IMO 2008 · Problem 1

An acute-angled triangle \(ABC\) has orthocentre \(H\). The circle passing through \(H\) with centre the midpoint of \(BC\) intersects the line \(BC\) at \(A_1\) and \(A_2\). Similarly, the circle passing through \(H\) with centre the midpoint of \(CA\) intersects the line \(CA\) at \(B_1\) and \(B_2\), and the circle passing through \(H\) with centre the midpoint of \(AB\) intersects the line \(AB\) at \(C_1\) and \(C_2\). Show that \(A_1\), \(A_2\), \(B_1\), \(B_2\), \(C_1\), \(C_2\) lie on a circle.

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

Solution

Let \(O\) and \(R\) be the circumcentre and circumradius of \(ABC\). We shall show that all six points have distance \[ \sqrt{\frac{OH^2+R^2}{2}} \] from \(O\).

Let \(M\) be the midpoint of \(BC\), and let \(D\) be the point diametrically opposite \(A\) on the circumcircle. Since \(AD\) is a diameter, \[ BD\perp AB,\qquad CD\perp AC. \] The altitude conditions give \(CH\perp AB\) and \(BH\perp AC\). Consequently, \(BHCD\) is a parallelogram. Its diagonals \(BC\) and \(HD\) bisect each other, so \(M\) is also the midpoint of \(HD\).

The midpoint identity therefore gives \[ OM^2+MH^2 =\frac{OH^2+OD^2}{2} =\frac{OH^2+R^2}{2}. \] For completeness, the first equality follows from \[ OH^2+OD^2 = \left\|\overrightarrow{OM}+\overrightarrow{MH}\right\|^2 + \left\|\overrightarrow{OM}-\overrightarrow{MH}\right\|^2 =2OM^2+2MH^2. \] Thus it remains valid when \(O,H,D\) are collinear or \(O=H\).

For either \(X=A_1\) or \(X=A_2\), the defining circle gives \(MX=MH\). Moreover, \(OM\perp BC\), because \(O\) lies on the perpendicular bisector of \(BC\). Hence \[ OX^2=OM^2+MX^2 =OM^2+MH^2 =\frac{OH^2+R^2}{2}. \]

Applying the same argument cyclically to \(CA\) and \(AB\) gives this same squared distance for \(B_1,B_2,C_1,C_2\). Since \(R>0\), the displayed radius is positive. All six points therefore lie on the circle with centre \(O\) and that radius. \(\square\)

The solution was checked against the official IMO shortlist solutions.