67th NMO Selection Tests for BMO and IMO

The required ratio equals $\sqrt{2}$. To prove this, let $S$ be the midpoint of the segment $PQ$, and let $B^{\prime }$ be the reflection of $B$ across $M$. Clearly, $ABC{B}^{\prime }$ is a parallelogram, $\angle AB{B}^{\prime }=\angle PQM$, and $\angle B{B}^{\prime }A=\angle B^{\prime }BC=\angle MPQ$, so the triangles $AB{B}^{\prime }$ and $MQP$ are similar. Since $AM$ and $MS$ are corresponding medians in these triangles, $$\angle SMP=\angle B^{\prime }AM=\angle BCA=\angle BTA.(1)$$ 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, $$\angle MTA=\angle TPS=\angle BQP=\angle BMP.(2)$$ 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 $B{M}^2=BS\cdot BT=B{T}^2/2$; 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 $B{T}^2=2B{M}^2$, and again $BT/BM=\sqrt{2}$.