Arkady M. Alt

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(2)

About one multidimensional sum with Fibonacci numbers

by Arkady M. Alt 100

Two refinements of power Lessels-Pelling inequality

by Mihály Bencze and Marius Dr˘agan 110 Problems

Elementary Problems: E59–E64 119

Easy–Medium Problems: EM59–EM64 120

Medium–Hard Problems: MH59–MH64 121

Advanced Problems: A59–A64 123

Mathlessons

Theorems and problems in classical geometry

by José Luis Díaz-Barrero and Marc Felipe Alsina 126 Contests

Problems and solutions from the 59th edition of the Interna- tional Mathematical Olympiad (IMO)

by Óscar Rivero Salgado 140

Solutions

Elementary Problems: E53–E58 149

Easy–Medium Problems: EM53–EM58 157

Medium–Hard Problems: MH53–MH58 168

Advanced Problems: A53–A58 178

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Articles

Arhimede Mathematical Journal aims to publish interesting and attractive papers with elegant mathematical exposition. Articles should include examples, applications and illustrations, when- ever possible. Manuscripts submitted should not be currently submitted to or accepted for publication in another journal.

Please, send submittals to José Luis Díaz-Barrero, Enginyeria Civil i Ambiental, UPC BARCELONATECH, Jordi Girona 1-3, C2, 08034 Barcelona, Spain, or by e-mail to

[email protected]

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About one multidimensional sum with Fibonacci numbers

Arkady M. Alt

Abstract

This note is motivated by the following problem: Let (fn)_n≥0 be the Fibonacci sequence defined by f0 = 0, f1 = 1 and, for all n ≥ 1, fn+1 = fn+ fn−1. Determine

hn = X

i,j,k

fifjfk,

where the sum is over i, j, k > 0 with i + j + k = n. This problem appeared in Mathematical Horizons [1], and is also motivated by a problem of Díaz-Barrero [2].

1 Main results

We will consider the general problem of computing the sum S_m(n) := X

i1,i2,...,im≥0 i1+i2+...+im=n

f_i₁f_i₂. . . f_i_m

for any non negative integers m and n. (Note that S0(0) = 0 as the sum is over the empty set.) It is easy to see that, in particular,

(5)

S_m(0) = 0 and S₁(n) = f_n. We have S_m(n) = X

i1,i2,...,im≥0 i1+i2+...+im=n

f_i₁f_i₂. . . f_i_m

= X

i1,i2,...,im−1≥0 i1+i2+...+im−1≤n

f_i₁f_i₂. . . f_i_m−1f_n−(ⁱ¹⁺ⁱ²^+...+im−1)

=

n−1

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+im−1=k

f_i₁f_i₂. . . f_i_m−1f_n−k

and

S_m(n + 1) =

n

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+im−1=k

f_i₁f_i₂. . . f_i_m−1f_n+1−k

=

n−1

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+im−1=k

fi1fi2. . . fi_m−1fn+1−k

+ X

i1,i2,...,im−1≥0 i1+i2+...+im−1=n

f_i₁f_i₂. . . f_i_m−1f₁

=

n−1

X

k=1

X

i1,i2,...,i_m−1≥0 i1+i2+...+im−1=k

f_i₁f_i₂. . . f_i_m−1f_n−k

+

n−1

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+i_m−1=k

f_i₁f_i₂. . . f_i_m−1f_n−1−k

+ X

i1,i2,...,i_m−1≥0 i1+i2+...+im−1=n

f_i₁f_i₂. . . f_i_m−1

=

n−1

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+im−1=k

f_i₁f_i₂. . . f_i_m−1f_n−k

(6)

+

n−2

X

k=1

X

i1,i2,...,im−1≥0 i1+i2+...+im−1=k

f_i₁f_i₂. . . f_i_m−1f_n−1−k

+ X

i1,i2,...,im−1≥0 i1+i2+...+im−1=n

f_i₁f_i₂. . . f_i_m−1.

Thus,

S_m(n + 1) = S_m(n) + S_m(n − 1) + S_m−1(n), n, m ∈ N. (1)

Using (1) we will constructively find explicit formulas for S2(n), S₃(n) and S4(n). Note that, since f0 = 0, then

h_n = S₃(n) = X

i,j,k≥0 i+j+k=n

f_if_jf_k =

n

X

k=0

X

i,j≥0 i+j=k

f_if_jf_n−k.

We will also use the notation g_n for S₂(n) and s_n for S₄(n). Namely, for m = 1, 2, 3, equation (1) becomes, respectively,

g_n+1 = g_n + g_n−1+ f_n, n ∈ N, (2) h_n+1 = h_n + h_n−1+ g_n, n ∈ N (3) and

s_n+1 = s_n + s_n−1 + h_n, n ∈ N. (4) Consider now the Fibonacci operator F defined by F (an) = a_n+1− a_n− a_n−1 for any sequence (a_n)_n≥0 of real numbers and, in particular, for any integer k consider two special applications of operator F. Namely,

F (a_nf_n+k) = a_n+1f_n+1+k − a_nf_n+k− a_n−1f_n−1+k

= a_n+1(f_n+1+k − f_n+k − f_n−1+k)

+ a_n+1f_n+k + a_n+1f_n−1+k − a_nf_n+k− a_n−1f_n−1+k

= (a_n+1− a_n)f_n+k + (a_n+1− a_n−1)f_n−1+k

= (a_n+1− a_n)f_n+k + (a_n+1− a_n−1)(f_n−1+k − f_n+k)

= (an+1− an−1)fn+1+k − (an − an−1)fn+k,

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so

F (a_nf_n+k) = (a_n+1 − a_n−1)f_n+1+k − (a_n − a_n−1)f_n+k, (5) and

F (anfn+k+1) = an+1fn+k+2− anfn+k+1 − an−1fn+k

= a_n+1(f_n+k+2− f_n+k+1− f_n+k)

+ a_n+1f_n+k+1+ a_n+1f_n+k − a_nf_n+k+1 − a_n−1f_n+k

= (a_n+1− a_n)f_n+k+1+ (a_n+1− a_n−1)f_n+k, so

F (a_nf_n+k+1) = (a_n+1− a_n)f_n+k+1+ (a_n+1− a_n−1)f_n+k. (6) Note that F (g_n) = f_n, F (h_n) = g_n and F (s_n) = h_n. Note also that

F (an) = 0 ⇐⇒ an = (a1 − a0)fn+ a0fn+1, as can be easily proven by induction.

