Hee-chan Lew Korea National University of Education Santiago, Chile 24 August, 2013

1. Working Backward Strategy for increasing students' deductive reasoning skills in solving geometric problems Hee-chanLew Korea National University of Education Santiago, Chile 24 August, 2013

2. Korea • Korea is a small country in East Asia. But, we have the long history of more than 4000 years. • It has been surrounded by China, Russia, Japan. • Korean Peninsula has been divided since 1950: Northand South.

3. Current Issues in Mathematics Education • Korean students have demonstrated a high level of achievement in TIMSS and PISA for a long time: • TIMSS 2003, 2007, 2011, • PISA 2003, 2006, 2009 • Despite of brilliant scores, the reports of TIMSS and PISA show the some negative results: • Lack of students’ mathematical attitude • Achievement gap induced by school locations.

4. Lack of Mathematics Attitude TIMSS 2011 Mathematical Attitude (4th Grade) TIMSS 2011 Mathematical Attitude (8th Grade)

5. Achievement Gap PISA surveys shows that there were a big gap among 5 areas of village, small town, town, city and large city classified by number of their populations.

6. Why? • Students think that mathematics is meaningless and is not important in their life, therefore we do not need to study mathematics hard. • But, how the city students’ score is so high? • It is because parents push their students to study hard.

7. Deductive Reasoning • Deductive reasoning (or proof) is a process to deduce a new result from assumptions existed in the problem, axioms, what was previously proven, etc. • Since the 6th century BC, it has been the flower of mathematics and a mark to distinguish mathematics from other science.

8. Euclid’s Elements • Euclid’s Elements written in BC 3rd century has been used as a textbook to develop students’ deductive reasoningover 2000 years. • But, the axiomatic method used in Euclid’s Elements has been criticized for a long time. • No mathematical activities such as imagination, experiment, guess, trial and error, and mistake etc. (Clairaut, 1741; Lakatos, 1976; Vincent, 2005).

9. Current textbook • The proof in current textbooks shows only final results of proving activities. • Teachers explain the results step by step very kindly but, do not guide students to find the solution and to explain why the solution works by themselves. • This is not honest

10. The Result of not honest way? • Students have few meaning in the proof and lose ability and confidence in constructing proof eventually. • I will show three examples that mathematics is treated imprudentlyfrom the thee levels of school: • One from the university level • One from Junior high school level • One from elementary level

11. Fermat Point • Fermat point is a point of a triangle in which the sum of lengths of three segments is minimum.

12. Procedure to make Fermat Point • Textbook introduce the procedure to find the Fermat Point.

13. Procedure to prove Fermat Point(1) • Then, textbook introduces the procedure to prove that the three lines make one point.

14. Procedure to prove Fermat Point(2) • Then textbook introduces the procedure that the sum of lengths of three segments is minimum

15. Tangent Line • Draw a tangent line of Circle A through point B

16. Procedure to draw the tangent line

17. Trisecting a Square Equally • Divide a square into three equal parts by using the method to divide a square into two equal parts and justify their way.

18. Procedure to trisect a square

19. Purpose of this talk • It is to provide an "honest" learning environment to teach deductive reasoningfor secondary students (9 & 10 graders). • This talkshows that the "analysis" method systemized by Greek mathematician Pappus in AD 3rd century might be the answer.

20. New strategy • In order to improve deductive reasoningability and confidence, an "active justification" to find the heuristics for solution and to explainthe reason by students themselves should be required rather than a "passive justification" to accept or to follow teacher’s explanation or persuasion • This talk introduces one possible strategy for active justification: "Analysis"with dynamic geometry software.

21. Analysis • The analysis method which is the oldest mathematics heuristics in the history of mathematics assumes “what is sought as if it were already done and inquire what it is from which this results and again what is the antecedent cause of the latter and so on, until by so retracing the steps coming up something already known or belonging to the class of first principles.” (Hearth, 1981, p.400)

22. Synthesis • The synthesis as the reverse of the analysis takes as already done that which was last arrived at in the analysis and arrives finally at the construction of what was sought by arranging in their natural order as consequences what before were antecedents and successively connecting them one with another.

23. Dialectic integration • Greek mathematicians thought the dialectic integration of analysis and synthesis as a substance of mathematical thought. • However, Euclid’s Elements considered only synthesis to reduce theorems from the foundation likeaxioms and the givens as a way to guarantee the truth of mathematics.

24. Design of an instructional scheme • Problem situation • There are two kinds of geometry problems: Proof problem or construction problem: • Prove something A under the condition(s) of B. • Construct something A under the condition(s) of B • In order to deductive reasoning ability, we need to start at the proper problem situation • Depending on the kind of problem, two kinds of different analysis are used.

