Here’s a problem for you:
Let $P\left(-\frac{1}{2},0\right)$ and $Q\left(\frac{1}{2},0\right)$ be adjacent vertices of a regular hexagon. If the hexagon is below the segment $\overline{PQ}$, find the coordinates of the four other vertices.
Let $P$, $Q$, $R$, $S$, $T$, and $U$ be the vertices of a regular hexagon. Two of the vertices, $P\left(-\frac{1}{2},0\right)$ and $Q\left(\frac{1}{2},0\right)$ are already given, so we need to find out the coordinates of the other four. Denote the other points as $R(r_{x},r_{y})$, $S(s_{x},s_{y})$, $T(t_{x},t_{y})$, and $U(u_{x},u_{y})$.
The length of side $\overline{PQ}$ is $s=\left|\overline{PQ}\right|=\sqrt{\left(\frac{1}{2}+\frac{1}{2}\right)^{2}+\left(0-0\right)^{2}}=1$. Recall that all of the sides of a regular hexagon are equal in length, and a regular hexagon can be split into six smaller equilateral triangles, with $s$ as the side of the triangle. From here, we can determine that the apothem $a$ is also equal to $s$, thus, $a=s=1$. By inspection, we can conclude that $p_{y}=q_{y}$, $p_{x}=t_{x}$, $q_{x}=s_{x}$, $r_{y}=u_{y}$, $t_{y}=s_{y}$, $u_{x}=t_{x}-k=-k-\frac{1}{2}$, and $r_{x}=s_{x}+k=k+\frac{1}{2}$ for some $k \in \mathbb{R}$. Thus, $P\left(-\frac{1}{2},0\right)$ and $Q\left(\frac{1}{2},0\right)$, $R(k+\frac{1}{2},r_{y})$, $S(\frac{1}{2},s_{y})$, $T(-\frac{1}{2},t_{y})$, and $U(-k-\frac{1}{2},u_{y})$.
Use the Pythagorean theorem to get the length of $\overline{QS}$ in order to get $s_{y}$.
$$\begin{eqnarray}
\left|\overline{QS}\right|^{2} & = & \left|\overline{TQ}\right|^{2}-\left|\overline{ST}\right|^{2} \nonumber \\\\\
& = & \left(2a\right)^{2}-s^2 \nonumber \\\\\
\left|\overline{QS}\right| & = & \sqrt{4a^{2}-s^2} \nonumber \\\\\
& = & \sqrt{4-1} \nonumber \\\\\
& = & \sqrt{3}
\end{eqnarray}$$
Now use the distance formula to find the value of $s_{y}$.
$$\begin{eqnarray*}
\left|\overline{QS}\right|=\sqrt{\left(\frac{1}{2}-\frac{1}{2}\right)^{2}+\left(s_{y}-0\right)^{2}} & = & \sqrt{3} \nonumber \\\\\
\sqrt{0+s_{y}^{2}} & = & \sqrt{3} \nonumber \\\\\
s_{y} & = & \sqrt{3}
\end{eqnarray*}$$
Since points $S$ and $T$ are below the $x$-axis, their $y$-coordinates must be negative. Thus, $S(\frac{1}{2},-\sqrt{3})$ and $T(-\frac{1}{2},-\sqrt{3})$.
Let $M\left(m_{x},m_{y}\right)$ be the midpoint of $\overline{PS}$ and also the center of the regular hexagon. Using the midpoint formula:
${\displaystyle m_{x}=\frac{-\frac{1}{2}+\frac{1}{2}}{2}=0}$ and
${\displaystyle m_{y}=\frac{0-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}}$,
thus $M\left(0,-\frac{\sqrt{3}}{2}\right)$
Since points $U$, $M$, and $R$ are collinear and the line passes through these points is parallel to the $x$-axis, their $y$-coordinates are equal. Therefore, $R(k+\frac{1}{2},-\frac{\sqrt{3}}{2})$ and $U(-k-\frac{1}{2},-\frac{\sqrt{3}}{2})$.
Now use the distance formula to determine the value of $k$, which in turn will be used to find out the $x$-coordinates of points $U$ and $R$.
$$\begin{eqnarray*}
\left|\overline{UP}\right|=\sqrt{\left(-\frac{1}{2}-\left(k+\frac{1}{2}\right)\right)^{2}+\left(0+\frac{\sqrt{3}}{2}\right)^{2}} & = & 1 \nonumber \\\\\
\sqrt{\left(-k-1\right)^{2}+\frac{3}{4}} & = & 1 \nonumber \\\\\
\sqrt{k^{2}+2k+\frac{7}{4}} & = & 1 \nonumber \\\\\
k^{2}+2k+\frac{3}{4} & = & 0 \nonumber \\\\\
\left(2k+1\right)\left(2k+3\right) & = & 0 \nonumber \\\\\
k & = & \left\{ -\frac{3}{2},-\frac{1}{2}\right\}
\end{eqnarray*}$$
There are two possible solutions for $k$, so our task now is to find out which solution is extraneous. We can test each solution by substituting it into each of the $x$-coordinates of $R$ and $U$. If $k=-\frac{3}{2}$, then the coordinates would be $R(-1,-\frac{\sqrt{3}}{2})$ and $U(1,-\frac{\sqrt{3}}{2})$. So far, so good. Now, If $k=-\frac{1}{2}$, then the coordinates would be $R(0,-\frac{\sqrt{3}}{2})$ and $U(0,-\frac{\sqrt{3}}{2})$. Points $R$ and $U$ can't be coincident! So we found out that the solution $k=-\frac{1}{2}$ is extraneous. Now, we're left with $k=-\frac{3}{2}$, thus the solution. Substituting $k$ with the $x$-coordinates of points $U$ and $R$, we get $R(-1,-\frac{\sqrt{3}}{2})$ and $U(1,-\frac{\sqrt{3}}{2})$.
Therefore, the coordinates of the other four vertices are $R(-1,-\frac{\sqrt{3}}{2})$, $S(\frac{1}{2},-\sqrt{3})$, $T(-\frac{1}{2},-\sqrt{3})$ and $U(1,-\frac{\sqrt{3}}{2})$.$\quad\blacksquare$
