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 $C$.)
IMO 2012: Day 1 Problem 1
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: IMO 2012: Day 1 Problem 1
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!!!
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: IMO 2012: Day 1 Problem 1
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.Nadim Ul Abrar wrote:Used pascel to say $AGLJKF$ is cyclic .
বড় ভালবাসি তোমায়,মা
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: IMO 2012: Day 1 Problem 1
had to provide an image for having no geogebra
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"
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"
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: IMO 2012: Day 1 Problem 1
Oh thank youTahmid Hasan wrote: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.Nadim Ul Abrar wrote:Used pascel to say $AGLJKF$ is cyclic .
$\frac{1}{0}$