Now we are ready to find gn, hn and, afterwards, sn. Applying (5) and (6) to an = n we obtain, for any integer k,

F (nf_n+k) = 2f_n+k− f_n+k−1 and F (nf_n+k+1) = f_n+k+1+ 2f_n+k. Since for k = 0 we have

F (nf_n+1) = f_n+1+ 2f_n and F (nf_n) = 2f_n+1 − f_n, thus

F (nf_n+1) + 2F (nf_n) = f_n+1 + 2f_n + 2(2f_n+1− f_n) = 5f_n+1 and, therefore,

f_n+1 = F 2f_n+1− nf_n 5

(7) using the fact F is linear.

We also have

2F (nf_n+1) − F (nf_n) = 2f_n+1+ 4f_n− (2f − n + 1 − f_n) = 5f_n,

(8)

from which

f_n = F 2nf_n+1− nf_n 5

. (8)

Hence, equation (2) is equivalent to F (g_n) = F 2nf_n+1− nf_n

5

⇐⇒ F

g_n − 2nf_n+1− nf_n 5

= 0,

from which we conclude that g_n = 2nf_n+1− nf_n

5 + c₁f_n+1 + c₂f_n.

From the fact that g₀ = 0 we get c₁ · 1 + c₂ · 0 = 0 and c1 = 0.

Likewise, from g1 = 0 we get c1· 1 + c₂· 1 +^2·1−1₅ = 0 and c2 = −¹₅. Substituting in the above expression, we obtain

gn = S2(n) = 2nf_n+1− (n + 1)f_n

5 , (9)

and now we can find hn. Indeed, applying (5) and (6) to an = n² we obtain

F (n²f_n+k) = ((n + 1)²− (n − 1)²)F_n+k+1− (n²− (n − 1)²)f_n+k

= 4nf_n+k+1 − (2n − 1)f_n+k, or

F (n²f_n+k) = 4nf_n+k+1 − (2n − 1)f_n+k, (10) and

F (n²f_n+k+1) = ((n + 1)² − n²)F_n+k+1 − (n + 1)²− (n − 1)²)f_n+k

= (2n + 1)f_n+k+1 + 4nf_n+k, or

F (n²f_n+k+1) = (2n + 1)f_n+k+1 + 4nf_n+k. (11) In particular, for k = 0 in (10) and (11) we obtain

F (n²f_n) = 4f_n+1− (2n − 1)f_n and

F (n²f_n+1) = (2n + 1)f_n+1+ 4nf_n.

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Hence,

F (2n²f_n)+F (n²f_n+1) = 8nf_n+1−(4n−2)f_n+(2n+1)f_n+1+4f_n

= 10nf_n+1+ f_n+1+ 2f_n and

10nf_n+1

= F (2n²f_n + n²f_n+1) − 2f_n − f_n+1

= F (2n²f_n + n²f_n+1)

− 2F 2nf_n+1− nf_n 5

− F nf_n+1+ 2nf_n 5

= F

n²fn+1+ 2n²fn − 2(2nf_n+1 − nf_n)

5 − nf_n+1+ 2nf_n 5

= F ((n² − n)f_n+1+ 2n²f_n), from which

nf_n+1 = F (n²− n)f_n+1+ 2n²f_n 10

. (12)

Likewise, from

F (2n²f_n+1) − F (n²f_n)

= (4n + 2)f_n+1+ 8nf_n − (4nf_n+1 − (2n − 1)f_n)

= 10nfn − fn+ 2fn+1

we get

nf_n = F 2n²f_n+1 − (n²+ n)f_n 10

= F ng_n 2

. (13) Then, using (13), (12) and (9) we obtain

5g_n = 2nf_n+1− (n + 1)f_n = F (5n² + 3n)f_n − 6nf_n+1 10

.

That is, g_n = F (5n² + 3n)f_n − 6nf_n+1 10

and, therefore, (2) is equivalent to F

h_n− (5n² + 3n)fn − 6nfn+1

50

= 0 and

h_n = (5n²+ 3n)f_n − 6nf_n+1

50 + c₁f_n+1 + c₂f_n.

(10)

Since h₀ = 0 = c₁ and h₁ = 0 = ₅₀² + c₂, we get c₂ = −₂₅¹ , from which it follows that

h_n = S₃(n) = (5n²+ 3n − 2)f_n− 6nf_n+1

50 .

