$SA \cdot SB=SC^2=SP^2$. Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$. Then $\angle{SPA}=\angle{SBP}$. Let $R, T$ be the intersections of $SP$ with $\Gamma$ ($R$ is in the $\widehat{LA}$, and $T$ is in the $\widehat{KB}$) $\frac{1}{2}(\widehat{RA}+\widehat{KT})=\angle{RPA}=\angle{LBA}=\frac{1}{2}\widehat{LA}=\frac{1}{2}(\widehat{LR}+\widehat{RA})$ Then $\widehat{KT}=\widehat{LR}$ Then $\angle{LPR}=\frac{1}{2}(\widehat{LR}+\widehat{TB})=\frac{1}{2}(\widehat{KT}+\widehat{TB})=\frac{1}{2}KB=\angle{PAB}$ $\angle{SPC}=\angle{LPR}+\angle{LPC}=\angle{PAB}+\angle{BPM}=\angle{PAB}+\angle{MCB}+\angle{PBC}$ $\angle{SCP}=\angle{SCA}+\angle{ACM}=\angle{SBC}+\angle{ACM}=\angle{SBL}+\angle{PBC}+\angle{ACM}$ But $\angle{SPC}=\angle{SCP}$. Then $\angle{SBL}+\angle{ACM}=\angle{MCB}+\angle{PAB}$ Then $\widehat{LM}=\widehat{KM}$ Then $LM=KM$

Solution (if it's correct please tell me) Call the circles $(S,SC) $ $\Omega$. Now the circles $\Omega$ and $ \Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $ \Omega$ and call this point $S'$. Assume that $A $ belong to $[S,B]$. Now the line $SP$ is tangent to the circumcircle of $ \triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2} $. Now $SC$ is the external symmedian of the vertex $C$. Calling the intersection of the angle bisector of $C$ with the line $ AB$, $P'$, we have the points $ (S',P',A,B) $ forming a harmonic range and this implies that $ SC = SP'$. So the circle $\Omega$ is is the appollonian circle of the vertex $ C $ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $ \angle LCM$. Let the external symmedian meet the the $S'C$ in $J$. We have $JN=JC$ and $ SC = SP$ so the traingle $ SCP $ and $ JCN $ are homothetic so the points $ A , P $ and $ N $ are collinear and we have $M$ being the midpoint of the arc $ \overarc{KMN}$.

here's my proof Since, angle BPC = angle BAK + angle CAK + angle ABL + angle ACL and angle SPC = angle SCP To see angle MLK = angle ACM + angle ABL = angle BCM + angle BAK = angle MKL it is sufficient to prove angle BPS = angle BPK. But we have SA * SB = SC^2 = SP^2, from this it is clear that triangle BPS is similar to triangle PAS this proves angle BPS = angle APS.

we know,as SC is tangent to the circle SC^2 =SA*SB so by the problem SP^2=SA*SB so SP/SA=SB/SP and as triangle SPA and triangle SPB have one angle in common,they are similar.so angleSPA and anglePBA(both equal to a) are equal. extending SP to meet the circle at x,we also have angleXPM=anglePAB(both equal to b) also,angleSCA=angleCBA(both equal to c) again ,being on the same arc, angleACM=angleABM(both equal to d) similarly,angleCAK=angleCMK(both equal to e) now in triangle APM, angleAPM=angleCPK=angleCAK+angleACM=e+d also, angleSPC=angleSCP=c+d,and angleSPA=a so adding,180=a+c+2d+e -(1) now, angleKLM=angleKLB+angleBLM=b+angleBLM also, angleBLM=angleBAM=180-b-c-d-e,after calculating from triangle CAM From (1)this turns out to be equal to a+d-b so angleKLM=a+d also angleLKM=angleLKA+angleAKM=a+d so in triangle KLM we have angleKLM=angleLKM so ML=MK................

