I think the key to this problem is to prove that $Q$ is on $AM$, where $M$ is the midpoint of $BC$.

A full solution: Indeed the radius of the circumcircle of triangle ABC is $\frac{2}{3}m$. Proof. The line Z is a symmetry axis of the isosceles triangle ABC; hence, it is an angle bisector, a median, an altitude and a diameter of the circumcircle simultaneously. Therefore, both the centroid G and the circumcenter O of triangle ABC lie on this line Z. Since the line Z is an angle bisector, actually the angle bisector of the angle BCA, we have < BCQ = < GCM. Also, < BQC = < GMC (since < BQC = < BMC because the point Q lies on the cirle through B, C and M). Therefore, the triangles BCQ and GCM are similar, and $\frac{CQ}{CM}=\frac{BC}{GC}$. Since CQ = m, we have $\frac{m}{CM}=\frac{BC}{GC}$. But the midpoint N of the side AB also lies on the line Z, and the centroid G of triangle ABC satisfies $GC=\frac{2}{3}\cdot NC$, so we obtain $\frac{m}{CM}=\frac{BC}{\frac{2}{3}\cdot NC}=\frac{3}{2}\cdot \frac{BC}{NC}$. Since the point O is the circumcenter of triangle ABC, it lies on the perpendicular bisector of the side AC, i. e. on the perpendicular to AC at the point M. Therefore, < CMO = 90°. Together with < CNB = 90°, we get < CNB = < CMO. Also, since the line Z bisects the angle BCA, we have < NCB = < MCO. Thus, the triangles CNB and CMO are similar, and $\frac{BC}{NC}=\frac{OC}{MC}$, so that $\frac{m}{CM}=\frac{3}{2}\cdot \frac{BC}{NC}=\frac{3}{2}\cdot \frac{OC}{MC}=\frac{3}{2}\cdot \frac{OC}{CM}$, and $m=\frac{3}{2}\cdot OC$, so $OC=\frac{2}{3}\cdot m$. But OC is just the radius of the circumcircle of triangle ABC. Hence we are done. Darij

Shortlists solution of Problem 2 Let P be the center of the circle $k_1$ through B, C and M, O be the circumcenter of ABC and K be the midpoint of MC. Since AC=BC, the center O lies on CH. Let KP intersect CH in point L. Since KP and OM are perpendicular to AC, then KP || OM. From MK = KC it follows that OL = CL. On the other hand OP is perpendicular to BC, hence angle LOP = angle COP = 90º - angle BCH. Also we have angle OLP = angle CLK = 90º - angle ACH. Since ABC isosceles and angle BCH = angle ACH, then angle LOP = angle OLP and LP = OP. Since CP = PQ we obtain that angle CLP = angle QOP and CL = OQ. Thus we have CL = LO = OQ, so CO = 2/3CQ. Finally for the radius R of the circumcircle of ABC we obtain R = 2/3m.

this is my solution for problem 2: the points M,Q,A,C are concycle so if we denote by ${P}=CQ\cap AM$ we get that $PM\cdot PA=PQ\cdot PC$ (*). We know that $PM=\displaystyle\frac 13 MA$, $PA=\displaystyle\frac23 MA$, $\displaystyle CP=\frac23 CC'$ where ${C'}=CQ\cap AB$. We know that $MA^2=m_a^2=\displaystyle\frac{2(b^2+c^2)-a^2}4$ (1) and $CC'^2=m_c^2=\displaystyle\frac{2(b^2+a^2)-c^2}4$ (2) If we denote by $\alpha=\angle ACQ$ and knowing that $\displaystyle R=\frac{abc}{4S}$ and using (*),(1),(2) and $CP+PQ=m$ with some easy computations we get the desired result.

This is a solution of this problem posted by vedran6 in a different thread: It is obvious that the BCMQ is cyclic and because of that is and we have that angle <CQM=<CBM (1) let $CZ\cap BM=\left(T\right)$ ; and $CZ\cap AB=\left(D\right)$ it is obvious that is $CT=\frac23\cdot CD$ and because of AC=BC we have <ACQ=<BCT (2) , from (1) and (2) we have that trinagle CQM is similar to triangle CBT. From similarity we have that is $\frac{CQ}{CB}=\frac{CM}{CT}$ which is equal to $CQ=\frac{3\cdot CB\cdot CB}{4\cdot CD}$ (3). It is well known that in every triangle is $R=\frac{a\cdot b\cdot c}{4P}$ so we use it at this triangle ABC and got $R=\frac{BC\cdot BC}{2\cdot CD}$ (4) . After dividing (3) and (4) we get $CQ=1,5\cdot R$ and the problem is finished