Before considering the computation of sn we present another way to obtain g_n. Note that F (F (g_n)) = F (f_n) = 0 and F (F (F (h_n))) = F (F (g_n)) = F (f_n) = 0. Since the characteristic polynomials of F (F (g_n)) and F (F (F (hn))) are (x² − x − 1)² and (x² − x − 1)³, respectively, then

g_n, h_n = P (n)φⁿ + Q(n)φⁿ,

where φ and φ are the roots of x² − x − 1 = 0 and P and Q are polynomials of degree at most 1 for g_n and at most 2 in the case of h_n. Since φⁿ and φⁿ can be represented as linear combinations of f_n+1 and fn, then we may also represent gn and hn in the form P (n)f_n+1+ Q(n)f_n, namely

g_n = (an + b)f_n+1+ (cn + d)f_n = anf_n+1(cn + d)f_n because g0 = 0 and

hn = (an² + bn + c)fn+1+ (pn²+ qn + r)fn

= (an² + bn)fn+1+ (pn² + qn + r)fn

because h₀ = 0, where g₀ = g₁ = 0, g₂ = 1, g₃ = 2, h₀ = h₁ = h₂ = 0, h3 = 1 and h4 = 3. Then, we get a = ²₅, c = −¹₅ and d = −¹₅ and, therefore,

gn = 2nf_n+1− (n + 1)f_n

5 .

The same expression may be obtained substituting gn = anfn+1 + (cn + d)f_n in gn+1− g_n − g_n−1 = f_n.

Now we consider the computation of sn = S4(n). Applying (5) and (6) to an = n³ for k = 0 we obtain

F (n³f_n) = ((n + 1)³ − (n − 1)³)f_n+1 − (n³ − (n − 1)³)f_n

= (6n²+ 2)fn+1− (3n² − 3n + 1)fn (14)

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and

F (n³f_n+1) = ((n + 1)³− n³)f_n+1− ((n + 1)³ − (n − 1)³)f_n

= (3n² + 3n + 1)fn+1+ (6n² + 2)fn. (15) Since gn = 2nf_n+1− (n + 1)f_n

5 , fn = F (g_n), nfn = F ng_n 2

and nf_n+1 = F (n² − n)F_n+1 + 2n²f_n

10

, then F (2n³f_n+1) − F (n³f_n)

= (6n² + 6n + 2)f_n+1+ (12n² + 4)f_n

− ((6n² + 2)f_n+1− (3n²− 3n + 1)f_n)

= 15n²f_n+ 6nf_n+1− (3n − 5)f_n

= 15n²f_n+ 6F (n²− n)f_n+1+ 2n²f_n 10

− 3F ng_n 2

+ 5F (g_n) and

15n²f_n = F (2n³f_n+1) − F (n³f_n) − 6F (n²− n)f_n+1+ 2n²f_n 10

+ 3F ng_n 2

− 5F (g_n)

= F (10 + 7n − 15n² − 10n³)fn + (20n³ − 14n)fn+1

10

, from which we conclude that

n²fn = F (10 + 7n − 15n²− 10n³)f_n + (20n³− 14n)f_n+1 150

.

Since S3(n) = h_n = (5n²+ 3n − 2)f_n− 6nf_n+1

50 ,

nf_n+1 = F (n² − n)fn+1+ 2n²fn

10

and

nf_n = F 2n²fn+1− (n² + n)fn

10

.

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Then, from the fact that F (S₄(n)) = S₃(n) we have F (s_n) = 1

10n²f_n + 3

50nf_n− 1

25f_n− 3

25nf_n+1

= F (11 + 5n − 30n²− 5n³)f_n+ (10n³ − 10n)f_n+1 750

. Hence,

s_n = (11+5n−30n²−5n³)f_n+(10n³−10n)f_n+1

750 +c₁f_n+1+c₂f_n. Since s₀ = s₁ = 0, then c₁ = 0 and c₂ = ₇₅₀¹⁹ and, therefore,

s_n = (11 + 5n − 30n²− 5n³)f_n+ (10n³− 10n)f_n+1

750 + 19

750f_n+1. Finally, rearranging terms we obtain

s_n = S₄(n) = (n − 1)(n + 1)(2nf_n+1− (n + 6)f_n)

150 . (16)

Remark. We claim that Sm(n) = Pm(n)fn+1 + Qm(n)fn, where P, Q are polynomials of degree at most m, because F^m(S_m(n)) = 0. Here, F^m = F ◦ F ◦ . . . ◦ F with characteristic polynomial (x²− x − 1)^m. Then, the polynomials P and Q can be determined

by substitution of Sm(n) in (1) assuming that

Sm−1(n) = Pm−1(n)fn+1+ Qm−1(n)fn

and that we know polynomials P_m−1 and Q_m−1. Using the fact that Sm(0) = Sm(1) = 0 we can determine both polynomials. Indeed, since

Pm(n + 1)fn+2+ Qm(n + 1)fn+1 − Pm(n)fn+1

− Q_m(n)f_n − P_m(n − 1)f_n− Q_m(n − 1)f_n−1

= (P_m(n + 1) − P_m(n) + Q_m(n + 1) − Q_m(n − 1))f_n+1 + (P_m(n + 1) − P_m(n − 1) − Q_m(n) + Q_m(n − 1))f_n,

then the fact that F (Sm(n)) = P_m−1(n)f_n+1 + Q_m−1(n)f_n implies P_m(n + 1) − P_m(n) + Q_m(n + 1) − Q_m(n − 1) = P_m−1(n), P_m(n + 1) − P_m(n − 1) + Q_m(n) + Q_m(n − 1) = Q_m−1(n).

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References

[1] Bloom, D. M. “Problem 55”. Mathematical Horizons (1996).

[2] Díaz-Barrero, J. L. “Problem E-58”. Arhimede math. j. 5.1 (2018), p. 28.

Arkady M. Alt

San Jose, California, USA.

[email protected]

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Two refinements of power Lessels-Pelling inequality

Mihály Bencze and Marius Dr˘ agan

Abstract

The Lessels-Pelling inequality states that, in a triangle, the sum of two angle bisectors and a median is at most √

3 of its semiperimeter. The purpose of this paper is to present refinements of the powered Lessels-Pelling inequality.