My proof: Because triangles $APC$ and $MPK$ are similar, we have that $\frac{MK}{MP}=\frac{AC}{AP}.$ Analogously we have that $\frac{ML}{MP}=\frac{BC}{BP}.$ So it's enough to prove that $\frac{AP}{BP}=\frac{AC}{BC}$. We have that $SP^2=SC^2=SA \cdot SB,$ so $\frac{SP}{SA}=\frac{SB}{SP}$. Using this result and because angle $CSA$ is common for triangles $SPA$ and $SBP$, we conclude that these triangles are similar. So $\frac{AP}{BP}=\frac{SA}{SP}=\frac{SA}{SC}.$ But triangles $SCA$ and $SBC$ are similar too. So $\frac{SA}{SC}=\frac{AC}{BC}.$ Therefore $\frac{AP}{BP} = \frac{AC}{BC},$ and we have done!

Let me add my solution too. I think it is a little different. Well by the powe point theorem we have that $SP^2=SA\cdot SB$ which means that the triangles SPB, PAS are similar. Using this fact a bit angle chasing shows that $\angle{SPL}=\angle{BAP}=\angle{BLK}$ so $SP//LK$ (1). But $OC\bot CS$ and $\angle{SPC}=\angle{SCP}$, so $OM\bot SP$ (2) From (1) and (2) we have that $OM\bot LK$ and we are done.

As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP. => angle BPS = angle BAK = angle BLK. Let KL intersect BC at R and SP intersect BC at Q. In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB. => SP is parallel to KL. As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA) So, In triangles MLC and MPD where D is the intersection point of ML and SR, angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL. As, SR is parallel to KL, angle MDP = angle MLK = angle MKL We Are Done!

KoolNerd wrote: As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP. => angle BPS = angle BAK = angle BLK. Let KL intersect BC at R and SP intersect BC at Q. In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB. => SP is parallel to KL. As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA) So, In triangles MLC and MPD where D is the intersection point of ML and SR, angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL. As, SR is parallel to KL, angle MDP = angle MLK = angle MKL We Are Done! My solution is quite similar to yours...It's really an easy problem...

mavropnevma wrote: Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$. We have $MK=ML\iff \angle PCA + \angle PBA = \angle PCB + \angle PAB\iff \angle PAB - \angle PBA = \angle PCA - \angle PCB$ From $SC=SP$ we conclude that $SP^2= SC^2=SA\cdot SB$ $\Longrightarrow SP$ is tangent to the circumcircle of triangle $ABP$ at $P$. Therefore $\angle PAB - \angle PBA = \angle PAB - \angle APS = \angle PSB\quad (1)$. On the other hand, $\angle PCA - \angle PCB=\angle SCP - \angle SCA-\angle PCB=\angle SPC-\angle SBC - \angle PCB=\angle PSB\quad (2)$. From $(1)$ and $(2)$ we have the conclusion.

Let the second tangent from $S$ meet $\Gamma$ at $C'$. Notice that $(A,B;C,C')$ is harmonic, because projecting from $S$ fixes $C$ and $C'$, and swaps $A$ and $B$. Now, let $C'P$ meet $\Gamma$ again at $M'$. Projecting from $P$ gives $(A,B;C,C') = (K,L;M,M') = -1$, thus $LM/MK = LM'/M'K$. Now, if we prove that $M$ and $M'$ are antipodal, we're done, because the only way for those fractions to be equal is that $M$ is the midpoint of arc $LK$ (if we move $M$, one fraction grows while the other is smaller). That is the same as saying $\angle C'CP + \angle CC'M' = 90^{o} - \angle C'BC$. But we know that $SC' = SP = SC$, thus $S$ is the circumcenter of $C'PC$, and $\angle C'CP + \angle CC'M = \angle C'SC / 2$. Now, if $O$ is the circumcenter of $ABC$, we have $\angle C'SC / 2 = (180^{o} - \angle C'OC)/2 = 90^{o} - \angle C'BC$, as we wanted.

