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Source: Iranian TST 2018, first exam day 2, problem 4

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164 posts

#1 h Apr 8, 2018, 11:00 AM • 7 Y

Y by mathenthusiastic, Mathuzb, itslumi, tiendung2006, Adventure10, Mango247, sami1618

Let $ABC$ be a triangle ( $\angle A\neq 90^\circ$ ). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$ . Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$ . Prove that $AP$ passes through the midpoint of $BC$ .

Proposed by Iman Maghsoudi, Hooman Fattahi

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157 posts

#2 h Apr 8, 2018, 11:26 AM • 3 Y

Y by Adventure10, Mango247, sami1618

Hey mate, could you pm me where i can find all tst iranian problems?

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1979 posts

#3 h Apr 8, 2018, 12:58 PM • 5 Y

Y by GGPiku, Wizard_32, Adventure10, Mango247, sami1618

Dadgarnia wrote:

Let $ABC$ be a triangle ( $\angle A\neq 90^\circ$ ). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$ . Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$ . Prove that $AP$ passes through the midpoint of $BC$ .

Proposed by Iman Maghsoudi, Hooman Fattahi

WLOG $AB<AC$ . Let $T=\overline{EF} \cap \overline{BC}$ ; $H=\overline{BE} \cap \overline{CF}$ and $X$ be the $A$ -HM point in $\triangle ABC$ . Then we claim that $X$ lies on $\odot(TMN)$ . Observe that ray $\overrightarrow{XM}$ bisects the arc $EAF$ of $\odot(AEF)$ since $\overline{AX}$ is a symmedian in $\triangle AEF$ and $X \in \odot(AEF)$ . Thus, $\angle FXM=90^{\circ}-\tfrac{1}{2}\angle A$ . Now $\tfrac{XB}{XC}=\tfrac{NB}{NC}$ hence $\angle BXN=90^{\circ}-\tfrac{1}{2}\angle A$ . Finally, $\angle BXF=\angle BXH+\angle FXH=\angle BCH+\angle FAH=180^{\circ}-2\angle B$ . Hence $\angle MXN=180^{\circ}-(\angle B-\angle C)$ so $X$ lies on $\odot(TMN)$ as desired.

Observe that $\angle PMN=\angle PNM=\tfrac{1}{2}(\angle B-\angle C)$ hence $P$ is the antipode of $T$ in $\odot(TMN)$ . Thus, $\overline{TX} \perp \overline{XP}$ . However $X$ lies on the $A$ -median which is also perpendicular to line $\overline{TX}$ ; thus $\overline{AP}$ bisects the side $\overline{BC}$ .

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698 posts

#4 h Apr 8, 2018, 1:13 PM • 3 Y

Y by Adventure10, Mango247, sami1618

Let $H_A$ be the A-Humpty point and $K$ be the second intersection of the circumcircle and the A-symmedian. Let $l$ be the perpendicular from $N$ to $BC$ . Note that $l$ is tangent to the A-Apollonius circle, which also happens to be the circumcircle of $AH_ANK$ . Let $AH_A\cap l=\{V\}$ and $AK\cap l=\{ W\}$ . Note that it is enough to prove that $AEF\cup\{V\}\sim ABC\cup\{W\}$ . Now we will use the well-know fact that $NH_A=NK$ . $\frac{AV}{AH_A}=\left(\frac{AN}{NH_A}\right)^2=\left(\frac{AN}{NK}\right)^2=\frac{AW}{AK} \implies AEFH_AV\stackrel{-}{\sim}ABCKW \ \ \ \blacksquare$

This post has been edited 2 times. Last edited by rmtf1111, Apr 8, 2018, 5:20 PM

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402 posts

GGPiku

#5 h Apr 8, 2018, 1:43 PM • 2 Y

Y by Adventure10, sami1618

Longer but we're gonna use "The old Switcharoo". Let $AB<AC$ and $X$ be the midpoint of $BC$ . Let $X_A$ be the $C(AEF)\cap C(BHC)$ point. $M,N$ the intersections of the angle bisector of $BAC$ and $BC,EF$ . $EF\cap BC=J$ . Then $X_A,M,J,N$ are conyclic.
$PROOF:$
We know that $J,H,X_A$ are collinear.
Claim 1
Claim 2
Now $JNX_A=90+\frac{A}{2}-\angle NFX_A=90+\frac{A}{2}-\angle EHX_A$ and $JMX_A=90+\frac{A}{2}-\angle MBX_A=-90+\frac{A}{2}+\angle FHX_A$ . Summing and using the fact that $\angle FHE=180-A$ , we get the conclusion. Now let $C(JNX_AM)$ intersect $AX$ in $P$ . Since
$JX_AP=90$ , we get that $JP$ is the diameter of the circle $JNX_AM$ . So $\angle JNP=\angle PMJ=90$ . But the perpendicular in $N$ on $EF$ and the perpendicular in $M$ on $BC$ have an unique point of intersection, so our $P$ is the required $P$ in the original problem and the conclusion follows.

This post has been edited 7 times. Last edited by GGPiku, Apr 15, 2018, 10:20 AM

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1475 posts

math90

#6 h Apr 8, 2018, 3:38 PM • 3 Y

Y by Adventure10, Mango247, sami1618

Where are problem 1 and problem 6?

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4233 posts

#7 h Apr 8, 2018, 4:01 PM • 3 Y

Y by Adventure10, Mango247, sami1618

hmm probably nothing new but

Let $H$ be the orthocenter of $ABC$ . Let $G$ be the intersection of $EF$ and $BC$ and let $X$ be the $A$ -HM point in $ABC$ . We will show that $GNXM$ is cyclic.

Because $AX$ is a symmedian in $AEF$ , $XM$ passes through the midpoint of arc $EAF$ in $(AEF)$ ; denote that by $Y$ . Then $\angle{GXM} = \angle{HXY} = \frac{1}{2}(\widehat{FY} + \widehat{FH})$ . But $\frac{1}{2}\widehat{FY} = 90 - \frac{1}{2}\angle{A}$ and $\frac{1}{2}\widehat{FH} = 90 - \angle{B}$ , so therefore $\angle{GXM} = 180 - \frac{1}{2}\angle{A} - \angle{B} = \angle{BNA}$ . Then $\angle{GXM} = \angle{GNM}$ so $GNXM$ is cyclic.

Next, it's pretty clear that $P$ is the antipode of $G$ on $(GMN)$ , so $GNPXM$ is cyclic. It follows that $\angle{GXP} = \angle{AXH} = 90$ and we're done.

