part a Four points are concyclic iff angle TCI=angle TYI iff angle TK1S=angle TK2X which is true. part b i only managed to prove the external bisector of angle C passes through I... can anyone guide me what to do next? 2*angle ACB = angle AK1B. 2*angle BCI = angle SK1B. Extend SK1 to meet AB at Z. Since SK1//K2X and K2X is perpendicular to AB, K1Z is perpendicular to AB. Since AK1B is isoceles, angle AK1Z=angle BK1Z. So angle AK1S=angle BK1S. So 2*Angle AK1S + Angle AK1B = 360 2*Angle BCI + Angle BCA = 180 So CI is the external bisector of angle C.

I did same as you siuhochung b) but to complete solution I think the following idea can be used: Let $YC$ and $YT$ intersect $k_1$ in $D$ and $Z$. $ZB$ intersects $XY$ in $J$. Using first idea $BJ$ is external bisector of $DBC$. From a) $BTJX$ are concyclic. <$XJB$ = < $BTX$ = <$SAB$ = <$BCI$ Hence $BCIJ$ are concyclic. And we need to prove < $ABD$ = <$IBJ$ sorry I didn't finish I found mistake.

First, clearly if the following statement(converse of the problem) is true, then the original conclusion is als otrue. $k_1$ be the circumcircle of a $\triangle ABC$. The external bisector of C meets $k_1$ at S. T is a point chosen on arc BC. Let $ST\cap AB=X$ and I be the excenter of ABC wrt BC. The circumcircle of TIC intersects XI again at Y. Then the circumcircle of XYT,$k_2$ is tangent to $k_1,BX,CY$. Proving BX and $k_1$ are tangent to $k_2$ is easy, so I will only prove for CY. Note that it is equivalent to $\angle TIS=\angle TXI\Leftrightarrow SI^2=ST\cdot SX$. By similar triangle, we also have $SB^2=ST\cdot SX$. Thus we only have to show that SI=SB, but this is trivial by angle chasing using IB is the external bisector of B. I know I have missed lots of details, but all of them can be obtained by angle chasing. Anyway, this problem really surprises me because it cannot be solved by the "ordinary" method.

b)Let CY meet AB in $T_1$ and AC meet XY in R. From Menelaos for $\triangle{AT_1C}$: $\frac{CY}{AX}=\frac{CR}{AR}$.(1) From Caysey applied to A,B,k2,C: $a=\frac{CY}{AX}\cdot{c}+\frac{BX}{AX}\cdot{b}$.(2) Using (1) and (2) and the converse of transversal theorem we obtain the conclusion.

Let me give a remark on where the problem (apparently) comes from. In the paper Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) pp. 225-229 (I have cited this paper in numerous threads, e. g. in http://www.mathlinks.ro/Forum/viewtopic.php?t=35337 ), a nice result - which is Lemma 1 in the paper - is shown: (1) Let ABC be a triangle, and D a point on its side BC. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the incenter I of triangle ABC. During the proof of Lemma 1, the following two additional facts are established: (2) If the line KE meets the circle $\mathcal{C}_2$ at a point M (apart from K), then the line AM passes through the point I. (3) The points A, K, F, I are concyclic. Now, let's "extravert" - i. e. state the analogous result for external points: (1') Let ABC be a triangle, and D a point on the extension of its side BC beyound C. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the B-excenter $I_b$ of triangle ABC. (2') If the line KE meets the circle $\mathcal{C}_2$ at a point M (apart from K), then the line AM passes through the point $I_b$. (3') The points A, K, F, $I_b$ are concyclic. Now, we change some notations: $A,B,C\to C,A,B$, $E\to X$, $F\to Y$, $K\to T$, $M\to S$, $\mathcal{C}_1\to k_2$, $\mathcal{C}_2\to k_1$, $I_b\to I_a$. These statements then become: (1'') Let ABC be a triangle, and D a point on the extension of its side AB beyound B. Let $k_2$ be a circle touching the segments DB and DC at two points X and Y and touching the circumcircle $k_1$ of triangle ABC at a point T. Then, the line XY passes through the A-excenter $I_a$ of triangle ABC. (2'') If the line TX meets the circle $k_1$ at a point S (apart from T), then the line CS passes through the point $I_a$. (3'') The points C, T, Y, $I_a$ are concyclic. But this is exactly the configuration of our problem, just seen from a slightly different viewpoint. Now, from (1'') and (2''), it follows that the point $I_a$ lies on the lines XY and CS, so that $I_a=XY\cap CS=XY\cap SC$. Comparing this with the problem which states $I=XY\cap SC$, we see that $I_a=I$. Thus, the point I coincides with the A-excenter $I_a$ of triangle ABC. This solves problem (b). Now, problem (a) follows from (3'') using $I_a=I$. Darij