An even simpler solution: $SA \cdot SB=SC^2=SP^2$. Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$. Then $\angle{SPA}=\angle{SBP}$. Let $R, T$ be the intersections of $SP$ with $\Gamma$ ($R$ is in $\widehat{LA}$, and $T$ is in $\widehat{KB}$) $\frac{1}{2}(\widehat{LR}+\widehat{RA})=\frac{1}{2}\widehat{LA}=\angle{LBA}=\angle{RPA}=\frac{1}{2}(\widehat{RA}+\widehat{KT})$. Then $\widehat{LR}=\widehat{KT}$ $\frac{1}{2}(\widehat{RC}+\widehat{RM})=\frac{1}{2}\widehat{CM}=\angle{SCP}=\angle{SPC}=\frac{1}{2}(\widehat{RC}+\widehat{TM})$. Then $\widehat{RM}=\widehat{TM}$ Then $\widehat{LM}=\widehat{KM}$. Then $LM=KM$

Yay. I liked this problem. I think that my solution is also a little bit different. Looks like there were quite a number of different solutions to this problem... Solution Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$. We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$. This yields that $MX \| PS$. Now consider the power of point $S$ with respect to $\Gamma$. \[SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}\] Hence $\triangle{SPA} \sim \triangle{SBP}$. Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$. Hence $PS \| LK$. This implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$. Hence $MK=ML$.

Let $SP$ cut the circumcircle at $X$ and $Y$, then $SC^2=SP^2=SX\cdot SY$ so if $Q$ is the reflection of $P$ over $S$ then $X$, $Y$ are harmonic conjugates of $P$, $Q$. Notice also that $\angle QCP=90^{\circ}$ then $\angle XCP=\angle PCY$, this means $MX=MY$, but since $SP$ is tangent to the circumcircle of $PAB$ ($SP^2=SC^2=SA\cdot SB$) we have $\angle SPA=\angle PBA=\angle PKL$ so $XY$ is parallel to $KL$ and the result follows.

We wish to show that $SA \cdot SB = SC^2 = SP^2$, that is, we want to show that $SP$ is tangent to the circumcircle of $\triangle ABP$. This is equivalent to $\angle APS = \angle ABP = \angle ABL = \angle AKL$, that is, $KL || PS$. Let $T = MA \cap CL$. By Pascal's theorem on $ABLCCM$, we see that $P$, $T$, and $S$ are collinear, so it is sufficient to show that $PT || KL$. By Pascal's theorem on $MMAKLC$, we see that $PT$, $KL$, and the tangent to $\Gamma$ at $M$ concur. But since $MK = ML$, the latter two lines are parallel, yielding $PT || KL$, as desired. EDIT: I appear to have misread the problem statement (my brain chose to move the word "from" so that it appeared after "follows.") Here is the revised proof: $SA \cdot SB = SC^2 = SP^2$, so $SP$ is tangent to the circumcircle of $\triangle ABP$. This is equivalent to $\angle APS = \angle ABP = \angle ABL = \angle AKL$, that is, $KL || PS$. Let $T = MA \cap CL$. By Pascal's theorem on $ABLCCM$, we see that $P$, $T$, and $S$ are collinear. By Pascal's theorem on $MMAKLC$, we see that $PT$, $KL$, and the tangent to $\Gamma$ at $M$ concur. But $PT || KL$, so we must have $MK = ML$. EDIT2: fixed some typoes.

Excuse me buddies someone can upload the problems in Spanish please¿? i don't understand some words =/ Thanks ^_^

http://www.imo-official.org/problems.aspx This link contains the problems in many different languages, including Spanish.

Dear everyone. $PS$ is tangent to Circle ($\Delta ABP$). Hence $~$ $\angle BAP=\angle SPL$. We need $~$ $\angle SPL+\angle PCB=\angle PCL\Longleftrightarrow \angle PCB+\angle LCS=\angle LPC$ , which is obvious.