1 Introduction

Let ABC be a triangle. We denote by a = BC , b = CA and c = AB the lengths of the sides; s = ¹₂(a+b+c)the semiperimeter;

m_a the median from A; w_b and w_c the bisectors from B and C , respectively.

In 1974, Garfunkel [5] conjectured, on basis of computer check, the following inequality:

m_a+ w_b+ h_c ≤ s√ 3,

where h_c represents the altitude from C . Gardner [4] proved it in 1976 by using a sequence of elementary transformations and a differentiable function. In 1977, Lessels and Pelling [6] gave the stronger inequality

m_a+ w_b+ w_c ≤ s√ 3

using a computer check. In 1980, Patuwo, Thomas, and Wang [9]

gave a proof of this inequality and some chains of inequalities. A

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new proof due to C. T˘an˘asescu appeared in the book of Mitrinovi´c, Peˇcari´c, and Volonec [7]. In 1981, Panaitopol [8] gave an elementary proof of the Lessels-Pelling inequality. Recently, Dr˘agan [2, 3]

gave a simple proof and some refinements of the Lessels-Pelling inequality, and Bencze and Dr˘agan [1] presented a weighted power version of the Lessels-Pelling inequality. Our goal in this paper is to give refinements of this weighted power inequality.

2 Main results

Hereafter we use the following notation:

x = a

s, y = b

s, z = c

s, u =p

1 − y, v = √

1 − z, s₁ = u + v, p₁ = uv, s₂ = βu + γv,

E = s

2(y² + z²) − x²

4 , t = 3α²− 2α + 1

2 , (1)

where α, β , γ are positive numbers such that (β − γ)(b − c) ≥ 0 and α + β + γ = 1.

Lemma 1. With the above notation, the following inequalities hold:

(i) E ≤ s

1 − s²₁ 2 . (ii) E ≤

s

1 − 2

(1 − α)² s²₂.

(16)

Proof. (i) From (1) we have

E = s

2(2 + u⁴+ v⁴− 2u² − 2v²) − (u² + v²)² 4

= s

2[2 + (u² + v²)² − 2p²₁ − 2(u²+ v²)] − (u² + v²)² 4

= s

4 + 2(s²₁ − 2p₁)² − 4p²₁ − 4(s²₁ − 2p₁) − (s²₁− 2p₁)² 4

= s

4 + s⁴₁ − 4s²₁ + 4p₁(2 − s²₁) 4

≤ s

4 − 4s²₁ + 2s²₁

4 =

s

1 − s²₁ 2 .

(ii) We have

s₂ = βu + γv ≤ 1

2 (β + γ)(u + v) = (1 − α)s₁

2 (2)

since, as (β − γ)(b − c) ≥ 0, it results that (β − γ)(u − v) ≤ 0.

The statement follows from (i) and (2).

Theorem 1. In any triangle ABC the following holds:

αm_a+ βw_b+ γw_c ≤ αm_a+ βp

s(s − b) + γp

s(s − c)

≤ α s

2s² − s √

s − b +√

s − c2

2 + 1 − α

2

ps(s − b) +p

s(s − c)

≤ s√ t, (3) where α, β , γ are positive numbers such that (β − γ)(b − c) ≥ 0 and α + β + γ = 1.

Proof. Since w_b ≤ ps(s − b) and w_c ≤ ps(s − c), to prove (3) it

(17)

will be sufficient to prove that αm_a + βp

s(s − b) + γp

s(s − c)

≤ α s

2S²−s √

s−b +√

s−c2

2 + 1−α

2

ps(s−b) +p

s(s−c)

≤ s√

t. (4)

On account of (1), inequality (4) may be written as

αE + S2 = α s

2(y²+ z²) − x²

4 + βp

1 − y + γ√ 1 − z

≤ α s

2 − S₁²

2 + 1 − α

2 S₁ ≤ √

t. (5)

From Lemma 1(i) and (2) we have that

αE + S₂ ≤ α s

2 − S₁²

2 + 1 − α 2 S₁. To prove (5), it will suffice to show that

α s

2 − S₁²

2 + 1 − α

2 S₁ ≤ √

t. (6)

We note that

S1 = p

1 − y +√

1 − z ≤ p

2(2 − y − z) =√ 2x =

s 2a

s < √ 2.

So, we have S1 < √ 2.

We consider the function f : 0,√

2

→ R defined by

f (S₁) = α s

2 − S₁²

2 + (1 − α)S₁

2 ,

(18)

for which the derivative is

f⁰(S₁) = − αS₁

p4 − 2S₁² + 1 − α 2 . We solve the equation f⁰(S₁) = 0 with the root

S₀ = 2(1 − α)

√6α²− 4α + 2.

Since f⁰(0) ≥ 0 and f⁰√ 2

≤ 0, it follows that S0 is a maximum point for f . So we have

f (S1) ≤ f (S0) = α s

1 2

2 − 4(1 − α)² 6α²− 4α + 2

+ (1 − α)²

√6α² − 4α + 2

= α s

4α²

6α² − 4α + 2 + (1 − α)²

√6α²− 4α + 2

= s

3α²− 2α + 1

2 =√

t,

which finishes the proof of (6).

Remark 1. Putting α = β = γ in (3) we obtain a refinement of the Lessels-Pelling inequality.

Theorem 2. In every triangle ABC the following inequalities hold:

(i) ma ≤

√t 2α · m²_a

s + α 2√

ts;

(ii)

√t 2α

m²_a s + α

2√

ts ≤ t + α² 2α√

ts−

√t α(1 − α)²

βp

s − b + γ√

s − c2

; (iii) ma ≤

s

s² − 2 (1 − α)²

βp

s − b + γ√ s − c

2

m_a ≤ t + α² 2α√

t s −

√t α(1 − α)²

βp

s − b + γ√

s − c2

,

(19)

where α, β , γ are positive numbers such that α + β + γ = 1 and (β − γ)(b − c) ≥ 0.

