Suppose convex hexagon $ \text{HEXAGN}$ has $ 120^\circ$-rotational symmetry about a point $ P$—that is, if you rotate it $ 120^\circ$ about $ P$, it doesn't change. If $ PX\equal{}1$, find the area of triangle $ \triangle{GHX}$.
1. **Assume the Hexagon is Regular:**
   Since the hexagon has $120^\circ$ rotational symmetry about point $P$, we can assume it is a regular hexagon. In a regular hexagon, all sides are equal, and all internal angles are $120^\circ$.

2. **Identify the Center and Side Length:**
   Let $P$ be the center of the hexagon. Given $PX = 1$, and since $P$ is the center, $PX$ is the radius of the circumscribed circle of the hexagon. In a regular hexagon, the radius of the circumscribed circle is equal to the side length of the hexagon. Therefore, each side of the hexagon is $1$.

3. **Determine the Shape of $\triangle GHX$:**
   Since $P$ is the center and the hexagon is regular, $\triangle GHX$ will be an equilateral triangle. This is because rotating the hexagon by $120^\circ$ about $P$ maps each vertex to another vertex, preserving the equilateral nature of the triangles formed by the vertices and the center.

4. **Calculate the Side Length of $\triangle GHX$:**
   The side length of $\triangle GHX$ is the same as the side length of the hexagon, which is $1$.

5. **Drop an Altitude in $\triangle GHX$:**
   To find the area of $\triangle GHX$, drop an altitude from one vertex to the opposite side. This altitude will split the equilateral triangle into two $30^\circ-60^\circ-90^\circ$ triangles.

6. **Calculate the Altitude:**
   In a $30^\circ-60^\circ-90^\circ$ triangle, the altitude (opposite the $60^\circ$ angle) is given by:
   \[
   \text{Altitude} = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2}
   \]

7. **Calculate the Area of $\triangle GHX$:**
   The area of an equilateral triangle with side length $s$ is given by:
   \[
   \text{Area} = \frac{s^2 \sqrt{3}}{4}
   \]
   Substituting $s = 1$:
   \[
   \text{Area} = \frac{1^2 \sqrt{3}}{4} = \frac{\sqrt{3}}{4}
   \]

The final answer is $\boxed{\frac{\sqrt{3}}{4}}$