Published in Mathematics & Informatics Quarterly, 6(3), 169-171, Sept 1996.
A dual to Kosnita's theorem
Michael de Villiers, Mathematics Education,
University of Durban-Westville, Durban 4000, South Africa
In the September 1995 issue of the Mathematics and Informatics Quarterly, the following interesting result was mentioned without proof in the column on Forgotten Theorems:
Kosnita's Theorem. The lines joining the vertices A, B, and C of a given triangle ABC with the circumcenters of the triangles BCO, CAO, and ABO (O is the circumcenter of triangle ABC), respectively, are concurrent.
On the basis of an often observed (but not generally true) duality between circumcenters and incenters (eg. see De Villiers, 1996), I immediately conjectured that the following dual to Kosnita's theorem might be true, namely:
Kosnita Dual. The lines joining the vertices A, B, and C of a given triangle ABC with the incenters of the triangles BCO, CAO, and ABO (O is the incenter of triangle ABC), respectively, are concurrent.
Investigation on the dynamic geometry program Sketchpad then confirmed that the conjecture was indeed true. I then found that both results could easily be proved by the following useful result for concurrency, a proof of which is given in De Villiers (1996). Interestingly, the point of concurrency for the special case with equilateral triangles on the sides is called the Fermat-Torricelli point.
Fermat-Torricelli Generalization. If triangles DBA, ECB and FAC are constructed on the sides of any triangle ABC so that angle DAB = angle CAF, angle DBA = angle CBE and angle ECB = angle ACF, then DC, EA and FB are concurrent (see Figure 1).
It should further be pointed out that the above result is also true when the triangles are constructed inwardly. As shown in Figure 2 we clearly have for Kosnita's dual that angle DAB = (angle A)/4 = angle CAF, angle DBA = (angle B)/4 = angle CBE and angle ECB = (angle C)/4 = angle ACF, and from the above result it therefore follows that DC, EA and FB are concurrent.
Kosnita's theorem follows a little less directly from the Fermat-Torricelli generalization. In this case, the "base triangle" is triangle DEF with A, B and C the outer vertices (see Figure 3). Since DBOA is a kite, we have angle BDO = angle ADO. But DBEO and DOFA are also kites. Therefore, angle BDE = (angle BDO)/2 and angle ADF = (angle ADO)/2 from which follows that angle BDE = angle ADF. In a similar fashion can be shown that angle BED = angle CEF and angle CFE = angle AFD. From the F-T generalization, it therefore follows that DC, EA and FB are concurrent.
De Villiers, M. (1995). A generalization of the Fermat-Torricelli point. The Mathematical Gazette, 79(485), July, 374-378.
De Villiers, M. (1996). Some Adventures in Euclidean Geometry. Durban: University of Durban-Westville.