Here is a short solution with harmonic quadrilaterals on hyperbolas Relabel $H$ as $D$. First note since the incircle of $\triangle PBC$ is tangent to $BC$ at $D$ we have $PB-PC=DB-DC$, so the locus of $P$ is a hyperbola with foci $B,C$ passing through $D$. Call this hyperbola $\mathcal{H}$. Next, note that if $I$ is the incenter of $\triangle PBC$ then clearly $(A,I;Q,D)=-1$. Projecting through $P$ onto $\mathcal{H}$, we know that $(PA\cap \mathcal{H}, PI \cap \mathcal{H}; PQ\cap \mathcal{H}, PD\cap \mathcal{H})=-1$. But since $PI$ bisects $\angle BPC$, we know $PI$ is tangent to $\mathcal{H}$ so $PI\cap \mathcal{H}=P$, and clearly $PD\cap \mathcal{H}=D$, so we know $(PA\cap \mathcal{H}, P;PQ\cap \mathcal{H},D)=-1$.But note if $R\neq D$ is the point on $\mathcal{H}$ with $AR$ tangent to $\mathcal{H}$ then $A = RR\cap DD \in AP$, so $(PA\cap \mathcal{H}, P;R,D)=-1$, hence $R = PQ\cap \mathcal{H}$ is the fixed point of line $PQ$, as desired.

MarkBcc168 wrote:

This is now my favorite "geometry fact that looks true but isn't."

wow prism lemma so op... The desired point is the point $X$ so that $A$ is the $X$-excenter of $\triangle XBC$. Let $T=k\cap\ell$. [asy][asy] //22slg6-0 //setup size(6cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn int i=0; real r=.3; pair A,B,C,H,T,Q; A=(3,12); B=(0,0); C=(14,0); H=foot(A,B,C); T=H+r*(A-H); Q=(T+r*A)/(1+r); pair A1,T1,H1,Y,P,X; A1=2*circumcenter(A,B,C)-A; T1=2*circumcenter(T,B,C)-T; H1=B+C-H; Y=extension(A,T1,B,C); P=extension(Q,Y,T,T1); X=extension(A,A1,Q,Y); //draw draw(H--A--X^^B--C,blu); draw(B--A1--C--X--B,blu+dotted); draw(B--T--C--P--B,purple+dotted); draw(P--T1--H1,purple); draw(P--X,magenta); draw(A--T1^^T--A1,red); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.15),a,linewidth(.3)); label(s,P,v);} string labels[]={"$A$","$B$","$C$","$H$","$T$","$Q$","$A'$","$T'$","$H'$","","$P$","$X$"}; //12 pair points[]={A,B,C,H,T,Q,A1,T1,H1,Y,P,X}; real dirs[]={60,-110,70,-130,180,20,-10,-160,50,0,80,-80}; pen colors[]={blu,blu,blu,blu,purple,magenta,blu,purple,purple, red,purple,blu}; for (i=0; i<12; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);} [/asy][/asy] To get rid of $E,F$, we may redefine (by Ceva-Menelaus) $Q$ as the point so that $(AT;QH)=-1$. Construct: $A',T'$ be the respective antipodes of $A,T$ in $(ABC)$, $(TBC)$; $H=\overline{AT}\cap\overline{BC}$ even though it can be any point on $\overline{AT}$. (This is sort of out of laziness.) Next, we spam prism lemma. Note that in any triangle $XYZ$ with $X$-incenter $I_x$ and incenter $I$, we have $(I,I_x;\overline{XI}\cap\overline{YZ},X)=-1$. $(T,T';P,\overline{TT'}\cap\overline{BC})=-1=(A,A',\overline{AA'}\cap\overline{BC})$ and $(AT;QH)=-1$... which we combine as follows: $(AT;QH)=(A,A',X,\overline{AA'}\cap\overline{BC})=-1$ $\implies$ $\overline{XQ},\overline{BC},\overline{A'T}$ concurrent; $(AT;QH)=(T',T;P,\overline{TT'}\cap\overline{BC})=-1$ $\implies$ $\overline{PQ},\overline{AT'},\overline{BC}$ concurrent; [asy][asy] //22slg6-1 //setup size(6cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn int i=0; real r=.3; pair A,B,C,H,T; A=(3,12); B=(0,0); C=(14,0); H=foot(A,B,C); T=H+r*(A-H); pair A1,T1,H1,Y; A1=2*circumcenter(A,B,C)-A; T1=2*circumcenter(T,B,C)-T; H1=B+C-H; Y=extension(A,T1,B,C); //draw draw(A--H^^B--C,blu); draw(A--B--A1,blu+dashdotted); draw(T--B--T1,purple+dashdotted); draw(A--T1^^T--A1,red); draw(T1--H1,purple); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.15),a,linewidth(.3)); label(s,P,v);} string labels[]={"$A$","$B$","$C$","$H$","$T$","$A'$","$T'$","$H'$",""}; //9 pair points[]={A,B,C,H,T,A1,T1,H1,Y}; real dirs[]={40,-110,90,60,140,0,0,90,0}; pen colors[]={blu,blu,blu,blu,purple,blu,purple,purple, red}; for (i=0; i<9; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);} [/asy][/asy] As a result it remains to prove that $\overline{AT'},\overline{TA'},\overline{BC}$ concurrent. By well-known configurations, $A',T'$ share a foot onto $\overline{BC}$, namely, $B+C-H$. Call it $H'$. To finish, define: $m=\overline{AT},m'=\overline{A'T'}$;* projective map $f:m\to m'$ map $P\in m$ to $P'\in m'$ such that $\angle P'BP=90^\circ$. (Think of this as a 90 degree rotation, then projected onto another line.) Using the notation $\infty_{RS}$ for the point at infinity along line $RS$ and similarly for $\infty_{\perp RS}$, we have \[(AT;H\infty_{\perp BC}) \overset f= (A'T';\infty_{\perp BC},H').\]One final application of prism lemma completes the needed concurrency. _______ * fun fact: I called these $\ell,\ell'$ on the MOP test, despite $\ell$ being used elsewhere in the problem. somehow, no docks!

