frac{BF}{FX}$. so we have $FX=\frac{PE\cdot BF}{BP}$. also,$\frac{AX}{AF}=\frac{AE}{AC}$. so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$. similarly we can get the other ratio $\blacksquare$solving the problemJoin $PE$
in $\triangle{CAB}$
applying lotu's ratio lemma we get $\frac{AQ}{QE}=\frac{AF}{AC}(1+\frac{CE}{EB}) \implies \frac{AQ}{QE}=\frac{3}{2}$.
set $[BPD]=A$, then $[PDE]=A.$
In $\triangle{EAB}$
again applying lotu's ratio lemma we get
$\frac{BP}{BQ}=\frac{BD}{DE}(1+\frac{EQ}{AQ})\implies \frac{BQ}{BQ}=\frac{5}{3}$.
now $[EPB]=\frac{5}{3}\cdot [EQP]$ , so $[EQP]=\frac{6A}{5}$, now similarly $[APQ]=\frac{3}{2}\cdot [PEQ]$, so we have $[APQ]=\frac{9A}{5}$ and hence we get $\frac{[APQ]}{[PDEQ]}=\boxed{\frac{9}{11}}$.
SolveForChocolate
07.11.2022 09:13
lifeismathematics wrote:
lotus ratio lemmaIn a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$, then $\frac{CP}{CF}=\frac{CE}{AE}(1+\frac{AF}{FB})$ and $\frac{BP}{PE}=\frac{BF}{AF}(1+\frac{AE}{EC})$.
proof of lemmaso construct $EX \parallel FC$. then we have $\triangle {BPF}\sim \triangle {BEX}$ and $\triangle{AXE} \sim \triangle {AFC}$. then we have $\frac{BP}=\frac{BF}{BX}$ and $\frac{AX}{AF}=\frac{AE}{AC}$. then also we have $\frac{BP}{PE}=\frac{BF}{FX}$. so we have $FX=\frac{PE\cdot BF}{BP}$. also,$\frac{AX}{AF}=\frac{AE}{AC}$. so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$. similarly we can get the other ratio $\blacksquare$solving the problemJoin $PE$
in $\triangle{CAB}$
applying lotu's ratio lemma we get $\frac{AQ}{QE}=\frac{AF}{AC}(1+\frac{CE}{EB}) \implies \frac{AQ}{QE}=\frac{3}{2}$.
set $[BPD]=A$, then $[PDE]=A.$
In $\triangle{EAB}$
again applying lotu's ratio lemma we get
$\frac{BP}{BQ}=\frac{BD}{DE}(1+\frac{EQ}{AQ})\implies \frac{BQ}{BQ}=\frac{5}{3}$.
now $[EPB]=\frac{5}{3}\cdot [EQP]$ , so $[EQP]=\frac{6A}{5}$, now similarly $[APQ]=\frac{3}{2}\cdot [PEQ]$, so we have $[APQ]=\frac{9A}{5}$ and hence we get $\frac{[APQ]}{[PDEQ]}=\boxed{\frac{9}{11}}$.
how in proof you get that $\frac{AC}{EC}$ divide $\frac{BF}{AF}$ is equal to $\frac{AE}{EC}$
lifeismathematics
14.11.2022 12:20
SolveForChocolate wrote: lifeismathematics wrote:
lotus ratio lemmaIn a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$, then $\frac{CP}{CF}=\frac{CE}{AE}(1+\frac{AF}{FB})$ and $\frac{BP}{PE}=\frac{BF}{AF}(1+\frac{AE}{EC})$.
proof of lemmaso construct $EX \parallel FC$. then we have $\triangle {BPF}\sim \triangle {BEX}$ and $\triangle{AXE} \sim \triangle {AFC}$. then we have $\frac{BP}=\frac{BF}{BX}$ and $\frac{AX}{AF}=\frac{AE}{AC}$. then also we have $\frac{BP}{PE}=\frac{BF}{FX}$. so we have $FX=\frac{PE\cdot BF}{BP}$. also,$\frac{AX}{AF}=\frac{AE}{AC}$. so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$. similarly we can get the other ratio $\blacksquare$solving the problemJoin $PE$
in $\triangle{CAB}$
applying lotu's ratio lemma we get $\frac{AQ}{QE}=\frac{AF}{AC}(1+\frac{CE}{EB}) \implies \frac{AQ}{QE}=\frac{3}{2}$.
set $[BPD]=A$, then $[PDE]=A.$
In $\triangle{EAB}$
again applying lotu's ratio lemma we get
$\frac{BP}{BQ}=\frac{BD}{DE}(1+\frac{EQ}{AQ})\implies \frac{BQ}{BQ}=\frac{5}{3}$.
now $[EPB]=\frac{5}{3}\cdot [EQP]$ , so $[EQP]=\frac{6A}{5}$, now similarly $[APQ]=\frac{3}{2}\cdot [PEQ]$, so we have $[APQ]=\frac{9A}{5}$ and hence we get $\frac{[APQ]}{[PDEQ]}=\boxed{\frac{9}{11}}$.
how in proof you get that $\frac{AC}{EC}$ divide $\frac{BF}{AF}$ is equal to $\frac{AE}{EC}$ where i have written that?
SolveForChocolate
01.12.2022 12:58
lifeismathematics wrote: SolveForChocolate wrote: lifeismathematics wrote:
lotus ratio lemmaIn a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$, then $\frac{CP}{CF}=\frac{CE}{AE}(1+\frac{AF}{FB})$ and $\frac{BP}{PE}=\frac{BF}{AF}(1+\frac{AE}{EC})$.
proof of lemmaso construct $EX \parallel FC$. then we have $\triangle {BPF}\sim \triangle {BEX}$ and $\triangle{AXE} \sim \triangle {AFC}$. then we have $\frac{BP}=\frac{BF}{BX}$ and $\frac{AX}{AF}=\frac{AE}{AC}$. then also we have $\frac{BP}{PE}=\frac{BF}{FX}$. so we have $FX=\frac{PE\cdot BF}{BP}$. also,$\frac{AX}{AF}=\frac{AE}{AC}$. so simplifying we get $\frac{BF}{AF}\cdot \frac{AC}{EC}=\frac{BP}{PE}$so we get $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$. similarly we can get the other ratio $\blacksquare$solving the problemJoin $PE$
in $\triangle{CAB}$
applying lotu's ratio lemma we get $\frac{AQ}{QE}=\frac{AF}{AC}(1+\frac{CE}{EB}) \implies \frac{AQ}{QE}=\frac{3}{2}$.
set $[BPD]=A$, then $[PDE]=A.$
In $\triangle{EAB}$
again applying lotu's ratio lemma we get
$\frac{BP}{BQ}=\frac{BD}{DE}(1+\frac{EQ}{AQ})\implies \frac{BQ}{BQ}=\frac{5}{3}$.
now $[EPB]=\frac{5}{3}\cdot [EQP]$ , so $[EQP]=\frac{6A}{5}$, now similarly $[APQ]=\frac{3}{2}\cdot [PEQ]$, so we have $[APQ]=\frac{9A}{5}$ and hence we get $\frac{[APQ]}{[PDEQ]}=\boxed{\frac{9}{11}}$.
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 $\frac{BF}{AF}\cdot (1+\frac{AE}{EC})=\frac{BP}{PE}$
lifeismathematics
01.12.2022 13:30
Yeah crct
SolveForChocolate
03.12.2022 09:57
lifeismathematics wrote: Yeah crct and? (AC\EC)\(BF\AF)=AE\EC why this is true?
Agrawal001
21.09.2024 10:59
Using menelaus , I got 9/11