IMO 2010 · Problem 2

Let \(I\) be the incentre of triangle \(ABC\) and let \(\Gamma\) be its circumcircle. Let the line \(AI\) intersect \(\Gamma\) again at \(D\). Let \(E\) be a point on the arc \(\widehat{BDC}\) and \(F\) a point on the side \(BC\) such that \[ \angle BAF=\angle CAE<\frac12\angle BAC. \] Finally, let \(G\) be the midpoint of the segment \(IF\). Prove that the lines \(DG\) and \(EI\) intersect on \(\Gamma\).

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

Solution

Put \[ \alpha=\frac12\angle BAC,\qquad \beta=\frac12\angle ABC,\qquad \gamma=\frac12\angle ACB. \] Thus \(\alpha+\beta+\gamma=90^\circ\).

Since \(AD\) bisects \(\angle BAC\), the point \(D\) is the midpoint of the arc \(BC\) not containing \(A\), and \(DB=DC\). Moreover, \[ \angle DBI=\angle DBC+\angle CBI=\alpha+\beta. \] As \(A,I,D\) occur in that order on their line, the exterior-angle theorem in triangle \(ABI\) gives \[ \angle BID=\alpha+\beta. \] Consequently, \[ DI=DB=DC. \]

Let \(J\) be the reflection of \(I\) in \(D\). The circle with centre \(D\) and radius \(DI\) contains \(B,C,I,J\), with \(IJ\) as a diameter. Hence \[ BI\perp BJ,\qquad CI\perp CJ. \] The point \(J\) lies beyond \(D\) on the ray \(AI\), and therefore on the opposite side of \(BC\) from \(A\). Since \(\angle ACI=\gamma\), it follows that \[ \angle ACJ=90^\circ+\gamma, \qquad \angle AJC=180^\circ-\alpha-(90^\circ+\gamma)=\beta. \] Thus triangles \(ABI\) and \(AJC\) are similar: their angles at \(A\) equal \(\alpha\), and \(\angle ABI=\angle AJC=\beta\). Therefore \[ AI\cdot AJ=AB\cdot AC. \tag{1} \]

Write \(t=\angle BAF=\angle CAE\), so \(0\le t<\alpha\). If \(t>0\), triangles \(ABF\) and \(AEC\) are similar, because \[ \angle BAF=\angle CAE, \qquad \angle ABF=\angle ABC=\angle AEC. \] Their corresponding sides give \(AF\cdot AE=AB\cdot AC\). If \(t=0\), then \(F=B\) and \(E=C\), so the same identity holds immediately. Combining this with (1), we obtain \[ AF\cdot AE=AI\cdot AJ, \qquad\text{hence}\qquad \frac{AI}{AF}=\frac{AE}{AJ}. \tag{2} \]

The angle condition also gives \[ \angle IAE=\angle FAJ=\alpha-t>0. \] Together with (2), this proves \[ \triangle AIE\sim\triangle AFJ. \] These triangles have the same orientation: the rays \(AF\) and \(AE\) lie on opposite sides of \(AI\), while \(AI\) and \(AJ\) are the same ray. Thus their corresponding directed angles are equal.

Let \(X\) be the second intersection of \(EI\) with \(\Gamma\). Since \(I\) is inside \(\Gamma\), this point exists and differs from \(E\). Also, \(E\ne D\) by the strict angle inequality, so \(X\ne A,D\); either equality would force \(EI=AD\).

Using directed angles modulo \(180^\circ\), cyclicity of \(A,D,E,X\), collinearity of \(E,I,X\), and the preceding similarity give \[ \angle(DA,DX) =\angle(EA,EX) =\angle(EA,EI) =\angle(JA,JF). \] Since \(DA\) and \(JA\) lie on the same line, this proves \[ DX\parallel JF. \]

Finally, \(D\) is the midpoint of \(IJ\), and \(G\) is the midpoint of \(IF\). The strict inequality \(t<\alpha\) ensures that \(F\notin IJ\), so the midpoint theorem in triangle \(IJF\) gives \[ DG\parallel JF. \] Therefore \(D,G,X\) are collinear. Since \(X\in EI\cap\Gamma\), the lines \(DG\) and \(EI\) intersect on \(\Gamma\), as required.

The solution was checked against the official IMO shortlist solutions.