A geometry problem by Rahil Sehgal

In the triangle $ABC$ the sum of the angles $\angle ABC+\angle BCA=180^\circ-40^\circ=140^\circ$

In the red quadrilateral $ABOC$ angle $BAC=40^\circ$ and $\angle ABO+\angle OCA=\frac12(\angle ABC+\angle BCA)=70^\circ$ .

That leaves $\angle BOC=360^\circ-40^\circ-70^\circ=250^\circ$

$\angle QOR=$ external angle $BOC=360^\circ-250^\circ=110^\circ$

$\angle QPR=\frac12\angle QOR=\dfrac{110}{2}=\boxed{55^\circ}$