“Can this happen only in Japan?”: mathematics teachers reflect on a videotaped lesson in a cross-cultural context

Abstract

This paper reports on a study conducted within VIDEO-LM (Viewing, Investigating and Discussing Environments of Learning Mathematics), a video-based professional development project for secondary mathematics teachers that aims to enhance reflection on practice. The study explored VIDEO-LM sessions where Israeli teachers watched an 8th grade geometry lesson from Japan. We analyzed teachers’ discussions around this lesson and characterized main themes and reflection types. Findings suggest that watching the Japanese lesson, using the VIDEO-LM framework, led to productive discussions among teachers. The fact that the videotaped lesson was culturally “far” from the teachers’ reality provoked them to re-see everyday aspects of their practice, revealing some dissonances which promoted reflective thoughts and actions.

Appendices

Appendix 1: The six-lens framework (SLF)

See Fig. 7.

Fig. 7

Description of SLF and its use in PD sessions (Karsenty 2018a)

Appendix 2: The lesson graph

See Fig. 8.

Fig. 8

Lesson graph of the Japanese lesson “Changing Shape without Changing Area” (adapted from https://www.timssvideo.com/67)

Appendix 3: Teachers’ proof by contradiction that the midline is not a solution

Let AC be the line through the midpoints of BE and DE (see Fig. 9); we shall prove that AC is not a solution for the task, that is,$S_{{\Delta {\text{BAF}}}} + S_{{\Delta {\text{CDG}}}} \ne S_{{\Delta {\text{EFG}}}}$ (we can think of it as compensation: if AC is a solution then the areas should “compensate” each other, hence it suffices to show that these areas are not equal). EH is a constructed parallel line to AB, through the point E. EH divides $\Delta {\text{EFG}}$ into two triangles: $\Delta {\text{EFH}}$ and $\Delta {\text{EGH}}$. The triangles $\Delta {\text{BAF}}$ and $\Delta {\text{EHF}}$ are congruent hence they have the same area: $S_{1} = S_{2}$. Now, we can draw a parallel line to EH through the point D, namely DI (see right-hand side of Fig. 9). These parallels delineate another pair of congruent triangles which have the same area: $T_{1} = T_{2}$. Therefore, the area of triangle $\Delta {\text{CDG}}$ is larger than the area of triangle $\Delta {\text{EGH}}$, thus $S_{{\Delta {\text{BAF}}}} + S_{{\Delta {\text{CDG}}}} > S_{{\Delta {\text{EFG}}}}$. It follows that the midline is the solution if and only if the top and bottom lines are parallel.

Fig. 9

Diagrams illustrating the teachers’ proof