IMO 2025 · Problem 2

Let \(\Omega\) and \(\Gamma\) be circles with centres \(M\) and \(N\), respectively, such that the radius of \(\Omega\) is less than the radius of \(\Gamma\). Suppose circles \(\Omega\) and \(\Gamma\) intersect at two distinct points \(A\) and \(B\). Line \(MN\) intersects \(\Omega\) at \(C\) and \(\Gamma\) at \(D\), such that points \(C\), \(M\), \(N\) and \(D\) lie on the line in that order. Let \(P\) be the circumcentre of triangle \(ACD\). Line \(AP\) intersects \(\Omega\) again at \(E\ne A\). Line \(AP\) intersects \(\Gamma\) again at \(F\ne A\). Let \(H\) be the orthocentre of triangle \(PMN\).

Prove that the line through \(H\) parallel to \(AP\) is tangent to the circumcircle of triangle \(BEF\).

(The orthocentre of a triangle is the point of intersection of its altitudes.)

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

Solution

Put \(r=AM\) and \(R=AN\), so \(r<R\). The distinct centres \(M,N\) both lie on the perpendicular bisector of \(AB\); therefore \(A\notin MN\), and triangle \(AMN\) is nondegenerate.

First observe that \(P\) is the excentre opposite \(A\) of triangle \(AMN\). Indeed, \(MA=MC\), so the perpendicular bisector of \(AC\) passes through \(M\) and bisects \(\angle AMC\). Since \(MC\) is opposite to \(MN\), this is the external angle bisector at \(M\) of triangle \(AMN\). Similarly, the perpendicular bisector of \(AD\) is the external angle bisector at \(N\). Their intersection \(P\) is consequently the excentre opposite \(A\).

Let \(I\) be the incentre of triangle \(AMN\). Thus \(A,I,P\) lie on the internal angle bisector at \(A\), and \[ MI\perp MP,\qquad NI\perp NP. \]

Let \(S\) be the midpoint of \(MN\), and let \(T\) and \(J\) be the images of \(A\) and \(I\), respectively, under the half-turn about \(S\). This half-turn interchanges \(M,N\), so \[ NJ\parallel MI\perp MP,\qquad MJ\parallel NI\perp NP. \] Hence \(J\) is the orthocentre of triangle \(PMN\), meaning \(J=H\). Since the half-turn maps \(AI\) to \(TH\), we obtain \[ TH\parallel AI=AP. \tag{1} \]

We will prove that \(T\) lies on the circumcircle of \(BEF\), with tangent parallel to \(AP\).

By its definition, \(T\) makes \(AMTN\) a parallelogram. In particular, \[ MT\parallel AN,\quad NT\parallel AM,\quad MT=R,\quad NT=r. \] Construct the rhombus with vertex \(A\), adjacent sides on the rays \(AM,AN\), and side length \(r\). Its opposite vertex lies on the internal angle bisector \(AP\) and is at distance \(r\) from \(M\); it is therefore \(E\). Likewise, the corresponding rhombus of side length \(R\) has opposite vertex \(F\).

Consequently, \(E\) lies on the ray \(MT\) with \(ME=r\), while \(F\) lies on the ray \(NT\) with \(NF=R\). Because \(r<R\), the point orders are \(M,E,T\) and \(N,T,F\). Thus \[ TE=R-r=TF,\qquad TE\parallel AN,\quad TF\parallel AM. \tag{2} \]

These facts show that \(E,F,T\) are distinct and noncollinear. Also, \(B\notin AP\): otherwise \(B\) would be the second intersection of \(AP\) with both original circles, giving \(E=F\). Finally, \(B\ne T\), since \(MB=r<R=MT\). Thus all circles and angles used below are nondegenerate.

Let \(t_\Omega,t_\Gamma\) be the tangents at \(A\) to \(\Omega,\Gamma\), respectively. Using directed angles modulo \(180^\circ\), the tangent–chord theorem gives \[ \begin{aligned} \angle(BE,BF) &=\angle(BE,BA)+\angle(BA,BF)\\ &=\angle(AP,t_\Omega)+\angle(t_\Gamma,AP)\\ &=\angle(t_\Gamma,t_\Omega)\\ &=\angle(AN,AM)\\ &=\angle(TE,TF). \end{aligned} \] Here the penultimate equality follows from \(t_\Gamma\perp AN\) and \(t_\Omega\perp AM\), and the last follows from (2). Therefore \(B,E,F,T\) are concyclic.

Let \(O\) be their circumcentre. By \(TE=TF\), both \(T\) and \(O\) lie on the perpendicular bisector of \(EF\). Hence \(OT\perp EF\), so the tangent at \(T\) is parallel to \(EF\), and thus to \(AP\). By (1), this tangent is precisely the line through \(H\) parallel to \(AP\), as required. \(\square\)

The solution was checked against the official IMO shortlist solutions.