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Source: Korea Final Round 2009

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#1 h Jul 24, 2010, 5:54 PM • 2 Y

Y by Adventure10, Mango247

$ABC$ is an acute triangle. (angle $C$ is bigger than angle $B$ ) Let $O$ be a center of the circle which passes $B$ and tangents to $AC$ at $C$ . $O$ meets the segment $AB$ at $D$ . $CO$ meets the circle $(O)$ again at $P$ , a line, which passes $P$ and parallel to $AO$ , meets $AC$ at $E$ , and $EB$ meets the circle $(O)$ again at $L$ . A perpendicular bisector of $BD$ meets $AC$ at $F$ and $LF$ meets $CD$ at $K$ . Prove that two lines $EK$ and $CL$ are parallel.

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#2 h Jul 24, 2010, 9:10 PM • 2 Y

Y by Adventure10, Mango247

$AO \; // \; PE \; \Longrightarrow \; |CA|=|EA|$

$|AC|^2=|AD|.|AB|=|EA|^2 \; \Longrightarrow \; \triangle AED \sim \triangle ABE$

$\angle AED=\angle ABE= \angle LCD$

Let $U$ be the midpoint of $[BD]$

By Pisagor's theorem

$|AU|^2-|BU|^2=|AF|^2-|BF|^2=|AF|^2-|DF|^2$

$|AC|^2=|AD|.|AB|=|AU|^2-|BU|^2$

$\Longrightarrow \; |AC|^2=|AF|^2-|DF|^2$

$\Longrightarrow \; |FD|^2=|AF|^2-|AC|^2=|FC|.|FE|$

$\Longrightarrow \; \triangle FDC \sim \triangle FED$

$\angle FDC = \angle DEF$

$\angle DLC = \angle DCE \; , \; \angle LCD = \angle DEC \; \Longrightarrow \; \triangle LDC \sim \triangle CDE$

$\Longrightarrow \; |DE|.|DL|=|CD|^2$

$\frac{|LD|}{|DE|}=\frac{|CD|^2}{|DE|^2}=\frac{|DF|^2}{|EF|^2}=\frac{|CF|}{|EF|}$

Let $V$ be the intersection point of the lines $KD$ and $EL$

$\angle LDV = \angle EDV \; \Longrightarrow \; \frac{|LD|}{|ED|}=\frac{|LV|}{|EV|}$

$\Longrightarrow \; \frac{|LV|}{|EV|}=\frac{|CF|}{|EF|}$

By Menelaus' theorem

$\frac{|KL|}{|KF|}.\frac{|FC|}{|CE|}.\frac{|EV|}{|VL|}=1$

$\Longrightarrow \; \frac{|KL|}{|KF|}=\frac{|CE|}{|FC|}.\frac{|VL|}{|EV|}=\frac{|CE|}{|EF|}$

$\Longrightarrow \; CL \; // \; KE$

Solution is done.

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#3 h Jul 28, 2010, 11:28 AM • 1 Y

Y by Adventure10

I have another idea

Let the radius of the circle is $1$

$OC\perp CA$ $\iff$ $\frac{o-c}{o-\frac{1}{c}}=c^{2}=\frac{c-a}{\overline a-\frac{1}{c}}$

$\implies$ $\frac{2c-a}{c^{2}}=\overline a$ $\cdots (1)$

$B-D-A$ are collinear $\iff$ $\frac{b-d}{\frac{1}{b}-\frac{1}{d}}=-bd=\frac{b-a}{\frac{1}{b}-\overline a}$

$\implies$ $\frac{b+d-a}{bd}=\overline a$ $\cdots (2)$

from $(1)$ and $(2)$ $\implies$ $(b+d-a)c^{2}=(2c-a)bd$

$\implies$ $a=\frac{bc^{2}+dc^{2}-2bcd}{c^{2}-bd}$

$C-O-P$ is collinear $\iff$ $p=-c$

$P-L-E$ is collinear $\iff$ $cl=\frac{l-e}{\frac{1}{l}-\overline e}$

$\implies$ $c-cl\overline e=l-e$ $\implies$ $\frac{c+e-l}{cl}=\overline e$ $\cdots (3)$

