IMO 2015 · Problem 4

Triangle \(ABC\) has circumcircle \(\Omega\) and circumcentre \(O\). A circle \(\Gamma\) with centre \(A\) intersects the segment \(BC\) at points \(D\) and \(E\), such that \(B\), \(D\), \(E\) and \(C\) are all different and lie on line \(BC\) in this order. Let \(F\) and \(G\) be the points of intersection of \(\Gamma\) and \(\Omega\), such that \(A\), \(F\), \(B\), \(C\) and \(G\) lie on \(\Omega\) in this order. Let \(K\) be the second point of intersection of the circumcircle of triangle \(BDF\) and the segment \(AB\). Let \(L\) be the second point of intersection of the circumcircle of triangle \(CGE\) and the segment \(CA\).

Suppose that the lines \(FK\) and \(GL\) are different and intersect at the point \(X\). Prove that \(X\) lies on the line \(AO\).

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

Solution

All angles between lines below are directed angles modulo \(180^\circ\).

Since \(AF=AG\) and \(OF=OG\), both \(A\) and \(O\) lie on the perpendicular bisector of \(FG\). Moreover, \(A\ne O\), because \(A\) lies on the nondegenerate circle \(\Omega\). Thus \(AO\) is precisely the perpendicular bisector of \(FG\).

Let \(t\) be the tangent to \(\Omega\) at \(A\). Both \(t\) and \(FG\) are perpendicular to \(AO\), so they are parallel. The tangent–chord theorem therefore gives \[ \angle(AB,BC)=\angle(t,AC)=\angle(FG,AC). \tag{1} \]

The concyclicity of \(B,D,F,K\), together with \(B,K,A\) collinear, gives \[ \angle(FK,FD)=\angle(BK,BD)=\angle(AB,BC). \] Consequently, by (1), \[ \begin{aligned} \angle(FK,FG) &=\angle(FK,FD)+\angle(FD,FG)\\ &=\angle(FG,AC)+\angle(FD,FG)\\ &=\angle(FD,AC). \end{aligned} \tag{2} \]

On the other hand, \(F,G,D,E\) lie on \(\Gamma\), so \[ \angle(FG,GE)=\angle(FD,DE). \] Also, \(C,G,E,L\) are concyclic, so \[ \angle(GE,GL)=\angle(CE,CL). \] Using the collinearities \(B,D,E,C\) and \(C,L,A\), we obtain \[ \begin{aligned} \angle(FG,GL) &=\angle(FG,GE)+\angle(GE,GL)\\ &=\angle(FD,DE)+\angle(CE,CL)\\ &=\angle(FD,BC)+\angle(BC,AC)\\ &=\angle(FD,AC). \end{aligned} \tag{3} \] Equations (2) and (3) yield \[ \angle(FK,FG)=\angle(FG,GL). \tag{4} \]

Now reflect the plane in \(AO\). This reflection interchanges \(F\) and \(G\). Since \(AO\perp FG\), equation (4) says that it also takes the direction of \(FK\) to the direction of \(GL\). Hence it takes the line \(FK\) to the line \(GL\), and conversely.

The two lines are distinct and meet at \(X\), so their unique intersection must be fixed by this reflection. The fixed points of reflection in \(AO\) are exactly the points on \(AO\). Therefore \(X\) lies on \(AO\), as required. \(\square\)

The solution was checked against the official IMO shortlist solutions.