Trac #29125: is_inscribed depends on order of vertices

Currently the inscription test for polyhedra depends on the order of
vertices:

{{{
sage: Polyhedron(vertices=[[-2,-1], [-2,1], [0,-1],
[0,1]]).is_inscribed()
True
sage: P = Polyhedron(vertices=[[-2,-1], [-2,1], [0,-1], [0,1]],
backend='field')
sage: P.is_inscribed()
False
sage: V = P.Vrepresentation()
sage: H = P.Hrepresentation()
sage: parent = P.parent()
sage: dic = {True: 0, False: 0}
sage: for V1 in Permutations(V):
....:     P1 = parent._element_constructor_(
....:         [V1, [], []], [H, []], Vrep_minimal=True,
Hrep_minimal=True)
....:     dic[P1.is_inscribed()] += 1
....:
sage: dic
{True: 18, False: 6}
}}}

The algorithm constructs a sphere around `dim + 1` vertices in general
position. The circumcenter is computed up to sign. Then, one vertex is
taken to determine, which sign to choose. However, up to `dim` vertices
might lie on the intersection of both spheres.

We fix this by checking distance from the circumcenter for all vertices
of that simplex.

URL: https://trac.sagemath.org/29125
Reported by: gh-kliem
Ticket author(s): Jonathan Kliem
Reviewer(s): Jean-Philippe Labbé

src/sage/geometry/polyhedron/base.py

@@ -2851,6 +2851,19 @@ def is_inscribed(self, certificate=False):
             ....:                          [+1, 0, 0, 0]])
             sage: P.is_inscribed()
             True
+
+        We check that :trac:`29125` is fixed::
+
+            sage: P = Polyhedron(vertices=[[-2,-1], [-2,1], [0,-1], [0,1]], backend='field')
+            sage: P.is_inscribed()
+            True
+            sage: V = P.Vrepresentation()
+            sage: H = P.Hrepresentation()
+            sage: parent = P.parent()
+            sage: for V1 in Permutations(V):
+            ....:     P1 = parent._element_constructor_(
+            ....:         [V1, [], []], [H, []], Vrep_minimal=True, Hrep_minimal=True)
+            ....:     assert P1.is_inscribed()
         """
 
         if not self.is_compact():
@@ -2895,8 +2908,8 @@ def is_inscribed(self, certificate=False):
         squared_circumradius = (sum(m**2 for m in minors) - 4 * a * c) / (4*a**2)
 
         # Checking if the circumcenter has the correct sign
-        test_vector = vertex.vector() - circumcenter
-        if sum(i**2 for i in test_vector) != squared_circumradius:
+        if not all(sum(i**2 for i in v.vector() - circumcenter) == squared_circumradius
+                   for v in vertices if v in simplex_vertices):
             circumcenter = - circumcenter
 
         is_inscribed = all(sum(i**2 for i in v.vector() - circumcenter) == squared_circumradius