We restate the problem as follow Proposition. We are given a triangle $\triangle ABC$ $(AC>AB).$ Circle $\omega$ with center $X$ and radius $\rho$ is tangent to the circumcircle $(O)$ and $\overline{BC}$ through $A,W,$ respectively. Perpendicular to $BC$ through $C$ cuts the external bisector of $ \angle BAC$ at $P.$ Then $B,X,P$ are colllinear. Obviously, $ AW$ bisects $ \angle BAC$ and $X \in AO,$ hence $ \angle XAW = \frac {_1}{^2}(\angle B -\angle C)$ $\Longrightarrow \ \rho = \frac {AW}{2 \cos(\frac {\angle B-\angle C}{2})} \ (1)$ Since $APCW$ is cyclic $\Longrightarrow$ $ \angle CPW= \angle CAW = \frac {_1}{^2}\angle A$ $\Longrightarrow$ $PC= CW \cdot \cot\frac {A}{2}$ On the other hand, $ CW = AW \cdot \frac {\sin\frac {A}{2}}{\sin C} \ (2).$ Then, combining $(1)$ and $(2)$ yields $\frac {PC}{\rho} = \frac {2\cos\frac {(\angle B -\angle C)}{2} \cdot \cos\frac {A}{2}}{\sin C}$ Combining this latter expression with the identities $\frac{\cos \frac {\angle B - \angle C}{2}}{\sin \frac {A}{2}}= \frac {b + c}{a}$ and $\sin A = 2\sin \frac {A}{2} \cdot \cos \frac {A}{2}$ $\Longrightarrow \ \frac {PC}{\rho}= \frac {b + c}{c}.$ But $ \frac {b + c}{c} = \frac {CB}{WB} \Longrightarrow \ \frac {PC}{\rho} = \frac {CB}{WB}$ $\Longrightarrow P,X,B$ are collinear.

Let $ K$ be, the center of the circle $ (K),$ instead of $ \omega$ and let $ O$ be, the center of the circle $ (O),$ instead of $ \Gamma$ and we denote as $ X,$ the midpoint of the arc $ AB,$ not containing the point $ S.$ From $ KT\parallel OX$ and $ \frac {SK}{SO} = \frac {KT}{OX},$ we conclude that the points $ S,\ T,\ X,$ are collinear. So, the line segment $ ST,$ is the angle bisector of the angle $ \angle S,$ of the triangle $ \bigtriangleup SAB.$ We denote the point $ P,$ as the intersection point of the line segments through vertices $ S,\ B$ and perpendicular to $ ST,\ AB$ respectively $ ($ $ SP$ is the external bisector of $ \angle S$ $ )$ and it is enough to prove that the line segment $ BP,$ passes through the point $ K.$ EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given with circumcircle $ (O)$ and let $ T,\ X$ be, the points of intersections of $ BC,\ (O)$ respectively, from the angle bisector of the angle $ \angle A.$ Through the vertex $ C,$ we draw the line perpendicular to $ BC,$ which intersects the external bisector of $ \angle A,$ at point so be it $ P$ and let be the point $ K\equiv OA\cap BP.$ Prove that $ KT\perp BC.$ PROOF. – [ t=265876 ] - Through the vertex $ B,$ we draw the line perpendicular to $ BC,$ which intersects the line segment $ AP,$ as the external bisector of $ \angle A,$ at point so be it $ Q$ and let be the point $ A'\equiv (O)\cap AO.$ It is easy to show that the line segments $ BQ,\ BA',$ are isogonal conjugates with respect to the angle $ \angle B,$ from $ QB\perp BC$ and $ AB\perp BA'.$ Similarly, the line segments $ CP,\ CA',$ are isogonal conjugates with respect to the angle $ \angle C.$ Because of now, the external angle bisector of $ \angle A,$ is a particular case of two lines through the vertex $ A,$ isogonal conjugates with respect to the angle $ \angle A,$ based on the Jacobi theorem ( I like better to say the Isogonic theorem ), we conclude that the line segments $ AA'\equiv AO,\ BP,\ CQ,$ are concurrent at one point so be it $ K$ and we will prove that $ KT\perp BC.$ $ \bullet$ From $ BQ\parallel CP$ $ \Longrightarrow$ $ \frac {KB}{KP} = \frac {BQ}{CP}$ $ ,(1)$ Because of the triangles $ \bigtriangleup AQB,\ \bigtriangleup APC$ have $ \angle BAQ = \angle CAP$ and $ \angle AQB + \angle APC = 180^{o},$ we conclude that $ \frac {BQ}{CP} = \frac {AB}{AC}$ $ ,(2)$ Because of $ AT$ is the angle bisector of $ \angle A,$ we have that $ \frac {AB}{AC} = \frac {TB}{TC}$ $ ,(3)$ From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {KB}{KP} = \frac {TB}{TC}$ $ ,(4)$ From $ (4)$ we conclude that $ KT\parallel PC$ $ \Longrightarrow$ $ KT\perp BC$ and the proof of the equivalent problem is completed. REMARK. - It is easy to show that the circle $ (K)$ centered at $ K$ with radius $ KA = KT$ ( easy to prove ), tangents internaly to $ (O)$ at point $ A$ and also tangents to $ BC,$ at point $ T$ and then, we have the configuration as the proposed problem states. REFERENCE. - http://www.mathlinks.ro/Forum/viewtopic.php?t=154396 Kostas Vittas.

