Problem 1637

1. Assume the vertices of the equilateral triangle are at specific coordinates:
- Let the vertices of the equilateral triangle be $A(1,0)$, $B(-1,0)$, and $C\left(0, \sqrt{3}\right)$.

2. Determine the regions where the vertices of the hexagon can be:
- The vertices of the regular hexagon $V_1, V_2, V_3, V_4, V_5, V_6$ can only be in the three triangular regions $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ formed by extending the sides of the equilateral triangle.

3. **Claim 0: There exists at least one vertex $V_k$ ( $1 \leq k \leq 6$ ) in each of $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$:**
- Assume otherwise, and let there be no vertex in region $\mathcal{A}$. Then the hexagon $V_1V_2V_3V_4V_5V_6$ is entirely to the right of line $BC$, and thus $\triangle ABC$ isn't inside $V_1V_2V_3V_4V_5V_6$. This is a contradiction.

4. **Claim 1: Points $V_i(x_1, y_1)$ and $V_{i+3}(x_2, y_2)$ cannot be in the same region:**
- Assume otherwise, and let them both be in $\mathcal{C}$. So $y_1 \geq \sqrt{3} \pm \sqrt{3}x_1$ and $y_2 \geq \sqrt{3} \pm \sqrt{3}x_2$.
- By complex numbers, $V_{i+2} \left( \frac{x_1 + 3x_2 - \sqrt{3}y_1 + \sqrt{3}y_2}{4}, \frac{\sqrt{3}x_1 - \sqrt{3}x_2 + y_1 + 3y_2}{4} \right)$.
- Since $\sqrt{3}x_1 - \sqrt{3}x_2 + y_1 + 3y_2 > y_1 + \sqrt{3}x_1 - \sqrt{3} + y_2 - \sqrt{3}x_2 - \sqrt{3} \geq 0$, $V_{i+2}$ is above the line $AB$ and therefore can only be in $\mathcal{C}$.
- We can similarly prove that each of $V_1, V_2, V_3, V_4, V_5, V_6$ is inside $\mathcal{C}$, which contradicts Claim 0.

5. **Claim 2: If $V_i(x_1, y_1)$ is in the same region as $V_{i+2}(x_2, y_2)$, then so are $V_{i+1}$ and $V_{i+4}$:**
- Let $V_i$ and $V_{i+2}$ be both in $\mathcal{C}$. So $y_1 \geq \sqrt{3} \pm \sqrt{3}x_1$ and $y_2 \geq \sqrt{3} \pm \sqrt{3}x_2$.
- By complex numbers, $V_{i+1} \left( \frac{\sqrt{3}x_1 + \sqrt{3}x_2 - y_1 + y_2}{2\sqrt{3}}, \frac{x_1 - x_2 + \sqrt{3}y_1 + \sqrt{3}y_2}{2\sqrt{3}} \right)$.
- Since $x_1 - x_2 + \sqrt{3}y_1 + \sqrt{3}y_2 > \frac{y_1 + \sqrt{3}x_1 - \sqrt{3}}{\sqrt{3}} + \frac{y_2 - \sqrt{3}x_2 - \sqrt{3}}{\sqrt{3}} \geq 0$, $V_{i+1}$ is above the line $AB$ and therefore can only be in $\mathcal{C}$.
- Additionally, $V_{i+4} \left( \frac{x_1 + x_2 + \sqrt{3}y_1 - \sqrt{3}y_2}{2}, \frac{y_1 + y_2 - \sqrt{3}x_1 + \sqrt{3}x_2}{2} \right)$, and since $y_1 + y_2 - \sqrt{3}x_1 + \sqrt{3}x_2 > y_1 - \sqrt{3}x_1 - \sqrt{3} + y_2 + \sqrt{3}x_2 - \sqrt{3} \geq 0$, $V_{i+4}$ is above the line $AB$ and therefore can only be in $\mathcal{C}$.

6. **Consider the multiset $S = \left\{ \#\{1 \leq i \leq 6 : V_i \in \mathcal{A}\}, \#\{1 \leq i \leq 6 : V_i \in \mathcal{B}\}, \#\{1 \leq i \leq 6 : V_i \in \mathcal{C}\} \right\}$:**
- Each element of $S$ has to be greater than or equal to one (by Claim 0) and less than or equal to three (otherwise by the pigeonhole principle, there exists a pair of opposite vertices both chosen, which yields a contradiction with Claim 1).
- We therefore have the following possibilities: $\{1, 2, 3\}$ and $\{2, 2, 2\}$.

7. Analyze the possibilities:
- For $\{1, 2, 3\}$, Claim 2 says that the "2" must be adjacent, say $V_1$ and $V_2$. But by Claim 2, the pairs $(V_4, V_6)$ and $(V_3, V_5)$ can't be in the same region; otherwise, the equilateral $\triangle V_1V_3V_5$ or $V_2V_4V_6$ yields a contradiction with Claim 2.
- So the only possibility is $S = \{2, 2, 2\}$, from which the solution is similar to StarChan's.

The final answer is False.