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2312 posts

#8 h Apr 8, 2018, 4:20 PM • 12 Y

Y by doxuanlong15052000, naw.ngs, MeineMeinung, mhq, Durjoy1729, fastlikearabbit, nguyendangkhoa17112003, char2539, enhanced, Adventure10, Mango247, sami1618

Let $T$ be the midpoint of $BC$ and let $I, J$ be the midpoint of arc $EF,$ arc $BC$ in $\odot (AEF), \odot (ABC),$ respectively. Note that $\triangle ABC \cup J \cup N \stackrel{-}{\sim} \triangle AEF \cup I \cup M,$ so $\frac{AI}{AM} = \frac{AJ}{AN}\ (\star).$ On the other hand, $TE, TF$ are tangent to $\odot (AEF),$ so by $( \star )$ we get $\triangle TIJ$ and $\triangle PMN$ are homothetic with center $A,$ hence we conclude that $T\in AP.$

This post has been edited 1 time. Last edited by TelvCohl, Apr 8, 2018, 10:14 PM

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412 posts

Yaghi

#9 h Apr 8, 2018, 4:35 PM • 3 Y

Y by Adventure10, Mango247, sami1618

My solution:
We begin with two well-known lemmas:
Let $K,L$ be the foot of perpendiculars from $N$ to $AB,AC$ and let $l$ be the line through $N$ perpendicular to $BC$ ( $N$ is the foot of angle bisector on $BC$ ),also,let $EF$ cut $BC$ at $X$ .then:
1- $X,K,L$ are collinear.
2- $l \cap KL$ is on the median of $A$ .
1) is easy since both $K,L$ are the foot of angle bisector of $X$ in $XFB,XEC$ .
2) is well-known and true for any point on the angle bisector of $A$ .proof is just law of sinus.
Now we prove the problem.Obviously, $X,N,P,M$ are concyclic and $\angle PXN=\angle PNM=\frac{B-C}{2} =\angle PXN$ so $XP$ is the angle bisector of $\angle MXN$ and this means that $P,K,L$ are collinear,so we are done by lemma 2.

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148 posts

#11 h Apr 9, 2018, 3:43 AM • 2 Y

Y by Adventure10, sami1618

Ohh wow I don't know how I got this solution but I got it anyways:

Let $EF \cap BC = D$ . It id easy to see that $MND$ is isosceles, and since $MNPD$ is cyclic with $P$ the antipode of $D$ , we get that MPD is actually a right kite, and so, it is a harmonic quadrilateral. Let $AP \cap (MNDP) = X$ , $AD \cap (MNDP) = Y$ . It is easy to see from the properties of cyclic harmonic quadrilaterals that $XMNY$ is also harmonic (say, after an inversion at $A$ ). Also, if $H$ is the orthocenter of the triangle, then we can see that $D(A,H; E, F) = -1 = D(X,M;N,Y)$ , which implies that $AP$ and $DH$ intersect at $X$ , or equivalently, they intersect at right angles, since $PD$ is a diameter of $MNPD$ . But then this implies that $AP$ passes through the midpoint of BC, as is can be proven that $H$ is the orthocenter of $APK$ , where $K$ is the midpoint of BC.

This post has been edited 1 time. Last edited by ManuelKahayon, May 28, 2018, 10:38 AM
Reason: (Whoops, typo)

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3070 posts

#12 h Apr 22, 2018, 3:30 AM • 3 Y

Y by Adventure10, Mango247, sami1618

$[asy]size(7cm); pair A=(1.5,7),B=(0,0),C=(10,0),Ee,F,M,Nn,P,X,H,Xa,D; Ee=foot(B,A,C);F=foot(C,A,B); M=extension(Ee,F,A,bisectorpoint(B,A,C)); Nn=extension(B,C,A,bisectorpoint(B,A,C)); P=extension(Nn,Nn+(B-Nn)*(0,1),M,M+(Ee-M)*(0,1)); H=extension(B,Ee,C,F);D=(B+C)/2; Xa=foot(H,A,D);X=extension(Ee,F,B,C); draw(circumcircle(X,M,Nn)^^circumcircle(A,Ee,F),green); D(MP("A",A,N)--MP("F",F,W)--MP("B",B,S)--MP("N",Nn,S)--MP("D",D,S)--MP("C",C,S)--MP("E",Ee,NE)--A--MP("M",M,NW)--F--MP("H",H,S)--C); D(B--H--Ee--M--Nn--MP("P",P,E)--M); D(A--MP("X_A",Xa,NE)--P--D,dotted); D(F--MP("X",X,S)--B); dot(A^^B^^C^^Ee^^F^^M^^Nn^^P^^Xa^^H^^D^^X); [/asy]$
Let $X=EF\cap BC$ and $X_A$ the HM-point w.r.t $A$ in $\triangle ABC$ . Clearly $P\in \odot (XMN)$ , and since $X_A$ is the center of the spiral similarity sending $BNC\mapsto EMF$ , we have $X_A\in\odot(XMN)$ as well. It's well-known that $X_A$ lies on the $A-$ Apollonius circle, therefore $\tfrac{XE}{XF}=\tfrac{XB}{XC}=\tfrac{AB}{AC}=\tfrac{AE}{AF}=\tfrac{EM}{FM}$ , so $X_AM$ bisects $\angle FX_AE$ .

We claim that $A,X_A,P$ are collinear. Assume the configuration shown above. We have $\angle MX_AA=\angle MX_AE-\angle AX_AE=\tfrac12 \angle FX_AE-\angle AFE=90^\circ-\tfrac12 \angle A-\angle C$ and also $\angle MX_AP=180^\circ-\angle MNC=180^\circ -\left(\tfrac12\angle A+\angle B-90^\circ\right)=90^\circ+\tfrac12 \angle A+\angle C$ which proves our claim. But since $AX_A$ passes through $D$ , the midpoint of $BC$ , the result is obvious. $\blacksquare$

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nguyenvanthien63

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nguyenvanthien63

#13 h Apr 22, 2018, 9:26 AM • 3 Y

Y by Adventure10, Mango247, sami1618

What is HM_point.?

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118 posts

Wictro

#14 h Apr 22, 2018, 1:32 PM • 3 Y

Y by Adventure10, Mango247, sami1618

nguyenvanthien63 wrote:

What is HM_point.?

Humpty point

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9775 posts

jayme

#15 h Apr 22, 2018, 2:09 PM • 2 Y

Y by Adventure10, sami1618

Dear Mathlinkers,

1. Reim theorem help for proving that A, Xa and P collinear
2. the tangents to (AEF) at E and F go through D (well known)
3. and we are done with the Boutin's theorem...

