Let $AH$ intersect $BC$ at $D$ and $(ABC)$ at $H_A$. We invert about $A$ with radius $\sqrt{AH\cdot AD}$, denoting the unnamed image of $X$ by $X'$. Then: $(H,D), (B,F), (C,E)$ swap, $A'$ gets sent to $A''$, the midpoint of $AH$, $H_A' = AH\cap EF$, finally $P' = A''C\cap EF$ and $Q' = EF \cap A''B$. By DDIT on quadrilateral $BECF$, we get \[(A''B,A''C), (A''E,A''F), (A''H,A''A) \overset{A''}{\to} (P',Q'), (E,F), (H_A', H_A')\]as pairs of an involution. Projecting again from $A$ to $(ABC)$, \[(B,C), (P,Q), (H_A, H_A)\]are pairs of an involution, implying that $RH_A$ is tangent to $(ABC)$. Hence \[\measuredangle RHD = \measuredangle DH_AR = \measuredangle ACH_A = \measuredangle AH_A'E = \measuredangle FH_A'H, \]implying $EF\parallel HR$ as desired. $\square$

Let $S$ be the second intersection of $AH$ and $(ABC)$, $T$ be the antipode of $A$ in $(ABC)$, $l$ be the line through $A'$ parallel to $BC$, $B'$ and $C'$ be the intersection of $l$ with $BE$ and $CF$ respectively, $U$ be the midpoint of $B'C'$, and $K$ be that well-known point (intersection point of $HT$ and $(ABC)$). Observe that by angle chasing it can be obtained that the tangent to $(HBC)$ at $H$ is parallel to $EF$. So the condition $EF||HR$ is equivalent to $RH$ tangent to $(HBC)$, or (after reflecting w.r.t. $BC$), $RS$ tangent to $(ABC)$. So it's suffices to prove that $RS$ is tangent to $(ABC)$. Let $P'$ be the intersection of $TB'$ and $(ABC)$. Since $\angle AA'B = \angle AEB' = \angle AP'T = \angle APB = 90^{\circ}$, therefore $P' \in (AA'E)$. So, we get $P'=P$. Similarly, $Q$ is the intersection of $TC'$ and $(ABC)$. Since $HT$ passes through midpoint of $BC$ (because $HBTC$ is a parallelogram), so $HT$ passes through $U$ as well. By projecting $l$ to $(ABC)$ we get: \[ -1 = (B',C';\infty,U) \overset{T}{=} (P,Q;S,K) \]But we know that $(B,C;S,K)$ is also harmonic (For example, by noting that $\frac{BS}{SC}=\frac{BH}{HC}=\frac{BF}{EC}=\frac{BK}{KC}$ from some similarities). Therefore, $PQ, BC$, and tangents to $(ABC)$ at $K,S$ are concurrent, and that means their intersection point is $R$. So, $RS$ is tangent to $(ABC)$, as desired.

Why did this take so long? ELMO SL 2019 G4 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $R$. Prove that $EF \parallel HR$ Solution: Assume, $AB < AC$. Let $\Psi$ be inversion at centered at $A$ with radius $\sqrt{AH \cdot AD}$, where $AH \cap BC=D$. Let $R$ be midpoint of $AH$ and $AH$ $\cap$ $\odot (ABC)$ $=$ $H'$ \begin{align*} \Psi(P) & \equiv \Psi(\odot (AA'E)) \cap \Psi(\odot (ABC)) \\ & \equiv CR \cap EF \equiv P' \in AP \end{align*}And similarly, \begin{align*} \Psi(Q) & \equiv \Psi(\odot (AA'F)) \cap \Psi(\odot (ABC)) \\ & \equiv BR \cap EF \equiv Q' \in AQ \end{align*}Let $EF \cap BC=T$ and $AT \cap \odot (ABC)=G$ $$-1=(T,D;B,C) \overset{R}{=} (T, RD ~ \cap ~ EF; Q', P') \overset{A}{=} (G,H';Q,P)$$But, $-1=(G,H';B,C)$,Let $K$ be such a point that, $KG, KH'$ tangent to $\odot (ABC)$, Then, $\overline{KQP}$ is the $P-$symmedian WRT $\Delta GH'P$ and $\overline{KBC}$ is the $C-$symmedian WRT $\Delta GH'C$ $\implies$ $HK$ tangent to $\odot (BHC)$. But tangent to $\odot (BHC)$ at $H$ is parallel to $EF$ $\implies$ $HK$ $||$ $EF$

Let $X$ be the midpoint of $\overline{AH}$, $T \equiv EF \cap BC$, $H'$ the reflection of $H$ about $BC$, and let $S$ be second intersection of $AT$ and $(ABC)$. The key claim is that $(S, H'; P, Q) = -1$. Invert at $A$ with radius $\sqrt{AH \cdot AD}$; then it's easy to see that the inverses $P'$ and $Q'$ of $P$ and $Q$ respectively are just the respective intersections of $CX$ and $BX$ with $EF$. Then $$-1 = (T, D; B, C) \overset{X}{=} (T, AH \cap EF; Q', P') \overset{A}{=} (S, H'; Q, P),$$so the tangents to $(ABC)$ at $S$ and $H$ pass through $R$. Then since $RH'$ is tangent to $\triangle ABC$, it follows that $RH'$ is tangent to $(BHC)$. But $RH$ and $BC$ are antiparallel in $\angle BHC$, so $RH \parallel EF$ as desired.

