Alternative Geometric Constructions: Promoting Mathematical Reasoning

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1 Eric. Pandiscio lternative Geometric onstructions: Promoting athematical Reasoning The use of multiple tools highlights the connections among ideas onstruction tools in most high school Euclidean geometry classes have typically been limited to a compass for drawing circular arcs and a straightedge for drawing segments. The strengths of these tools include both mathematical precision and a long history of use. However, alternatives can provide fresh possibilities for engaging students in the mathematical reasoning that lies at the heart of traditional geometry (Gibb 1982; Robertson 1986). This article proposes that a single task completed with a variety of construction tools fosters a greater sense of mathematical contemplation than multiple tasks done with the same tool. The premises are simple: each tool fosters different mathematical ideas, and using multiple tools not only requires understanding of a greater breadth and depth of geometric concepts but also highlights the connections that exist among different ideas. For constructions in this article, we do not use compass and straightedge but instead use the following: a ira for constructing images of a reflection; a three-by-five-inch card for drawing right angles and transferring lengths; and a twoedged straightedge (TESE) for drawing parallel s a fixed distance apart. Each tool has different strengths and weaknesses. construction that is easy with one tool may be difficult with another. Since techniques that work well with one tool often work poorly with another tool, the user should find strategies that take advantage of a tool s strengths. given construction may vary in complexity depending on the tool used and the techniques required. This article presents three construction tasks that can be assigned to secondary geometry students and then gives solutions that use each tool. Such mathematical tasks accomplish many goals, among them encouraging students to construct representations of... geometric objects using a variety of tools (NT 2000, p. 308) and to link mathematical ideas as a way to develop robust understandings of problems (NT 2000, p. 354). The reader is welcome to try each challenge with all three tools and to consider how to support students as different geometric relationships become apparent when each tool is used. We begin by giving a brief description of how each tool works. THE THREE ONSTRUTION TOOLS ira The ira is a commercially available piece of semitransparent plastic that generates reflected images of geometric objects. onstructions that are straightforward with this tool include right angles, perpendicular bisectors, and angle bisectors. The ira helps show the power of geometric transformations and throws a different twist on construction problems. The ira can also be used to draw a, called the ira, along its beveled straight edge. The ira is becoming popular in middle-level classrooms, but it also has legitimate uses in a high school geometry course. One compelling reason for encouraging students to use the ira is that it is a good tool for exploratory construction activities. The ira generates reflective images that move when the ira is moved, thereby stimulating students visual imagination and suggesting possible solution strategies. The advantage of the ira over a mirror is that, given point P, the student can reach around the ira and mark the apparent location of P', the reflection image of P, on the paper. The ira should be the perpendicular bisector of segment PP' when the beveled edge of the ira is touching the paper and facing the user. The optical center of the ira is then on the ira. Right-angle source (a 3 5 index card) n index card is useful for drawing right angles and testing for perpendicularity. With this rightangle source, a student can construct altitudes and segments and can inscribe right triangles in Eric Pandiscio, eric.pandiscio@umit.maine.edu, teaches at the University of aine, Orono, E He is interested in geometric problem solving, understanding proof, and proportional reasoning. 32 THETIS TEHER opyright 2002 The National ouncil of Teachers of athematics, Inc. ll rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NT.

