IMO 2018 · Problem 1

Let \(\Gamma\) be the circumcircle of acute-angled triangle \(ABC\). Points \(D\) and \(E\) lie on segments \(AB\) and \(AC\), respectively, such that \(AD = AE\). The perpendicular bisectors of \(BD\) and \(CE\) intersect the minor arcs \(AB\) and \(AC\) of \(\Gamma\) at points \(F\) and \(G\), respectively. Prove that the lines \(DE\) and \(FG\) are parallel (or are the same line).

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

Solution

Let \(O\) and \(R\) be the circumcenter and circumradius, and put \[ t=AD=AE,\qquad \theta=\frac12\angle BAC. \] Here \(t>0\), since otherwise \(D=E=A\) and the line \(DE\) is undefined.

Choose Cartesian coordinates with origin \(O\), whose positive horizontal direction is parallel to the internal angle bisector from \(A\). Orient the vertical axis so that the unit vectors along \(AB\) and \(AC\) are \[ u=(\cos\theta,\sin\theta), \qquad v=(\cos\theta,-\sin\theta). \] Consequently, \[ D-E=t(u-v)=(0,2t\sin\theta), \] so \(DE\) is vertical. We shall prove that \(F\) and \(G\) have equal horizontal coordinates.

Let \(M\) be the midpoint of \(AB\), and \(P\) the midpoint of \(BD\). Since \(OM\perp AB\), we have \(M\cdot u=0\). Moreover, \[ P=M+\frac t2u. \] The perpendicular bisector of \(BD\) passes through \(P\) perpendicular to \(u\). Thus \[ F\cdot u=P\cdot u=\frac t2. \] Applying the same argument to \(AC\) and \(CE\) gives \[ G\cdot v=\frac t2. \]

The outward unit normals to the sides \(AB\) and \(AC\), respectively, are \[ n=(-\sin\theta,\cos\theta), \qquad m=(-\sin\theta,-\cos\theta). \] Because \(ABC\) is acute, \(O\) lies strictly inside the triangle. A minor arc lies on the side of its chord opposite the center; hence the specified arc choices imply \[ F\cdot n>0,\qquad G\cdot m>0. \] This determines which of the two circle intersections is being used.

Since \(|F|=|G|=R\), and each of \((u,n)\) and \((v,m)\) is an orthonormal pair, it follows that \[ F=\frac t2u+hn,\qquad G=\frac t2v+hm, \qquad h=\sqrt{R^2-\frac{t^2}{4}}. \] Here \(h>0\): indeed, \(t\le\min\{AB,AC\}<2R\), since neither side is a diameter in an acute triangle. Expanding, \[ F=\left( \frac t2\cos\theta-h\sin\theta,\, \frac t2\sin\theta+h\cos\theta \right), \] \[ G=\left( \frac t2\cos\theta-h\sin\theta,\, -\frac t2\sin\theta-h\cos\theta \right). \] Their horizontal coordinates are equal, and their vertical coordinates differ by \[ t\sin\theta+2h\cos\theta>0. \] Therefore \(FG\) is a well-defined vertical line. Since \(DE\) is also vertical, the two lines are parallel or coincide, as required.

For completeness, if \(D=B\) or \(E=C\), the corresponding zero-length segment has no ordinary perpendicular bisector. If such endpoints are admitted using the natural limiting construction—the perpendicular to the containing side through the endpoint—the same proof applies. Its formulas give \(F=D=B\) or \(G=E=C\), respectively, and the two lines then coincide.

The solution was checked against the official IMO shortlist solutions.