The required ratio equals $\sqrt{2}$. To prove this, let $S$ be the midpoint of the segment $PQ$, and let $B'$ be the reflection of $B$ across $M$. Clearly, $ABCB'$ is a parallelogram, $\angle ABB' = \angle PQM$, and $\angle BB'A = \angle B'BC = \angle MPQ$, so the triangles $ABB'$ and $MQP$ are similar. Since $AM$ and $MS$ are corresponding medians in these triangles,
$$\angle SMP = \angle B'AM = \angle BCA = \angle BTA. \qquad (1)$$
Next, $\angle ACT = \angle PBT$ and $\angle TAC = \angle TBC = \angle BTP$, so the triangles $TCA$ and $PBT$ are similar. Since $TM$ and $PS$ are corresponding medians in these triangles,
$$\angle MTA = \angle TPS = \angle BQP = \angle BMP. \qquad (2)$$
If $S$ does not lie on the segment $BM$, we may and will assume that $S$ and $A$ both lie on the same side of the line $BM$, since the configuration is symmetric in $A$ and $C$. By (1) and (2), $\angle BMS = \angle BMP - \angle SMP = \angle MTA - \angle BTA = \angle MTB$, so the triangles $BSM$ and $BMT$ are similar, whence $BM^2 = BS \cdot BT = BT^2/2$; that is, $BT/BM = \sqrt{2}$.

If $S$ lies on the segment $BM$, then (2) shows that $\angle BCA = \angle MTA = \angle BMP = \angle BQP$, so $(PQ, AC)$ and $(PM, AT)$ are pairs of parallel lines. Consequently, $BS/BM = BP/BA = BM/BT$, so $BT^2 = 2BM^2$, and again $BT/BM = \sqrt{2}$.