Johnson Circles - Proofs

Proofs

Property 1 is obvious from the definition. Property 2 is also clear: for any circle of radius r, and any point P on it, the circle of radius 2r centered at P is tangent to the circle in its point opposite to P; this applies in particular to P=H, giving the anticomplementary circle C. Property 3 in the formulation of the homothety immediately follows; the triangle of points of tangency is known as the anticomplementary triangle.

For properties 4 and 5, first observe that any two of the three Johnson circles are interchanged by the reflection in the line connecting H and their 2-wise intersection (or in their common tangent at H if these points should coincide), and this reflection also interchanges the two vertices of the anticomplementary triangle lying on these circles. The 2-wise intersection point therefore is the midpoint of a side of the anticomplementary triangle, and H lies on the perpendicular bisector of this side. Now the midpoints of the sides of any triangle are the images of its vertices by a homothety with factor −½, centered at the barycenter of the triangle. Applied to the anticomplementary triangle, which is itself obtained from the Johnson triangle by a homothety with factor 2, it follows from composition of homotheties that the reference triangle is homothetic to the Johnson triangle by a factor −1. Since such a homothety is a congruence, this gives property 5, and also the Johnson circles theorem since congruent triangles have circumscribed circles of equal radius.

For property 6, it was already established that the perpendicular bisectors of the sides of the anticomplementary triangle all pass through the point H; since that side is parallel to a side of the reference triangle, these perpendicular bisectors are also the altitudes of the reference triangle.

Property 7 follows immediately from property 6 since the homothetic center whose factor is -1 must lie at the midpoint of the circumcenters O of the reference triangle and H of the Johnson triangle; the latter is the orthocenter of the reference triangle, and its nine-point center is known to be that midpoint. Since the central symmetry also maps the orthocenter of the reference triangle to that of the Johnson triangle, the homothetic center is also the nine-point center of the Johnson triangle.

There is also an algebraic proof of the Johnson circles theorem, using a simple vector computation. There are vectors, and, all of length r, such that the Johnson circles are centered respectively at, and . Then the 2-wise intersection points are respectively, and, and the point clearly has distance r to any of those 2-wise intersection points.