WLOG $AS > BS$. By Power of a Point, $SP^2 = SC^2 = SB \cdot SA$, so $\overline{SP}$ is tangent to the circumcircle of $\triangle ABP$. Thus, $\angle KPS = 180 - \angle SPA = \angle ABP$. It follows that \begin{align*} \widehat{ML} = \widehat{MA} + 2\angle AKL = \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL = 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} = 2\angle PCS - \widehat{KC} = \widehat{MK}. .\end{align*} [asy][asy] import graph; size(10.71cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(9); defaultpen(dps); pen ds=black; real xmin=-2.22,xmax=4.48,ymin=-1.99,ymax=3; pen ccqqqq=rgb(0.8,0,0), qqzzqq=rgb(0,0.6,0), evevff=rgb(0.9,0.9,1); filldraw(arc((0.08,1.23),0.37,-17.48,12.32)--(0.08,1.23)--cycle,evevff,blue); filldraw(arc((0,0),0.37,0,29.8)--(0,0)--cycle,evevff,blue); filldraw(arc((1.42,0.81),0.37,-179.98,-150.2)--(1.42,0.81)--cycle,evevff,blue); draw((2,2.55)--(4,0),linewidth(1.6)+ccqqqq); draw((2,2.55)--(0,0),linewidth(1.6)+ccqqqq); draw(circle((2,0.49),2.06),linewidth(1.2)); draw((0.76,-1.16)--(3.43,1.97)); draw((3.43,1.97)--(0.08,1.23)); draw((0.08,1.23)--(0.76,-1.16)); draw((3.43,1.97)--(xmin,0.22*xmin+1.22)); draw((0.08,1.23)--(4,0)); draw((1.42,0.81)--(-1.85,0.81),linewidth(1.6)+qqzzqq); draw((0,0)--(4,0)); draw(circle((1.42,3.15),2.34),linetype("4 4")); draw(arc((0.08,1.23),0.37,-17.48,12.32),blue); draw(arc((0.08,1.23),0.31,-17.48,12.32),blue); draw(arc((0,0),0.37,0,29.8),blue); draw(arc((0,0),0.31,0,29.8),blue); draw(arc((1.42,0.81),0.37,-179.98,-150.2),blue); draw(arc((1.42,0.81),0.31,-179.98,-150.2),blue); draw((2,2.55)--(0.76,-1.16)); draw((0,0)--(3.43,1.97)); draw((-1.85,0.81)--(xmax,-0.75*xmax-0.58)); draw((-1.85,0.81)--(0.76,-1.16),linewidth(1.6)+qqzzqq); dot((0,0),ds); label("$K$",(-0.30,0.06),NE*lsf); dot((4,0),ds); label("$L$",(4.2,0.09),NE*lsf); dot((2,2.55),ds); label("$M$",(2.05,2.62),NE*lsf); dot((1.42,0.81),ds); label("$P$",(1.27,0.96),NE*lsf); dot((3.43,1.97),ds); label("$A$",(3.30,2.17),NE*lsf); dot((0.76,-1.16),ds); label("$C$",(0.91,-1.19),NE*lsf); dot((0.08,1.23),ds); label("$B$",(-0.11,1.49),NE*lsf); dot((-1.85,0.81),ds); label("$S$",(-1.8,0.89),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Solved with a hint from cxyerl. Let $CM\cup LK=X$. Since $SP^2=SC^2=(SA)(SB)$, we have $\triangle SBP\sim\triangle SPA$, so $\angle SPB=\angle PAB=\angle KLB$, so $SP\parallel KL$. Now $\angle CAM=\angle MCS=\angle SPC=\angle PXL$. Therefore, \[\angle MKL=\angle MCL=\angle MXL-\angle KLC=\angle MAC-\angle KAC=\angle MAK=\angle MLK\;\blacksquare\]

Let $D=LK\cap MC$; it's obvious that $SP^2=SC^2=SA\cdot SB\implies SPB=PAB=KLB\implies SP\parallel KL$. Then $$MKL=MCL=MDL-CLK=DPC-KLC=SCM-KLC=CAM-CAK=KAM=KLM,$$as desired. $\blacksquare$

