what is $T$?

maybe I'm not right, but I proved that $M$ is incenter of $\triangle ABC$. But then the statement is not true???

Outline: Use inversion with center $P$ and arbitrary radius. Let $O_1$ and $O_2$ be the circumcenters of $T'PK'$ and $A'PM'$ respectively. So $O_1,P,O_2$ are collinear, which implies that $P$ lies between the perpendicular bisectors of $T'K'$ and $A'M'$. If $(PT'A')$ is bigger than $(PK'M')$, it is easy to see that $P$ does not lie between the perpendicular bisectors of $T'K'$ and $A'M'$. Similarly, $(PK'M')$ cannot be bigger than $(PT'A')$. Thus $(PK'M')$ and $(PT'A')$ have equal radii. By considering the translation which maps $(T'PA')$ to $(K'PC')$, we conclude that $T'K'C'A'$ is a parallelogram, which implies that $T'A'\parallel K'C'$, hence the result follows.

A generalisation: In triangle $ABC$, let $T,K,M$ be on $AB,BC,BC$ such that $BT=BK$ and $BA=BM$. Let $Q$ be on $AB$. The circumcircles of $AMC$ and $TKQ$ intersect at $P$. Then the circumcircles of triangles $KPC$ and $QPA$ are tangent. Note that $\angle PCM+\angle PAT=\angle PAM+\angle PAT=\angle MAT=\angle KTB=\angle QPK$, thus the circumcircles of triangles $CPK$ and $APQ$ are tangent.

TK AM so that angle KPT = TKB = AMB = 180 - AMC =180 - APC Let I and J and R and S is center of (ATP) and (KCP) and(TPK) and (ACP) so that IRJS are concyclic thus angle TPK = RSJ but angle PIS = PKT ang angle PJS = PTK so that PIS +PJS + ISJ=180 hence I P J are conllinear hence APT and KPC are tangent

Sorry to revive, but i don't see how, alive_ghost_vn wrote: Let I and J and R and S is center of (ATP) and (KCP) and(TPK) and (ACP) so that IRJS are concyclic Is it true that the cirucm centers of these must form a concyclic quad? Also, oneplusone wrote: and $BA=BM$. how is this so?

first came with the fact that $ISJ=APC , IRJ=TPK$ modulo $\pi$ . second caused by paralleling of $TK,AM$ and the fact $BT=BK$ which comes from tangency of $W.$

As above, $AT=BK, BM=AB$. Let's note $I, J, R$ the circumcenters of triangles $\Delta APT, \Delta KPC,\Delta PTK$. See that $\angle KPT=\angle BRT=\frac{\widehat{A}+\widehat{C}}{2}\ (\ 1\ )$. Also, easily, $P$ is the incenter of $\Delta ABC$, hence $\angle PAT=\angle BAP=\frac{\widehat{A}}{2}\ (\ 2\ ), \angle KCP=\angle BCP=\frac{\widehat{C}}{2}\ (\ 3\ )$. Consequently $\angle IPT=90^\circ-\angle PAT\ (\ 4\ ), \angle JPC=90^\circ-\angle PCK\ (\ 5\ )$. With $(1), (4), (5)$, keeping into account $(2), (3)$ we get $\angle IPJ=\angle IPT+\angle KPT+\angle KPJ=180^\circ$, done. (Here $\hat A,\hat B,\hat C$ means the measures of the angles of the triangle $\Delta ABC$). Best regards, sunken rock

Suppose $\angle{PAC}=\theta,\angle{PCM}=\alpha,\angle{KPC}=x$. Now certainly $P$ lies on the angular bisector of $\angle{ABC}$.Now so from ceva we've $Sin(\theta)Sin(\alpha)=Sin(A-\theta)Sin(C-\alpha)$. And also we've $\theta-\alpha=90^2-C-\frac {B}{2}=\frac {A}{2}-\frac {C}{2}$. So solving two equations we get $P$ is in center of triangle $ABC$.Now $\angle{PKC}=\pi-(x+\frac {C}{2})=\angle{TKC}\implies \angle{PTA}=x-\frac{\pi}{2}+\frac {B+C}{2}$ and so finally that implies $\angle{TPA}=\pi-x$ ,where we also had $\angle{APC}+\angle{TKP}=\pi$, and hence we're done.

Let l be the tangent to circle APT at P which meets circle AMC at S different from P.Let l' be the common tangent at P to the circles PTK and APC meeting AB and AC at X and Y respectively. Now <SPA=<XTP=<XPT=<PKT. Then <AMC=<TKY=<PKT+<PKY=<SPA+<PKY But <APC=<SPA+<SPC=<AMC=<SPA+<PKY Then <SPC=<PKY=<PKC Therefore,circle KPC is tangent to l at P and we are done.