Sincerely
Jean-Louis

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tuannghia18.09MNP

24 posts

tuannghia18.09MNP

#16 h May 16, 2018, 1:51 PM • 6 Y

Y by mautilus1, luizp, nguyendangkhoa17112003, rashah76, Adventure10, sami1618

Let $O$ be the intersection of $NP$ and the line through $A$ and perpendicular to $EF$ , $(O,OA)$ cuts $CA,AB$ at $K,L$ . Since $AO$ is perpendicular to $EF$ , $KL$ is parallel to $BC$ .
First, $\triangle AEF \sim \triangle ABC$ then $\frac {AM} {AN}= \frac {AE} {AB}$ .
Then, by Thales theorem, $\frac {OP}{ON}=\frac {AM}{AN}=\frac {AE}{AB}$ .
This leads to $\triangle OPL \sim \triangle AEB (s.a.s)$ .
So $\angle OPL=90^\circ$ , or $P$ is the mid-point of $KL$ .
Then, by Thales theorem, $AP$ bisects $BC$ .

Attachments:

This post has been edited 1 time. Last edited by tuannghia18.09MNP, May 16, 2018, 1:57 PM
Reason: .

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851 posts

#17 h May 19, 2018, 2:33 AM • 2 Y

Y by Adventure10, sami1618

Define $Z\equiv \overline{EF} \cap \overline{BC}$ ; since $\overline{EF}$ and $\overline{BC}$ are antiparallel we obtain that $\angle ZMN = \angle ZNM$ . In particular, $\triangle ZMN$ is isosceles, and since $P$ is the antipode of $Z$ in $\odot(ZMN)$ , we also obtain that $PM = PN$ . Additionally, define $\ell \perp \overline{AP}$ with $A \in \ell$ , $X \equiv \ell \cap \overline{EF}$ and $Y \equiv \ell \cap \overline{BC}$ ; notice that that $P \in \odot(XYZ)$ via Simson's theorem, and so by spiral similarity we obtain that $PX = PY$ and hence $AX = AY$ .

The line $\ell$ through $A$ has the property that its intersections with $\overline{BC}$ and $\overline{EF}$ are equidistant to $A$ ; such a line is unique by construction, and so it suffices to show that the line through $A$ perpendicular to the $A$ -median satisfies this property. Let $S$ be the midpoint of $\overline{BC}$ , $T\equiv AS\cap EF$ , and $X_A$ the $A$ -HM point of $\triangle ABC$ ; via an inversion about $A$ with radius $\sqrt{AF \cdot AB}$ , we see that $(TS; AX_A)$ forms a harmonic bundle, and so projecting through $Z$ yields the desired conclusion.

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475 posts

#18 h May 20, 2018, 6:21 AM • 5 Y

Y by Ankoganit, H.HAFEZI2000, hutu683, Adventure10, sami1618

I know there are many nice properties in this diagram, but I wonder if my solution fails or not, since it's too short...

Take the midpoint of $BC$ say $T$ , and consider $P_1, P_2$ be the points on $AT$ s.t $NP_1\perp BC, MP_2\perp EF$ . Then take $N', M'$ be midpoint of arcs opposite to $A$ in $(ABC), (AEF)$ , then $\frac{AP_1}{AT}=\frac{AN}{AN'}=\frac{AM}{AM'}=\frac{AP_2}{AT}$ So $P_1=P_2$

Doesn't this work?

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102 posts

#19 h May 26, 2018, 1:44 PM • 3 Y

Y by Adventure10, Mango247, sami1618

For an arbitrary point $X$ , let $Y$ and $Z$ be its projections onto $BC$ and $EF$ respectively. Consider the locus of $X$ such that $\dfrac{BY}{CY}=\dfrac{EZ}{FZ}$ . By standard linearity argument, it’s a line. Finally, it’s easy to verify that $A$ , $P$ and the midpoint of $BC$ belong to it, so we’re done.

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25 posts

jozuch

#21 h Mar 26, 2019, 2:22 AM • 2 Y

Y by Adventure10, sami1618

new solution:
Let $T=EF \cap BC$ , $S$ the midpoint of $BC$
$K, L$ the intersection point of the line passing through $A$ perpendicular to $AS$ with $BC, EF$ resp.
By Butterfly we have $AL=AK$ . Easy angle chase to see $\triangle TMN$ isosceles. So redefine $P$ as the intersection point of the angle bisector of $\angle MTN$ (denote by $\gamma$ ) with $AS$
then we have $P \in (LKT)$ . so by simson we get that the projections of $P$ on line $BC, EF$ and $A$ collinear (denote by $\ell$ this line)
and its easy to see $\ell$ is perpendicular to $\gamma$ thus not hard to see it concides with $A-$ angle bisctor
so $P$ concides with the original $P$ , done.

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1618 posts

#22 h Mar 26, 2019, 3:04 AM • 2 Y

Y by Adventure10, sami1618

Here is my solution for this problem
Solution
Let $Q$ be $A$ - Humpty point of $\triangle$ $ABC$ ; $G$ $\equiv$ $EF$ $\cap$ $BC$ ; $H$ be orthocenter of $\triangle$ $ABC$
We have: $\dfrac{QE}{QF}$ = $\dfrac{AE}{AF}$ = $\dfrac{ME}{MF}$
Then: $QM$ is internal bisector of $\widehat{EQF}$
So: $\widehat{MQG}$ = $\widehat{MQF}$ + $\widehat{FQH}$ = $\widehat{MQF}$ + $\widehat{FQH}$ = $180^o$ $-$ $\widehat{BAN}$ $-$ $\widehat{ABC}$ = $\widehat{MNG}$ or $Q$ $\in$ ( $MGNP$ )
Hence: $\widehat{GQP}$ = $90^o$ or $A$ , $Q$ , $P$ are collinear
Therefore: $AP$ passes through midpoint of $BC$

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2125 posts

#24 h Jun 6, 2019, 1:55 PM • 3 Y

Y by Adventure10, Mango247, sami1618

Let $G$ be midpoint of $\overline{BC}$ . Let $EF \cap BC=D$ and $AG \cap \odot (AFHE)=L$ $\implies$ $L$ is $A-$ humpty point $\implies$ $D-H-L$ . $AL$ is the $A-$ symmedian WRT $\Delta AFE$ and $H$ is $A-$ antipode in $\odot (AFE)$ , Let $Q$ be midpoint of minor arc $EF$ in $\odot (AFE)$ . Let $ML \cap \odot (AFE)$ $=$ $R$ , then $ARHQ$ is rectangle
$\angle LDN=\angle HAL=\angle HRL=\angle NML$ $\implies$ $MDLN$ is cyclic. Let $\odot (MDLN)$ $\cap$ $AG$ $=$ $P'$
$\implies \angle DLP'=90^{\circ}=\angle DMP'=\angle DNP' \implies P' \equiv P$

This post has been edited 3 times. Last edited by AlastorMoody, Jun 6, 2019, 4:15 PM

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924 posts

#25 h Nov 30, 2019, 1:44 PM • 4 Y

Y by amar_04, AlastorMoody, Adventure10, sami1618

Beautiful. Here goes my solution.

