SOLUTIONS

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149. [2004, 129] Proposed by Joe Howard, Portales, New Mexico and Les Reid, Southwest Missouri State University, Springfield, Missouri.

Let A, B, C be the angles of a triangle. Show

3 + cos A + cos B + cos C ≥ 2(sin A sin B + sin B sin C + sin C sin A)
≥ 9(cos A + cos B + cos C − 1)

with equality if and only if the triangle is equilateral.

Solution by the proposers. Let r, R, and s denote the inradius, circumradius, and semiperimeter, respectively. Euler’s inequality (R ≥ 2r) is well known. The following can be found in Crux Mathematicorum, [2029] 22 (3) (1996), p. 130 and on pages 45 and 55–56 of D. S. Mitrinović, J. E. Pec̆urić, and V. Volenec, Recent Advances in Geometric Inequalities, Kluwer Academic Pub., 1989.

sin A sin B + sin B sin C + sin C sin A = \frac{s^{2} + 4Rr + r^{2}}{4R^{2}}  (1)

cos A + cos B + cos C = 1 + \frac{r}{R}  (2)

4R^{2} + 4Rr + 3r^{2} ≥ s^{2} ≥ 16Rr − 5r^{2}  (3)

Using (1) and (2), the given inequality is equivalent to

4 + \frac{r}{R} ≥ 2(\frac{s^{2} + 4Rr + r^{2}}{4R^{2}} ) ≥ 9(\frac{r}{R})

which is equivalent to

8R^{2} − 2Rr − r^{2} ≥ s^{2} ≥ 14Rr − r^{2}.

For the first inequality: from Euler’s inequality (R ≥ 2r); R^{2} ≥ 4r^{2} and 3R^{2} ≥ 6Rr which implies 4R^{2} ≥ 6Rr + 4r^{2}. By this and (3),

8R^{2} − 2Rr − r^{2} ≥ 4R^{2} + 4Rr + 3r^{2} ≥ s^{2}.

For the second inequality: from Euler’s inequality (R ≥ 2r); 2Rr ≥ 4r^{2} and with (3) implies

s^{2} ≥ 16Rr − 5r^{2} ≥ 14Rr − r^{2}.

The equality is clear when the triangle is equilateral. Conversely, if 8R^{2} − 2Rr − r^{2} = s^{2}, then the first part of (3) yields

4R^{2} + 4Rr + 3r^{2} ≥ 8R^{2} − 2Rr − r^{2}

or

0 ≥ 4R^{2} − 6Rr − 4r^{2} = 2(R − 2r)(2R + r).

This implies that 2r ≥ R, which along with Euler’s inequality forces 2r = R, hence the triangle must be equilateral. A similar argument works for the other inequality.

Also solved by Scott H. Brown, Auburn University, Montgomery, Alabama and Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan.

150. [2004, 130] Proposed by Ovidui Furdui, Western Michigan University, Kalamazoo, Michigan.

Evaluate

∫∫_{D} \{\frac{1}{x + y} \}dx dy,

where D = [0, 1] × [0, 1] and {a} is the fractional part of a.

Solution by Huizeng Qin, Shandong University of Technology, Zibo, People’s Republic of China. Let

x = \frac{1}{2}(u − v), y = \frac{1}{2}(u + v).  (1)

Then,

∫∫_{D} \{\frac{1}{x + y} \}dx dy = \frac{1}{2}(∫∫_{D_{1}} + ∫∫_{D_{2}})\{\frac{1}{u}\}du dv = I_{1} + I_{2},  (2)

where

D_{1} = {−u ≤ v ≤ u, 0 ≤ u ≤ 1} and D_{2} = {u − 2 ≤ v ≤ 2 − u, 1 ≤ u ≤ 2}. Now, we evaluate I_{1} and I_{2}, separately. Since {1/u} = 1/u when u > 1, we first have

I_{2} = \frac{1}{2} ∫∫_{D_{2}} \frac{1}{u}du dv = \frac{1}{2} ∫_{1}^{2} ∫_{u−2}^{2−u} \frac{1}{u}dv du = ∫_{1}^{2} \frac{2 − u}{u} du = 2 ln 2 − 1. (3)

To evaluate I_{1}, we rewrite {a} as a − ⌊a⌋, where ⌊a⌋ is the integer part of a. Now, we can use the well known result

∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6}  (4)

to obtain

I_{1} = \frac{1}{2} ∫_{0}^{1} ∫_{−u}^{u}(\frac{1}{u} − ⌊\frac{1}{u}⌋)dv du = 1 − ∫_{0}^{1} u⌊\frac{1}{u}⌋du  (5)

= 1 − ∫_{1}^{∞} \frac{⌊v⌋}{v^{3}}dv = ∑_{n=1}^{∞} ∫_{n}^{n+1} \frac{n}{v^{3}} dv

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

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

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

= \frac{1}{2} − \frac{1}{2}(\frac{π^{2}}{6} − 1) = 1 − \frac{π^{2}}{12} .

