A dyadic supertriangle contains the sites and excludes every input circumcircle

For each noncollinear triple $(a,b,c)$ from $P$, its center $o$ is uniquely determined by $2o\cdot (b-a)=\mathrm{\parallel }b{\mathrm{\parallel }}^2-\mathrm{\parallel }a{\mathrm{\parallel }}^2,2o\cdot (c-a)=\mathrm{\parallel }c{\mathrm{\parallel }}^2-\mathrm{\parallel }a{\mathrm{\parallel }}^2\mathrm{.}$ The coefficient matrix has nonzero determinant because $a,b,c$ are noncollinear, so this system has a unique real solution; put $r^2=\mathrm{\parallel }o-a{\mathrm{\parallel }}^2$. There are finitely many such triples.

Let $B$ be the maximum of $1$, the absolute values of all input coordinates, the absolute values of all center coordinates from step 1.1, and all corresponding values $r^2$. This is a finite maximum in the total order of $\mathbb{R}$. By [F3], choose an integer $M>B$.

Take baseline vertices $v_1=(-8M,-4M)$, $v_2=(8M,-4M)$, and $v_3=(0,8M)$. At height $y\in [-M,M]$, their triangle has horizontal half-width $\frac{2}{3}(8M-y)\ge \frac{14}{3}M>M$. Since every input site has $\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }<M$, all sites lie strictly inside this triangle.

Step 2.1 gives $\mathrm{\mid }{o}_1\mathrm{\mid },\mathrm{\mid }{o}_2\mathrm{\mid }<M$ and $r^2<M$, while $M>1$. For either lower baseline vertex $v_1=(-8M,-4M)$ or $v_2=(8M,-4M)$, $\mathrm{\parallel }{v}_i{\mathrm{\parallel }}^2=80M^2$ and $\mathrm{\mid }{v}_i\cdot o\mathrm{\mid }<12M^2$, so $\mathrm{\parallel }{v}_i-o{\mathrm{\parallel }}^2=\mathrm{\parallel }{v}_i{\mathrm{\parallel }}^2-2v_i\cdot o+\mathrm{\parallel }o{\mathrm{\parallel }}^2>56M^2>M>r^2\mathrm{.}$ For the top vertex $v_3=(0,8M)$, $\mathrm{\parallel }{v}_3{\mathrm{\parallel }}^2=64M^2$ and $\mathrm{\mid }{v}_3\cdot o\mathrm{\mid }<8M^2$, giving $\mathrm{\parallel }{v}_3-o{\mathrm{\parallel }}^2>48M^2>M>r^2$. Thus each vertex is strictly outside every candidate circle, using only squared distances. The input containment and all circle exclusions have positive margins.

The strict site-containment, strict circle-exclusion, and nonzero orientation of the supertriangle persist under sufficiently small changes in its three baseline vertices, because they are finitely many strict inequalities in continuous polynomial or squared-distance expressions. Choose open disks $U_1,U_2,U_3$ about the baseline vertices small enough to preserve these conditions for every choice $v_i\in U_i$.

Dyadic rationals are dense: for $a<b$, use [F3] to choose an integer $N>1\mathrm{/}(b-a)$, then choose $k>N$; induction gives $2^k\ge k+1>N$, hence $2^{-k}<b-a$. The least integer $m>2^{k}a$ exists by the Archimedean property and the well-ordering of the naturals after translating by an integer lower bound; minimality gives $m\le 2^{k}a+1<2^{k}b$, so $m{2}^{-k}\in (a,b)$. Hence dyadic points are dense in every open disk: place an open rectangle inside the disk, choose a dyadic first coordinate away from the finitely many vertical lines among the forbidden set, and then choose a dyadic second coordinate away from the finitely many intersections with that vertical line. If [A1] holds, choose $v_1\in U_1$ outside the finitely many lines through pairs and circles through triples of input sites. Next choose $v_2\in U_2$ outside all lines through pairs and circles through noncollinear triples of $P\cup \{v_1\}$, and choose $v_3\in U_3$ outside those determined by pairs and triples of $P\cup \{v_1,v_2\}$. A finite union of such lines and circles cannot contain an open disk by the rectangle argument. These choices make the augmented set have no collinear triple or cocircular quadruple. Without [A1], any dyadic choices in the disks still satisfy the first part.

The selected vertices are dyadic, preserve strict containment and exclusion by [step 4.1], and satisfy the additional general-position properties by [step 5.1]. This proves both claims.