Since $L$ and $M$ are the mid points of arc $FDE$ and $FE$ , hence $MX$ $MY$ , are their Simson lines ... it is well known that simson lines of 2 diametrically opposite points with respect to a triangle are perpendicular... The problem is overkilled ..

It is well known that $NE,NF$ are tangent to $(AEF).$ (EGMO lemma ) So $L,M,N$ are collinear and $LN \perp EF$ at $M.$ Thus, $LMFX$ is cyclic. So $\angle MLF=\angle MXF.$ Similarly, $\angle MNF=\angle MYF.$ Thus, $\angle XMY=\angle LFN=90^\circ$ and done. (here, $\angle LFN=90^\circ$ since $LN$ is a diameter of the nine-point circle.)

Someone good at LaTeX post P4. I don't know how to make arrays, and it's not even comfortable on phone

Math-wiz wrote: Someone good at LaTeX post P4. I don't know how to make arrays, and it's not even comfortable on phone You can use this https://webdemo.myscript.com/views/math/index.html#

thewitness wrote: Math-wiz wrote: Someone good at LaTeX post P4. I don't know how to make arrays, and it's not even comfortable on phone You can use this https://webdemo.myscript.com/views/math/index.html# But still it isn't comfortable in phone, plus in a moving car

I just proved $XM || BC$, $YM || AD$ by some angle chasing (a nice diagram helped in that observation ) hence done.

It's well known $NF,NE$ are tangent to $\odot(AFHE)$. Claim:-$L-M-N$ are collinear As $L,N$ are the centers of $\odot(AFHE)$ and $\odot(BFEC)$, So as $LN\perp EF$ as EF is the radical axis of $\odot(AFHE)$ and $\odot(BFEC)$, hence, $LN$ must pass through the midpoint of $EF$ hence, $L-M-N$ are collinear. So, $\angle BAC=\angle EFN=\angle MYN=\angle NDY\implies MY\perp BC\implies AH\|MY$. Now $\angle FDB=\angle FAE=\angle FLM=\angle MXY\implies XM\|BC\implies\angle XMY=90^\circ$. This was perhaps the easiest problem in today's RMO, @below, OH YES, I didn't notice P2.

no @above it cant defeat p1 and p2 in being easy.

If u guys ever did this synthetically(not sure if bashy techniques exist) then u will understand its beauty... First,$AL=LE=LH$, $\angle LEM = \angle LFN = 90$ since it is the diameter of the nine point circle. Now, LEFN cyclic and $NE=NF$(as midlines onto BC) so, $NM$ is angle bisector by angle bisector as $M$ is midpoint of $EF$. So, $L-M-N$ collinear, and, $\angle LEF=\angle MNF=\angle MYF=\angle XMF$ and after two more, you get $LE \mid \mid XM$ and $EN \mid \mid MY$ so, $\angle LEN =90 = \angle XMY$. Done!

Let $X'$, $Y'$ be the feet of altitude from $L $, $N $ onto the line $DE $. Since, $L $ is the circumcenter of $(AEF) $, and $LM \perp EF$. Applying Simson's theorem $MXX'$ is collinear. Similarly $MYY'$ is collinear. Only show that the Simson lines of $L$,$N $ are perpendicular. $L $, $N $ are antipodes of $(AEF) $. $X $, $Y $ are antipodes of $\triangle ABC $. Simson lines of $X $, $Y $ are pendicular. $\text {Another proof} $ $LMN $ is the Newton line of $\{AF,FH,HE,EA\} $. Apply Simson line of $L $ wrt $(DEF) $ and $N $ wrt $(DEF) $.

Anyways, a more detailed solution I suppose [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.912480306830471, xmax = 0.9180304974667128, ymin = -2.826658307249468, ymax = 7.789436798224485; /* image dimensions */ pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); pen ffwwzz = rgb(1,0.4,0.6); draw(arc((-3.9508270833034356,4.3439889081743965),0.3526941895506291,-89.198712669893,-57.439962575555136)--(-3.9508270833034356,4.3439889081743965)--cycle, linewidth(1.6) + ffwwzz); draw(arc((-5.225462047177262,3.5300779214655362),0.3526941895506291,-68.5590062233107,-36.80025612897283)--(-5.225462047177262,3.5300779214655362)--cycle, linewidth(1.6) + ffwwzz); draw(arc((-4.257400015216454,4.824101087976465),0.3526941895506291,-89.198712669893,-57.439962575555136)--(-4.257400015216454,4.824101087976465)--cycle, linewidth(1.6) + ffwwzz); 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If $X$ is a point on $AC$, such that $\angle MXA = 90^{\circ}$, then $LM$ and $LX$ are isogonal with respect to $\angle ALC.$ PROOF: Let $\angle BAC = 2A, \angle ABC = 2B, \angle ACB = 2C.$ Then not that by angle chasing, $\angle ALM = (A + 2B) - 90^{\circ}$ and $\angle XLM = 90^{\circ} - (A + 2C)$. Hence, the conclusion follows. $\blacksquare$ Applying this claim with reference triangle as $DEF,$ we obtain that $LF$ and $LD$ are isogonal with respect to $\angle XLM$. Hence, $$\angle XMF = \angle XLF = \angle DLM = \angle NFD = \angle NFY = \angle YMN.$$

