Uniqueness of midpoints in inner product spaces

Question

Does anyone have an elegant proof of the following fact?

Let $V$ be a real inner product space and let $x$ and $y$ be two elements of $V$. If $z\in V$ is such that $\lVert x-z\rVert=\lVert y-z\rVert=\tfrac12\lVert x-y\rVert$, then $z=\frac12(x+y)$.

Answer 1

Let $v=z-x$ and $w=z-y.$ Then, $\|v\|+\|w\|=\|v-w\| \stackrel{\text{traingle ineq.}}{\leq} \|v\| + \|(-w)\|,$ so the equality (Cauchy-Schwarz equality) occurs when $v=\alpha (-w)$ for $\alpha > 0$ (assuming non of them are zero). Thus, from norm equality, we get $\alpha=1,$ hence $v+w=0.$

Answer 2

Let $\bar{z} = \frac{1}{2}(x+y)$. Then $$ \|x-z\|^2 = \|x-\bar{z}\|^2 + \|\bar{z}-z\|^2 + 2\langle x-\bar{z},\bar{z}-z\rangle $$ and similarly for $\|y-\bar{z}\|$. Since $\|x-\bar{z}\| = \frac{1}{2}\|x-y\| = \|y-\bar{z}\|$ and $\|x-z\|=\|y-z\|$, we have $$ \|x-\bar{z}\|^2+\|\bar{z}-z\|^2 + 2\langle x-\bar{z},\bar{z}-z\rangle = \|y-\bar{z}\|^2+\|\bar{z}-z\|^2 + 2\langle y-\bar{z},\bar{z}-z\rangle \\ \implies \langle x-\bar{z},\bar{z}-z\rangle = \langle y-\bar{z},\bar{z}-z\rangle \\ \implies \langle x-y,\bar{z}-z\rangle = 0.$$ Since $\|x-z\| = \|x-\bar{z}\|$, we have $$ 0 = \|\bar{z}-z\|^2 + 2\langle x-\bar{z},\bar{z}-z\rangle = \|\bar{z}-z\|^2 + \langle 2x-(x+y),\bar{z}-z\rangle = \|\bar{z}-z\|^2.$$ This is probably the standard way to do it.

Answer 3

By a translation and a dilatation by a factor of two, this is equivalent to showing that for every $y$ there is a unique $z$ with $z-y\perp y$ and $|z|=|y|$. But $|z|=|y|$ gives $z-y\perp z$ too, so that $z-y\perp z-y$ and $z=y$.