**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 *AC ^{p}* =

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 *AC ^{p}* =

The value of *p* can easily be found by drawing the graphs of *g*(*x*) = log(*AB ^{x}* +

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.

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 *AC ^{p}* =

The value of *p* can easily be found by drawing the graphs of *g*(*x*) = log(*AB ^{x}* +

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 90^{o} or *p* is greater than or equal to *p*(*ABC*) as angle *ABC* is less than or equal to 90^{o}.

**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.*

This page uses

Michael de Villiers, 6 February 2011.