Iranian TST 2018 ,first exam day 2, problem 4 wrote:

Let $ABC$ be a triangle ( $\angle A\neq 90^\circ$ ). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$ . Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$ . Prove that $AP$ passes through the midpoint of $BC$ .

Proposed by Iman Maghsoudi, Hooman Fattahi

Solution:Let $I=EF \cap BC$ and $H_A$ be the $A-\text{humpty}$ point.
Since $\angle IMN=\angle C+\frac{\angle A}{2}=\angle MNI\implies IM=IN$ . Now so we can have a circle centred at $P$ with radius $PM=PN$ . Call this $\gamma$ . We perform an inversion about $\gamma$ . Clearly we have that $IP \perp MN \implies I \overset{\gamma}{\mapsto} IP \cap MN=T$ . Its well known that $IH_A \perp AH_A \implies AH_ATI$ concyclic.Also since the images of $I$ remains on $\odot(AH_ATI) \implies \odot(AH_ATI)$ and $\gamma$ are orthogonal. Let $S$ be the midpoint of $AI$ .Let $P_1=AN \cap KL$ where $\gamma \cap \odot(AH_ATI)=K,L$ . Clearly $P_1$ lies on polar of $I$ and $S$ w.r.t $\gamma \implies AI$ is the polar of $P_1 \implies PP_1 \perp AI \implies P_1$ is the orthocentre w.r.t $\Delta API \implies IP_1 \perp AP$ . Let $K=IP_1 \cap AP$ . Since $IH_A \perp A-\text{median}$ we can easily notice $K \equiv H_A$ . Done $\blacksquare$ .

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1740 posts

#26 h Dec 20, 2019, 7:56 AM • 5 Y

Y by char2539, SPACE123JK, Adventure10, sami1618, MS_asdfgzxcvb

Cool problem.
$[asy] size(6cm); defaultpen(fontsize(10pt)); pair A,B,C,O,EE,F,NN,M,SS,T,P,H; A=dir(120); B=dir(210); C=dir(330); O=(B+C)/2; EE=foot(B,C,A); F=foot(C,A,B); NN=extension(A,incenter(A,B,C),B,C); M=extension(A,NN,EE,F); SS=extension(B,C,EE,F); T=foot(SS,A,O); P=2*circumcenter(SS,M,NN)-SS; H=A+B+C; draw(circumcircle(SS,M,NN)); draw(A--NN,gray); draw(B--EE,gray); draw(C--F,gray); draw(circumcircle(A,EE,F),gray); draw(arc(reflect(B,C)*(0,0),1,-10,190),gray); draw(B--A--C--SS); draw(SS--EE); draw(A--O); dot("$A$",A,N); dot("$B$",B,SE); dot("$C$",C,E); dot("$O$",O,S); dot("$E$",EE,NE); dot("$F$",F,dir(110)); dot("$N$",NN,dir(220)); dot("$M$",M,dir(250)); dot("$S$",SS,dir(210)); dot("$T$",T,dir(75)); dot("$P$",P,E); dot("$H$",H,S); [/asy]$
Let $O$ be the midpoint of $\overline{BC}$ , let $H$ be the orthocenter, let $T$ be the $A$ -Humpty point, and denote $S=\overline{BC}\cap\overline{EF}$ . By construction, $T\in\overline{AO}$ and $\overline{AO}\perp\overline{ST}$ .

Since $T$ lies on $(HEF)$ and $(HBC)$ , $T$ is the Miquel point of $BCFE$ . Let $\Psi$ be the spiral similarity at $T$ sending $\overline{BC}$ to $\overline{EF}$ . Note that $\frac{NB}{NC}=\frac{AB}{AC}=\frac{AE}{AF}=\frac{ME}{MF},$ so $\Psi$ sends $N$ to $M$ . It follows that $SMTN$ is cyclic, say with circumcircle $\Gamma$ , but by definition, $\overline{SP}$ is a diameter of $\Gamma$ . Thus $\angle STP=90^\circ$ , so $P$ lies on $\overline{ATO}$ , as desired.

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6871 posts

#27 h Feb 18, 2020, 1:50 AM • 11 Y

Y by NJOY, stroller, fukano_2, Hamper.r, Kanep, v4913, CrazyMathMan, MeowX2, Adventure10, sami1618, MS_asdfgzxcvb

Solution with Aahan, Aatman Supkar, Aibek, Anshul, Archit, Arindam, Ethan Liu, G, Grant Yu, Maximus Lu, Naruto. D. Luffy, Paul Hamrick, R. Correaa , Rohan Goyal, yuanfeng:

Claim: $PM = PN$ .

Proof. Since $\angle MAE = \angle NAB$ and $\angle AEM = \angle AEF = \angle ABC = \angle ABN$ , it follows $\angle AME = \angle ANB$ . Thus $\triangle PMN$ is isosceles. $\blacksquare$

Let $T = \overline{EF} \cap \overline{BC}$ and let $\overline{AD}$ be an altitude. Thus $PMTN$ is a kite inscribed in some circle $\gamma$ with diameter $\overline{PT}$ . Let $Q$ denote the foot from $T$ to $\overline{AP}$ (also on $\gamma$ ).

