2.1 Create visual proofs similar to the one in Figure 2.1 to show that (a)p3 and (b)p5 are irrational.
2.2 Prove thatp2 is irrational by showing that there are no solutions to the congruence m2 2n2.mod 3/ with m and n relatively prime.
2.3 Prove thatpn2 is irrational for all n 3. [Hint: use Fermat’s Last The- orem: xnC ynD znhas no solutions in positive integers for n 3.] 2.4 Prove that the length of the diagonal of a regular pentagon with side
2.13. Challenges 37 2.5 An equilateral triangle with side 1 is partitioned into three equilateral triangles and a trapezoid by line segments parallel to the sides, as shown in Figure 2.15a. Show that the shaded trapezoids in Figures 2.15b and 2.15c are similar if and only if x D 1='.
x
(a) (b) (c)
Figure 2.15.
2.6 Let A denote the area of an annulus of outer radius a and inner radius b and E the area of an ellipse with semi-major and semi-minor axes of lengths a and b, respectively, as illustrated in Figure 2.16. Find the ratio of a to b if A and E are equal [Rawlins, 1995].
b a b a A E Figure 2.16.
2.7 Three identical rectangles measuring a b with a < b fit together as shown in Figure 2.17a, each perpendicular to the other two and inter- secting at their centers. Their vertices determine a convex polyhedron with triangular faces as shown in Figure 2.17b. Show that the polyhe- dron is a regular icosahedron if and only if b=a D ', i.e., only golden rectangles are sections of a regular icosahedron.
(a) (b)
2.8 Let ŒŒx denote the nearest integer function, i.e., the integer nearest to x (when x is a half-integer, it is the nearest even integer). Show that for n 1,
FnD ŒŒ'n=
p 5:
2.9 A sequence of the form fa; b; a C b; a C 2b; 2a C 3b; g with ab ¤ 0 in which each term after the second is the sum of the preceding two is called a generalized Fibonacci sequence. Are any of them geometric progressions?
2.10 Prove that < 22=7. [Hint: Prove 22=7 D Z 1
0
Œx4.1 x/4= .1 C x2/dx. This was Problem A-1 on the 29th William Lowell Putnam Mathematical Competition in 1968.]
2.11 (a) Show that for 0 a < b and any positive integer n, we have .n C 1/an< b
nC1 anC1
b a :
(b) Use (a) to show that the sequence f.1 C 1=n/nC1g is decreasing and bounded below.
(c) Show that the sequences f.1 C 1=n/ng and f.1 C 1=n/nC1g have the same limit, and consequently .1 C 1=n/n< e < .1 C 1=n/nC1for
all n 1. 2.12 Show that lim
n!1 n p nŠ n D 1 e. 2.13 Which is larger, e or e?
2.14 (a) Show that a rational number to an irrational power may be irra- tional.
(b) Show that a rational number to an irrational power may be rational. (c) Why do we not ask about an irrational number to a rational power? 2.15 If fang is as in (2.10), show that an > 1=2.n C 1/. As a conse-
quence, it converges very slowly to . 2.16 Show that for k 2,
r k C
q
k Cpk Cpk C D .1 Cp1 C 4k/=2. [Hint: Consider the sequence fxng defined by x1 D
p
k and xnC1 D
p
CHAPTER
3
Points in the Plane
Mighty is geometry; joined with art, resistless.
Euripedes
Geometry is the art of correct reasoning on incorrect figures.
George P´olya
In this chapter we present some intriguing results, and their delightful proofs, about some of the simplest geometric configurations in the plane. These include figures consisting solely of points and lines, including those con- structed from the lattice points in the plane. We will deal with structures such as triangles, quadrilaterals, and circles in later chapters.
3.1
Pick’s theorem
Pick’s theorem is admired for its elegance and its simplicity; it is a gem of elementary geometry. Although it was first published in 1899, it did not at- tract much attention until seventy years later when Hugo Steinhaus included it in the first edition of his lovely book Mathematical Snapshots [Steinhaus, 1969]. Georg Alexander Pick (1859–1942) was born in Vienna but lived much of his life in Prague. Pick wrote many mathematical papers in the ar- eas of differential equations, complex analysis, and differential geometry. Sadly, Pick was arrested by the Nazis in 1942 and sent to the concentration camp at Theresienstadt, where he perished.
A lattice point in the plane is a point with integer coordinates, and a lattice polygon is a polygon whose vertices are lattice points. A polygon is simple if it has no self-intersections. Pick’s theorem gives the area A.S / of a simple lattice polygon S in terms of the number i of interior lattice points and the number b of lattice points on the boundary: A.S / D i Cb=21. For example, for the lattice polygon S2in Figure 3.1, i D 15, b D 8, and so A.S2/ D 18.
S2 S1
Figure 3.1.
Today elementary school students often study Pick’s theorem. Using a geoboard (a wooden board with a rectangular array of nails driven part way into the board) and rubber bands to form polygons, students can easily cal- culate areas.
There are a variety of ways to prove Pick’s theorem. Our proof [Varberg, 1985] is elegant in its simplicity; it is direct and intuitive. First note that any lattice polygon can be partitioned into a union of lattice triangles (this can be easily proved by induction on the number of sides, using an interior diagonal of the polygon).
We begin by defining a visibility angle k(in degrees) for each lattice point
Pkof the polygon, the angle with which one can “see” into the polygon and
a weight wk D k=360ı. For example, wk D 1 for an interior lattice point,
wk D 1=2 for a boundary point that is not a vertex, and wk D 1=4 for a
right-angled vertex. Now let W .S / DPP
k2Swkdenote the total weight of
the polygon S . Surprisingly, we have
Lemma 3.1. If S is a simple lattice polygon, thenW .S / D A.S /.
Proof. Like area, the weight function W is additive: if S1and S2are disjoint
(except for perhaps a shared boundary) and S D S1[ S2(as in Figure 3.1),
then W .S / D W .S1/ C W .S2/. This follows from the observation that the
visibility angles in S1and S2at a common lattice point add to give the visi-
bility angle in S for that point. We now examine several cases.
Consider Case 1, a lattice rectangle whose sides are horizontal and verti- cal, as illustrated in Figure 3.2a. Each lattice point corresponds to a square or rectangle whose area equals the weight of the lattice point, so the weight and area of the rectangle are equal.
In Figure 3.2b we have Case 2, a right triangle S with legs horizontal and vertical, and W .S / D A.S / follows by division by 2 from Case 1.
In Case 3, Figures 3.2c and 3.2d, we surround a general lattice triangle S by Case 2 triangles and Case 1 rectangles, and again W .S / D A.S / by the
3.2. Circles and sums of two squares 41