2 circles. This tool encourages students to use right triangles creatively. The card may be marked, thereby allowing the transfer of lengths, but students should not fold, tear, or otherwise alter it. Two-edged straightedge (TESE) This tool draws parallel s a fixed distance, the width of the straightedge, apart. For the constructions in this article, the corners of this tool may not be used as a right-angle source. This tool is probably the least familiar to students and readers and its use may require creative thinking to be valuable. traditional ruler, without the use of any measures, can serve as a TESE. ONSTRUTIONS USING EH TOOL onstruction 1 Given a segment, construct its perpendicular bisector. This construction is useful any time that we need to find the midpoint of a segment. With a ira. Given segment, we place the ira across segment so that the image ' of is on top of, as shown in figure 1. ira m is the perpendicular bisector of. Fig. 2 =. Line is perpendicular to at midpoint. = ' edges, and repeat with the role of the edges reversed, as shown in figure 3. This process produces a rhombus with diagonal. The other diagonal,, determines the perpendicular bisector of. The reader is asked to justify that is indeed a rhombus. The task can be used to introduce fundamentally important properties within a problemsolving context ira = ' Fig. 1 ira is the perpendicular bisector of. Fig. 3 is a rhombus. iagonal is the perpendicular bisector of diagonal. With a 3 5 card. Given segment, we place a mark on one edge of the card and use the marked card to construct congruent, perpendicular segments and on opposite sides of, as illustrated in figure 2. Line segment intersects at its midpoint,. Triangles and are congruent by the S congruence test. We use the card to draw the perpendicular to at, calling it l. With a TESE. Given segment, we place the TESE so that one edge goes through and the other edge goes through, draw s along the two In this construction, the ira solution is based on a simple reflection, the index card uses congruent triangles, and the TESE highlights the properties of a rhombus. The solution using the rhombus can be used either to reinforce ideas that students have previously learned or to introduce fundamentally important properties within a problem-solving context. This latter approach is consistent with recommendations given by the National ouncil of Teachers of athematics for connecting problem solving with content (NT 1989, 2000). Vol. 95, No. 1 January

3 The three tools and the solution strategies still work if angle is obtuse onstruction 2 Given two adjacent sides, construct a parallelogram. With a ira. Given segments and with common point, we let be opposite the missing vertex, as shown in figure 4. We know that the intersection of the diagonals is the center of the parallelogram and that the diagonals bisect each other. The construction first finds the midpoint of using ira m 1. We next reflect onto its extension on the other side of across ira m 2. The reflection ' of is the required fourth vertex. ' copy of triangle, with at and ' along. Point, which corresponds to, is the fourth vertex of the parallelogram. n alternative and perhaps simpler but not necessarily as creative method would be to construct a parallel to through by constructing a perpendicular to through and a perpendicular to the perpendicular through. We repeat to construct a parallel to through. With a TESE. We use the TESE to construct three parallel s, with the first two going through and. We let be the intersection of ray with the third parallel and note that is the midpoint of. We repeat the process to find E on ray, with the midpoint of E, as shown in figure 6. We use the TESE to construct the midpoint,, of E using the midpoint technique shown in construction 1. Quadrilateral is the required parallelogram, and is parallel to. We note that,, and are the midpoints of the sides of triangle E. Fig. 4 is the midpoint of both and '. iagonals of a parallelogram bisect each other. Thus ' is the fourth vertex. 1 2 E E With a 3 5 card. One approach is a cut-andpaste method that turns a parallelogram into a rectangle. We use our index card to drop a perpendicular from, with the shorter side of the parallelogram, to side, as shown in figure 5. We let be the foot of the perpendicular and mark the locations of vertices and of right triangle on the sides of a card. We draw triangle ', a E Fig. 6 ecause of the triangle midpoint theorem, is parallel to E and is parallel to. card position 1 card position 2 L ' Fig. 5 Transferring of triangle to a new location highlights the relationship between parallelograms and rectangles. The solutions given for construction 2 all assume that the original angle is acute. However, the three tools and all the solution strategies still work if angle is obtuse. The most significant adaptation is that with the index card, the sides of the original angle must be extended as auxiliary s past the vertex angle to find the correct placement of the card. The reader may wish to try such an example. onstruction 3 Given a side, construct an equilateral triangle. 34 THETIS TEHER

