SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the problem editor.

47. [1992, 88; 1993, 94] Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Find all the solutions of

(x − 1)x(x + 1)(x + 2) = −1.

Comment by the editor.

Due to an error by the editor, a remark by the proposer was omitted from the solution to the problem. At the time the problem was proposed, the proposer noted that if x is an integer, the factorization provided in the solution indicates that the product of four consecutive integers is one less than the square of an integer.

57. [1993, 90] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Let s and k be positive integers. Evaluate

lim_{n→∞} ∏_{i=1}^{k} ∑_{j=1}^{n} j^{i}(\frac{s}{n})^{i+1}.

Solution I by Joseph E. Chance, University of Texas–Pan American, Edinburg, Texas; Seung–Jin Bang, Albany, California; and the proposer.

∑_{j=1}^{n} j^{i}(\frac{s}{n})^{i+1} = ∑_{j=1}^{n}(\frac{js}{n})^{i} \frac{s}{n}

is an upper Riemann sum for

∫_{0}^{s} x^{i}dx.

Since the product of sums is finite and each sum has a limit,

lim_{n→∞} ∏_{i=1}^{k} ∑_{j=1}^{k} j^{i}(\frac{s}{n})^{i+1} = ∏_{i=1}^{k} lim_{n→∞} ∑_{j=1}^{n}(\frac{js}{n})^{i} \frac{s}{n}

= ∏_{i=1}^{k} ∫_{0}^{s} x^{i}dx

= ∏_{i=1}^{k} \frac{s^{i+1}}{i + 1}

= \frac{s^{k(k+3)/2}}{(k + 1)!} .

Solution II by N. J. Kuenzi, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin and Frank J. Flanigan, San Jose State University, San Jose, California.

It is known that

∑_{j=1}^{n} j^{i}

can be represented as a polynomial in n of degree i + 1 with leading coefficient 1/(i + 1).

(See for example: R. S. Luthar, “A Simple Way of Evaluating ∑_{i=1}^{k} i^{n}”, Pi Mu Epsilon Journal, 6 (1976), 282–284.) Using this fact we can represent ∑_{j=1}^{n} j^{i} in the form

∑_{j=1}^{n} j^{i} = (\frac{1}{i + 1})n^{i+1} + P (n),

where P (n) is a polynomial of degree i. Since

\frac{P (n)}{n^{i+1}} → 0 as n → ∞,

it follows that

∑_{j=1}^{n} \frac{j^{i}}{n^{i+1}} → \frac{1}{i + 1} as n → ∞.

Hence,

\begin{aligned}lim_{n→∞} ∏_{i=1}^{k} ∑_{j=1}^{n} j^{i}(\frac{s}{n})^{i+1} & = ∏_{i=1}^{k} s^{i+1}( lim_{n→∞} ∑_{j=1}^{n} \frac{j^{i}}{n^{i+1}} ) \\ & = ∏_{i=1}^{k} \frac{s^{i+1}}{i + 1} = \frac{s^{k(k+3)/2}}{(k + 1)!} . \end{aligned}

Also solved by Jayanthi Ganapathy, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

58. [1993, 90] Proposed by Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

Let G be a proper subgroup of ℝ, the reals under addition. Prove that ℝ and the complement of G have the same cardinality.

Solution I by the proposer.

Let |G^{c}| denote the cardinality of the complement of G and H denote the set of all finite linear combinations of elements from G^{c} with integer coefficients (that is, an element of H is of the form

∑_{i=1}^{m} n_{i}x_{i},

where m is a positive integer, the n_{i}’s are integers and the x_{i}’s are elements from G^{c}). Then G^{c} ⊆ H and H is a subgroup of ℝ. Suppose |G^{c}| < |ℝ|. Then G is not a subset of H, because if G ⊆ H, then ℝ = G ∪ G^{c} ⊆ H and consequently |H| = |ℝ|, which contradicts the fact that |G^{c}| < |ℝ|. Since G is not a subset of H and H is not a subset of G, there exist elements g ∈ G and h ∈ H such that g ∉ H and h ∉ G. If g +h ∈ G, then h = g +h−g ∈ G, but h ∉ G. Hence g + h ∉ G. Similarly g + h ∉ H. Hence g + h ∉ G ∪ H. But g + h ∈ ℝ. This contradicts the fact that ℝ = G ∪ H. Thus |G^{c}| = |ℝ|.

Solution II by Frank J. Flanigan, San Jose State University, San Jose, California.

We will deduce the above assertion after establishing the following theorem. \underline{Theorem}. Let G be a subset of an infinite set S whose complement G^{c} = S −G satisfies

|G^{c}| ≥ |G|. Then |G^{c}| = |S|.

