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

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== See also ==
 
== See also ==
 
*http://www.artofproblemsolving.com/community/c6h14020p99445
 
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{{IMO box|year=2004|before=First Problem|num-a=2}}

Latest revision as of 23:51, 18 November 2023

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 \[\angle{CNR} + \angle{RMB} = a + k + b + k = a + b + 180^\circ - a - b = 180^\circ.\]

The problem is solved.

$\textbf{NOTE:}$ We have $\angle{RKB} + \angle{KBA} + \angle{BAK} = 180^\circ \implies \angle{BAK} = 180^\circ - (180^\circ - \angle{RMB}) - \angle{KBA}$. Noting that $180^\circ - \angle{RMB} = 180^\circ - b - k = 180^\circ - (90^\circ - \tfrac{a}{2} + \tfrac{b}{2}) = 90^\circ + \tfrac{a}{2} - \tfrac{b}{2}$, we then have \[\angle{BAK} = 180^\circ - (90^\circ + \tfrac{a}{2} - \tfrac{b}{2}) - b = 90^\circ - \tfrac{a}{2} - \tfrac{b}{2}\] which is indeed the measure of $\angle{BAR}$. This implies that $K$ lies on the bisector of $\angle{BAC}$, and from this, it is clear that $K$ must lie on the interior of segment $BC$. Not proving that $K$ had to lie in the interior of $BC$ was a reason that a large portion of students who submitted a solution received a 1-point deduction.

See also

2004 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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