Problem Posing for and Through Investigations in a Dynamic Geometry Environment

Abstract

This chapter analyzes three different types of problem posing associated with geometry investigations in school mathematics, namely (a) problem posing through proving; (b) problem posing for investigation; and (c) problem posing through investigation. Mathematical investigations and problem posing which are central for activities of professional mathematicians, when integrated in school mathematics, allow teachers and students to experience meaningful mathematical activities, including the discovery of new mathematical facts when posing mathematical problems. A dynamic geometry environment (DGE) plays a special role in mathematical problem posing. I describe different types of problem posing associated with geometry investigations by using examples from a course with prospective mathematics teachers. Starting from one simple problem I invite the readers to track one particular mathematical activity in which participants arrive at least at 25 new problems through investigation in a DGE and through proving.

Appendices

Appendix A

Proof for Problem 5h (Figure 18.5)

• (1) $ET\left|\right|CA$ thus triangles CBQ and MEQ are similar;

• (2) $\frac{QE}{BQ}=\frac{5}{4}$ (2b, Figure 18.4);

• (3) From (1) and (2) $\frac{EM}{BC}=\frac{5}{4}$;

• (4) $\frac{AC}{BC}=\frac{5}{4}$ (1b, Figure 18.4) thus $CA=EM$.

• (5) Hence $EM\left|\right|CA$ and $EM=CA$; that is CEMA is a parallelogram.

Appendix B

Proof for Problems 7a, 7b (Figure 18.7)

Construction outline:

CAME is a rhombus, T-midpoint on EM, $F=AT\cap CM$

DK on $AT:D:DA=AF,\kern0.24em K:KF=FA$

Prove:

• 7a. $DC=DF\left(\iff DK=2DC\right)$

• 7b. $CAME\;\mathrm{is}\;\mathrm{a}\;\mathrm{square}\;\mathrm{when}\angle CDK=36.9{}^{\circ}$

• 7c. ${K}^{\prime}\mathrm{on}\;CE\;:{K}^{\prime }E=EC\iff {K}^{\prime}\mathrm{coincides}\;\mathrm{with}\;K$

Proof

• 1. According to the construction: $AC=CE=EM= AM=x$, $ET=TM=\frac{1}{2}x$, $FA= AD=2y$, $FK=DF=4y\Rightarrow DK=8y$;

• 2. $CAME\;\mathrm{is}\;\mathrm{a}\;\mathrm{rhombus}\Rightarrow \varDelta CAF\cong \varDelta CEF\Rightarrow FE=2y$;

• 3. MF-bisects angle $AMT,\kern0.24em AM=2MT\Rightarrow AF=FT$; $FT=y$; $TK=3y$

• 4.
  $$ \begin{array}{l}\varDelta TEK\cong \varDelta TMA;\kern0.22em AT=TK,\kern0.24em ME\left|\right|CA\Rightarrow TE\;\mathrm{midline}\;\mathrm{on}\;\\ {}\kern4.2em \varDelta\;ACK\Rightarrow EK=CE\end{array} $$

  (18.7c)
• 5.
  $$ \begin{array}{l}TE\;\mathrm{midline}\;\mathrm{on}\;\varDelta\;ACK\Rightarrow CD=2EF\Rightarrow DC=4y\\ {}\kern6em \Rightarrow DC=AF\left(\iff DK=2DC\right)\end{array} $$

  (18.7a)

• 6. $\varDelta FEK\sim \varDelta DCK\sim \varDelta DAC\Rightarrow \angle DCA=\angle CKD$

• 7.
  $$ \begin{array}{l}\mathrm{If}\; CAME\;\mathrm{is}\;\mathrm{a}\;\mathrm{square}\;\alpha =45{}^{\circ};\kern0.24em \tan \beta =\frac{1}{2}\iff \beta =26.57{}^{\circ}\iff \angle CDK\\ {}\kern5.52em =71.57{}^{\circ}\angle CDK=36.9{}^{\circ}\end{array} $$

  (18.7b)

Appendix C

Problems posed for and through investigations by a PMT who participated in the study

Rasha’s Problem

Initial problem: Midline in a triangle

The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long. (Given: $\varDelta\;ABC,\;AE= EB,\;AP=PC$; Prove: $EP\left|\right|BC,\kern0.24em EP=\frac{1}{2}BC$)

Posed problems:

Given:

$$ \varDelta ABC,\kern0.22em AE= EB,\kern0.22em AP=PC $$

$$ \begin{array}{l}PD\ \mathrm{is}\ \mathrm{a}\ \mathrm{continuation}\ \mathrm{of}EP,\kern0.24em ED=2EP,\kern0.24em F=EC\cap BP,\kern0.22em G=BD\cap AC,\\ {}\kern8.88em S=EC\cap BD,\kern0.24em O=FG\cap SP\end{array} $$

Prove:

$$ \frac{ED}{FG}=3;\kern0.36em \frac{BA}{SP}=4,\kern0.22em FO=OG,\kern0.22em OP=2 OS $$