G4. Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triangle $A B C$. Determine all possible values of $B T / B M$.
Answer. $\sqrt{2}$.
Solution 1. Let $S$ be the center of the parallelogram $B P T Q$, and let $B^{\prime} \neq B$ be the point on the ray $B M$ such that $B M=M B^{\prime}$ (see Figure 1). It follows that $A B C B^{\prime}$ is a parallelogram. Then, $\angle A B B^{\prime}=\angle P Q M$ and $\angle B B^{\prime} A=\angle B^{\prime} B C=\angle M P Q$, and so the triangles $A B B^{\prime}$ and $M Q P$ are similar. It follows that $A M$ and $M S$ are corresponding medians in these triangles. Hence,
$$
\angle S M P=\angle B^{\prime} A M=\angle B C A=\angle B T A .
$$

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
$$
\angle M T A=\angle T P S=\angle B Q P=\angle B M P \text {. }
$$

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 (1) and (2), 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 (2) that $\angle B C A=\angle M T A=\angle B Q P=\angle B M P$ (see Figure 2). Thus, $P Q \| A C$ and $P M \| A T$. Hence, $B S / B M-B P / B A-B M / B T$, so $B T^{2}=2 B M^{2}$ and $B T=\sqrt{2} B M$.
Comment 1. Here is another way to show that the triangles BSM and $B M T$ are similar. Denote by $\Omega$ the circumcircle of the triangle $A B C$. Let $R$ be the second point of intersection of $\omega$ and $\Omega$, and let $\tau$ be the spiral similarity centered at $R$ mapping $\omega$ to $\Omega$. Then, one may show that $\tau$ maps each point $X$ on $\omega$ to a point $Y$ on $\Omega$ such that $B, X$, and $Y$ are collinear (see Figure 3). If we let $K$ and $L$ be the second points of intersection of $B M$ with $\Omega$ and of $B T$ with $\omega$, respectively, then it follows that the triangle $M K T$ is the image of $S M L$ under $\tau$. We now obtain $\angle B S M=\angle T M B$, which implies the desired result.
Solution 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 $\triangle X P M-\angle B Q M$, 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 $B^{\prime} \neq B$ be the point on the ray $B M$ such that $B M=M B^{\prime}$. Then $\angle M B^{\prime} A=$ $\angle M B C=\angle M X B$ and $\angle C B^{\prime} M-\angle A B M=\angle B Y M$. This means that the triangles $B M X$, $B A B^{\prime}, Y M B$, and $B^{\prime} C B$ are all similar; hence $B A \cdot B X-B M \cdot B B^{\prime}-B C \cdot B Y$. Thus there exists an inversion centered at $B$ which swaps $A$ with $X, M$ with $B^{\prime}$, and $C$ with $Y$. This inversion then swaps $\Omega$ with the line $X Y$, and hence it preserves $T$. Therefore, we have $B T^{2}-B M \cdot B B^{\prime}=2 B M^{2}$, and $B T=\sqrt{2} B M$.

Solution 3. 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
$$
B Q \cdot B C-B J \cdot B T \text {. }
$$

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,
$$
T J \cdot B T=T Q \cdot B A-B P \cdot B A \text {. }
$$

Combining (3) and (4) now yields the desired result.
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
$$
B P \cdot B A+B Q \cdot B C=B T^{2} \text {. }
$$

Since $B A-2 B X$ and $B C=2 B Y$, we have $B T^{2}-2 B M^{2}$, and so $B T=\sqrt{2} B M$.
Comment 2. Here we give another proof of the lemma using Ptolemy's theorem. We readily have
$$
T C \cdot B A+T A \cdot B C=A C \cdot B T \text {. }
$$

The lemma now follows from
$$
\frac{B P}{T C}=\frac{B Q}{T A}=\frac{B T}{A C}=\frac{\sin \angle B C T}{\sin \angle A B C} .
$$