SOLUTIONS

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

104. [1997, 35; 1998, 59–61] Proposed by Kenneth Davenport, Box 491, Frackville, Pennsylvania.

Show that

1 · sin \frac{π}{2n} + 3 · sin \frac{3π}{2n} + 5 · sin \frac{5π}{2n} + ⋯ + (2n − 1) sin \frac{(2n − 1)π}{2n} = n csc \frac{π}{2n}.

Solution III by Paul S. Bruckman, Edmonds, Washington.

Let

o_{k} = \frac{1 − (−1)^{k}}{2} .

Then,

∑_{k=0}^{2n−1} k sin \frac{kπ}{2n} · o_{k} = n · csc \frac{π}{2n}, n = 1, 2, …

is easily evaluated using complex variables. That is,

sin x = \frac{1}{2i}(e^{ix} − e^{−ix}),

and then we can use geometric series sum formulas to evaluate the sum. Letting θ = exp(iπ/2n) and using the fact that θ^{2n} = −1, we have

∑_{k=0}^{2n−1} k sin \frac{kπ}{2n} · o_{k} = \frac{1}{2i} ∑_{k=0}^{2n−1} ko_{k}(θ^{k} − θ^{−k})

= \frac{1}{2i} ∑_{k=0}^{2n−1} ko_{k}θ^{k} − \frac{1}{2i} ∑_{k=1}^{2n}(2n − k)o_{2n−k}θ^{k−2n}

= \frac{1}{2i} ∑_{k=0}^{2n−1} ko_{k}θ^{k} + \frac{1}{2i} ∑_{k=0}^{2n−1}(2n − k)o_{k}θ^{k} = \frac{n}{i} ∑_{i=0}^{2n−1} o_{k}θ^{k}

= \frac{n}{2i}(\frac{θ^{2n} − 1}{θ − 1} − (\frac{(−θ)^{2n} − 1}{−θ − 1})) = \frac{n}{2i}(\frac{−2}{θ − 1} + \frac{−2}{θ + 1})

= \frac{−n}{i} · \frac{2θ}{θ^{2} − 1} = n · \frac{2i}{θ − θ^{−1}} = n csc \frac{π}{2n}.

109. [1997, 184] Proposed by Kenneth Davenport, Box 491, Frackville, Pennsylvania.

Let n be a positive integer and a ≥ 2 be a positive integer. Show that

\begin{matrix}∫_{0}^{∞} \frac{dx}{1^{a} + x^{a}} + ∫_{0}^{∞} \frac{dx}{2^{a} + x^{a}} + ⋯ + ∫_{0}^{∞} \frac{dx}{n^{a} + x^{a}} \\ = [\frac{1}{1^{a−1}} + \frac{1}{2^{a−1}} + ⋯ + \frac{1}{n^{a−1}} ]\frac{π/a}{sin(π/a)}. \end{matrix}

Solution I by Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri.

Consider

∫_{C} f(z)dz

where f(z) = 1/(m^{a} + z^{a}), m = 1, 2, …, n and C is the boundary of the circular sector described by 0 ≤ θ ≤ 2π/a and 0 ≤ r ≤ R.

The only singularity of f(z) that is inside C is a simple pole at z_{0} = me^{πi/a} and

Res_{z=z_{0}}f(z) = −e^{πi/a}/(am^{a−1}).

By the Residue Theorem,

∫_{0}^{R} \frac{dx}{m^{a} + x^{a}} + ∫_{0}^{2π/a} \frac{Re^{iθ}idθ}{m^{a} + R^{a}e^{iaθ}} + ∫_{R}^{0} \frac{e^{2πi/a}dr}{m^{a} + r^{a}} = \frac{−2πie^{πi/a}}{am^{a−1}} .  (1)

Let I = I(R) represent the second integral on the left hand side of (1). It is straightforward to show that

|I| ≤ \frac{2πR}{a|R^{a} − m^{a}|}.

Since a ≥ 2, lim_{R→∞} I(R) = 0. Then (1) becomes

(1 − e^{2πi/a}) ∫_{0}^{∞} \frac{dx}{m^{a} + x^{a}} = \frac{−2πie^{πi/a}}{am^{a−1}} .  (2)

Equating real parts in (2) gives

(1 − cos \frac{2π}{a} ) ∫_{0}^{∞} \frac{dx}{m^{a} + x^{a}} = \frac{2π sin \frac{π}{a}}{am^{a−1}}

and so

∫_{0}^{∞} \frac{dx}{m^{a} + x^{a}} = \frac{2π sin(π/a)}{2am^{a−1} sin^{2}(π/a)} = \frac{π}{am^{a−1} sin(π/a)}.