in the pictures an "ordinary" solution. pic.1 <TCI = <TYI ==> TCYI CIclico ==> <TXY = <TYC = < TIC ==> SI^2 = ST*SX <SAT = <XYT = <AXT ==> SA^2 = ST*SX ==>SA=SI <SAT = <AXT ==> <SBA = <ATX = <SAB ==>SA=SB pic.2 SC _|_ CF ==> SC bisettrice esterna di <C <ASK = < AEC = 2<AIK ==> EC =EI <BEA = <BSA = 2<BIA ==> BE =EI CE = BE ==> <BAI = <CAI

Attachments:

darij grinberg wrote: (I have cited this paper in numerous threads, e. g. in http://www.mathlinks.ro/Forum/viewtopic.php?t=35337 ), a nice result - which is Lemma 1 in the paper - is shown: (1) Let ABC be a triangle, and D a point on its side BC. Let $\mathcal{C}_1$ be a circle touching the segments DC and DA at two points E and F and touching the circumcircle $\mathcal{C}_2$ of triangle ABC at a point K. Then, the line EF passes through the incenter I of triangle ABC. Darij The following argument is also applicable to give a proof of this Lemma <TCI = <TYI ==> TCYI ciclico ==> <TXY = <TYC = < TIC ==> SI^2 = ST*SX <SAT = <XYT = <AXT ==> SA^2 = ST*SX ==>SA=SI <SAT = <AXT ==> <SBA = <ATX = <SAB ==>SA=SB SC _|_ CF ==> SC bisettrice esterna di <C <ASK = < AEC = 2<AIK ==> EC =EI <BEA = <BSA = 2<BIA ==> BE =EI CE = BE ==> <BAI = <CAI To complete the proof of the Thebault theorem, instead of that used in the paper, one can use the following argument. [I refer to the picture in the paper Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003).] If I = GF/\EH let U and V be the intersection of c(PEI) with GF and GE respectively. Since GF _|_ EH, then UP// GE therefore VE=UP. Now, as GF//DP, UP=GD so we have that VE=GD. Since QGD ~ GIE, is easy to deduce that IGQ ~ IEV, then IQGV is cyclic which means that <QIV = 90° and then I lies on PQ.

Somewhere here in Bulgaria-2 also 'hiding' a nice proof that the radical axis of circle ABC and it's A-excircle is parallel to the line through traces of B- and C-bisectors on sides AC and AB. T.Y. M.T.

armpist wrote: Somewhere here in Bulgaria-2 also 'hiding' a nice proof that the radical axis of circle ABC and it's A-excircle is parallel to the line through traces of B- and C-bisectors on sides AC and AB. Is perhaps this? if Ia is the excenter of ABC and K and L the ends of a chord of circumcircle of ABC through the feet of B- and C-bisector on AC and AB respectively, then c(IKL), where I is the incenter of ABC, is homotetic with c(ABC) with center Ia and ratio 2 (*). Therefore, if P=IaK/\c(ABC), FP//KI so <IKIa = < IaAK then IaK is tangent at K to the circumcircle c(AIK) from which IaK^2 = IaI*IaA. Similarly we have that IaL^2 = IaI*IaA, then IaK = IaL. This implies that IaO is othogonal to LK which is equivalent to the claim. (*) this is a lemmaRL22022006.

Hello RL ! Initially I made a naming mistake in my drawing, but after overdue correction I think that this very well may be it. Nice idea! Hope Darij likes it, otherwise back to the drawing board. T.Y. M.T.

Ciao MT. I'm not sure to understand exactly what you are referring to. What was the naming mistake? Why you are asking if this like to Darij?

Sailor wrote: converse of transversal theorem What is it?

N.T.TUAN wrote: Sailor wrote: converse of transversal theorem What is it? http://www.mathlinks.ro/Forum/viewtopic.php?t=20289

Very nice! Thank you very much, tdl and darji.