Let $X = k \cap l$, $PC = b$, $PB = c$, $BC = a$, and $p=\dfrac{a+b+c}{2}$. Note that $X$ is the incenter of $\triangle PBC$, so, $CH-BH=(p-c) - (p-b) = b-c = CP-BP$. Therefore, the locus of $P$ is the "half hyperbola" with foci $B, C$ passing through $H$. Let $h$ be the complete hyperbola. Let $D \in h$, $D \neq H$, such that $AD$ is tangent to $h$ (not hard to see this point exists), and $R = AP \cap h$, $R \neq P$. Note that $PHRD$ is a harmonic quadrilateral in the hyperbola, so, $(H,D;P,R) = -1$. Also, it's a known fact that $(H,Q;X,A) = -1$ (Ceva and Menelaus). $PX$ is bisector of $\angle{BPC}$, so, a property of hyperbola says that $PX$ is tangent to $h$. Then, by projecting in $AH$ through $P$ we have that $$-1 = (H,D;P,R) = (H, PD \cap AH; X, A)$$ So, $(H,Q;X,A) = -1 = (H, PD \cap AH; X, A) \therefore Q = PD \cap AH$, a fixed point. $_\blacksquare$

CantonMathGuy wrote: MarkBcc168 wrote:

This is now my favorite "geometry fact that looks true but isn't." Can't you just handle that case by continuity?

Some notations used in the solution: Let $I, \Gamma$ be the incenter and incircle of $\triangle PBC$, respectively. $G = PB \cap \Gamma$, $J = PC \cap \Gamma $. $\Omega = (A, AH)$. Let $X$ be the point such that $\Omega$ is the $X-excircle$ of $\triangle BXC$. $M = XB \cap \Omega$, $N = XC \cap \Omega$. $\omega_{B} = (B, BH)$, $\omega_{C} = (C, CH)$. Solution. Note that $X$ is a fixed point. I'll show that $PQ$ passes through $X$ and we'll be done. Clearly, $M, G \in \omega_{B}$, $J, N \in \omega_{C}$. Moreover, $AM, AH, IG$ are tangent to $\omega_{B}$ and $AN, AH, IJ$ are tangent to $\omega_{C}$. Let $L$ be the second intersection point of circle $\omega_{B}$ with line $AG$ and $Q' = MG \cap AH$. $-1 = (M, H, G, L) = (Q', H, I, A)$ by projecting through $G$ onto $AH$. Therefore, $Q' \in EF$ implying $Q' = Q$. A similar argument shows that $NJ$ passes through $Q$. $K_{1} = GJ \cap BC$. It is well-known that $-1 = (B, C, H, K_{1})$. Similar relation holds for $K_{2} = MN \cap BC$, therefore $K_{1} = K_{2}$. Note that $\triangle MBG$ and $\triangle JCN$ are perspective since $GJ, MN, BC$ concur. Now apply Desargue's theorem for these two triangles, which gives us that the points $MB \cap NC = X$, $BG \cap CJ = P$, $MG \cap NJ = Q$ are lying on the same line and we are done.

This took quite a while to solve, even with the aid of geogebra.

We use Cartesian coordinates. Let $A=(0,1)$, $H=(0,0)$, $B=(-b,0)$, $C=(c,0)$, and $X$, the intersection of $k$ and $\ell$, be $(0,t)$. We can find $E$ as the intersection of the two lines $k$ and $AC$, which are represented by \begin{align*} y&=\frac tb(x+b)\\ y&=-\frac 1c(x-c) \end{align*}respectively, which can easily be solved obtaining \[ E=\left(\frac{bc(1-t)}{b+ct},\frac{t(b+c)}{b+ct}\right).\]Similarly \[ F=\left(-\frac{bc(1-t)}{c+bt},\frac{t(b+c)}{c+bt}\right).\]The line $EF$ has slope \[\frac{t(b+c)\left(\frac{1}{b+ct}-\frac{1}{c+bt}\right)}{bc(1-t)\left(\frac{1}{b+ct}+\frac{1}{c+bt}\right)}=\frac{t(b+c)(c-b)(1-t)}{bc(1-t)(b+c)(1+t)}=\frac{t(c-b)}{bc(1+t)},\]and so $Q$ is \begin{align*} Q&=\left(0,\frac{t(b+c)}{b+ct}-\left(\frac{bc(1-t)}{b+ct}\right)\left(\frac{t(c-b)}{bc(1+t)}\right)\right)\\ &=\left(0,\frac{t}{(b+ct)(1+t)}((b+c)(1+t)-(1-t)(c-b))\right)\\ &=\left(0,\frac{2t}{1+t}\right). \end{align*}Now note that since $k$ and $\ell$ have slopes $t/b$ and $-t/c$, the lines $PB$ and $PC$ have, by tangent double angle, slopes \[\frac{2(t/b)}{1-(t/b)^2}=\frac{2bt}{b^2-t^2}\]and \[-\frac{2(t/c)}{1-(t/c)^2}=-\frac{2ct}{c^2-t^2}.\]Thus the $x$-coordinate of $P$ satisfies \begin{align*} \frac{2bt}{b^2-t^2}(x+b)&=-\frac{2ct}{c^2-t^2}(x-c)\\ \left(\frac{b}{b^2-t^2}+\frac{c}{c^2-t^2}\right)x&=\frac{c^2}{c^2-t^2}-\frac{b^2}{b^2-t^2}\\ x&=\frac{c^2(b^2-t^2)-b^2(c^2-t^2)}{b(c^2-t^2)+c(b^2-t^2)}\\ x&=\frac{t^2(b^2-c^2)}{(b+c)(bc-t^2)}\\ x&=\frac{t^2(b-c)}{bc-t^2}. \end{align*}We can then easily find \[P=\left(\frac{t^2(b-c)}{bc-t^2},\frac{2bct}{bc-t^2}\right).\]The line $PQ$ is \begin{align*} y&=\frac{\frac{2bct}{bc-t^2}-\frac{2t}{1+t}}{\frac{t^2(b-c)}{bc-t^2}}x+\frac{2t}{1+t}\\ y&=\frac{2(bc+t)x}{(1+t)(b-c)}+\frac{2t}{1+t}. \end{align*}The fixed point \[\left(\frac{c-b}{1-bc},-\frac{2bc}{1-bc}\right),\]is clear from special values of $t$, and indeed this is correct since plugging it in we obtain \begin{align*} -\frac{2bc}{1-bc}&=\frac{2(bc+t)\frac{c-b}{1-bc}}{(1+t)(b-c)}+\frac{2t}{1+t}\\ -bc(1+t)&=-(bc+t)+t(1-bc)\\ -bc-bct&=-bc-bct \end{align*}and we are done.

