= Solution
{c}
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,
$$
\mathbf b-\mathbf c=\tfrac12(\mathbf z-\mathbf y).
$$
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 <1/a/iii>, the <medial triangle> has a unique <orthocenter>, so \b[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 $2\mathbf q\cdot(\mathbf z-\mathbf y)=\|\mathbf z\|^2-\|\mathbf y\|^2$. Applying this to the three sides shows that the common <point> is equally far from $X,Y,Z$. It is the <circumcenter>.
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