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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* = 3*OG*).

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 3*PG* = *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 4*PG* = *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.

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Created by Michael de Villiers, August 2012; updated 15 September 2012; 18 August 2020.