Sniped for Coordinate bash... btw why G6 is coordinate bash-able? Choose coordinates so that $D(0,0)$, $A(0,1)$, $B(b,0)$, $C(c,0)$ and $M(m,0)$ for $c,m > 0$ and $b<0$, where $M$ is the common intersection point of $k$, $\ell$ and $AD$. It suffices to find a point on $PQ$ with coordinates depending only on $b$ and $c$. Recall that the equation of a line through $(A_1,B_1)$ and $(A_2,B_2)$ is $y = \frac{B_2-B_1}{A_2-A_1}{x} + \frac{A_2B_1 - A_1B_2}{A_2-A_1}$. We automatically have the following equations of lines \[ AC: y=-\frac{x}{c} + 1, \ \ AB: y=-\frac{x}{b}+1, \ \ BM: y = m\left(-\frac{x}{b}+1\right), \ \ CM: y = m\left(-\frac{x}{c}+1\right).\] Hence the intersection of $AC$ and $BM$ is $E\left(\frac{bc(1-m)}{b-cm}, \frac{m(b-c)}{b-cm}\right)$ and analogously (by swapping $b$ and $c$) the intersection of $CM$ and $AB$ is $F\left(\frac{bc(1-m)}{c-bm},\frac{m(c-b)}{c-bm}\right)$. From here we compute \[ EF: y = \frac{\frac{m(c-b)}{c-bm} - \frac{m(b-c)}{b-cm}}{\frac{bc(1-m)}{c-bm} - \frac{bc(1-m)}{b-cm}}x + \frac{\frac{bc(1-m)}{c-bm}\frac{m(b-c)}{b-cm} - \frac{bc(1-m)}{b-cm}\frac{m(c-b)}{c-bm}}{\frac{bc(1-m)}{c-bm} - \frac{bc(1-m)}{b-cm}} = \mbox{irrelevant coefficient} \cdot x + \frac{2m}{m+1} \](note firstly the cancellation of $bc(1-m)$ and then the $b-c$ factor in $(b-cm) - (c-bm)$ after clearing denominators) and in particular $Q\left(0, \frac{2m}{m+1}\right)$. To compute $P$ via $BP$ and $CP$, note that the slope of $BM$ is $-\frac{m}{b}$, so that of $BP$ is (using $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2 \theta}$) $\frac{\frac{-2m}{b}}{1 - \frac{m^2}{b^2}} = -\frac{2bm}{b^2-m^2}$, hence $BP: y = -\frac{2bm}{b^2-m^2}(x-b)$; analogously (by just replacing $b$ with $c$) $CP: y = -\frac{2cm}{c^2-m^2}(x-c)$. Solving the equation $-\frac{2bm}{b^2-m^2}(x-b) = -\frac{2cm}{c^2-m^2}(x-c)$ yields $P\left(\frac{m^2(b+c)}{bc+m^2}, \frac{2bcm}{bc+m^2}\right)$. Therefore the equation of the desired line is \[ PQ: y = \frac{\frac{2bcm}{bc+m^2} - \frac{2m}{m+1}}{\frac{m^2(b+c)}{bc+m^2}}x + \frac{2m}{m+1} = \frac{2(bc-m)}{(m+1)(b+c)}x + \frac{2m}{m+1} = \frac{2bcx - 2mx + 2mb + 2mc}{(m+1)(b+c)}.\]To obtain a constant point from here, it makes sense to aim to cancel $m+1$, so it is enough to have $2bcx + 2x - 2b - 2c = 0 $, i.e. take $x = \frac{b+c}{1+bc}$, The corresponding point has coordinates $\left(\frac{b+c}{1+bc}, \frac{2bc}{1+bc}\right)$. Remark. Actually, p.lazarov06 noticed after seeing this solution that $Q$ can be computed instantly if one observes $(A, M; Q, H) = -1$, wow!

