Johnson circles

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Illustration of the Johnson circles theorem which states that if the three blue circles in the picture have equal radius and intersect as shown, then the resulting red circle has the same radius as the blue circles.

In geometry, a set of Johnson circles comprise three circles of equal radius $r$ sharing one common point of intersection $H$ . In such a configuration the circles usually have a total of four intersections (points where at least two of them meet): the common point $H$ that they all share, and for each of the three pairs of circles one more intersection point (referred here as their 2-wise intersection). If two of the circles happen to lie directly opposite to each other they only have $H$ as a common point, and it will then be considered their 2-wise intersection as well; it they should coincide we declare their 2-wise intersection to be their point diametrically opposite to $H$ .

1 Properties
2 Proofs
3 Further properties
4 References

[edit] Properties

The centers of the Johnson circles lie on a circle of the same radius $r$ as the Johnson circles centered at $H$ . These centers form the Johnson triangle.
The circle centered at $H$ with radius $2 r$ is tangent to each of the Johnson circles in its point opposite to $H$ .
Those points of tangency form another triangle, similar to the Johnson triangle, and in fact its image by the homothety with factor 2 centered at $H$ .
The Johnson circles theorem: the three points of 2-wise intersection, forming what is called the reference triangle, lie on a circle of the same radius $r$ as the Johnson circles.
The reference triangle is in fact congruent to the Johnson triangle, and homothetic to it by a factor −1.
The point $H$ is the orthocenter of the reference triangle.

[edit] Proofs

Property 1 is obvious from the definition. Property 2 is also clear: a circle centered in a point P of another circle C and of twice its radius is tangent to C in its point opposite to P. Property 3 in the formulation of the homothety immediately follows; call the triangle of points of tangency the doubled Johnson 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 doubled Johnson triangle lying on these circles. The 2-wise intersection point therefore is the midpoint of a side of the doubled Johnson 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 doubled Johnson 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 doubled Johnson 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.

There is also a more pedestrian proof of the Johnson circles theorem, using a simple vector computation. There are vectors $\vec{u}$ , $\vec{v}$ , and $\vec{w}$ , all of length r, such that the Johnson circles are centered respectively at $H+\vec{u}$ , $H+\vec{v}$ , and $H+\vec{w}$ . Then the 2-wise intersection points are respectively $H+\vec{u}+\vec{v}$ , $H+\vec{u}+\vec{w}$ , and $H+\vec{v}+\vec{w}$ , and the point $H+\vec{u}+\vec{v}+\vec{w}$ clearly has distance r to any of those 2-wise intersection points.

[edit] Further properties

The center of the homothety of property 5, which according to the given proof can be computed from H and the barycenter of the Johnson triangle and is collinear with them, is the nine-point center of the reference triangle. The point H is both the orthocenter of the reference triangle and the circumcenter of the Johnson triangle. Furthermore, under the the reflections in the sides of the reference triangle, its circumscircle is respectively interchanged with each of the Johnson circles, its orthocenter H with certain points of that circumcircle, and its circumcenter with the vertices of the Johnson triangle.

The Johnson triangle and its reference triangle share the same nine-point center, the same Euler line and the same nine-point circle. The six points formed from the vertices of the reference triangle and its Johnson triangle all lie on the Johnson circumconic that is centred at the nine-point center and that has the point $X (216)$ of the reference triangle as its perspector.

Finally there are two interesting and documented circumcubics that pass through the six vertices of the reference triangle and its Johnson triangle as well as the circumcenter, the orthocenter and the nine-point center. The first is known as the first Musselman cubic – $K 026$ . This cubic also passes through the six vertices of the medial triangle and the medial triangle of the Johnson triangle. The second cubic is known as the Euler central cubic – $K 044$ . This cubic also passes through the six vertices of the orthic triangle and the orthic triangle of the Johnson triangle.

The $X (i)$ point notation is the Clark Kimberling ETC classification of triangle centers.