Cross ratio solution. Let $H'$ be the reflection of $H$ across $BC$. Let $M_a, M_b, M_c, M$ be midpoints of $BC$, $CA$, $AB$, $AH$ respectively. Now, notice that $X=MM_c\cap BC$ and $Y=MM_b\cap BC$ are centers of $\odot(AA'E)$ and $\odot(AA'F)$ respectively. Thus if we let line through $O$ parallel to $BC$ meet $MM_c, MM_b$ at $U,V$ respectively. Then $$\begin{array}{rll} A(P,H';B,C) & = O(X,\infty_{BC};M_c,M_b) & (\text{Rotate } 90^{\circ}) \\[8pt] & = (X, U; M_c, \infty_{MM_c}) & \\[8pt] & = (Y, V; M_b, \infty_{MM_b}) & \\[8pt] & = O(Y, \infty_{BC}; M_c, M_b) & \\[8pt] & = A(Q,H';C,B) & (\text{Rotate } 90^{\circ}) \end{array}$$This implies that $PQ, H'H', BC$ are concurrent at $R$ i.e. $H'R$ is tangent to $\odot(BH'C)$. By symmetry, $HR$ is tangent to $\odot(BHC)$ so $HR\parallel EF$.

Observe that $\angle BAC\neq 90^\circ$ because then there's no circle $AA'E$. Also $AB\neq AC$ (otherwise pair $(A,P)$ is symmetric to $(B,Q)$ with respect to $A$-angle bisector and there's no point of intersection of $BC$ and $PQ$). When you read this solution at it's middle you will probably think this is one of the roads to hell. Luckily ending is very rewarding. Let $a,b,c\in\mathbb C$ be vertices of triangle $ABC$. We can take coordinate system such that it's circumscribed circle is the unit circle, i.e. $|a|=|b|=|c|=1$. Then also $|p|=1=|q|$. By the observation $b+c\neq 0\wedge a^2\neq bc$. $$a'=b+c-\frac{bc}{a}$$$$2e=a+b+c-\frac{ac}{b}\wedge 2f=a+b+c-\frac{ab}{c}$$Using these formulas $$\frac{a-p}{e-p}\cdot\frac{e-a'}{a-a'}=\frac{a-p}{2(e-p)}\cdot\frac{(a-b)(ab-ac-2bc)}{b(a-b)(a-c)}=\frac{a-p}{2(e-p)}\cdot\frac{ab-ac-2bc}{b(a-c)}$$Points $A,A',E,P$ are concyclic $$\frac{a-p}{e-p}\cdot\frac{e-a'}{a-a'}=\overline{\left(\frac{a-p}{e-p}\cdot\frac{e-a'}{a-a'}\right)}\iff$$$$\frac{a-p}{2(e-p)}\cdot\frac{ab-ac-2bc}{b(a-c)}=\frac{a-p}{2a(1-p\overline e)}\cdot\frac{2a+b-c}{a-c}\iff$$$$p=\frac{a(ab-ac-2bc)-be(2a+b-c)}{a\overline e(ab-ac-2bc)-b(2a+b-c)}\iff$$$$p=\frac{-(b+c)(b^2+3ab+ac-bc)}{-(b+c)(b^2+3bc+ac-ab)}\cdot\frac{bc}{a}$$$$p=\frac{b^2+3ab+ac-bc}{b^2+3bc+ac-ab}\cdot\frac{bc}{a}$$Similarly $$q=\frac{c^2+3ac+ab-bc}{c^2+3bc+ab-ac}\cdot\frac{bc}{a}$$Because $$|b|=|c|=|p|=|q|=1$$we have $$r=\frac{pq(b+c)-bc(p+q)}{pq-bc}$$Auxiliary calculations: $$pq(b+c)-bc(p+q)=\frac{b^2+3ab+ac-bc}{b^2+3bc+ac-ab}\cdot\frac{bc}{a}\cdot\frac{c^2+3ac+ab-bc}{c^2+3bc+ab-ac}\cdot\frac{bc}{a}\cdot (b+c)-bc\cdot\left(\frac{b^2+3ab+ac-bc}{b^2+3bc+ac-ab}\cdot\frac{bc}{a}+\frac{c^2+3ac+ab-bc}{c^2+3bc+ab-ac}\cdot\frac{bc}{a}\right)$$$$pq(b+c)-bc(p+q)=\frac{b^2c^2}{a^2}\cdot\frac{(b+c)(b^2+3ab+ac-bc)(c^2+3ac+ab-bc)-a(b^2+3ab+ac-bc)(c^2+3bc+ab-ac)-a(c^2+3ac+ab-bc)(b^2+3bc+ac-ab)}{(b^2+3bc+ac-ab)(c^2+3bc+ab-ac)}$$$$pq(b+c)-bc(p+q)=\frac{b^2c^2(a-b)(c-a)(c-b)^2(c+b+2a)}{a^2(b^2+3bc+ac-ab)(c^2+3bc+ab-ac)}$$$$pq-bc=\frac{b^2+3ab+ac-bc}{b^2+3bc+ac-ab}\cdot\frac{bc}{a}\cdot \frac{c^2+3ac+ab-bc}{c^2+3bc+ab-ac}\cdot\frac{bc}{a}-bc$$$$pq-bc=\frac{bc}{a^2}\cdot\frac{bc(b^2+3ab+ac-bc)(c^2+3ac+ab-bc)-a^2(c^2+3bc+ab-ac)(b^2+3bc+ac-ab)}{(c^2+3bc+ab-ac)(b^2+3bc+ac-ab)}$$$$pq-bc=\frac{bc}{a^2}\cdot\frac{(b-c)^2(c-a)(a-b)(bc-a^2)}{(c^2+3bc+ab-ac)(b^2+3bc+ac-ab)}$$Hence $$r=\frac{bc(2a+b+c)}{bc-a^2}$$Auxiliary calculations: $$r-h=\frac{bc(2a+b+c)-(a+b+c)(bc-a^2)}{bc-a^2}=\frac{a(a+b)(a+c)}{bc-a^2}$$$$e-f=\frac{a(c-b)(b+c)}{bc}$$Hence $$\frac{r-h}{e-f}=\frac{bc(a+b)(a+c)}{(c-b)(b+c)(bc-a^2)}$$and finally $$\overline{\left(\frac{r-h}{e-f}\right)}=\frac{\frac{1}{bc}\cdot\frac{a+b}{ab}\cdot\frac{a+c}{ac}}{\frac{b-c}{bc}\cdot\frac{b+c}{bc}\cdot\frac{-bc+a^2}{a^2bc}}=\frac{r-h}{e-f}$$which is equivalent to $$RH||EF$$

