G. Información
VII. Conclusiones
A circular helix (or coil) is a curve that lies on the surface of a right circular cylinder with the property that tangents to the curve intersect the elements of the cylinder at a constant angle. Its parametric equations are x D r cos , y D r sin , z D k for in Œ0; 2T , where is the parameter and r, k, and T are constants: r is the radius of the cylinder, 2k the vertical separation between the loops of the helix, and T the number of revolutions around the cylinder. In Figure 9.28 we see a helix with r D 1, k D 1, and T D 2.
The projection of a circular helix onto a plane parallel to its axis (the axis of the cylinder) is a sinusoidal curve.
9.7. Helices 155
Theorem 9.12. The shortest path between two points on a cylinder is either
a straight line or a fractional turn of a helix.
Proof. If the two points lie on the same element of the cylinder, then the shortest path is the line segment joining them. Otherwise we can cut the cylinder along an element not passing through either point, and lay it out flat, as shown in Figure 9.29. The shortest path between two points in the rectangle is a portion (i.e., a fractional turn) of some helix on the cylinder. A curious fact is that if one cuts the cylinder along the helix (rather than on an element of the cylinder) and lays it out flat, the resulting plane figure is a parallelogram.
Figure 9.29.
Figure 9.29 also aids us in finding the length of a circular helix. The length of the helix once around the cylinder is the length of the hypotenuse of a right triangle with legs 2 r and 2k, that is, 2pr2C k2, thus the total length
for given values of r, k, and T is 2Tpr2C k2.
Helices everywhere
Helices abound in nature—from the structure of the DNA molecule to the shapes of some animal horns to the paths taken by squirrels as they chase one anther around a tree trunk. Helices are also common in man-made objects such as screws, bolts, springs, staircases, and helical antennas.
9.8
Challenges
9.1 In the Cartesian plane let O be the origin, C the circle of radius a centered at (0,a/, and T the line tangent to C at (0,2a/. The cissoid of Diocles is the locus of points P such thatjOP j D jABj where A and B are the points of intersection of OP with C and T , respec- tively. See Figure 9.30. Find the equation of the cissoid, and show how it can be used to duplicate the cube, that is, to represent the cube root of 2. O A B P a C T 2a Figure 9.30.
9.2 Conic sections can also be used to duplicate the cube. In Figure 9.31 we see two parabolas with a common vertex and perpendicular axes. Find a focus and directrix for each parabola so that the x-coordinate of the point of intersection is p3 2. 2 3 0 x y Figure 9.31.
9.3 Show that a lune constructed on the side of an equilateral triangle is not squarable.
9.8. Challenges 157 9.4 Show that the shaded portions of the objects in Figure 9.32 are squarable
(the curves are arcs of circles).
(a) (b)
Figure 9.32.
9.5 Leonardo’s claw is the gray portion of the circle in Figure 9.33a re- maining when a smaller circle and a lens, defined by an isosceles right triangle whose legs are radii of the circle, have been removed. Show that Leonardo’s claw is squarable, and that its area is the same as the area of the square in its grasp (see Figure 9.33b).
(a) (b)
Figure 9.33.
9.6 Prove the broken chord theorem of Archimedes: If AB and BC are two chords in a circle withjABj > jBC j and M is the midpoint of arc ABC , then the foot F of the perpendicular line from M to BC is the midpoint of the broken chord ABC
9.7 Show that a Cartesian equation of the quadratrix of Hippias is given by x D y cot.y=2/ for y 2 .0; 1 and x D 2= for y D 0.
9.8 Consider a circular disk partitioned into four regions by semicircular arcs, as illustrated in Figure 9.34. The boundaries of the regions par- tition the horizontal diameter into four intervals of equal width. Prove that the regions have equal area [Esteban, 2004].
Figure 9.34.
9.9 Let be the graph of a differentiable function concave on the interval [a; b]. Prove that the point at which the tangent line to minimizes the shaded area in Figure 9.35 is the midpoint of [a; b]. (Hint: no calculus required!)
a b
Γ
CHAPTER
10
Adventures in Tiling
and Coloring
“What’s the color got to do with it?”
