Expanding Contexts for Teaching Upper Secondary School Geometry

Abstract

This chapter describes how the theatrical performance based on the history of mathematics—‘An Amazing Story: The Measurement of the Earth by Eratosthenes’—created the opportunity of a ‘third,’ expanding learning space, which allowed for new practices and tools to emerge. It also permitted students to approach mathematical concepts in an experiential way and (re)negotiate their own learning processes, their conceptions of mathematical Discourse, and the nature of mathematics. We analyze a one semester-long, interdisciplinary, didactical intervention for 10th grade students in a public school in Athens, where different funds of knowledge and Discourses expanded the boundaries of the official school Discourse. Our aim is to show how an experiential way of integrating the history of mathematics—a theatrical performance based on history—can create a ‘third,’ expanded learning space, where new tools and new Discourses are applied.

Appendix

In this appendix, we provide the solutions of some of the given problems above, with an analysis of students’ worksheets.

• (i) Concurrent line segments in a triangle

Let ΑΔ, ΒΕ, ΓΖ be concurrent at K. Imagine weights β _Α , β _Β and β _Γ suspended at A, B and Γ, respectively. The main idea for this solution is to choose the weights β _Α , β _Β and β _Γ so that the center of gravity of the triangle ΑΒΓ they define is on K. To this end we note that,

• (1) Since the center of gravity of ΑΒΓ lies on ΓΖ, the center of gravity of AB is at Z.

• (2) For the same reason the center of gravity of ΒΓ is at Δ.

• (3) Similarly, the center of gravity of ΑΓ lies on E.

By (1), (2), (3) and the definition of the center of gravity of two point masses, we have, respectively $\frac{{\beta_{ A} }}{{\beta_{B} }} = \frac{{\gamma_{2} }}{{\gamma_{1} }}.$, $\frac{{\beta_{B} }}{{\beta_{\varGamma } }} = \frac{{\alpha_{2} }}{{\alpha_{1} }}$, $\frac{{\beta_{\varGamma } }}{{\beta_{A} }} = \frac{{\beta_{2} }}{{\beta_{1} }}.$

Multiplying these three equations gives $\frac{{\beta_{A} }}{{\beta_{B} }} \cdot \frac{{\beta_{B} }}{{\beta_{\varGamma } }} \cdot \frac{{\beta_{\varGamma } }}{{\beta_{A} }} = \frac{{\gamma_{2} }}{{\gamma_{1} }} \cdot \frac{{\alpha_{2} }}{{\alpha_{1} }} \cdot \frac{{\beta_{2} }}{{\beta_{1} }} = 1 \Leftrightarrow {\mkern 1mu} \gamma_{2} \beta_{2} \alpha_{2} = \gamma_{1} \beta_{1} \alpha_{1}$, that is, Ceva’s theorem.

The converse theorem holds: Ιn a triangle ΑΒΓ let Δ, E and Ζ be given on its sides ΒΓ, ΑΓ and AB respectively, so that the following relation holds

$$\frac{{\gamma_{2} }}{{\gamma_{1} }} \cdot \frac{{\alpha_{2} }}{{\alpha_{1} }} \cdot \frac{{\beta_{2} }}{{\beta_{1} }} = 1$$

Then the three line segments ΑΔ, ΒΕ and ΓΖ are concurrent at a point K.

Indeed, if we choose the weights β _Α , β _Β and β _Γ so that the center of gravity of ΑΒ is at Ζ, and the center of gravity of ΒΓ is at Δ, then we have that $\frac{{\beta_{A} }}{{\beta_{B} }} = \frac{{\gamma_{2} }}{{\gamma_{1} }}$ and $\frac{{\beta_{B} }}{{\beta_{\varGamma } }} = \frac{{\alpha_{2} }}{{\alpha_{1} }}$ and the center of gravity of ΑΒΓ should be on both ΓΖ and ΑΔ; that is, at their point of intersection K.

Then by the hypothesis, we have that $\frac{{\beta_{A} }}{{\beta_{\varGamma } }} = \frac{{\gamma_{2} }}{{\gamma_{1} }} \cdot \frac{{\alpha_{2} }}{{\alpha_{1} }} = \frac{{\beta_{1} }}{{\beta_{2} }}$, which shows that the center of gravity of β _Α and β _Γ is at E. Consequently, K, the center of gravity of ΑΒΓ must be on ΒΕ as well, and therefore the three lines are concurrent at K.

The greatest difficulty faced by students during the proof of the theorem was the change of the context of the solution of the problem. The students have learnt to think of a mathematical problem only in a mathematical framework. Students were not able to prove neither the converse nor the theorem of the median, neither in the frame of mathematics (as an application of the Ceva’s theorem) nor in the physics frame (i.e. to apply to the vertices of the triangle equal weights).

• (ii) Finding the area of an irregular shape

Here are two solutions: To calculate the area of the irregular shape indirectly we can initially weigh the entire square ABΓΔ and measure its area. Then we have to cut the irregular shape of the frame and weigh it. The ratio of the weights is the ratio of the areas. We can also calculate the area by counting the number of squares contained in the irregular surface. To find the area of this irregular shape all our students chose to use millimeter-grid paper or created a grid of squares and counted the number of squares contained in the irregular surface. In some cases, students chose to split the irregular shape to other known shapes, as triangles or squares.

• (iii) Using geometry to solve a problem in Physics

To solve the above equilibrium problem students should work in a frame of geometry, after linking the mass of solids with their volumes through density. The solution is as follows: Since the objects are made from the same homogeneous material, it means that they have the same density and due to the formula d = m/V, which connects density (d), mass (m), and volume (V) of a solid, we conclude that the ratio of the volumes of the two objects is equal to the ratio of their masses.

In this equilibrium problem of solids, students were led to a geometric solution, i.e. the determination of the relation of the volumes, because the problem was given during the geometry class. Then, students calculated the volumes of the geometrical solids; they found the ratio of volumes and thus, they found the number of solids needed for the equilibrium. They were led automatically to this solution without proving it through the formula of the density of a solid body.