A Generalization of the Theorem of Pythagoras

Introduction: One of the many aspects related to the 'processes' of mathematics and developing genuine 'mathematical thinking' in the classroom I've focused on consistently over the years has been to try and develop my students' 'problem posing' abilities. By encouraging them to think divergently all the time, to reflect critically on problems and to ask 'what-if' questions themselves, has often lead to some interesting, novel investigations. This is an example of one such an investigation started by a question by a Grade 11 student of mine, Hannes du Plooy, at Diamantveld High School, Kimberley, South Africa way back in 1981.

We were dealing with a similarity proof of the Pythagorean theorem in class when Hannes asked whether in general for any triangle ABC, a number might exist so that ACp = ABp + BCp. I replied saying that it was a great question, but that I didn't know and encouraged him to go and investigate. This is the result of that investigation, with some assistance by myself (see De Villiers & Du Plooy, 1982). It should be noted that the original investigation was done in the days before we had access to graphic calculators or any graphing programs on computer: instead a computer program was written for the necessary iterative calculations using an HP 33E. This episode always serves to remind me how creative students can be if encouraged and allowed to think divergently!

Please install Java (version 1.4 or later) to use JavaSketchpad applets.

A Generalization of the Theorem of Pythagoras: Theorem 1

Theorem 1: For any triangle ABC with sides AC > BC >= AB there exists a number p > 1 so that ACp = ABp + BCp, and p is smaller than or equal to 2 as angle ABC is greater than or equal to 90o or p is greater than or equal to 2 as angle ABC is smaller than or equal to 90o. (Proof follows from the Intermediate Value Theorem).

The value of p can easily be found by drawing the graphs of g(x) = log(ABx + BCx)/log(AC) and h(x) = x as shown, then the x-value of the intersection of the two graphs equals p.

1) Drag point D to the intersection of the two graphs to find p = x[D] as shown by the measurement in green. 2) Drag point C to change the triangle. 3) Drag point E to change angle ABC while maintaining the condition of the theorem, and repeat Step 1.

Please install Java (version 1.4 or later) to use JavaSketchpad applets.

A Generalization of the Theorem of Pythagoras: Theorem 2

Theorem 2: For any triangle ABC with sides AC < BC =< AB there exists a number p < 0 so that ACp = ABp + BCp. (Proof follows from the Intermediate Value Theorem).

The value of p can easily be found by drawing the graphs of g(x) = log(ABx + BCx)/log(AC) and h(x) = x as shown, then the x-value of the intersection of the two graphs equals p.

1) Drag point D to the intersection of the two graphs to find p = x[D] as shown by the measurement in green. 2) Drag point A to change the triangle, and repeat Step 1. 3) Drag point E to change angle ABC while maintaining the condition of the theorem, and repeat Step 1.

Theorem 3a: If AB = BC in Theorems 1 or 2 and only angle ABC is given, then p can be solved in closed form from: p(ABC) = log 4/log (2(1 - cos (ABC)). More-over, if AB is not equal to BC, then p is smaller than or equal to p(ABC) as angle ABC is greater than or equal to 90o or p is greater than or equal to p(ABC) as angle ABC is less than or equal to 90o.

Theorem 3b: If AB = BC in Theorems 1 or 2 and AB and BC is given, then p can be solved in closed form from: p = log 2/log (AC/AB).

Reference: De Villiers, M. & Du Plooy, H. (1982). 'n Veralgemening van Pythagoras se Stelling. Spectrum, 20(3), Okt, pp. 27-30.

Note: A Sketchpad 5 version of the above Java sketches, in which the intersection of the graphs is accurately constructed, and Theorem 3 is also illustrated, is available for downloading at Sketchpad Exchange at: A generalization of Pythagoras.


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Michael de Villiers, 6 February 2011.