Let \(D=k\cap\ell\), and let \(Z\) be the intersection of the reflection of \(\overline{BC}\) over \(\overline{AB}\) and the reflection of \(\overline{BC}\) over \(\overline{CA}\). Then \(A\) is an incenter or excenter of \(\triangle ZBC\) and \(D\) is an incenter or excenter of \(\triangle PBC\). We assume both are incenters without loss of generality (the other cases are analogous). We show \(Z\) is the fixed point. If we define \(Q'=\overline{ZP}\cap\overline{AD}\), we wish to show \(Q'\) is collinear with \(E=\overline{AC}\cap\overline{DB}\) and \(F=\overline{AB}\cap\overline{DC}\). By Ceva-Menelaus, we wish to show: Quote: Let \(ZBPC\) be a quadrilateral. Let \(\omega_A\) and \(\omega_D\) be the incircles and \(A\) and \(D\) the incenters of \(\triangle ZBC\) and \(\triangle PBC\), respectively. If \(\omega_A\) and \(\omega_D\) are tangent at a point \(H\) on \(\overline{BC}\), then \(Q'=\overline{ZP}\cap\overline{AD}\) satisfies \((AD;HQ')=-1\). [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=lightred; pen sec=lightblue; pen tri=purple+pink; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair Z,B,C,A,H,D,P,J,Q; Z=dir(140); B=dir(200); C=dir(340); A=incenter(Z,B,C); H=foot(A,B,C); D=H+(H-A)*.6; P=extension(B,reflect(B,D)*C,C,reflect(C,D)*B); J=extension(Z,A,P,D); Q=extension(Z,P,A,D); draw(A--Q,tri+dashed); draw(Z--Q,tri); filldraw(circle(J,abs(J-foot(J,Z,B))),sfil,sec+dashed); filldraw(Z--B--P--C--cycle,fil,pri); draw(B--C,pri); filldraw(incircle(Z,B,C),fil,pri); filldraw(incircle(P,B,C),fil,pri); dot("\(Z\)",Z,N); dot("\(B\)",B,W); dot("\(C\)",C,E); dot("\(P\)",P,SW); dot("\(A\)",A,E); dot("\(D\)",D,E); dot("\(Q'\)",Q,S); dot("\(H\)",H,NE); [/asy][/asy] Note that \[\frac{ZB+BC-ZC}2=BH=\frac{PB+BC-PC}2 \implies ZB-ZC=PB-PC,\]so \(ZBPC\) has an incircle \(\omega\) by Pitot. Note that \(H\) is the insimilicenter of \(\omega_A\) and \(\omega_D\). By Monge on \(\omega\), \(\omega_A\), \(\omega_D\), the exsimilicenter is \(Q'\). The desired harmonic bundle follows. Remark: Another proof is to draw the hyperbola \(\mathcal H\) with foci \(B\) and \(C\) through \(Z\), \(H\), \(P\), and note that \(A=\overline{ZZ}\cap\overline{PP}\) and \(D=\overline{HH}\cap\overline{PP}\). Then the desired harmonic bundle follows from duality.

This problem is proposed by Alireza Danaie from Iran.

during mop i was strongly considering coordinate bashing this problem. I realized everything computed quite easily if I parametrized the problem in terms of the intersection of $k$ and $\ell$. The only thing I had to figure out was how to reflect the x-axis over another line, which is just tangent double angle. then i forgot the tangent double angle formula existed, did not think there existed a formula expressing $\tan(2\theta)$ solely in terms of $\tan(\theta)$ in any reasonable form, and then gave up and did nothing

you couldve derived it from sin and cos double angle

We claim that the fixed point is a point $D,$ such that $A$ is a $D-$excenter of $\triangle BCD.$ Let $\omega,\Omega$ denote incircle of $\triangle PBC$ and $D-$excircle of $\triangle BCD$ - since these circles are tangent to $BC$ at the same point, by Pivot there exist a circle $\Gamma$ which is tangent to extensions of rays $PB,PC,BD,CD.$ By property of complete quadrilateral we obtain $(A,BE\cap CF,Q,H)=-1,$ hence $Q$ is the insimilicenter of $\omega ,\Omega .$ Then by Monge $P,Q,D$ are collinear, done.

Omg amazing problem I haven't felt excitement like this after solving a geo problem in quite a long time. Let the reflections of line $\overline{BC}$ over $\overline{AB}$ and $\overline{AC}$ intersect at $T$. We claim that $T$ is the desired fixed point. Let $R$ be the incenter of $\triangle PBC$ which lies on $\overline{AH}$ by the problem statement. Let $\omega$ be the incircle of $\triangle PBC$ and $\Gamma$ be the $T$-excircle of $\triangle TBC$. Then the two circles are tangent to each other internally at $H$, and $(Q, H; A, R) = -1$, so $Q$ is the internal homothetic center of $\omega$ and $\Gamma$. Note that \[ PC + BC - PB = 2CD = TB + BC - TC \Longrightarrow PC + TC = PB + TB. \]Therefore, by Pitot's theorem, there exists a circle $\Omega$ which is tangent to rays $TB$, $TC$, $BP$ and $CP$. Monge's theorem on $\omega$, $\Gamma$ and $\Omega$ shows that $\overline{PQ}$ passes through $T$.