$[asy]size(8cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair E = foot(B, C, A); pair F = foot(C, A, B); pair I = incenter(A, B, C); pair M = extension(E, F, A, I); pair N = extension(B, C, A, I); pair Z = N+dir(270); pair P = extension(A, midpoint(B--C), N, Z); draw(M--P--N, red); filldraw(unitcircle, invisible, blue); draw(A--B--C--cycle, blue); draw(A--P, blue); draw(A--N, blue); draw(E--F, blue); pair T = extension(E, F, B, C); draw(F--T--B, lightblue); draw(circumcircle(P, M, N), deepgreen); pair Q = foot(T, A, midpoint(B--C)); draw(T--Q, deepgreen); pair D = foot(A, B, C); pair S = extension(Q, T, M, N); draw(A--D, blue); draw(B--E, blue+dotted); draw(C--F, blue+dotted); pair X = extension(E, F, A, D); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(45)); dot("$F$", F, dir(F)); dot("$M$", M, dir(190)); dot("$N$", N, dir(N)); dot("$P$", P, dir(45)); dot("$T$", T, dir(T)); dot("$Q$", Q, dir(45)); dot("$D$", D, dir(D)); dot("$S$", S, dir(260)); dot("$X$", X, dir(X)); /* TSQ Source: A = dir 130 B = dir 210 C = dir 330 E = foot B C A R45 F = foot C A B I := incenter A B C M = extension E F A I R190 N = extension B C A I Z := N+dir(270) P = extension A midpoint B--C N Z R45 M--P--N red unitcircle 0.1 lightcyan / blue A--B--C--cycle blue A--P blue A--N blue E--F blue T = extension E F B C F--T--B lightblue circumcircle P M N deepgreen Q = foot T A midpoint B--C R45 T--Q deepgreen D = foot A B C S = extension Q T M N R260 A--D blue B--E blue dotted C--F blue dotted X = extension E F A D */ [/asy]$

Claim: $\overline{QT}$ passes through the orthocenter $H$ of $\triangle ABC$ .

Proof. Let $S = \overline{QT} \cap \overline{AMN}$ , and $H' = \overline{TSQ} \cap \overline{AD}$ . Since $PMTN$ is a cyclic kite, $-1 = (PT;MN)_\gamma \overset{Q}{=} (AS; MN) \overset{T}{=} (AH';XD).$ On the other hand, $BFEC$ is a complete quadrilateral, so $-1 = (BC;TD) \overset{F}{=} (AH;XD)$ . We conclude $H = H'$ . $\blacksquare$

Thus we see $Q$ is the foot from $A$ to $\overline{TH}$ . So $Q$ is the $A$ -HM point, and thus $\overline{AQP}$ bisects $\overline{BC}$ .

This post has been edited 1 time. Last edited by v_Enhance, Feb 18, 2020, 1:50 AM

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#28 h Feb 18, 2020, 2:05 AM • 2 Y

Y by Adventure10, sami1618

20 min length bash

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399 posts

#29 h May 20, 2020, 8:02 AM • 2 Y

Y by lilavati_2005, sami1618

A homothety is all that it takes!

Let $\measuredangle ABC=B$ and etc.Also assume WLOG $B>C$ .
$[asy] size(6cm); pair A,B,C,E,F,N,R,M,G,O,I,H,S,T,K,X,L,Y,Z,P; A=dir(130); B=dir(210); C=dir(330); H=orthocenter(A,B,C); R=midpoint(B--C); O=circumcenter(A,B,C); I=incenter(A,B,C); S=midpoint(A--H); E=foot(B,C,A); F=foot(C,A,B); X=foot(S,A,I); Y=foot(O,A,I); G=extension(S,R,E,F); K=2*X-A; L=2*Y-A; M=extension(A,K,B,C); N=extension(A,K,E,F); T=extension(B,C,E,F); Z=circumcenter(T,M,N); P=2*Z-T; draw(A--B--C--A); draw(A--L,gray); draw(A--H,gray); draw(F--E,gray); draw(S--R,gray); draw(circumcircle(A,F,E),gray); draw(unitcircle); draw(B--E,gray); draw(A--R,dashed); draw(C--F,gray); draw(R--L,gray); draw(P--N--M--P,gray); dot("$A$",A,dir(120)); dot("$B$",B,dir(180)); dot("$C$",C,dir(0)); dot("$E$",E,NE); dot("$F$",F,dir(200)); dot("$H$",H,dir(270)); dot("$R$",R,dir(50)); dot("$S$",S,dir(0)); dot("$K$",K,2*dir(240)); dot("$L$",L,dir(270)); dot("$N$",N,dir(80)); dot("$M$",M,dir(250)); dot("$P$",P,dir(0)); [/asy]$

Let $H$ be the orthocenter of $\triangle ABC$ .Let $R$ and $S$ be the midpoints of $\overline{BC}$ and $\overline{AH}$ repectively.Also let $K$ and $L$ be the arc midpoints of $\widehat{BC}$ in $(ABC)$ and $\widehat{EF}$ in $(AFHE)$ respectively.Now a simple angle chase shows that $\triangle PMN$ is isosceles and $\measuredangle PNM = B-C$

Claim: We have $\triangle RLK \sim \triangle PMN$

Proof: Now observe that $K$ lies on $\overline{RS}$ by the incenter-excenter lemma.Now angle chase to get $\measuredangle SAK = \measuredangle AKS =\measuredangle LKR = \measuredangle RLK = B-C$ where the last equality holds due to $\overline{AH} \parallel \overline{RL}$ and whence the claim holds.

Now remark that $\overline{RL} \parallel \overline{PM}$ and $\overline{KR} \parallel \overline{PN}$ .Whence $\triangle PMN$ and $\triangle RLK$ are homothetic with center $A$ and whence $\overline{APR}$ are collinear and we are done $\blacksquare$

This post has been edited 1 time. Last edited by char2539, May 20, 2020, 9:22 AM

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#30 h Jun 25, 2020, 12:04 PM • 1 Y

Y by sami1618

Let $K=EF\cap BC$ , also let $M_{BC}$ and $M_{EF}$ be the midpoints of $BC$ and $EF$

Claim : $PM=PN$ .

proof : Note that, since $\angle PMK=\angle PNK=90^{\circ}$ , so $PMKN$ is cyclic. Now since $\angle ANB=\angle AME=\angle KMN$ it follows that $KM=KN$ which means that $PM=PN$

Hence $EF,BC$ are tangent to $\odot(P,PM)=\omega$ at $M,N$ respectively,

Now we invert around $A$ with radius $\sqrt{AF\cdot AB}$ to get the following diagram

Inverted diagram wrote:

$E,F$ are the feet of the altitudes of $\triangle ABC$ from $B,C$ respectively. $M^*$ is the midpoint of the minor arc $\widehat{BC}$ of $(ABC)$ and $N^*$ is the midpoint of the minor arc $\widehat{EF}$ of $(AEF)$ . $\omega^*$ is the circle tangent to $(ABC)$ and $(AEF)$ at $M^*$ and $N^*$ respectively.