Let the tangents at $ S,T$ of $ \omega$ meet at $ Q$ For any point $ X$ denote by $ h_X$ the distance of the point $ X$ from line $ QO$ We have $ QT^2 = QA\cdot QB\implies \frac {QB}{QT} = \frac {QT}{QA}\implies \frac {QB}{QT} = \frac {QT}{QA} = \frac {QB - QT}{QT - QA} = \frac {TB}{TA}$ And since $ \frac {h_T}{h_A} = \frac {QT}{QA}$ we have $ \frac {h_T}{h_A} = \frac {TB}{TA}$ Since $ QO\perp TS$ $ PS\perp TS\iff PS||QO\iff h_P = h_S\iff h_A\cdot \frac {PO}{AO} = h_T\iff$ $ \frac {PO}{AO} = \frac {h_T}{h_A}\iff \frac {PO}{AO} = \frac {TB}{TA}\iff\Delta PAB\sim \Delta OAT\iff PB\perp AB$ $ QED$

Dear Mathlinkers, 1. (0) the circle going through S, A and B 2. (1) the circle tangent to AB and internally to 0 2. X the midpoint of the arc AB, not containing the point S 3. Tx the tangent to 0 at X 4. According to Reim's theorem applied to (0) and (1), Tx // ATB 5. 2 the circle with diameter PT 6. According to a converse of Reim's theorem aplied to (0) and (2), X, T and the second intersection of 0 and 2à (other than B) are collinear. 7. We are done... Sincerely Jean-Louis

Dear Jean-Louis, Can you please state the Reim's theorem? (I searched in google and almost all the hits refer to ML-AOPS) I know and can prove (without Reim's theorem) that $ X,T,S$ are collinear. But could not use it to solve this problem when I tried. So I want to understand your solution. Thank you very much Akashnil

Dear Mathlinkers, for the Reim's theorem (general case and particular case), for the two converses, see http://perso.orange.fr/jl.ayme then: à propos Sincerely Jean-Louis

I do not understand the language ...

vittasko wrote: Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$ and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$ respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$ which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ . Proof. Suppose w.l.o.g. $ b > c$ . Observe that $ \|\begin{array}{c} m(\angle KAB) = 90^{\circ} - C \\ \\ m(\angle KAP) = 90^{\circ} - \frac {B - C}{2} \\ \\ m(\angle APC) = 90^{\circ} - \frac {B - C}{2}\end{array}\|$ . Denote $ \|\begin{array}{c} D\in BC \\ \ AD\perp BC\end{array}\|$ . Therefore, $ \frac {KB}{KP} = \frac {AB}{AP}\cdot\frac {\sin\widehat {KAB}}{\sin\widehat {KAP}} =$ ${ |\frac {c\cdot\sin (90^{\circ} - C)}{AP\cdot \cos\frac {B - C}{2}}}| =$ $ \frac {c\cdot |\cos C|}{DC} = \frac cb$ . In conclusion, $ \frac {KB}{KP} = \frac {TB}{TC}$ $ \implies$ $ KT\parallel PC$ $ \implies$ $ KT\perp BC$ .

vittasko wrote: Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$ and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$ respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$ which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ . Proof (proiectively). Denote $ L\in BC\cap AP$ , the diameter $ [XY]$ , the line $ d$ for which $ K\in d$ , $ d\parallel PC$, i.e. $ d\perp BC$ and the points $ U\in AP\cap d$ , $ V\in PT\cap LK$ , $ W\in LK\cap PC$ , $ T_1\in d\cap AX$ , $ T_2\in d\cap BC$ . Observe that $ KU=KT_1$ because $ K\in AO$ - the $ A$-median in $ \triangle AXY$ and $ \overline {UKT_1}\parallel \overline {XOY}$ . Thus, the division $ \{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the pencil $ P\{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the divison $ \{K,W;L,V\}$ is harmonically. Therefore, $ KT_2\parallel WC$ $ \Longrightarrow$ $ KT_2=KU$ $ \Longrightarrow$ $ KT_1=KT_2$ $ \Longrightarrow$ $ T_1\equiv T_2\equiv T$ $ \Longrightarrow$ $ KT\perp BC$ .

