SOLUTIONS

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141. [2003, 200] Proposed by Kenneth B. Davenport, Frackville, Pennsylvania.

(a) Show that

∑_{k=1}^{n−1} \frac{1}{1 − tan^{2}(\frac{kπ}{2n} )} = \frac{n − 1}{2},

for n = 3, 5, 7, 9, . . ..

(b) Show that

∑_{k=1}^{n−1} \frac{1}{1 + tan^{2}(\frac{kπ}{2n} )} = \frac{n − 1}{2},

for n = 2, 3, 4, 5, . . ..

Solution by Joe Howard, Portales, New Mexico. Note that cos θ = − cos(π −θ) and sec θ = − sec(π − θ). By pairing in this way (Ex: cos \frac{π}{3} + cos \frac{2π}{3} = 0) it follows that

∑_{k=1}^{n−1} cos \frac{kπ}{n} = 0 and ∑_{k=1}^{n−1} sec \frac{kπ}{n} = 0 for n = 3, 5, 7, 9, ….

Also, cos \frac{π}{2} = 0 so

∑_{k=1}^{n−1} cos \frac{kπ}{n} = 0 for n = 2, 4, 6, 8, ….

(a) For n = 3, 5, 7, 9, …

∑_{k=1}^{n−1} \frac{1}{1 − tan^{2} \frac{kπ}{2n}} = ∑_{k=1}^{n−1} \frac{cos^{2} \frac{kπ}{2n}}{cos^{2} \frac{kπ}{2n} − sin^{2} \frac{kπ}{2n}} = \frac{1}{2} ∑_{k=1}^{n−1} \frac{1 + cos \frac{kπ}{n}}{cos \frac{kπ}{n}}

= \frac{1}{2} ∑_{k=1}^{n−1}(1 + sec \frac{kπ}{n} ) = \frac{n − 1}{2} + \frac{1}{2} ∑_{k=1}^{n−1} sec \frac{kπ}{n} = \frac{n − 1}{2}.

(b) For n = 2, 3, 4, 5, …

∑_{k=1}^{n−1} \frac{1}{1 + tan^{2} \frac{kπ}{2n}} = ∑_{k=1}^{n−1} cos^{2} \frac{kπ}{2n} = \frac{1}{2} ∑_{k=1}^{n−1}(1 + cos \frac{kπ}{n} )

= \frac{n − 1}{2} + \frac{1}{2} ∑_{k=1}^{n−1} cos \frac{kπ}{n} = \frac{n − 1}{2}.

Also solved by Joe Flowers, Texas Lutheran University, Sequin, Texas; Don Redmond, Southern Illinois University, Carbondale, Illinois; Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri; Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan; and the proposer.

Comment by the proposer. For a related problem, see Problem H-566 [2000, 377; 2001, 474–476] in The Fibonacci Quarterly.

142. [2003, 201] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Solve the differential equation

y′ + \frac{y}{x^{x}} (x^{x} ln y)^{n} + y(ln x ln y) = 0,

where n is any real number.

Solution by Joe Flowers, Texas Lutheran University, Sequin, Texas. The substitution v = x^{x} ln y leads in straightforward fashion to the Bernoulli equation

v′ − v = −v^{n}.

If n = 1, then v′ = 0, so v = c, hence

y = e^{\frac{c}{x^{x}}} .

For n ≠ 1, the substitution v = w^{\frac{1}{1−n}} yields the linear equation

w′ + (n − 1)w = n − 1.

Applying the integrating factor e^{(n−1)x}, we obtain the solution

w = 1 + ce^{(1−n)x}

which then gives

v = (1 + ce^{(1−n)x})^{\frac{1}{1−n}}

and finally

y = exp(\frac{(1 + ce^{(1−n)x})^{\frac{1}{1−n}}}{x^{x}} ).

Also solved by Kenneth B. Davenport, Frackville, Pennsylvania, Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan, J. D. Chow, Edinburg, Texas; and the proposer.

144. [2003, 201] Proposed by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan.

Prove that in any triangle the following inequality holds:

∑ \frac{b + c}{a} tan \frac{B}{2} tan \frac{C}{2} ≥ 2,

where the notations are usual.

Solution I by Mangho Ahuja, Southeast Missouri State University, Cape Girardeau, Missouri. Using the identity

tan \frac{A}{2} = \sqrt{\frac{(s − b)(s − c)}{s(s − a)}},

where s = (a + b + c)/2, we have

tan \frac{B}{2} tan \frac{C}{2} = \sqrt{\frac{(s − a)(s − c)}{s(s − b)}}\sqrt{\frac{(s − b)(s − a)}{s(s − c)}} = \frac{s − a}{s}.

The given expression

∑ \frac{b + c}{a} tan \frac{B}{2} tan \frac{C}{2} = ∑ \frac{b + c}{a} · \frac{s − a}{s} = ∑ \frac{b + c}{a}(1 − \frac{a}{s})

= ∑ \frac{b + c}{a} − ∑ \frac{b + c}{s} = \frac{b}{a} + \frac{c}{a} + \frac{c}{b} + \frac{a}{b} + \frac{a}{c} + \frac{b}{c} − \frac{2a + 2b + 2c}{s}

= (\frac{a}{b} + \frac{b}{a}) + (\frac{b}{c} + \frac{c}{b}) + (\frac{c}{a} + \frac{a}{c}) − \frac{4s}{s}

= (\frac{a}{b} + \frac{b}{a}) + (\frac{b}{c} + \frac{c}{b}) + (\frac{c}{a} + \frac{a}{c}) − 4.

But using the AM-GM inequality, the quantity

\frac{a}{b} + \frac{b}{a} ≥ 2.

Hence, the given expression is greater than or equal to

2 + 2 + 2 − 4 = 2.

Solution II by Joe Howard, Portales, New Mexico. We use the formula that

∑_{cyclic} tan \frac{B}{2} tan \frac{C}{2} = 1,

the inequality

u + \frac{1}{u} ≥ 2,

and Chebyshev’s Inequality

∑_{i=1}^{n} x_{i}y_{i} ≥ \frac{1}{n}(∑_{i=1}^{n} x_{i})(∑_{i=1}^{n} y_{i}),

where x_{1} ≥ x_{2} ≥ … ≥ x_{n} > 0 and y_{1} ≥ y_{2} ≥ … ≥ y_{n} > 0. Without loss of generality assume a ≥ b ≥ c. Then

\frac{a + b}{c} ≥ \frac{a + c}{b} ≥ \frac{b + c}{a}

and

tan \frac{A}{2} tan \frac{B}{2} ≥ tan \frac{A}{2} tan \frac{C}{2} ≥ tan \frac{B}{2} tan \frac{C}{2} .

Then

∑_{cyclic} \frac{b + c}{a} tan \frac{B}{2} tan \frac{C}{2}

≥ \frac{1}{3}(\frac{b}{a} + \frac{c}{a} + \frac{a}{b} + \frac{c}{b} + \frac{a}{c} + \frac{b}{c})( ∑_{cyclic} tan \frac{B}{2} tan \frac{C}{2} )

≥ \frac{1}{3}(6)(1) = 2.

Also solved by the proposer.