An equilateral triangle $ABC$ with side $a$ and two rhombuses $ACMN$ and $ABFE$ are arranged such that points $M$ and $B$ lie on opposite sides of line $AC$, and points $F$ and $C$ lie on opposite sides of line $AB$. Find the distance between the centers of the rhombuses, if $\angle EAB = \angle ACM = \alpha \left(\alpha < 90^{\circ}\right)$.
Let $P$ and $Q$ be the centers of the rhombi. Using the Law of Cosines, find $P Q$ in triangle $P A Q$.

## Solution

Let $P$ and $Q$ be the centers of the rhombi $A C M N$ and $A B F E$. Then

$$
\begin{gathered}
\angle P A Q=\angle P A C+\angle C A B+\angle B A Q=\left(90^{\circ}-\frac{\alpha}{2}\right)+60^{\circ}+\frac{\alpha}{2}=150^{\circ} \\
\qquad A Q=A B \cos \angle Q A B=a \cos \frac{\alpha}{2}, A P=A C \sin \angle P C A=a \sin \frac{\alpha}{2}
\end{gathered}
$$

By the Law of Cosines in triangle $P A Q$, we find that

$$
\begin{gathered}
P Q^{2}=A Q^{2}+A P^{2}-2 A Q \cdot A P \cos 150^{\circ}= \\
\left(a \cos \frac{\alpha}{2}\right)^{2}+\left(a \sin \frac{\alpha}{2}\right)^{2}+2 a^{2} \cos \frac{\alpha}{2} \sin \frac{\alpha}{2} \cdot \frac{\sqrt{3}}{2}= \\
=a^{2}+a^{2} \cdot \frac{\sqrt{3}}{2} \cdot \sin \alpha=a^{2}\left(1+\frac{\sqrt{3}}{2} \sin \alpha\right)
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-42.jpg?height=557&width=598&top_left_y=1271&top_left_x=730)

## Answer

$a \sqrt{\left(1+\frac{\sqrt{3}}{2}\right) \sin \alpha}$