Therefore,

∑_{m=1}^{n} ∫_{0}^{∞} \frac{dx}{m^{a} + x^{a}} = \frac{π/a}{sin(π/a)} ∑_{m=1}^{n} \frac{1}{m^{a−1}} .

Solution II by Joseph E. Chance, University of Texas–Pan American, Edinburg, Texas.

For i = 1, 2, …, n,

∫_{0}^{∞} \frac{dx}{i^{a} + x^{a}} = \frac{1}{i^{a}} ∫_{0}^{∞} \frac{dx}{1 + (x/i)^{a}} .

Let y = x/i and the integral is transformed to

\frac{1}{i^{a−1}} ∫_{0}^{∞} \frac{dy}{1 + y^{a}} .

The required sum can now be written as

[\frac{1}{1^{a−1}} + \frac{1}{2^{a−1}} + ⋯ + \frac{1}{n^{a−1}} ] ∫_{0}^{∞} \frac{dy}{1 + y^{a}} .

In this integral let

u = \frac{1}{1 + y^{a}} , or y = (\frac{1 − u}{u})^{\frac{1}{a}} , dy = \frac{1}{a}(\frac{1 − u}{u})^{\frac{1}{a} −1}(\frac{−1}{u^{2}})

which transforms the integral to

\frac{1}{a} ∫_{0}^{1}(1 − u)^{\frac{1}{a} −1}u^{− \frac{1}{a}} du.

In terms of the famous beta function, the integral is

\frac{1}{a}B(1 − \frac{1}{a}, \frac{1}{a}),

but in terms of the more famous gamma function and its identities, this beta function is

\frac{1}{a} \frac{Γ(\frac{a−1}{a})Γ(\frac{1}{a})}{Γ(\frac{a−1}{a} + \frac{1}{a})} = \frac{1}{a}Γ(1 − \frac{1}{a})Γ(\frac{1}{a}) = \frac{1}{a} \frac{π}{sin(π/a)}.

Solution III by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; Carl Libis, University of Alabama, Tuscaloosa, Alabama, Jerry Masuda, Metropolitan Community College, Blue Springs, Missouri; and Ice B. Risteski, Skopje, Macedonia.

According to Formula 402 on p. 303 of the Eleventh Edition of the C. R. C. Standard Mathematical Tables for 0 < m < a,

∫_{0}^{∞} \frac{t^{m−1}dt}{1 + t^{a}} = \frac{π}{a sin \frac{mπ}{a}} .  (∗)

Let k be an arbitrary fixed positive integer and let a ≥ 2 be a positive integer. Then, letting m = 1 and t = x/k in (∗), we have

∫_{0}^{∞} \frac{dx}{k^{a} + x^{a}} = \frac{1}{k^{a}} ∫_{0}^{∞} \frac{dx}{1 + (\frac{x}{k})^{a}} = \frac{k}{k^{a}} ∫_{0}^{∞} \frac{\frac{1}{k} dx}{1 + (\frac{x}{k})^{a}}

= \frac{1}{k^{a−1}} ∫_{0}^{∞} \frac{d(\frac{x}{k})}{1 + (\frac{x}{k})^{a}} = \frac{1}{k^{a−1}} \frac{π}{a sin \frac{π}{a}}

= \frac{1}{k^{a−1}} \frac{\frac{π}{a}}{sin(\frac{π}{a} )}.

Thus,

∫_{0}^{∞} \frac{dx}{1^{a} + x^{a}} + ∫_{0}^{∞} \frac{dx}{2^{a} + x^{a}} + ⋯ + ∫_{0}^{∞} \frac{dx}{n^{a} + x^{a}}

= \frac{1}{1^{a−1}} \begin{matrix}( & \frac{π}{a} & ) \\ sin & ( & \frac{π}{a} & ) \end{matrix} + \frac{1}{2^{a−1}} \begin{matrix}( & \frac{π}{a} & ) \\ sin & ( & \frac{π}{a} & ) \end{matrix} + ⋯ + \frac{1}{n^{a−1}} \begin{matrix}( & \frac{π}{a} & ) \\ sin & ( & \frac{π}{a} & ) \end{matrix}

= [\frac{1}{1^{a−1}} + \frac{1}{2^{a−1}} + ⋯ + \frac{1}{n^{a−1}} ]\begin{matrix}( & \frac{π}{a} & ) \\ sin & ( & \frac{π}{a} & ) \end{matrix}.