25. Twokinds of analysis • The first is for proof problem. It is to find justification process by getting a series of previous sufficient conditions of the conclusion to be deduced under the assumption that what is required to be deduced is already done. • The second is for construction problem. This is to find the construction process by getting a series of necessary conditions from the assumption that what is required to be constructed is already constructed.

26. Design of an instructional scheme • Four phase problem solving process • First is “understanding” phase • Second is “analysis” phase to assume what to be solved is done and to find the construction ideas by a series of necessary conditions • Third is “synthesis” phase to construct a deductive proof as a reverse of the analysis • Finally, “reflection” phase to reflect on whole problem solving process.

27. Design of an instructional scheme • Dynamic geometry • Analysis method is very difficult to be applied in the paper and pencil environment because various dynamic operations are required. • It might be because of the lack of proper dynamic tools that the analysis known well by Greek mathematicians has not emphasized in schools since Greek era.

28. Teacher’s roles • Technical assistant: provide students information about mathematical knowledge and computer environment • Counselor: advise some ideas in solving process • Collaborator: cooperate with students in solving process

29. Draw a tangent line of Circle A through point B Understanding

30. Assume that line BC is a tangent line of the circle A Analysis

31. Draw segments AC and AB Triangle ABC is a right triangle Let D is a midpoint of the segment AB DC = DA = DB Analysis

32. Draw a segment AB and let D is a midpoint of AB Draw a circle D with radius DA and let C is an intersection point of the two circles BC is the tangent line to find Synthesis

33. Reflection • Let students to find another method. • Let them to write the construction process on the paper • Let them to make an encapsulization of the whole process. • Let them to test the construction process by dragging the constructed object • Let them to discuss something difficult during their problem solving process

34. Construct an inscribed circle to the fan shaped figure OAB Understanding

35. Assume that circle O is the inscribed circle. Analysis

36. Let point F be a tangent point of the circle E and the arc AB. Draw a perpendicular line to the radius OF at Point F Analysis

37. Extend line OB to make line OB1 Extend line OA to make line OA1 Then the circle E is an inscribed circle of the triangle OA1B1 Now, students can find the way to draw the inscribed circle Analysis

38. Draw line OC to bisect angle AOB. Let C be the intersection point of line OC and arc AB At C, draw the perpendicular line to line OC Extend line OB Let D be the intersection point of line OB and the perpendicular line Synthesis

39. At D draw the line to bisect angle ODC. Let E be the intersection point of line OC and the angle bisector at D Draw a circle with radius CE Synthesis

40. Reflection

41. There is a triangle ABC. Make three equilateral triangles ABD, BCE, AFC by using each side of the given triangle ABC. Then prove that quadrilateral BEFD is a parallelogram Understanding

42. Analysis • To prove that quadrilateral BEFD is a parallelogram we have to show DF=BE and BD=FE • To prove that DF=BE, we have to find two congruent triangles of which one side is DF and BE respectively. But, only △ADF and △BCE are triangle having DF and BE as a side. But, these are not congruent

43. Analysis • How to find the two triangles? • Here teachershave to remind students that BE=BC • Students can find the two triangles, △ADF and △ABC, in which DF=BC=BE. • We have to prove that DF=BC

44. Analysis • To prove that DF=BC, we have to show that △ADF and △ABC are congruent. • Why? • Students can find the reason based on the fact that DA=BA, AF=AC, ∠DAF=∠BAC. The reason comes from that △ADF and △ABC are equilateral.

45. 1. △ABD, △BCE, △AFC are equilateral triangles 2. ∠FCA and ∠ECB are 60° and ∠FCB is common (∠DAB and ∠FAC are 60° and ∠FAB is common) 3.∠ACB = ∠FCE and AC = FC, BC = EC (∠ BAC =∠ DAF and AC = AF, AB = AD) Synthesis

46. 4. △ABC ≡ △FEC (△ABC ≡△ADF) 5. AB = FE (DF=BC) 6. BD = FE (DF=BE) 7. Quadrilateral BEFD is a parallelogram Synthesis

47. Trisecting a Square Equally • Divide a square into three equal parts by using the easy method to divide a square into two equal parts and justify their way.

48. Analysis • Let’s assume that E is the trisecting point of AD. It is important that E is the assumed point rather than a final mathematical solution.

49. Analysis • The ratio value of EG and GFG is 1: 2. Intuitively this is clear to students. They can use 180 degree rotation of △MEG and translation of the rotated triangle.

50. Analysis • On the other hand, let H be a point of the intersection of EF and DB. As a similar reason as the previous activity, students can understand that H is a point to divide EF is the ration of 1:2.