Proof. (i) From (1), the inequality in (i) may be written as E ≤

√t

2αE²+ α 2√

t or, equivalently, √

tE − α

2

≥ 0.

(ii) From Lemma 1(ii), we have

√t

2αE² + α 2√

t ≤

√t 2α −

√t

α(1 − α)² S₂² + α 2√

t

= t + α² 2α√

t −

√t

α(1 − α)² S₂². (7) Taking in (7) the notation from (1) we obtain the equality from the statement.

(iii) The first inequality is just the inequality in Lemma 1(ii). We will prove the second inequality, which is equivalent to

s

1 − 2

(1 − α)² S₂² ≤ t + α² 2α√

t −

√t

α(1 − α)² S₂², or



 s

t − 2t

(1 − α)²S₂²y − α





2

≥ 0.

Theorem 3. In every triangle ABC the following chains of in- equalities are true:

(i) ≤αm_a+ βw_b+ γw_c

≤ αma+ βp

s(s − b) + γp

s(s − c)

≤

√t 2

ma²

s + α² 2√

ts + βp

s(s − b) + γp

s(s − c)

≤ t + α² 2√

t s −

√t (1 − α)²

βp

s − b + γ√

s − c2

≤+ βp

s(s − b) + γp

s(s − c) ≤ s√ t;

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(ii) ≤αm_a+ βw_b+ γw_c

≤ α s

s²− 2 (1 − α)²

βp

s − b + γ√

s − c2

+ βw_b+ γw_c

≤ t + α² 2√

t s −

√t (1 − α)²

βp

s − b + γ√

s − c2

+ βw_b+ γw_c

≤ t + α² 2√

t s −

√t (1 − α)²

βp

s − b + γ√

s − c2

≤+ βp

s(s − b) + γp

s(s − c) ≤ s√ t,

where α, β , γ are positive numbers such that α + β + γ = 1 and (β − γ)(b − c) ≥ 0.

Proof. (i) The first inequality follows from the inequalities wb ≤ ps(s − b) and wc ≤ ps(s − c). The second inequality results from Theorem 2(i). The third follows from Theorem 2(ii). It remains to prove that

t + α² 2√

t s −

√t 2(1 − α)²

βp

s − b + γ√

s − c2

+ βp

s(s − b) + γp

s(s − c) ≤ s√

t. (8)

Using (1), inequality (8) may be written as t + α²

2√ t −

√t

(1 − α)² S₂²+ S₂ ≤ √ t or, equivalently,

tS₂²− 2√

t(1 − α)²S₂ + (t − α²)(1 − α)² ≥ 0, or

tS₂² − 2√

t(1 − α)²S₂ + (1 − α)⁴ 2 ≥ 0, or

2√

tS₂ − 1 + α2

≥ 0.

(ii) The first inequality from this chain follows from Theorem 2(iii).

The second also results from Theorem 2(iii). The third is true since w_b ≤ ps(s − b) and wc ≤ ps(s − c), and the last inequality from the chain is the inequality from (i).

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References

[1] Bencze, M. and Dr˘agan, M. “A weighted power Lessels-Pelling inequality”. Math. Reflection 6 (2017), pp. 1–4.

[2] Dr˘agan, M. “Some geometric inequalities”. Arhimede (Roma- nian) 7-12 (2008), pp. 1–4.

[3] Dr˘agan, M. “Some extensions and refinements of Lessels- Pelling inequality”. Gazeta Matematica Seria B (Romanian) 6- 7-8 (2013), pp. 281–284.

[4] Gardner, S. “Solution to Problem E2504”. Amer. Math. Monthly 83 (1976), pp. 289–290.

[5] Garfunkel, J. “Problem E2504”. Amer. Math. Monthly 81 (1974), p. 1111.

[6] Lessels, G. S. and Pelling, M. J. “An inequality for the sum of two angle bisectors and a median”. Univ. Beograd, Publ.

Electrotehn. Fak. Ser. Mat. Fiz. 577-598 (1977), pp. 59–62.

[7] Mitrinovi´c, D. S., Peˇcari´c, J. E., and Volonec, V. Recent Ad- vances in Geometric inequalities. 1st. Kluwer, 1989.

[8] Panaitopol, L. “Some about geometric inequalities”. Gazeta Matematica Seria B (Romanian) (1981), pp. 6–8.

[9] Patuwo, B. E., Thomas, R. S. D., and Wang, C.-L. “The triangle inequality of Lessels and Pelling”. Univ. Beograd, Publ.

Electrotehn. Fak. Ser. Mat. Fiz. 678-715 (1980), pp. 45–47.

Mihály Bencze

H˘armanului 6, 505600 S˘acele-Négyfalu, Bra ¸sov, Romania.

[email protected] Marius Dr˘agan

61311 Timi ¸soara 35,

Bl. 0D6, Sc. E, et. 7, Ap. 176, Sect. 6, Bucure ¸sti, Romania

[email protected]

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Problems

This section of the Journal offers readers an opportunity to ex- change interesting and elegant mathematical problems. Proposals are always welcome. Please observe the following guidelines when submitting proposals or solutions:

1. Proposals and solutions must be legible and should appear on separate sheets, each indicating the name and address of the sender. Drawings must be suitable for reproduction.

2. Proposals should be accompanied by solutions. An asterisk (*) indicates that neither the proposer nor the editor has supplied a solution.

[email protected]

The section is divided into four subsections: Elementary Problems, Easy–Medium High School Problems, Medium–Hard High School Problems, and Advanced Problems mainly for undergraduates.