AlastorMoody wrote: Why did this take so long? ELMO SL 2019 G4 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $R$. Prove that $EF \parallel HR$ Solution: Assume, $AB < AC$. Let $\Psi$ be inversion at centered at $A$ with radius $\sqrt{AH \cdot AD}$, where $AH \cap BC=D$. Let $R$ be midpoint of $AH$ and $AH$ $\cap$ $\odot (ABC)$ $=$ $H'$ \begin{align*} \Psi(P) & \equiv \Psi(\odot (AA'E)) \cap \Psi(\odot (ABC)) \\ & \equiv CR \cap EF \equiv P' \in AP \end{align*}And similarly, \begin{align*} \Psi(Q) & \equiv \Psi(\odot (AA'F)) \cap \Psi(\odot (ABC)) \\ & \equiv BR \cap EF \equiv Q' \in AQ \end{align*}Let $EF \cap BC=T$ and $AT \cap \odot (ABC)=G$ $$-1=(T,D;B,C) \overset{R}{=} (T, RD ~ \cap ~ EF; Q', P') \overset{A}{=} (G,H';Q,P)$$But, $-1=(G,H';B,C)$,Let $K$ be such a point that, $KG, KH'$ tangent to $\odot (ABC)$, Then, $\overline{KQP}$ is the $P-$symmedian WRT $\Delta GH'P$ and $\overline{KBC}$ is the $C-$symmedian WRT $\Delta GH'C$ $\implies$ $HK$ tangent to $\odot (BHC)$. But tangent to $\odot (BHC)$ at $H$ is parallel to $EF$ $\implies$ $HK$ $||$ $EF$ Where can I learn these notations, and how to use them? By that I meant Inversion $\sqrt{bc}$ and harmonics. I tried EGMO but... It was painful for me

Yet another non-inversive solution. We shall replace $R$ by $X$ for our convenience. Here’s the restated question. tastymath75025 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $X$. Prove that $EF \parallel HX$. Proposed by Daniel Hu Let $T = (AH) \cap (ABC)$, $A_1 = AH \cap (ABC)$ and let $S$ be the $A$-antipode and $M$ the midpoint of $BC$. We shall use the following results without proof. Lemma 1: $T, H, S, M$ are collinear. Lemma 2 : $(T,B;C,S) = -1$. Here’s the key claim for the solution. Claim : $(T, P; Q, S) = -1$ Proof : Let $Y = (AEA’) \cap BH$, $Z = (AFA’) \cap CH$. Note that $Y, A’, Z$ are collinear and $\overline{YA’Z} \parallel BC$. Note that $$ \angle{AQS} + \angle{AQY} = 90^{\circ} + 90^{\circ} = 180^{\circ}$$So, $\overline{YQS}$, $\overline{ZPS}$ are collinear. Let $YZ \cap HM = N$. Then, $N$ is the midpoint of $YZ$. Let $P_{\infty}$ be the point at infinity on $YZ$. $$-1 = (Y, M; Z, P_{\infty}) \stackrel{S}{=} (T, P; Q, S)$$ Combining the above and Lemma 1, we have $XA_1$ is tangent to $(ABC)$.So, by our Claim $XH$ is tangent to $(HBC)$. So, $$\angle{XHB} = \angle{HCB} = \angle{HED} = \angle{HEF}$$So, $XH \parallel EF$. $\blacksquare$

