Argentina_2018

Let the parallel to $AD$ through $P$ intersect $BD$ at $O$. We show that $O$ is the circumcenter of $DPQ$.

One has $\overrightarrow{ADP}=\overrightarrow{BDP}$ ($DP$ bisects $\overrightarrow{ADB}$) and $\overrightarrow{ADP}=\overrightarrow{OPD}$ (as $PO\parallel AD$). Hence $\overrightarrow{ODP}=\overrightarrow{OPD}$ and so $OP=OD$. Also $DO=AP$, $BO=BP$ as triangle $ADB$ is isosceles, then $\frac{DO}{BO}=\frac{AP}{BP}=\frac{DA}{DB}=\frac{b}{a}$ by the bisector theorem in triangle $ABD$.

On the other hand the same theorem in triangle $BCD$ gives $\frac{CQ}{BO}=\frac{DC}{DB}=\frac{b}{a}$. In summary,

$$\frac{DO}{BO}=\frac{b}{a}=\frac{CQ}{BO}.$$

Now the converse of Thales' theorem implies $QQ\parallel CD$. Then



$B\widehat{DQ}=C\widehat{DQ}=O\widehat{QD}$, hence $OQ=OD$. We obtained $OP=OQ=OD$, meaning that $O$ is the circumcenter of $DPQ$.

In addition $OD=AP$ from the isosceles triangle $ADB$. Since $AP+BP=a$ and $\frac{AP}{BP}=\frac{b}{a}$, standard calculus lead to $AP=\frac{ab}{a+b}$, so triangle $DPQ$ has circumradius $\frac{ab}{a+b}$.