SOLUTIONS

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137. [2002, 210] Proposed by José Luis Díaz, Universidad Politécnica de Cataluña, Barcelona, Spain.

Find all non-negative integers a, b, and c such that a + b + c and abc are consecutive integers.

Solution by Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri. Assume without loss of generality that a ≤ b ≤ c. If 2 ≤ a ≤ b ≤ c, then writing b = a + m, c = a + n and abc = a + b + c + 1 implies

a^{3} + (m + n)a^{2} + amn = 3a + m + n + 1.

This is a contradiction as the left-hand side is larger than the right-hand side when a ≥ 2 and m and n are positive integers. We get a similar contradiction when we write abc = a + b + c − 1. It follows that a has to be either 0 or 1.

\underline{Case I}. If a = 0, then abc = a + b + c + 1 gives a contradiction. But abc + 1 = a + b + c implies 1 = b + c which implies b = 0 and c = 1. So the solution in this case is a = b = 0 and c = 1.

\underline{Case II}. If a = 1, then either bc = b + c + 2 or bc = b + c.

(i) If bc = b + c, then b = b/c + 1 and so we must have b = c. This implies that b = 2 and c = 2.

(ii) If bc = b + c + 2, then b = (b + 2)/c + 1. This implies that b + 2 = c. By substitution, b(b + 2) = 2b + 4 or b^{2} = 4 and so b = 2 and c = 4.

The solutions are (0, 0, 1), (1, 2, 2), and (1, 2, 4).

Also solved by J. D. Chow, Edinburg, Texas; Joe Howard, Portales, New Mexico; James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri; and the proposer.

138. [2002, 210] Proposed by Joe Howard, Portales, New Mexico. Suppose an acute triangle ABC has inradius r and area △. Prove

cot A + cot B + cot C ≥ \frac{△}{3r^{2}} .

Solution I by José Luis Díaz, Universidad Politécnica de Cataluña, Barcelona, Spain. From the Law of Cosines and Sines, we have the cyclic identities

cot A = \frac{cos A}{sin A} = \frac{b^{2} + c^{2} − a^{2}}{2bc sin A} = \frac{R}{abc}(b^{2} + c^{2} − a^{2}),

where R is the circumradius. We must show

cot A + cot B + cot C

= \frac{R}{abc}[(b^{2} + c^{2} − a^{2}) + (c^{2} + a^{2} − b^{2}) + (a^{2} + b^{2} − c^{2})] ≥ \frac{△}{3r^{2}}

or

3(a^{2} + b^{2} + c^{2}) ≥ \frac{△abc}{Rr^{2}}.  (1)

Taking into account that

R = \frac{abc}{4△} and △ = sr,

where s is the semiperimeter, inequality (1) becomes

3(a^{2} + b^{2} + c^{2}) ≥ (a + b + c)^{2}.  (2)

Finally, we need to prove (2). In fact,

\begin{aligned}& (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) \\ ≤ a^{2} + b^{2} & + c^{2} + (a^{2} + b^{2}) + (b^{2} + c^{2}) + (c^{2} + a^{2}) = 3(a^{2} + b^{2} + c^{2}). \end{aligned}

Note that equality holds for an equilateral triangle. This completes the proof.

Solution II by James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri. Assume sides a, b, c are opposite angles A, B, C, respectively. Let h_{a}, h_{b}, h_{c} be the altitudes on sides a, b, c, respectively. Since ABC is an acute triangle,

\frac{a}{h_{a}} = cot B + cot C, \frac{b}{h_{b}} = cot A + cot C, and \frac{c}{h_{c}} = cot A + cot B.

Thus,

cot A + cot B + cot C = \frac{1}{2}(\frac{a}{h_{a}} + \frac{b}{h_{b}} + \frac{c}{h_{c}} )

= \frac{1}{2}( \frac{a}{\frac{2△}{a}} + \frac{b}{\frac{2△}{b}} + \frac{c}{\frac{2△}{c}} )

= \frac{a^{2} + b^{2} + c^{2}}{4△} .

