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The Euler line of a triangle is mostly valued, not for any practical application, but purely as a beautiful, esoteric example of post-Greek geometry. However, the following theorem by the British mathematician James Joseph Sylvester (1814-1897) involves an interesting application of forces that relate to the Euler line (segment).
Theorem of Sylvester: The resultant of three equal forces OA, OB and OC acting on the circumcentre O of a triangle ABC, is the force represented by OH, where H is the orthocentre of the triangle (hence, the resultant OH = 3OG).
The result generalizes to forces acting similarly on any point as follows and as shown in the interactive sketch below:
Generalization: The resultant of three forces PA, PB and PC acting on any point P of a triangle ABC, is the force represented by 3PG = PQ, where G is the centroid of the triangle.
A generalization of Sylvester's theorem to any point in a triangle or quadrilateral
Can you explain why the result is true? (I.e. prove it?) First try yourself, but if stuck, click on the button in the sketch to show some constructions that might assist you in using the parallelogram law of forces to prove the result.
The result generalizes further to any point in a quadrilateral as follows and as shown in the LINK in the sketch above.
Quadrilateral generalization: The resultant of four forces PA, PB, PC and PD acting on any point P of a quadrilateral ABCD, is the force represented by 4PG = PQ, where G is the centroid of the quadrilateral.
Can you explain why this result is true? (I.e. prove it?) Can you generalize further to other polygons or to 3D? See Sylvester's theorem for a tetrahedron.
See my 2012 Mathematical Gazette paper Generalizing a problem of Sylvester for more details.
Created by Michael de Villiers, August 2012; updated 15 September 2012; 18 August 2020.