A Tiling of the Plane With Triangles

A Tiling of the Plane with TrianglesAuthor(s): Paul T. MielkeSource: The Two-Year College Mathematics Journal, Vol. 14, No. 5 (Nov., 1983), pp. 377-381Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/3026759 .Accessed: 31/10/2014 16:55

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A Tiling of the Plane with Triangles Paul T. Mielke

Paul Mielke received his Ph.D. in mathematics from Purdue L3 EUniversity in 1951. He has been on the staff of Wabash College

/;* ETw;Sf ~ for 29 years, having served for 15 of those years as chairman of its mathematics department. He has been a member of the MAA for 35 years, having served as Executive Director of

.: C 9 JutCUPM in 1970-71, as a member of its Board of Governors from 1972 to 1975, and as associate editor for mathematics educa- tion of the "American Mathematical Monthly" from 1974 to 1978.

9

Figure 1.

Figure 1 might be used as a problem without words for a geometry class. If that proves too cryptic, the student might be led along by the following suggestions:

1. Count the equilateral triangles and observe their sizes (of form n2 _ 1). 2. Establish the pattern of formation of these triangles (starting with side 3,

successive enlargment by isosceles trapezoids of slant height 2n + 1 in a counterclockwise pattern). Extend this pattern to cover an arbitrarily large area, thus tiling the plane.

3. Consider the diagonals of these trapezoids. Use the Rule of Pythagoras (or the cosine law) to find their lengths (of form n2 + n + 1).

4. Discover relationships between the sides of the triangles, e.g.,

72 = 32 + 52 + 3 X 5 = 52 + 82 5 x 8= 82 + 32 8 x 3.

Generalize these equations.

377


5. Show that the map of the plane resulting from the tiling needs only three colors.

6. Find the similar triangles among the tiles. (There are only two pairs. Proof? Note that among the equilateral triangles, only the smallest one is actually a tile.)

My discovery of this problem resulted from a colloquium talk that was given at Wabash College in the Spring of 1980 by Professor Rodney Hood of Franklin College. Hood's talk centered on the Diophantine equation

a2 + b2- ab = c2 (1)

that results from setting 9 = 7r/3 in the cosine law. L. E. Dickson reports (see [1], p. 214) that this equation was solved in 1625 by Albert Girard. Hood provides an independent solution [2]. In either case, the solutions constitute a two-parameter family of pairs of number triples of form (a, b, c) and (a', b, c) with

a + a' = b. (2)

Each triple represents a triangle having integral sides and one 600 angle, and the paired triangles fit together to form an equilateral triangle of side b. Figure 2 represents the pair (3,8,7) and (5, 8,7).

8 7 8

3 5 Figure 2.

Girard also noted that the triple (a, a', c) represents a triangle in which the angle is 1200, that is,

2 ~2 2 a + a + aa= c (3)

For Girard's parametrization of the solutions, let m and n be positive integers with m < n. Then all solutions will be given by the following, together with their integral multiples:

2 2 ~~~~~~2 2 a = 2mn + m2, b =2mn + n2, c=m +mn+n, (4) and

a' = m It is an easy exercise in algebra to show that equations (1), (2), and (3) are all satisfied. To obtain Hood's parametrization, replace m by n - v.

378


In thinking about Hood's solution to the equation, I could not resist preparing a computer-generated list of the triples for possible future classroom use, their appeal being comparable to that of the Pythagorean triples. Scanning the list provided several surprises, all of which are associated with the subset of triples for which m= 1. Since there is then only one parameter, let us denote the relationships with this parameter by writing equations (4) as

an= 2n + 1,

bn=2n+n2 =n(n+2)=(n+ 1)21 (5)

cn =n2 +n+ 1,

and

n=a = 1. Referring once more to Figure 1, one sees that the above formulas for an and bn are easily guessed, and the form for cn can be derived easily from the cosine law or, with a bit more exercise, from the Rule of Pythagoras. Now come two delightful surprises that lead to the tiling. What I had noticed from scanning the list was that

= a (6) and this is now algebraically obvious from equations (5). The most appealing surprise, however, is a kind of Pascal's Law that arises from applying equation (6) to equation (2) in the form

an + an =n+

thus,

an + 1+ bn = bn+l *(7)