[asy][asy] size(15cm); pair A, B, C, H, P, Q, E, F, X; A=(0,8); B=(-2,0); C=(6,0); H=(0,0); X=(0,2); E=extension(B,X,A,C); F=extension(C,X,A,B); Q=extension(E,F,A,H); P=extension(foot(C,B,X)*2-C,B,foot(B,C,X)*2-B,C); draw(P--extension(foot(C,B,A)*2-C,B,foot(B,C,A)*2-B,C)--C-- extension(foot(C,B,A)*2-C,B,foot(B,C,A)*2-B,C)--B); draw(A--B--C--A--H--B--P--C--X--B--E--Q--F--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$H$", H, S); label("$P$", P, NW); label("$Q$", Q, W); label("$E$", E, NE); label("$F$", F, W); label("$X$", X, NW); [/asy][/asy] Let $k$ and $\ell$ intersect at $X$. Let $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, and $X=(0,d)$. Then, $H=(0,0)$. The reflection of $H$ over $BX$ has coordinates $\left(\frac{2bd^2}{b^2+d^2},\frac{2b^2d}{b^2+d^2}\right)$, so line $BP$ has equation $$y=\frac{2bd}{d^2-b^2}(x-b).$$Similarly, line $CP$ has equation $$y=\frac{2cd}{d^2-c^2}(x-c).$$This implies that the coordinates of $P$ satisfy \begin{align*} \frac{2bd}{d^2-b^2}(x-b)&=\frac{2cd}{d^2-c^2}(x-c)\\ \frac{x-b}{x-c}&=\frac{c(d^2-b^2)}{b(d^2-c^2)}\\ \frac{c-b}{x-c}&=\frac{cd^2-cb^2-bd^2+bc^2}{b(d^2-c^2)}\\ x-c&=\frac{b(d^2-c^2)}{bc-d^2}\\ x&=\frac{(b-c)d^2}{bc-d^2}. \end{align*}Therefore, $P=\left(\frac{(b-c)d^2}{bc-d^2},\frac{2bcd}{bc-d^2}\right)$ and the reflections of $BC$ over $AB$ and $AC$ intersect at $\left(\frac{(b-c)a^2}{bc-a^2},\frac{2abc}{bc-a^2}\right)$. Since $(A,X;Q,H)=-1$, if $Q=(0,q)$, then we get $$(a-d)^2=(a+d-2q)(a+d)$$so $q=\frac{a+d}2-\frac{(a-d)^2}{2(a+d)}=\frac{2ad}{a+d}$. Therefore, line $PQ$ has equation $$y=\frac{2(bc+ad)}{(a+d)(b-c)}x+\frac{2ad}{a+d},$$so $\left(\frac{(b-c)a^2}{bc-a^2},\frac{2abc}{bc-a^2}\right)$ lies on this line. Therefore, $PQ$ passes through a fixed point.

We claim that the fixed point is the point $T$ such that $A$ is the $T-$excenter of $\triangle TCD.$ Let $I_P$ be the intersection of $k$ and $l$. $I_P$ is the therefore the incenter of $\bigtriangleup{PBC}$. Let $I_T$ be the incenter of $\bigtriangleup{TBC}$ and let $D = BC \cap TI_T$. It is well known that $A$ lies on the line $TI_T$ and that $-1 = (A, I_T; D, T)$. By Ceva-Menelaus, $-1 = (A, I_P; Q, H) \Rightarrow -1 = (A, I_P; H, Q)$ so: $-1 = (A, I_T; D, T) = (A, I_P, H, Q) \Rightarrow I_TI_P$, $DH$, $TQ$ concur $\Rightarrow I_TI_P \cap BC = K$ then $TKQ$ collinear. So it suffices to show $TKP$ collinear. We will proceed with Trig Ceva. Let $b = \angle{CBI_T}$, $c = \angle{BCI_T}$, $m = \angle{CBI_P}$, $n = \angle{BCI_P}$, $s = \angle{TPB}$, $t = \angle{TPC}$, $x = \angle{I_TI_PB}$, $y = \angle{I_TI_PC}$. By Trig Ceva on $\bigtriangleup{PBC}$ and the concurrency at $T$: $\sin{\angle{TPC}}\sin{\angle{TCB}}\sin{\angle{TBP}} = \sin{\angle{TCP}}\sin{\angle{TBC}}\sin{\angle{TPB}} \Rightarrow$ $\sin{t}\sin{2c}\sin{2b + 2m} = \sin{2c + 2n}\sin{2b}\sin{s} \Rightarrow \frac{\sin{s}}{\sin{t}} = \frac{\sin{2b + 2m}\sin{2c}}{\sin{2c + 2n}\sin{2b}}$ By Trig Ceva on $\bigtriangleup{I_PBC}$ and the concurrency at $I_T$: $\sin{\angle{I_TI_PC}}\sin{\angle{I_TCB}}\sin{\angle{I_TBI_P}} = \sin{\angle{I_TCI_P}}\sin{\angle{I_TBC}}\sin{\angle{I_TI_PB}} \Rightarrow$ $\sin{y}\sin{c}\sin{b + m} = \sin{c + n}\sin{b}\sin{x} \Rightarrow \frac{\sin{y}}{\sin{x}} = \frac{\sin{c + n}\sin{b}}{\sin{b + m}\sin{c}}$. $\frac{\sin{s}}{\sin{t}} \cdot \frac{\sin{y}}{\sin{x}} = \frac{\cos{b + m}\cos{c}}{\cos{c + n}\cos{b}}$ after using $\sin{2x} = 2\sin{x}\cos{x}$. Using additional formula, we have: $\frac{\sin{s}}{\sin{t}} \cdot \frac{\sin{y}}{\sin{x}} = \frac{\cos{b}\cos{c}\cos{m} - \sin{b}\sin{m}\cos{c}}{\cos{b}\cos{c}\cos{n} - \sin{c}\sin{n}\cos{b}}$. Dividing through by $\cos{b}\cos{c}$ gives: $\frac{\sin{s}}{\sin{t}} \cdot \frac{\sin{y}}{\sin{x}} = \frac{\cos{m} - \sin{m}\tan{b}}{\cos{n}-\sin{n}\tan{c}} = \frac{\cos{m}(1 - \tan{m}\tan{b})}{\cos{n}(1 - \tan{n}\tan{c})}$ If $S$ is the foot of the perpendicular from $I_T$ to $BC$, then it is well known that $CS = BH$ since the touchpoints of the incircle and excircle are reflections of each other about the midpoint of $BC$. In particular, $\frac{BH}{HC} = \frac{CS}{SB} \Rightarrow \frac{\tan{n}}{\tan{m}} = \frac{\tan{b}}{\tan{c}} \Rightarrow \tan{n}\tan{c} = \tan{m}\tan{b}$ so the above fractions simplifies to $\frac{\cos{m}}{\cos{n}} \Rightarrow \frac{\sin{x}}{\sin{y}} = \frac{\sin{s}\cos{n}}{\sin{t}\cos{m}}.$ Let $K' = PT \cap BC$. Now, $\frac{BK'}{K'C}$ = $\frac{\frac{BK'}{PK'}}{\frac{K'C}{PK'}} = \frac{\sin{s}\sin{2n}}{\sin{t}\sin{2m}}$ and $\frac{BK}{KC} = \frac{\frac{BK}{PK}}{\frac{KC}{PK}} = \frac{\sin{x}\sin{n}}{\sin{y}\sin{m}} = \frac{\sin{s}\cos{n}\sin{n}}{\sin{t}\cos{m}\sin{m}}$ by above. $=\frac{\sin{s}\sin{2n}}{\sin{t}\sin{2m}} = \frac{BK'}{K'C} \Rightarrow K = K'$ so $TKP$ is collinear as needed and we are done.