Akashnil, Reim's theorem is stated here:http://alt1.mathlinks.ro/viewtopic.php?t=35337

Thank you dgreenb801. But I still can't understand the way it is used here. Can anyone help me?

This problem have many good geometry method，http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=15&Id=15330 see one of then：

It is sufficient to show that the perpendicular to $ ST$ at $ S$, the perpendicular to $ AB$ at $ B$, and $ AO$ concur. We show this by trig Ceva on triangle $ SAB$: Let the perpendicular to $ ST$ at $ S$ meet $ AO$ at $ P$, and the perpendicular to $ AB$ at $ B$ meet $ AO$ at $ P'$. Then it suffices to show that \[ \frac{\sin{\angle BSP}}{\sin{\angle ASP}} \frac{\sin{\angle SAP}}{\sin{\angle BAP}} \frac{\sin{\angle ABP}}{\sin{\angle SBP}} = 1\] Let $ \angle SAB = \alpha, \angle SBA = \beta$. We prove a lemma: Lemma. $ \angle TSA = \angle TSB$. This is fairly well-known; extend $ ST$ to hit the outer circle at $ R$, and let $ AS$ intersect $ \omega$ at $ A'$. Then, $ \angle ATA' = \angle TSA'$ because $ AT$ is tangent to $ \omega$. However, by the homothety centered at $ S$ taking $ \omega$ to the outer circle, $ A'T || AR$. Hence, $ \angle ATA' = \angle TAR$, which equals $ \angle RSB = \angle TSB$ because they intercept the same arc. $ \blacksquare$ So, we may let $ \angle TSB = \angle TSA = \gamma$. Now, $ \angle BSP = 90+\gamma, \angle aSP = 90-\gamma$ because $ SP \perp TS$. Hence, $ \displaystyle\frac{\sin{\angle BSP}}{\sin{\angle ASP}} = 1$. Also, $ \angle SAP = 90-\alpha$, $ \angle BAP = 90$. It suffices to show that $ \displaystyle \frac{\sin{\angle ABP}}{\sin{\angle SBP}} = \frac{1}{\sin{90-\alpha}$. But this is evident from considering triangles $ BTO, BSO$: $ \sin \angle ABP = \frac{TO}{BO} = \frac{SO}{BO} = \frac{\sin{\angle SBP}}{\sin \angle OSB}$, but $ \angle OSP$ = $ 90-\angle OSM = 90-\angle SAB = 90-\alpha$, where $ M$ is an arbitrary point on the common tangent to the circles at $ S$ on the same side of $ OS$ as $ B$. So, we are done.

First, I will show that if $PB \perp PA$ implies that $SP \perp ST$, then $SP \perp ST$ implies $PB \perp PA$. If $SP \perp ST$, define $Q$ on $AO$ such that $P'B \perp BA$. By assumption, we must have $SQ \perp ST$. Hence, both $P$ and $Q$ lie on the intersection of $OS$ and the perpendicular to $ST$ through $S$, yielding $P = Q$. The result follows. Consider the homothety centered at $S$ that maps $\omega$ to $\Gamma$. Let the images of $A$, $B$, $O_1$, $P_1$, and $T$ under this homothety be $A_1$, $B_1$, $O_1$, $P_1$, and $T_1$, respectively. The tangent to $\Gamma$ at $T_1$ is parallel to $AB$, so $T$ is the midpoint of arc $AB$, whence $\angle B_1 S T_1 = \angle T _1 S A_2$. Let $P_1'$ be the intersection of the perpendicular to $ST_1$ through $S$ and the perpendicular to $A_1 B_1$ through $B_1$. It is sufficient to show that $P_1'$, $O_1$, and $A_1$ are collinear, for then both $P_1$ and $P_1'$ must lie on $AO$ and the perpendicular to $B_1 A_1$ through $B$, which implies they are equal. Perform an inversion centered at $T_1$ with arbitrary radius; let the images of $P_1'$, $S$, $A_1$, $O_1$, and $B_1$ be $P_2$, $S_2$, $A_2$, $O_2$ and $B_2$, respectively. $\angle T_1 S_1 B_1 = \angle T_1 S_1 A_1$, yielding $\angle T_1 B_2 S_2 = \angle T_1 A_2 S_2$, whence $S_2 A_2 = S_2 B_2$. $O_1$ is sent to the reflection of $T_1$ across the line through $S_2$ parallel to $A_2 B_2$. $P_2$ lies on $B_2 S_2$ and $P_2 T_1 \perp B_2 S_2$ (since $T _1 B_1 P_1 S$ is cyclic, and $T_1 B_1 \perp P_1 B_1$.) We wish to show that $T_1 P_2 O_2 A_2$ is cyclic. Let $X$ be the intersection of $B_2 S_2$ and the line through $A_2$ perpendicular to $A_2 B_2$. Note that $XO_2 || A_2 T_1$. It follows that $XP_2 T_1 A_2$ and $T_1 P_2 O_2 X$ are cyclic, so $O_2 P_2 T_1 A_2$ is cyclic, as desired.

