Soluzione del problema n°3388

(1)

SOLUTIONS

No problemis ever permanently losed. Theeditorisalways pleased

to onsiderforpubli ationnewsolutionsornewinsightsonpastproblems.

We belatedlya knowledge a orre t solutionto#3340by \Solver X",

dedi atedtothememoryofJim Totten,whi hwehadpreviously lassied

asin orre t.Ourapologies.

3376. [2008:430,432℄ ProposedbyD.J.Smeenk,Zaltbommel,the

Neth-erlands.

Theverti esofquadrilateral

ABCD

lieona ir le. Let

K

,

L

,

M

,and

N

bethein entresof

△ABC

,

△BCD

,

△CDA

,and

△DAB

,respe tively. Showthatquadrilateral

KLM N

isare tangle.

CommentbyFran is oBellotRosado,I.B.EmilioFerrari,Valladolid,Spain.

This problem has already appeared in Crux. The proof of the result

and the proof of its onverse was published together with omments and

referen esin[1980:226-230℄.

Solutions, omments,andotherreferen esweresentbyGEORGEAPOSTOLOPOULOS,

Messolonghi, Gree e;

SEFKET ARSLANAGI

C , University of Sarajevo, Sarajevo, Bosnia and

Herzegovina; RICARDO BARROSOCAMPOS, Universityof Seville, Seville, Spain; MICHEL

BATAILLE, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh

Long,Vietnam;CHIPCURTIS,MissouriSouthernStateUniversity,Joplin,MO,USA;OLIVER

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

JANOUS,Ursulinengymnasium,Innsbru k,Austria;V

ACLAVKONE

CN

Y,BigRapids,MI,USA;

TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; MADHAV R. MODAK, formerlyof Sir

ParashurambhauCollege,Pune,India;DUNGNGUYENMANH,HighS hoolofHUS,Hanoi,

Vietnam;JOELSCHLOSBERG,Bayside,NY,USA;ALBERTSTADLER,Herrliberg,Switzerland;

PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;andtheproposer.

Geupel provides the link

www.mathlinks.ro/viewtopic.php?t=137589

. Janous pro-vides the link

http://www.gogeometry.com/problem/p035_incenter_cyclic_quadrilateral

. Kone ny and S hlosberg found a solution at

www.cut-the-knot.org/Curriculum/Geometry/

CyclicQuadrilateral.shtml

S hlosberg also gives two other links,

forumgeom.fau.edu/

FG2002volume2/FG200223.ps

and

mathworld.wolfram.com/CyclicQuadrilateral.html

.

3377. [2008:430,432℄ ProposedbyToshioSeimiya,Kawasaki,Japan.

Let

ABC

beatrianglewith

∠B = 2∠C

. Theinteriorbise torof

∠BAC

meets

BC

at

D

. Let

M

and

N

bethemidpointsof

AC

and

BD

,respe tively. Supposethat

A

,

M

,

D

,and

N

are on y li . Provethat

∠BAC = 72

◦

.

SolutionbyMi helBataille,Rouen,Fran e.

Wewillusethefamiliarnotationsfortheelementsofthetriangle

ABC

. First,wenotethatfrom

∠B = 2∠C

,wehave

(2)

ItthensuÆ estoprovethat

a

= b

,forthenwewillhave

B

= A

,andthen

A

=

2

5 A

+ A +

A

2 =

2

5 (A + B + C) = 72

◦

. From

DB

c

=

DC

b

=

a

b

+ c

,weobtain

DB

=

ac

b

+ c

and

DC

=

ab

b

+ c

,sothat

CN

=

ab

b

+ c

+

ac

2(b + c)

=

a(2b + c)

2(b + c)

.

Sin e

A

,

M

,

D

,and

N

are on y li , wehave

CM

· CA = CD · CN

,or, afterasimple al ulation,

c

2 b

= c

a

2 − 2b

2 + b

2a

2 − b

2

. (2)

Therelation(1),rewrittenas

c

2 _b

_{= b}

3 _{− abc}

,togetherwith(2)yields

c

(a − b)(2b + a) = 2b(b − a)(b + a)

.

Now, if

a

6= b

, then

c <

0

,whi h isimpossible. Therefore,

a

= b

,whi h ompletestheproof.

