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3

Let the points be $(a1,a2,a3,a4),(b1,b2,b3,b4),(c1,c2,c3,c4),(d1,d2,d3,d4)$. The circumcentre is in the hyperplane, so it is a convex combination of the points $$CC = wA+xB+yC+(1-w-x-y)D$$ The distances to the points are the same. $$|(1-w)AD-xBD-yCD|^2\\=|-wAD+(1-x)BD-yCD|^2\\=|-wAD-xBD+(1-y)CD|^2\\=|-wAD-xBD-yCD|^2$$ Subtract the last expression from the ...

2

I suggest a family of metrics wich may be combined (by weighted means, for example) to yield a number. Then finding a good threshold for this number (i.e. if $\mathrm{mydist}(A,B) \le \epsilon$, the clouds $A$ and $B$ are deemed similar) will complete the algorithm. d_i(A,B) = \frac1{|A|+|B|} \left( \sum_{a\in A} \mathrm{dist}_i^i(a,B) + \sum_{b\in B} ...

2

Here is a somewhat efficient approach: Build a matrix $P$ of distances, $P_{ij} = P_{ji} = d(p_i, p_j)$ Keep an array $d_i = \min_{q\in Q} d(p_i, q)$ and an array of lists $Q_i = \{q\in Q \mid d(p_i,q) = d_i\}$. We'll build these on-the-fly For $i = 1$ perform the normal search. For each $i \ge 2$: Find $\bar d = \min_{j = 1}^{i-1} P_{ij} + d_j$ and ...

2

The radius of the circumsphere is three times the radius of the inscibed sphere. Hence one tetrahedron with the sphere of radius $1$ around the origin would be given by the vertices $(0,0,3)$, $(\sqrt 8,0,-1)$, $(-\sqrt 2,\sqrt 6,-1)$, $(-\sqrt 2,-\sqrt 6,-1)$.

2

Assuming that each tile has uniform surface density (mass per unit of area) then the centre of mass of any one tile is located at its geometrical centre. Then, it's easy to prove that the CM of an object made of multiple tiles is simply the weighted average of the centres of the constituent tiles, where the weights are the masses of the tiles. In equations, ...

1

If the radius of inscribed sphere is $r$ then the edge length of the tetrahedron is $2r\sqrt{6}$ & radius of circumscribed sphere is $3r$ Then in general form, for inscribed sphere with a radius $r$ centered at the origin, the vertices of the tetrahedron are $(0, 0, 3r)$,$\left(2r\sqrt{2},0, -r\right)$, $\left(-r\sqrt{2},r\sqrt{6}, -r\right)$, ...

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Here is one method I can think of. Since your point clouds form a curve, create a function from $[0,1]$ to $\Bbb R^2$ out of both point clouds. In essence, "connect the dots". Let's call the function of the first cloud $A$ and the function of the second cloud $B$. So basically, $A(0)$ would be the first point in cloud $A$ and $A(1)$ would be the final point ...

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As far as I know there is no practical tool which provide automatically a readable proofs of the same kind as high school proofs for many geometry theorems. As Mitch pointed there are algebraic methods such as Grobner bases and Wu's method, but they do not provide readable proofs. They have been implemented and are available in software such as: ...

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This goes well beyond your question, but this paper was posted to the arXiv just last week: Jean-Daniel Boissonnat, Ramsay Dyer, Arijit Ghosh. "A probabilistic approach to reducing the algebraic complexity of computing Delaunay triangulations." (arXiv abstract). Computing Delaunay triangulations in $\mathbb{R}^d$ involves evaluating the ...

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