by Hannes du Plooy (1981)

**Introduction**: One of the many aspects related to the '*processes*' of mathematics and developing genuine '*mathematical thinking*' in the classroom I've tried focusing 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. Below is a *Facebook* picture of Hannes from 2018.

We were dealing with a similarity proof of the Pythagorean theorem in class when Hannes asked the provoking question whether in general for any triangle *ABC*, a number might exist so that *AC ^{p}* =

A Generalization of the Theorem of Pythagoras by Hannes du Plooy

Basically, the generalization states that for any non-equilateral triangle *ABC* (assuming *AC* ≠ 1) there always exists a number *p* so that *AC ^{p}* =

**Theorem 1**: For any triangle *ABC* with sides *AC* > *BC* ≥ *AB* and θ opposite *AC*, there exists a number *p* > 1 so that *AC ^{p}* =

The value of

(Click on the '*Link to Theorem 2*' button in the sketch above to navigate to it.)

**Theorem 2**: For any triangle *ABC* with sides *AC* < *BC* ≤ *AB* and θ opposite *AC*, there exists a number *p* < 0 so that *AC ^{p}* =

Again the value of

(Click on the '*Link to Theorem 3 - case 1*' or '*Link to Theorem 3 - case 2*' buttons in the sketch above to navigate to two illustrative cases of this theorem.)

**Theorem 3a**: If *AB* = *BC* in Theorems 1 or 2, and only ∠*ABC* = θ is given, then *p* can be solved in closed form from: *p*(*θ*) = log 4/log (2(1 - cos (*θ*)). More-over, if *AB* ≠ *BC*, then *p* ≤ *p*(*θ*) as *θ* ≥ to 90^{o} or *p* ≥ *p*(*θ*) as *θ* ≤ 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 *WebSketchpad* sketches, with **accurate** constructions of the intersections of the given graphs, can be downloaded at: Zipped Sketchpad file.

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Michael de Villiers, created with *JavaSketchpad* on 6 February 2011; updated to *WebSketchpad* on 16 July 2022.