“It’s got everything to do with it. Illinois is green, Indiana is pink.
“Indiana PINK? Why, what a lie!”
“It ain’t no lie; I’ve seen it on the map, and it’s pink.”
Mark Twain
Tom Sawyer Abroad There are many lovely proofs in which the tiling of regions and the coloring of objects appear. We have seen examples in earlier chapters, such as tiling in the proof of Napoleon’s theorem in Section 6.5, and coloring in the proof of the art gallery theorem in Section 4.6. Tiles enable us to com- pare areas without calculation, and color enables us to distinguish relevant parts of figures easily.
In this chapter we continue our exploration of elegant proofs that employ these techniques. We begin with a brief survey of basic properties of tilings, followed by some results using tilings with triangles and quadrilaterals, in- cluding the Pythagorean theorem. We also discuss the lovely tilings found in the Alhambra in Granada, Spain, and in the work of M. C. Escher. After visiting the seven frieze patterns, we examine the use of color in proofs, and proofs about colorings. Our examples include tiling chessboards with polyominoes, packing calissons into boxes, map coloring, and Hamiltonian circuits on dodecahedra.
10.1
Plane tilings and tessellations
A tiling of the plane is any countable family of closed sets (the tiles) that cover the plane without gaps or overlaps [Gr¨unbaum and Shephard, 1987]. In this section and ones to follow, our tiles will be polygons. An edge-to- edge tiling is a polygonal tiling in which each edge of each polygon co- incides with an edge of another polygon. Non-edge-to-edge tilings lead to some unexpected proofs, as we shall see in the next section. A vertex of an edge-to-edge tiling is a vertex of one of the tiles. An edge-to-edge tiling in which all the tiles are regular polygons is called a tessellation. A monohedral tiling is one in which all the tiles are congruent, and a uniform tessellation is one with identical vertices, i.e., one with the same arrangement of polygons at each vertex.
We begin by considering regular tilings, the monohedral tessellations— edge-to-edge tilings with congruent polygons. Clearly (as Figure 10.1 shows) equilateral triangles, squares, and regular hexagons yield regular tilings.
Figure 10.1.
Are there others? The negative answer is given in the next theorem.
Theorem 10.1. The only regular tilings consist of equilateral triangles,
squares, and regular hexagons.
Proof. Suppose k n-gons meet at a vertex. Since each interior vertex angle of an n–gon measures .1 2=n/ 180ı, we have k .1 2=n/ 180ıD 360ı, or .n 2/.k 2/ D 4. The only solutions to this equation in positive integers are .n; k/ D .3; 6/, (4,4), and (6,3).
The regular tilings are clearly uniform tessellations. We can generalize regular tilings in several ways, two of which are (i) use more than one kind of regular polygon (a tessellation, but not monohedral), or (ii) use identi- cal nonregular polygons (monohedral, but not a tessellation). One class of non-monohedral tessellations is the class of uniform non-monohedral tessel- lations, called semiregular (or Archimedean) tilings. To find all the semireg- ular tilings, we need first the following lemma.
Lemma 10.1. The number of regular polygons that may share a common
10.1. Plane tilings and tessellations 161 Proof. Suppose k polygons share a common vertex in a given edge-to-edge tiling. Clearly k 3. Let ˛1; ˛2; : : : ; ˛k denote the vertex angles of the
polygons. Then each ˛iis at least 60ıso that 360ıD ˛1C ˛2C C ˛k
k 60ı, or k 6.
In Figure 10.2 we see eight semiregular tilings. In Theorem 10.2 we prove that there are no others.
Figure 10.2.
Theorem 10.2. There are eight classes of semiregular tilings.