Proposals that appeared in Math Contests around the world and most appropriate for Math Olympiads training are always welcome.

The source of these proposals will appear when the solutions are published.

Solutions to the problems stated in this issue should be posted before

April 30, 2019

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Elementary Problems

E–59.

Proposed by Oriol Baeza Guasch, Institut de Terrassa, Ter- rassa, Spain. Let ABC be a triangle with circumcentre O , and let P be the point of intersection of ray AO with side BC . Denote by D and E the feet of the perpendiculars from P to sides AB and AC, respectively. Show that BD = CE if and only if AB = AC .

E–60.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce- lona, Spain. Find all real numbers a, b such that

(a² +p

1 + a⁴) (b² +p

1 + b⁴) = 1.

E–61.

Proposed by Mihaela Berindeanu, Bucharest, România. If bac is the integer part of the real number a, solve in R the equation

x²− x x² − x + 1

+ 2x² + 1 3x = 0.

E–62.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce- lona, Spain. On a 2019 × 2019 board symmetrical with respect to the main diagonal, the numbers from 1 to 2019 are placed so that in each row and column each number appears once and only once.

Show that all the numbers appear on the main diagonal.

E–63.

Proposed by Mihaela Berindeanu, Bucharest, România.

Find all pairs (p, q) of prime numbers for which p p²+ p + 3 = q(q + 4).

E–64.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce- lona, Spain. Let a, b, c be three positive real numbers such that ab + bc + ca = 2abc. Prove that

√1

ab + 1

√bc + 1

√ca ≤ 2.

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Easy–Medium Problems

EM–59.

Proposed by Pedro Henrique O. Pantoja, University of Campina Grande, Brazil. Find all functions f : R → R such that

f (x²y + y²z + z²x) = xf (x) + f (f (y) + z) + f (zx) for all real numbers x, y , z .

EM–60.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar- celona, Spain. Two hundred students participated in a mathematical contest. They had six problems to solve. It is known that each problem was correctly solved by at least 120 participants. Prove that there must be two participants such that every problem was solved by at least one of these two students.

EM–61.

Let x, y, z be real numbers. Show that X

cyclic

1

(1 + 2^y+1−x)(1 + 2^z+1−x) ≥ 1 3.

EM–62.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar- celona, Spain. Let a1 ≤ a₂ ≤ . . . ≤ a_n be n ≥ 1 real numbers lying in the interval (0, π/2). Show that

a₁ ≤ arctan sin a₁ + sin a2 + . . . + sin an

cos a1 + cos a2 + . . . + cos an

≤ a_n.

EM–63.

Proposed by Oriol Baeza Guasch, Institut de Terrassa, Terrassa, Spain. Let ABC be a triangle with usual notation. Show that

1 − cos β sin β

! 1 − cos γ sin γ

!

= 1 − 2a a + b + c.

EM–64.

If z ∈ C and |z²− 1| = |4z − i|, then show that |z| ≤ p

9 +√ 85.

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Medium–Hard Problems

MH–59.

Proposed by Arkady Alt, San Jose, California, USA. Prove that 2nFn+1 − (n + 1)F_n is divisible by 5 for any integer n ≥ 1.

Here, F_n is the n-th Fibonacci number defined by F₀ = 0, F₁ = 1 and, for all n ≥ 2, Fn = Fn−1+ Fn−2.

MH–60.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar- celona, Spain. Prove that in any acute triangle ABC with the usual notations the following inequality holds:

sin A sin B

cos C + sin B sin C

cos A + sin C sin A cos B ≥ 9

4.

MH–61.

Proposed by Mihály Bencze, Bra ¸sov, Romania. Suppose that ABCDA1B₁C₁D₁ is a rectangle parallelepiped with sides a, b, c and diagonal d. Prove that

a + b + c

d + 1

abc ≥ 4√ 3.

MH–62.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar- celona, Spain. Find all integer solutions of the equation

(x + 1)(y − 1) = x²y².

MH–63.

Proposed by Pedro Henrique O. Pantoja, University of Campina Grande, Brazil. Let 0 ≤ a, b, c < ^π₂. Prove that

8 tan a 2 tanb

2tan c

2 ≤ tan(a + b)

2 tan(b + c)

2 tan(c + a) 2

< (1 + tan^{2 a}₂)(1 + tan^{2 b}₂)(1 + tan^{2 c}₂) (1 − tan^{2 a}₂)(1 − tan^{2 b}₂)(1 − tan^{2 c}₂).

MH–64.

Proposed by Marc Felipe Alsina, BarcelonaTech, Barce- lona, Spain. Let ABC be an isosceles triangle with ∠ABC > 90^◦. Let D be the projection of C onto the line AB . Let Γ₁ be the

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circle centered at A that passes through C and let Γ₂ be the circle centered at D that passes through A. Let Γ1 and Γ2 intersect at X and Y . Prove that X and Y belong to the perpendicular to AB through B .

(27)

Advanced Problems

A–59.

Proposed by Víctor Martín Chabrera, FME, BarcelonaTech, Spain. Compute

∞

X

s=2

(ζ(s) − 1), where ζ(s) is the Riemann zeta function.

A–60.

Proposed by Mihály Bencze, Bra ¸sov, Romania. Let A ∈ M₂(C) with Tr(A) = √

2/2. Show that

det A² + 3√ 2

2 A + 3I₂

!

− det A²−

√2 2 A

!

= 15.

A–61.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar- celona, Spain. If the coefficients of the power series P∞

n=0a_nzⁿ are given by the recurrence a0 = a₁ = 1 and for all n ≥ 2, 14a_n + 3a_n−1− a_n−2 = 0, then find the radius of convergence of the series and the function to which it converges in its disc of convergence.

