IMO 2002 · Problem 2

\(BC\) is a diameter of a circle center \(O\). \(A\) is any point on the circle with \(\angle AOC > 60^\circ\). \(EF\) is the chord which is the perpendicular bisector of \(AO\). \(D\) is the midpoint of the minor arc \(AB\). The line through \(O\) parallel to \(AD\) meets \(AC\) at \(J\). Show that \(J\) is the incenter of triangle \(CEF\).

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

Solution

First suppose \(A\ne B\), so that the stated construction is nondegenerate. Let the circle have radius \(R\), and let \(I\) be the incenter of triangle \(CEF\). We will prove that \(OI\parallel AD\).

Since \(E\) and \(F\) lie on the perpendicular bisector of \(AO\), \[ AE=OE=R,\qquad AF=OF=R. \] Thus \(AOE\) and \(AOF\) are equilateral triangles on opposite sides of \(AO\). Consequently, \(A\) is the midpoint of the minor arc \(EF\), and the minor arcs \(EA\) and \(AF\) each measure \(60^\circ\).

The hypothesis \(\angle AOC>60^\circ\) places \(C\) on the open major arc \(EF\). Hence \[ \angle ECA=\angle ACF=\angle AEF=30^\circ. \] In particular, \(CA\) is the internal angle bisector at \(C\). Since \(A\) and \(C\) lie on opposite sides of \(EF\), the points \(C,I,A\) are collinear in that order.

Write \(\beta=\angle CEF\). Because \(EI\) bisects this angle and \(A\) lies on the opposite side of \(EF\) from \(I\), \[ \angle AEI=\angle AEF+\angle FEI =30^\circ+\frac{\beta}{2}. \] On the other hand, \(IA\) is the extension of \(CI\), so the exterior-angle theorem in triangle \(CEI\) gives \[ \angle EIA=\angle ECI+\angle CEI =30^\circ+\frac{\beta}{2}. \] Therefore triangle \(AEI\) is isosceles, and \[ AI=AE=R. \tag{1} \]

Now let \(M\) be the midpoint of \(AB\). Since \(D\) is the midpoint of the minor arc \(AB\), \(O,M,D\) are collinear, with \(M\) between \(O\) and \(D\). Also, because \(O\) and \(M\) are the midpoints of \(BC\) and \(BA\), \[ \overrightarrow{OM}=\frac12\overrightarrow{CA}. \] It follows that \(\overrightarrow{DO}\) has the same direction as \(\overrightarrow{AC}\). Its length is \(R\); by (1), and because \(I\) lies on segment \(AC\), we obtain \[ \overrightarrow{AI}=\overrightarrow{DO}. \] Thus \(ADOI\) is a parallelogram, so \(OI\parallel AD\). Since \(I\in AC\), the defining intersection \(J\) is precisely \(I\). This proves the result.

For completeness, \(A=C\) and \(C\in\{E,F\}\) are excluded by the strict angle hypothesis. If \(A=B\), then \(D=A\), so the line \(AD\) is undefined; this coincident case is implicitly excluded by the construction. Under its natural limiting interpretation, \(AD\) becomes the tangent at \(B\), giving \(J=O\). In that case all three sides of \(CEF\) have length \(\sqrt3R\), so \(CEF\) is equilateral with incenter \(O\), consistent with the conclusion.

The solution was checked against Evan Chen's published IMO solution notes.