Parallel lines

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syk0526

202 posts

syk0526

#1 h Aug 19, 2012, 9:06 AM • 3 Y

Y by Ikeronalio, Adventure10, Mango247

Let $w$ be the incircle of triangle $ABC$ . Segments $BC, CA$ meet with $w$ at points $D, E$ . A line passing through $B$ and parallel to $DE$ meets $w$ at $F$ and $G$ . ( $F$ is nearer to $B$ than $G$ .) Line $CG$ meets $w$ at $H ( \ne G )$ . A line passing through $G$ and parallel to $EH$ meets with line $AC$ at $I$ . Line $IF$ meets with circle $w$ at $J (\ne F )$ . Lines $CJ$ and $EG$ meets at $K$ . Let $l$ be the line passing through $K$ and parallel to $JD$ . Prove that $l, IF, ED$ meet at one point.

This post has been edited 3 times. Last edited by syk0526, Sep 1, 2012, 1:05 PM

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yunxiu

571 posts

yunxiu

#2 h Aug 20, 2012, 4:17 AM • 2 Y

Y by Adventure10, Mango247

syk0526 wrote:

Segments $BC, CA$ meet with $O$ at $D, E$ .

What does meet mean?

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syk0526

202 posts

syk0526

#3 h Aug 20, 2012, 7:43 AM • 2 Y

Y by Adventure10, Mango247

yunxiu wrote:

syk0526 wrote:

Segments $BC, CA$ meet with $O$ at $D, E$ .

What does meet mean?

Umm,,, $O$ is the incircle, so $OD \bot BC$ and $OE \bot CA$ .

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Vo Duc Dien

341 posts

Vo Duc Dien

#4 h Sep 3, 2012, 9:53 PM • 2 Y

Y by Adventure10, Mango247

Hint:

Let $CF$ intersect $w$ at $N$ , $DJ$ intersect $EK$ at $M$ , $FG$ intersect $MD$ at $W$ and $IF$ intersect $EK$ at $S$ . We will have $HJ || KE$ , $EN || IF$ . $G$ is then the midpoint of $EM$ . Triangle $IFG$ is similar to triangle $ENH$ . $W$ is the midpoint of $MD$ . The rest involves the steps to come up with the equal ratios and finally $\frac {MS}{MJ} = \frac {KS}{KP}$ and you're done. Needless to say, it's a very difficult problem.

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polya78

105 posts

polya78

#5 h Jun 4, 2013, 7:32 PM • 3 Y

Y by cauterez, Adventure10, Mango247

Indeed it is a difficult problem. Let $R$ be center of $w$ , $f$ be the dilation with center $C$ such that $f(H)=G$ ,and $g$ denote the reflection about $CR$ .

Since $IG \parallel EH, f(E)=I$ and since $GF \parallel ED, g(E)=D, g(G)=F$ . If $f(D)=T$ , then $T=g(I)$ as well, and $IT\parallel GF$ . So $\angle GFJ =\angle TIF =\angle GTI =\angle HDE$ , which means that $JH\parallel GE$ . As a result $f(J)=K \Rightarrow KT \parallel JD$ .

It only remains to prove that if $Y = KT \cap IF$ , then $E,D,Y$ are collinear. First we note that since $f$ transforms $E,H,D,J$ into $I,G,T,K$ ,then $I,G,T,K$ lie on the circle $f(w)$ , and $g(G)=F$ must lie on this circle as well. So $\angle TYF = \angle KGI -\angle IKG = \angle JHE -\angle EJH = \angle GDE =\angle FED =\angle FDT$ , so $F,T,Y,D$ are concyclic. So $\angle TDY =180-\angle TFI = \angle IKT =\angle EJD = \angle EDC$ , which proves that $T,D,E$ are collinear.

Attachments:

korea.pdf (451kb)

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Mosquitall

571 posts

Mosquitall

#6 h Jun 6, 2013, 11:17 AM • 2 Y

Y by Adventure10, Mango247

Easy it takes me 3 minutes
Easy see that circle $(E'FGI)$ is tangent to $E'C$ and $IC$ , at points $E',C$ . Let $GE$ cuts $(E'FGI)$ at $P$ , then $KI||JE$ , so from homotecy $P$ is on $JC$ , so $P=K$ . Let $IF\cap DE'=T$ , $\angle JFH=\angle EDC$ , so $(FE'TD)$ is circle, so $T$ is on $KE'$ .

