Example 3 modified

For a right triangle, the circumcenter lies on one side.

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(x,y) circumcenter, and (a,b) is perpendicular to (a-e, b).

H :
x^2+y^2-(x-e)^2-y^2,
x^2+y^2-(x-a)^2-(y-b)^2
(a-e)a+b^2

T : y=0

Use R::=Q[txyabe], Lex;

t..y
[t, x, y]
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Theorem is false :

NF(1, Ideal(x^2+y^2-(x-e)^2-y^2, x^2+y^2-(x-a)^2-(y-b)^2, (a-e)a+b^2, yt-1))

1
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Non-Degeneracy conditions :

Elim(t..y, Ideal(x^2+y^2-(x-e)^2-y^2, x^2+y^2-(x-a)^2-(y-b)^2, (a-e)a+b^2, yt-1))

Ideal(a^2 - ae + b^2, -be)

So theorem is true if triangle is not rectangular OR be­0
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Discovery conditions :

Elim(t..y, Ideal(x^2+y^2-(x-e)^2-y^2, x^2+y^2-(x-a)^2-(y-b)^2, (a-e)a+b^2, y))

Ideal(a^2 - ae + b^2)

So theorem could be true if triangle is rectangular, nothing really new.
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