Also partially solved by the proposer.

110. [1997, 184] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Let α, β, and γ be the three angles of any triangle. Show that

\frac{sin α}{1 + sin β sin γ} + \frac{sin β}{1 + sin α sin γ} + \frac{sin γ}{1 + sin α sin β} < 2.

Solution I by the proposer. Without loss of generality, we may assume that sin α ≤ sin β ≤ sin γ. Now,

from the fact that 0 ≤ (1 − sin α)(1 − sin β), we have

sin α + sin β ≤ 1 + sin α sin β < 1 + 2 sin α sin β.

On the other hand,

sin α + sin β + sin γ ≤ 1 + sin α + sin β < 2 + 2 sin α sin β = 2(1 + sin α sin β). Therefore,

\frac{sin α}{1 + sin β sin γ} + \frac{sin β}{1 + sin α sin γ} + \frac{sin γ}{1 + sin α sin β}
≤ \frac{sin α}{1 + sin α sin β} + \frac{sin β}{1 + sin α sin β} + \frac{sin γ}{1 + sin α sin β}
= \frac{sin α + sin β + sin γ}{1 + sin α sin β} < \frac{2(1 + sin α sin β)}{1 + sin α sin β}
= 2.

Solution II by Joe Howard, New Mexico Highlands University, Las Vegas, New Mexico.

A generalization is given as Problem 25, p. 34 of Leningrad Math Olympiads (1987–1991) by D. Fomin and A. Kirichenko, MathPro Press, 1994.

\underline{Problem}. If 0 ≤ A, B, C ≤ 1, prove

\frac{A}{1 + BC} + \frac{B}{1 + AC} + \frac{C}{1 + AB} ≤ 2.

\underline{Solution}. Without loss of generality, assume 0 ≤ A ≤ B ≤ C ≤ 1. Since 0 ≤ (1−A)(1−B), we have A+B ≤ 1+AB ≤ 1+2AB. Furthermore, A+B +C ≤ A + B + 1 ≤ 2 + 2AB = 2(1 + AB). Hence,

\frac{A}{1 + BC} + \frac{B}{1 + AC} + \frac{C}{1 + AB} ≤ \frac{A}{1 + AB} + \frac{B}{1 + AB} + \frac{C}{1 + AB} = \frac{A + B + C}{1 + AB} ≤ 2.

Note that in the original problem, only one sin can be 1 (angle = π/2) so we have strict inequality.

111. [1997, 185] Proposed by Herta T. Freitag, Roanoke, Virginia.

D is a 3 by 3 determinant whose elements are polygonal numbers P_{n,k} such that

a_{i,j} = P_{n+3i+j−4,k}, k ≥ 3,

where P_{n,k} is the nth polygonal number of k “dimensions” (P_{5,3} is the 5th triangular number). Show that D is a cube independent of n.

Solution by Ice B. Risteski, Skopje, Macedonia.

The formula for the n–th polygonal number of k “dimensions” is given by P_{n,k} = nr/2, where r = pn + q, p = k − 2 and q = −k + 4, (see [1]). Now, the 3 × 3 determinant D whose entries are a_{i,j} = P_{n+3i+j−4,k}, (k ≥ 3) has a form

D = \frac{1}{8} det ( \begin{matrix}nr & (n + 1)(r + p) & (n + 2)(r + 2p) \\ (n + 3)(r + 3p) & (n + 4)(r + 4p) & (n + 5)(r + 5p) \\ (n + 6)(r + 6p) & (n + 7)(r + 7p) & (n + 8)(r + 8p) \end{matrix} ) .  (1)

Let R_{1}, R_{2}, and R_{3} denote the rows of determinant (1). Then, we easily obtain the linear combination

R_{1} − 2R_{2} + R_{3} = 18p ( 1 1 1 ) .