The inequality (a − b)^{2} + (a − c)^{2} + (b − c)^{2} ≥ 0 can be used to derive the inequality

a^{2} + b^{2} + c^{2} ≥ \frac{(a + b + c)^{2}}{3} .

It can also be shown that △ = sr, where s is the semiperimeter of the triangle. Applying these above gives

cot A + cot B + cot C = \frac{a^{2} + b^{2} + c^{2}}{4△} ≥ \frac{(a + b + c)^{2}}{12△} = \frac{(2s)^{2}}{12sr} = \frac{s}{3r} = \frac{sr}{3r^{2}} = \frac{△}{3r^{2}} .

This completes the proof.

Solution III by Mangho Ahuja, Southeast Missouri State University, Cape Girardeau, Missouri. Let E denote the expression

E = cot A + cot B + cot C.

Then,

E = \frac{2bc cos A}{2bc sin A} + \frac{2ca cos B}{2ca sin B} + \frac{2ab cos C}{2ab sin C} .

Let s denote the semiperimeter of the triangle ABC. We will use the identities

△ = \frac{1}{2}ab sin C, b^{2} + c^{2} − a^{2} = 2bc cos A, and △ = rs.

Then,

E = \frac{1}{4△}[(b^{2} + c^{2} − a^{2}) + (c^{2} + a^{2} − b^{2}) + (a^{2} + b^{2} − c^{2})] = \frac{1}{4△}(a^{2} + b^{2} + c^{2}).

Using

\sqrt{\frac{a^{2} + b^{2} + c^{2}}{3}} ≥ \frac{a + b + c}{3},

we get

E ≥ \frac{1}{4△} · \frac{(a + b + c)^{2}}{3} = \frac{1}{4△} · \frac{4s^{2}}{3} = \frac{△}{3r^{2}} .

Solution IV by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan. In [1], the following two lemmas are proved.

\underline{Lemma 1}. In any triangle the following indentity holds.

cot A + cot B + cot C = \frac{s^{2} − r(4R + r)}{2sr}.

Here s is the semiperimeter and R is the circumradius.

\underline{Lemma 2}. In any triangle the following inequality holds.

s^{2} ≥ 3r^{2} + 12Rr.

The inequality we wish to prove now reads

\frac{s^{2} − r(4R + r)}{2sr} ≥ \frac{△}{3r^{2}} = \frac{sr}{3r^{2}} ,

since △ = sr. But this inequality is equalivalent to

3[s^{2} − r(4R + r)] ≥ 2s^{2}

which is equivalent to

s^{2} ≥ 3r^{2} + 12Rr.

This last inequality holds according to Lemma 2. This completes the proof.

\underline{Observation}. The above inequality holds in any triangle (not necessarily an acute one).

\underline{Reference}

1. Constantin C. Florea, Abordave globala a geometriei triunghiului cu implicatii creative.

Also solved by J. D. Chow, Edinburg, Texas; Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri; and the proposer.

139. [2002, 210] Proposed by José Luis Díaz, Universidad Politécnica de Cataluña, Barcelona, Spain.

Let F_{n} denote the nth Fibonacci number (F_{0} = 0, F_{1} = 1, and F_{n} = F_{n−1} + F_{n−2} for n ≥ 2) and let L_{n} denote the nth Lucas number (L_{0} = 2, L_{1} = 1, and L_{n} = L_{n−1} + L_{n−2} for n ≥ 2). Show that

∑_{k=1}^{n} L_{k}^{3} = \frac{1}{2}(F_{3n+3} + F_{3n+1} + 12(−1)^{n}F_{n} + 6(−1)^{n−1}F_{n−1} + 3).

Solution by James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri. Let

α = \frac{1 + \sqrt{5}}{2} and β = \frac{1 − \sqrt{5}}{2}.

It is known that the nth Fibonacci number can be represented as

F_{n} = \frac{α^{n} − β^{n}}{\sqrt{5}}

and the nth Lucas number can be represented as

L_{n} = α^{n} + β^{n}.