Wow beautiful problem I'll prove the well known facts just for completeness. $\triangle AFE \sim \triangle ACB$ $\implies \angle FAM = \angle CAN$ (corresponding parts of similar triangles) $\implies AM$ and $AN$ are isogonal, in fact $AN$ is symmedian in $\triangle AFE$ $\implies NF = NE$ (they are tangents to $(AFE)$) $\implies NM \perp FE$ In fact $L-M-N$ are collinear since $LN$ is diameter of nine point circle and midpoint of $LN$ is the center of nine point circle $\in MN$ ($\triangle FNE$ is isosceles) $\implies \angle NMF = 90^\circ = \angle LXF \implies LXFM$ is cyclic. Also $\angle NMF = 90^\circ = \angle FYN \implies NMFY$ is cyclic. $\angle FXM = \angle FLM$ $\angle FLM = 90^\circ - FNM = 90^\circ - \angle FYM$ $\implies \angle FXM + \angle FYM = 90^\circ$ and we are done.

To Prove: $YLMF$ and $FMNX$ are concyclic. Notice that $M$ is midpoint of $FE$ and $\angle FMN=\angle LMF=90^{\circ}$ and because $YL$ and $XN$ are perpendicular to $XY$ we can say that $\angle LMF=90^\circ$ and $\angle FMN=90^\circ$ so $YLMF$ and $FMNX$ are concyclic. Note that $\triangle AFL$ and $\triangle NFC$ are \textbf{isosceles} so we have $$\angle FAL = \angle AFL =\angle NFC = \angle NCF=90^\circ-\angle ABC$$Also, $$\angle LFN=\angle LFC + 90^\circ-\angle ABC=\angle LFC+\angle LFA=90^\circ$$So we have, $$\angle YXM=\angle LNF$$and $$\angle FYM=\angle MLF \Rightarrow \angle YMX=90^\circ$$

Lol I remember being stuck on this for months when I started olympiads. Today basically headsolved when I looked at it again after having forgotten my original solution. Note that the problem can be restated entirely in terms of $\Delta DEF$ which allows us to delete $\Delta ABC$ from the picture and construct a non-convoluted diagram. Restated Problem wrote: In $\Delta ABC$, $N_a$ and $M_a$ are midpoints of arc $BC$ containing and not containing $A$ respectively. Their perpendiculars to $AC$ are $Y$ and$X$ respectively. Prove that the midpoint $M$ of $BC$ lies on the circle with diameter $XY$. Now angle chasing with cyclic quadrilaterals $AN_aCM$ and $CXMM_a$ gives us that $AN_a$ and $XM$ are parallel which is obviously enough to conclude.

Lol, this is better than INMO ‘19 P5

We have $\angle{XMY} = \angle{XMF} + \angle{FMY} = \angle{XLF} + \angle{FNY} = \angle{XLN} - \angle{FLN} + \angle{YNL} - \angle{FNL} = 1880^{\circ} - 90^{\circ} = 90^{\circ}$

Aryan-23 wrote: Since $L$ and $M$ are the mid points of arc $FDE$ and $FE$ , hence $MX$ $MY$ , are their Simson lines ... it is well known that simson lines of 2 diametrically opposite points with respect to a triangle are perpendicular... The problem is overkilled .. I think their is a typo can you please rewrite it

Let the line through $E$ parallel to $BC$ cut $DF$ at $G$ and let the line through $E$ parallel to $AH$ cut $DF$ at $I$. Then$$\measuredangle FGE=\measuredangle FDB=\measuredangle FAC=\measuredangle FAE$$and$$\measuredangle EIF=\measuredangle HDF=\measuredangle HBF=\measuredangle EBF,$$so $AEHFG$ and $BFECI$ are cyclic. Since $L$ is the circumcentre of $AEHFG$ and $N$ is the circumcentre of $BFECI$, we get that $X$ is the midpoint of $FG$ and $Y$ is the midpoint of $FI$. Then $XM\parallel EG\parallel BC$ and $MY\parallel EI\parallel AH$, so $XM\perp MY$, as desired.