In the first example below, each vertex is joined to the point (1/4) along the opposite side (measured say anti-clockwise), and in the second example (1/5) along the opposite side.

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Feynman triangle: 1/4 division

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Feynman triangle: 1/5 division

1. Can you now generalize to find a formula for a triangle *ABC* in the plane, when each vertex is joined to the point (1/*p*), (*p* > 2) along the opposite side (measured say anti-clockwise)?

2. Read my article at *Feynman's triangle: Some feedback and more*

3. The result actually generalizes further for sides divided into different ratios as shown by *Routh's Theorem (1896)*

4. Recommended reading: Y.K. Man. (2009). On Feynman's Triangle problem and the Routh Theorem. *Teaching Mathematics & its Applications*, 28(1):16-20.

5. Other references:

*One-seventh area triangle*

*Feynman's and Steiner's triangle*

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Michael de Villiers, Sept 2009.