Otis Walkthrough: Let $D$ denote the second tangency point from $S$. Let $DP$ meet $(ABC)$ at $N \neq D$. Now note that, $-1 = (AB,DC) \overset{P}{=} (KL,MN)$ and thus we have $KMLN$ is harmonic. Now it suffices to show $KMLN$ is a kite, which is equivalent to showing that $MN$ is a diameter. We will now use phantom points. Let $N'$ be the $M$-antipode and let $D' = N'P \cap (ABC)$. It then suffices to show $D' \equiv D$. Begin by noting that from the Three Tangents Lemma applied to $MNX$, where $X = MD' \cap NC$, we find that $OD'$ and $OC'$ are tangents to $(CD'P)$ where $O$ is the center of $(ABC)$. Therefore $(CD'P)$ and $(ABC)$ are orthogonal. Now by noting $SC = SP$ and $SC$ tangent to $(ABC)$ we can conclude $S$ is the circumcenter of $CD'P$. From the orthogonality we then must have $SD'$ tangent to $(ABC)$ from which $D' = D$. Thus $MN$ is a diameter and we must have $NL = NK$ from harmonic $KMLN$.

Our given condition implies $\frac{SC}{SB} = \frac{SA}{SC} \implies \frac{SP}{SB} = \frac{SA}{SP}$, or $\triangle SPB \sim \triangle SAP$. Thus we get \[\angle SPB = \angle PAS = \angle KLB,\]or $LK \parallel PS$. Notice the desired implies that $LK$ is parallel to the tangent at $M$, so we want to show this tangent is also parallel to $PS$. Suppose the tangent at $M$ meets the tangent at $C$ at point $T$. Since $TM = TC$ and $SP = SC$, we know $PS \parallel MT$, which concludes our proof. $\blacksquare$

let $MC$ and $KL$ intersect at $Q$, and let $SP$ intersect $MK$ at $R$ we get that $MCK$ and $MKQ$ are similar, and that $MLQ$ and $MCL$ are similar we have $BAC=BLC=BCS$ extend $CK$ to meet $SP$ at $T$ we easily get $TKR=MLC$, and that $TKR$ is similar to $TPC$ we then get that $SP||KL$, and that $SPQ=PQL=MLC=MCS$, so $SC=SP$

Solved with deduck. Claim. The desired condition is equivalent to $\angle PBA + \angle ACP = \angle PCB + \angle PAB$. Proof. $MK=ML$ is equivalent to $\angle MLK = \angle MKL$, so by angle chasing, we have \begin{align*} \angle MKL &= \angle MKA + \angle AKL \\ &= \angle MCA + \angle ABL \\ &= \angle PCA + \angle ABP \end{align*}Further, we also have, \begin{align*} \angle MLK &= \angle MLB + \angle BLK \\ &= \angle MCB + \angle BAK \\ &= \angle PCB + \angle PAB \end{align*}so the claim follows. $\blacksquare$ Thus, we could remove points $M,L,K$ from the diagram. Now, we note that$$SB \cdot SA = SC^2 = SP^2$$which leads to $\Delta SAP \sim \Delta SPB$, and $\angle SAP = \angle SPB$. To finish, we have,\begin{align*} \angle PCB + \angle PAB &= \angle PCB + \angle BPS \\ &= \angle PCS + \angle BPS - \angle BCS \\ &= \angle BPC - \angle BCS \\ &= \angle BPC - \angle BAC \end{align*}Therefore, it is left to show that $\angle PBA + \angle ACP + \angle BAC = \angle BPC$. However, this is true since, \begin{align*} \angle PBA + \angle ACP + \angle BAC &= \angle PBA + \angle ACP + \angle BAP + \angle PAC \\ &= (180^{\circ} - \angle APB) + (180^{\circ} - \angle APC) \\ &= 360^{\circ} - \angle APB - \angle APC \\ &= \angle BPC \end{align*}as desired. $\blacksquare$

We use PoP to get $SA \cdot SB = SC^2$ and also $SC^2 = SP^2$ from which we get that $SP$ is tangent to $(PAB)$. This then implies that $\angle PAB = \angle BPS$ and $ABKL$ cyclic implies $\angle PAB = \angle KAB = \angle KLB \implies KL \parallel SP$. So then we apply Pascal on $BMCCKA$ to get that $BM \cap CK$ lies on $SP$. Then apply the same on $MMCKLB$ to get that $MM \cap KL$ lies on $SP$ which implies $MM$ is parallel to $KL$, so $MK = ML$ as desired.