This problem is amazing! Even though it seems really simple, it has a lot of structure going through it. tastymath75025 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $R$. Prove that $EF \parallel HR$. Proposed by Daniel Hu Define $T=(ABC) \cap (AEF), S=EF \cap BC, H'=AH \cap (ABC)$ and $D$ to be the projection of $A$ on $BC.$ Redefine $R$ to be the point on line $BC$ such that $RH \parallel EF.$ We start off by a claim: Claim: $R$ is the intersection of the tangents to $(ABC)$ at $T,H'.$ Proof: Since $BTCH'$ is harmonic, hence the tangents to $(ABC)$ at $T,H'$ meet on $BC.$ So it suffices to show that $RH'$ is tangent to $(ABC),$ for which it is enough to show that $RH$ is tangent to $(HBC).$ We can do this by angle chasing, but I will provide a different method. Let $U$ be the midpoint of $EF,$ and let $K=UH \cap BC.$ Clearly $HK$ is a symmedian in $\triangle BHC,$ since $\triangle HBC \overset{-}{\sim} \triangle HFE.$ Hence $$-1=(E,F;U,\infty_{EF}) \overset{H}{=} ( B,C;K, R )$$and so $RH$ is tangent to $(HBC),$ as desired. $\square$ It suffices to show that the line $PQ$ and the tangents to $(ABC)$ at $T,H'$ are concurrent, for which it suffices to show that $TQH'P$ is a harmonic quadrilateral. Consider the inversion $\mathcal{I}$ about $A$ that swaps $H,D.$ Then if $N$ is the midpoint of $AH,$ then $\mathcal{I}: A' \leftrightarrow N, S \leftrightarrow T.$ Now $P=(AEA') \cap (ABC),$ hence $P \mapsto P'=CN \cap EF.$ Similarly $Q \mapsto Q'=BN \cap EF.$ We can easily use the inversion length formula to show that cross ratios remain invariant under inversion. 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mufree wrote: By that I meant Inversion $\sqrt{bc}$ and harmonics. I tried EGMO but... It was painful for me EGMO is the best. Give it some time and it will be completely worth it. Otherwise, you can also try reading these two documents:

A slightly different ending with the above inversion solutions The main part is to perform the $\sqrt{bc}$ inversion at $A$. Let us denote the image of point $X$ by $X'$. Let $T=AH' \cap B'C'$, $U=B'C' \cap E'F'$, $S$ be the midpoint of $TU$. Then it is easy to see that $EF$ is parallel to $HR$ iff $\angle(AH', H'R')=\angle(AH', A'R')$, and iff $(AR' \cap UT, R', H', T)$ is concyclic. A key observation is that $AR' \cap UT$ coincides with $S$. This could be checked since both $(UT; B'C'), (UT,Q'P')$ is harmonic, which leads to $$SB' \cdot SC'=ST^2=SP' \cdot SQ',$$or $S$ lies on the radical axis of $\odot(A'B'C'), \odot(A'P'Q')$, i.e. $S$ lies on $AR'$. Now note that $ST^2=SH'^2=SR'\cdot SA'$, hence $\angle SR'H'=\angle SH'T'=\angle ST'H'$, so $(SR' T'H')$ is concyclic, as desired. $\square$

A short and cute solution. Call $E',F',H'$ symmetric of $E,F,H$ wrt $BC$; $X=EE'\cap BC, Y=FF'\cap BC, W=AA'\cap BC, Z=EF\cap BC$. Let $\omega=\left(BFECE'F'\right)$. Radical center of $\omega,\left(ABQPC\right),\left(AEPE'A'\right)$ is $EE'\cap BC=X$, so $P=AX\cap\left(ABC\right)$. Similarly $Q=AY\cap\left(ABC\right)$. Define $\left(AEFH\right)\cap\left(ABC\right)=S\ne A$. It's well known that $SA$ passes through $Z$, and that $\left(S,H';B,C\right)=-1$ so tangents to $\left(ABC\right)$ through $S$ and $H'$ intersect in $R'$ on $BC$. Defining $Y'=YA\cap FC$ and $X'=XA\cap BE$ we get that $Z,Y',X'$ are collinear by Desargues on $EFH$ and $XYA$. Now, putting $W'=HW\cap FE$ we get $\left(S,H';P,Q\right)\stackrel{A}{=}\left(Z,W';X',Y'\right)\stackrel{H}{=}\left(Z,W;B,C\right)=-1$. From this $R'$ lies also on $QP$, so $R'=BC\cap QP=R$ and we are done because by symmetry $R'H$ is tangent to $\left(BHC\right)$ hence $\angle R'HB=\angle R'CH=\angle FEB\Rightarrow R'H\parallel FE$.

