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SOLUTIONS

Au unprobleme n'estimmuable.L'editeur esttoujoursheureux

d'en-visager la publi ation de nouvelles solutions oude nouvelles perspe tives

portantsurdesproblemes anterieurs.

We have re eived alate bat h of orre t solutions to problems 3478,

3479,3480,3481,3482,3483,and3486fromWaltherJanous,

Ursulinengym-nasium,Innsbru k,Austria.

3488. [2009:515,517℄ ProposedbyPhamHuuDu ,Ballajura,Australia.

Let

a

,

b

,and

c

bepositiverealnumbers. Provethat

a

2a

2

+ bc

+

b

2b

2

+ ca

+

c

2c

2

+ ab

Ê

a

−1

+ b

−1

+ c

−1

a + b + c

.

Solution by Paolo Perfetti, Dipartimento di Matemati a, Universita degli

studidiTorVergataRoma,Rome,Italy.

Let

t = (t

1

, t

2

, . . . , t

n

)

and

s = (s

1

, s

2

, . . . , s

n

)

bearbitrary

n

-tuples ofnonnegativerealnumbers. Wewillwrite

t ≻ s

if

(i)

t

1

≥ · · · ≥ t

n

and

s

1

≥ · · · ≥ s

n

, (ii)

k

P

i=1

t

i

k

P

i=1

s

i

forall

k = 1

,

2

,

. . .

,

n

,withequalitywhen

k = n

.

Let

R

+

denotethesetofpositiverealnumbers,let

P

bethesetofall permutationsof

{1, 2

,

. . .

,

n}

,andde ne

[t] : R

n

+

→ R

by

[t](x) =

X

σ∈P

x

t

1

σ(1)

x

t

2

σ(2)

· · · x

t

n

σ(n)

forall

x = (x

1

, x

2

, . . . , x

n

)

.

Muirhead's inequality states that if

t ≻ s

, then

[t] ≥ [s]

. Here, as usual,

[t] ≥ [s]

meansthat

[t](x) ≥ [s](x)

forall

x ∈ R

n

+

. Now,bysquaring andsimplifying,thegiveninequalityisequivalentto

A ≥ B

,where

A

= 12[8, 5, 1] + 23[7, 4, 3] + 16[6, 6, 2] + 12[8, 4, 2] + 4[7, 6, 1]

+ 4[9, 3, 2] + 8[7, 7, 0]

,

B

= 12[7, 5, 2] + 22[6, 5, 3] + 26[6, 4, 4] +

5

2

[8, 3, 3] +

33

2

[5, 5, 4]

. Butthislastinequalityholdsbytheseappli ationsofMuirhead'sinequality:

[8, 5, 1]

≥ [7, 5, 2]

,

[7, 4, 3]

≥ [6, 5, 3] and [7, 4, 3] ≥ [6, 4, 4]

,

[8, 4, 2]

≥ [8, 3, 3] and [8, 4, 2] ≥ [6, 4, 4]

,

(2)

[7, 6, 1] ≥ [5, 5, 4]

,

[9, 3, 2] ≥ [5, 5, 4]

,

[7, 7, 0] ≥ [5, 5, 4]

.

AlsosolvedbyOLIVERGEUPEL,Bruhl,NRW,Germany;WALTHERJANOUS,

Ursulinen-gymnasium,Innsbru k,Austria;ALBERTSTADLER,Herrliberg,Switzerland;andtheproposer.

Onein ompletesolutionwassubmitted.

3489. [2009:515, 517℄ ProposedbyJose LuisDaz-Barrero,Universitat

Polite ni a deCatalunya,Bar elona,Spain.

Let

n

beanonnegativeinteger. Provethat

1

2

n−1

n

X

k=0

k



2n

k



Ê

n



2

2n

+



2n

n

 ‹ .

A omposite of similar solutions by George Apostolopoulos, Messolonghi,

Gree e;andAlbertStadler,Herrliberg,Switzerland.

By usingtheelementaryfa tsthat

2n

k



=

2n−k

2n

 for

0 ≤ k ≤ 2n

and

k

2n

k



= 2n

2n−1

k−1



for

1 ≤ k ≤ 2n

,andalsotheCau hy{S hwarzInequality, wehavethat "

n

X

k=0

k



2n

k

 #

2

=

"

n

X

k=0

Ê 

2n

k



k

Ê 

2n

k

 #

2

"

n

X

k=0



2n

k

 #"

n

X

k=0

k



2n

k

 #

=

1

2

" 

2n

n



+

2n

X

k=0



2n

k

 #"

2n

n

X

k=1



2n − 1

k − 1

 #

=

1

2

•

2n

n



+ 2

2n

˜ "

2n

n−1

X

k=0



2n − 1

k

 #

=

1

2

•

2n

n



+ 2

2n

˜ "

2n ·

1

2

2n−1

X

k=0



2n − 1

k

 #

=

1

2

•

2

2n

+



2n

n

(n · 2

2n−1

)

