Belorusija 2012

Since $AB=BC$, $CD=DE$ we have $\sim AB=\sim BC$, $\sim CD=\sim DE$.



Therefore, $$\angle BPA=\angle DPE=\frac{1}{2}(\sim AB+\sim DE)=\frac{1}{2}(\sim AB+\sim CD)=\angle BQA.(1)$$

So, points $A$, $B$, $Q$, $P$ belong to the same circle and we have $$\angle QAB=\angle QPB\text{ (subtended by the same chord BQ)}(2)$$ and $$\angle QAB=\angle CAB=\angle CEB\text{ (subtended by the same chord BC).}(3)$$ Thus from (2) and (3) we have $\angle QPB=\angle CEB$, so $PQ\parallel CE$. Therefore, $$\angle TQP=\angle DTE=\frac{1}{2}(\sim BC+\sim DE)=\frac{1}{2}(\sim AB+\sim CD)=\angle AQB.(4)$$ Similarly, points $P$, $T$, $D$, $E$ belong to the same circle because $$\angle EPD=\frac{1}{2}(\sim AB+\sim DE)=\frac{1}{2}(\sim BC+\sim DE)=\angle ETD.$$ So, $\angle PTE=\angle PDE$ (subtended by the same chord $PE$). Since $\angle ACE=\angle ADE$ (subtended by the same chord $AE$) we have $\angle ACE=\angle PTE$. Therefore, $PT\parallel AC$ and hence $\angle PTQ=\angle AQB$. Now from (4) it follows that $\angle PTQ=\angle TQP$, thus triangle $PTQ$ is isosceles.