SOLUTIONS

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

25. [1990, 140] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Let P be the free product with amalgamations of any collection {C_{γ}}_{γ∈Γ} of infinite cyclic groups, where Γ is an indexing set of cardinality greater than one.

A. Construct an element h ∈ P and a subgroup T < P such that |P : T | is infinite but |P :< h, T > | is finite.

B. Is h uniquely determined? Is T uniquely determined?

Solution by the proposer.

Let C_{γ} =< c_{γ} >, let H_{γ} =< c_{γ}^{m_{γ}} >, where m_{γ} are natural numbers, greater than one, and let

P = ⋆_{γ∈Γ}(C_{γ}; H_{γ})

with amalgamated subgroup H.

A. Consider an arbitrary element

h = c_{γ}^{qm_{γ}} (γ ∈ Γ)

of H, where q is a natural number. Now, let

T =< c_{γ}^{p_{γ}} : γ ∈ Γ > ,

where p_{γ} are prime numbers greater than one and (p_{γ}, qm_{γ}) = 1, for all γ ∈ Γ. We claim that this h is the required element and also the subgroup T defined as above is the required subgroup. (Note that if Γ is finite and p is any prime number greater than one, such that (p, qm_{γ}) = 1, then

T =< c_{γ}^{p} : γ ∈ Γ >) .

Proof of claim: (p_{γ}, qm_{γ}) = 1 implies that there are integers r_{γ} and s_{γ} such that

r_{γ}p_{γ} + s_{γ}qm_{γ} = 1 .

Thus,

c_{γ} = c_{γ}^{r_{γ}p_{γ}+s_{γ}qm_{γ}}

is an element of < T, h >, so that < T, h >= P. Therefore, it only remains to prove that T is of infinite index in P. To establish this fact, we choose α, β, …, η distinct in Γ, and we show that g_{1}, g_{2}, …, g_{r}, … are incongruent (mod T), where

g_{r} = (c_{α}c_{β} ⋯ c_{η})^{r}

and r is a natural number. For r and s different natural numbers, we wish to verify that

g_{r}^{−1}g_{s} ∉ T .

If not, then there exists an element t ∈ T such that

g_{r}^{−1}g_{s} = (c_{α}c_{β} ⋯ c_{η})^{s−r} = t = ∏_{i=1}^{n} c_{γ_{i}}^{k_{i}p_{γ_{i}}} ,

where k_{i} is a non–zero integer. Now if r > s, then

1 = c_{α}c_{β} ⋯ c_{η} ⋯ c_{α}c_{β} ⋯ c_{η}t .

If γ_{1} ≠ η, no cancellations or simplifications are possible. Also, if γ_{1} = η, then

c_{η}c_{γ_{1}}^{k_{1}p_{γ_{1}}} = c_{η}^{1+k_{1}p_{γ_{1}}}

is not 1. Thus, we have a non–trivial expression for 1, a contradiction. A similar argument will apply for r < s. Therefore, r ≠ s implies that

g_{r} ≢ g_{s} (mod T) .

This completes the proof.

B. Obviously, neither h nor T is unique. Because a different q will produce a different h and also, as primes p_{γ} change so does T.

26*. [1990, 140] Proposed by Stanley Rabinowitz, Westford, Massachusetts.

Prove that

∑_{k=1}^{38} sin \frac{k^{8}π}{38} = \sqrt{19} .

Comment by the editor.

No solutions have been received on this problem to date.

27. [1990, 141] Proposed by Don Redmond, Southern Illinois University, Carbondale, Illinois.

Show that

∫_{1}^{∞} \frac{t − [t]}{t^{2}(t + 1)^{2}} (2t + 1)dt = log 2 − \frac{1}{2} ,

where log 2 denotes the natural logarithm of 2 and [t] is the greatest integer less than or equal to t.

Solution I by Jayanthi Ganapathy, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

First note that, for any integer i ≥ 2, ‘integration by parts’ yields

∫_{i−1}^{i} \frac{(t − [t])(2t + 1)}{t^{2}(t + 1)^{2}}dt = −\frac{t − [t]}{t(t + 1)}|_{i−1}^{i} + log(\frac{t}{t + 1})|_{i−1}^{i}

= −\frac{1}{i(i + 1)} + log(\frac{i}{i + 1}) − log(\frac{i − 1}{i})

= −(\frac{1}{i} − \frac{1}{i + 1}) + log(\frac{i}{i + 1}) − log(\frac{i − 1}{i}) ,

since, for any integer k ≥ 1, t − [t] → 0 as t → k^{+} and t − [t] → 1 as t → k^{−}. Thus for any integer n > 1,

∫_{1}^{n} \frac{(t − [t])(2t + 1)}{t^{2}(t + 1)^{2}}dt = − ∑_{i=2}^{n}(\frac{1}{i} − \frac{1}{i + 1}) + ∑_{i=2}^{n}(log(\frac{i}{i + 1}) − log(\frac{i − 1}{i}))

= −\frac{1}{2} + \frac{1}{n + 1} + log(\frac{n}{n + 1}) − log(\frac{1}{2}) .

Let n → ∞. It now follows that

\begin{aligned}∫_{1}^{∞} \frac{(t − [t])(2t + 1)}{t^{2}(t + 1)^{2}}dt & = −\frac{1}{2} − log(\frac{1}{2}) \\ & = log 2 − \frac{1}{2} . \end{aligned}

Solution II by Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

It can be easily shown that

\frac{(t − n)(2t + 1)}{t^{2}(t + 1)^{2}} = \frac{1}{t} − \frac{n}{t^{2}} − \frac{1}{t + 1} + \frac{n + 1}{(t + 1)^{2}} .