Proof. Since semiregular tilings are uniform, at each vertex we have k reg- ular polygons with n1; n2; : : : ; nk sides, 3 k 6 and each ni 3. Since
the interior vertex angles of the polygons are .1 2=ni/ 180ı, we have
Pk i D1.1 2=ni/ 180ıD 360ı, or equivalently k X i D1 1 ni D k 2 2 :
The solutions to these four equations (k D 3, 4 5, 6) are given in Table 10.1.
The six numeric solutions marked✗ do not yield tilings, since in a tiling if one of fn1; n2; n3g is odd, then the other two must be equal, as those tiles
alternate around the tile with an odd number of sides. Also, the solution (3,3,3,4,4) corresponds to two semiregular tilings, one with the two squares adjacent and one where they are not adjacent.
Table 10.1. k n1 n2 n3 n4 n5 n6 Conclusion 3 7 42 ✗ 3 8 24 ✗ 3 9 18 ✗ 3 10 15 ✗ 3 3 12 12 semiregular (1) 4 5 20 ✗ 4 6 12 semiregular (2) 4 8 8 semiregular (3) 5 5 10 ✗ 6 6 6 regular 3 3 4 12 not uniform 4 3 3 6 6 semiregular (4) 3 4 4 6 semiregular (5) 4 4 4 4 regular 5 3 3 3 3 6 semiregular (6) 3 3 3 4 4 semiregular (7 & 8) 6 3 3 3 3 3 3 regular
We now turn our attention to monohedral tilings that are not tessella- tions. We begin with triangles and quadrilaterals. It is convenient to consider quadrilaterals first.
Theorem 10.3. Any (convex or concave) quadrilateral tiles the plane.
Proof. See Figure 10.3.
10.1. Plane tilings and tessellations 163
Figure 10.4.
Since any triangle can be doubled to form a parallelogram (see Figure 10.4) and parallelograms are quadrilaterals, the following corollary is immediate
Corollary 10.1. Any triangle tiles the plane.
In the next section we explore the use of these tilings in the proofs of some theorems about triangles and quadrilaterals.
What about pentagons? Clearly regular pentagons do not tile the plane (although they do form the surface of a regular dodecahedron in three di- mensions), but some non-regular pentagons do. For example, one can par- tition each hexagon in the regular hexagonal tiling in Figure 10.1 into two or three congruent pentagons to obtain the monohedral pentagonal tilings in Figure 10.5.
Figure 10.5.
Another pentagonal tiling can be created by overlaying two non-regular hexagonal tilings illustrated in Figure 10.6. This rather attractive monohedral
pentagonal tiling is sometimes called the Cairo tiling, for its reported use as a street paving design in that city.
Thus we arrive at the question: How many different convex pentagons tile the plane? The answer to this question has an interesting history [Klarner, 1981]. Five different pentagons that tile (including the three above) were described by K. Reinhardt in 1918 and three more by R. B. Kerschner in 1968. A ninth pentagon that tiles was found by R. James in 1975, and four more by Marjorie Rice, an amateur mathematician and homemaker, in 1976– 77. A fourteenth pentagon that tiles the plane was found by R. Stein in 1985.
That is where things stand now. Since 1985 no new pentagons that tile the plane have been discovered, nor has a proof appeared that the classification is complete. An answer to this open problem would close the study of mono- hedral tilings, since, as we shall see, the classification for n-gons with n 6 is known.
As with pentagons, not all convex hexagons tile the plane. In 1918, K. Reinhardt described the three classes of convex hexagons that tile the plane [Gardner, 1988]. Let A, B, C , D, E, and F denote the vertices, and a, b, c, d , e, and f the sides, as illustrated in Figure 10.7. Then the hexagon tiles the plane if and only if it belongs to one of three classes: I. A C B C C D 360ı; II. A C B C D D 360ı, a D d , c D e; III. A D C D E D 120ı, a D b, c D d , e D f .
Finally, any convex n-gon with n 7 cannot tile the plane [Kerschner, 1969; Niven, 1978], which concludes our brief survey of tilings and tessel- lations. A B C D E F a b c d e f A B C D E F a b c d f e A B C D E F a b c d f e I. II. III. Figure 10.7.