I was judged rather strongly by Mains of Coordinate System Proofs for not finishing this solution is a reasonable amount of time (whatever that means), so here is a solution with unconventional notation and every other line left as an exercise to the reader.

1, we have that each of $BP$ and $CP$ move with degree $2$, so by (Weak2) Zack's Lemma, $\deg P \le 2$. It's easy to check that $\deg P = 2$, and that for almost every $\triangle ABC$ (I think for $AB \ne AC$), the locus of $P$ is a conic. We can easily throw out the other cases. Finally, $E$ and $F$ have degree $1$, by Zack's Lemma, so $\deg EF = 1 + 1 - 1$, by Zack's Lemma, so $\deg Q = 1$, by Zack's Lemma. By $R \to D$ and $R \leftrightarrow \text{reflection of } R \text{ over BC}$, we see that the locus of $P$ is a conic passing through $D$ and symmetric over $BC$; it follows that it is tangent to $AD$. To finish, note that as $R \to D$, the cross ratio $(A, R; Q, D) = -1$ implies that $QD = 2RD + o(RD)$3, and the angle condition implies that $d(P, BC) = 2RD + o(RD)$4. It follows that, \[ P|_{R \to D} = Q|_{R \to D} \quad \text{and} \quad P'|_{R \to D} = Q'|_{R \to D}, \]which implies that $\deg PQ = 1 + 2 - 2 = 1$, by Strong5 Zack's Lemma, as desired. $\blacksquare$

Yes, I coordinate bashed it. No, I don't feel ashamed. After 2023 TSTST 6, I've learned that shame is only a social construct. Let $R = k \cap \ell$, $H'$ be the reflection of $H$ over the midpoint of $BC$, $D = AH \cap (ABC)$, and $L$ be the reflection of $H$ over $R$. Then by Ceva-Menelaus theorem, we have $\frac{AQ}{QR} = \frac{AH}{HR}$ so if $LQ = x$, we have $\frac{AH - 2RH+x}{RH - x} = \frac{AH}{RH}$ so $x = \frac{2RH^2}{RH+AH}$. Furthermore, we have $QH = 2RH - x = \frac{2RH \cdot AH}{RH + AH}$. Notice that $H$ is the $P$-intouch point and $H'$ is the $P$-extouch point of $\triangle PBC$ so it follows that if $A'$ is the antipode of $A$ on $(ABC)$ and $Q' = PQ \cap H'A'$ then by homothety between incircle and $P$-excircle, we have that $\frac{H'Q'}{x} = \frac{r_P}{r} = \frac{BH \cdot CH}{RH^2} = \frac{AH \cdot HD}{RH^2}$ so $H'Q' = \frac{2 AH \cdot HD}{RH + AH}$. Now setting up a coordinate system where $BC$ is parallel to the $x$-axis, one may notice that $\frac{HD \cdot Q - HA \cdot Q'}{HD - HA}$ is in fact a fixed expression which does not depend on the value of $RH$. Thus we have proven the existence of a fixed point lying on line $PQ$ which does not depend on $RH$, finishing.