Claim : $M_{BC}$ is the center of $\omega^*$

proof : Simply note that $M_{BC}M^*$ is perpendicular to the common tangent of $\omega^*$ and $(ABC)$ . $M_{BC}M_{EF}\perp EF$ and so $M_{BC}N^*$ is perpendicular to the common tangent of $\omega^*$ and $(AEF)$ . Hence $M_{BC}$ is the center of $\omega^*$ .

Inverting back, this implies that $P$ , the center of $\omega$ , lies on the $A-$ median of $\triangle ABC$ and we are done. $\blacksquare$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -9.723561704743021, xmax = 11.15124640898547, ymin = -7.721700374224664, ymax = 4.4607071733965205; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-3.0371269115540978,2.4580737591692805)--(-4.12,-3.666363636363637), linewidth(0.5)); draw((-4.12,-3.666363636363637)--(4.607272727272718,-3.6481818181818184), linewidth(0.5)); draw((4.607272727272718,-3.6481818181818184)--(-3.0371269115540978,2.4580737591692805), linewidth(0.5)); draw((-3.8523178858599465,-2.152425976034979)--(-0.711663246950657,0.6005210221578313), linewidth(0.5)); draw((-3.8523178858599465,-2.152425976034979)--(4.607272727272718,-3.6481818181818184), linewidth(0.5)); draw((-0.711663246950657,0.6005210221578313)--(-4.12,-3.666363636363637), linewidth(0.5)); draw((-3.0371269115540978,2.4580737591692805)--(0.24363636363635877,-3.6572727272727277), linewidth(0.5) + dotted); draw(circle((-0.7320655081523083,-1.8385633255371434), 1.8207381627180734), linewidth(0.5) + ccqqqq); draw(circle((0.23868179362544875,-1.2790791220359572), 4.969631206682314), linewidth(0.5)); draw(circle((-3.032172341543187,0.07988015393250965), 2.3781987662415993), linewidth(0.5)); draw((-3.0371269115540978,2.4580737591692805)--(-0.7282723118783992,-3.659297537013383), linewidth(0.5)); draw((-0.7282723118783992,-3.659297537013383)--(-0.7320655081523083,-1.8385633255371434), linewidth(0.5)); draw((-0.7320655081523083,-1.8385633255371434)--(-1.9322330206358465,-0.4693718125788347), linewidth(0.5)); draw(circle((-3.157505747597731,-2.753987383020227), 2.5924431643027126), linewidth(0.5)); draw((-0.7282723118783992,-3.659297537013383)--(0.2490351695044648,-6.248699543963433), linewidth(0.5)); draw((-1.4645461139418077,-1.7085206351215922)--(0.24363636363635877,-3.6572727272727277), linewidth(0.5)); draw((0.24363636363635877,-3.6572727272727277)--(0.2490351695044648,-6.248699543963433), linewidth(0.5)); draw((-2.2819905664053017,-0.7759524769385739)--(-1.4645461139418077,-1.7085206351215922), linewidth(0.5)); draw(circle((0.24363636363635877,-3.6572727272727277), 2.5914324404407134), linewidth(0.5) + dotted + ccqqqq); draw((-3.8523178858599465,-2.152425976034979)--(-5.582945987043152,-3.66941144050331), linewidth(0.5)); draw((-4.12,-3.666363636363637)--(-5.582945987043152,-3.66941144050331), linewidth(0.5)); draw((-3.0371269115540978,2.4580737591692805)--(-3.0272177715322774,-2.2983134513042605), linewidth(0.5)); /* dots and labels */ dot((-3.0371269115540978,2.4580737591692805),dotstyle); label("$A$", (-2.9718659554589624,2.617853019606435), N * labelscalefactor); dot((-4.12,-3.666363636363637),dotstyle); label("$B$", (-4.048223248823088,-3.4978134199624327), NW * labelscalefactor); dot((4.607272727272718,-3.6481818181818184),dotstyle); label("$C$", (4.676794204961867,-3.4815049761235826), E * labelscalefactor); dot((-0.711663246950657,0.6005210221578313),linewidth(4pt) + dotstyle); label("$E$", (-0.6397584865033575,0.7260735342997986), NE * labelscalefactor); dot((-3.8523178858599465,-2.152425976034979),linewidth(4pt) + dotstyle); label("$F$", (-3.7872881474014815,-2.0300534744659045), NW * labelscalefactor); dot((-3.0272177715322774,-2.2983134513042605),linewidth(4pt) + dotstyle); label("$H$", (-2.9555575116201123,-2.160521025176707), NW * labelscalefactor); dot((-1.9322330206358465,-0.4693718125788347),linewidth(4pt) + dotstyle); label("$M$", (-1.765041111384034,-0.12196554532041776), NE * labelscalefactor); dot((-0.7282723118783992,-3.659297537013383),linewidth(4pt) + dotstyle); label("$N$", (-0.656066930342208,-3.5304303076401333), NE * labelscalefactor); dot((-0.7320655081523083,-1.8385633255371434),linewidth(4pt) + dotstyle); label("$P$", (-0.6723753741810583,-1.7038845976888983), NE * labelscalefactor); dot((0.24363636363635877,-3.6572727272727277),linewidth(4pt) + dotstyle); label("$M_{BC}$", (0.3061312561499647,-3.5304303076401333), NE * labelscalefactor); dot((-5.582945987043152,-3.66941144050331),linewidth(4pt) + dotstyle); label("$K$", (-5.515983194319622,-3.546738751478984), N * labelscalefactor); dot((0.2490351695044648,-6.248699543963433),linewidth(4pt) + dotstyle); label("$M^*$", (0.3061312561499647,-6.123472878017333), NE * labelscalefactor); dot((-1.4645461139418077,-1.7085206351215922),linewidth(4pt) + dotstyle); label("$N^*$", (-1.4062553469293255,-1.5734170469780957), E * labelscalefactor); dot((-2.2819905664053017,-0.7759524769385739),linewidth(4pt) + dotstyle); label("$M_{EF}$", (-2.825089960909309,-0.6112188604859272), S * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 3 times. Last edited by Greenleaf5002, Jun 25, 2020, 12:08 PM

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1051 posts

#31 h Jul 6, 2020, 3:07 AM • 1 Y

Y by sami1618

darn iran tst has some good geo

Let $H_A$ be the $A$ -humpty point, let $X$ be the midpoint of $BC$ , and let $T = BC \cap EF$ . It is well-known that $T,H,H_A$ are collinear and $\angle TH_AX = 90$ . Furthermore, $BHH_AC$ is cyclic. We prove that $P$ lies on $\overline{AH_AX}$ , or equivalently, $\angle TH_AP = 90$ .