=

n



2

2n

+



2n

n

· 2

2n−2

,

(3)

AlsosolvedbyARKADYALT,SanJose,CA,USA;DIONNECAMPBELL,ELSIE

CAMP-BELL,andCHARLESR.DIMINNIE,AngeloStateUniversity,SanAngelo,TX,USA;MICHEL

BATAILLE, Rouen, Fran e; CHIPCURTIS,Missouri SouthernState University, Joplin,MO,

USA; OLIVER GEUPEL, Bruhl, NRW, Germany; WALTHER JANOUS, Ursulinengymnasium,

Innsbru k,Austria;PAOLOPERFETTI,DipartimentodiMatemati a,Universitadeglistudidi

TorVergataRoma,Rome,Italy;andtheproposer.

3490. [2009:515,518℄ ProposedbyMi haelRozenberg,Tel-Aviv,Israel.

Let

a

,

b

,and

c

benonnegativerealnumberssu hthat

a + b + c = 1

. Provethat (a)

9 − 32ab +

9 − 32ac +

9 − 32bc ≥ 7

; (b)

1 − 3ab +

1 − 3ac +

1 − 3bc ≥

6

.

Solution to part (a) by Oliver Geupel, Bruhl, NRW, Germany; solution to

part (b) by George Apostolopoulos, Messolonghi,Gree e, modi edby the

editor.

(a) Fornonnegativeintegers

,

m

,and

n

,let

[ℓ, m, n] =

P

symm.

a

b

m

c

n

.

Thefollowinginequalityisa onsequen eofMuirhead'sTheorem,

27

Y y li €

9(a + b + c)

2

− 32ab

Š

€

11(a + b + c)

2

+ 16(ab + bc + ca)

Š

3

= 9176 [6, 0, 0] + 34320 [5, 1, 0] − 36336 [4, 2, 0] + 50184 [4, 1, 1]

− 54352 [3, 3, 0] + 100320 [3, 2, 1] − 103312 [2, 2, 2]

≥ 0

.

We put

a + b + c = 1

in theabove, andwe observe thatby theAM-GM Inequality P y li È

(9 − 32ab)(9 − 32bc) ≥ 3

 Q y li

(9 − 32ab)



1/3

. Itfollows that P y li È

(9 − 32ab)(9 − 32bc) ≥ 11 + 16(ab + bc + ca)

,andwededu e

„ X y li p

9 − 32ab

Ž

2

≥ 49

,

fromwhi htheinequalityin(a)follows.

(b) Let

x = 3a

,

y = 3b

,and

z = 3c

. Then

x

,

y

,and

z

arenonnegative realnumberssu hthat

x + y + z = 3

,andwearetoshowthat

X

p

(4)

Note rstthat P y li É

(3 + z)

2

8

=

1

8

P y li

(3 + z) =

1

8

(9 + 3) = 3

2

. Also,

(3 + z)

2

8



3 −

(x + y)

2

4



=

(3 + z)

2

8

− 3 +

(3 − z)

2

4

=

1

8

€

9 + 6z + z

2

− 24 + 18 − 12z + 2z

2

Š

=

3

8

(z − 1)

2

. (2) Hen e,

(3 + z)

2

8

≥ 3 −

(x + y)

2

4

, (3) and(1)isequivalentto X y li p

3 − xy −

Ê

3 −

(x + y)

2

4

!

X y li Ê

(3 + z)

2

8

Ê

3 −

(x + y)

2

4

! . (4)

Let

H

and

K

denotetheleftandrightsideof(4),respe tively. Then

H =

1

4

X y li

(x − y)

2

3 − xy +

È

3 −

(x+y)

4

2

1

4

X y li

(x − y)

2

3 +

3

=

1

8

3

X y li

(x − y)

2

. (5)

Ontheotherhand,using(2)and(3),wehave

K =

X y li

(3+z)

2

8

− 3 +

(x+y)

2

4

È

(3+z)

2

8

+

È

3 −

(x+y)

4

2

=

3

8

X y li

(z − 1)

2

È

(3+z)

2

8

+

È

3 −

(x+y)

4

2

3

8

X y li

(z − 1)

2

2

È

3 −

(x+y)

4

2

3

8

X y li

(z − 1)

2

2

È

3 −

9

4

=

3

8

X y li

(z − 1)

2

. (6) Finally, X y li

(x − y)

2

= 3

X y li

x

2

!

X y li

x

!

2

= 3

X y li

x

2

!

− 9

= 3

X y li

x

2

!

− 6

X y li

x

!

+ 9 = 3

X y li

(z − 1)

2

! . (7)

(5)

Part(b)wasalsosolvedbyCHIPCURTIS,MissouriSouthernStateUniversity,Joplin,

MO,USA;OLIVERGEUPEL,Bruhl, NRW,Germany;andtheproposer. Twoin omplete

solu-tionsweresubmitted.