Hence, (1) will become

\begin{aligned}D & = \frac{9p}{4} det ( \begin{matrix}1 & 1 & 1 \\ (n + 3)(r + 3p) & (n + 4)(r + 4p) & (n + 5)(r + 5p) \\ (n + 6)(r + 6p) & (n + 7)(r + 7p) & (n + 8)(r + 8p) \end{matrix} ) \\ & = \frac{9p}{4} det ( \begin{matrix}np + r + 7p & np + r + 9p \\ np + r + 13p & np + r + 15p \end{matrix} ) = (6 − 3k)^{3}. \end{aligned}

\underline{Reference}

1. J. T. Bruening, Solution to Problem 97, Missouri Journal of Mathematical Sciences, 9 (1997), 189–191.

Also solved by James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri; Carl Libis, University of Alabama, Tuscaloosa, Alabama; Kenneth B. Davenport, Box 491, Frackville, Pennsylvania; and the proposer.

112. [1997, 185] Proposed by Mathew Timm, Bradley University, Peoria, Illinois.

Let Y be a connected, first countable Hausdorff space. Then Y is h–connected if and only if where p: X → Y is a finite–sheeted covering projection from a connected space X onto Y, it follows that X is homeomorphic to Y. Y is trivially h–connected if and only if whenever p: X → Y is a connected finite–sheeted covering projection of X onto Y, it follows that p is a homeomorphism of X onto Y.

Note that for non–trivially h–connected spaces, the covering projection p: X → Y is not required to be a homeomorphism, only that some homeomorphism exist between X and Y. Examples of non–trivially h–connected spaces include the circle S^{1}, the torus S^{1} × S^{1}, and, more generally, the n–tori S^{1} × ⋯ × S^{1}. Examples of trivially h–connected spaces include any simply connected finite simplicial complex or, more generally, any finite simplicial complex whose fundamental group has no proper finite index subgroups.

Recall that a topological space Y has the fixed point property if and only if, for every continuous function f: Y → Y, there is a y ∈ Y such that f(y) = y.

Now assume that Y is a first countable, Hausdorff, connected, locally path connected, semi–locally 1–connected space. Show that if Y is h–connected and has the fixed point property, then Y is trivially h–connected.

Solution by the proposer.

Since Y is connected, locally path connected, and semi–locally 1–connected, it follows from [2, 2.5.13] that for each subgroup H ≤ π_{1}(Y), there is a covering projection p: X → Y of a connected, locally path connected, semi–locally 1–connected X onto Y such that p_{∗}(π_{1}(X)) = H.

Now assume that Y is h–connected and has the fixed point property. In addition, assume that Y is non–trivially h–connected. Then there is an n–to–1 covering projection p: X → Y for some 1 < n < +∞. The space X satisfies the conditions listed in the first paragraph of the solution. In particular, X is connected and [π_{1}(Y) : p_{∗}(π_{1}(X))] = n > 1 is finite. Therefore, by standard results in group theory, e.g., [1, 3.3.5], there is a normal subgroup N ≤ π_{1}(Y), called the normal core of p_{∗}(π_{1}(X)), such that N ≤ p_{∗}(π_{1}(X)) and n ≤ [π_{1}(Y) : N] = k < +∞.

By the remarks in the first paragraph of this solution, there is a covering projection q: Ỹ → Y such that Ỹ is connected and q_{∗}(Ỹ) = N. Since N is normal in π_{1}(Y), it follows from [2, 2.6.2] that q: Ỹ → Y is a regular covering and has a group of covering transformations Aut_{Y} Ỹ ≅ π_{1}(Y)/N.

Now let f ∈ Aut_{Y} Ỹ. Since Ỹ is a connected k–to–1 covering of Y and Y is h–connected it follows that Ỹ is homeomorphic to Y. Since Y has the fixed point property, so does Ỹ. Therefore, f has a fixed point, y_{0}. So, by [2, 2.6.5] and the paragraph following [2, 2.6.6], it follows that f = idỸ. Thus Aut_{Y} Ỹ ≅ π_{1}(Y)/N = 1. So N = π_{1}(Y). So, since N ≤ p_{∗}(π_{1}(X)), it also follows that p_{∗}(π_{1}(X)) = π_{1}(Y). Thus, p is a 1–1 covering projection and so it follows that p is a homeomorphism. Thus, Y is trivially h–connected.

An inspection of the above proof points out that the full power of the fixed point property is not needed. All that is required is that homeomorphisms from Y to itself have the fixed point property.

\underline{References}

1. W. R. Scott, Group Theory, Dover, 1987.

2. E. H. Spanier, Algebraic Topology, McGraw–Hill, 1996.