Trisecting an Angle and Doubling the Cube Using Origami Method

広島経済大学研究論集第38巻第 4 号 2016年 3 月 Note Trisecting an Angle and Doubling the Cube Using Origami Method Kenji Hiraoka* and Laura Kokot** meaning to fold, and kami meaning paper, refers 1. Introduction to the traditional art of making various attractive Laura Kokot, one of the authors, Mathematics and decorative figures using only one piece of and Computer Science teacher in the High school square sheet of paper. This art is very popular, of Mate Blažine Labin, Croatia came to Nagasaki, not only in Japan, but also in other countries all Japan in October 2014, for the teacher training over the world and everyone knows about the program at the Nagasaki University as a MEXT paper crane which became the international (Ministry of education, culture, sports, science symbol of peace. Origami as a form is continu- and technology) scholar. Her training program ously evolving and nowadays a lot of other was conducted at the Faculty of Education, possibilities and benefits of origami are being Nagasaki University in the field of Mathematics recognized. For example in education and other. Education under Professor Hiraoka Kenji. The goal of this paper is to research and For every teacher it is important that pupils learn more about geometric constructions by in his class understand and learn the material as using origami method and its properties as an easily as possible and he will try to find the best alternative approach to learning and teaching pedagogical approaches in his teaching. It is not high school geometry. Use of origami in learning always easy to motivate pupils and to get them Mathematics allows pupils much more visualiza- interested in the materials, especially when teach- tion and requires logical thinking and creativity. ing Mathematics.Unfortunately, in Croatia when There are variety of origami geometric construc- teaching Mathematics it is uncommon to use a tions, however in this paper are shown two par- paper as a teaching tool. Therefore, applying ticulary appealing examples. origami method in teaching and learning geom- Trisecting an angle and Doubling the cube, etry, as a new, creative method, can increase two ancient Greek problems, are chosen because pupil’s motivation and their interest in the lec- throughout the history many mathematicians tures. tried to solve them, however in the end it was Origami, meaning paper folding, from oru proven that they can not be solved using only a compass and an unmarked straightedge. These * Professor, Hiroshima University of Economics, geometric constructions are interesting due to Hiroshima, Japan their simplicity and practical use in classroom. It ** Mathematics and Computer Science Teacher, High School of Mate Blažine Labin, Croatia is fascinating that these constructions can be 168 広島経済大学研究論集第38巻第 4 号 easily achieved using origami method therefore vertex D and mark that point as G. This crease the procedure of both geometric constructions is intersects the side BC in the point H. See Fig. 1 b). included in this paper. Now, let us see why is ∠GDE = 60° and ∠CDG = 30°. 2. Trisecting An Angle See Fig. 1 b). In the DEG we can see that Trisecting angles by Euclidean methods is ∠DEG is a right angle and since DE is half of the one of the unsolved problems of Greek antiquity. side DA (because of the first fold) and GD = CD, It is proven that angle trisection is impossible the hypotenuse GD is twice the lenght of the side using only an unmarked straightedge and a DE. Therefore, ∠GDE = 60° and ∠CDG = ∠CDE – compass, but we can trisect angles using paper ∠GDE = 90° – 60° = 30°. folding (origami method). 