http://olympiads.mccme.ru/vmo/

@Mathuz : Do you mean K is the intersection point of BC and MN?

$K$ is the point of intersection $MN$ and $AC$.

War-Hammer wrote: $K$ is the point of intersection $MN$ and $AC$. the intersection of $MN$ and $AC$ is $N$....

There is a typo in the problem statement. $N$ should lie on $BC$, under the same condition $CN=AM$.

Solution: Let $(P)$ be tangent to $MN,AC$ at $D,E$, and let $Q$ be tangent to $MN,AC$ at $F,G$. Let $X,Y,Z$ be the midpoints of $AC,MN,PQ$, and let $S,T$ be the midpoints of $EG,DF$. Note that since $PQ$ passes through $K$ and $DP,FQ\perp MN$, $TZ\perp MN$ as well. Similarly $ZS\perp AC$. Now \[ KD=KE=\frac{1}{2}(KM+KA-MA) \] and \[ KF=KG=\frac{1}{2}(KN+KC+NC) \] so \[ 2KT=KD+KF=\frac{1}{2}(KM+KA+KN+KC) \] \[=\frac{1}{2}(KM+KN)+\frac{1}{2}(KA+KC)=KX+KY \] Thus \[ KT=KS=\frac{1}{2}(KX+KY)\Rightarrow YT=SX \] $Z$ lies on $PQ$, the angle bisector of $MKA$, so $ZT=ZS$. Then $\triangle ZTY\cong \triangle ZSX$, yielding $ZY=ZX$. So $Z$, the intersection of the angle bisector of $YKX$ and the perpendicular bisector of $XY$, is the midpoint of arc $XY$ on circle $(KXY)$. But $RA=RC,MA=NC$, and $\angle MAR=\angle NCR$, so $\triangle MAR\cong \triangle NCR$ implying $RM=RN$. Thus $RY\perp YK$, and of course $RX\perp XK$. So $RXYK$ is cyclic with diameter $RK$. Then $Z$ also lies on this circle, so $RZ\perp ZK$, and $R$ lies on the perpendicular bisector of $PQ$, giving $RP=RQ$.

Can anyone make all of it into the page contest?

Domination1998 wrote: Can anyone make all of it into the page contest? It is here.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=587501

OK thanks for the page

weirdly worded.

Lol...rotation and done...for the foll sol we do not need angle chase at all. Just note two things,$1)RM=RN,RC=RA$ and that $2)\odot MPA,\odot NQC$ are congruent.Now $1)$ imples $R$ is the centre of rotation sending $\odot MPA $to$\odot NQC$.$2)$ implies it is equidistant from their centres,and so present on line of symmetry between the circles.Since $PQ$ passes thru centres of those circles,,by symmetry $RP=RQ$ is obvious.

All Russia MO 2014/4 wrote: Given a triangle $ABC$ with $AB>BC$, let $ \Omega $ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$. M. Kungodjin

Deleted.

I appreciate your concern and some genuine points you raise. I'd appreciate it even more though, if the discussion was less public.

The above post is definitely uncalled for and should have been communicated privately, not sprang like a trap in a wild geometry thread. But since you decided to make it public, let me address some of the points raised. First, you seem to be of the impression that writing up concise and short solutions are of no use since that does not earn you any extra marks in IMO. But how concise your solution is is a benchmark of how well you understand it and attempting to be concise will definitely push your understanding. A solution does not mark the end of a problem, but is a call for further reflection, further distillation of its essence and in trying to make it concise, you are forcing yourself to differentiate what is important and what are just unnecessary details that can be filled up at will. In math olympiad, the vast majority of problems reuse the same bunch of key ideas but cleverly disguised differently and it is via this process that will enable you to see through the smoke screens. Secondly, I believe that for olympiad problems, especially those which are not on the easy side (which may be subjective), any solution is welcome even if it is a rehash of the ideas already presented in previous posts. Positing a solution, all the moreso a complete one, is an invitation for scrutiny and criticism and should always be encouraged. Furthermore even if the outline is the same, different people have different perspectives on certain steps and these show in their presentation of the solution. For instance, they may find different steps trivial and complement each other by explaining parts that are left out by each other (I have seen this in action with advanced texts), or have different explanations for intermediate steps that may click for different groups of people. Finally, unless you know anantmudgal09 personally, I do not see why you have an issue with him focusing solely on geometry and NT. Different people have different priorities in life and different tastes in problems. For many people, the color of their medal, or even getting one, is ultimately secondary to the enjoyment gained through problem solving.