my solution: consider $M$ and $M'$ are midpoint of arc $AB$ and $ASB$, respectively. as we know $S,T,M$ are colliner(a useful lemma!). i want to prove that if $P$ is a point on $AO$ such that $PS \bot ST$ then $ PB \bot BT$ . $PS\bot ST$ so $P,S,M'$ are colliner. so it's sufficient to prove that$ STBP$ is cyclic. for this i will prove that $\angle BSP=\angle BTP$ but this is equivalence to $TP \| AM'$. so i define $L$ as a line through $T$ parall to $AM'$. so i will prove that $L,AO,SM'$ are concurrent. prove this by using ceva(sinus form) in triangle $STA$.

I have proved this problem on this site yet , but i don't remember where Let TH is diameter of omega Let point P such that angle PBT = TSP = 90 let's prove that P is on line AO Let line AS intersect w at points J and S Let line AS intersect (BTS) at points S and I Well known that TS is inner angle bissector of BSA , so angle BSP = PSI , so BP = IP After Reim's Theorem easy to see that TJ || BI TO || BP and TO = JO , so A is Homotety center of triangles TJO and BIP . done

Let $O_1,r_1,r$ be the center of $\Gamma$, radius of $\Gamma$, and radius of $\omega$, respectively. Let $M=ST\cap \Gamma$; it's well-known that $M$ is the midpoint of arc $AB$ not containing $S$. Let $Q$ be the intersection of tangent $BB$ and $(STB)$; then $QB\perp AB$ and $QS\perp TS$, and we want to show that $A,O,Q$ are collinear. Now $\angle{BQT}=\angle{BST}=\alpha$, so \[ \frac{AT}{OT}\frac{AB}{QB} = \frac{AT}{r}\frac{TB\cot\alpha}{2r_1\sin{2\alpha}} = \frac{TS\cdot TM}{4rr_1\sin^2\alpha}. \]But $\triangle{MTA}\sim\triangle{MAS}$ by simple angle chasing, so $MT\cdot MS = MA^2 = 4r_1^2\sin^2\alpha$. We thus obtain \[ \frac{AT}{OT}\frac{AB}{QB} = \frac{TS}{MS} \frac{MT\cdot MS}{4rr_1\sin^2\alpha} = \frac{OS}{O_1S} \frac{r_1}{r} = 1, \]whence $A,O,Q$ are collinear, as desired.

My solution: Let another tangent of $ \omega $ through $ A $ cut $ \Gamma $ at $ C $ and tangent to $ \omega $ at $ R $ . From incenter of triangle we get the midpoint $ H $ of $ TR $ lie on $ (TBS) $ , so $ AB \perp BP \Longleftrightarrow S, H, T, B, P $ are concyclic $ \Longleftrightarrow TS \perp SP $ . Q.E.D

Dear Mathlinkers, with harmonic division we have also a proof... Sincerely Jean-Louis

Take an inversion with center T and any circle. Then $\omega$ circle transforms into line $\ell$ and the $\Gamma$ circle tranforms into a circle tangent to line $\ell$.And the line $\ell$ bisects $OT$ (which is under inversion) .The line AO tranforms into the circle passing through points $A,O,T$(which are under inversion). The circle that passes through points $S,T,B$ tranforms into the line that passes through $S$ and $B$(which are under inversion) . Point $P$ under inversion transforms into the intersection between $SB$ and the circle that passes through points $A,O,T$. Our condition is equvalent to showing $ TP\perp SB.$ Let line through $A,S$ intersect the circle that passes through points $A,O,T$ the second time at point K.Let $OP$ intersect $\ell$ at point $R$.Then since $ \ell\parallel AB$ and $ \triangle ABS$ is isosceles(which is obvious) we find that $ \angle OAK = \angle PAT=\angle TOP$. $\angle SPO = \angle SRO-\angle RSB = 90^{\circ}-\angle TOP -\angle SAT= 90^{\circ}-\angle OAT=\angle AOT=\angle APT $ .And because $\angle ATO = \angle APO=90^{\circ}$, we find $\angle TPS=90^{\circ} $. So $ TP\perp SB$ and we are done.

