IMO 2015 · Problem 3

Let \(ABC\) be an acute triangle with \(AB > AC\). Let \(\Gamma\) be its circumcircle, \(H\) its orthocentre, and \(F\) the foot of the altitude from \(A\). Let \(M\) be the midpoint of \(BC\). Let \(Q\) be the point on \(\Gamma\) such that \(\angle HQA = 90^\circ\), and let \(K\) be the point on \(\Gamma\) such that \(\angle HKQ = 90^\circ\). Assume that the points \(A, B, C, K\) and \(Q\) are all different, and lie on \(\Gamma\) in this order.

Prove that the circumcircles of triangles \(KQH\) and \(FKM\) are tangent to each other.

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

Solution

Let \(D\ne Q\) be the second intersection of \(HQ\) with \(\Gamma\). Since the triangle is acute, \(H\) lies strictly inside \(\Gamma\), so \(H\) lies between \(D\) and \(Q\). The condition \(\angle HQA=90^\circ\) gives \(\angle DQA=90^\circ\), hence \(AD\) is a diameter of \(\Gamma\).

Consequently, \[ BD\perp AB,\qquad CD\perp AC. \] Since \(CH\perp AB\) and \(BH\perp AC\), the quadrilateral \(BHCD\) is a parallelogram. Its diagonals \(BC\) and \(HD\) bisect each other, so \(M\) is the midpoint of \(HD\). In particular, \(D,M,H,Q\) are collinear, in that order.

Let \(E\) be the reflection of \(H\) in \(BC\). Then \(F\) is the midpoint of \(HE\). Moreover, \(E\) and \(A\) lie on opposite sides of \(BC\), and \[ \angle BEC=\angle BHC=180^\circ-\angle BAC, \] so \(E\in\Gamma\).

Notice that \(F\ne M\), since otherwise \(AF\) would be the perpendicular bisector of \(BC\), giving \(AB=AC\). Also \(K\notin BC\), because \(K\in\Gamma\) and \(K\ne B,C\). Thus the circumcircle of \(FKM\) is well-defined.

Now perform an inversion with centre \(H\) and any positive radius. Write \(X'\) for the inverse of a point \(X\ne H\). Since \(H\notin\Gamma\), the image of \(\Gamma\) is a circle \(\Gamma'\).

Because \(M\) and \(F\) are the midpoints of \(HD\) and \(HE\), respectively, inversion gives \[ D'\text{ is the midpoint of }HM', \qquad E'\text{ is the midpoint of }HF'. \tag{1} \]

The circle with diameter \(HM\) contains \(F\), since \(HF\perp FM\). Its inverse is the line through \(M'\) perpendicular to \(HM\). Therefore \[ F'M'\perp HM. \tag{2} \] Likewise, the circumcircle of \(KQH\) has diameter \(HQ\), so its inverse is the line \(Q'K'\), perpendicular to \(HQ\). As \(H,M,Q\) are collinear, \[ Q'K'\parallel F'M'\perp HM. \tag{3} \] These two parallel lines are distinct: they intersect \(HM\) at the distinct points \(Q'\) and \(M'\). Hence \(F',M',K'\) are noncollinear, and the inverse of the circumcircle of \(FKM\) is the circle \[ \sigma'=(F'K'M'). \]

By (1), the midpoint theorem in triangle \(HF'M'\) gives \[ D'E'\parallel F'M'. \] Since \(D',Q'\) lie on \(HM\), equation (2) yields \[ \angle E'D'Q'=90^\circ. \] Thus \(E'Q'\) is a diameter of \(\Gamma'\). The lines \(D'E'\) and \(Q'K'\) are distinct parallel lines, so \(K'\ne E'\). As \(K'\in\Gamma'\), we obtain \[ \angle E'K'Q'=90^\circ, \qquad\text{and therefore}\qquad E'K'\parallel HM. \tag{4} \]

Let \(T\) be the midpoint of \(F'M'\). Because \(E'\) is the midpoint of \(HF'\), the midpoint theorem gives \(E'T\parallel HM'\). Together with (2) and (4), this shows that \(E'K'\) is the perpendicular bisector of \(F'M'\).

The centre of \(\sigma'\) therefore lies on the line \(E'K'\), which also contains \(K'\). Its tangent at \(K'\) is consequently perpendicular to \(E'K'\), and hence is exactly \(Q'K'\).

Thus the inverses of the two circles are tangent at \(K'\). Inversion preserves tangency away from its centre, so the original circumcircles of \(KQH\) and \(FKM\) are tangent at \(K\).

The solution was checked against the official IMO shortlist solutions.