IMO 1982 · Problem 5

The diagonals \(AC\) and \(CE\) of the regular hexagon \(ABCDEF\) are divided by the inner points \(M\) and \(N\), respectively, so that \[ \frac{AM}{AC}=\frac{CN}{CE}=r. \] Determine \(r\) if \(B,M,\) and \(N\) are collinear.

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

Answer

\[ \boxed{r=\frac{1}{\sqrt3}}. \]

Solution

A similarity preserves both collinearity and ratios along segments, so we may place the regular hexagon on the unit circle with \[ A=(1,0),\quad B=\left(\frac12,\frac{\sqrt3}{2}\right), \quad C=\left(-\frac12,\frac{\sqrt3}{2}\right), \quad E=\left(-\frac12,-\frac{\sqrt3}{2}\right). \] Write \(h=\sqrt3/2\). Since \(M\) and \(N\) are inner points, \(0<r<1\), and the given ratios imply \[ M=(1-r)A+rC=\left(1-\frac{3r}{2},hr\right), \] \[ N=(1-r)C+rE=\left(-\frac12,h(1-2r)\right). \] Consequently, \[ \overrightarrow{BM}=\left(\frac12-\frac{3r}{2},h(r-1)\right), \qquad \overrightarrow{BN}=(-1,-2hr). \] The points \(B,M,N\) are collinear exactly when the determinant of these vectors vanishes. That determinant equals \[ \left(\frac12-\frac{3r}{2}\right)(-2hr) -h(r-1)(-1) =h(3r^2-1). \] As \(h\ne0\), collinearity is therefore equivalent to \(3r^2=1\). The restriction \(0<r<1\) selects \[ r=\frac1{\sqrt3}. \] Conversely, this value lies in \((0,1)\) and makes the determinant zero, so it indeed gives inner points \(M,N\) collinear with \(B\).

The solution was checked against John Scholes' IMO solution archive (Kalva).