These representations can be used to prove the following identities:

L_{n−1} = F_{n} + F_{n−2},

∑_{k=1}^{n}(−1)^{k}L_{k} = (−1)^{n}L_{n−1} + 1,

2 · ∑_{k=1}^{n} L_{3k} = L_{3n+2} − L_{2} = L_{3n+2} − 3.

Thus,

∑_{k=1}^{n} L_{k}^{3} = ∑_{k=1}^{n}(α^{k} + β^{k})^{3}

= ∑_{k=1}^{n}(α^{3k} + 3α^{2k}β^{k} + 3α^{k}β^{2k} + β^{3k})

= ∑_{k=1}^{n}(α^{3k} + β^{3k} + 3α^{k}β^{k}(α^{k} + β^{k}))

= ∑_{k=1}^{n}(α^{3k} + β^{3k} + 3(−1)^{k}(α^{k} + β^{k}))

= ∑_{k=1}^{n}(L_{3k} + 3(−1)^{k}L_{k}) = ∑_{k=1}^{n} L_{3k} + 3 ∑_{k=1}^{n}(−1)^{k}L_{k}

= \frac{L_{3n+2}}{2} − \frac{3}{2} + \frac{6}{2}((−1)^{n}L_{n−1} + 1)

= \frac{1}{2}(L_{3n+2} + 6(−1)^{n}L_{n−1} + 6 − 3)

= \frac{1}{2}(F_{3n+3} + F_{3n+1} + 6(−1)^{n}(F_{n} + F_{n−2}) + 3)

= \frac{1}{2}(F_{3n+3} + F_{3n+1} + 6(−1)^{n}(F_{n} + F_{n} − F_{n−1}) + 3)

= \frac{1}{2}(F_{3n+3} + F_{3n+1} + 6(−1)^{n}(2F_{n} + F_{n−1}) + 3)

= \frac{1}{2}(F_{3n+3} + F_{3n+1} + 12(−1)^{n}F_{n} + 6(−1)^{n−1}F_{n−1} + 3).

This completes the proof.

\underline{Note}. The proposer notes that

∑_{k=1}^{n} F_{k}^{3} = \frac{1}{10}F_{3n+2} + \frac{3}{5}(−1)^{n−1}F_{n−1} + \frac{1}{2}

was a problem in [1].

\underline{Reference}

1. C. Cooper and R. Kennedy, “Problem 3,” Missouri Journal of Mathematical Sciences, 0 (1988), 29.

Also solved by Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri; Joe Howard, Portales, New Mexico; Don Redmond, Southern Illinois University at Carbondale, Carbondale, Illinois; Kenneth B. Davenport, Frackville, Pennsylvania; and the proposer.

140. [2002, 211] Proposed by José Luis Díaz, Universidad Politécnica de Cataluña, Barcelona, Spain.

The numbers a, b and c are in geometric progression if and only if

(ab + bc + ca)^{3} = abc(a + b + c)^{3}.

Prove this.

Solution by Mangho Ahuja, Southeast Missouri State University, Cape Girardeau, Missouri. Let E denote the expression

E = abc(a + b + c)^{3} − (ab + bc + ca)^{3}.

If we substitute b^{2} = ac in E, the expression reduces to zero, hence E has a factor (b^{2} − ac). But since E is symmetric in a, b, c; E must have the other two factors (c^{2} − ab) and (a^{2} − bc) as well. Thus,

E = (b^{2} − ac)(c^{2} − ab)(a^{2} − bc)k.

On comparing the powers of the polynomial expressions on both sides of this equation, we conclude that k must be a constant. On comparing (on both sides) the coefficients of one of the terms, say a^{4}bc, we see that k must be 1.

We have thus established that

E = (b^{2} − ac)(c^{2} − ab)(a^{2} − bc).

It follows that E = 0 if and only if one of the three factors is zero, which means the three numbers a, b, c are in geometric progression.

Also solved by James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri; J. D. Chow, Edinburg, Texas; Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri; Joe Howard, Portales, New Mexico; and the proposer.