First observe that since circles APMC and PTK are tangent at P <APT=<TKP+<ACP, it can be easily proven that this condition is a sufficient and necessary condition for the two circles to be tangent. Our aim is to show that this condition exists for circles APT and KPC. <BTK=<BAM (AM||TK), <BAM=<TAP+<PAM=<TAP+<PCK, but <TPK=<BTK=<BAM=<TAP+<PCK, so we are done. But the question that came to my mind is that the fact that circles AMC and PTK are tangent wasn't any help for the solution (except that I used it to introduce my observation initially), it seems for this to be true the condition that these two circles intersect at two points (with one of them being P). Can just someone show me if I have mistaken.

This problem is a bit too easy for a TST. Note that $TK \parallel AM \implies AB=BM$ since $TB=BK$.Next observation is that if $I$ is the incenter of $\triangle{ABC}$ then $BI$ lies on the bisector of $TK$ and $AM$ which is just another way of saying that $B,O_W,O_{APMC},I$ are collinear. We also have $\angle{APC}=\angle{AMC}=90+\frac{B}{2}$ and $\angle{TPK}=\angle{BTK}=90-\frac{B}{2}$.Thus $\angle{TPK}+\angle{APC}=180^{\circ},\angle{ATP}+\angle{PKC}=180-\angle{TPK}$ which means that $\angle{TAP}+\angle{PCK}=90-\frac{B}{2}=\angle{TPK}$ or circles $(APT),(PCK)$ are tangent at $P$ as desired.

let circumcircle of KPC and circumcircle of APT meet at P and L. APMC is cyclic so <APC=<AMC and TK is parallel to AM so <AMC=<TKC KPLC is cyclic so <LPC=<LKC TPLA is cyclic so <LPA=<LTA we have <LKC+<LKT=<LKC+<LTA so <LTA=<LKT so circumcircle of TKL is tangent to AB let circumcircle of TKL meet BC at K and Q so TB^2=BK*BQ but we have TB^2=BK^2 so we have BQ=BK so circumcircle of TLK is tangent to AB and BC so circumcircle of TKL and circumcircle TKP is tangent ogether at T and K so this two circle are the same and L is P and we are done

easy solution : all we have to do is to prove that $\angle{TAP} + \angle{PCK} =\angle{TPK} $ now $\angle{PCK} = \angle{PAM}$ so now we have to say that $\angle{TAM} =\angle{TPK}$ which is obvious by considering the parallel lines .

Excuse me if the name of points in the attachment doesn't match with the names in the problem. First we should note that BD=AB. By Sawayama's lemma, T,G and the excenter of triangle ADC are collinear(In the picture its name is Ic.); and also by homothety it is obvious that T,P and the midpoint of the arc DAC are collinear. So IcA and TP and circle W are concurre. Let L be the intersection of GT and circle PTC. <IcAP=<AMP+<APM and M is the midpoint of arc DAC so <IcAP=<PTC so A,Ic,G and P are concyclic. Now we have <LPC=<APIc; <AMC=<ADB=<LTC=<LPC, so L,P and A are collinear and Ic,P and C are collinear too. By computing angles it is obvious that CL||AIc. Now if we say IcACL is a cyclic trapzoide we are done. <IcMC=<GTB => TIcMC is cyclic => <PMC=<PAC=<TIcC => IcACL is cyclic and we are done

SDEIBAR wrote: let circumcircle of KPC and circumcircle of APT meet at P and L. APMC is cyclic so <APC=<AMC and TK is parallel to AM so <AMC=<TKC KPLC is cyclic so <LPC=<LKC TPLA is cyclic so <LPA=<LTA we have <LKC+<LKT=<LKC+<LTA so <LTA=<LKT so circumcircle of TKL is tangent to AB let circumcircle of TKL meet BC at K and Q so TB^2=BK*BQ but we have TB^2=BK^2 so we have BQ=BK so circumcircle of TLK is tangent to AB and BC so circumcircle of TKL and circumcircle TKP is tangent ogether at T and K so this two circle are the same and L is P and we are done use Latex please!

This is too easy!! Just note that $$\angle TPK=\angle BTK=\angle BAM=\angle BAP+\angle PAM=\angle TAP+\angle PAK$$which directly gives the desired result. Done

jgnr wrote: Outline: Use inversion with center $P$ and arbitrary radius. Let $O_1$ and $O_2$ be the circumcenters of $T'PK'$ and $A'PM'$ respectively. So $O_1,P,O_2$ are collinear, which implies that $P$ lies between the perpendicular bisectors of $T'K'$ and $A'M'$. If $(PT'A')$ is bigger than $(PK'M')$, it is easy to see that $P$ does not lie between the perpendicular bisectors of $T'K'$ and $A'M'$. Similarly, $(PK'M')$ cannot be bigger than $(PT'A')$. Thus $(PK'M')$ and $(PT'A')$ have equal radii. By considering the translation which maps $(T'PA')$ to $(K'PC')$, we conclude that $T'K'C'A'$ is a parallelogram, which implies that $T'A'\parallel K'C'$, hence the result follows. Could someone elaborate this