Easy Claim 1 The circumcircles of $\triangle KAM$ and $\triangle KCN$ are congruent Proof Note that $MA\div sin\angle AKM=NC\div sin\angle NKC$ Claim 2 $RM=RN$ Proof Let $X$ be the second intersection of the circumcircles of $\triangle BMN$ and $\triangle ABC$ By spiral similarity $\triangle XMA\equiv \triangle XNC$ (because $MA=NC$) Thus we get that $X=R$ and consequently $RM=RN$ Main problem Let the angle bisector of $\angle CKN$ intersect $(AKM)$ at $Y$ and $(CKN)$ at $Z$.Since $(AKM)$ and $(CKN)$ are congruent,by Incenter-Excenter Lemma we get that $YP$=$ZQ$. Also by spiral similarity we get that $RY=RZ$. Moreover $\angle RYP=\angle RZQ$.Thus we get that $\triangle RYP\equiv \triangle RZQ$ and the result follows $\blacksquare$

interesting

yikes this problem has two threads. I think this one is the more popular one so I'm posting here too. Sorry for the double post [asy][asy] unitsize(1.5inches); pair B=dir(125); pair A=dir(215); pair C=dir(-35); pair I=incenter(A,B,C); pair L=circumcenter(A,I,C); pair R=-L; pair M=0.73*B+(1-0.73)*A; pair N=2*foot(circumcenter(B,R,M),B,C)-B; pair K=extension(M,N,A,C); pair P=incenter(A,M,K); pair Q=extension(P,K,C,I); pair PP=((C-R)/(A-R))*(P-R)+R; pair QQ=((A-R)/(C-R))*(Q-R)+R; draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(circumcircle(B,M,N)); draw(C--K); draw(M--K); draw(circle(R,abs(R-P)),dotted); draw(QQ--M--P--A--cycle); draw(PP--N--Q--C--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(-60)); dot("$R$",R,dir(R)); dot("$M$",M,dir(30)); dot("$N$",N,1.5*dir(80)); dot("$K$",K,dir(0)); dot("$P$",P,dir(-75)); dot("$Q$",Q,dir(-110)); dot("$P'$",PP,dir(0)); dot("$Q'$",QQ,dir(70)); [/asy][/asy] Claim: $R\in(BMN)$. Proof: Let $R'=(BMN)\cap(BAC)$. We have a spiral similarity $\psi$ centered at $R'$ sending $MA\mapsto NC$. Since $MA=NC$, this has to be a rotation, so $R'A=R'C$, so $R'=R$, as desired. We also see that $RM=RN$ since $\psi(M)=N$. $\blacksquare$ Let $P'=\psi(P)$ and $Q'=\psi^{-1}(Q)$. Since $Q$ is a rotation, we see that \[P'NQC=\psi(PMQA),\]so $P'NQC\cong PMQ'A$. Note that $\angle MPA=\pi/2+\frac{1}{2}K$ and $\angle NQC=\pi/2-\frac{1}{2}K$, so $\angle MPA+\angle NQC=\pi$. Thus, $\angle NP'C+\angle NQC=\pi$, so actually the two congruent quadrilaterals $P'NQC$ and $PMQ'A$ are cyclic. We claim the quadrilateral $P'QPQ'$ is cyclic. Since $Q'P=PQ'$, it suffices to show that it is an isosceles trapezoid, or that $\angle P'QP=\angle Q'PQ$. We see that \begin{align*} \angle Q'PQ &= \angle Q'PM+\angle +\angle MPQ \\ &=\angle Q'AM+\angle MPK \\ &=\angle QCN+\angle (\pi/2+A/2) \\ &=\pi/2+A/2+C/2, \end{align*}and by symmetry then, we also have $\angle P'QP=\pi/2+A/2+C/2$, so $P'QPQ'$ is cyclic, as desired. Since $RP=RP'$ and $RQ=RQ'$, we see that the circumcenter of $(P'QPQ')$ is $R$, so $RP=RQ$, as desired.