4 With a ira. We know that the perpendicular bisectors of the sides of an equilateral triangle are axes of reflective symmetry. First, we construct the perpendicular bisector m 1 of. Then we place the ira on and rotate it until the image ' of lies on m 1. The second ira is m 2, as shown in figure 7. Since m 1 and m 2 are perpendicular bisectors of the segments and ', respectively, triangle ' is isosceles with base and with base ', so ' = ' and = '. Thus, triangle ' is equilateral. ' 1 lateral triangle, we construct the perpendicular bisector of, as demonstrated in construction 1, figure 3. The third vertex of the equilateral triangle lies on this perpendicular bisector; the challenge is to find the specific location. To accomplish that task, we must create a segment that forms a 60 degree angle with segment. The point of intersection of this new segment and the perpendicular bisector of, m, gives the third vertex of the equilateral triangle. We construct a perpendicular to at point by first constructing three parallel s to find points and N along <> with as the midpoint. We construct the perpendicular bisector of N, which gives, perpendicular to, as shown in figure 9. 2 Fig. 7 m 1 is the perpendicular bisector of. m 2 is the perpendicular bisector of '. With a 3 5 card. We construct the perpendicular bisector, k, of segment, as in construction 1 and mark a copy of on our index card s edge. We place the card so that one mark is at and the other mark is at, a point on the perpendicular bisector. See figure 8. Triangle is isosceles, with its base congruent to. The triangle is therefore equilateral. Line N Fig. 9 is the perpendicular bisector of ; is perpendicular to. With a TESE. This task is a challenging one. Given segment as the side of the desired equi- k Fig. 8 k is the perpendicular bisector of ; = ;. Using the straightedge, we mark two points on this new perpendicular, one of which is the rule s width away from and the other of which is twice the width of the rule away from, and call these points and, respectively. We notice that is the midpoint of segment and use the perpendicularbisector technique to construct rhombus FE. ngle E is 60 degrees. We extend E> until it intersects m at G, as shown in figure 10. Triangle G is the equilateral triangle. The big question is, Why is the measure of angle E 60 degrees? This question forms the crux of the construction. In some ways, this proof is deceptively simple and elegant. We construct segment EF, producing triangle FE, which is equilateral because the three altitudes of FE are all equal, since they were constructed to be exactly one rule s width in length. Ray bisects angle EF; therefore, the measure of angle E equals 30 degrees. Since angles E and E are complementary, the This proof is deceptively simple and elegant Vol. 95, No. 1 January

5 G onstructions promote a spirit of exploration and discovery E Fig. 10 easure of angle E = 60 degrees; =. F E I Fig. 11 ltitudes FH, EI, and are all one rule width in length; triangle FE is equilateral; the measure of angle E is 30 degrees; and the measure of angle E is 60 degrees. H F measure of angle E equals 60 degrees. See figure 11. This construction is by no means simple or straightforward. However, numerous powerful geometric ideas can be either learned or reinforced in such a legitimate problem-solving context. ISUSSION The use of constructions in a geometry class can be justified on many grounds. mong the arguments is the position that [c]onstructions can reinforce proof and lend visual clarity to many geometric relationships (Sanders 1998, p. 554). onstructions, according to Robertson (1986, p. 380) give the secondary school student, starved for a Piagetian concrete-operational experience, something tangible. In addition, unless constructions simply ask students to mimic a given example, they promote true problem solving through the use of reasoning. Finally, constructions promote a spirit of exploration and discovery and can be guided to the extent that the teacher desires. This article has presented a detailed mathematical look at constructions through taking seemingly routine tasks and making them challenging and motivating by requiring that students solve them in three unique ways with three distinct tools. y demanding that students expand their outlook on solution strategies, we are helping them to both broaden and deepen their mathematical perspective. REFERENES Gibb, llan. Giving Students an dded Edge in onstructions. athematics Teacher 75 (pril 1982): National ouncil of Teachers of athematics (NT). urriculum and Evaluation Standards for School athematics. Reston, Va.: NT, Principles and Standards for School athematics. Reston, Va.: NT, Robertson, Jack. Geometric onstructions Using Hinged irrors. athematics Teacher 79 (ay 1986): Sanders, athleen V. Sharing Teaching Ideas: Geometric onstructions: Visualizing and Understanding Geometry. athematics Teacher 91 (October 1998): THETIS TEHER