\underline{Proof}. The theorem follows from

|S| = |G ∪ G^{c}| = |G| + |G^{c}| = |G^{c}|,

since |G^{c}| is necessarily infinite. (The second equality follows since the sets G and G^{c} are disjoint.)

Now if G is a proper subgroup of ℝ, then the standard Lagrange decomposition of ℝ as a union of cosets x + G shows that the complement G^{c} contains at least one such coset x + G. Thus

|G^{c}| ≥ |x + G| = |G|,

so the theorem applies.

\underline{Comment}. This approach does not involve uncountability, intermediate cardinalities, the continuum hypothesis, Hamel bases, etc. The assertion holds for any proper subgroup of any infinite group.

59. [1993, 91] Proposed by Ollie Nanyes, Bradley University, Peoria, Illinois.

Find a topology τ_{1} for the real line ℝ^{1} such that:

1) (ℝ^{1}, τ_{1}) is a second countable, metrizable space and

2) there is a homeomorphism

f: (ℝ^{1}, τ_{1}) → (ℝ^{2}, τ_{2}),

where τ_{2} is the product topology τ_{1} × τ_{1}.

Solution by the proposer.

Let τ_{1} be the topology generated by [q, x) where q and x are rational. Notice that these basis elements are closed and open. Clearly (ℝ^{1}, τ_{1}), denoted by ℝ_{q}, is Hausdorff, regular and second countable. Therefore, ℝ_{q} is metrizable by the Urysohn metrization theorem.

However, we can find a basis B for ℝ_{q},

B = ⋃_{i=1}^{∞} B_{i},

where:

1) for all i, B_{i} is a disjoint, countable cover of ℝ^{1} by non-empty closed-open sets, 2) for all i, if B ∈ B_{i}, there is a countable, disjoint cover B(B) of B such that

B = ⋃_{B′∈B(B)} B′

and

B_{i+1} = ⋃_{B∈B_{i}}{B(B)}

and

3) for every collection {B_{i}} where, for all i, B_{i} ∈ B_{i} and B_{i+1} ⊂ B_{i}, then ∩_{i}B_{i} is a single point set. Then, ℝ_{q} is homeomorphic to ℤ^{ℤ} in the product (Tychonoff) topology by the map g: ℝ_{q} → ℤ^{ℤ} by

g(x) = (n_{1}, n_{2}, …, n_{k}, …)

where

x ∈ ⋂_{j=n_{k}, k∈ℤ^{+}} B_{i}^{j}

(with B_{i}^{k} ∈ B_{i} for all i). So, let τ_{2} be the product topology for ℝ_{q} × ℝ_{q}; since ℤ^{ℤ} is homeomorphic to ℤ^{ℤ} × ℤ^{ℤ},

ℝ_{q} is homeomorphic to ℝ_{q} × ℝ_{q}.

We can show that the necessary basis B exists by induction: let C = {[a, b) | a, b are rational } and enumerate the rationals by q_{i}. Our induction hypothesis is that B_{n} ∈ C. If we let:

B_{1} = ⋃_{i}[i, i + 1),

then our hypothesis is satisfied for i = 1. Now assume that B_{n} ∈ C. We will describe the construction of B_{n+1}. If B ∈ B_{n}, then B = [a, b) by hypothesis. Let B′(B) = {[c, d) ∈ C | a ≤ c < d < b with |d − c| < 1/n}. Certainly if q_{n} ∈ [a, b) there is a rational d_{n} such that [q_{n}, d_{n}) ∈ B′(B). Choose a countably infinite family of disjoint members of B′(B) whose union covers B such that [q_{n}, d_{n}) ∈ B(B) if q_{n} ∈ B. Call this collection B(B). Define

B_{n+1} = ⋃_{B∈B_{n}}{B(B)}.

It is now easy to check that

B = ⋃_{i=1}^{∞} B_{i}

is our basis which meets requirements 1, 2, and 3. (It turns out that ℝ_{q} is homeomorphic to the irrationals in the normal Euclidean subspace topology since the irrationals are homeomorphic to ℤ^{ℤ}. (See F. Willard, General Topology, Addison-Wesley, 1970, Reading, Massachusetts, exercise 24K.)

Acknowledgement by the proposer. I would like to thank an anonymous referee and John Duncan who explained the solution to me.

One incorrect solution was also received.

60. [1993, 91] Proposed by Alvin Tinsley, Central Missouri State University, Warrensburg, Missouri.