Hence for every positive integer n,

∫_{n}^{n+1} \frac{t − [t]}{t^{2}(t + 1)^{2}} (2t + 1)dt = ∫_{n}^{n+1} \frac{(t − n)}{t^{2}(t + 1)^{2}} (2t + 1)dt

= (log t + \frac{n}{t} − log(t + 1) − \frac{n + 1}{t + 1})|_{n}^{n+1}

= log(n + 1) − log n − \frac{1}{n + 1} − log(n + 2) + log(n + 1) + \frac{1}{n + 2} .

Hence the integral in the given problem equals

lim_{n→∞} ∑_{r=1}^{n}(log(r + 1) − log r + log(r + 1) − log(r + 2) + \frac{1}{r + 2} − \frac{1}{r + 1})

= lim_{n→∞}(log(n + 1) + log 2 − log(n + 2) + \frac{1}{n + 2} − \frac{1}{2})

= lim_{n→∞}(log(\frac{n + 1}{n + 2}) + log 2 + \frac{1}{n + 2} − \frac{1}{2})

= log 2 − \frac{1}{2} .

This completes the solution.

Solution III by the proposer. Let a be a positive real number and b be a nonnegative real number. Then

(1)  ∑_{k=1}^{∞} \frac{1}{(ak + b)(a(k + 1) + b)} = \frac{1}{a} ∑_{k=1}^{∞}(\frac{1}{ak + b} − \frac{1}{a(k + 1) + b}) = \frac{1}{a(a + b)} .

If we write this sum as an integral we have

(2)  \begin{aligned}∑_{k=1}^{∞} & \frac{1}{(ak + b)(a(k + 1) + b)} = ∫_{1^{−}}^{∞} \frac{d[t]}{(at + b)(a(t + 1) + b)} \\ & = ∫_{1}^{∞} \frac{dt}{(at + b)(a(t + 1) + b)} + ∫_{1^{−}}^{∞} \frac{d([t] − t)}{(at + b)(a(t + 1) + b)} \\ & = I_{1} + I_{2} . \end{aligned}

We have

(3)

I_{1} = \frac{1}{a} ∫_{1}^{∞}(\frac{1}{at + b} − \frac{1}{at + b + a})dt

= \frac{1}{a^{2}} ( lim_{t→∞} log \frac{at + b}{at + b + a} − log \frac{a + b}{2a + b})

= −\frac{1}{a^{2}} log \frac{a + b}{2a + b} .

To deal with I_{2} we integrate by parts to get

(4) \begin{aligned}I_{2} & = \frac{[t] − t}{(at + b)(at + b + a)}|_{1^{−}}^{∞} − ∫_{1}^{∞} \frac{(t − [t])}{(at + b)^{2}(at + a + b)^{2}} (2a^{2}t + a^{2} + 2ab)dt \\ & = \frac{1}{(a + b)(2a + b)} − ∫_{1}^{∞} \frac{(t − [t])(2a^{2}t + a^{2} + 2ab)}{(at + b)^{2}(at + a + b)^{2}}dt . \end{aligned}

If we now combine (1)–(4) we obtain

\begin{aligned}∫{∞ \atop 1} & \frac{(t − [t])(2a^{2}t + a^{2} + 2ab)}{(at + b)^{2}(at + a + b)^{2}}dt = −\frac{1}{a^{2}(a + b)} − \frac{1}{a^{3}} log \frac{a + b}{2a + b} + \frac{1}{a(a + b)(2a + b)} \\ & = −\frac{1}{a^{3}} log \frac{a + b}{2a + b} − \frac{1}{a^{2}(2a + b)} . \end{aligned}

If we take a = 1 and b = 0 we get the desired result.

28. [1990, 141] Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Let ABC be an equilateral triangle with segment lengths as indicated in the diagram. Determine s as a function of a, b and c.

Solution by the proposer. Using the law of sines in triangle BCO gives

\frac{sin(60^{∘} − ϕ)}{b} = \frac{sin(60^{∘} − θ)}{c} ,

which simplifies to

(1)  b sin θ − c sin ϕ = \sqrt{3}(b cos θ − c cos ϕ) .

However, from triangles ABO and ACO,

(2)  cos θ = \frac{−a^{2} + b^{2} + s^{2}}{2bs} ,

and

(3)  cos ϕ = \frac{−a^{2} + c^{2} + s^{2}}{2cs} .

Substituting (2) and (3) into (1) yields

(4)  b sin θ − c sin ϕ = \frac{\sqrt{3}(b^{2} − c^{2})}{2s} .

Since the area of triangle ABC is the sum of the areas of triangles ABO, ACO and BCO,

(5)  b sin θ + b sin(60^{∘} − θ) + c sin ϕ = \frac{\sqrt{3}s}{2} .

Adding equations (4) and (5) and simplifying with the use of (2) leads to

(6)  \sqrt{3}b sin θ = \frac{s^{2} + a^{2} + b^{2} − 2c^{2}}{2s} .

Now, square equation (6), replace sin^{2} θ with 1 − cos^{2} θ and use (2) to arrive at

s^{4} − (a^{2} + b^{2} + c^{2})s^{2} + a^{4} + b^{4} + c^{4} − a^{2}b^{2} − a^{2}c^{2} − b^{2}c^{2} = 0 ,

which has

s = \sqrt{\frac{a^{2} + b^{2} + c^{2} + \sqrt{−3(a^{4} + b^{4} + c^{4} − 2a^{2}b^{2} − 2a^{2}c^{2} − 2b^{2}c^{2})}}{2}}

as the desired root.