12. Given a regular hexagon with side length 1. Consider an equilateral triangle with side length 1 inside the hexagon, with two vertices constrained to the perimeter of the hexagon. Move the equilateral triangle so that one of the two constrained vertices completes exactly one full turn along the perimeter of the hexagon. What is the length of the path traveled by the unconstrained vertex?
(A) $\pi / 2$
(B) $\frac{\pi \sqrt{3}}{3}$
(C) $8 \sqrt{3}-12$
(D) $(4 \sqrt{3}-6) \pi$
(E) None of the above.

## Numerical answer problems - 5 points
12. The answer is (C). We show that, in moving the first constrained vertex of the triangle from one end to the other of a single side of the hexagon, the unconstrained vertex travels back and forth along a segment of length $\frac{2 \sqrt{3}}{3}-1$.

Indeed, let $A B C D E F$ be the hexagon, and $O$ its center; let $G$ and $H$ be on segments $A B$ and $B C$, respectively, such that $G H=A B$, and let $R$ be the point on the same side of $O$ relative to the line $G H$ such that $G H R$ is an equilateral triangle. We want to prove that $R$ lies on the line $O B$.
![](https://cdn.mathpix.com/cropped/2024_04_17_9f6e2f6c8f4b083321c1g-12.jpg?height=502&width=1370&top_left_y=148&top_left_x=390)

Notice that the quadrilateral $G B H R$ is cyclic, since the sum of the angles $\widehat{G B H}=120^{\circ}$ and $\widehat{G R H}=60^{\circ}$ is a straight angle. It follows that $\widehat{G B R}=\widehat{G H R}=60^{\circ}$, hence $\widehat{G B R}=\widehat{G B O}$ as desired. We also calculate the distance of $R$ from $O$ in the case where $G H$ is perpendicular to $B O$ (i.e., $H$ is the reflection of $G$ across this line). In this case, let $K$ be the intersection point of $B O$ and $G H$, the triangles $K G B$ and $K H B$ are congruent and right-angled, each being half of an equilateral triangle, so $B K=\frac{1}{\sqrt{3}} G K=\frac{\sqrt{3}}{6}$. Since $R K=\frac{\sqrt{3}}{2}$ and $O B=1$, we have $R O=R K-(O B-B K)=2 \frac{\sqrt{3}}{3}-1$.

Thus, when a constrained vertex of the right triangle moves from $A$ to $B$ and the second constrained vertex from $B$ to $C$, the unconstrained vertex travels back and forth along a segment $O X$ of length $2 \frac{\sqrt{3}}{3}-1$. In total, the distance traveled by the unconstrained vertex is $12 \cdot O X=8 \sqrt{3}-12$.