Combining Geometrical Transformations: A Meta-mathematical Narrative

Abstract

A fictional narrative format between a teacher and a mathematics educator is used to introduce pedagogical considerations around the theme of combining geometrical transformations. Questions about scalings from different centres, scaling from and rotating about different centres, and rotations about different centres are brought up and used to illustrate ways of working with students. The mathematical results are applied to other geometrical situations.

Appendix

Theorem

In the diagram, $\frac{\overline{AR}}{\overline{AB}}\times \frac{\overline{CX}}{CR}=\frac{\overline{QX}}{QB}$ . Here Band C are the centres for scalings of A.

Proof

The idea is to use the fact that areas of triangles which have a common vertex and whose edges opposite that vertex lie on a common line have their areas proportional to their ‘bases’. This makes it possible to convert ratios of lengths into ratios of areas, and since areas can be added and subtracted, ratios can be manipulated. For manipulating ratios, the following lemma is crucial.

Lemma

If $\frac{a}{b}=\frac{p}{q}$ then $\frac{\lambda a+\mu p}{\lambda b+\mu q}=\frac{a}{b}=\frac{p}{q}$ as long as λb + μq ≠ 0.

Proof

Let the common ratio be t. Then a = bt and p = qt so λa + μp = (λb + μq)t from which the lemma follows.//

In order to get started it is important to become familiar with picking out the sub-configurations that are needed in the reasoning. The key sub-configuration is shown in the first diagram in Fig. 11. Looking at the second diagram, which is the diagram for the theorem, the sub-configurations are only available when extra lines are drawn in. These are ‘present in the imagination even though not present in the actual diagram (Fig. 12).

Fig. 12

(a) is the configuration of two triangles with a common vertex; (b) is the diagram for the theorem; (c) shows the extra lines needed

Preparatory Task

How many different versions of the configuration in the first diagram can be found in the third diagram?

Of course what matters is the experience of looking, not the actual count.

Proof of Theorem

$$ \frac{\overline{AR}}{\overline{AB}}=\frac{\varDelta ARC}{\varDelta ABC}=\frac{\varDelta ARX}{\varDelta ABX}=\frac{\varDelta ARC-\varDelta ARX}{\varDelta ABC-\varDelta ABX}=\frac{\varDelta AXC}{\varDelta ABC-\varDelta ABX} $$

and

$$ \frac{\overline{CX}}{\overline{CR}}=\frac{\varDelta AXC}{\varDelta ARC}=\frac{\varDelta XBC}{\varDelta RBC}=\frac{\varDelta AXC+\varDelta XBC}{\varDelta ARC+\varDelta RBC}=\frac{\varDelta ABC-\varDelta ABX}{\varDelta ABC} $$

But

$$ \frac{\overline{QX}}{\overline{QB}}=\frac{\varDelta AXQ}{\varDelta ABQ}=\frac{\varDelta QXC}{\varDelta QBC}=\frac{\varDelta AXQ+\varDelta QXC}{\varDelta ABQ+\varDelta QBC}=\frac{\varDelta AXC}{\varDelta ABC}=\frac{\overline{AR}}{\overline{AB}}\times \frac{\overline{CX}}{\overline{CR}} $$

//.

A similar style of reasoning can be used with this diagram to prove that $\frac{\overline{AR}}{RB}+\frac{\overline{AQ}}{\overline{QC}}=\frac{\overline{AX}}{\overline{XP}}$ and that$\frac{\overline{AR}}{\overline{RB}}\times \frac{\overline{BP}}{\overline{PC}}\times \frac{\overline{CQ}}{\overline{QA}}=1$ Which is Ceva’s theorem. How many other similar ratio sums and ratio products can you find in the diagram ?