We'll prove the statement for a general point $A'$ on $\overline{AH}$. Pretty similar to everybody else, we invert about $A$ with radius $\sqrt{AH \cdot AD}$. Let $O$ be the circumcenter of $\triangle ABC$ and let $AO$ meet $\overline{BC}$ at $T$. Then the inverse of the line through $H$ parallel to $\overline{EF}$ is the circle through $A$ and $D$, tangent to $\odot(ABC)$ which is the circumcircle of $\triangle ADT$. Moreover, if $X = BA' \cap EF$ and $Y = CA' \cap EF$ are the inverses of $P$ and $Q$, respectively, then it will suffice to prove that the second intersection of $\triangle AXY$ and $\triangle ADT$ is fixed. (We may then prove the intersection of the line through $H$ parallel to $EF$ and $BC$ is this desired fixed point under inversion by considering the case $A' = H$ or $A' = D$.) In order to prove the existence of the fixed point, we'll resort to projective, setting $M = AH \cap EF$ to obtain $$-1 = (B, C; M, EF \cap BC) \stackrel{A'}{=} (X, Y; M, EF \cap BC)$$which combined with $AA' \perp BC$ gives $\angle XDH = \angle YDH$ so the circumcircles of $\triangle DXY$ and $\triangle DEF$ are tangent. It isn't too hard to prove that the nine-point circle $\equiv \odot(DEF)$ is tangent to $\odot(ADT)$ so $\odot(DXY)$ is tangent to $\odot(ADT)$. Let $Q$ denote the second intersection of the circumcircles of $\triangle AXY$ and $\triangle ADT$. Then, by the radical axis theorem, $AQ, EF$ and the tangent to $\odot(DEF)$ at $D$ concur so the intersection of $AQ$ and $EF$ is fixed so inverting back we are done. $\blacksquare$

ELMOSL 2019 G4 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $R$. Prove that $EF \parallel HR$. Proposed by Daniel Hu Solution with Crystal1011 and RAMUGAUSS :- Let $M$ be the midpoint of $AH$. Perform a $\sqrt{AH\cdot AD}$ inversion. Let this map be $\Psi$ and $\Psi$ swaps $\{\odot(AA'E),MC\},\{\odot(AA'F),MB\},\{\odot(ABC),EF\}$. Then $$\Psi(P)\leftrightarrow MC\cap FE=P'$$Hence, $\overline{A-P'-P}$. Similarly we get that $\overline{A-Q'-Q}$. So now the Problem can be reduced as follows. Reduced Problem wrote: $ABC$ be a triangle with altitudes $BE$ and $CF$ and Orthocentre $H$. Let $M$ be the midpoint of $AH$ and let $\{MB\cap EF,MC\cap EF\}=\{Q',P'\}$ and let $AQ'\cap \odot(ABC)=Q$ and let $AP'\cap\odot(ABC)=P$ and let $R=QP\cap BC$. Then prove that $RH\|EF$. Let $AH\cap\odot(ABC)=H_A'$. Now let's project. $$-1=(Q_A,H_A';B,C)\overset{M}{=}(Q_A,H_A';Q',P')\overset{A}{=}(Q_A,H_A';Q,P)$$So from this we directly get that $RH_A'$ is tangent to $\odot(ABC)$. Now notice that $\{EF,BC\}$ and $\{BC,HR\}$ are antiparallel. Hence, $HR\|EF$. $\blacksquare$

tastymath75025 wrote: Let triangle $ABC$ have altitudes $BE$ and $CF$ which meet at $H$. The reflection of $A$ over $BC$ is $A'$. Let $(ABC)$ meet $(AA'E)$ at $P$ and $(AA'F)$ at $Q$. Let $BC$ meet $PQ$ at $R$. Prove that $EF \parallel HR$. Proposed by Daniel Hu Nice and hard problem Solution- Let $\Gamma=(ABC)$ and $AH$ meet $\Gamma$ again at $H'$ Lemma- $RH'$ is tangent to $\Gamma$ Assume that the lemma is true. Reflecting about $BC$ we get that $RH$ is tangent to $(BHC)$ and hence $\angle HRB=\angle HBC-\angle RHB=\angle HBC-\angle HCB=(90-C)-(90-B)=B-C=\angle ABC- \angle AFE = \angle (FE,BC)$ . And hence $FE\parallel HR$ Hence proved Now we prove the lemma Lemma- $RH'$ is tangent to $\Gamma$ Proof- Invert the figure in $A$ and let there new names be same as there old names (Yeah I am lazy) Note that here $A,B,F;A,C,E;A,A'H';F,P,A';E,Q,A'$ are all triads of collinear points. Now why would one not use projective geometry Let $l=\overleftrightarrow{BC}$ and $m=\overleftrightarrow{AA'}$ ,consider projectivity $\tau:l\rightarrow l$ which is composition of $l\xrightarrow{E}m$ and $m\xrightarrow{F}l$ Note that $AH',BE,CF$ are concurrent at the inverse of the foot of perpendicular from $A$ in original diagram. Hence we have $\tau(Q)=P, \tau(H')=H', \tau(B)=C, \tau (C)=B$ and hence we get that $\tau$ is an involution and hence an inversion from a point $S$ Note that $A,R,P,Q ; A,R,B,C$ are sets of concyclic points Let $AR\cap l=S'$ and hence we have $S'P\times S'Q=S'R\times S'A=S'B\times S'C$ and hence $S\equiv S'$ And also $SH'^2=SP\times SQ=SR\times SA$ and hence $BC$ is tangent to $(ARH')$ and this is exactly the required result in the original figure. Hence proved !!!!

