BMO 2019 Shortlist

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 $△BKQ$, then $T_{1}B=T_{1}Q$. Due to $\angle DPO=90^{\circ }$, it suffices to show that $\angle OP{T}_1=90^{\circ }$ – indeed, $T\equiv T_1$ and $TB=TQ$ would follow.

Denote by $Y$ the midpoint of $BQ$. Then $\angle OXB=\angle T_{1}YB=90^{\circ }$. The quadrilateral $QKB{T}_1$ is inscribed in a circle, hence $\angle B{T}_{1}Q=180^{\circ }-\angle BKQ=\angle AKB$. Then $\angle XBO=\frac{1}{2}\angle AKB=\frac{1}{2}\angle B{T}_{1}Q=\angle B{T}_{1}Y$ and thus $△OXB\sim △BY{T}_1$. The quadrilaterals $PXBY$

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

Consequently, $$\frac{OX}{XP}=\frac{OX}{BY}=\frac{XB}{T_{1}Y}=\frac{PY}{T_1{Y}^{\prime }}$$ which along with $\angle PXB=\angle PYB$ and $\angle OXB=\angle T_{1}YB$ gives $\angle OXP=\angle PY{T}_1$ and $△OXP\sim △PY{T}_1$. Thus $\angle XPO=\angle Y{T}_{1}P$ and $\angle POX=\angle T_{1}PY$.

In conclusion, $$\angle OP{T}_1=\angle XPY+\angle XPO+\angle YP{T}_1=\angle PXA+\angle XPO+\angle XOP=90^{\circ }$$ $\square$