navi_09220114 wrote: Let $ABC$ be an actue triangle with $AB<AC$. Let $\Gamma$ be its circumcircle, $I$ its incenter and $P$ is a point on $\Gamma$ such that $\angle API=90^{\circ}$. Let $Q$ be a point on $\Gamma$ such that $$QB\cdot\tan \angle B=QC\cdot \tan \angle C$$Consider a point $R$ such that $PR$ is tangent to $\omega$ and $BR=CR$. Prove that the points $A, Q, R$ are colinear. What is $\omega$? Incircle? Btw, nice to see another SD-point involved problem (as $P$ satisfies both the properties of lying on the circumcircle, along with $\angle API = 90^{\circ}.$)

Yes, that point that you mentioned is exactly this $P$ in this problem

A quick solution using Involution. navi_09220114 wrote: Let $ABC$ be an actue triangle with $AB<AC$. Let $\Gamma$ be its circumcircle, $I$ its incenter and $P$ is a point on $\Gamma$ such that $\angle API=90^{\circ}$. Let $Q$ be a point on $\Gamma$ such that $$QB\cdot\tan \angle B=QC\cdot \tan \angle C$$Consider a point $R$ such that $PR$ is tangent to $\textcolor{red}{\Gamma}$ and $BR=CR$. Prove that the points $A, Q, R$ are colinear. [asy][asy] defaultpen(fontsize(9pt)); size(7cm); pair A,B,C,M,N,O,A1,I,P,D,Q,V,R,A2; A=dir(120); B=dir(210); C=dir(330); I=incenter(A,B,C); D=foot(A,B,C); O=circumcenter(A,B,C); M=dir(90); A1=-A+2O; N=-M+2*foot(O,O,M); P=-A1+2*foot(O,A1,I); Q=-N+2*foot(O,N,D); V=extension(A,I,B,C); R=extension(Q,A,N,M); A2=extension(B,C,N,A1); draw(A--B--C--A); draw(A--D); draw(circumcircle(A,B,C)); draw(A1--P); draw(circumcircle(A,I,P)); draw(N--Q); draw(A--V,linewidth(0.3)); draw(A1--Q); draw(N--A2); draw(V--N,linewidth(0.3)); draw(Q--R,linewidth(0.3)); draw(C--A2); draw(N--R,linewidth(0.3)); draw(R--P,linewidth(0.3)); draw(arc(circumcenter(A,D,N),circumradius(A,D,N),-10,-230),dotted); draw(arc(circumcenter(B,I,C),circumradius(B,I,C),0,180)); dot("$A$" , A , dir(A)); dot("$B$" , B , dir(B)); dot("$C$" , C , dir(C)); dot("$M$" , M , dir(M)); dot("$N$" , N , dir(N)); dot("$A^*$" , A1 , dir(A1)); dot("$I$" , I , dir(270)); dot("$P$" , P , dir(P)); dot("$D$" , D , dir(270)); dot("$Q$" , Q , dir(Q)); dot("$V$" , V , dir(V)); dot("$R$" , R , dir(R)); dot("$A'$" , A2 , dir(A2)); [/asy][/asy] Let $D$ be the foot of $A$ on $\overline{BC}$. Let $M$ be the midpoint of the arc $\widehat{BAC}$ and $N$ be the antipode of $M$ WRT $\odot(ABC)$. We characterize $Q$ by showing that $Q=\overline{ND}\cap\odot(ABC)$. Notice that $\frac{QB}{QC}=\frac{BD}{DC}=\frac{\tan C}{\tan B}$. Let $A^*$ be the $A-$ antipode WRT $\odot(ABC)$ and let $\overline{AI}\cap\overline{BC}=V$. We show that $Q\in\overline{VA^*}$. Consider an Inversion $\Psi$ around $(BIC)$. $\Psi(V)=A$ , $\Psi(Q)=D$ and $\Psi(A^*)=\overline{NA^*}\cap\overline{BC}=A'$. It's obvious to see $A,D,N,A'$ concyclic $\overset{\Psi}{\implies} Q,V,A^*$ are collinear. Now Consider the Involution $\Psi$ on $\overline{AI}$. $\Psi$ swaps $(I,I),(A,V),(N,\infty_{\overline{AI}})$. Now Project these pairs from $A^*$ on $\odot(ABC)$ to get that $(P,P),(M,N),(A,Q)$ are pairs of Involution on $\odot(ABC)$. Hence, $\overline{PP},\overline{MN},\overline{AQ}$ are concurrent. $\quad\blacksquare$