i thought i was not terrible at projective triangle geo but apparently i'm bad at that too Let $AD$ be an altitude of $\triangle ABC$, $T = (ABC) \cap (AEF)$, $L= EF \cap BC$, $H'$ be the reflection of $H$ over $BC$. Note that $A,T,L$ are collinear. Finally let $Z$ be the midpoint of $AH$. Note that by antiparallel and tangents it suffices to show that $(T,H'Q,P) = -1$. (There's a bit more to it but I'm tired) Perform the classic $\sqrt{AH \cdot AD}$ inversion at $A$, and let the inverses $P' = CX \cap EF$ and $Q' = BX \cap EF$. Note that $A,P',P$ and $A,Q',Q$ are collinear, and let $N = AH \cap EF$. so $$-1 = (L,D;B,C) \stackrel{Z}{=} (L,N;Q',P') \stackrel{A}{=} (T,H';Q,P)$$

I couldn't solve this a long time ago, and blamed it on my lack of DDIT knowledge. Now, having learned DDIT, I find myself a fool when I realise there is no need for DDIT in this problem.

Why no one told me it's from ELMOSL 2019 ? , anyways nice problem .

khina wrote: I couldn't solve this a long time ago, and blamed it on my lack of DDIT knowledge. What is DDIT?

Johann Peter Dirichlet wrote: khina wrote: I couldn't solve this a long time ago, and blamed it on my lack of DDIT knowledge. What is DDIT? Dual of Desargues' Involution Theorem

Nice problem,we are going to use inverzion and a bit of $DDIT$ to solve this problem. Since a lot of circles pass through $A$ we are going to use an inverzion centered at $A$, and since we get an orthocenter style problem we are going to use $\sqrt{AH\cdot AD}$ -inverzion.Now we know that $(ABC)<->EF$, so $\phi P$ will lie on $EF$,Also $A'$ get swapet with $N$-the midpoint of $AH$,so from all of this we can say that $\phi(P)$ Will be $AP\cap CN$ , similarly $\phi(Q)$ will be $AQ\cap BN$. easier to note ,let $\phi (P)$=$Y$, and $\phi(Q)$=$X$.Also $AH\cap (ABC)=R$ Now since $N$ is the midpoint of $AH$ (and the center of $(AEF)$)we know that $NE=NF$, this gives us that $(E,F)$ isogonal W.R.T $\angle BNC$. So $(NE,NF)$,$(NB,NC)$,$(NA,NH)$ -reciprocal pairs,let $AH\cap EF$ at $L$, so "projecting" from $N$ to $EF$ we get that $(E,F)$,$(X,Y)$,$(L,L)$.Now projecting from $A$ to $(ABC)$ we get that $(B,C)$,$(Q,P)$,$(R,R)$, so there exist a point $T$ on $BC$ such that $TR^2=TB*TC$,this tell that the intersection of the tangent at $L$ and $PQ$,$BC$ are collinear. So we are done ,actually from here is a very known result. $\blacksquare$

Another non-inversive solution. Let $X$ and $Y$ be foots of perpendiculars from $E$ and $F$ to $BC$. $M$ is the miquel point of quadrilateral $BCEF$, $Z=AM \cap EF \cap BC$, and $A_{1}=AD\cap (ABC)$. Let $AX$ meet $(ABC)$ at $P'$, and let $E'$ be the reflection of $E$ over $BC$. Then by power of a point, $$XE\cdot XE'=XE^2=XB\cdot XC=XA\cdot XP'$$and so $AEP'E'$ is cyclic. However as $A,A',E,E'$ and $P$ are also concyclic, it follows that $P=P'$. Similarly, $A,Y,Q$ are collinear. Then projecting through $A$ and onto line $ZE$ shows that $$(M,A_{1};Q,P)\stackrel{A}{=}(Z,D;Y,X)=(Z,AD\cap EF;F,E)\stackrel{H}{=}(Z,D;C,B)=-1$$and so $PQ$ passes through the intersection of tangents at $M$ and $A_{1}$. But $$(M,A_{1};B,C)\stackrel{A}{=}(Z,D;B,C)=-1$$so $BC$ also passes through that point. Hence $R$ must be the intersection of tangents at $M$ and $A_{1}$. Consequently, $$\measuredangle RHB=-\measuredangle RA_{1}B=-\measuredangle A_{1}CB=\measuredangle HCB=\measuredangle FEB$$and so $EF \parallel RH$.

wait what i just realized this is pretty fast from ratio lemma lol. We do the generalized form where $A'$ is arbitrary on $AH$. Note that by ratio lemma, since the line through $H$ parallel to $EF$ is tangent to $(BHC)$, the final conclusion we want is \[\frac{BP}{CP} \cdot \frac{BQ}{CQ}=\left(\frac{BH}{CH}\right)^2.\]Let $D$ be $AH \cap BC$, and let $(AEA')$ meet $AB$ at $Y$, and $BE$ at $Z$. Angles gives $A'Z || BC$. We have that by spiral similarity, \[\frac{BP}{CP}=\frac{BY}{CE}=\frac{BY \cdot AB}{AB \cdot CE}=\frac{BE \cdot BZ}{AB \cdot CE}=\frac{BE \cdot d(B,A'Z)}{AB \cdot CE\cos(\angle BZA')}=\frac{BE \cdot d(A',BC)}{AB \cdot CE\sin(90-C)}\]\[=\frac{1}{2R\sin(C)} \cdot \tan(C) \cdot \frac{d(A',BC)}{\cos(C)}=\frac{d(A',BC)}{2R} \cdot \frac{1}{\cos^2(C)}\]and similar for $Q$, implying the result.