2.2 Trisecting an acute angle1) 2.1 Trisecting a right angle1) This paper method of trisecting an acute The paper method of trisecting a right angle angle was developed by Hisashi Abe. is possible by folding angles of 30° and 60°. In Let us take a square piece of paper and mark this trisection we will use the fact that any altitude the folding square as ABCD. Then we will trans- divides an equilateral triangle into two right fer the acute angle θ such that θ = ∠EAB, where angled triangles with angles 30°, 60° and 90° and E is a point on the side BC or on the side CD the hypotenuse of each such triangle is twice the different from B and different from D. Look at lenght of its shortest leg. the Fig. 2 a). Let us take a square piece of paper and mark We will make two creases FF and GG paral- the folding square as ABCD. We have a right lel to side AB, such that the points F and G lie angle θ such that θ = ∠CDA. Let us fold upper on the side DA (F ≠ A), the point G is a midpoint edge to lower edge (a book fold) and unfold it. of FA and the points F and G lie on the side BC See Fig. 1 a). Let us mark the middle point of the (Fig. 2 b)). sides DA and BC as E and F respectively. Now Now let us fold the square such that the point we will fold the vertex C and place it onto the F comes onto point F′ on the AE and the point middle line EF such that crease passes through A comes onto point A′ on the GG (Fig. 3 a)). The a) b) Fig. 1 Trisecting an Angle and Doubling the Cube Using Origami Method 169 a) b) Fig. 2 a) b) Fig. 3 point G is simultaneously folded onto point G′ nition of the fold. We also know that GA′ folds to and the lines AA′ and AG′ arerisectors of θ. See G′A so AG′ is a third altitude and passes through J. Fig. 3 b). Let us see why. We have to prove that θ = 3α. We know that the gained crease cannot be Since IA = A′I therefore, AA′I is a isosce- parallel to AB since F′ would then lie on AD and les triangle and ∠HIA = ∠A′IH. If we mark angle that contradicts our assumption that E is a point ∠HIA and ∠A′IH with α we have ∠G′AB = ∠A′IA = on the side CD different from D. Similary, the 2α (because we know that AG′ is orthogonal to crease cannot be parallel to AD since A′ would A′I and AB to IA. then lie on AB and that would mean AB = GG, therefore AB = FF which contradicts F ≠ A. Let us mark the point of intersection of the crease with the side DA as I and the point of intersection of the crease with the side AB as H. See Fig. 4. If we mark the point of intersection of the crease with GG as J, we see that J is the orthocenter of AA′I, since GA′ lies on the same line with GG wich is orthogonal to IA and AA′ is orthogonal to the crease IH because of the defi- Fig. 4 170 広島経済大学研究論集第38巻第 4 号 Let us denote the foot of A′ on AB as A. Let us take a square piece of paper and mark There are two triangles AAA′ and AA′G′. the folding square as ABCD. Then we will trans- Since ∠AAA′ = ∠A′G′A = 90°, A′A is a common fer the obtuse angle θ such that θ = ∠YOX. Look side and AA′ = A′G′ (equal to the lenght of GA), at the Fig. 5 a). AAA′ and AA′G′ are congruent triangles. First we have to extend the lines XO and YO. Therefore, ∠G′AA′ = ∠AA′ A = α. Let us mark the point of intersection of the line We can also look at two triangles AG′F′ XO with the side DA as X′ and the the point of and AA′G′. They are also congruent triangles intersection of the line YO with the side AB as because both of them are right angled triangles Y′. See Fig. 5 b). with a common side AG′ and G′F′ = A′G′ (G is Now let us determine and mark the points a midpoint of FA, so G′F′ = FG = GA = A′G′). E and E′ on the lines YO and Y′O respectively at Therefore, ∠F′AG′ = ∠G′AA′ = α. So we have equal distances from point O (E is symmetric to ∠F′AG′ = ∠G′AA′ = ∠AA′ A = α and the lines E′ with respect to O). We will fold the line XO AA′ and AG′ are trisectors of θ. onto the line X′O through the point O and we will get the line perpendicular to the line XO through 2.3 Trisecting an obtuse angle1,3) the point O and let us mark the intersection of This paper method of trisecting an obtuse that line with the side AB as F. See Fig. 6 a). angle was developed by Jacques Justin. Let us make a first fold such that the points a) b) Fig. 5 a) b) Fig. 6 Trisecting an Angle and Doubling the Cube Using Origami Method 171 a) b) Fig. 7 E and E′ lie on the lines X′O and FO respectively we know now that ∠OGO′ = ∠O′OG = α. and unfold it. Mark this obtained points as G and Now let us look at the angles in the GO′O. G′. See Fig. 6 b). The final step of this trisection ∠GO′OO+∠ ′OG+∠ OGO′ =°180 is to fold a line perpendicular to the first crease 180°−θ +++ααα=°180 ⇒ θα=⋅3 through point O and this crease trisects the original angle θ (Fig.

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