AlsosolvedbyMIGUELAMENGUALCOVAS,CalaFiguera,Mallor a,Spain;GEORGE

APOSTOLOPOULOS, Messolonghi, Gree e;

SEFKET ARSLANAGI

C , University of Sarajevo,

Sarajevo,BosniaandHerzegovina;CHIPCURTIS,MissouriSouthernStateUniversity,Joplin,

MO,USA;OLIVERGEUPEL,Bruhl, NRW,Germany;JOHNG.HEUVER,GrandePrairie,AB;

WALTHERJANOUS,Ursulinengymnasium,Innsbru k,Austria;V ACLAVKONE CN Y ,BigRapids,

MI, USA; KEE-WAI LAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka,

Japan;SOUTHEASTMISSOURISTATEUNIVERSITYMATHCLUB;MADHAVR.MODAK,

for-merlyofSirParashurambhauCollege,Pune,India;ALBERTSTADLER,Herrliberg,Switzerland;

HAOHAOWANGandJERZYWOJDYLO,SoutheastMissouriStateUniversity,CapeGirardeau,

Missouri, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU,

Comane sti,Romania;andtheproposer.

AmengualCovasnotedthattherelation(1)hadalreadyappearedinseveralotherissues

ofCRUX,namely,[1976:74℄,[1984:278℄,and[1996:265-267℄.

3378. [2008:430,432℄ ProposedbyMihalyBen ze,Brasov,Romania.

Let

x

,

y

,and

z

bepositiverealnumbers. Provethat X y li

xy

+ x + y

≤

X y li

x

2x + z

.

CounterexamplebyGeorgeApostolopoulos,Messolonghi,Gree e.

Theinequalityisfalseingeneral. Forexample,if

x

= 2

and

y

= z = 1

, thentheinequalitybe omes

2

5 +

1

3 +

2

5 ≤

2

5 +

1

4 +

1

3

,or

2

5 <

1

4

,whi his learlyfalse. Alsodisprovedby SEFKETARSLANAGI

(3)

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

Ursulinengymnasium,Innsbru k,Austria;RICHARDI.HESS,Ran hoPalosVerdes,CA,USA;

PAOLOPERFETTI,DipartimentodiMatemati a,Universita deglistudidiTorVergataRoma,

Rome, Italy; STAN WAGON,Ma alesterCollege, St.Paul, MN, USA;andTITUZVONARU,

Comane sti,Romania.

Let

f(x, y, z) =

P y li

x

2x + z

−

P y li

xy

+ x + y

. Then Curtis showed that

f(x, x, x) < 0

forall

x > 1

. Perfettishowedthat,ingeneral,

f(x, y, z) < 0

if

x

,

y

,

z

allliein

(1, ∞)

,while

f(x, y, z) ≥ 0

if

x

,

y

,

z

allliein

(0, 1]

.

Janousreportsthatasimilarinequality, P y li

xy

+ x

2 _{+ y}

2 ≤

P y li

x

2x + z

,where

x

,

y

,

z

arepositive,wasproblem4ofthe2009MediterraneanMathemati sCompetition.

Let

a

1

,

a

2

,

. . .

,

a

n

bepositiverealnumbers. Provethat

n

X

i=1

a

i

a

i

+ (n − 1)a

i+1

≥ 1

,

wherethesubs riptsaretakenmodulo

n

. SolutionbyMi helBataille,Rouen,Fran e.

Set

x

i

=

(n − 1)a

i+1

a

i

forea h

i

;theproblemisthentoprovethat

n

X

i=1

1 1 + x

i

≥ 1

(1)

subje ttothe onstraint

x

1 x

2 · · · x

n

= (n − 1)

n

.

Let

y

i

=

1 1 + x

i

forea h

i

,sothat

x

i

=

1 − y

i

y

i

.