Storage Iran TST 2010 wrote: Let $M$ be an arbitrary point on side $BC$ of triangle $ABC$. $W$ is a circle which is tangent to $AB$ and $BM$ at $T$ and $K$ and is tangent to circumcircle of $AMC$ at $P$. Prove that if $TK||AM$, circumcircles of $APT$ and $KPC$ are tangent together. Let $\omega$ be the circle tangent to $AB,BM$ and $\odot(APC)$ and denote $O_1,O_2$ as the centers of $\omega$ and $\odot(APC)$ respectively. Notice that $TK\|AM\implies AB=BM$ as $TB=BK$. Let $\angle BAP=\alpha$ and $\angle PCK=\theta$. Now draw a tangent $(\ell)$ to $\odot(APT)$ through $P$ so it suffices to show that $\ell$ is tangent to $\odot(KPC)$ as well. So we just need to prove that $\angle TAP+\angle PCK=\angle TPK$ (This comes by Angle Segment Theorem). \begin{align*} \angle TAP+\angle PCK &=\theta+\alpha=\angle TAP+\angle PAM=\angle BAM=\angle AMB\\ &\implies\angle ABM=180^\circ-2\theta-2\alpha\\ &\implies\angle TO_1K=2\theta+2\alpha\\ &\implies\angle TPK. \end{align*} So, $\odot(APT)$ and $\odot(KPC)$ are tangent to each other at their Exsimillicenter $P$. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.901527400284496, xmax = 11.837771332732142, ymin = -4.187471033493936, ymax = 6.022314009338907; /* image dimensions */ pen qqzzqq = rgb(0,0.6,0); pen ccqqqq = rgb(0.8,0,0); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); /* draw figures */ draw((0,3.4589473764443337)--(-4,-3), linewidth(2) + qqzzqq); draw((-4,-3)--(6,-3), linewidth(2) + qqzzqq); draw((6,-3)--(0,3.4589473764443337), linewidth(2) + qqzzqq); draw((0,3.4589473764443337)--(2,-3), linewidth(2) + linetype("2 2") + ccqqqq); draw(circle((4,1.1584176244867497), 4.614372886930793), linewidth(2) + blue); draw(circle((-1.4126640339838517,-1.559961879580064), 1.4452321002485076), linewidth(2) + red); draw((-2.583426339957532,-0.7126063187321385)--(-1.5350814973055948,-3), linewidth(2) + linetype("2 2") + ccqqqq); draw((-2.583426339957532,-0.7126063187321385)--(-0.08860944398479287,-0.9806732932164655), linewidth(2) + yqqqyq); draw((-0.08860944398479287,-0.9806732932164655)--(0,3.4589473764443337), linewidth(2) + yqqqyq); draw((-1.5350814973055948,-3)--(-0.08860944398479287,-0.9806732932164655), linewidth(2) + yqqqyq); draw((-0.08860944398479287,-0.9806732932164655)--(6,-3), linewidth(2) + yqqqyq); /* dots and labels */ dot((0,3.4589473764443337),dotstyle); label("$A$", (-0.25666879794920533,3.558906571678657), NE * labelscalefactor); dot((-4,-3),dotstyle); label("$B$", (-4.144697404376829,-2.8667284916158504), NE * labelscalefactor); dot((6,-3),dotstyle); label("$C$", (6.065087638456014,-2.8518886877745233), NE * labelscalefactor); dot((2,-3),dotstyle); label("$M$", (1.9841415820911426,-3.4009614299036155), NE * labelscalefactor); dot((-2.583426339957532,-0.7126063187321385),dotstyle); label("$T$", (-2.8536344701813965,-0.6110783077341756), NE * labelscalefactor); dot((-1.5350814973055948,-3),dotstyle); label("$K$", (-1.4735327129380036,-3.3267624106969813), NE * labelscalefactor); dot((-0.08860944398479287,-0.9806732932164655),dotstyle); label("$P$", (-0.0340717403293032,-0.8336753653540777), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] This is cute Angle Chasing. To prove: $\angle TAP+\angle PCK=\angle TPK$. Notice that by Angle Segment Theorem, we have $\angle TPK=\angle BTK$. Now, since $TK ||AM$, we get $\angle BTK=\angle BAM=\angle BAP+\angle PAM$. But, $\angle BAP=\angle TAP$. Now, by the property of chord subtends equal angle from two points, we get $\angle PAM=\angle PCM=\angle PCK$. Hence, we proved that $\angle TPK=\angle TAP+\angle PCK. \blacksquare$

Extremely easy for an Iran TST. Same angle chase solution as the above people.