Suppose a unit square has its left–hand corner at the origin and its sides along the x– & y–axes. Initially, place the base of an equilateral triangle with unit sides on the x–axis between 0 and 1. Slide the triangle to the left and up, always keeping the two vertices of the base in contact with the x– & y–axes until the base of the equilateral triangle is on the y–axis between 0 and 1. What is the equation of the locus of points determined by the third vertex of the triangle?

Solution I by N. J. Kuenzi, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; Leon Hall, University of Missouri–Rolla, Rolla, Missouri; and Rhonda McKee, Central Missouri State University, Warrensburg, Missouri.

Let t be the y–coordinate of the vertex A on the y–axis. Then \sqrt{1 − t^{2}} is the x– coordinate of the vertex B on the x–axis. For 0 < t < 1, the slope of the line through A and B is −t/\sqrt{1 − t^{2}}, the slope of the perpendicular bisector of side AB is \sqrt{1 − t^{2}}/t, and the coordinates of the midpoint M of side AB are (\sqrt{1 − t^{2}}/2, t/2). (See figure 1.) Now the third vertex C lies on the perpendicular bisector of side AB at a distance of \sqrt{3}/2 units from M. So if (x, y) are the coordinates of vertex C we have

(1)  y − \frac{t}{2} = \frac{\sqrt{1 − t^{2}}}{t} (x − \frac{\sqrt{1 − t^{2}}}{2} )

and

(2)  (y − \frac{t}{2})^{2} + (x − \frac{\sqrt{1 − t^{2}}}{2} )^{2} = \frac{3}{4}.

Substituting the right hand side of (1) for (y − t/2) in (2) and simplifying yields

(3)  (x − \frac{\sqrt{1 − t^{2}}}{2} )^{2} = \frac{3t^{2}}{4} .

Suppose the initial position for vertex C was quadrant I. Then from equations (3) and (1) we get the following parametric equations:

x = \frac{\sqrt{3}}{2}t + \frac{\sqrt{1 − t^{2}}}{2}

and

y = \frac{t}{2} + \frac{\sqrt{3}}{2}\sqrt{1 − t^{2}}, 0 ≤ t ≤ 1.

(See figure 2.) Using these equations we can eliminate the radical and solve for t,

t = \sqrt{3}x − y.

Substituting \sqrt{3}x − y for t in the parametric equation for y and simplifying yields,

\sqrt{3}y − x = \sqrt{1 − (\sqrt{3}x − y)^{2}}.

Squaring both sides and simplifying yields

4x^{2} − 4\sqrt{3}xy + 4y^{2} = 1

where 1/2 ≤ x ≤ 1 and 1/2 ≤ y ≤ 1. Next suppose the initial position for vertex C was quadrant IV. Then from equations (3) and (1) we get the following parametric equations:

x = −\frac{\sqrt{3}}{2}t + \frac{\sqrt{1 − t^{2}}}{2}

and

y = \frac{t}{2} − \frac{\sqrt{3}}{2}\sqrt{1 − t^{2}}, 0 ≤ t ≤ 1.

(See figure 3.) Using the same technique as in the previous case we get

\sqrt{3}x + y = −t

and

4x^{2} + 4\sqrt{3}xy + 4y^{2} = 1

where −\sqrt{3}/2 ≤ x ≤ 1/2, −\sqrt{3}/2 ≤ y ≤ 1/2, and (x < 0 or y < 0).

107 108

Solution II by the proposer; Joseph E. Chance, University of Texas–Pan American, Edinburg, Texas; and J. Sriskandarajah, University of Wisconsin Center–Richland, Richland Center, Wisconsin.

Consider the figure

The triangle slides as θ moves from 0 to 1. The equations of the non–base sides of the triangle are

y = tan(\frac{2π}{3} − \frac{π}{2} θ)(x − cos \frac{π}{2} θ)

and

y − sin \frac{π}{2} θ = tan(\frac{π}{3} − \frac{π}{2} θ)x.

Solving both equations simulataneously (and using some trig identities), the locus of the third vertex of the triangle is

(x, y) = (cos(\frac{π}{3} − \frac{π}{2} θ), sin(\frac{π}{3} + \frac{π}{2} θ)).

Finding θ in terms of x, we have

\frac{π}{2} θ = \frac{π}{3} − arccos x.

Thus

y = sin(\frac{2π}{3} − arccos x)

= \frac{\sqrt{3}}{2}x + \frac{1}{2}\sqrt{1 − x^{2}}.

Hence

2y − \sqrt{3}x = \sqrt{1 − x^{2}}

so

4x^{2} − 4\sqrt{3}xy + 4y^{2} − 1 = 0.