Difference between revisions of "2004 IMO Problems/Problem 1"

Revision as of 22:14, 8 September 2015

Problem

Let $ABC$ be an acute-angled triangle with $AB\neq AC$ . The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$ . The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$ . Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$ .

Solution

$[asy] unitsize(2.5cm); size(60); pair A,B,C,O,K,M,N,R,D; C=(0,0); B=(3.2,0); A=(1.1,3.35); O=(1.6,0); K=(1.51,0); R=(1.42,0.73); M=(2.3,1.44); N=(0.31,0.95); D=(1.1, 2.02); draw(arc((1.6,0),1.6,0,180)); draw(A--B--C--cycle); draw(M--N); draw(A--K); draw(O--D); draw(R--M); draw(C--R); draw(B--R); draw(N--R); draw(N--O); draw(O--M); draw(N--D); draw(M--D); draw(A--D); draw(circumcircle(A,M,N)); draw(circumcircle(C,R,N)); draw(circumcircle(B,M,R)); dot(A); dot(B); dot(C); dot(O); dot(K); dot(M); dot(N); dot(R); dot(D); label("$A$",A,N); label("$B$",B,SE); label("$O$",O,SE); label("$K$",(1.38,-0.1)); label("$C$",C,SW); label("$M$",M,NE); label("$N$",(0.24,1.13)); label("$R$",(1.3,0.59)); markscalefactor=0.025; draw(anglemark(R,N,M),blue); draw(anglemark(N,M,R),blue); draw(anglemark(N,A,R),blue); draw(anglemark(R,A,M),blue); draw(anglemark(O,C,N),green); draw(anglemark(C,N,O),green); draw(anglemark(A,M,N),green); draw(anglemark(M,N,A),red); draw(anglemark(M,B,O),red); draw(anglemark(O,M,B),red); draw(anglemark(O,N,R),yellow); draw(anglemark(R,M,O),yellow); markscalefactor=0.01; draw(rightanglemark(N,(1.305,1.195),O)); [/asy]$

Let $\angle{ACB} = a$ , $\angle{CBA} = b$ , and $\angle{ONR} = k$ . Call $\omega$ the circle with diameter $BC$ and $\odot{AMN}$ the circumcircle of $\triangle{AMN}$ .

Our ultimate goal is to show that $\angle{CNR} + \angle{RMB} = 180^\circ$ . To show why this solves the problem, assume this statement holds true. Call $K$ the intersection point of the circumcircle of $\triangle{CNR}$ with side $BC$ . Then, $\angle{RKC} = 180^\circ - \angle{CNR}$ , and $\angle{RKB} = \angle{CNR}$ . Since $\angle{RMB} = 180^\circ - \angle{CNR}$ , $\angle{RKB} + \angle{RMB} = 180^\circ$ , implying $K$ also lies on the circumcircle of $\triangle{BMR}$ , thereby solving the problem.

We now prove that $\angle{CNR} + \angle{RMB} = 180^\circ$ . Note that $ON$ and $OM$ are radii of $\omega$ , so $\triangle{MON}$ is isosceles. The bisector of $\angle{MON}$ is thus the perpendicular bisector of $MN$ . Since $R$ lies on the bisector of $\angle{MON}$ , $\angle{ONR} = \angle{RMO} = k$ . Angle computations yield that $\angle{RMN} = 180^\circ - \angle{BMO} - \angle{RMO} - \angle{AMN}$ $= 180^\circ - b - k - a,$ from $\angle{AMN} = 180^\circ - \angle{BMN} = 180^\circ - (180^\circ - \angle{ACB}) = \angle{ACB}$ and from $\angle{BMO} = \angle{MBO} =$ $\angle{ABC}$ .

The bisector of $\angle{BAC}$ hits $\odot{AMN}$ at the midpoint of the arc $MN$ not containing $A$ . This point must lie on the perpendicular bisector of segment $MN$ , which is the bisector of $\angle{MON}$ . It follows that $R$ is indeed the midpoint of arc $MN$ , so $A$ , $M$ , $R$ , $N$ , are concyclic. Since $\angle{RAN}$ and $\angle{RMN}$ subtend the same arc $NR$ , $\angle{RAN}$ = $\angle{RMN}$ . With $AR$ being the bisector of $\angle{BAC}$ , we have $\angle{RAN} = \angle{RMN} = \tfrac{\angle{CAB}}{2} = \tfrac{180^\circ - a - b}{2} = 90^\circ - \tfrac{a}{2} - \tfrac{b}{2}.$ We know that $\angle{RMN} = 180^\circ - b - k - a$ . so we have $180^\circ - b - k - a = 90^\circ - \tfrac{a}{2} - \tfrac{b}{2}$ . Since $\angle{CNR} = \angle{CNO} + \angle{ONR} = a + k$ , and $\angle{RMB} = \angle{OMB} + \angle{RMO} = b + k$ , we have $\angle{CNR} + \angle{RMB} = a + k + b + k = a + b + 180^\circ - a - b = 180^\circ.$