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The circumcircle of a nondegenerate triangle is the unique circle through its three vertices. Its center is the circumcenter, found by intersecting two perpendicular bisectors; equality of distances to all three vertices proves existence, and subtracting the squared-distance equations proves uniqueness. For a right triangle, the midpoint of its hypotenuse is equidistant from all vertices, so its circumradius is half the hypotenuse length.

The radius of a triangle's circumscribed circle, whose center is the circumcenter. For side lengths $a,b,c$ and area $A$, it satisfies $R=abc/(4A)$.

The medial triangle joins the three midpoints of a triangle's sides. Each side is parallel to the corresponding original side and has half its length: $((\mathbf{x}+\mathbf{z})/2)-((\mathbf{x}+\mathbf{y})/2)=(\mathbf{z}-\mathbf{y})/2$. Each altitude of a triangle in the medial triangle is a perpendicular bisector of an original side, so its orthocenter is the original triangle's circumcenter.

Put $q=|z{|}^2$ and retain the domain restriction $z\ne 0$. Then $z+1/\bar{z}=(q+1)/\bar{z}$, so the continued fraction simplifies toThe other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomesMultiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first geometric locus in the complex plane is the circleIts center is $-2$ and radius is $\sqrt{2}$; it does not contain the excluded point $z=0$ on its circumference.

The principal complex logarithm has imaginary part equal to the complex argument, so its equation gives the open rayIf one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi /3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.

These latter loci bound a right triangle with verticesThe bounded open region is $0<x<1/\sqrt{3}$, $\sqrt{3}x<y<1$. Its closure is the closed triangle. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier circle is a separate locus and is not a side of this triangle.

The right angle is at $P$, so the circumcenter is the midpoint of hypotenuse $OQ$:Thus its circumcircle has the complex equationTo find the incircle, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt{3}x=0$ is $(1-r-\sqrt{3}r)/2$. Requiring this distance to be $r$ givesThereforeThis tangent circle is indeed the largest contained circle. For any contained disk of radius $\rho$, its center has distances $d_j\ge \rho$ from the three side lines. Splitting the triangle into triangles with that center as vertex gives area $\Delta =\frac{1}{2}{\sum }_j{\ell }_j{d}_j\ge s\rho$, where $s$ is its semiperimeter. Here $\Delta =1/(2\sqrt{3})$ and $s=(1+\sqrt{3})/2$, so $\rho \le \Delta /s=r$. Our circle attains this upper bound, proving maximality rather than just tangency.

Expanding the squared Euclidean distances givesThus one may takeThe normal is parallel to $\overrightarrow{AB}$, so $L_{AB}$ is its perpendicular bisector. If $\mathbf{x}$ lies on both $L_{AB}$ and $L_{BC}$, then $|\mathbf{x}-\mathbf{a}|=|\mathbf{x}-\mathbf{b}|=|\mathbf{x}-\mathbf{c}|$, hence it also lies on $L_{CA}$. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its circumcenter.

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.

Write $z=x+iy$, with $y\ne 0$ because the triangle is nondegenerate. Let $w=u+iv$ be its circumcenter. Equality of the complex modulus values $|w|=|w-1|$ gives $u=1/2$. Equality of $|w|=|w-z|$, after expanding the squares, givesThus the real part and imaginary part determineUsing the complex conjugate $\bar{z}=x-iy$, we also haveThereforeThe nonzero denominator is exactly the nondegeneracy condition. If $z$ is real, the three distinct collinear points do not have a finite circumcenter.

At the north pole $\mathbf{r}=+R\hat{z}$, so the velocities reverse:The triangle still has side lengths 20, 48 and 52 mm/year. Across AB, A moves toward positive $y$ relative to B; across AC, A moves toward positive $x$ relative to C. Both boundaries therefore open and are spreading ridges, with full opening speeds 20 and 48 mm/year. Together with BC they form a ridge-ridge-transform triple junction.

In the symmetric normal-spreading model, a ridge's normal velocity is the mean normal velocity of its two plates. AB gives $U_y=(0-20)/2=-10$, and AC gives $U_x=(48+0)/2=24$. Their intersection lies on the BC transform velocity line:This northern junction is also instantaneously kinematically stable. The diagrams use the same fixed geocentric tangent axes at both poles; reversing velocities is not the same operation as reflecting a map drawn from an exterior observer's viewpoint.

Not every ridge-ridge-transform geometry works. For pure normal spreading, each ridge velocity line is the perpendicular bisector of the corresponding side of the plate velocity triangle. Their intersection is the triangle's circumcenter. A transform junction must also lie on the third side's line. A nondegenerate triangle's circumcenter lies there only at that side's midpoint, so the opposite angle is a right angle. Equivalently, $({\mathbf{v}}_B-{\mathbf{v}}_A)\cdot ({\mathbf{v}}_C-{\mathbf{v}}_A)=0$. The two ridge directions are then perpendicular, as they are here. RRF junctions are conditionally stable, with the right-angle ridge geometry in this symmetric pure-normal model; they are not all stable. Oblique or asymmetric accretion changes the velocity-line assumptions and must be specified separately.