CWMO 2010, Day 2, Problem 6

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CWMO 2010, Day 2, Problem 6

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#1 h Oct 30, 2010, 1:10 PM • 3 Y

Y by meowme, Adventure10, Mango247

$\Delta ABC$ is a right-angled triangle, $\angle C = 90^{\circ}$ . Draw a circle centered at $B$ with radius $BC$ . Let $D$ be a point on the side $AC$ , and $DE$ is tangent to the circle at $E$ . The line through $C$ perpendicular to $AB$ meets line $BE$ at $F$ . Line $AF$ meets $DE$ at point $G$ . The line through $A$ parallel to $BG$ meets $DE$ at $H$ . Prove that $GE = GH$ .

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lssl

#2 h Oct 30, 2010, 4:45 PM • 2 Y

Y by Adventure10 and 1 other user

Let us denote the intersection of $AB$ and $DE$ be $K$ , Asume $EL$ be the diameter of $\odot B$ , denote the intersection of $AB$
and $FC$ be $M$ .

Join $AL$ , if we can prove $GB // AL$ then since $EB=BL$ ,so $EG=EH$ ,we can finish the proof .

Prove : $BG//AL$ :
Easy to get : $BC^2=BM*BA=BE^2$
hence : $\triangle BEN \sim \triangle BAE$
because $E,F,M,K$ are concyclic ,
hence $\angle BEM=\angle FHM=\angle EAB$
hence $FH//EA$ $\Longrightarrow$ $\frac{BF}{EF}=\frac{BH}{HA}$ (#)
to $\triangle AFB$ and line $KGE$ , by Menelaus Th. , we get :
$\frac{EF}{EB}*\frac{BH}{HA}*\frac{AG}{GF}=1$ ,
By (#) , hence : $\frac{AG}{GF}=\frac{BE}{FB}=\frac{BL}{BF}$
$\Longrightarrow$ $BG//AL$

QED.

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#3 h Nov 20, 2010, 12:19 PM • 2 Y

Y by Adventure10, Mango247

i know this may sound stupid, but i used coordinate geometry to solve this problem in 1 hour during the competition

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vslmat

#4 h Sep 22, 2012, 1:39 PM • 2 Y

Y by Adventure10, Mango247

Because I am a little confused with Issl's proof, for example "Let us denote the intersection of $AB$ and $DE$ be $K$ " (???) and " $\Delta BEN\sim BAE$ " (???) ..., I will post my solution here. However, I see that we both may have some same ideas:

Let $FB$ cut the circle with center $B$ , radius $BC$ again at $K$ , what we have to prove is equivalent to prove that $AK\parallel GB$
Let $BC$ cut the circle with diameter $DB$ again at $I$ , as $\angle DIB = 90^{\circ}, DIBE$ is cyclic, hence $\angle EIB = \angle BDE = \angle CFB$ , thus $ICEF$ is cyclic, but as $BCE$ is isosceles, it is obvious that $CE\parallel IF$
Thus $\frac{IC}{CB} = \frac{FE}{EB}$ , notice that $\frac{IC}{CB} = \frac{DA}{AB}$ .
Using Melenaus in $\Delta AFB$ we get
$\frac{GF}{GA}.\frac{DA}{DB}.\frac{EB}{EF} = 1$ , but $\frac{DA}{DB} = \frac{EF}{EB}$ , so $\frac{GF}{GA} = \frac{FB}{EB} = \frac{FB}{BK}$ hence $AK\parallel GB$

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#5 h Sep 28, 2012, 4:29 PM • 1 Y

Y by Adventure10

Dear lssl:
I think ∆ $BEM$ ~∆ $BAE$

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jred

#6 h May 29, 2014, 2:43 PM • 1 Y

Y by Adventure10

vslmat wrote:

Because I am a little confused with Issl's proof, for example "Let us denote the intersection of $AB$ and $DE$ be $K$ " (???) and " $\Delta BEN\sim BAE$ " (???) ..., I will post my solution here. However, I see that we both may have some same ideas:

Let $FB$ cut the circle with center $B$ , radius $BC$ again at $K$ , what we have to prove is equivalent to prove that $AK\parallel GB$
Let $BC$ cut the circle with diameter $DB$ again at $I$ , as $\angle DIB = 90^{\circ}, DIBE$ is cyclic, hence $\angle EIB = \angle BDE = \angle CFB$ , thus $ICEF$ is cyclic, but as $BCE$ is isosceles, it is obvious that $CE\parallel IF$
Thus $\frac{IC}{CB} = \frac{FE}{EB}$ , notice that $\frac{IC}{CB} = \frac{DA}{AB}$ .
Using Melenaus in $\Delta AFB$ we get
$\frac{GF}{GA}.\frac{DA}{DB}.\frac{EB}{EF} = 1$ , but $\frac{DA}{DB} = \frac{EF}{EB}$ , so $\frac{GF}{GA} = \frac{FB}{EB} = \frac{FB}{BK}$ hence $AK\parallel GB$

Obviously you must have misunderstood the statement of this problem. According to the statement, D should be on side AC, not on AB. BTW, I make a solution with projective geometry.

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#7 h Apr 18, 2020, 8:37 AM

Y by

Let $X=DE\cap AB$ , $Y=BD\cap CE$ , $Z=AB\cap CF$
Notice that $DC$ is also tangent to the circle hence $CE\perp BD$ . Now,
$\underline{CLAIM.}$ $\frac{GX}{XE}=\frac{BX}{BA}$
Proof.Applying Menelaus Theorem to $\triangle BEX$ , we have
$\frac{XG}{GE}=\frac{AX}{AB}\cdot\frac{FB}{FE}$
It hence suffices to prove
$\frac{FB}{FE}=\frac{BX}{AX}$
since $\angle ECF=\angle YCZ=\angle YBZ=\angle DBA$
$\begin{align*} \frac{FB}{FE}&=\frac{\sin\angle BCF}{\sin\angle ECF}\cdot\frac{\sin\angle BEC}{\sin\angle CBE}\\ &=\frac{\sin\angle DAB}{\sin\angle ECF}\frac{\sin\angle{BDX}}{\sin\angle ADX}\\ &=\frac{BX}{AX} \end{align*}$ which is exactly what we want to prove.
Hence the claim is proved and
$\frac{GH}{GE}=\frac{GX}{GE}\cdot\frac{GH}{GX}=\frac{BX}{BA}\cdot\frac{BA}{BX}=1$

This post has been edited 1 time. Last edited by mathaddiction, Apr 18, 2020, 8:38 AM

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#8 h Jul 18, 2023, 8:37 AM

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Let $P = \overline{AD} \cap \overline{BC}$ . Notice that $EDCO$ is cyclic, and furthermore it is the nine-point circle of $ABX$ . But $E$ is the $O$ -antipode in $(DOCE)$ , hence as $F$ is the orthocenter, $M$ is the foot of the $X$ -altitude, and thus lies on the circle too.

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