$OC \perp CE$ $\iff$ $c^{2}=\frac{c-e}{\overline e-\frac{1}{c}}$

$\implies$ $\overline e=\frac{2c-e}{c^{2}}$ $\cdots (4)$

from $(3)$ and $(4)$ $\implies$ $\frac{c+e-l}{cl}=\frac{2c-e}{c^{2}}$

$\implies$ $e=\frac{3cl-c^{2}}{l+c}$

Let midpoint of $BD$ is $R$ , $r=\frac{b+d}{2}$

$R-O-F$ is collinear $\iff$ $bd=\frac{f}{\overline f}$

$\implies$ $bd\overline f =f$ $\cdots(5)$

$OC\perp CF$ $\iff$ $c^{2}=\frac{c-f}{\overline f-\frac{1}{c}}$

$c^{2}\overline f-c=c-f$ $\implies$ $\frac{2c-f}{c^{2}}=\overline f$ $\cdots (6)$

from $(5)$ and $(6)$ $\implies$ $\frac{2c-f}{c^{2}}=\frac{f}{bd}$

$f=\frac{2bcd}{c^{2}+bd}$

$LF \cap DC=\{K\}$ $\implies$

$D-C-K$ collinear $\iff$ $-cd=\frac{k-c}{\overline k-\frac{1}{c}}$

$\implies$ $d-cd\overline k=k-c$ $\implies$ $\overline k=\frac{c+d-k}{cd}$ $\cdots(7)$

$L-F-K$ collinear $\iff$ $\frac{lc^{2}+lbd-2bcd}{c^{2}+bd-2cl}=\frac{k-l}{l\overline k -1}$

$\implies$ $\overline k=\frac{kc^{2}+kbd-2kcl+2l^{2}c-2bcd}{l^{2}c^{2}+l^{2}bd-2bdcl}$ $\cdots (8)$

from $(7)$ and $(8)$ $\implies$ $\frac{c+d-k}{cd}=\frac{kc^{2}+kbd-2kcl+2l^{2}c-2bcd}{l^{2}c^{2}+l^{2}bd-2bdcl}$

$\implies$ $k=\frac{l^{2}c^3+l^2c^2db-2bc^2dl+l^2dc^2+l^2d^2b-2b^2dcl-2l^2c^2d+2bc^2d^2}{c^3d+bd^2c-2c^2dl+l^2c^2+l^2bd-2bcdl}$

$EK // CL$ $\iff$ $\frac{e-k}{\overline e-\overline k}=-cl$

$e=\frac{3cl-c^{2}}{l+c}$ ,

$k=\frac{l^{2}c^3+l^2c^2db-2bc^2dl+l^2dc^2+l^2d^2b-2b^2dcl-2l^2c^2d+2bc^2d^2}{c^3d+bd^2c-2c^2dl+l^2c^2+l^2bd-2bcdl}$

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#4 h Sep 5, 2012, 6:56 PM • 2 Y

Y by Adventure10, Mango247

I did not understand why P-L-E are collinear.
Aslinda B-L-E dogrusal. Burada bir hata gozukuyor gibi.
O kismini aciklayabilir misin ..
Can you explain..

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leader

#5 h May 20, 2013, 8:04 PM • 1 Y

Y by Adventure10

let $H$ be on $CD$ such that $EH||CL$ then $\angle EHD=\angle DCL=\angle DBE$ so $BDEH$ is cyclic. now since by power of $A$ to $(O)$ $AC^2=AD*AB$ notice that $CA/AE=CO/OP=1$ so $AE^2=AC^2=AD*AB$ and $AD/AE=AE/AB$ giving $AED\sim AEB$ so $\angle AED=\angle DBE$ so $AE$ is tangent to circle $BDEH$ meaning that $AC$ is the common tangent of circles $BDEH$ and $(O)$ while the perpendicular bisector of $BD$ passes through the centers of these circles so $F$ is the center of psitive homothethy that pictures circle $(O)$ to circle $BDEH$ and $C$ to $E$ but since $CL||EH$ $L$ goes to $H$ so $F,L,H$ are collinear and $H=FL\cap CD=K$ and $EK||CL$

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#6 h Mar 13, 2016, 1:07 AM • 2 Y

Y by Adventure10, Mango247

Lemma. Let there be two circles with center $O_1$ and $O_2$ . Let $T_1T_2$ be a common external tangent of the two circles.
$T_1$ lies on circles $O_1$ and $T_2$ lies on circle $O_2$ . Let the two circles meet each other at $U, V$ .
Let $T_1V \cap O_2$ be $P$ and $T_2U \cap O_1$ be $Q$ . Let $T_1T_2 \cap O_1O_2$ be $R$ . Then $P,Q,R$ are colinear.

Proof of Lemma. $\angle QT_2P = \angle PVU = \angle T_1VU = \angle T_1QU$ , so $QT_1$ and $PT_2$ are parallel.
Since $R$ is the center of homothety of the two circles, clearly $P, Q, R$ are colinear as required.

Denote the circumcircle of $\triangle BDE$ as $\omega$ .
Now since $AC^2 = AE^2 = AD \cdot AB$ , we have $AE$ tangent to $\omega$ .
Also, note that the center of $\omega$ lies on the perpendicular bisector of $BD$ .
This implies that $F$ is the center of homothety of circle $O$ and $\omega$ .
Now $L=EB \cap O$ , $CD \cap \omega$ and $F$ are colinear by our Lemma, so $K=FL \cap CD$ lie on $\omega$ .
It is clear that $EK \parallel CD$ , as required. $\blacksquare$

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