IMO 2019 · Problem 2

In triangle \(ABC\), point \(A_1\) lies on side \(BC\) and point \(B_1\) lies on side \(AC\). Let \(P\) and \(Q\) be points on segments \(AA_1\) and \(BB_1\), respectively, such that \(PQ\) is parallel to \(AB\). Let \(P_1\) be a point on line \(PB_1\), such that \(B_1\) lies strictly between \(P\) and \(P_1\), and \(\angle PP_1C = \angle BAC\). Similarly, let \(Q_1\) be a point on line \(QA_1\), such that \(A_1\) lies strictly between \(Q\) and \(Q_1\), and \(\angle CQ_1Q = \angle CBA\).

Prove that points \(P\), \(Q\), \(P_1\), and \(Q_1\) are concyclic.

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

Solution

Write \(\alpha=\angle BAC\) and \(\beta=\angle CBA\).

First, we address possible endpoint configurations. The angle conditions exclude \(P=A\), \(Q=B\), \(B_1=C\), and \(A_1=C\): each would make one of the prescribed angles undefined, zero, or straight. If \(B_1=A\), then \(A\) lies strictly between \(P\) and \(P_1\). The collinear case is again impossible; otherwise, the exterior-angle theorem in triangle \(AP_1C\) gives \[ \angle PP_1C<\angle PAC\leq\alpha, \] contrary to the hypothesis. Symmetrically, \(A_1\ne B\). Thus \(A_1,B_1\) lie in the interiors of their respective sides. Also, \(P\ne Q\), since the line \(PQ\) is specified.

Let \(\omega\) be the circumcircle of \(ABC\). Let \(D\) be the second intersection of \(BB_1\) with \(\omega\), and let \(E\) be the second intersection of \(AA_1\) with \(\omega\). Since \(A_1,B_1\) lie strictly inside \(\omega\), the orders on these lines are \[ B,\ B_1,\ D \qquad\text{and}\qquad A,\ A_1,\ E. \] In particular, \(D\) lies on the open arc \(AC\) not containing \(B\), while \(E\) lies on the open arc \(BC\) not containing \(A\). Consequently, \(D\ne E\), and the inscribed-angle theorem gives \[ \angle B_1DC=\angle BDC=\alpha, \qquad \angle CEA_1=\angle CEA=\beta. \]

The points \(D\) and \(P_1\) are on the same side of \(AC\), opposite \(B\): this follows for \(P_1\) because \(P\) is on the same side as \(B\), and \(B_1\) lies strictly between \(P\) and \(P_1\). Therefore \[ \angle B_1P_1C=\angle PP_1C=\alpha=\angle B_1DC \] shows that \(C,B_1,D,P_1\) are concyclic. Symmetrically, \(E,Q_1\) are on the same side of \(BC\), and \(C,A_1,E,Q_1\) are concyclic. Denote these conclusions by \[ C,B_1,D,P_1\text{ are concyclic}, \qquad C,A_1,E,Q_1\text{ are concyclic}. \tag{1} \]

For the remainder of the proof, all angles are directed modulo \(180^\circ\).

Let \(\Sigma\) be the circle through \(P,Q,D\). These points are noncollinear: the line \(QD\) is \(BB_1\), which is not parallel to \(PQ\). Using \(A,P,E\) collinear, \(B,Q,D\) collinear, and \(PQ\parallel AB\), we obtain \[ \angle PED =\angle AED =\angle ABD =\angle PQD. \] Hence \(E\in\Sigma\).

If \(P_1=D\), then \(P_1\in\Sigma\) already. Otherwise, the first cyclicity in (1) yields \[ \begin{aligned} \angle PP_1D &=\angle B_1P_1D\\ &=\angle B_1CD\\ &=\angle ACD\\ &=\angle ABD\\ &=\angle PQD. \end{aligned} \] Thus \(P_1\in\Sigma\).

Similarly, if \(Q_1=E\), its membership is immediate. Otherwise, the second cyclicity in (1) yields \[ \begin{aligned} \angle QQ_1E &=\angle A_1Q_1E\\ &=\angle A_1CE\\ &=\angle BCE\\ &=\angle BAE\\ &=\angle QPE. \end{aligned} \] Therefore \(Q_1\) lies on the circle through \(Q,P,E\), which is \(\Sigma\).

Consequently, \(P,Q,P_1,Q_1\) are concyclic. \(\square\)

The solution was checked against the official IMO shortlist solutions.