I think I've got a slightly different proof for (b), proving that I is the A-excenter of ABC. From a quick angle chase, we can see S is the midpoint of the arc AB, in the same segment as B; and IC is the external angle bisector of <ACB. Thus it suffices to prove that, the E, the A-excenter of ABC lies on XY. Let R be the midpoint of arc AC in circle k1, that lies in the same segment as B. It's easy to see RB and SC intersect at E. Applying Pascal's Theorem to A,R,B,S,C,T - {RB,SC},{AB,ST},{RT,AC} are collinear - i.e. E lies on XZ where Z is where RT intersects intersects AC. In order to prove E lies on XY, it is now sufficient to prove X,Y,Z are collinear. Let the tangents to k2 at X,Y intersect at P. Using Menelaus in triangle ACP, and noting that XP=YP, now reduces the problem to proving AZ/CZ=AX/CY. But noticing that RT is the external bisector of <ATC, we have AZ/CZ=AT/CT. We now have to prove AT/CT = AX/CY. Unfortunately, for this seemingly simple problem I don't have a Euclidean proof, but a trig-approach is simple enough: forgetting about all the unnecessary points, angle chase the angles in quadrilateral ACXY and triangles ACT,ATX,TXY,TCY. Using the quadrilateral extension of trig-ceva (for lines AT,XT,YT,CT) gives an equation, which after some manipulation, reduces to the sine-rule equivalent of AT/CT=AX/CY.

Dear Mathlinkers, for a) we can use a converse of the pivot theorem applied to the triangle IST with C on IS, T on SX, Y on IX, pivot being T. Sincerely Jean-Louis

Dear Mathlinkers, for b) we can use the result of Protassov in an extraversion, I being the A-excenter of ABC. http://perso.orange.fr/jl.ayme vol. 2 Protassov Sincerely Jean-Louis

yes!thanks!

Here's my solution: Let $CT\cap k_2=D$, since $T$ is the center of homothetic of the two circles, hence $CS\parallel DX\Rightarrow \angle XYT=180-\angle XDT=180-\angle SCT$ and a) follows. For b) let $O_1,O_2$ denote the circumcenters of the two circles. Since $(SO_1\parallel XO_2)\perp AB$, hence $SB=SA\Rightarrow CI$ externally bisects $\angle C$. Let $(IYCT)\cap CB=E$, Since $\angle BTS=\angle BCS=180-\angle XYE$, therefore $BT\cap YE=F\in k_2$ $\Rightarrow \angle BFE=\angle CYT=\angle TEB\Rightarrow BE^2=BT*BF=BX^2$ $\Rightarrow BE=BX$, since $\angle XTC=\angle IEB=\angle IXB$, we can deduce that $BI$ externally bisects $\angle B$ and we done.

Bulgaria MO 2005/P2 wrote: Consider two circles $k_{1},k_{2}$ touching externally at point $T$. a line touches $k_{2}$ at point $X$ and intersects $k_{1}$ at points $A$ and $B$. Let $S$ be the second intersection point of $k_{1}$ with the line $XT$ . On the arc $\widehat{TS}$ not containing $A$ and $B$ is chosen a point $C$ . Let $\ CY$ be the tangent line to $k_{2}$ with $Y\in k_{2}$ , such that the segment $CY$ does not intersect the segment $ST$ . If $I=XY\cap SC$ . Prove that : (a) the points $C,T,Y,I$ are concyclic. (b) $I$ is the excenter of triangle $ABC$ with respect to the side $BC$. Here's an elementary and most important , complete solution $\textbf{(a)}$ Consider the homothety $\mathcal{H}$ taking $k_2$ to $k_1$ . Then $\mathcal{H} : X\leftrightarrow{S}$ . So $\mathcal{H} : \overarc{XT}\leftrightarrow{\overarc{TS}}$ and $\angle TAS=\angle TXA=\angle TCI=\angle TYI$ , as desired . $\textbf{(b)}$ It's easy to see $SA$ is tangent to $(ATX)\implies{SA^{2}=ST.SX}$ , but also $SI$ is tangent to $(ITX)\implies{SI^2=ST.SX}$ , similarly $SB$ is tangent to $(TXB)\implies{SB^2=ST.SX}$ , hence $SA=SI=SB$ . \[\angle BCI=180-\angle BCS=90-\frac{\angle BCA}{2}\], so $CI$ is the external bisector of $\angle ACB$ . Analogously , since $\angle SBI=\angle SBI=90-\frac{\angle BAC}{2}$ , we get \[\angle CBI=90-\frac{\angle ABC}{2}\]. Thus $I$ is the $A$-excenter of $\triangle ABC$ . $\blacksquare$

This problem is same as Curvilinear Incircle configuration but here circles are externally tangent. The proof is same as the proof of curvilinear incircle. By homothety, $S$ is the midpoint of arc $AB$ containing $T$ and $\widehat{ST}=\widehat{XT}$. So \[ \angle TCI=\frac{1}{2}\widehat{ST}=\frac{1}{2}\widehat{XT}=\angle TYI \]So $CTIY$ is cyclic. Notice that $\triangle SCT\sim\triangle SIY$. So $SA^2=SB^2=SI^2=ST\cdot SX$ implies $I$ is the $A-$excenter of $\triangle ACB$ because $CI$ is the angle bisector of exterior angle $\angle ACB$.