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Source: RMO 2019 P5

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MathPassionForever

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MathPassionForever

#1 h Oct 20, 2019, 10:54 AM • 6 Y

Y by Some_one_, centslordm, HWenslawski, math_comb01, Adventure10, Rounak_iitr

In an acute angled triangle $ABC$ , let $H$ be the orthocenter, and let $D,E,F$ be the feet of altitudes from $A,B,C$ to the opposite sides, respectively. Let $L,M,N$ be the midpoints of the segments $AH, EF, BC$ respectively. Let $X,Y$ be the feet of altitudes from $L,N$ on to the line $DF$ respectively. Prove that $XM$ is perpendicular to $MY$ .

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#2 h Oct 20, 2019, 10:59 AM • 6 Y

Y by Pluto1708, Some_one_, centslordm, HWenslawski, Entrepreneur, Adventure10

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 ..

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#3 h Oct 20, 2019, 11:03 AM • 6 Y

Y by Aryan-23, Some_one_, AmirKhusrau, Purple_Planet, centslordm, Adventure10

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.)

This post has been edited 2 times. Last edited by Wizard_32, Oct 20, 2019, 11:05 AM

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#4 h Oct 20, 2019, 11:08 AM • 4 Y

Y by Some_one_, centslordm, Adventure10, Mango247

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

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#5 h Oct 20, 2019, 11:10 AM • 3 Y

Y by Some_one_, The_Maitreyo1, Adventure10

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#

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#6 h Oct 20, 2019, 11:16 AM • 3 Y

Y by Some_one_, centslordm, Adventure10

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

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#7 h Oct 20, 2019, 11:18 AM • 4 Y

Y by Some_one_, AmirKhusrau, centslordm, Adventure10

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

) hence done.

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#8 h Oct 20, 2019, 11:47 AM • 6 Y

Y by Some_one_, Pakistan, AmirKhusrau, Purple_Planet, centslordm, Adventure10

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.

This post has been edited 6 times. Last edited by amar_04, Nov 12, 2019, 7:45 PM

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#9 h Oct 20, 2019, 11:48 AM • 3 Y

Y by Some_one_, AmirKhusrau, Adventure10

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

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#10 h Oct 21, 2019, 5:23 AM • 4 Y

Y by Some_one_, centslordm, Adventure10, Mango247

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!

This post has been edited 1 time. Last edited by o_i-SNAKE-i_o, Oct 21, 2019, 5:24 AM

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#12 h Oct 31, 2019, 8:18 AM • 3 Y

Y by AmirKhusrau, centslordm, Adventure10

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 $(DEF)$ .

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Dr_Vex

#13 h Jul 20, 2020, 9:34 PM • 2 Y

Y by centslordm, Rounak_iitr

Anyways, a more detailed solution I suppose
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CLAIM: Let $I$ be the incenter of $\Delta ABC$ and $M$ be the midpoint of $BC.$ $AI$ intersects $\odot(ABC)$ at $L$ . 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, $\angle XMF = \angle XLF = \angle DLM = \angle NFD = \angle NFY = \angle YMN.$

This post has been edited 1 time. Last edited by Dr_Vex, Sep 25, 2020, 5:45 AM

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#15 h May 5, 2021, 9:25 AM • 2 Y

Y by centslordm, PRMOisTheHardestExam

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.

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#17 h Sep 15, 2021, 9:20 AM

Y by

Very nice problem

I think my solution is a little different. Solution

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L13832

#18 h Jul 5, 2024, 11:29 AM • 1 Y

Y by CRT_07

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 $\angle FYM=\angle MLF \Rightarrow \angle YMX=90^\circ$

This post has been edited 3 times. Last edited by L13832, Sep 13, 2024, 6:57 PM

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Eka01

#19 h Sep 13, 2024, 7:23 AM • 1 Y

Y by alexanderhamilton124

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.

This post has been edited 1 time. Last edited by Eka01, Sep 13, 2024, 7:24 AM

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L13832

#20 h Sep 13, 2024, 6:57 PM

Y by

Lol, this is better than INMO ‘19 P5

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#21 h Sep 14, 2024, 1:35 PM

Y by

We have $\angle{XMY} = \angle{XMF} + \angle{FMY} = \angle{XLF} + \angle{FNY} = \angle{XLN} - \angle{FLN} + \angle{YNL} - \angle{FNL} = 180^{\circ} - 90^{\circ} = 90^{\circ}$

This post has been edited 1 time. Last edited by khanhnx, Feb 21, 2025, 4:17 AM

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ANBB

#22 h Sep 20, 2024, 1:24 PM

Y by

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

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Ianis

#23 h Sep 20, 2024, 2:00 PM

Y by

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.

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