We WLOG assume $CA<CB$. Draw the other tangent to $\Gamma$ from $S$, and let the tangency point be $D$. Let the intersection of $DP$ and $\Gamma$ distinct from $D$ be $N$. Then we have: $-1=(AB;CD) \overset{P}{=} (KL;MN)$. Now we angle chase to show that $MN$ is a diameter. Since $C$, $D$, and $P$ lie on a circle centered at $S$, we have that $\angle CPD=180-\frac12 \angle CSD$, and thus $\angle MPD = \frac12 \angle CSD$. Then we can calculate that the measure of arc $MN$ equals $\angle COD+2\angle MPD = 180 - \angle CSD + 2\left(\frac12 \angle CSD \right) = 180$. Since $MN$ is a diameter of $\Gamma$ and $LMKN$ is harmonic, by symmetry we must have that $MK=ML$, as desired.

Since in shortlist says iff I will show that from $MK=ML$ $\implies$ $SP = SC$ Let $Q=CL\cap MA$, from Pascals theorem on $CCMABL$ we see that $S$ ,$P$ and $Q$ are coliniear. $\angle QSP = \angle LCM = \angle LKM = \angle KLM = \angle MAK = \angle MAP$ $\implies CQAP$ is cyclic. Then $\angle SPA = \angle QPA = \angle QCA = \angle LCA = \angle SBP$ Which implies that $SP$ is tangent to $(APB)$ $\implies SP\cdot SP= SA\cdot SB=SC\cdot SC$ $\implies SC=SP$ You can also go backwards to get $SP = SC$ $\implies$ $MK=ML$

We will show that $SC=SP \iff MK=ML$. Notice that $SC=SP \iff P$ lies on the $C$-Apollonian circle so $SC=SP \iff \frac{PA}{PB}=\frac{CA}{CB}$. Now $MK=ML \iff \angle LCP=\angle KCP$. From LOS in $\triangle LCP$ and $\triangle KCP$ we have $$\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{LP\sin(\angle LCP)}{KP\sin(\angle CKP)}$$ From PoP we have $\frac{PA}{PB}=\frac{PL}{PK}$ and from LOS we have $\frac{\sin(\angle LCP)}{\sin(\angle CKP)}=\frac{CB}{CA}$. Hence we get $$\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{CB}{CA}\cdot \frac{PA}{PB}$$Now just plug in either condition in this equation to get the other one. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -8.202896110170563, xmax = 19.213125730938394, ymin = -10.497987045605893, ymax = 3.8696816774635145; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); pen qqzzqq = rgb(0.,0.6,0.); draw((5.,-6.)--(13.,-6.)--(5.738284588040686,1.9324528510591215)--cycle, linewidth(0.8) + qqzzcc); draw((5.738284588040686,1.9324528510591215)--(3.8138853212849435,-0.8654634846866291)--(8.117756722334056,-3.271092593269745)--cycle, linewidth(0.8) + qqzzqq); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749)--(8.117756722334056,-3.271092593269745)--cycle, linewidth(0.8) + qqzzqq); /* draw figures */ draw((5.,-6.)--(13.,-6.), linewidth(0.8) + qqzzcc); draw((13.,-6.)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzcc); draw((5.738284588040686,1.9324528510591215)--(5.,-6.), linewidth(0.8) + qqzzcc); draw(circle((9.,-2.3717026281975766), 5.4004205223508635), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(-4.729370032483704,-6.), linewidth(0.8)); draw((-4.729370032483704,-6.)--(5.,-6.), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(10.12208915873399,-7.65426429258731), linewidth(0.8)); draw((5.,-6.)--(13.126060669122644,1.112571352554749), linewidth(0.8)); draw((3.8138853212849435,-0.8654634846866291)--(13.,-6.), linewidth(0.8)); draw((3.8138853212849435,-0.8654634846866291)--(5.738284588040686,1.9324528510591215), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749), linewidth(0.8)); draw((5.738284588040686,1.9324528510591215)--(3.8138853212849435,-0.8654634846866291), linewidth(0.8) + qqzzqq); draw((3.8138853212849435,-0.8654634846866291)--(8.117756722334056,-3.271092593269745), linewidth(0.8) + qqzzqq); draw((8.117756722334056,-3.271092593269745)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzqq); draw((5.738284588040686,1.9324528510591215)--(13.126060669122644,1.112571352554749), linewidth(0.8) + qqzzqq); draw((13.126060669122644,1.112571352554749)--(8.117756722334056,-3.271092593269745), linewidth(0.8) + qqzzqq); draw((8.117756722334056,-3.271092593269745)--(5.738284588040686,1.9324528510591215), linewidth(0.8) + qqzzqq); /* dots and labels */ dot((5.,-6.),dotstyle); label("$A$", (4.7006540756095925,-6.6791480835613585), NE * labelscalefactor); dot((13.,-6.),dotstyle); label("$B$", (13.211848649521938,-6.169988292520035), NE * labelscalefactor); dot((5.738284588040686,1.9324528510591215),dotstyle); label("$C$", (5.694720674902673,2.299868060960004), NE * labelscalefactor); dot((-4.729370032483704,-6.),linewidth(4.pt) + dotstyle); label("$S$", (-5.0011423857687545,-5.864293213305793), NE * labelscalefactor); dot((8.117756722334056,-3.271092593269745),dotstyle); label("$P$", (8.143746020443698,-3.000412997509203), NE * labelscalefactor); dot((13.126060669122644,1.112571352554749),linewidth(4.pt) + dotstyle); label("$K$", (13.195759434826453,1.2471396820992215), NE * labelscalefactor); dot((3.8138853212849435,-0.8654634846866291),linewidth(4.pt) + dotstyle); label("$L$", (3.345695329361647,-0.9249043017914502), NE * labelscalefactor); dot((10.12208915873399,-7.65426429258731),linewidth(4.pt) + dotstyle); label("$M$", (10.18707628677048,-7.421482326940897), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

