, 2006

The problem may be rephrased as follows: line segments $PA$, $PB$, $PC$ are given. Let $l_1$, $l_2$, $l_3$ denote the perpendicular bisectors of $PA$, $PB$, $PC$ respectively. Given that $l_1$, $l_3$ meet on $PB$ and $l_2$, $l_3$ meet on $PA$, we have to show that $l_1$, $l_2$ intersect on $PC$.

Let $l_1$, $l_3$ intersect in $R$ on $PB$; and let $l_2$, $l_3$ intersect in $H$ on $PA$. If we consider $l_1$ and $l_2$, we observe that $l_1$ is perpendicular to $PH$, $l_2$ is perpendicular to $PR$, and $PC$ is perpendicular to $l_3$, a part of which is $HR$. It follows that $l_1$, $l_2$, $PC$ are concurrent. The other propositions may be proved similarly.

Let $A'$, $B'$, $C'$ denote the mid-points of the line segments $PA$, $PB$, $PC$ respectively. We have to show that the circumcentre of $ABC$ passes through the circumcentres of $PAB$, $PBC$ and $PCA$. Note that $PBC$ is a homothetic expansion of $PB'C'$. If $O$, $O'$ denote the circumcentres of $PBC$ and $PB'C'$, then we get $PO' = PO/2$. However, $O'$ is the mid-point of $HP$. (This can be seen easily from right-triangle $PHB'$.) It follows that $O = H$. Thus $H$ is the circumcentre of $PBC$.

We also observe that $PRA'C'$ is a cyclic quadrilateral. Hence the circumcentre of $PC'A'$ is the same as that of $PC'R$ and this is the mid-point of $PR$ since $PC'R$ is a right-triangle. But $PCA$ is the homothetic expansion of $PC'A'$ with centre $P$ and ratio $2$. Hence the circumcentre of $PCA$ is along $PR$ and moved by a factor of $2$ from the circumcentre of $PC'A'$; we get the point $R$. Hence $R$ is the circumcentre of $PCA$. Similarly, we may prove that the circumcentre of $PBA$ is $Q$.

Now $\angle CHB = 2\angle P$ since $H$ is the circumcentre of $PBC$ and
$$\angle BAC = \angle BAP + \angle PAC = \angle B'A'P + \angle PA'C.$$
But the cyclic quadrilateral $B'HA'R$ shows that $\angle B'A'P = \angle PRH = 90^\circ - \angle P$. Similarly, $\angle PA'C' = 90^\circ - \angle P$. Thus $\angle BAC = 180^\circ - 2\angle P$. Thus $A$, $B$, $H$, $C$ are concyclic. It follows that $H$ lies on the circumcircle of $ABC$.

We also observe that $\angle BQC = 2\angle B'QC = 2(90^\circ - \angle P)$ and similarly $\angle BRC = 2(90^\circ - \angle P)$. Thus $BC$ subtends the same angles at $A$, $Q$, $R$.

It follows that $A$, $B$, $C$, $Q$, $R$ all lie on the same circle. Since $H$ is already on the circumcircle of $ABC$, we conclude that $Q$, $R$, $H$ lie on the circumcircle of $ABC$.