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$S=(AEHF)\cap (ABC), G=(ABC)\cap AH$. $DEF$ is the orthic triangle. $\textbf{Claim:}$ $RS$ is tangent to $(ABC)$. Proof: $\frac{RB}{RC}=\frac{QB}{QC} \cdot \frac{PB}{PC}$. We will use inversion centered at $A$ with $r^2=AH\cdot AD$ to find the last product. If $L$ is the midpoint of$ AH$, inverse of $P$ is $M=CL \cap EF$ and inverse of $Q$ is $N=BL\cap EF$. $CM$ and $BN$ are symmedians in triangles $ECF$ and $EBF$. Thus $\frac{PB}{PC}=\frac{\frac{r^2}{AM\cdot AF}MF}{\frac{r^2}{AM \cdot AE}ME}=\frac{MF}{ME}\cdot \frac{AE}{AF}=\frac{CF^2}{CE^2}\cdot \frac{AE}{AF}$. Similarly $\frac{QB}{QC}=\frac{BF^2}{BE^2}\cdot \frac{AE}{AF}$. $$\frac{RB}{RC}=\frac{QB}{QC} \cdot \frac{PB}{PC}=\frac{CF^2}{CE^2}\cdot \frac{AE^2}{AF^2}\cdot \frac{BF^2}{BE^2}=\frac{BK^2}{CK^2}\cdot \frac{CF^2}{BE^2}=\frac{BH^2}{BC^2}\cdot \frac{BC^2}{CH^2}=\frac{BF^2}{CE^2}=\frac{SB^2}{SC^2}$$as needed.$\square$ $-1=(G,S;B,C) \implies RH^2=RG^2=RQ\cdot RP \implies \angle RHB=\angle RCH=\angle FEH$ as desired.

Actually, $A'$ can be any point on $AH$. https://artofproblemsolving.com/community/c6h2650502_generalization_of_2019_elmo_g4

Let $AB$ and $BE$ meet $AEA'$ at $K$ and $K'$. $AC$ and $CF$ meet $AFA'$ at $T$ and $T'$. Let $R'$ be point on $BC$ such that $R'H || EF$. Claim $: T'A' || BC || A'K'$. Proof $:$ Note that $\angle FT'A' = \angle FAA' = \angle FCB$ so $T'A' || BC$. we prove the other one with same approach. Claim $:CPE$ and $PBK$ are similar as well as $QBF$ and $QCT$. Proof $:$ Note that $\angle ECP = \angle PBK$ and $\angle PEC = \angle PKA = \angle PKB$ so $CPE$ and $PBK$ are similar. we prove the other one with same approach. Now we wanted to prove $R',P,Q$ are collinear. we instead prove $\frac{R'B}{R'C} = \frac{BQ.BP}{CQ.CP}$ Note that $\angle BHR' = \angle HEF = \angle HCB$ so $HR'$ is tangent to $BHC$ so $\frac{R'B}{R'C} = \frac{BH}{CH}^2$. Also $\frac{BQ.BP}{CQ.CP} = \frac{BF}{CT} . \frac{BK}{CE}$. $\frac{BF}{CT} = \frac{BF.AC}{AC.CT} = \frac{BF.AC}{CF.CT'} = \frac{BF.AC.\sin{90+C-B}}{CF.A'T'.\sin{C}}$ $\frac{BK}{CE} = \frac{BK.AB}{CE.AB} = \frac{BE.BK'}{AB.CE} = \frac{BE.A'K'.\sin{B}}{CE.AB.\sin{90+B-C}}$ so we have $\frac{BQ.BP}{CQ.CP} = \frac{BF}{CF}.\frac{BE}{CE}.\frac{AC}{AB}.\frac{A'K'}{A'T'}.\frac{\sin{B}}{\sin{C}} . \frac{\sin{90+C-B}}{\sin{90+B-C}} = \frac{BH}{HC}.\frac{AB}{AC}.\frac{AC}{AB}.\frac{\sin{C}.HK'}{\sin{B}.HT'}.\frac{\sin{B}}{\sin{C}} = \frac{BH}{CH}.\frac{HK'}{HT'} = \frac{BH}{CH}^2$ as wanted.

Time to full triple down!. (wlog $AB \ge AC$) Make an inversion wtih center $A$ and radius $\sqrt{AH \cdot AD}$, let $N$ the midpoint of $AH$ so the inversion sends $A' \to N$, $P \to P'=CN \cap EF$, $Q \to Q'=BN \cap EF$ and let $AH \cap (ABC)=H_A$ and $EF \cap BC=G$ then $H_A \to H_A'=AD \cap EF$ so by projecting $$-1=(B, C; G, D) \overset{N}{=} (Q', P'; G, H_A') \overset{A}{=} (Q, P; AG \cap (ABC), H_A)$$Its known that $AG \cap (ABC)$ is the A-queque point so $-1=(B, C; AG \cap (ABC), H_A)$ and with the later result this gives that $RH_A$ is tangent to $(ABC)$ so reflecting over $BC$ gives $RH$ tangent to $(BHC)$ which mean $\angle HRC=\angle ABC-\angle ACB$ but also since $BFEC$ is cyclic we get that $\angle EGC=\angle ABC-\angle ACB$ so $EF \parallel HR$ thus we are done