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harrumit

3 posts

harrumit

#7 h Nov 9, 2013, 2:23 PM • 1 Y

Y by Adventure10

Mosquitall wrote:

Easy it takes me 3 minutes
Easy see that circle $(E'FGI)$ is tangent to $E'C$ and $IC$ , at points $E',C$ . Let $GE$ cuts $(E'FGI)$ at $P$ , then $KI||JE$ , so from homotecy $P$ is on $JC$ , so $P=K$ . Let $IF\cap DE'=T$ , $\angle JFH=\angle EDC$ , so $(FE'TD)$ is circle, so $T$ is on $KE'$ .

I can't understand your solution. What's $E'$ ?

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jayme

9775 posts

jayme

#8 h Nov 10, 2013, 9:14 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
an outline of an idea of a synthetic proof...
This problem accepts fully the Reim's theorem... and the proof finishs with the Pascal's theorem.
Sincerely
Jean-Louis

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kormidscoler

64 posts

kormidscoler

#9 h Oct 7, 2015, 12:22 PM • 2 Y

Y by Adventure10, Mango247

why don't you just use the fact EH=JG (which can be proved easily), and let IJ and ED meet at T, and let the line l' be a line that is parallel to line JD and passes through a point T, and let K' be a point that is on EG and l' at same time?

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cauterez

1 post

cauterez

#10 h Jan 20, 2017, 1:57 PM • 1 Y

Y by Adventure10

polya78 wrote:

Indeed it is a difficult problem. Let $R$ be center of $w$ , $f$ be the dilation with center $C$ such that $f(H)=G$ ,and $g$ denote the reflection about $CR$ .

Since $IG \parallel EH, f(E)=I$ and since $GF \parallel ED, g(E)=D, g(G)=F$ . If $f(D)=T$ , then $T=g(I)$ as well, and $IT\parallel GF$ . So $\angle GFJ =\angle TIF =\angle GTI =\angle HDE$ , which means that $JH\parallel GE$ . As a result $f(J)=K \Rightarrow KT \parallel JD$ .

It only remains to prove that if $Y = KT \cap IF$ , then $E,D,Y$ are collinear. First we note that since $f$ transforms $E,H,D,J$ into $I,G,T,K$ ,then $I,G,T,K$ lie on the circle $f(w)$ , and $g(G)=F$ must lie on this circle as well. So $\angle TYF = \angle KGI -\angle IKG = \angle JHE -\angle EJH = \angle GDE =\angle FED =\angle FDT$ , so $F,T,Y,D$ are concyclic. So $\angle TDY =180-\angle TFI = \angle IKT =\angle EJD = \angle EDC$ , which proves that $T,D,E$ are collinear.

The greatest solution I've ever seen. Congrats!

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parmenides51

30630 posts

parmenides51

#11 h Jun 3, 2020, 3:18 PM

Y by

another solution here

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IDMasterz

1412 posts

IDMasterz

#12 h Jun 23, 2022, 9:05 PM • 1 Y

Y by Mango247

Here's a solution. First we start with a bit of motivation:
Suppose $T:=IF \cap ED$ , and let $\cal{C}$ be the circle through $KTE$ ; suppose $\cal{C}$ meets the circle $w$ at a point $M$ other that $E$ . We see that the statement is true, by the converse of Reim's theorem, if and only if $M$ lies on $CJ$ , so we try to prove this. Brief angle chasing yields that in fact, if this holds then $I$ lies on $\cal{C}$ and too that $IK$ is parallel to $EJ$ .

The solution:
So, we want to show that $IK \parallel EJ$ . First, we reduce this to $KG$ being parallel to $HJ$ , which is equivalent to showing that $JGEH$ is an isosceles trapezoid, which is equivalent to showing that $\angle IFG = \angle CEH = \angle CIG$ , or that the circle through $IFG$ is tangent to $CI$ , which is because $\angle DEC = \angle FIE = 180 - \angle FGI$ as desired. We then directly deduce from this that $I$ lies on $\cal{C}$ . The converse of Reim's theorem now applies to show that $M$ lies on $CJ$ .

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