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

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 $\overline{BC}$.

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 $NH_A=NK$. $$\frac{AV}{AH_A}=\left(\frac{AN}{NH_A}\right)^2=\left(\frac{AN}{NK}\right)^2=\frac{AW}{AK} \implies AEFH_AV\stackrel{-}{\sim}ABCKW \ \ \ \blacksquare$$

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.

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.

Where are problem 1 and problem 6?

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.

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. $

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.

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.

[asy][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][/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. $\blacksquare$

What is HM_point.?

nguyenvanthien63 wrote: What is HM_point.? Humpty point

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

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 $BC$.

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.

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?

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.

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.

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 $BC$

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 $\odot (MDLN)$ $\cap$ $AG$ $=$ $P'$ $$\implies \angle DLP'=90^{\circ}=\angle DMP'=\angle DNP' \implies P' \equiv P$$

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 $\blacksquare$.

Cool problem. [asy][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][/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.

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][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][/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 $\overline{BC}$.

A homothety is all that it takes! Let $\measuredangle ABC=B$ and etc.Also assume WLOG $B>C$. [asy][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][/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 $\blacksquare$

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][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)); 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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.

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 $AN \perp BC$.

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 $BC$

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.

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 $\measuredangle XH_AP=90=\measuredangle XH_AA.$ $\square$

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. $\blacksquare$

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.

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

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 $TH_m \perp H_m P$.

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.