Proving coincidence in Euclidean geometry by using finitely many constellations Two polynomials $f(x)$ and $g(x)$ of degree $n$ are equal if they are equal for $n+1$ different $x$. 
Is anything like this true for Euclidean geometry? Say, I have three arbitrary points in the plane (called $A$, $B$, $C$), and I use $k$ construction steps (like adding a circle or a line, or intersecting existing one-dimensional objects) to get a point $P$. From the same three points, I could also do $l$ construction steps getting a point $Q$.
Now I conjecture that $P$ and $Q$ coincide. Is it possible to check this conjecture by looking at a finite number of constellations for $A$, $B$, $C$? These finitely many constellations of course need to be "independent" in the sense that they could not be constructed from each other through rotations or other simple maps.
 A: No, there is no number $m$ such that two constructions agreeing on $m$ independent inputs must agree on all inputs. Consider:


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*Let $P$ be the point on line $AB$, distinct from $A$, such that $CP=CA$.

*Let $Q$ be the point on line $AB$, distinct from $A$ and $B$, such that $CQ=\max(CA,CB)$.
These are easy to construct, potentially with clauses for degenerate cases, depending on your definition of "construction step". They obviously agree infinitely often when $CA>CB$, and disagree sometimes when $CB>CA$.
So the agreement of constructions would only follow from agreement on different regions of inputs. Here are some facts about polynomials which provide better analogies:


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*If $f$ and $g$ are piecewise polynomial functions of $x$, with the same pieces of definition, and each piece of degree at most $n$, and any overlapping pieces of $f$ and $g$ agreeing on $n+1$ points, then $f=g$ always.

*If $f$ and $g$ are piecewise polynomial functions of $(x_1,\ldots,x_k)$, with the same pieces of definition, and each piece of degree at most $n$, and any overlapping pieces of $f$ and $g$ agreeing on $(n+1)^k$ points, then $f=g$ always.
