Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is
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The incenter I of the triangle ABC formed by intersections A, B and C is equidistant from lines AB, BC and CA.
Also the three centers of three excircles are equidistant.
The distance is = r = inradius. Or r1, r2 ,r3: exradii .
In 2 dim answer = 4.
Draw a straightline L Perpendicular To the plane of ABC and passing through Incenter I or ex centers . All points on L are equidistant from lines AB BC and CA.
In 3dimensions there infinite such points.
Also the three centers of three excircles are equidistant.
The distance is = r = inradius. Or r1, r2 ,r3: exradii .
In 2 dim answer = 4.
Draw a straightline L Perpendicular To the plane of ABC and passing through Incenter I or ex centers . All points on L are equidistant from lines AB BC and CA.
In 3dimensions there infinite such points.
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