The aseofequalitywasnotrequested,thoughGeupel laimedequalitypre iselywhen

a = b = c = 1/3

,buttheproposernotedthatequalityalsoo urswhen

a = b = 1/2

,

c = 0

.

3491. [2009:515,518℄ProposedbyDorinMarghidanu, ColegiulNational

\A.I.Cuza",Corabia,Romania.

Let

a

1

,

a

2

,

. . .

,

a

n+1

bepositiverealnumberswhere

a

n+1

= a

1

. Prove that

n

X

i=1

a

4

i

(a

i

+ a

i+1

)(a

2

i

+ a

2

i+1

)

1

4

n

X

i=1

a

i

.

SolutionbyGeorgeApostolopoulos,Messolonghi,Gree e.

Let

A

=

n

X

i=1

a

4

i

(a

i

+ a

i+1

)(a

2

i

+ a

2

i+1

)

,

B

=

n

X

i=1

a

4

i+1

(a

i

+ a

i+1

)(a

2

i

+ a

2

i+1

)

. Then

A − B =

n

X

i=1

a

4

i

− a

4

i+1

(a

i

+ a

i+1

)(a

2

i

+ a

2

i+1

)

=

n

X

i=1

a

i

− a

i+1

= 0

, andhen e

A = B

.

Wenowshowthatforallpositiverealnumbers

a

and

b

wehave

a

4

+ b

4

(a + b)

2

(a

2

+ b

2

)

4

.

Indeed, usingtheinequality

(x + y)

2

≤ 2(x

2

+ y

2

)

twi eweobtain

(a + b)

2

(a

2

+ b

2

) ≤ 2(a

2

+ b

2

)

2

≤ 4(a

4

+ b

4

)

. Hen e,

2A = A + B

=

n

X

i=1

a

4

i

+ a

4

i+1

(a

i

+ a

i+1

)(a

2

i

+ a

2

i+1

)

1

4

n

X

i=1

(a

i

+ a

i+1

) =

1

2

n

X

i=1

a

i

.

Equalityholdsifandonlyif

a

1

= a

2

= · · · = a

n

.

AlsosolvedbyARKADYALT,SanJose,CA,USA;CHIPCURTIS,MissouriSouthernState

University,Joplin,MO,USA;OLIVERGEUPEL,Bruhl, NRW, Germany;WALTHERJANOUS,

Ursulinengymnasium, Innsbru k, Austria; PAOLO PERFETTI, Dipartimento di Matemati a,

(6)

3492

. [2009 : 515, 518℄ Proposed by Ovidiu Furdui, Campia Turzii, Cluj,Romania.

Let

P

beapointintheinterioroftetrahedron

ABCD

su hthatea h of

∠P AB

,

∠P BC

,

∠P CD

,and

∠P DA

isequal to

arccos

È

2

3

. Provethat

ABCD

isaregulartetrahedronandthat

P

isits entroid.

Theproblemremainsopen. Theonlysubmission,fromPeter Y.Woo,

BiolaUniversity,LaMirada,CA,USA,gavea ounterexamplewhere

ABCD

isadegeneratetetrahedron. Inparti ular,heprovidedanelegantproofthat

if

P

is the entreofaparallelogram

ABCD

withsides

AD = BC = 3

2

and

AB = CD =

6

,anddiagonals

AC = 2

3

and

AC = 6

,then

∠P AB = ∠P BC = ∠P CD = ∠P DA = arccos

r

2

3

.

This ertainly addresses the question that was asked, and itsuggests that

there are in nitely many tetrahedra with an interiorpoint

P

that satis es thegivenanglerequirement,butitfailstoprovideanexpli itnondegenerate

example.

3494. [2009:516,518℄ ProposedbyMi helBataille,Rouen,Fran e.

Let

n > 1

beanintegerandforea h

k = 1

,

2

,

. . .

,

n

let

σ(n, k) =

X

1≤i

1

<···<i

k

≤n

i

1

i

2

· · · i

k

. Provethat

n

X

k=1

ln n

n + 1 − k

· σ(n, k) ∼ (n + 1)! ∼

n

X

k=1

n + 1 − k

ln n

· σ(n, k)

,

where

f (n) ∼ g(n)

meansthat

f (n)

g(n)

→ 1

as

n → ∞

.

Solutionbytheproposer.

Let

P

n

(x)

=

(x + 1)(x + 2) · · · (x + n)

=

x

n

+ σ(n, 1)x

n−1

+ · · · + σ(n, n − 1)x + σ(n, n)

. If

U

n

denotes

n

P

k=1

σ(n, k)

n + 1 − k

,then

U

n

=

 Z

1

0

P

n

(x) dx



1

n + 1

.

