IMO 2005 · Problem 1

Six points are chosen on the sides of an equilateral triangle \(ABC\): \(A_1, A_2\) on \(BC\), \(B_1, B_2\) on \(CA\) and \(C_1, C_2\) on \(AB\), such that they are the vertices of a convex hexagon \(A_1A_2B_1B_2C_1C_2\) with equal side lengths. Prove that the lines \(A_1B_2\), \(B_1C_2\) and \(C_1A_2\) are concurrent.

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

Solution

We first prove that \(A_1B_1C_1\) is equilateral.

Let \(s>0\) be the common side length of the hexagon. By convexity, its stated cyclic order agrees with the order of its vertices around the boundary of \(ABC\). Choose the counterclockwise orientation for this order, and let \(R\) denote rotation of vectors counterclockwise through \(120^\circ\).

Write \[ \begin{aligned} u_1&=\overrightarrow{A_1A_2},& u_2&=\overrightarrow{B_1B_2},& u_3&=\overrightarrow{C_1C_2},\\ v_1&=\overrightarrow{A_2B_1},& v_2&=\overrightarrow{B_2C_1},& v_3&=\overrightarrow{C_2A_1}. \end{aligned} \] The vectors \(u_1,u_2,u_3\) have equal lengths and follow the three sides of the equilateral triangle in order. Hence \[ u_2=Ru_1,\qquad u_3=R^2u_1,\qquad u_1+u_2+u_3=0. \] The sum of all six edge vectors is zero, so \[ v_1+v_2+v_3=0. \] Moreover, \(|v_1|=|v_2|=|v_3|=s\). Therefore \[ s^2=|v_1+v_2|^2 =2s^2+2v_1\cdot v_2, \] giving \(v_1\cdot v_2=-s^2/2\). The same calculation applies to the other two pairs. Thus the directions of \(v_1,v_2,v_3\) are spaced by \(120^\circ\).

As a convex polygon is traversed counterclockwise, its edge directions occur in counterclockwise circular order: its exterior turns are nonnegative and sum to \(360^\circ\). Consequently \(v_1,v_2,v_3\) occur in that order, and \[ v_2=Rv_1,\qquad v_3=R^2v_1. \] This reasoning permits zero exterior turns, so it also covers any collinear consecutive edges.

Now \[ \begin{aligned} \overrightarrow{A_1B_1}&=u_1+v_1,\\ \overrightarrow{B_1C_1}&=u_2+v_2=R(u_1+v_1),\\ \overrightarrow{C_1A_1}&=u_3+v_3=R^2(u_1+v_1). \end{aligned} \] These three vectors have equal lengths. Since the vertices are distinct, \(A_1B_1C_1\) is a nondegenerate equilateral triangle.

Let \(O\) be its circumcenter. Because \[ B_1B_2=B_2C_1, \] the point \(B_2\) lies on the perpendicular bisector of \(B_1C_1\). Since \(A_1B_1=A_1C_1\), this perpendicular bisector also contains \(A_1\), and hence it is precisely the line \(A_1B_2\). It contains \(O\).

Similarly, \(B_1C_2\) is the perpendicular bisector of \(C_1A_1\), and \(C_1A_2\) is the perpendicular bisector of \(A_1B_1\). Both contain \(O\). Thus the three required lines are concurrent at \(O\).

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