Suppose onthe ontrary that

n

P

i=1

y

i

<

1

. Then

1 −

n

P

i=1

y

i

>

0

, and hen eforea h

i

wehavebytheAM{GMInequalitythat

1 − y

i

>

X

j6=i

y

j

≥ (n − 1)

Y

j6=i

y

j

1 n−1

. (2)

Multiplying a rosstheinequalitiesin(2)overall

i

,wethenobtain

n

Y

i=1

(1 − y

i

) > (n − 1)

n

Y

i=1

y

i

, or

n

Q

i=1

1−y

i

y

i

>

(n − 1)

n

(4)

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

Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long,

Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER

GEUPEL,Bruhl, NRW,Germany;JOHNG.HEUVER,GrandePrairie,AB;WALTHERJANOUS,

Ursulinengymnasium, Innsbru k, Austria; SALEM MALIKI

C, student, Sarajevo College,

Sarajevo, Bosnia and Herzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi,

Vietnam;CRISTINELMORTICI,ValahiaUniversityofTargovi^ ste,Romania;PAOLOPERFETTI,

DipartimentodiMatemati a,UniversitadeglistudidiTorVergataRoma,Rome,Italy;ALBERT

STADLER,Herrliberg,Switzerland;PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;and

theproposer.

Forpositive

α

Janousprovedthegeneralization

n

X

i=1

a

i

a

i

+ αa

i+1

≥ min

n

1 + α

, 1

o

.

Let

a

,

b

,

c

,

x

,

y

,and

z

berealnumbers. Showthat

(a

2 + x

2 )(b

2 + y

2 )

(c

2 _{+ z}

2 _{)(a − b)}

2 +

(b

2 + y

2 )(c

2 + z

2 )

(a

2 _{+ x}

2 _{)(b − c)}

2 +

(c

2 + z

2 )(a

2 + x

2 )

(b

2 _{+ y}

2 _{)(c − a)}

2 ≥

a

2 _{+ x}

2 |(a − b)(a − c)|

+

b

2 _{+ y}

2 |(b − a)(b − c)|

+

c

2 _{+ z}

2 |(c − a)(c − b)|

. Similar solutions by Arkady Alt, San Jose, CA, USA; and Mi hel Bataille,

Rouen,Fran e. Let

u

=

Ê

(a

2 _{+ x}

2 _)(b

2 _{+ y}

2 ₎

(c

2 _{+ z}

2 _{)(a − b)}

2

;

v

=

Ê

(b

2 _{+ y}

2 _)(c

2 _{+ z}

2 ₎

(a

2 _{+ x}

2 _{)(b − c)}

2

;

w

=

Ê

(c

2 _{+ z}

2 _)(a

2 _{+ x}

2 ₎

(b

2 _{+ y}

2 _{)(c − a)}

2

. Then

uv

=

Ê

(b

2 _{+ y}

2 ₎

2 (a − b)

2 _{(b − c)}

2 =

b

2 + y

2 |(a − b)(b − c)|

, andsimilarly

uw

=

a

2 _{+ x}

2 |(a − b)(c − a)|

;

vw

=

c

2 + z

2 |(b − c)(c − a)|

.

Theoriginalinequalitynowfollowsfromthewell-knowninequality

u

2 + v

2 + w

2 ≥ uv + uw + vw

.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e;

SEFKET

(5)

Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW,

Germany; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; SALEMMALIKI

C,

student,SarajevoCollege,Sarajevo,BosniaandHerzegovina;DUNGNGUYENMANH,High

S hoolofHUS,Hanoi,Vietnam;PAOLOPERFETTI,DipartimentodiMatemati a,Universita

deglistudidiTorVergataRoma,Rome,Italy;CAOMINHQUANG,NguyenBinhKhiemHigh

S hool,VinhLong,Vietnam;TITUZVONARU,Comane sti,Romania;andtheproposer.

3381

⋆

. [2008 :431, 433℄ ProposedbyShiChangwei,XianCity, Shaan XiProvin e,China.

Let

n

beapositiveinteger. Provethat (a)

1 −

1 ₆

1 −

₆

1

2 · · ·

1 −

₆

1 n

>

4

5

; (b) Let

a

n

= aq

n

,where

0 < a < 1

and

0 < q < 1

. Evaluate

lim

n→∞

n

Y

i=1

(1 − a

i

)

.

Solutiontopart(a)byPeterY.Woo,BiolaUniversity,LaMirada,CA,USA.