Combining (2), (3), and (5), we get

∫∫_{D} \{\frac{1}{x + y} \}dx dy = 2 ln 2 − \frac{π^{2}}{12} .

Also solved by the proposer.

151. [2004, 130] Proposed by José Luis Diaz-Barrero, Universidad Politécnica de Cataluña, Barcelona, Spain.

Solve the differential equation

\sqrt{1 + y^{2}}x^{2}e^{arctan x}dx + \sqrt{1 + x^{2}}dy = 0.

Solution by the proposer. Dividing both sides of the given equation by \sqrt{(1 + x^{2})(1 + y^{2})} and rearranging terms yields

−\frac{dy}{\sqrt{1 + y^{2}}} = \frac{x^{2}e^{arctan x}}{\sqrt{1 + x^{2}}} dx.  (1)

Integrating both sides of (1), we get

− ln(y + \sqrt{1 + y^{2}}) = − sinh^{−1}(y) = F (x),

where

F (x) = ∫ \frac{x^{2}e^{arctan x}}{\sqrt{1 + x^{2}}} dx.

To obtain F (x), we call

I_{1} = ∫ \frac{e^{arctan x}}{\sqrt{1 + x^{2}}} dx

and

I_{2} = ∫ \frac{xe^{arctan x}}{\sqrt{1 + x^{2}}} dx.

Then,

I_{1} = ∫ \frac{e^{arctan x}}{\sqrt{1 + x^{2}}} dx = ∫ \frac{\sqrt{1 + x^{2}}e^{arctan x}}{1 + x^{2}} dx

= ∫ \sqrt{1 + x^{2}}d(e^{arctan x}) = \sqrt{1 + x^{2}}e^{arctan x} − I_{2}

and so,

I_{1} + I_{2} = \sqrt{1 + x^{2}}e^{arctan x}.  (2)

On the other hand,

I_{2} = ∫ \frac{xe^{arctan x}}{\sqrt{1 + x^{2}}} dx = ∫ \frac{x\sqrt{1 + x^{2}}e^{arctan x}}{1 + x^{2}} dx

= ∫ x\sqrt{1 + x^{2}}d(e^{arctan x})

= x\sqrt{1 + x^{2}}e^{arctan x} − I_{1} − 2 ∫ \frac{x^{2}e^{arctan x}}{\sqrt{1 + x^{2}}} dx.

That is,

F (x) = \frac{x\sqrt{1 + x^{2}}e^{arctan x} − (I_{1} + I_{2})}{2}.

Taking into account (2), we obtain

F (x) = \frac{1}{2}(x − 1)e^{arctan x}\sqrt{1 + x^{2}}.

Therefore,

sinh^{−1}(y) = \frac{1}{2}(1 − x)e^{arctan x}\sqrt{1 + x^{2}} + C

or

y = sinh(\frac{(1 − x)e^{arctan x}\sqrt{1 + x^{2}}}{2} + C), C ∈ ℝ

and we are done.

Also solved by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan and Kenneth Davenport, Dallas, Pennsylvania.

152. [2004, 130] Proposed by Joe Flowers and Doug Martin (student), Texas Lutheran University, Seguin, Texas.

Let

F (s) = L[f(t)] = ∫_{0}^{∞} e^{−st}f(t) dt

denote the Laplace transform of f(t). Find L[sin^{n} bt], where b is any real constant and n is any non-negative integer.

Solution by the proposers. Let

F_{n}(s) = L[sin^{n} bt].

The results

F_{0}(s) = \frac{1}{s} and F_{1}(s) = \frac{b}{s^{2} + b^{2}}

are well-known. For n > 1, let

f(t) = sin^{n} bt.

Then,

f′(t) = nb · cos bt · sin^{n−1} bt

and

f′′(t) = nb^{2}((n − 1) sin^{n−2} bt − n · sin^{n} bt)

and therefore,

L[f′′(t)] = b^{2}(n(n − 1)F_{n−2}(s) − n^{2}F_{n}(s)).

Also, from the theory of Laplace transforms,

L[f′′(t)] = s^{2}F_{n}(s) − s · f(0) − f′(0) = s^{2}F_{n}(s).

From these we obtain the second degree recurrence formula

F_{n}(s) = \frac{b^{2} · n(n − 1)}{s^{2} + n^{2}b^{2}} · F_{n−2}(s)

and then by iteration,

F_{n}(s) = \frac{b^{n} · n!}{s · ∏_{k=1}^{n/2}(s^{2} + 4k^{2}b^{2})}

for even values of n and

F_{n}(s) = \frac{b^{n} · n!}{∏_{k=0}^{(n−1)/2}(s^{2} + (2k + 1)^{2}b^{2})}

for odd values of n.

Also solved by Thomas Dence, Ashland University, Ashland, Ohio and Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan. A partial solution was also received.