Note that $H_A$ is the Miquel Point of complete quadrilateral $BCFE$ , so there exists a spiral similarity at $H_A$ sending $BC$ to $EF$ . Furthermore, note that by similarity and angle bisector theorem $\frac{ME}{MF} = \frac{AE}{AF} = \frac{AB}{AC} = \frac{NB}{NC}.$ Thus the aforementioned spiral similarity also takes $M$ to $N$ , and thus $TMH_AN$ is cyclic. Note that $TMPN$ is obviously cyclic, so the two cyclic quadrilaterals have the same circumcircle. Now $\angle TH_AP = 90$ , which is what we wanted, and we are done.

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mira74

#32 h Jul 6, 2020, 7:26 AM • 1 Y

Y by sami1618

Given today's world, I'm surprised that I don't see this solution:

We make the following more general claim:

If we replace $M,N$ with any points on $EF$ and $BC$ with $\angle BAM = \angle CAN$ , then $P$ lies on $AM$ . Also, let $T=EF \cap BC$

But this is simple by moving points - if we move $M$ linearly, since $AEMF \sim ABNC$ , $N$ moves linearly, meaning that the circumcircle of $TMN$ moves through some fixed point (call it $X$ ), and thus since $\angle TXP=90^{\circ}$ , $P$ moves on a fixed line. Now, we can just check the cases where $N$ is the midpoint of $BC$ and $AN \perp BC$ .

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259 posts

#33 h Dec 25, 2020, 8:50 AM • 1 Y

Y by sami1618

Let $G$ be $A$ -Humpty. then $FE$ , $HG$ , $BC$ concur at $D$ and $G$ is the spiral center of ${E,M,F}\rightarrow{B,N,C}$ . Easy angle chase gives $DM=DN$ . Since $AG$ bisects $BC$ at $K$ and $KF,KE$ are tangent to $(AH)$ , $AFHE$ is harmonic and so, $GM$ bisects $\angle FGE$ . So, $\measuredangle MGF=\measuredangle EGM=\measuredangle BGN$ Also, $\measuredangle FGH=\measuredangle FAH=\measuredangle HCB=\measuredangle HGB$ $\implies \measuredangle MGD=\measuredangle DGN \implies (MGND)$ -cyclic. So, $G\in(DP)\implies DG\perp GP$ But also, $DG \perp AG \implies$ $A,G,P$ -collinear. Since $AG$ bisects $BC$ , $AP$ bisects $BC$

This post has been edited 3 times. Last edited by lahmacun, Dec 25, 2020, 8:51 AM

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173 posts

#34 h Jan 2, 2021, 10:05 AM • 1 Y

Y by sami1618

Claim. Let $ABC$ be a triangle and let $P_{1}$ and $P_{2}$ be points on $BC$ such that $(P_{1},P_{2};B,C)=-1$ . Define $Q_{1}$ and $Q_{2}$ similarly. Let $T$ be a point such that $AP_{1}\parallel TQ_{1}$ and $AQ_{2}\parallel TP_{2}$ . Then $T$ lies on the $A$ - median.
Proof. Let $M$ be the midpoint of $BC$ , then as $(P_{1},P_{2};B,C)=-1$ , we see that $P_{1}$ and $P_{2}$ are inverses with respect to the circle centered at $M$ with radius $MB$ . Similarly, $Q_{1}$ and $Q_{2}$ are inverses too. Therefore, $MP_{1}\cdot MP_{2}=MQ_{1} \cdot MQ_{2}\Rightarrow \frac{MP_{1}}{MQ_{1}}=\frac{MQ_{2}}{MP_{2}}$ . where the lengths are directed. If the lines through $Q_{1}$ and $P_{2}$ parallel to $AP_{1}$ and $AQ_{2}$ meet $AM$ at $T_{1}$ and $T_{2}$ ,
$\frac{MA}{MT_{1}}=\frac{MP_{1}}{MQ_{1}}=\frac{MQ_{2}}{MP_{2}}=\frac{MA}{MT_{2}}$ and so $T_{1}=T_{2}=T$ as desired.
Now we're ready to solve the problem. Let the external angle bisector of $ABC$ meet $BC$ at $T$ , and let $S=EF\cap BC$ . Also let $D$ be the foot of perpendicular from $A$ to $BC$ . Then $\measuredangle SMN=\measuredangle BAM+\measuredangle EFA=\measuredangle NAC+\measuredangle ACB=\measuredangle MNS$ which shows that $SMPN$ is a kite. In particular, $SP\perp AM$ , so $SP\parallel TA$ . Now letting $P_{1}=T,P_{2}=N, Q_{1}=S, Q_{2}=D$ and applying the claim shows that $P$ lies on $A$ - median as desired.

This post has been edited 1 time. Last edited by KST2003, Jan 2, 2021, 10:06 AM

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1080 posts

#35 h Apr 16, 2022, 7:10 PM • 2 Y

Y by centslordm, sami1618

Let $H_A$ be the $A$ -HM point of $\triangle ABC$ and let $X=\overline{EF}\cap\overline{BC}.$

Claim: $XMH_AN$ is cyclic.
Proof. As $H_A=(AEF)\cap (HBC),$ we know $H_A$ is the center of the spiral similarity $\overline{BC}\mapsto\overline{EF}.$ Notice $ME/MF=AE/AF=AB/AC=NB/NC$ so $\overline{NB}\mapsto\overline{ME}.$ Hence, $H_A$ is the Miquel point of self-intersecting quadrilateral $MEBN$ and $H_A\in (MNX).$ $\blacksquare$

Notice $(PMNX)$ is cyclic with diameter $\overline{PX}$ so $\measuredangle XH_AP=90=\measuredangle XH_AA.$ $\square$

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302 posts

#36 h Nov 28, 2022, 4:27 AM • 1 Y

Y by sami1618

I think I found the worst possible solution to this problem

Lemma: If a point $X$ moves along a line $\ell_X$ linearly and a point $Y$ moves along a line $\ell_Y$ linearly, then the point $Z$ defined as the intersection of the line passing through $X$ perpendicular to $\ell_X$ and the line passing through $Y$ perpendicular to $\ell_Y$ lies on a fixed line $\ell_Z$