Let $RC $ interesects the circumcircle of $\triangle CNQ $ at $I$. $RA $ intersects the circumcircle of $\triangle APM $ at $J $. It is easy to show that $\triangle RCN =\triangle RAM $. We have, $\angle CIN=90^{\circ}+\frac {1}{2}\angle MKA=\angle MIA $. Note that $\triangle CIN=\triangle AIN $, so the radius of the circumcircle of $\triangle CIN $ is equal to the radius of the circumcircle of $\triangle AIN $. $R $ lies on the radical axis of $(CIN) $ and $(AJM) $, so $ R$ also lies on the perpendicular bisector of $PQ $. $PQ $ is the line, which passes through the circumcircle of $\triangle CIN $ and $\triangle AJM $. $\blacksquare $

ARMO 2014 Grade 10 P4 wrote: Given a triangle $ABC$ with $AB>BC$, let $ \Omega $ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$. Claim 1:- $\triangle RMA\cong\triangle RNC$. Easy to see that $RC=RA$, $MA=NC$ and $\angle NCR=\angle MAR$. So, $MR=RN$, After a bit of angle chasing we find that $RBNM$ is a cyclic quadrilateral. Hence, $R$ is the Miquel Point of $CMNA$. Hence, $R\in$ {$\odot(KAM),\odot(KCN)$}. Now notice that as triangles $RMA,RNC$ were congruent, their Circumcircles will also be congruent. Let $KP\cap\odot(KAM)=D$ and $KP\cap\odot(KCN)=E$. Claim 2:- $\triangle REC\cong\triangle RDA$. $RC=RA$. $\angle ECR=\angle EKR=\angle DKR=\angle DAR$. Now notice that $\frac{DA}{\sin DKA}=\frac{EC}{\sin DKA}\implies DA=EC$. Note that we took this step just because their Circumcircles are congruent. Hence, $RE=RP$. Claim 3:- $\triangle REQ\cong\triangle RDP$. $RE=RP$, $\angle RFQ=\angle RDP$ and $QE=EC=DA=DP$ by IEL or Fact $5$. Hence, $\boxed{RP=RQ}$.

I find a quite interesting proof(though it's a little complex ). First, let $I'$ be the incentre of the triangle $RAM$ , let $I''$ be the incentre of the triangle $RCN$. $R$ is the midpoint of the arc $ABC \implies AR=CR$. Now notice that $AM=CN, \angle RAB= \angle RCB \implies \triangle RAM\cong\triangle RCN$ $\implies RM=RN \implies \angle RMN=\angle RNM=\angle RAC=\angle RCA$. Thus, we can claim that $(R,M,A,K),(R,N,C,K)$ are concyclic. Obviously, $\angle APM=\angle AI'M$, so $(A,P,I',M)$ are concylic. $\angle RI'P=90°+\frac{1}{2}\angle RMA-\angle PI'A=90°+\frac{1}{2}\angle RMA-\angle PMB$ $=90°+\frac{1}{2}(\angle RMA-\angle KMA)=180°-\frac{1}{2}\angle RAC$. Similarly, $\angle NI''C=90°+\frac{1}{2}\angle K,$$\angle NQC=90°-\frac{1}{2}\angle K\implies (N,I'',C,Q)$are concyclic. Similarly, we can find that $\angle RI''Q=180°-\frac{1}{2}\angle RCA \implies \angle RI''Q=\angle RI'P$. In addition, we can find $RI'=RI''$ obviously. As a result, all we need to do is to prove that $I'P=I''Q.$ If so, we can claim that $\triangle RI'P\cong\triangle RI''Q$, which $\implies RP=RQ$.Applying the Ptolemy Theorem, we can easily find that it is true.