Tangent double angle formula: We have $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$. Let $H$ be on $\overline{BC}$ (WLOG). Use coordinates with $A=(1,0),B=(b,0),C=(c,0),H=(0,0)$. Let $k \cap \ell \cap \overline{AH}:=I$ be the incenter of $\triangle PBC$, and suppose $I=(0,t)$. The slope of $\overline{BI}$ is $-\tfrac{t}{b}$, hence the slope of $\overline{BP}$ is $\tfrac{2bt}{t^2-b^2}$ by the tangent double angle formula. Therefore, the equation of $\overline{BP}$ is $y=\tfrac{2bt}{t^2-b^2}(x-b)$. Likewise, the equation of $\overline{CP}$ is $y=\tfrac{2ct}{t^2-c^2}(x-c)$. We first find $P$, whose $x$-coordinate satisfies $$\frac{b}{t^2-b^2}(x-b)=\frac{c}{t^2-c^2}(x-c) \implies \left(\frac{b}{t^2-b^2}-\frac{c}{t^2-c^2}\right)x=\frac{t^2}{t^2-b^2}-\frac{t^2}{t^2-c^2} \implies (b-c)(bc+t^2)x=t^2(b-c)(b+c) \implies x=\frac{(b+c)t^2}{bc+t^2}.$$By plugging this into either equation we then get $P=(\tfrac{(b+c)t^2}{bc+t^2},\tfrac{2bct}{bc+t^2})$. Line $\overline{AC}$ has equation $y=-\tfrac{1}{c}x+1$, so we may compute $E=\overline{BI} \cap \overline{AC}$ as $(\tfrac{bc(t-1)}{ct-b},\tfrac{(c-b)t}{ct-b})$ and similarly $F$ as $(\tfrac{bc(t-1)}{bt-c},\tfrac{(b-c)t}{bt-c})$. The slope of $\overline{EF}$ is then $$\frac{\frac{(b-c)t}{bt-c}+\frac{(b-c)t}{ct-b}}{\frac{bc(t-1)}{bt-c}-\frac{bc(t-1)}{ct-b}}=\frac{(b-c)t(ct-b+bt-c)}{bc(t-1)(ct-b-bt+c)}=-\frac{(b+c)t}{bc(t+1)},$$hence its $y$-intercept is (using $F$ and so-called "point-slope form") $$\frac{(b-c)t}{bt-c}+\frac{(b+c)t}{bc(t+1)}\cdot \frac{bc(t-1)}{bt-c}=\frac{t}{(t+1)(bt-c)}((b-c)(t+1)+(b+c)(t-1))=\frac{2t}{t+1} \implies Q=\left(0,\frac{2t}{t+1}\right).$$I then claim that the fixed point on $\overline{PQ}$ as $t$ varies is $T:=(\tfrac{b+c}{bc+1},\tfrac{2bc}{bc+1})$. We have $$\frac{\frac{2bc}{bc+1}-\frac{2t}{t+1}}{\frac{b+c}{bc+1}}=\frac{2bc-\frac{2t}{t+1}(bc+1)}{b+c}=\frac{\frac{2}{t+1}bc-\frac{2t}{t+1}}{b+c}=\frac{2bc-\frac{2}{t+1}(bc+t^2)}{(b+c)t}=\frac{\frac{2bct}{bc+t^2}-\frac{2t}{t+1}}{\frac{(b+c)t}{bc+t^2}},$$hence the slopes of $\overline{PQ}$ and $\overline{TQ}$ are the same, so we are done. $\blacksquare$ Remark: Guessing $T$ can be done as follows. Send $t \to -1$ to get that $T$ lies on the line joining $(\tfrac{b+c}{bc+1},-\tfrac{2bc}{bc+1})$ to "$(0,\pm \infty)$", which is basically the vertical line $x=\tfrac{b+c}{bc+1}$. Plug in $t=1$ to get that $T$ lies on the line joining $(\tfrac{b+c}{bc+1},\tfrac{2bc}{bc+1})$ to $(0,1)$. Hence we should expect $T=(\tfrac{b+c}{bc+1},\tfrac{2bc}{bc+1})$.

Let $D$ be the intersection of $k$ and $\ell$. Note that $(A,D;Q,H)=-1$ so so $Q$ is the insimilicenter of the circle with center $D$ passing through $H$ and the circle with center $A$ passing through $H$. Let the tangent from $B$ and $C$ to circle $A$, neither of which is $BC$, intersect at $R$. We have $BP-PC=BH-HC=RC-BR$ so by Pitot's there exists a circle $\omega$ externally tangent to all four sides of $BPCR$. Thus, $R$ is the exsimilicenter of circle $A$ and $\omega$ and $P$ the insimilcenter of circle $D$ and $\omega$. By Monge's we are done.

oo hyperbolas!

Sad Problem

Let $M,N,H$ be the incircle touchpoints, $L,J$ be foot of altitudes, $K$ be $AO \cap (BOC)$. By Prism Lemma, $EF,LJ,BC,MN$ concur, $A',T'$ be antipodes of $A,T$ in $(ABC),(TBC)$. Then keep using Prism lemma to get $PQ,BC,IA'$ concur, $PQ,AT'BC$ concur. It duffices to prove $AT',BC,IA'$ concur which follows by another application of prism lemma.