A–62.

Proposed by Nicolae Papacu, Slobozia, Romania. Assume that A(+, ◦) is a ring such that 1 + 1 + 1 is invertible. If x, y ∈ A verify that x + y = 1 and x³ = x, then prove that the elements 1 − xy and 1 − yx are invertible.

A–63.

Proposed by Óscar Rivero Salgado, BarcelonaTeach, Bar- celona, Spain. Let k ≥ 1 be a fixed positive integer, and let n ≥ 0 be a non-negative integer. Show that

n

X

j=0

kn kj

= 2^kn

k + 2^kn+1 k ·

bn/2c

X

j=1

(−1)^jn · cosπj n

kn

.

A–64.

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce- lona, Spain. Assume that polynomial A(z) with leading coefficient

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one and degree n has distinct zeros α₁, α₂, . . . , α_n. Prove that

|∆(A)| ≤ 2ⁿ⁽ⁿ⁻¹⁾

n

Y

k=1

max{1, |α_k|}

!2n−2

,

where ∆(A) is the discriminant of A(z).

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Mathlessons

This section of the Journal offers readers an opportunity to ex- change interesting and elegant mathematical notes and lessons with material useful to solve mathematical problems.

[email protected]

(30)

Theorems and problems in classical geometry

José Luis Díaz-Barrero and Marc Felipe Alsina

1 Introduction

In classical elementary plane geometry, there are two kinds of lines that usually are most drawn, straight lines and circles. In this note, three well-known results involving the above mentioned elements are presented. These results may be applied to solve problems that frequently appear in mathematical competitions, similar to the ones given in this Mathlesson.

2 Theorems

We begin with a result that establishes a relationship between the lengths of the sides and the length of a cevian in a triangle due to the Scottish mathematician Matthew Stewart, who published it in 1746 [3].

Theorem 1 (Stewart). Let D be a point in the side BC of triangle ABC. Let m = BD , n = DC , and p = AD . Then,

b²m + c²n = a(p²+ mn).

Proof. Since ∠ADB+∠ADC = π, then cos ∠ADB+cos ∠ADC = 0. Taking into account the law of cosines, we have

m² + p² − c²

2mp + n²+ p² − b² 2np = 0,

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D C B

A

a

c b

m n

p

Figure 1: Scheme for Stewart’s theorem.

or equivalently,

n(m² + p² − c²) + m(n² + p² − b²) = 0, from which it follows that

b²m + c²n = (m + n)(p² + mn) = a(p² + mn), as claimed.

Corollary 1 (Length of the median). In any triangle ABC the length of the median is given by the expression

m_a = 1 2

p2b² + 2c² − a² (cyclic).

Proof. Applying Stewart’s theorem with m = n = a/2 and p = m_a, we have

b²a

2 + c²a 2 = a

m²_a + a² 4

,

from which we get 4m²_a = 2b² + 2c² − a², and the statement follows.

The preceding expressions are also known in the literature as Apollonius formulae.

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Next we present a fascinating result in classical Euclidean geometry, the Butterfly theorem.

Theorem 2 (Butterfly). Let M be the midpoint of a chord P Q of a circle, through which two other chords AB and CD are drawn.

Chords AD and BC intersect P Q at points X and Y , respectively.

Then, M is the midpoint of XY .

Q

P

y

2

y

1

x

2

x

1

Y'' X'' Y'

X'

M Y

X

D

C

B A

Figure 2: Scheme for Butterfly’s theorem

Proof. We begin by dropping perpendiculars x1 = XX⁰⁰ and y1 = Y Y⁰⁰ from X and Y to AB , and x2 = XX⁰ and y2 = Y Y⁰ from X and Y to CD , as shown in Figure 2. Letting a = P M = M Q, x = XM, y = M Y , and observing the pairs of similar triangles 4M XX⁰⁰ ∼ 4M Y Y⁰⁰, 4M XX⁰ ∼ M Y Y⁰, 4AXX⁰⁰ ∼ CY Y⁰ and 4DXX⁰ ∼ BY Y⁰⁰, yields

M X

M Y = XX⁰⁰

Y Y⁰⁰ ⇐⇒ x y = x₁

y₁, M X

M Y = XX⁰

Y Y⁰ ⇐⇒ x y = x₂

y₂, AX

CY = XX⁰⁰

Y Y⁰ ⇐⇒ AX CY = x₁

y₂, DX

BY = XX⁰

Y Y⁰⁰ ⇐⇒ DX

BY = x2

y1

.

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From the preceding, and taking into account the power of points X and Y respect to the circle, we get

x²

y² = x₁x₂ y₁y₂ = x₁

y₂ · x₂

y₁ = AX · XD

CY · Y B = P X · XQ P Y · Y Q

= (a − x)(a + x)

(a + y)(a − y) = a²− x²

a² − y² = (a²− x²) + x²

(a² − y²) + y² = a² a² = 1, from which x = y follows, as we wanted to prove.

According to Coxeter and Greitzer [1], one of the proofs to the But- terfly theorem was submitted in 1815 by W. G. Horner of Horner’s method fame [2]. However, two recent discoveries have made it clear that Wallace’s proof came before Horner’s proof by about ten years. The first of those discoveries is an 1803 publication of The Gentlemen’s Mathematical Companion containing a generalization of the problem by Wallace. The second is a correspondence from Sir William Herschel to Wallace in 1805. In the letter, Herschel presents the problem of proving that the two line segments XM and Y M in the theorem are equal in length. In turn, he is asking Wallace for a proof of the generalized butterfly theorem.

Finally, we present and give three proofs of a classical result of Wittaker appeared in 1820 [4].

