Romanian Mathematical Olympiad

Triangle $ABC$ is isosceles, so $\angle ABC=\angle ACB=70^{\circ }$.

Notice that $\angle TAC=40^{\circ }-10^{\circ }=30^{\circ }$ and $\angle ACS=70^{\circ }-10^{\circ }=60^{\circ }$, hence $\angle APC=90^{\circ }$.

Triangles $ABT$ and $BSC$ are similar, whence $\frac{BS}{BC}=\frac{BT}{AB}$.

Also, triangles $BST$ and $BCA$ are similar, therefore $TB=TS$ and $\angle TSB=70^{\circ }$.

Since $\angle CSA=\angle SBC+\angle SCB=70^{\circ }+10^{\circ }=80^{\circ }$, it follows that $\angle PST=180^{\circ }-80^{\circ }-70^{\circ }=30^{\circ }$.

Now, triangle $STP$ has a right angle in $P$ and $\angle PST=30^{\circ }$, so $BT=2PT$.