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 $B T / B M$. (Russia) Answer. $\sqrt{2}$.
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 
$$ \angle M T A=\angle T P S=\angle B Q P=\angle B M P. $$
Now we deal separately with two cases. Case 1. $\quad S$ 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. $\quad S$ 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 $B T=\sqrt{2} B M$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-49.jpg?height=960&width=730&top_left_y=1713&top_left_x=246) 
Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-49.jpg?height=749&width=729&top_left_y=1938&top_left_x=1092) 
Figure 2 
Comment 1. Here is another way to show that the triangles $B S M$ 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. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-50.jpg?height=758&width=757&top_left_y=752&top_left_x=204) 
Figure 3 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-50.jpg?height=954&width=846&top_left_y=551&top_left_x=1002) 
Figure 4