DottedCaculator wrote: [asy][asy] size(15cm); pair A, B, C, H, P, Q, E, F, X; A=(0,8); B=(-2,0); C=(6,0); H=(0,0); X=(0,2); E=extension(B,X,A,C); F=extension(C,X,A,B); Q=extension(E,F,A,H); P=extension(foot(C,B,X)*2-C,B,foot(B,C,X)*2-B,C); draw(P--extension(foot(C,B,A)*2-C,B,foot(B,C,A)*2-B,C)--C-- extension(foot(C,B,A)*2-C,B,foot(B,C,A)*2-B,C)--B); draw(A--B--C--A--H--B--P--C--X--B--E--Q--F--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$H$", H, S); label("$P$", P, NW); label("$Q$", Q, W); label("$E$", E, NE); label("$F$", F, W); label("$X$", X, NW); [/asy][/asy] Let $k$ and $\ell$ intersect at $X$. Let $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, and $X=(0,d)$. Then, $H=(0,0)$. The reflection of $H$ over $BX$ has coordinates $\left(\frac{2bd^2}{b^2+d^2},\frac{2b^2d}{b^2+d^2}\right)$, so line $BP$ has equation $$y=\frac{2bd}{d^2-b^2}(x-b).$$Similarly, line $CP$ has equation $$y=\frac{2cd}{d^2-c^2}(x-c).$$This implies that the coordinates of $P$ satisfy \begin{align*} \frac{2bd}{d^2-b^2}(x-b)&=\frac{2cd}{d^2-c^2}(x-c)\\ \frac{x-b}{x-c}&=\frac{c(d^2-b^2)}{b(d^2-c^2)}\\ \frac{c-b}{x-c}&=\frac{cd^2-cb^2-bd^2+bc^2}{b(d^2-c^2)}\\ x-c&=\frac{b(d^2-c^2)}{bc-d^2}\\ x&=\frac{(b-c)d^2}{bc-d^2}. \end{align*}Therefore, $P=\left(\frac{(b-c)d^2}{bc-d^2},\frac{2bcd}{bc-d^2}\right)$ and the reflections of $BC$ over $AB$ and $AC$ intersect at $\left(\frac{(b-c)a^2}{bc-a^2},\frac{2abc}{bc-a^2}\right)$. Since $(A,X;Q,H)=-1$, if $Q=(0,q)$, then we get $$(a-d)^2=(a+d-2q)(a+d)$$so $q=\frac{a+d}2-\frac{(a-d)^2}{2(a+d)}=\frac{2ad}{a+d}$. Therefore, line $PQ$ has equation $$y=\frac{2(bc+ad)}{(a+d)(b-c)}x+\frac{2ad}{a+d},$$so $\left(\frac{(b-c)a^2}{bc-a^2},\frac{2abc}{bc-a^2}\right)$ lies on this line. Therefore, $PQ$ passes through a fixed point. I think there might be some mistake with $P$ in your answer. In the case of $\angle{BXC}=90$, which happens if $a^2=-bc,$ you can't find $P.$ So I think $P=?/(bc+a^2)$ This is my answer (but in Chinese): https://hackmd.io/@sine/22G6

$ABC$ is acute

solved with hint on fixed point Let $R$ be the incenter of $\triangle BPC$, let $D$ be the point such that $\triangle DBC$ has $D$-excenter of $A$. I claim that $D$ is the fixed point. Consider hyperbola $\mathcal{H}$ with foci at $B$ and $C$ which passes through $D,P,H$. Since $PR$ bisects $\angle BPC$ then $PP\cap HH=R$. Then since $DD\cap HH=A$ we can write \[-1=(D,H;P,AP\cap \mathcal{H})\stackrel{P}{=}(DP\cap AH,H;R,A)\]hence $DP\cap AH=Q$ and we're done.

asdf334 wrote: solved with hint on fixed point (the green mop test gave us the fixed point for free)

funny problem. Let $\mathcal H$ be the hyperbola with foci $B,C$ through $H$; as $H$ is the intouch point in $PBC$ we find $H$ lies on $\mathcal H$. To finish, if $I$ is the incenter of $PBC$ then $-1=(AI;QD)$ by ceva-menelaus and projecting this through P onto the hyperbola we get $-1=(PA\cap \mathcal H, P; PQ\cap \mathcal H, D )$ since $PI$ bisects $\angle BPC$ so its tangent to $\mathcal H$. We conclude that $PQ\cap H$ is then where the other tangent from $A$ hits $\mathcal H$ which is fixed.

Very Beautiful Problem : Let $\mathcal{H}$ be the hyperbola with focii $B$ and $C$ that passes through $H$. Since $PB-PC=HB-HC$, $P$ lies on one branch of $\mathcal{H}$. Let $PQ$ intersect $\mathcal{H}$ at $R$. Let $AP$ intersect the $\mathcal{H}$ at $T$. Let $I$ be the incenter of $BCP$. Notice that $PI$ is tangent to $\mathcal{H}$. Notice that $(A,I;Q,H)$ is a harmonic bundle. Projecting the bundle through $P$ onto $\mathcal{H}$ we get that $(T,P;R,H)$ is also a harmonic bundle. Projecting through $A$ onto $\mathcal{H}$ we get that $AR$ must be tangent to $\mathcal{H}$. So $R$ is uniquely defined, concluding the proof.

Whoa, this problem is really cool. Also, we really don't need hyperbolas. We claim the fixed point is the intersection point of the reflections of $BC$ across $AB$ and $AC$. Note that if $k \cap l = R$, we can encode $Q$ as the unique point such that $(A, R; Q, H) = -1$. We can hence reformulate the problem as follows. reformulated problem wrote: Consider a triangle $ABC$ with incenter $I$ and foot $D$ from $I$ to $BC$. Consider a variable point $A'$ with incenter $I'$ lying on $ID$. Let $K = AA' \cap ID$. then prove $(I, I'; K, D) = -1$. To prove the reformulated statement, let $\omega_1$ be the incircle of $ABC$ and $\omega_2$ the incircle of $A'BC$. Note that $AB - AC + BC = 2BD = A'B - A'C + BC$, so $AB + A'C = A'B + AC$. So $ABA'C$ inscribes an incircle $\omega$. Now $D$ is the exsilimicenter of $\omega_1, \omega_2$ so it suffices to prove $K$ is the insilimicenter. But this follows from Monge on $\omega, \omega_1, \omega_2$.