Wehave

(1 − aq)

1 − aq

2 >

_{1 − a}

q

+ q

2

. Ingeneral,ifweknow that

n

Q

k=1

1 − aq

k

>

1 −

n

P

k=1

aq

k

holdsforsome

n

≥ 2

,then

n+1

Y

k=1

1 − aq

k

>

1 − aq

n+1

1 −

n

X

k=1

aq

k

!

= 1 −

n+1

X

k=1

aq

k

+ aq

n+1

n

X

k=1

aq

k

!

>

_{1 −}

n+1

X

k=1

aq

k

.

Byindu tion,theinequalityholdsforea h

n

≥ 2

,andfurthermorewehave

n

Q

k=1

1 − aq

k

>

_{1 −}

n

P

k=1

aq

k

_>

_{1 −}

aq

(1 − q)

. Taking

a

= 1

and

q

=

1

6

yields

1 −

1 ₆

1 −

₆

1

2 · · ·

1 −

₆

1 n

>

1 −

1 ₅

=

4

5

.

AlsosolvedbyARKADYALT,SanJose,CA,USA(part(a));GEORGEAPOSTOLOPOULOS,

Messolonghi, Gree e (part(a)); PAUL BRACKEN, University of Texas, Edinburg, TX, USA;

OLIVERGEUPEL,Bruhl, NRW,Germany;RICHARDI.HESS,Ran hoPalosVerdes,CA,USA

(part(a));WALTHERJANOUS,Ursulinengymnasium,Innsbru k,Austria;ALBERTSTADLER,

Herrliberg,Switzerland;andSTANWAGON,Ma alesterCollege,St.Paul,MN,USA.Therewas

onein orre tsolutiontopart(b)submitted.

(6)

Bra kennotesthatthe produ tisrelatedtoRamanujan's

q

-extensionoftheGamma fun tion, the ustomarynotationsbeing

(a; q)

k

= (1 − a)(1 − aq) · · · 1 − aq

k−1

and

(a; q)

∞

=

∞

Q

k=0

1 − aq

k

.The

q

-binomialtheoremimpliesthatfor

|x| < 1

,

|q| < 1

wehave

1 (a;q)

∞

=

∞

P

n=0

x

n

(q;q)

n

,whi hleadstoanexpressionfortherequiredprodu t.

Geupelgivestheexpansion

∞

Y

k=1

1 − q

k

=

√

1

3

·

1

24 √

q

· ϑ

2 π

6 ,

6 √

_q

,

where

ϑ

2

isaJa obithetafun tion. Forfurtherinformationhereferstotheonlinesurvey

http://mathworld.wolfram.com/q-PochhammerSymbol.html

.

Stadlerprovidedsixdierentexpansionsrelatedtotherequiredprodu t,oneinvolving

the Dedekindetafun tionandtheothersduetoEulerandJa obi. Hereferstheinterested

readertoMarvinI.Knopp,ModularFun tionsinAnalyti NumberTheory(2

nd

ed.),Chelsea,

NewYork(2003).

Wagon alsoreferstothe Wolfram website, aswellasB. Gordon andR.J.M Intosh,

\Some EighthOrderMo kThetaFun tions", J.LondonMath.So . 62,pp.321-335(2000).

Thereisgiventheasymptoti estimate

(q; q)

∞

=

p

π

t

exp

t

12 −

π

2 12t

+ o(1)

;

t = −

1

2 ln q

, whi hyieldsavaluefor

(1/6; 1/6)

∞

thatiswithin

2 · 10

−10

ofthetruevalue.

3382. [2008 : 431, 433℄ Proposed by Jose Luis Daz-Barreroand Josep

Rubio-Masseg u, UniversitatPolite ni a deCatalunya,Bar elona,Spain.

Let

n

beapositiveinteger. Provethat

sin

P

n+2

4P

n

P

n+1

+ cos

P

n+2

4P

n

P

n+1

<

3

2 sec

3P

n

+ 2P

n−1

4P

n

P

n+1

, where

P

n

isthe

n

th

Pellnumber, whi his denedby

P

0 = 0

,

P

1 = 1

,and

P

n

= 2P

n−1

+ P

n−2

for

n

≥ 2

.

SolutionbyOliverGeupel,Bruhl, NRW,Germany.

