Triangle Points

Explanation

Each point $C$ can be described as a vector from the midpoint of $\overrightarrow{AB}$ to the point $C$. Let's call this vector $\vec{C}$. $\vec{C}$ has 2 important properties; it is orthogonal to $\overrightarrow{AB}$ and it's magnitude is constant for all terminal points $C$. It shouldn't be that hard to convince yourself of this, try solving the problem in $\mathbb{R}^2$ and it should become apparent fairly quickly.

Here's the intuition that I used. Given $\overrightarrow{AB}$ in $\mathbb{R}^n$ space, we can draw an infinite amount of planes that contain $\overrightarrow{AB}$ (assuming $n>2$). For example, in $\mathbb{R}^3$ space, imagine rotating a plane about an axis along $\overrightarrow{AB}$. That's the set of all planes that contain $\overrightarrow{AB}$.

On each of these planes, we can draw the problem in 2 dimensions. You have probably noticed that there are always 2 solutions in $\mathbb{R}^2$. Simply draw a square with the diagonal being $\overrightarrow{AB}$ and calculate where the last 2 corners are. In fact, because the solutions in $\mathbb{R}^2$ are the secondary diagonal points of a square, this tells us that the magnitude of $\vec{C}$ is half that of the diagonal ($\|\vec{C}\|=\frac{1}{2}\|\overrightarrow{AB}\|$).

Similarly, we can determine that $\vec{C}$ must be orthogonal to $\overrightarrow{AB}$ using the idea of square diagonals. The diagonals of a square are by definition orthogonal to each other. This means that $\vec{C} \cdot \overrightarrow{AB} = 0$.

To combine these conditions, we can write the first condition as $\|\vec{C}\|-\frac{1}{2}\|\overrightarrow{AB}\|=0$. Then we equate this to the second condition. Of course, we still need to specify that both of these expressions are equal to zero.