(7)

Clearly,

P

n

(x)

P

n

(x)

=

1

x + 1

+

1

x + 2

+ · · · +

1

x + n

,sothatforall

x ∈ [0, 1]

,

1

2

+

1

3

+ · · · +

1

n + 1

P

n

(x)

P

n

(x)

≤ 1 +

1

2

+ · · · +

1

n

. (1)

Multiplying by

P

n

(x)

andintegratingover

[0, 1]

leadsto

(H

n+1

− 1)



U

n

+

1

n + 1

‹

≤ P

n

(1) − P

n

(0) ≤ H

n



U

n

+

1

n + 1

‹ , where

H

n

= 1 +

1

2

+ · · · +

1

n

denotesthe

n

th

harmoni number. Sin ewe

alsohave

P

n

(1) − P

n

(0) = (n + 1)! − n! =

n

n + 1

· (n + 1)!

,weobtain

n

n + 1

·

(n + 1)!

H

n

1

n + 1

≤ U

n

n

n + 1

·

(n + 1)!

H

n+1

− 1

1

n + 1

forallpositiveintegers

n

. Re allingthat

H

n

∼ ln(n)

,theSqueezeTheorem forlimitsyields

lim

n→∞

U

n

ln(n)

(n + 1)!

= 1

,thatis,

n

X

k=1

ln(n)

n + 1 − k

· σ(n, k) ∼ (n + 1)!

. Let

V

n

=

n

P

k=1

(n + 1 − k)σ(n, k)

. From(1)and

P

n

(1) = (n + 1)!

,wededu e that

(H

n+1

− 1)(n + 1)! ≤ P

n

(1) ≤ H

n

(n + 1)!

. Also,for

n > 1

,

V

n

=

n

X

k=1

(n − k)σ(n, k) +

n

X

k=1

σ(n, k)

=

P

n

(1) − n + (n + 1)! − 1

=

P

n

(1) + (n + 1)! − (n + 1)

, sothat

H

n+1

− 1

ln(n)

+

1

ln(n)

1

n! ln(n)

V

n

(n + 1)! ln(n)

H

n

ln(n)

+

1

ln(n)

1

n! ln(n)

.

Again, theSqueezeTheoremyields

(n + 1)! ∼

n

P

k=1

n + 1 − k

ln(n)

· σ(n, k)

,and theproofis omplete.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; and ALBERT

(8)

3495. [2009:516, 518℄ Proposedby CosminPohoata, TudorVianu

Na-tionalCollege,Bu harest,Romania.

Let

a

,

b

,

c

bepositiverealnumberswith

a + b + c = 2

. Provethat

1

2

+

X y li

a

b + c

X y li €

a

2

+ bc

Š

b + c

1

2

+

X y li

a

2

b

2

+ c

2

.

A ombinationofsolutionsbyGeorgeApostolopoulos,Messolonghi,Gree e

and Paolo Perfetti, Dipartimento diMatemati a, Universita degli studi di

TorVergataRoma,Rome,Italy,modi edbytheeditor.

Forve tors

a = (a

1

, a

2

, . . . , a

n

)

and

(b

1

, b

2

, . . . , b

n

)

withrealentries, thenotation

a ≺ b

meansthat

a

1

+ a

2

+ · · · + a

n

= b

1

+ b

2

+ · · · + b

n

and

a

1

+ a

2

+ · · · + a

j

≤ b

1

+ b

2

+ · · · + b

j

holdsforea h

j = 1

,

2

,

. . .

,

n − 1

. Sin e

a + b + c = 2

,theinequalityontheleftisequivalentto

„

1

2

+

X y li

a

b + c

Ž

a + b + c

2

X y li

a

2

+ bc

b + c

or X symmetri

(a

4

+ a

2

bc) ≥

X symmetri

(a

4

+ a

2

bc)

.

S hur'sInequalityyields

X symmetri

(a

4

+ a

2

bc) ≥ 2

X symmetri

(a

3

b)

.

NowusingMuirhead'sinequalityfor

(2, 2, 0) ≺ (3, 1, 0)

weobtain X symmetri

(a

3

b) ≥

X symmetri

(a

2

b

2

)

,

whi hprovestheinequalityontheleft.

Nowtheinequalityontherightisequivalentto

X y li

a

2

+ bc

b + c

„

1

2

+

X y li

a

2

b

2

+ c

2

Ž

a + b + c

2

, or X symmetri

(2a

9

b + 4a

8

bc + 7a

7

b

2

c + a

7

b

3

+ 2a

4

b

4

c

2

)

X symmetri

(2a

6

b

4

+ a

5

b

5

+ 5a

5

b

3

c

2

+ a

4

b

3

c

3

+ 5a

5

b

4

c + 2a

6

b

3

c)

.

