### 2015 IMO SL #G4

Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.

The only answer is $\boxed{\sqrt{2}}$.

Let $N$ be the midpoint of $PQ$, $BT\cap (BPQ)=X$, $BD\cap (ABC)=Y$. Then, $PNQXM$ and $AMCTY$ are similar, so we must have$$\angle\widehat{NX, MT}=\angle\widehat{NM, MY}\implies \angle MTB=\angle BMN\implies BMN\sim BTM$$Hence, $BM^2=BN\cdot BT=\frac{BT^2}{2}\implies\frac{BT}{BM}=\sqrt{2}$. We are done. $\square$