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dot((-6.5476277224685795,-1.8881695128468945),linewidth(4pt) + dotstyle); label("$K$", (-6.4630355474703665,-1.7234200304833067), NE * labelscalefactor); dot((-4.630578998328487,-1.4870167599308866),linewidth(4pt) + dotstyle); label("$P$", (-4.536459019617474,-1.3255400953832532), NE * labelscalefactor); dot((2.7931377967971787,0.06643589452331444),linewidth(4pt) + dotstyle); label("$Q$", (2.8766724027730057,0.2240975465853765), NE * labelscalefactor); dot((-2.9247888620673153,-1.1300710024019107),linewidth(4pt) + dotstyle); label("$D$", (-2.8402340331382963,-0.9695422587147843), NE * labelscalefactor); dot((1.0873476605360077,-0.29050986300566106),linewidth(4pt) + dotstyle); label("$E$", (1.1804474162938283,-0.13190029008309248), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Notice that $$\angle RAM=\angle RCN$$$$\frac{RA}{AM}=\frac{RC}{CN}$$Hence $R$ is the center of spiral sim. sending $\overline{AM}$ to $\overline{CN}$, as well as the miquel point of $AMNC$. Let $D$ be the arc midpoint of arc $AM$ not containing $K$ in $(KAMR)$, and $E$ the arc midpoint of arc $NC$ not containing $K$ in $(NCAR)$. Then by spiral similarity, $RADM\cong RCEN$. In particular, $RD=RE$. Meanwhile, by Fact 5, $$DP=DM=EN=EQ$$Hence $\triangle RDP\cong REQ$ and we are done.

A different solution (it appears long, but is very straightforward and easily motivated): Let $S = \odot(KQN) \cap \odot(KPM) \ne K$ and $T = \odot(KCQ) \cap \odot(KAP) \ne K$ ; $X,Y,Z$ be midpoints of segments $AC,MN,PQ$. Claim: $R$ is center of spiral sim taking $\overline{MN}$ to $\overline{AC}$. Proof: Let $R'$ be the center of spiral sim taking $\overline{MN}$ to $\overline{AC}$. Then $R' \in \odot(ABC)$. Also, $$\frac{RA}{RC} = \frac{AM}{CN} = 1$$so $R'$ lies on the perpendicular bisector of segment $AC$. So $R'$ must be one of the midpoints of arc $\widehat{AC}$ of $\odot(ABC)$, and because of the confi we obtain $R = R'$. $\square$ Claim: The four points $K,R,S,T$ lie on some circle $\omega$ with diameter $\overline{KR}$. Proof: Note that $R = \odot(KCN) \cap \odot(KAM) \ne K$. Now invert at $K$. Then, $Q^*$ is incenter of $\triangle KN^*C^*$. $P^*$ is $K$-excenter of $\triangle KA^*M^*$. $R^* = \overline{A^*M^*} \cap \overline{C^*N^*}$. $S^*$ is incenter of $\triangle K^*M^*N^*$. $T^*$ is $R^*$-excenter of $\triangle R^*A^*C^*$. $\overline{R^*K}$ is external angle bisector of $\angle A^*R^*C^*$ as $$\angle KR^*C^* = \angle KCR = 180^\circ - \angle KAR = 180^\circ - \angle KR^*A^*$$ Now as all the points $R^*,S^*,T^*$ lie on the internal angle bisector of $\angle A^*R^*C^*$ (which is perpendicular to the external angle bisector), so we are done. $\square$ Claim: The midpoint of segments $AC,MN,PQ$ also lie on $\omega$. Proof: We already have $\angle RXK = 90^\circ$, i.e. $X \in \omega$. Now because of spiral sim, $Y \in \odot(KXR) ~ \implies ~ Y \in \omega$. $Z \in \odot(KXT) ~ \implies ~ Z \in \omega$. $\square$ Now we basically have that $\angle RYK = 90^\circ$, which implies $RP = RQ$. This completes the proof of the problem.