(9)

UsingMuirhead'sinequalityrepeatedlyweobtain:

(6, 4, 0) ≺ (9, 1, 0) =⇒

X symmetri

2a

9

b ≥

X symmetri

2a

6

b

4

(6, 2, 2) ≺ (7, 2, 1) =⇒

X symmetri

2a

7

b

2

c ≥

X symmetri

2a

6

b

2

c

2

(6, 3, 1) ≺ (8, 1, 1) =⇒

X symmetri

2a

8

bc ≥

X symmetri

2a

6

b

3

c

(5, 4, 1) ≺ (8, 1, 1) =⇒

X symmetri

2a

8

bc ≥

X symmetri

2a

5

b

4

c

(5, 5, 0) ≺ (7, 3, 0) =⇒

X symmetri

a

7

b

3

X symmetri

a

5

b

5

(5, 3, 2) ≺ (7, 2, 1) =⇒

X symmetri

a

7

b

2

c ≥

X symmetri

a

5

b

3

c

2

(4, 3, 3) ≺ (7, 2, 1) =⇒

X symmetri

a

7

b

2

c ≥

X symmetri

a

4

b

3

c

3

(5, 4, 1) ≺ (7, 2, 1) =⇒

X symmetri

a

7

b

2

c ≥

X symmetri

a

5

b

4

c

Also,bytheAM-GMInequality,wehave

X symmetri

(2a

6

b

2

c

2

+ 2a

4

b

4

c

2

) ≥

X symmetri

4a

5

b

3

c

2

.

Weaddalltheseinequalities,andwearedone.

Also solved by 

SEFKET ARSLANAGI 

C , University of Sarajevo, Sarajevo, Bosnia and

Herzegovina; OLIVER GEUPEL, Bruhl, NRW, Germany; WALTHER JANOUS,

Ursulinen-gymnasium,Innsbru k,Austria;TITUZVONARU,Comanesti,Romania;andtheproposer.One

in ompletesolutionwassubmitted.

ZvonaruobservedthatthisproblemappearedinthebookOldAndNewInequalities,

Vol.2,byVoQuo BaCanandCosminPohoata,GilPublishingHouse,2008.

3496. [2009:516, 519℄ Proposed by EliasC.BuissantdesAmorie,

Cas-tri um,theNetherlands.

Provethefollowingequations:

(a)

tan 72

= tan 66

+ tan 36

+ tan 6

.

(b)

tan 84

= tan 78

+ tan 72

+ tan 60

;

[Ed.: Theproposergavesixmorerelationsoftheform

f (θ) =

4

P

i=1

tan k

i

θ = 0

(10)

CompositeofsolutionsbyKee-WaiLau,HongKong,ChinaandD.J.Smeenk,

Zaltbommel,theNetherlands.

With thehelp ofappropriatetrigonometri identities, bothequations

anberedu edtopropertiesofthegoldense tion

τ =

1 +

5

2

,whi histhe positiverootofthequadrati equation

τ

2

= τ + 1

. (1)

Be ause

τ

istheratioofadiagonaltoasideofaregularpentagon,itsatis es

cos 36

=

τ

2

and

cos 72

=

1

. (2)

(a) Thefollowingequationsareequivalent.

tan 72

=

tan 66

+ tan 36

+ tan 6

,

tan 72

− tan 36

=

tan 66

+ tan 6

,

sin(72

− 36

)

cos 72

cos 36

=

sin(66

+ 6

)

cos 66

cos 6

,

2 sin 36

cos 66

cos 6

=

2 sin 72

cos 72

cos 36

= sin 144

cos 36

,

2 sin 36

cos 66

cos 6

=

sin 36

cos 36

,

2 cos 66

cos 6

=

cos 36

,

cos 72

+ cos 60

=

cos 36

,

cos 72

− cos 36

+

1

2

=

0

,

andthelastequalityfollowsimmediatelyfromequations(1)and(2).

(b) Thefollowingequationsareequivalent.

tan 84

= tan 78

+ tan 72

+ tan 60

,

tan 84

− tan 60

= tan 78

+ tan 72

,

sin(84

− 60

)

cos 84

cos 60

=

sin(78

+ 72

)

cos 78

cos 72

=

1

2 cos 78

cos 72

,

cos 84

= 4 sin 24

cos 72

cos 78

,

sin 6

= 2(sin 96

− sin 48

) cos 78

,

sin 6

= (sin 174

+ sin 18

) − (sin 126

− sin 30

)

,

sin 6

= sin 6

+ cos 72

− cos 36

+

1

2

,

andthelastequalityfollowsimmediatelyfromtheequations(1)and(2)just

asinpart(a).

BothpartswerealsosolvedbyGEORGEAPOSTOLOPOULOS,Messolonghi,Gree e;ROY

(11)

CAMPBELL,ELSIECAMPBELL,andCHARLESR.DIMINNIE,AngeloStateUniversity,San

An-gelo,TX,USA;CHIPCURTIS,MissouriSouthernStateUniversity,Joplin,MO,USA;OLIVER

GEUPEL, Bruhl, NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD,

Portales,NM, USA;WALTHER JANOUS,Ursulinengymnasium,Innsbru k,Austria; ALBERT

STADLER,Herrliberg,Switzerland;EDMUNDSWYLAN,Riga,Latvia;JANVERSTER,Kwantlen

UniversityCollege,BC;andTITUZVONARU,Comanesti,Romania.STANWAGON,Ma alester

College,St.Paul,MN,USAgavea omputerveri ation.

