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 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 calculator. This episode always serves to remind me how creative students can be if encouraged and allowed to think divergently!
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 ACp = ABp + BCp. In the event that AC = 1, the triangle can simply be scaled up to a triangle of similar shape so that AC ≠ 1. More specific details below.
Theorem 1: For any triangle ABC with sides AC > BC ≥ AB and θ opposite AC, there exists a number p > 1 so that ACp = ABp + BCp, and p ≤ 2 as θ ≥ 90o or p ≥ 2 as θ ≤ 90o. (Existence 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 P of the two graphs equals the required value of p.
(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 ACp = ABp + BCp. (Existence proof follows from the Intermediate Value Theorem).
Again 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 P of the two graphs equals the required value of p.
(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 90o or p ≥ p(θ) as θ ≤ 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).
De Villiers, M. & Du Plooy, H. (1982). 'n Veralgemening van Pythagoras se Stelling. Spectrum, 20(3), Okt, pp. 27-30.
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.
Michael de Villiers, created with JavaSketchpad on 6 February 2011; updated to WebSketchpad on 16 July 2022.