IMO 2012 · Problem 1

Given triangle \(ABC\) the point \(J\) is the centre of the excircle opposite the vertex \(A\). This excircle is tangent to the side \(BC\) at \(M\), and to the lines \(AB\) and \(AC\) at \(K\) and \(L\), respectively. The lines \(LM\) and \(BJ\) meet at \(F\), and the lines \(KM\) and \(CJ\) meet at \(G\). Let \(S\) be the point of intersection of the lines \(AF\) and \(BC\), and let \(T\) be the point of intersection of the lines \(AG\) and \(BC\). Prove that \(M\) is the midpoint of \(ST\).

(The excircle of \(ABC\) opposite the vertex \(A\) is the circle that is tangent to the line segment \(BC\), to the ray \(AB\) beyond \(B\), and to the ray \(AC\) beyond \(C\).)

Official text of the problem. Official problem paper (PDF)

Solution

Choose points \(X,Y\) on the line \(BC\), with \(X\) beyond \(B\) and \(Y\) beyond \(C\), such that \[ BX=BA,\qquad CY=CA. \] We will show that \(S=X\) and \(T=Y\).

Equal tangent lengths from \(A,B,C\) give \[ AK=AL,\qquad BK=BM,\qquad CL=CM. \] Since the orders of the relevant points are \(A,B,K\), \(A,C,L\), and \(X,B,M,C,Y\), we obtain \[ XM=XB+BM=AB+BK=AK \] and \[ MY=MC+CY=CL+CA=AL. \] Thus \(XM=MY\), so \(M\) is the midpoint of \(XY\).

Next, \(BM=BK\) and \(JM=JK\) show that \(BJ\) is the perpendicular bisector of \(MK\). Reflection in \(BJ\) therefore interchanges \(K\) and \(M\), and hence the rays \(BK\) and \(BM\). The rays \(BA\) and \(BX\) are opposite to these respective rays, and \(BA=BX\); consequently, this reflection maps \(A\) to \(X\). Thus \(BJ\) is also the perpendicular bisector of \(AX\).

Similarly, \(CJ\) is the perpendicular bisector of \(ML\). Reflection in \(CJ\) interchanges the rays \(CL\) and \(CM\), and therefore maps \(A\) to \(Y\). Hence \(CJ\) is also the perpendicular bisector of \(AY\).

Let \(P,Q\) be the midpoints of \(AX,AY\), respectively. The preceding observations give \[ P\in BJ,\qquad Q\in CJ,\qquad AX\perp BJ,\qquad AY\perp CJ. \] Applying the midpoint theorem in the nondegenerate triangle \(AXY\), we have \[ MP\parallel AY,\qquad MQ\parallel AX. \] But \(LM\perp CJ\) and \(KM\perp BJ\). Therefore \[ MP\parallel LM,\qquad MQ\parallel KM. \] Since these lines pass through \(M\), it follows that \(P\in LM\) and \(Q\in KM\). Consequently, \[ F=P,\qquad G=Q. \] These intersections are unique: \(J\notin BC\) and \(M\) lies strictly between \(B\) and \(C\), so \(M\notin BJ\cup CJ\); hence \(LM\ne BJ\) and \(KM\ne CJ\).

Finally, \(P\) and \(Q\) are interior points of \(AX\) and \(AY\), so \(AF=AX\) and \(AG=AY\) as lines. Since \(A\notin BC\), their intersections with \(BC\) are precisely \(X\) and \(Y\). Thus \(S=X\) and \(T=Y\), and the already established equality \(XM=MY\) proves that \(M\) is the midpoint of \(ST\). \(\square\)

The solution was checked against the official IMO shortlist solutions.