Let $\ell_X$ and $\ell_Y$ be number lines where $\ell_X \cap \ell_Y$ is $0$ on both lines, and $X$ moving one unit on $\ell_X$ corresponds to $Y$ moving on unit along $\ell_Y$ . Now, consider the linear transformation taking this to the cartesian plane, ie: if $\ell_X \to \ell_X^*$ and $\ell_Y \to \ell_Y^*$ then $\ell_X^* \perp \ell_Y^*$ and $\ell_X^*$ and $\ell_Y^*$ are scaled equally. Let $\ell_X^* \cap \ell_Y^*=O$ . Notice that there are fixed angle $\theta_X$ and $\theta_Y$ where $\angle OX^*Z^* = \theta_X$ and $\angle OY^*Z^* = \theta_Y$ which implies all such $Z^*$ must lie on a line. Undoing the linear transformation means that $Z$ lies on a fixed line. $\square$

Now, we will define points $M_1$ and $N_1$ of $EF$ and $BC$ where $\angle M_1AE = \angle N_1AB$ . Since $AFE$ and $ABC$ are similar, it must follow that as $M_1$ moves linearly along $EF$ , $N_1$ moves linearly along $BC$ since they are the same points wrt to the two similar triangles. Thus, by our lemma, $P_1$ also lies on a fixed line ( $P_1$ is just the generalized $P$ ). Setting $N_1$ to the midpoint of $BC$ gives $P_1 = N_1$ . So, the midpoint of $BC$ lies on this line. Likewise, setting $M_1$ to be the foot of the altitude from $A$ to $EF$ gives $P_1=A$ , so $A$ is part of the line as well. Thus, our line passes through both $A$ and the midpoint of $BC$ . Finally, setting $M_1=M$ and $N_1=N$ gives $P_1=P$ which lies on that line. $\blacksquare$

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7586 posts

#37 h Oct 12, 2023, 11:06 PM • 1 Y

Y by sami1618

here's a solution. (I really need to draw everything in my diagrams oops)
also I used Reim's twice, there's probably a way to just use it once but oh well.

Let $EF$ intersect $BC$ at $D$ . Let $\ell$ be the line through $A$ parallel to $BC$ and let $DP\cap \ell=Q$ . Also let $\overline{AMN}\cap \overline{DPQ}=R$ . Finally let $DP$ intersect $AB$ and $AC$ at $U$ and $V$ . Oh also let $K$ be the midpoint of $BC$ . Oh let's also let $H$ be the orthocenter and let $L=AH\cap BC$ .

Claim: $A,U,N,H,V$ are concyclic.
Proof: By Reim's. By Law of Sines (you can technically turn this into a synthetic argument) we have $NV\parallel BE$ and since $ABLE$ is cyclic this implies $ANLV$ is cyclic. Similarly $ANLU$ is cyclic, done.

Claim: $AMPQ$ is cyclic.
Proof: By Reim's. Just notice that $PNDM$ is cyclic and $AQ\parallel ND$ .

Now we are almost done. Notice that $RU\cdot RV=RA\cdot RN=RA\cdot RM=RP\cdot RQ$ , therefore:
$(U,V;P,Q)\stackrel{A}{=}(B,C;AP\cap BC,\infty_{BC})$ implying that $AP\cap BC=K$ , done.

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176 posts

#38 h Jan 19, 2024, 3:29 AM • 1 Y

Y by sami1618

Let $P'$ be the $A$ -Humpty point.
Lemma: We know that the Miquel point of quadrilateral $BCEF$ is the $A$ - Humpty point $(P')$
Now since $AP'$ is $A$ -symmedian in $(AEF)$ , we have $\frac{AF}{AE}=\frac{FP}{PE}$
Let $M'$ be the foot of perpendicular from $P$ to $FE.$
We know that $\frac{FM'}{M'E}=\frac{FP}{PE}=\frac{AF}{AE}$ . Now therefore we have that $AM'$ is angle bisector. Therefore $M'=M$
Similarly proceeding like that we get $P=P'$
We are done

:cool:

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815 posts

#39 h May 6, 2024, 7:33 PM • 1 Y

Y by sami1618

Invert about $A$ with radius $\sqrt{AE\cdot AC}$ . Some trivial angle chasing show $M,N$ go to the arc midpoints $G,H$ of $BC$ and $EF$ on $(ABC)$ and $(AEF)$ . Let $Q$ be the queue point; I claim that $QGHD$ where $D$ is the midpoint is cyclic. This is true by more angle chasing since $DG \perp BC$ and $DH \perp EF$ . Thus inverting back the humpty point lies on $(TMN)$ with $T=EF\cap BC$ which finishes as if $TH_m \perp H_m P$ .

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4178 posts

#40 h Jul 6, 2024, 10:31 PM • 1 Y

Y by sami1618

The point of the problem is really the following claim:

Claim: In $\triangle ABC$ , if a line $\ell$ perpendicular to $BC$ intersects the A-median at $P_m$ and the A-symmedian at $P_s$ , then $\frac{AP_m}{AP_s}=\cos\alpha.$
Let $M$ be the midpoint of $BC$ , $L$ be the midpoint of arc $BC$ on $(ABC)$ , and let $T$ be the intersection of the tangents to $B$ and $C$ . Then, $\frac{AP_m}{AP_s}=\frac{AM}{AT}=\frac{ML}{LT}=\frac{BM}{BT}=\cos\alpha$ since $AL$ bisects $\angle TAM$ and $BL$ bisects $\angle TBM$ .

This kills the original problem. Let the perpendicular to $EF$ at $M$ intersect the A-median in $\triangle ABC$ (and thus $A$ -symmedian in $\triangle AEF$ ) at $P_1$ , and let the perpendicular to $BC$ at $N$ intersect the A-median at $P_2$ . Note that $AEMF$ and $ABNC$ are similar. When going from $P_1$ to $P_2$ , the triangle was scaled by a factor of $\frac{1}{\cos\alpha}$ as this is the scale factor between $\triangle ABC$ and $\triangle AEF$ , but the symmedian was replaced by a median, which by our claim multiplies the distance by $\cos\alpha$ . Thus, $AP_1=AP_2$ and we are done.

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#41 h Apr 2, 2025, 10:05 AM

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IDK my solution is just really simple; let $P'$ be the intersection of perpendicular in $N$ to $BC$ and $AS$ . now Let $K$ be midpoint of $EF$ . Now let $R$ be on $AK$ such that $P'R$ is perpendicular to $EF$ . we know $AEF$ and $ACB$ are similar. We know $SK$ and $P'R$ are parallel(because $SE=SF$ ) so we have $\frac{AR}{AK} = \frac{AP'}{AS}$ and because of similarity of 2 triangles we get $R$ is intersection of perpendicular from $M$ and $AK$ so $P'M$ is perpendicular to $EF$ either so $P'=P$ and $P'$ lies on $AS$ .

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