Part(a)wasalsosolvedbyARKADYALT,SanJose,CA,USA;PANOSE.TSAOUSSOGLOU,

Athens,Gree e;PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;andtheproposer.

WagonusedMathemati ato on rmthatthereareseven

4

-tuples

(a, b, c, d)

ofdistin t

integersbetween

0

and

90

(otherthanthepairfeaturedinourproblem)thatsatisfytherelation

tan a

= tan b

+ tan c

+ tan d

,namely

(60; 42, 36, 6),

(72; 60, 42, 24),

(78; 66, 60, 36),

(78; 72, 42, 36)

(60; 50, 20, 10),

(70; 60, 40, 10),

(80; 70, 60, 50)

.

The rstfourare learlyrelatedtothegoldense tionasinourfeaturedpair,whilethe nal

threeseemtoberelatedtotheregularenneagon(ornonagon,ifyouprefer)asdis ussedin

\TrigonometryandtheNonagon"byAndrewJobbings(seewww.arbelos. o.uk/papers.html).

Itisamusingtonotethattheproposerthoughtthathehadfoundonethatfailsto teitherof

thetwopatterns,butitturnsoutthat

tan 62

di ersfrom

tan 48

+ tan 24

+ tan 18

by

about

10

−5

.Wagonfurtherprodu edalistof

49

su hequationsallowingrepeatedangles,and

determinedthattherewerenosu h

3

-termequationsandnosu h

5

-termequations.

3497. [2009 :516, 519℄ Proposed by Salem Maliki , student, Sarajevo

College, Sarajevo,BosniaandHerzegovina.

Let

P

be a point in the interior of triangle

ABC

, and let

r

be the inradiusof

ABC

. Provethat

max{AP

,

BP

,

CP } ≥ 2r

.

I.SolutionbyRoyBarbara,LebaneseUniversity,Fanar,Lebanon.

Re allthatthe onvexhullofatriangle

T

istheunionofitsinteriorand boundary.If

C

isa ir lewithradius

r

inthe onvexhullofatriangle

T

1

with inradius

r

1

,then

r ≤ r

1

. (Hereisaproofofthissimplefa t: Considerthe threetangentsto

C

thatareparalleltothesidesof

T

1

andseparatethe entre of

C

fromthe orrespondingsides;theyformatrianglethatissimilarto

T

1

forwhi h

C

isthein ir le.Sin eallpointsof

C

areinsideoron

T

1

,theratioof thesidesofthenewtriangletothesidesof

T

1

|whi hisalsotheratioofthe inradii| ouldbeatmost

1

;thatis,

r ≤ r

1

.)Let

T = △ABC

beanarbitrary triangle with in ir le

C

and inradius

r

, and let

P

bea point in the onvex hull of

T

. Withoutloss ofgenerality, we mayassume thatmax

{AP

,

BP

,

CP } = AP

andshow that

AP ≥ 2r

. Extend (ifne essary) thesegments

P B

to

P B

1

and

P C

to

P C

1

su h that

P B

1

= P C

1

= P A

. Then

P

is the ir um entreoftriangle

T

1

= △AB

1

C

1

,and

P A

its ir umradius;let

r

1

denoteitsinradius. Notethatbe ause

P

isassumedtolieinthe onvexhull of

T

,

T

mustlieinthe onvexhullof

T

1

; onsequentlythein ir leof

T

also liesinthat onvexhull,sothat(fromoursimplefa t)

r

1

≥ r

.

(12)

II.SolutionbyMi helBataille,Rouen,Fran e.

Generalization: Thefollowingresultholdsforanypoint

P

intheplane of

△ABC

. Let

R

,

O

,

a

,

b

, and

c

be the ir umradius, ir um entre, and sidesof

△ABC

,andlet

M = max{AP

,

BP

,

CP }

;then

(a) if

△ABC

is a ute,

M ≥ R ≥ 2r

,with

M = 2r

ifandonlyif

P = O

andthetriangleisequilateral;

(b) if

△ABC

isnota ute,

M ≥

max{a, b, c}

2

≥ 2r

, with

M = 2r

ifand onlyif

P

isthemidpointofthelongestside.

