Ratios in length of a triangle

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Ratios in length of a triangle

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Source: Indonesian MO (INAMO) 2009, Day 1, Problem 3

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#1 h Aug 8, 2009, 8:38 AM • 2 Y

Y by Adventure10, Mango247

For every triangle $ABC$ , let $D,E,F$ be a point located on segment $BC,CA,AB$ , respectively. Let $P$ be the intersection of $AD$ and $EF$ . Prove that:
$\frac{AB}{AF}\times DC+\frac{AC}{AE}\times DB=\frac{AD}{AP}\times BC$

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#2 h Aug 9, 2009, 3:32 AM • 2 Y

Y by Adventure10, Mango247

wangsacl wrote:

For every triangle $ABC$ , let $D,E,F$ be a point located on segment $BC,CA,AB$ , respectively. Let $P$ be the intersection of $AD$ and $EF$ . Prove that:
$\frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB = \frac {AD}{AP}\times BC$

It is nice but easy!.
In the case $AB<AC$
Let $BI$ ; $CJ$ be parallel to $EF$ where $I;J$ belong to $AD$ . Thales Theorem shows us $\frac{AB}{AF}=\frac{AI}{AP}$ ; $\frac{AC}{AE}=\frac{AJ}{AP}$ ; $\frac{ID}{DB}=\frac{JD}{DC}=\frac{IJ}{BC}$ . (1)
Notice that $AI=AD-ID=AD-\frac{DB.IJ}{BC}$ thus $\frac{AI.DC}{AP}=\frac{AD.DC}{AP}-\frac{IJ.DB.DC}{BC.AP}$
A same thing is $\frac{AJ.DB}{AP}=\frac{AD.DB}{AP}+\frac{IJ.DB.DC}{BC.AP}$ .
Now , it is immediate that $\frac{AI.DC}{AP}+\frac{AJ.DB}{AP}=\frac {AD}{AP}\times BC$ since the fact $DC+DB=BC$ .
Now, the proof can be completed after we look at again (1)

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#3 h Aug 9, 2009, 5:50 AM • 2 Y

Y by Adventure10, Mango247

Yup. The official solution only uses ratios in length and area, which is very-very beautiful (and algebraic

)

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IW@IT

#4 h Aug 11, 2009, 9:53 AM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Here is my solution
$\frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}+\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}=\frac{[ABD]}{[AFP]}.\frac{[ADC]}{[ABC]}+\frac{[ADC]}{[APE]}.\frac{[ABD]}{[ABC]}=\frac{[ABD].[ADC]}{[ABC]}(\frac{1}{[AFP]}+\frac{1}{[APE]})=\frac{[ABD].[ADC]}{[AFP].[APE]}.\frac{[AFE]}{[ABC]}=\frac{AB.AD}{AF.AP}.\frac{AD.AC}{AP.AE}.\frac{AF.AE}{AB.AC}=\frac{AD^2}{AP^2}$

So, $\frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}+\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}=\frac{AD^2}{AP^2}$ , which is equivalent with the problem

$QED$

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shoki

#5 h Aug 11, 2009, 9:57 AM • 1 Y

Y by Adventure10

it's (almost) Cristea's Theorem: let ABC be a scalene triangle and D a point on the segment BC. M belongs to the segment AD. E and F belong to the segment AB and AC ,respectively. Then EF passes through M if and only if:
DC(EB/EA) + BD(FC/FA) = BC(MD/MA)

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dyd

#6 h Aug 13, 2009, 12:17 PM • 2 Y

Y by Adventure10, Mango247

BTW, that Cristea's Theorem is not well-known to me. But this problem is just an old result from http://www.mathlinks.ro/viewtopic.php?p=1143691#1143691

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#7 h Aug 13, 2009, 12:31 PM • 2 Y

Y by Adventure10, Mango247

A simple proof is using vector.
We have $BD. \vec{AC}+DC.\vec{AB}=BC.\vec{AD}$
$\Rightarrow \frac{AC}{AE}.BD.\vec{AE}+\frac{AB}{AF}.DC.\vec{AF}=\frac{AD}{AP}.BC.\vec{AP}$ or $x.\vec{AE}+y.\vec{AF}=z.\vec{AP}$
On the other side, $FP.\vec{AE}+EP.\vec{AF}=EF.\vec{AP}$
So $\frac{x}{FP}=\frac{y}{EP}=\frac{z}{EF}=\frac{x+y}{FP+EP}=\frac{x+y}{EF}$
$\Rightarrow x+y=z$ (QED)

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#8 h Oct 12, 2009, 3:02 PM • 2 Y

Y by Adventure10, Mango247

wangsacl wrote:

For every triangle $ABC$ , let $D,E,F$ be a point located on segment $BC,CA,AB$ , respectively. Let $P$ be the intersection of $AD$ and $EF$ . Prove that:
$\frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB = \frac {AD}{AP}\times BC$

I am giving a solution using Menelaus Theorem.
Construct $F',E'$ on $AB, AC$ resp. such that $F'E' \parallel BC$
Now Menelaus on $\triangle AF'E'$ and transversal $FPE$ gives,
$\begin{align*} & \frac{AF}{FF'}\cdot \frac{F'P}{PE'} \cdot \frac{EE'}{EA}=1 \\ \iff & \frac{PE'}{PF'}=\frac{AF}{FF'} \cdot \frac{EE'}{EA} \\ \iff & \frac{DC}{DB}\cdot \frac{AB}{AC}=\frac{EE'}{FF'}\cdot \frac{AF}{AE} \cdot \frac{AF'}{AE'}=\frac{\frac{AE'-AE}{AE\times AE'}}{\frac{AF-AF'}{AF\times AF'}}\\ \iff & AB\times DC \left[\frac{1}{AF'}-\frac{1}{AF}\right]=AC\times DB \left[\frac{1}{AE}-\frac{1}{AE'}\right] \\ \iff & AB\times DC \times \frac{1}{AF'}+AC\times DB \times \frac{1}{AE'} = AB\times DC \times \frac{1}{AF}+AC\times DB \frac{1}{AE} \end{align*}$
But $\begin{align*} \frac{AD}{AP}\times BC=& \frac{AD}{AP}\times (BD+DC)=\frac{AB}{AF'}\times DC+\frac{AC}{AE'}\times BD\\ =&\frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB \end{align*}$

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