IMO 2007 · Problem 2

Consider five points \(A, B, C, D\) and \(E\) such that \(ABCD\) is a parallelogram and \(BCED\) is a cyclic quadrilateral. Let \(\ell\) be a line passing through \(A\). Suppose that \(\ell\) intersects the interior of the segment \(DC\) at \(F\) and intersects line \(BC\) at \(G\). Suppose also that \(EF = EG = EC\). Prove that \(\ell\) is the bisector of angle \(DAB\).

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

Solution

Take \(A\) as the origin, and write \(B=\mathbf b\), \(D=\mathbf d\), so that \(C=\mathbf b+\mathbf d\). The vectors \(\mathbf b,\mathbf d\) are linearly independent because \(B,C,D\) are distinct points on a circle.

Since \(F\) lies inside \(DC\), we have \[ F=\mathbf d+t\mathbf b,\qquad 0<t<1. \] The point \[ \mathbf b+\frac1t\mathbf d=\frac1tF \] lies on both \(BC\) and \(AF\), so it is \(G\). In particular, \(B,C,G\) occur in this order.

Put \[ x=CF,\qquad y=CG,\qquad k=\frac{t}{1-t}>0. \] The expressions above give \[ DF=kx,\qquad DC=(k+1)x,\qquad BC=ky,\qquad BG=(k+1)y. \]

Let \(r=EC>0\). By hypothesis, the circle \(\Gamma\) with centre \(E\) and radius \(r\) passes through \(C,F,G\). Applying power of a point to \(D\) and \(B\), using the orders \(D,F,C\) and \(B,C,G\), yields \[ ED^2-r^2=DF\cdot DC=k(k+1)x^2, \] \[ EB^2-r^2=BC\cdot BG=k(k+1)y^2. \] Consequently, \[ ED^2-EB^2=k(k+1)(x^2-y^2). \tag{1} \]

Now let \(R\) be the radius of the circle through \(B,C,E,D\), and set \[ \alpha=\angle ECF,\qquad \beta=\angle ECG. \] Neither triangle \(ECF\) nor \(ECG\) is degenerate: otherwise \(E\) would lie on \(CD\) or \(BC\), contradicting the distinctness of the four concyclic vertices.

The cosine rule, together with \(EC=EF=EG=r\), gives \[ \cos\alpha=\frac{x}{2r},\qquad \cos\beta=\frac{y}{2r}. \] Since rays \(CF,CD\) coincide and rays \(CG,CB\) are opposite, the extended sine rule in triangles \(ECD\) and \(ECB\) gives \[ ED=2R\sin\alpha,\qquad EB=2R\sin\beta. \] Therefore \[ \begin{aligned} ED^2-EB^2 &=4R^2(\sin^2\alpha-\sin^2\beta)\\ &=4R^2(\cos^2\beta-\cos^2\alpha)\\ &=-\frac{R^2}{r^2}(x^2-y^2). \end{aligned} \tag{2} \] Comparing (1) and (2), we obtain \[ \left(k(k+1)+\frac{R^2}{r^2}\right)(x^2-y^2)=0. \] The coefficient is positive, so \(x=y\). Hence \[ DF=kx=ky=BC=AD. \]

Thus triangle \(ADF\) is isosceles, and, because \(DF\parallel AB\), \[ \angle DAF=\angle AFD=\angle FAB. \] Finally, \(\overrightarrow{AF}=\mathbf d+t\mathbf b\) with \(t>0\), so ray \(AF\) lies inside \(\angle DAB\). Therefore \(\ell=AF\) is its internal bisector. \(\square\)

The solution was checked against the official IMO shortlist solutions.