$Z$ is the antipode of $T$ at $\omega$ and $SZ \cap AB \equiv X \Rightarrow (A,B,X,T)=-1$ $PS \perp TS \Longleftrightarrow P \in SZ \Longleftrightarrow P(Z,T,A,B)=-1 \Longleftrightarrow PB \parallel ZT \Longleftrightarrow PB \perp AB$

Let $ST$ cut $\Gamma$ for a second time at $M$, and let the perpendicular to $AB$ through $B$ cut $AO$ at $Q.$ Let the tangent to $\omega$ at $S$ meet $AB$ at $Z$, and denote $O_1 \equiv OZ \cap QT.$ We will prove that $QS \perp TS.$ First, note that the homothety with center $S$ that takes $\omega \mapsto \Gamma$ also takes $T \mapsto M.$ It follows that this homothety takes $AB$ to the line tangent to $\Gamma$ at $M.$ Therefore, the tangent to $\Gamma$ at $M$ is parallel to $AB$, which implies that $M$ is the midpoint of arc $\widehat{AB}.$ Hence, $ST$ bisects $\angle ASB.$ Now, from the Menelaus Theorem for the transversal $ZOO_1$ w.r.t. $\triangle AQT$, we obtain \[1 = \frac{ZT}{ZA} \cdot \frac{OA}{OQ} \cdot \frac{O_1Q}{O_1T}.\] Observe that $\triangle AOT \sim \triangle AQB$ and $\triangle ZAS \sim \triangle ZSB.$ It follows from these two relations and the Angle Bisector Theorem that $OA : OQ = TA : TB = SA : SB = ZA : ZS.$ Because $ZS = ZT$ by equal tangents, we obtain $O_1Q = O_1T.$ Hence, $O_1$ is the midpoint of $\overline{QT}$ in right triangle $\triangle BQT.$ It follows that $O_1B = O_1Q = O_1T.$ Moreover, since $O_1$ lies on $OZ$, the perpendicular bisector of $\overline{ST}$, we have $O_1S = O_1T.$ Thus, $B, Q, T, S$ are concyclic, and $QT$ is a diameter of their circumcircle. Hence, $QS \perp TS$ as desired. $\square$

My solution : Considering the equivalent problem :set intersection of $BP$ and circumcircle of $ABC$ $X$ , and set $\angle XAC=x$. Now we can calculate that angles $\angle KAO=(B-C)/2$$\angle OAP=90-(b-c)/2$$\angle AKn=(B-C)/2$ $\angle APB=x-(B-C)/2$$\angle BP K=90-x-a/2$$\angle nKP=a/2$( $n$ is perpendicular to $BC$ at $K$ ) which now implies that Cevas theorem hold for triangle $AKP$. So we are done.

My solution, take $P$ such that $\angle TSP=\angle PBA=90$. Then, we will prove that $A$, $O$ and $P$ are collinear,. Set $SP \cap AB=X$ and $SP \cap \omega=Y$. Considering homotethy centred at $S$ which brings $\omega$ to $\Gamma$ gives us that $ST$ bissects $\angle ASB$. Notice that $\angle XST$ is adjacent to $\angle TSP$, so $\angle XST=90$ meaning that $(X,T,A,B)$ are harmonic quadruple. Also, $\angle YST \equiv \angle TSP=90$. So $O$ is the midpoint of $YT$. we will prove that $PA$ bissects $YT$, which completes solution. Indeed, consider harmonic pencil $(PX,PT,PA,PB)$ intersecting line $YT$. $PB$ is perpendicular to $AB$ from our choice of $P$, $YT$ is perpendicular to $AB$ from is diametr of $\omega$ and $AB$ is tangent to $\omega$ at $T$, so $PB$ and $YT$ are parallel, $PX \cap YT=Y$, $PT \cap YT=T$, so $PA$ intersects $YT$ at the midpoint, which is $O$, $QED$.