Let $AH$ meet $BC$ at $D$. It's obvious that $H,B,C$ and $A'$ are concyclic. We claim that the circles $(HBC)$ and $(HPQ)$ are tangent to each other. This claim would solve our problem. Indeed, if that were true, by radical axis on $(ABC), (HPQ)$ and $(HBC)$, we get that $RH$ is tangent to $(HBC)$, so $\angle RHB = \angle HCB = \angle HEF$. Now we will proceed with the proof. We will use inversion. Let $M$ be the midpoint of $AH$. Consider the inversion with center $A$ and radius $\sqrt{AH \cdot AD}$. We have $AM \cdot AA' = AH \cdot AD$ so the image of $A'$ is $M$. Thus the circle $(AA'E)$ goes to the line $CM$. Similarly, the circle $(AA'F)$ goes to the line $BM$. We also know that the circle $(ABC)$ goes to the line $EF$. From these, we infer that the images of $P$ and $Q$ are precisely the points $P' = CM \cap EF$ and $Q' = BM \cap EF$, respectively. Thus, it suffies to prove that the circles $(DEF)$ and $(DP'Q')$ are tangent to each other. This is equivalent to showing that $\angle FDQ' = \angle EDP'$, which is further equivalent to (by steiners theorem) $\frac{FQ'}{Q'E} \cdot \frac{FP'}{P'E} = (\frac{DF}{DE})^2$. A simple angle chasing shows that $MF, ME$ are tangent to $(BFEC)$. Hence, in triangle $FBE$, the line $MB$ coincides with the $B$ symmedian. Thus, we have $\frac{FQ'}{Q'E} = (\frac{FB}{BE})^2$. Similarly for triangle $CFE$ we get $\frac{FP'}{P'E} = (\frac{CF}{CE})^2$. Finally, $\frac{FQ'}{Q'E} \cdot \frac{FP'}{P'E} = (\frac{FB}{BE} \cdot \frac{CF}{CE})^2 = (\frac{HD}{DC} \cdot \frac{BD}{HD})^2 = (\frac{BD}{CD})^2 = (\frac{FD}{DE})^2$, which is what we wanted.

Take the inversion $\sqrt{bc}$. $A'$ swaps with $O$, $P^*,Q^*$ are intersections of $OE^*,OF^*$ with $BC$. New Problem Statement: $ABC$ is a triangle whose orthocenter is $H$ and $D,E,F$ are altitudes. $N$ is the midpoint of $AH$ and $NB,NC$ intersect $EF$ at $P,Q$ respectively. The tangent to $(BHC)$ at $H$ intersects $BC$ at $K$ and $AK$ intersects $(AEFH)$ at $R$. Then, $A,P,Q,R$ are cyclic. Note that $KH\perp AO$. Invert the diagram from $H$ with radius $\sqrt{-HA.HD}$. New Problem Statement: $ABC$ is a triangle with orthocenter $H$ and $D$ is the altitude from $A$ to $BC$. $AH$ intersects $(ABC)$ at $L$. $(ELH)$ and $(FLH)$ intersect $(BHC)$ at $P,Q$ respectively. $AO$ and $BC$ intersect at $T$. Then, $P,Q,D,T$ are cyclic. Circumecenters of $(FHL)$ and $(EHL)$ are on $BC$ and the reflection of $(BHC)$ with respect to $BC$ is $(ABC)$. Reflect $D,T,P,Q$ to $BC$. If $(FHL)\cap (ABC)=S$ and $(EHL)\cap (ABC)=R,$ then we want to show that $S,R,D,T$ are cyclic. Denote $B',C'$ as the antipodes of $B,C$ on $(ABC)$. \[\angle ASF=90+\angle B-\angle C-\angle FSL=90-\angle C=\angle ASB'\]Thus, $S,F,B'$ are collinear. Similarily, $R,E,C'$ are collinear. Now invert from $A$ with radius $\sqrt{bc}$ (This inversion is not in the same triangle with the initial diagram so inverting twice does not have to be unnecessary). $B'$ swaps with the intersection of $AB'$ with $BC$ and $C'$ swaps with the intersection of $AC'$ with $BC$. $\angle F^*S^*B'^*=90=\angle C'^*R^*E^*$. New Problem Statement: $ABC$ is a triangle with orthocenter $H$ and $D,E,F$ are the altitudes. The perpendicular feet from $B,C,H$ to $EF,EF,AO$ are $R,S,K$ respectively. Then $R,H,K,S$ are cyclic. Let $AK\cap EF=M$. $FHKE$ is an isosceles trapezoid and we want to show that $RHKS$ is an isosceles trapezoid, too. Hence we will show that $FR\overset{?}{=}ES$. \[MF.\frac{FB}{FA}=FR\overset{?}{=}ES=EM.\frac{EC}{EA}\]\[\frac{ME}{MF}.\frac{EC}{EA}.\frac{FA}{FB}=\frac{DB}{DC}.\frac{EC}{EA}.\frac{FA}{FB}=1\]As desired.$\blacksquare$