Nice solution using involutions. My original method (two of them) is as below: Let $P'R$ be another tangent line of $\Gamma$ different from $PR$, then since $BR=CR$, $P$ and $P'$ must be symmetric to the perpendicular bisector of $BC$. It is sufficient to prove that the quadrilateral $APQP'$ is harmonic. Now, we present $2$ ways to prove this fact, each uses the following lemma: \textbf{Lemma.} Given a triangle $\triangle ABC$ and a point $X$ on arc $BC$ opposite to $A$. Suppose $AX$, $BC$ intersect at $D$, then $$\frac{AB}{AC}\cdot\frac{XB}{XC}=\frac{BD}{DC}$$\textbf{Proof.} By Generalized Angle Bisector Theorem, we have $$\frac{BD}{DC}=\frac{AB}{AC}\cdot \frac{\sin \angle BAD}{\sin\angle CAD}=\frac{AB}{AD}\cdot \frac{\sin \angle BCX}{\sin\angle CBX}=\frac{AB}{AD}\cdot \frac{XB}{XC}$$which proves this lemma. \textbf{Method 1.} From above paragraph, it is enough to show that $\displaystyle \frac{AP}{AP'}=\frac{QP}{QP'}$. Let $X$ be the antipodal point of $A$ with respect to $\Gamma$. Since $\angle API=90^{\circ}$, then $P, I, X$ are colinear. Denote $I_a$ be the $A$-excenter of $\triangle ABC$, we claim that $P', X, I_a$ are also colinear. Suppose $P'X$ intersect $AI$ at $J$, and $M$ is the midpoint of arc $BC$ not contaning $A$, then since $PP'\parallel BC$, $M$ is also the midpoint of arc $PP'$ not containing $A$. So $\angle PAI=\angle QAJ$. Since we also have $\angle API=\angle AQJ=90^{\circ}$, we have $\triangle API\sim \triangle AQJ$, which implies $\angle XIJ=\angle AIP=\angle AJQ=\angle XJQ \Rightarrow XI=XJ$. Since we know that $MX\perp IJ$, then $MI_a=MI=MJ \Rightarrow J=I_a$. So $\triangle API\sim \triangle AQI_a$, and we have $$\frac{AP}{AP'}=\frac{PI}{P'I_a}$$On the other hand, if $D=QX\cap BC$, then by the lemma, $$\frac{BD}{DC}=\frac{QB}{QC}\cdot\frac{XB}{XC}=\frac{\tan \angle C}{\tan \angle B}\cdot \frac{\sin\angle BCX}{\sin\angle CBX}=\frac{\tan \angle C}{\tan \angle B}\cdot \frac{\sin (90^{\circ}-\angle C)}{\sin (90^{\circ}-\angle B)}$$$$=\frac{\tan \angle C}{\tan \angle B}\cdot \frac{\cos\angle C}{\cos\angle B}=\frac{\sin\angle C}{\sin\angle B}=\frac{AB}{AC} $$So $D$ is the intersection of the internal angle bisector of $\angle BAC$ with $BC$, which means that the lines $II_a, BC, QX$ concur at $D$. Since $\angle IBI_a=\angle ICI_a=90^{\circ}$, so $BICI_a$ is cyclic. By Power Of Point Theorem, $$DI\cdot DI_a=DB\cdot DC=DQ\cdot DX$$so $QIXI_a$ is cyclic. So we have $\angle QIP=\angle QIX=\angle QI_aP'$. Together with the fact that $\angle QPI=\angle QPX=\angle QP'X=\angle QP'I_a$, we have $\triangle QPI\sim \triangle QP'I_a$. This implies that $$\frac{AP}{AP'}=\frac{PI}{P'I_a}=\frac{QP}{QP'}$$and we are done. $\blacksquare$ \textbf{Comment.} Alternatively, once proving that $P'X \cap AI=I_a$ and $D$ is the intersection of the internal angle bisector of $\angle BAC$ with $BC$ , one can finish the solution immidiately by noting that $(A,Q;P,P')=(A,D;I,I_a)=-1$. Sine rule also works directly by noting $$\frac{AP}{AP'}=\frac{AI}{AI_a}=\frac{DI}{DI_a}=\frac{\sin \angle DXI}{\sin \angle DXI_a}=\frac{\sin \angle QP'P}{\sin \angle QPP'}=\frac{QP}{QP'} $$ \textbf{Method 2.} This time we let $X$ be the midpoint of arc $BC$, and we prove that $(A,Q;P,P')=-1$. Denote $H$ be the feet of altitude from $A$ to $BC$, $AX\cap BC=J$, and $D$, $E$, $F$ be the touch points of the incircle to of $BC$, $AB$, $AC$ respectively. Note that $$ \frac{BH}{CH}=\frac{BH}{AH}\cdot \frac{AH}{CH}=\frac{\tan \angle C}{\tan \angle B}$$By the lemma, if $QX\cap BC=H'$, then $$\frac{BH'}{CH'}=\frac{QB}{QC}\cdot\frac{XB}{XC}=\frac{\tan \angle C}{\tan \angle B}=\frac{BH}{CH}$$Since the quantity $\displaystyle f(Y)=\frac{BY}{CY}$ is strictly increasing as $Y$ moves from $B$ to $C$ in segment $BC$, we must have $H=H'$. Now we claim that $P, D, X$ are colinear. Since $\angle API=\angle AEI=\angle AFI=90^{\circ}$, then $APEIF$ is cyclic. Note that $\angle PBE=\angle PBA=\angle PCA=\angle PCF$ and $\angle PEB=180^{\circ}-\angle PEA =180^{\circ}-\angle PFA=\angle PFC$, so $\triangle PBE\sim \triangle PFC$. Note that $PX$ is the angle bisector of $\angle BPC$, so it is sufficient to prove that $DX$ bisects $\angle BPC$. But this is true because $BD=BE$, and $CD=CF$, so $$\frac{BP}{PC}=\frac{BE}{CF}=\frac{BD}{CD}$$which proves the claim. So if $G$ is the touch point of the $A$-excircle and $BC$, then $G$ is symmetric to $D$ with respect to segment $BC$, so $P', G, X$ are also colinear. Consider the antipodal point $D'$ of $D$ with respect to the incircle, then since the tangent to incircle at $D'$ is parallel to $BC$, the homothety that maps the incircle to $A$-excircle maps $D'$ to $G$. So $A, D', G$ are colinear. Finally, we obtain $$(A,Q;P,P')=(AX,AQ;AP;AP')=(J,H;D,G)=(AJ,AH;AD,AG)=(I,\infty;D,D')=-1$$and the problem is solved. $\blacksquare$