By Michael Sewell
21 July 2012
I have been devising such material since 2000, and teaching it to selected pupils at Bisham School near Maidenhead. There is a privately printed version (100 A4 pages), and a publisher will be sought. I spoke briefly about a few of these topics at The Royal Institution Mathematics Masterclasses Organisers Conference on 22 June 2012. This note describes two of the items.
Pythagorus Proof
The following proof of Pythagoras’ Theorem was novel to me when I learned it (during a recent public lecture by John Ockendon), but I do not know who was the originator. Any right-angled triangle ABC with the right angle at B can be subdivided into two triangles AHB and BHC “similar” to it by “dropping” the perpendicular BH from the right angle at B to the hypotenuse at H. All three triangles ABC, AHB and BHC are “similar” because they contain the same trio of angles. Because of this similarity, either one of these acute angles (say HAB =HBC =γ) induces an (otherwise unspecified) non-zero function f(γ) which allows the areas of the three triangles to be expressed as (AC)2f(γ), (AB)2f(γ) and (BC)2f(γ) (i.e. the square of the hypotenuse multiplied by this same function in all three cases). Thus (AC)2f(γ) = (AB)2f(γ) + (BC)2f(γ) because triangle BAC contains HAB with HBC. Pythagoras’ Theorem follows because f(γ) ≠ 0.
The second item is a proof that a photograph of an arc of a rainbow is not necessarily an arc of a circle. I used a photograph (by Robert Peel of South West News Service, published in The Daily Telegraph) of a ground-to-ground arc over Malmesbury. Three points are marked on the yellow arc, and the perpendicular bisectors of their joins intersect in a point which serves as the centre of a circle passing through those three points on the yellow arc. But that circle can be seen to deviate from the yellow arc towards the two ends of the rainbow. That proves that this photographic image of a rainbow is not a circular arc.