IMO(1995-1) Collinearity Of AM, DN and XY

Discussion on International Mathematical Olympiad (IMO)
User avatar
Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Location: Dhaka, Bangladesh.

IMO(1995-1) Collinearity Of AM, DN and XY

Unread post by Labib » Tue Jan 31, 2012 4:24 pm

Let $A, B, C,$ and $D$ be distinct points on a line, in that order. The circles
with diameters $AC$ and $BD$ intersect at $X$ and $Y$ . $O$ is an arbitrary point
on the line $XY$ but not on $AD$. $CO$ intersects the circle with diameter
$AC$ again at $M$, and $BO$ intersects the other circle again at $N$. Prove that
the lines $AM, DN,$ and $XY$ are concurrent.
Last edited by Labib on Tue Jan 31, 2012 11:40 pm, edited 1 time in total.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

User avatar
Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Location: Dhaka, Bangladesh.

Re: IMO(1995-1) Colliniarity Of AM, DN and XY

Unread post by Labib » Tue Jan 31, 2012 8:12 pm

Here's my solution::
Let me denote the circles with the diameters $AC$ and $BD$ as $\Gamma _1$ and $\Gamma _2$ respectively.
IMO 1995-1.png
Figure:: IMO(1995-1) Colliniarity Of AM, DN and XY
IMO 1995-1.png (381.3 KiB) Viewed 1482 times
As, $O$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$, so,
$BO.NO=CO.OM$ which implies that $BCNM$ is a cyclic quadrangle.
Let, $AM\cap DN = K$ and $AD\cap XY = G$ .
As, $\angle ONK=\angle OMK=90^{\circ}$.
$OMKN$ is cyclic as well.
$OK$ is the diameter.

Now, We need to prove that, $K,O,G$ are collinear.
As, $\Delta MKO \sim \Delta OCG$,
$\frac{MO}{GO}=\frac{KO}{CO}=\frac{MK}{CG}$
$\Rightarrow \frac{CG.MO}{MK.CO}=\frac{GO}{CO}$.

Now, $\Delta AGM \sim \Delta AKD$ as $\angle AGM= \angle AKD$.(As $MKDG$ is cyclic)
So, $\frac{AG}{AK}=\frac{AM}{AC}$.

But, $\Delta AMC \sim \Delta COG$.
So,$\frac{CO}{GO}=\frac{AC}{AM}$.

Therefore,
$\frac{AK}{AG}=\frac{CO}{GO}$.

So,
$\frac{AK.MO.CG}{AG.MK.CO}=1$.
Using the converse of Menelaus's theorem for $\Delta AMC$, we conclude that, $K,O,G$ are collinear.
[Proved]
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

User avatar
bristy1588
Posts: 92
Joined: Sun Jun 19, 2011 10:31 am

Re: IMO(1995-1) Collinearity Of AM, DN and XY

Unread post by bristy1588 » Wed Feb 01, 2012 12:00 am

my solution might be wrong, if it is, i request people to correct it:
I have denoted the point $O$ as $P$ here. Sorry for the inconvinience.

Let $T$ be the intersection of $AM$ and $XY$(extended). Now $TQ$ is the perpendicualr to $AD$.$Q$ is the foot of perpendicular from $T$ to $AD$. Now, $\angle TMC=\angle TQC=90^{\circ}.$ So, the pointd $T,M,Q,C$ are concyclic. so $\angle MTQ=\angle MCQ=\angle MCB$.
Using power of point:
$PM.PC=PX.PY=BP.BN$. Therefore, $B,C,M,N$ are concylic. Now, $\angle BCM=\angle BNM , so \angle MTP=\angle MTQ=\angle BCM= \angle BNM=\angle PNM$. Therefore, points $M,T,P,N$ are concyclic. So, $\angle PNT= 180^{\circ}-\angle PMT=90^{\circ}$. So $\angle DNP+\angle PNT=90^{\circ}+90^{\circ}=180^{\circ}$. So, $D,N,T$ are collinear, Therefore $T$ is also the point of intersection of $DN$ and $XY$.
Hopefully, proved
Bristy Sikder

Post Reply