IMO 2005 · Problem 5

Let \(ABCD\) be a fixed convex quadrilateral with \(BC = DA\) and \(BC\) not parallel with \(DA\). Let two variable points \(E\) and \(F\) lie of the sides \(BC\) and \(DA\), respectively and satisfy \(BE = DF\). The lines \(AC\) and \(BD\) meet at \(P\), the lines \(BD\) and \(EF\) meet at \(Q\), the lines \(EF\) and \(AC\) meet at \(R\).

Prove that the circumcircles of the triangles \(PQR\), as \(E\) and \(F\) vary, have a common point other than \(P\).

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

Solution

We use vectors in the Euclidean plane, with \(P\) as the origin. Since the quadrilateral is convex, there are positive numbers \(a,b,c,d\) and linearly independent unit vectors \(\mathbf u,\mathbf v\) such that \[ A=a\mathbf u,\qquad C=-c\mathbf u,\qquad B=b\mathbf v,\qquad D=-d\mathbf v. \] Put \[ S=a+c,\qquad T=b+d,\qquad \Delta=ab-cd. \] We have \(\Delta\ne0\). Indeed, if \(ab=cd\), then \[ \overrightarrow{BC} =-c\mathbf u-b\mathbf v =-\frac ca(a\mathbf u+d\mathbf v) =-\frac ca\,\overrightarrow{DA}, \] contrary to the hypothesis that \(BC\) and \(DA\) are not parallel.

Because \(BC=DA\) and \(BE=DF\), we may write \[ t=\frac{BE}{BC}=\frac{DF}{DA},\qquad 0\le t\le1. \] First suppose \(0<t<1\). Then \[ E=-tc\mathbf u+(1-t)b\mathbf v,\qquad F=ta\mathbf u-(1-t)d\mathbf v. \] A general point \(E+\lambda(F-E)\) of their line has vector \[ t(-c+\lambda S)\mathbf u +(1-t)(b-\lambda T)\mathbf v. \] Taking \(\lambda=c/S\) gives its intersection with \(BD\), while taking \(\lambda=b/T\) gives its intersection with \(AC\). Consequently, \[ Q=\frac{(1-t)\Delta}{S}\mathbf v, \qquad R=\frac{t\Delta}{T}\mathbf u. \tag{1} \] In particular, both intersections exist, neither equals \(P\), and \(P,Q,R\) are noncollinear.

Define the fixed vector \[ \mathbf w=T\mathbf u+S\mathbf v \] and the fixed point \(X\) by \[ \overrightarrow{PX}=\mathbf x =\frac{\Delta}{\lVert\mathbf w\rVert^2}\mathbf w. \tag{2} \] Linear independence of \(\mathbf u,\mathbf v\) gives \(\mathbf w\ne0\), and \(\Delta\ne0\) therefore gives \(X\ne P\). We will prove that every required circle contains \(X\).

Let \(\mathbf z_t\) be the vector of the centre of the circumcircle of \(PQR\). Its radius is \(\lVert\mathbf z_t\rVert\), since it passes through the origin. Using its passage through \(R\) and \(Q\), respectively, and (1), we obtain \[ 2\mathbf z_t\cdot\mathbf u=\frac{t\Delta}{T}, \qquad 2\mathbf z_t\cdot\mathbf v=\frac{(1-t)\Delta}{S}. \] For example, the first identity follows by expanding \[ \left\lVert \frac{t\Delta}{T}\mathbf u-\mathbf z_t\right\rVert^2 =\lVert\mathbf z_t\rVert^2 \] and dividing by the nonzero number \(t\Delta/T\).

It follows that \[ 2\mathbf z_t\cdot\mathbf w =T\frac{t\Delta}{T} +S\frac{(1-t)\Delta}{S} =\Delta. \] Thus, by (2), \[ 2\mathbf z_t\cdot\mathbf x =\frac{\Delta^2}{\lVert\mathbf w\rVert^2} =\lVert\mathbf x\rVert^2. \] Hence \[ \lVert\mathbf x-\mathbf z_t\rVert^2 =\lVert\mathbf z_t\rVert^2, \] which says precisely that \(X\) lies on the circumcircle of \(PQR\).

Finally, at \(t=0\), the line \(EF\) equals \(BD\), so \(Q\) is not uniquely defined; at \(t=1\), it equals \(AC\), so \(R\) is not uniquely defined. These endpoint choices do not define triangles or circumcircles. Every other choice has been covered above. If endpoint circles are understood as limits, the displayed equations for \(\mathbf z_t\) give finite, nonzero limiting centres, and the same calculation shows that those circles also contain \(P\) and \(X\).

Therefore all the stated circumcircles have the fixed common point \(X\ne P\). \(\square\)

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