The applet below illustrates a configuration of three equal circles concurrent in a point. The discovery of the configuration that has many engaging properties is often credited to R. A. Johnson, although a century earlier the problem in a different guise has been solved by Mary Fairfax Somerville (1780--1872).
Here we just summarize several properties of the configuration that have been addressed elsewhere on various pages at the site. The the circles are denoted by their centers: C(Oa), C(Ob), C(Oc), with O being their common point. A is the intersection (other than O) of C(Ob) and C(Oc), etc.
If three equal circles pass through a point, the circle through their other three intersections is equal to them.
and several related ones have been proved elsewhere.
The conditions of the theorem are enforced in the applet by simply restricting the centers of the circles to a circle centered at O.
5 March 2015, Created with GeoGebra
The circumcircle of ΔABC has the same radius as any of C(Oa), C(Ob), C(Oc).
O is the orthocenter of ΔABC so that the four points O, A, B, C form an orthocentric system. The circumscribes of the four triangles formed by the points of an orthocentric system taken three at a time have the same radius.
The seven points A, Ob, C, Oa, B, Oc, and O form three adjacent rhombi so that their sidelines are naturally split into triples of parallel lines: AOc||ObO||COa, etc.
As a consequence of #3, ABOaOb is a parallelogram so that AB||OaOb and
AB = OaOb.This is true of any position of Oc, provided he other centers and O are kept fixed. In other words, the length and direction of AB is the same for all positions of Oc.
As a consequence of #3, triangles ABC and OaObOc are equal and homothetic with the coefficient -1. If Q is the orthocenter of ΔOaObOc then the midpoint of OQ serves as the center of homothety. Since O is the circumcenter of ΔOaObOc, OQ is its Euler line and N is the 9-point center.
O is the orthocenter and Q the circumcenter of ΔABC so that OQ is the Euler line of both triangles. (The centers of the circumcircles of the orthocentric system A, B, C, O form another orthocentric system Oa, Ob, Oc, Q.)
The anticomplementary ΔA'B'C' of ΔABC has the property that OA', OB', and OC' are diameters of C(Oa), C(Ob), C(Oc), its circumcircle is twice as big as C(O) and is tangent to C(Oa) at A', to C(Ob) at B' and to C(Oc) at C'.
ΔA'B'C' has the largest perimeter among all triangles with vertices on the three given circles whose sides pass through the points of intersection A, B, C. This is because (#7) its sides are parallel to the lines of centers OaOb, etc.
For the same reason as in #8, triangles A'B'C' and OaObOc are homothetic with the coefficient 2 and center O.
- N. Altshiller-Court, College Geometry, Dover, 1980, p. 95
- F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, p. 142
- R. Honsberger, Mathematical Gems II, MAA, 1976, pp. 18-22
- R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 17-18
- R. A . Johnson, Am Math Monthly, Vol. 23, No. 5. (May, 1916), pp. 161-162
- R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960, p. 75, pp. 165-166
- T. Lalesco, La Géométrie du Triangle, Éditions Jacques Gabay, sixiéme édition, 1987, 1.11
- G. Polya, Mathematical Discovery, John Wiley and Sons, 1981, Ch. 10.
- M. F. Somerville, Mathematical Repository, volume IV (1819), 91-92
3 Equal and Concurrent Circles
- Johnson Circles
- Three equal circles
- 3 circles having the same radius
- Inversionn in the Incircle
- Orthocenter and Three Equal Circles
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