Let

A

,

B

, and

C

bethemidpointsofthesidesoppositeverti es

A

,

B

,and

C

,respe tively. Forpart(a)we xpoints

D

,

E

,and

F

onthe per-pendi ularbise torsofthesidessothattherays

[OD)

,

[OE)

,and

[OF )

are oppositetherays

[OA

)

,

[OB

)

,and

[OC

)

,respe tively. Thewholeplaneis

theunionofthenonoverlappingangles

∠EOF

,

∠F OD

,and

∠DOE

. With-out loss ofgenerality we an assume that

P

is in or on thesides of angle

∠EOF

(boundedbytherays

[OE)

and

[OF )

)sothat

M = P A

. Let

E

0

on

AB

and

F

0

on

AC

besu hthat

OE

0

||AC

and

OF

0

||AB

. Notethatbe ause

O

isintheinteriorof

△ABC

,

E

0

and

F

0

belongtotherays

[AB)

and

[AC)

, while

OE

0

⊥ OE

and

OF

0

⊥ OF

. Sin e

A

isintheinteriorof

∠E

0

OF

0

,the angle

∠P OA

isobtuse,hen e

M = P A ≥ OA = R

,withequalityexa tly when

P = O

. The inequality

R ≥ 2r

is Euler's inequality, with

R = 2r

exa tlywhen

△ABC

isequilateral,sotheproofofpart(a)is omplete.

For part (b) we rstsuppose that

∠BAC

, say, is obtuse. Then

O

is exteriorto

△ABC

withline

BC

separating

O

from

A

,andtheplaneisthe union ofthe three angles

∠EOF

,

∠EOA

, and

∠F OA

. If

P

is in

∠EOF

then

M = P A ≥ R >

a

2

(mu hasin part(a)). Otherwise, withoutlossof generality,we ansupposethat

P

is in

∠EOA

,in whi h ase

M = P C ≥

A

C =

a

2

. To he kthattheminimumvalueof

M

,namely

a

2

,o urswhen

P = A

, note that

A

and

A

are on the same side of the perpendi ular

bise torofthesegment

AC

,sothat

A

A < A

C

;thatis, if

P = A

,then

M = A

C = A

B =

a

2

. If

∠BAC = 90

, this argument an easily be

adaptedtoshowthat

M ≥

a

2

= R

. To ompletetheproofweshowthatin thepresent asewehave

a

2

≥ 2r

. Let

h = AH

bethealtitudefrom

A

,and let

A

0

bethepointontheray

[HA)

su hthat

∠BA

0

C = 90

. Wewantto

showthat

ah ≥ 4rh

;thatis,that

a + b + c ≥ 4h

(sin e

ah

2

=

r(a + b + c)

2

=

area

(△ABC)

). But

h ≤ HA

0

=

HB · HC ≤

HB + HC

2

=

a

2

,

when e

a ≥ 2h

;moreover,

b

,

c ≥ h

,sothat

a + b + c ≥ 4h

,asdesired. Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,

Messolonghi, Gree e (2solutions); 

SEFKETARSLANAGI 

(13)

BosniaandHerzegovina;JOEHOWARD,Portales,NM,USA;WALTHERJANOUS,

Ursulinen-gymnasium, Innsbru k, Austria; V 

ACLAV KONE 

CN 

Y , BigRapids, MI, USA;KEE-WAI LAU,

HongKong,China;VICTORPAMBUCCIAN,ArizonaStateUniversityWest,Phoenix,AZ,USA;

ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; GEORGE

TSINTSIFAS,Thessaloniki,Gree e;PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;and

theproposer.Thereweretwoin ompletesubmissions.

Tsintsifasextendedtheresultto

n

-dimensionalEu lideanspa e: Forapoint

P

inthe

interiorofthesimplex

A

1

A

2

. . . A

n+1

,

max{A

1

P

,

A

2

P

,

. . .

,

A

n+1

P } ≥ nr

.

Janous pointedout thatthe inequalityfollowsfrom themore generalassertion that

AP + BP + CP ≥ 6r

,whi hisitem12.14ofO.Bottema etal., Geometri Inequalities, Wolters-Noordho Publ.,Groningen,1969.

3498. [2009:517, 519℄ ProposedbyJose LuisDaz-Barrero,Universitat

Polite ni a deCatalunya,Bar elona,Spain.

Let

F

n

be the

n

th

Fibona i number, that is,

F

0

= 0

,

F

1

= 1

, and

F

n

= F

n−1

+ F

n−2

for

n ≥ 2

. Forea hpositiveinteger

n

,provethat Ê

F

n+3

F

n

+

Ê

F

n

+ F

n+2

F

n+1

> 1 + 2

‚Ê

F

n

F

n+3

+

Ê

F

n+1

F

n

+ F

n+2

Π.

Solution by Chip Curtis, Missouri Southern State University, Joplin, MO,

USA. Let

x =

É

F

n+3

F

n

and

y =

É

F

n

+ F

n+2

F

n+1

. The laimed inequality is

su essivelyequivalentto

x + y

>

1 + 2



1

x

+

1

y

‹ , 

1 −

2

xy

‹

(x + y) >

1

.

