BMO Shortlist, 2019

As $DC$ is tangent to $k$ at $C$ then $\angle OCD = 90^\circ$. Denote by $X$ the midpoint of $AB$. Then $\angle OXA = 90^\circ$ because $OX$ is the perpendicular bisector of the side $AB$. The pentagon $PXOCD$ is inscribed in the circle with diameter $OD$, hence $\angle PXA = \angle PXD = \angle PCD = \angle QCD = \angle QBA$ (the latter is due to $QBCD$ being cyclic). We deduce that $PX \parallel QB$ and that $P$ is the midpoint of $AQ$, so $AP = PQ$.

Figure 5: G5

Now let $T_1$ be the midpoint of the arc $BQ$, not containing $K$, from the circumcircle of $\triangle BKQ$, then $T_1B = T_1Q$. Due to $\angle DPO = 90^\circ$, it suffices to show that $\angle OPT_1 = 90^\circ$ – indeed, $T \equiv T_1$ and $TB = TQ$ would follow.

Denote by $Y$ the midpoint of $BQ$. Then $\angle OXB = \angle T_1YB = 90^\circ$. The quadrilateral $QKBT_1$ is inscribed in a circle, hence $\angle BT_1Q = 180^\circ - \angle BKQ = \angle AKB$. Then $\angle XBO = \frac{1}{2}\angle AKB = \frac{1}{2}\angle BT_1Q = \angle BT_1Y$ and thus $\triangle OXB \sim \triangle BYT_1$. The quadrilaterals $PXBY$

and $AXYP$ are parallelograms, since $XY$ and $PY$ are middle lines of the triangle $AQB$.

Consequently,
$$\frac{OX}{XP} = \frac{OX}{BY} = \frac{XB}{T_1Y} = \frac{PY}{T_1Y'}$$
which along with $\angle PXB = \angle PYB$ and $\angle OXB = \angle T_1YB$ gives $\angle OXP = \angle PYT_1$ and
$\triangle OXP \sim \triangle PYT_1$. Thus $\angle XPO = \angle YT_1P$ and $\angle POX = \angle T_1PY$.

In conclusion,
$$\angle OPT_1 = \angle XPY + \angle XPO + \angle YPT_1 = \angle PXA + \angle XPO + \angle XOP = 90^\circ$$
$\Box$