Theorem 3 (Napoleon’s theorem). If equilateral triangles AGB , BHC and CKA are constructed on the sides of any triangle ABC , then the centers of the circumcircles of those equilateral triangles D , E, F themselves form an equilateral triangle.

Proof 1. We begin with the following well-known result.

Lemma 1 (Fermat’s point). Lines KB , HA and GC in Figure 3 meet at point P . (This point is called Fermat’s point.)

Proof. Let P be the intersection of AH and KB . Triangles KCB and ACH are congruent because the second is a 60^◦ rotation of the first about point C . So, ∠CKP = ∠CAP and

∠CHP = ∠CBP . From the preceding, it immediately follows that quadrilaterals KAP C and HBP C are concyclic. Thus,

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K

H

G C

A B

P

Figure 3: Scheme for Fermat’s theorem.

∠AP C = ∠AP B = ∠BP C = 120^◦. Since ∠AP B and ∠AGC add up to 180^◦, then AP BG is also concyclic. Hence, ∠AP G =

∠ABG = 60^◦ (both have the common arc AG). Finally, since

∠AP G+∠AP C = 180^◦, then points G, P and C are collinear.

S

R PQ H

G K

F

E D

C

B A

Figure 4: Scheme for proof 1 of Napoleon’s theorem.

Now we give the first proof of the theorem. Indeed, let P be the intersection point of the circumcircles of the equilateral triangles ABG, BCH and CAK . Then, by the inscribed angle theorem

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(capacious arc), ∠AP C = 180 − ∠AKC (cyclic). That is,

∠AP C = ∠BP C = ∠AP B = 120^◦.

Since P A, P B and P C are, respectively, radical axis of two circles, then P Q⊥EF , P R⊥DF and P S⊥DE , as is well-known. Now, considering the quadrilateral F RP Q, we have ∠RF Q + ∠AP C = 180^◦ or, equivalently, ∠RF Q + 120^◦ = 180^◦, from which it follows that ∠RF Q = 60^◦. Likewise, we obtain that ∠RDS = ∠QSD = 60^◦, and the proof is complete.

Proof 2. In Figure 5 we have that ∠ACK = ∠BCH = 60^◦. Adding to each ∠ACB , we have ∠BCK = ∠ACH . Since sides AC = KC and CH = CB , then 4BKC and 4AHC are congruent and AH = BK.

M

L K

H

G F

E

D

C

A B

Figure 5: Scheme for proof 2 of Napoleon’s theorem.

Extend BD and BE to L and M , respectively. Then, since D and E are the centers of the equilateral triangles ABG and CBH, then ∠ABL = ∠CBM = 30^◦. Moreover, BD = ²₃ BL and BE = ²₃ BM as is well-known. Triangles BCM and ABL are

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similar. Therefore, AB

BC = BL

BM = 3/2 · BD

3/2 · BE = BD BE.

Since ∠CBE + ∠ABD = ∠CBH . then adding ∠ABC to each, we get ∠DBE = ∠ABH and triangles DBE and ABH are similar.

Likewise, triangles AKB and ADF are similar too. Hence, AB

AH = BD

DE and AB

BK = AD DF.

Since we have already seen that BK = AH and AD = BD , then we get DE = DF . A similar procedure leads us to obtain that DF = F E. So, 4DEF is equilateral and the proof is complete.

Remark 1. The relation AB

AH = BD

DE can also be obtained from the right triangles BM H and ALB . Indeed, in 4BM H we have

cos 30^◦ = BM

BH = 3/2 BE BH , and in 4ALB we have

cos 30^◦ = BL

AB = 3/2 BD AB , from which it follows that AB

AH = BD DE.

Finally, we give a third proof using complex numbers. We need the following lemma.

Lemma 2. Let z₁, z₂, z₃ be the coordinates of the vertices A₁, A₂, A₃ of a positively oriented triangle. Then, the following holds:

4A₁A₂A₂ is equilateral ⇐⇒ z1+ e^i2π/3z₂ + e^i4π/3z₃ = 0.

Proof. It is well-known that 4A₁A₂A₂ is equilateral and positively oriented if and only if A3 is obtained from A2 by rotation about A1

through an angle of π/3. Namely,

z₃ − z₁ = e^iπ/3(z₂ − z₁),

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from which it follows that z3 = z1+ 1

2 + i

√3 2

!

(z2− z1) = 1 2 − i

√3 2

!

z1+ 1 2 + i

√3 2

! z2. Therefore,

z₁ + e^i2π/3z₂+ e^i4π/3z₃

= z₁ + −1 2 + i

√3 2

! z₂

+ −1 2 − i

√3 2

!"

1 2 − i

√3 2

!

z1 + 1 2 + i

√3 2

! z2

#

= z₁ + −1 2 + i

√3 2

!

z₂ − z₁+ 1 2 − i

√3 2

!

z₂ = 0, and the proof is complete.

Proof 3. Let zA, zB, zC be the affixes of vertices A, B and C , respectively (see Figure 3). On account of Lemma 2, we have

z_A + z_C + ²z_k = 0, z_G + z_B + ²z_A = 0, z_B + z_H + ²z_C = 0, where = e^i2π/3. The centers of gravity of triangles GAB , HBC and KAC are, respectively,

z_D = 1 3

z_A+ z_B + z_G , z_F = 1

3

z_A+ z_C + z_K

, z_E = 1

3

z_B+ z_C + z_H . Finally, we have

3(z_D + z_E + ²z_F)

= (z_A + z_B + z_G) + (z_B + z_C + z_H) + ²(z_A + z_C + z_K)

= (z_A + z_c+ ²z_K) + (z_B + z_H + ²z_C) + (z_G+ z_B + ²z_A)

= 0,

and 4DEF is equilateral.