nice problem

[asy][asy] unitsize(4cm); import geometry; point B = dir(110); point C = dir(210); point A = dir(330); point O = (0,0); point Rp = intersectionpoint(line(B,incenter(triangle(A,B,C))), perpendicular(O,line(A,C))); point R = scale(-1, O) * Rp; point N = (B-C) * 0.75 + C; point M = reflect(line(R, circumcenter(triangle(R,B,N)))) * N; point K = intersectionpoint(line(M,N), line(A,C)); point P = incenter(triangle(A,K,M)); point Q = excenter(triangle(C,K,N).AC); point T = (P+Q)/2; point D = intersectionpoint(line(P,Q), perpendicular(circumcenter(triangle(A,K,M)),line(A,M))); point E = intersectionpoint(line(P,Q), perpendicular(circumcenter(triangle(C,K,N)),line(C,N))); dot("$A$", A, dir(0)); dot("$B$", B, dir(110)); dot("$C$", C, dir(180)); dot("$R$", R, dir(90)); dot("$M$", M, dir(45)); dot("$N$", N, dir(135)); dot("$P$", P, dir(45)); dot("$Q$", Q, dir(135)); dot("$K$", K, dir(90)); dot("$T$", T, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(270)); draw(A--B--C--cycle); draw(R--M); draw(R--N); draw(R--D, green); draw(R--E, green); draw(K--C); draw(K--N); draw(K--Q); draw(E--N, dotted+green); draw(D--M, dotted+green); draw(R--T); draw(N--E); draw(M--D); draw(circle(R,N,C), dashed+blue); draw(circle(R,M,A), dashed+blue); draw(circle(A,B,C), red); [/asy][/asy] First, Claim: $R$ is the center of spiral similarity sending $MN$ to $AC$. In particular, then $\triangle RNC \cong \triangle RMA$, and $(RNCK)$ and $(RMAK)$ are cyclic. Proof. First, note that $RA = RC$ and $MA = NC$. But then directing angles, \[ \measuredangle RAM = \measuredangle RAB = \measuredangle RCB = \measuredangle RCN \]hence we find $\triangle RAM \cong \triangle RCN$. By the definition of spiral similarity, then $(RNCK)$ and $(RMAK)$ are cyclic. $\blacksquare$ Now, we note that $K$, $P$ and $Q$ are collinear on the angle bisector of $\angle MKA$. Claim: Define $D$ and $E$ as the intersections of $PQ$ with $(RMAK)$ and $(RNCK)$ respectively. Then $D$ is the midpoint of minor arc $AM$, $E$ is the midpoint of minor arc $CN$ and $RD = RE$ and $NE = MD$. Proof. The fact that $D$ and $E$ are arc midpoints is immediate from the fact that they lie on the angle bisector of $\angle MKA$. To prove $RD = RE$, note that directing angles, \[ \measuredangle REN = \measuredangle RCN = \measuredangle RCB = \measuredangle RAB = \measuredangle RAM = \measuredangle RDM \]and moreover directing angles, \[ \measuredangle RNE = \measuredangle RKE = \measuredangle RKD = \measuredangle RMD \]since $RN = RM$ then we find $\triangle RNE \cong \triangle RMD$, which is enough. $\blacksquare$ We now prove that Claim: We have $PD = QE$. Proof. Using Incenter/Excenter Lemma, note that \[ PD = MD = NE = QE \]as needed. $\blacksquare$ Now, let $T$ be the midpoint of $DE$. Then since $PD = QE$, we see that $T$ is also the midpoint of $PQ$. But $\angle RTQ = \angle RTE = 90$, so in fact $RP = RQ$ as needed. $\square$