ItthussuÆ estoshowthatthefollowingtwoinequalitieshold:

1 −

2

xy

1

3

, (1)

x + y

> 3

. (2) Set

λ =

F

n+1

F

n

. Then

xy

=

Ê

F

n+3

F

n

·

F

n+2

+ F

n

F

n+1

=

s

(2F

n+1

+ F

n

) (F

n+1

+ 2F

n

)

F

n

F

n+1

=

Ê

(2λ + 1)



1 +

2

λ

‹ .

(14)

Hen e,(1)isequivalenttoea hof Ê

(2λ + 1)



1 +

2

λ

‹

≥ 3

,

2 (λ − 1)

2

λ

≥ 0

,

andthelatteris learlytrue.

BytheAM{GMInequality,

x + y

>

2 ·

4

s

F

n+3

(F

n

+ F

n+2

)

F

n

F

n+1

=

2 ·

4

Ê

(2λ + 1)



1 +

2

λ

‹ .

For(2),itthussuÆ estoshowthat

(2λ + 1)



1 +

2

λ

‹

>

81

16

, whi hisequivalentto

32λ

2

− λ + 32

16λ

> 0

, whi his learlytrue.

Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,

Messolonghi,Gree e;MICHELBATAILLE,Rouen,Fran e;BRIAND.BEASLEY,Presbyterian

College, Clinton, SC, USA; CHARLES R. DIMINNIE, AngeloState University, San Angelo,

TX,USA;OLIVERGEUPEL,Bruhl,NRW,Germany;WALTHERJANOUS,Ursulinengymnasium,

Innsbru k,Austria;ALBERTSTADLER,Herrliberg,Switzerland;andtheproposer.Two

in om-pletesolutionsweresubmitted.

3499

. [2009 : 517, 519℄ Proposed by Bernardo Re aman, Instituto AlbertoMerani,Bogota, Colombia.

Abuildinghas

n

oorsnumbered

1

to

n

andanumberofelevatorsall ofwhi hstopatboth oors

1

and

n

,andpossiblyother oors. Forea h

n

, ndtheleastnumberofelevatorsneededinthisbuildingifbetweenanytwo

oorsthereisatleastoneelevatorthat onne tsthemnon-stop.

Forexample,if

n = 6

,nineelevatorssuÆ e:

(1, 6)

,

(1, 5, 6)

,

(1, 4, 6)

,

(1, 3, 4, 6)

,

(1, 2, 4, 5, 6)

,

(1, 2, 5, 6)

,

(1, 2, 6)

,

(1, 3, 5, 6)

,and

(1, 2, 3, 6)

.

SolutionbyGeorgeApostolopoulos,Messolonghi,Gree e.

Theansweris ›

n

2

4

ž .

(15)

Toseethatatleastthismanyelevatorsareneeded, onsidertheset

P =

n

(x, y) ∈ Z

2

: 1 ≤ x, y ≤ n, x ≤

n

2

, y >

n

2

o .

Any elevator an onne t at most one pair of oors in the set

P

, and the ardinalityof

P

is

›

n

2

4

ž

,soatleastthismanyelevatorsareneeded.

Toshowthat ›

n

2

4

ž

elevatorssuÆ e,wegivea onstru tionintwo ases.

Case 1:

n = 2k

. Here

k

2

elevators are needed. Let integers

i

and

j

be restri tedsothat

1 ≤ i ≤ k

and

k + 1 ≤ j ≤ 2k

,anddes ribeea helevator bythetupleof oorsitstopsat. Theelevatorsarethen

8 < :

(1, i, j, 2k)

, if

i + j = 2k + 1

,

(1, 2k + 1 − j, i, j, 2k)

, if

i + j > 2k + 1

,

(1, i, j, 2k + 1 − j, 2k)

, if

i + j < 2k + 1

. Case2:

n = 2k + 1

. Here

k

2

+ k

elevatorsareneeded. Letintegers

i

and

j

berestri tedsothat

1 ≤ i ≤ k

and

k + 1 ≤ j ≤ 2k + 1

,anddes ribeea h elevatorbythetupleof oorsitstopsat. Theelevatorsarethen

8 < :

(1, i, j, 2k + 1)

, if

i + j = 2k + 2

,

(1, 2k + 2 − j, i, j, 2k + 1)

, if

i + j > 2k + 2

,

(1, i, j, 2k + 2 − j, 2k + 1)

, if

i + j < 2k + 2

.

This ompletestheproof.

AlsosolvedbyOLIVERGEUPEL,Bruhl, NRW,Germany;D.P.MEHENDALE(Dept.of

Ele troni s)andM.R.MODAK,(formerlyofDept.Mathemati s), S.P.College,Pune,India;

MISSOURISTATEUNIVERSITYPROBLEMSOLVINGGROUP,Spring eld,MO,USA;MORTEN

H.NIELSEN,UniversityofWinnipeg,Winnipeg,MB;andPETERY.WOO,BiolaUniversity,La

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