 Solution

Let $A=(Y+Z)/2$, $B=(Z+X)/2$, and $C=(X+Y)/2$, interpreting these expressions as position vectors. The resulting medial triangle is nondegenerate: for example,Thus $BC$ is parallel to $YZ$, and the same calculation holds cyclically. The altitude of a triangle through $A$ is consequently the line through the midpoint of $YZ$ that is orthogonal to $YZ$, namely its perpendicular bisector. Each remaining altitude of a triangle similarly gives one of the other two perpendicular bisectors.

By iii, the medial triangle has a unique orthocenter, so the three perpendicular bisectors meet in one point. Every point on the perpendicular bisector of $YZ$ is equally far from $Y$ and $Z$: expanding the squares of the two Euclidean distances gives $2\mathbf{q}\cdot (\mathbf{z}-\mathbf{y})=\Vert \mathbf{z}{\Vert }^2-\Vert \mathbf{y}{\Vert }^2$. Applying this to the three sides shows that the common point is equally far from $X,Y,Z$. It is the circumcenter.