Here is my solution (I think that it differs from the other preceding solutions): Let $w=C(O,R)$ be a circumcircle of the triangle $ABC$ and let $U$ be the middlepoint of the segment $[AM]$. From $\widehat {QCM}\equiv \widehat {QCB}\Longrightarrow QM=QB=QA$ (the point $Q$ is the circumcenter for the triangle $ABM$). From $QU\perp AC,\ MO\perp AC\Longrightarrow QU\parallel MO\Longrightarrow$ $\frac 23 =\frac{CM}{CU}=\frac{CO}{CQ}=\frac{R}{m}\Longrightarrow R=\frac 23 m$.

My Solution Let $MC=AM=x$ Then $BC=2x$ From sinus we have $\longrightarrow$ $\frac{2x}{cos\alpha}=2R$ Where $R$ be the circumradius of $\triangle{ABC}$ and $\angle{ACQ=BCQ=\alpha}$. Then from $\triangle{BCQ}$ we have by sinus $\frac{m}{sin(\alpha+\beta)}=\frac{2Rcos\alpha}{sin(2\alpha+\beta)}$ Where $\angle{MBC}=\beta$. Then we have $m=\frac{2Rcos\alpha\cdot sin(\alpha+\beta)}{sin(2\alpha+\beta)}$ By some trigonometric identies we have $m=\frac{Rsin(2\alpha+\beta)+Rsin\beta}{sin(2\alpha+\beta)}$ $\longrightarrow$ $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}$ Loking at $\triangle{MBC}$ by sinus we have $\longrightarrow$ $\frac{x}{sin\beta}=\frac{2x}{sin(2\alpha+\beta)}$ $\longrightarrow$ $\frac{sin\beta}{sin(2\alpha+\beta)}=\frac{1}{2}$ so $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}=\frac{3R}{2}$ $\longrightarrow$ $R=\frac{2m}{3}$ Anar Abbas Istek Lyceum

Let $K$ $\in [CA$ , $|MA|=|AK|$ , $M \neq K$ and $O$ is $\triangle ABC$'s circumcircle center $|CM|=|MA|=|AK|$ $\Rightarrow$ $|MK|=|AC|=|BC|$ $.$ $\dotsm$ $(i)$ $C,M,Q,B$ is circularly $\Rightarrow$ $\angle QMA = \angle QBC$ $\dotsm$ $(ii)$ $ABC$ is isoscles triangle and $D$ is $BA$'s midpoint $\Rightarrow$ $\angle ACD = \angle BAD$ $.$ Also $B,Q,M,C$ is circularly $\Rightarrow$ $|MQ|=|BQ|$ $\dotsm$ $(iii)$ Combine $(i),(ii),(iii)$ $\Rightarrow$ $\triangle QMK \equiv \triangle QBC$ İn here $\angle QKM=\angle QCB=\angle OCA=\angle OAC$ $\Rightarrow$ $OA \parallel QK$ $\Rightarrow$ $\frac{AC}{KC} = \frac{OC}{QC}$ $.$ Also $|AC|=2|MC|$ , $|KC|=3|MC|$ $\Rightarrow$ $\frac{OC}{QC} = \frac{2}{3}$ $\Rightarrow$ $|OC|=\frac{2m}{3}$ $Q.E.D$

Let length of side $CB$ = $x$ and length of $QM = a$. We shall first prove that $QM = QB$. Let $O$ be the circumcenter of $\triangle ACB$ which must lie on line $Z$ as $Z$ is a perpendicular bisector of isosceles $\triangle ACB$. So, we have $\angle ACO = \angle BCO = \angle C/2$. Now $MQBC$ is a cyclic quadrilateral by definition, so we have: $\angle QMB = \angle QCB = \angle C/2$ and, $\angle QBM = \angle QCM = \angle C/2$, thus $\angle QMB = \angle QBM$, so $QM = QB = a$. Therefore in isosceles $\triangle QMB$ we have that $MB = 2 QB \cos C/2 = 2 a \cos C/2$. Let $R$ be the circumradius of $\triangle ACB$. So we have $CM = x/2 = R \cos C/2$ or $x = 2R \cos C/2$ Now applying Ptolemy's theorem in cyclic quadrilateral $MQBC$, we get: $m . MB = x . QM + (x/2) . QB$ or, $m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2$ or, $ R = (2/3)m$

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