simple one, posting for storage

swaqar wrote: Solution (if it's correct please tell me) Call the circles $(S,SC) $ $\Omega$. Now the circles $\Omega$ and $ \Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $ \Omega$ and call this point $S'$. Assume that $A $ belong to $[S,B]$. Now the line $SP$ is tangent to the circumcircle of $ \triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2} $. Now $SC$ is the external symmedian of the vertex $C$. Calling the intersection of the angle bisector of $C$ with the line $ AB$, $P'$, we have the points $ (S',P',A,B) $ forming a harmonic range and this implies that $ SC = SP'$. So the circle $\Omega$ is is the appollonian circle of the vertex $ C $ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $ \angle LCM$. Let the external symmedian meet the the $S'C$ in $J$. We have $JN=JC$ and $ SC = SP$ so the traingle $ SCP $ and $ JCN $ are homothetic so the points $ A , P $ and $ N $ are collinear and we have $M$ being the midpoint of the arc $ \overarc{KMN}$. bringing modern artillery to a snowball fight Works though

pascal on \(CCMABL\) gives \(S,P,T\) collinear where T = \(MA \cap CL\) now i claim that the quadrilateral \(PATC\) is cyclic , that ends the problem because, it means \(\angle PCA = \angle MAK\) which means arc \(MK\) = arc \(ML\) so it finishes. now i prove \(\angle TPL = \angle BAK\) . notice \(\angle TPL = \angle BPS = \angle BAK\) where the last one follows from the fact that \(SP\) is tangent to the circumcircle of \(\triangle BPA\) since \(BS * AS = SC^2 = SP^2\) from \(\angle TPL = \angle BAK = \angle BLK\) we obtain \(TL \parallel ST\) which indicates \(\angle KAC = \angle KLC = \angle PTC\) giving \(PATC\) cyclic so we're done!

My first IMO Upload.