IMO 2013, Day 2-P4

[Masum]

Let $\triangle ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes drawn from $B$ and $C$, respectively. $\omega_1$ is the circumcircle of $\triangle BWN$, and $X$ is a point such that $WX$ is a diameter of $\omega_1$. Similarly, $\omega_2$ is the circumcircle of $\triangle CWM$, and $Y$ is a point such that $WY$ is a diameter of $\omega_2$. Show that the points $X, Y$, and $H$ are collinear.

[SANZEED]

Let $V=\omega_1\cap\omega_2$. Now according to the definitions of $X,Y, \angle XVW=\angle YVW=90^{\circ}$ and thus $V\in XY$. Notice that $W,M,N$ are three points on the sides $BC,CA,AB$ of $\triangle ABC$, and $\bigodot BWN$ and $\bigodot CWM$ pass through $V$. So $V$ is the Miquel point. So, $M\in \bigodot AMN$. Also $H\in \bigodot AMN$ since $H$ is the orthocenter. Thus $\angle AVH=\angle AMH=90^{\circ}$. this means that $H\in XV$ i.e. $H,X,Y$ are collinear.

[asif e elahi]

Let $AH$ meets $BC$ at $L$. So $AL\perp BC$
Let $W$ lies between $L$ and $C$
We have $\measuredangle XBW=\measuredangle XNW=90$. So $\measuredangle XNB=\measuredangle =90-\measuredangle BNW=\measuredangle CNW$
$\measuredangle XBN=90-\measuredangle B=\measuredangle NCW$
So $\triangle XBN$ and $\triangle NWC$ are similar

SO $\displaystyle \frac{BX}{BN}=\frac{CW}{CN},BX=\frac{CW.BN}{CN}=CW\frac{BL}{AL}$

Similarly $CY=BW\dfrac{CL}{AL}$
Again as $\triangle CHL$ and $\triangle ABL$ are similar

$\displaystyle \frac{HL}{CL}=\frac{BL}{AL},HL=CL$. $\dfrac{BL}{AL}$ as we assumed that $CW<CL$
so $CW\cdot \frac{BL}{AL}<CL\cdot \dfrac{BL}{AL}$
so $BX<HL$. Similarly as $BW>BL,CY>HL$
Let $XP\perp HL$ and $HQ\perp CY$. Now $XBLP$ and $HLCQ$ are rectangle

so $\displaystyle \frac{PH}{XP}=\frac{HL-BX}{BL} =\frac{HL}{BL}-\frac{BX}{BL}=\frac{CL}{AL}-\frac{CW.BL}{AL.BL}$

$\displaystyle =\frac{CL}{AL}-\frac{CW}{AL}=\frac{LW}{AL}$. So $\triangle XPH$ and $\triangle ALW$ are similar.

similarly $\triangle ALW$ and $\triangle HQY$ are similar. So $\triangle XPH$ and $\triangle HQY$ are similar
so $\measuredangle HYQ=\measuredangle XHP$
$\measuredangle XHY=\measuredangle XHP+90+\measuredangle YHQ$
$=\measuredangle HYQ+90+\measuredangle YHQ=180$
so $X,H,Y$ are collinear.

Notes for the poster:
1. Never make such a mess again using so much dollar signs. Only put a \$ at the beginning and put another at the end. Also when you write a long mathematical line, use $\backslash [...\backslash ]$ instead of \$...\$.

2. If you don't like large latex fonts, then you can change its size. On the bottom-right corner of the screen there is a button named jsMath. Go to "jsMath>Options>Chang math font size" to change it.

3. Use \triangle instead of \Delta, \dfrac instead of \frac.