IMO 2012: Day 1 Problem 1

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Moon
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IMO 2012: Day 1 Problem 1

Unread post by Moon » Tue Jul 10, 2012 11:28 pm

Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)
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Re: IMO 2012: Day 1 Problem 1

Unread post by SANZEED » Wed Jul 11, 2012 1:55 am

First we can see that $\angle BFM=\frac{1}{2}\angle A$. This implies that $\angle JFL=\angle JAL$. Thus the quadrilateral $AFJL$ is cyclic.So $\angle AFJ=180^{\circ}-\angle ALJ=90^{\circ}$. We can now easily notice that $AB=BS$. Similarly $AC=CT$. Now, $MS=SB+BM=AB+BM=AK=AL=AC+CM=CT+CM=MT$.It is proved!!! :P :P :P :P :P
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Re: IMO 2012: Day 1 Problem 1

Unread post by Nadim Ul Abrar » Wed Jul 11, 2012 7:41 pm

Used pascel to say $AGLJKF$ is cyclic :D .
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Re: IMO 2012: Day 1 Problem 1

Unread post by Tahmid Hasan » Tue Jul 17, 2012 12:46 pm

Nadim Ul Abrar wrote:Used pascel to say $AGLJKF$ is cyclic :D .
I'm afraid you can't conclude that.The converse of Pascal implies the points lie on a conic which in some cases may be a parabola or hyperbola etc.See this link for more details.
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Re: IMO 2012: Day 1 Problem 1

Unread post by nafistiham » Wed Jul 18, 2012 12:42 am

:oops: :oops: had to provide an image for having no geogebra
DSC01365.JPG
it's paint
DSC01365.JPG (294.22KiB)Viewed 6155 times
Let us join $A,O$ and, $F,G$

$\angle BOC= \frac {\angle B + \angle C} {2}$
$\angle COJ= \frac {\angle C} {2}$
$\angle FCO= \frac {\angle A + \angle C} {2}$
So, $\angle OFJ= \frac {\angle A} {2}= \angle JAO$
which means $AFOJ$ is cyclic
So, $\angle AFO= \frac {\pi} {2}= \angle AHO $
$\angle BHO= \frac {\pi} {2}$
$\angle ABF= \angle HBO$
$\angle FAB= \angle HOB= \angle BHI $
So, $AF$ is parallel to $GH$
similarly $AG$ is parallel to $FC$
$AFIG$ is a parallelogram
$BF$ is perpendicular to $AD$ and angle bisector of $\angle ABD$
So, $FD=AF=GI$
similarly, $GE=AG=FI$
$\angle DFI= \angle DAE= \angle IGE$
so, $\triangle FDI$ and $\triangle GIE$ are congruent
Summing up, $DI=IE$

for this solution, I am really grateful to Fahim Ferdous for the hint "Try Angle Chasing" ;)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: IMO 2012: Day 1 Problem 1

Unread post by Nadim Ul Abrar » Sat Jul 21, 2012 3:07 pm

Tahmid Hasan wrote:
Nadim Ul Abrar wrote:Used pascel to say $AGLJKF$ is cyclic :D .
I'm afraid you can't conclude that.The converse of Pascal implies the points lie on a conic which in some cases may be a parabola or hyperbola etc.See this link for more details.
Oh thank you :D
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