This can also be checked algebraically from equations (5). Equations (6) and (7) now permit a fresh geometrical interpretation that leads to

the tiling; given any equilateral triangle constructed from a pair of triples (an, bn, C), (an, bn, cn), one can imbed it suggestively in the equilateral triangle associated with n + 1. Figure 3 explains the imbedding more eloquently than words can:

a +

n+

Figue 3. 39 a an+ In an

Figure 3. 379


Incidentally, equation (3) for Figure 3 becomes 2 ,2 =2

an+ I+ an+ I+ an+,an+ 1 = Cn+ 1I

Notice also how equations (6) and (7) apply to the figure. Mathematical induction now suggests that one can tile the plane with triangles of

this kind beginning at any place in their sequence. Furthermore, if n = 1 is permitted, the construction can begin with the equilateral triangle of side a, = b, = = 3, in which case a' = 0. Figure 4 is an exploded view of the first five steps of one such construction. Beginning with an equilateral triangle of side 3, join isosceles trapezoids (a), (b), (c), and (d), respectively, in the counterclockwise pattern shown. These trapezoids have tops of length n2 _ 1, sides of length 2n + 1, bases of length (n + 1)2 - 1, and diagonals of length n2 + n + 1, and they have base angles of 60 degrees. How to continue the tiling should be clear. There are other interesting ways to join the trapezoids, some leading to tilings and others not. For young children, cutouts based on Figure 4 could serve as jigsaw puzzles. For purposes of arithmetic instruction, the back of each trapezoid could be imprinted with its related equations, e.g., for trapezoid (d),

312= 112+242+11 x24

= 242 + 352 - 24 x 35 = 352 + 112-35 x 11.

Colors may be added. If the starting equilateral triangle and the bottom triangle of each trapezoid are colored one color, and if alternate top triangles of the trapezoids are colored with a second and third color, then the tiling will become a three-color map of the plane.

What is not so simple is that there are only two pairs of similar triangles in the tiling. Keep in mind that there is really only one equilateral tile. It is useful in this connection to recall that one defines a triple to be primitive if its three elements have no common factor. Hood proves that each of a pair of triples is primitive iff m and n are relatively prime and n - m is not a multiple of 3. Since m = 1 for the tiling triples, the only nonprimitives among them are those for which n = 3k + 1. Setting m = 1 and n = 3k + I in equations (4) yields

a =3(2k + 1), b = 3(k + 1)(3k + 1), c = 3(3k2 + 3k + 1), (8)

and

a' = 3k(3k + 2).

Each such triple is therefore a multiple of 3. But a further scan of the computer- generated list revealed that if one lets m = k and n = k + 1, in equations (4), then

a = k(3k + 2), b = (k + 1)(3k + 1), c = 3k2 + 3k + 1, (9) and

a' = 2k + 1.

380


(c) 24

21

(b) 15 - ~~~~8 13 15

9

(a)/

(d) Figure 4.

Furthermore, since k and k + 1 are relatively prime for any k, these triples are primitive according to Hood. Equations (8) and (9) make it clear that the only non-primitive tiling triples are multiples of the primitives where n = m + 1 and where a and a' have been reversed, and this correspondence is clearly one-to-one. One can now use this information to discover that there are only two pairs of similar tiles.

I conclude by thanking Murray Klamkin for the reference to Girard's work.

REFERENCES

1. Leonard Eugene Dickson, History of the Theory of Numbers, vol. II, Diophantine Analysis, Chelsea, 1966.

2. Rodney T. Hood, On equilateral and related triangles with integral sides, paper in preparation.

381


Article Contentsp. 377p. 378p. 379p. 380p. 381

Issue Table of ContentsThe Two-Year College Mathematics Journal, Vol. 14, No. 5 (Nov., 1983), pp. i+370-444Front Matter [pp. ]Shiing-shen Chern: A Man and His Times [pp. 370-376]A Tiling of the Plane with Triangles [pp. 377-381]On the Radii of the Inscribed and Escribed Circles of Right Triangles-A Second Look [pp. 382-389]Products of Sets of Complex Numbers [pp. 390-397]What Makes Mathematics Lessons Easy to Follow, Understand, and Remember? [pp. 398-406]The Address Problem [pp. 407-415]Book ReviewsReview: untitled [pp. 415]

Computer and CalculatorsComputer Simulations to Clarify Key Ideas of Statistics [pp. 416-421]Integrating Writing into the Mathematics Curriculum [pp. 421-424]Another Way to Introduce Natural Logarithms and e [pp. 424-426]Approximation of Square Roots [pp. 427-431]

Classroom CapsulesMean Inequalities [pp. 431-434]Antisubmarine Warfare: Passive vs. Active Sonar [pp. 434-435]

Ellipses from a Circular and Spherical Point of View [pp. 436-438]Problems [pp. 438-444]Back Matter [pp. ]

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A Tiling of the Plane With Triangles