IMO Shortlist

Let $S$ be the center of the parallelogram $B P T Q$, and let $B' \neq B$ be the point on the ray $B M$ such that $B M = M B'$ (see Figure 1).
It follows that $A B C B'$ is a parallelogram. Then, $\angle A B B' = \angle P Q M$ and $\angle B B' A = \angle B' B C = \angle M P Q$, and so the triangles $A B B'$ and $M Q P$ are similar. It follows that $A M$ and $M S$ are corresponding medians in these triangles. Hence,
$$\angle S M P = \angle B' A M = \angle B C A = \angle B T A .$$
Since $\angle A C T = \angle P B T$ and $\angle T A C = \angle T B C = \angle B T P$, the triangles $T C A$ and $P B T$ are similar. Again, as $T M$ and $P S$ are corresponding medians in these triangles, we have
$$\angle M T A = \angle T P S = \angle B Q P = \angle B M P .$$
Now we deal separately with two cases.

Case 1. $S$ does not lie on $B M$. Since the configuration is symmetric between $A$ and $C$, we may assume that $S$ and $A$ lie on the same side with respect to the line $B M$.
Applying the previous results, we get
$$\angle B M S = \angle B M P - \angle S M P = \angle M T A - \angle B T A = \angle M T B,$$
and so the triangles $B S M$ and $B M T$ are similar. We now have $B M^2 = B S \cdot B T = B T^2 / 2$, so $B T = \sqrt{2} B M$.

Case 2. $S$ lies on $B M$. It follows from the previous results that $\angle B C A = \angle M T A = \angle B Q P = \angle B M P$ (see Figure 2). Thus, $P Q \parallel A C$ and $P M \parallel A T$. Hence, $B S / B M = B P / B A = B M / B T$, so $B T^2 = 2 B M^2$ and $B T = \sqrt{2} B M$.

Figure 1

Figure 2

Again, we denote by $\Omega$ the circumcircle of the triangle $A B C$.
Choose the points $X$ and $Y$ on the rays $B A$ and $B C$ respectively, so that $\angle M X B = \angle M B C$ and $\angle B Y M = \angle A B M$ (see Figure 4). Then the triangles $B M X$ and $Y M B$ are similar. Since $\angle X P M = \angle B Q M$, the points $P$ and $Q$ correspond to each other in these triangles. So, if $\overrightarrow{B P} = \mu \cdot \overrightarrow{B X}$, then $\overrightarrow{B Q} = (1-\mu) \cdot \overrightarrow{B Y}$. Thus
$$\overrightarrow{B T} = \overrightarrow{B P} + \overrightarrow{B Q} = \overrightarrow{B Y} + \mu \cdot (\overrightarrow{B X} - \overrightarrow{B Y}) = \overrightarrow{B Y} + \mu \cdot \overrightarrow{Y X},$$
which means that $T$ lies on the line $X Y$.
Let $B' \neq B$ be the point on the ray $B M$ such that $B M = M B'$. Then $\angle M B' A = \angle M B C = \angle M X B$ and $\angle C B' M = \angle A B M = \angle B Y M$. This means that the triangles $B M X$, $B A B'$, $Y M B$, and $B' C B$ are all similar; hence $B A \cdot B X = B M \cdot B B' = B C \cdot B Y$. Thus there exists an inversion centered at $B$ which swaps $A$ with $X$, $M$ with $B'$, and $C$ with $Y$. This inversion then swaps $\Omega$ with the line $X Y$, and hence it preserves $T$. Therefore, we have $B T^2 = B M \cdot B B' = 2 B M^2$, and $B T = \sqrt{2} B M$.

Figure 4

We begin with the following lemma.

Lemma. Let $A B C T$ be a cyclic quadrilateral. Let $P$ and $Q$ be points on the sides $B A$ and $B C$ respectively, such that $B P T Q$ is a parallelogram. Then $B P \cdot B A + B Q \cdot B C = B T^2$.

Proof. Let the circumcircle of the triangle $Q T C$ meet the line $B T$ again at $J$ (see Figure 5). The power of $B$ with respect to this circle yields
$$B Q \cdot B C = B J \cdot B T.$$
We also have $\angle T J Q = 180^{\circ} - \angle Q C T = \angle T A B$ and $\angle Q T J = \angle A B T$, and so the triangles $T J Q$ and $B A T$ are similar. We now have $T J / T Q = B A / B T$. Therefore,
$$T J \cdot B T = T Q \cdot B A = B P \cdot B A.$$
Combining the above, we get the desired result. $\square$

Let $X$ and $Y$ be the midpoints of $B A$ and $B C$ respectively (see Figure 6). Applying the lemma to the cyclic quadrilaterals $P B Q M$ and $A B C T$, we obtain
$$B X \cdot B P + B Y \cdot B Q = B M^2$$
and
$$B P \cdot B A + B Q \cdot B C = B T^2.$$
Since $B A = 2 B X$ and $B C = 2 B Y$, we have $B T^2 = 2 B M^2$, and so $B T = \sqrt{2} B M$.

Figure 5

Figure 6