Forallrealnumbers

x

and

y

with

0 < y <

π

2

,wehave

sin x + cos x =

√

2 sin

x

+

π

4 ≤

√

2 <

3

2 ≤

3

2 sec y

. Setting

x

=

P

n+2

4P

n

P

n+1

and

y

=

3P

n

+ 2P

n−1

4P

n

P

n+1

givesthedesired result, ifwe

anshowthat

0 < y <

π

2

.

Be ausethePellnumbersformastri tlyin reasingsequen e,weobtain

forpositive

n

that

0 <

3P

n

+ 2P

n−1

4P

n

P

n+1

<

5P

n

4P

2 n

≤

5

4 <

π

2

,

(7)

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

Messolonghi, Gree e;

SEFKET ARSLANAGI

C , University of Sarajevo, Sarajevo, Bosnia and

Herzegovina; MICHELBATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State

University, Joplin,MO,USA;WALTHER JANOUS,Ursulinengymnasium, Innsbru k,Austria;

SALEM MALIKI

C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; ALBERT

STADLER, Herrliberg,Switzerland; PETER Y. WOO, BiolaUniversity, La Mirada, CA,USA;

TITUZVONARU,Comanesti,Romania;andtheproposers.

Itis learfromthefeaturedsolutionthatontherightsideoftheinequality

3

2

anbe repla edbythesmaller oeÆ ient

√

2

.

3383. [2008:431,433℄ ProposedbyMi helBataille,Rouen,Fran e.

Let

ABC

be atrianglewith

∠BAC

6= 90

◦

,let

O

beits ir um entre and let

M

bethemidpoint of

BC

. Let

P

beapoint ontheray

M A

su h that

∠BP C = 180

◦

_{− ∠BAC}

. Let

BP

meet

AC

at

E

andlet

CP

meet

AB

at

F

. If

D

istheproje tionofthemidpointof

EF

onto

BC

,showthat

(a)

AD

isasymmedianof

△ABC

;

(b)

O

,

P

,andtheortho entreof

△EDF

are ollinear.

A omposite of solutions by Madhav R. Modak, formerly of Sir

Parashu-rambhauCollege,Pune,Indiaandtheproposer.

Werstshowthatthereisauniquepoint

P

ontheray

M A

su hthat

∠BP C = 180

◦

−∠BAC

,and

P

anbe onstru tedbyextendingthemedian

AM

towhereitmeetsthe ir um ir le

Γ

of

△ABC

at

P

′

. Claim: Theline

through

B

parallelto

P

′

_C

meets

AM

at

P

and

AC

at

E

. Toseethis,note that

△P BM ∼

=

△P

′

_CM

(

BM

= M C

and orrespondinganglesareequal), when e

P BP

′

_C

isaparallelogram. Itfollowsthat

∠BP C = ∠CP

′

_B

. But

P

′

,

A

lieonoppositear s

BC

of

Γ

,so

∠BP C = ∠CP

′

_B

_{= 180}

◦

_{− ∠BAC}

,

when e

P

istheuniquepointontheray

M A

thatformstherequiredangle. Moreover,sin e

∠BAC

6= 90

◦

,

P

isdierentfrom

A

. Thehomothetywith entre

A

thattakes

P

′

to

P

willtake

C

to

E

,

B

to

F

,and

Γ

tothe ir um ir le

Γ

′

ofthequadrilateral

AF P E

. Notethat

EF

isparallelto

BC

;itsmidpoint, allit

N

,lieson

AP

.

We turn now to part (a). Sin e

AP

and

F E

interse t at

N

, by the lassi al onstru tion

N

isthepoleoftheline

BC

withrespe tto

Γ

′

(be ause

B

and

C

arethediagonalpointsotherthan

N

ofquadrilateral

AF P E

).The polarof

D

,therefore,passesthrough

N

and,sin e

DN

passesthroughthe entre of

Γ

′

and is perpendi ular to

EF

, this polar is a tually

EF

. As a result,

DE

and

DF

are tangent to

Γ

′

at

E

and

F

. It follows that

AD

is the symmedian from

A

in

△EAF

. (See, for example, Roger A. Johnson, Advan edEu lideanGeometry,(Dover,2007),page215,no.347,orNathan

AltshillerCourt,CollegeGeometry,(Dover,2007),page248,Theorem560.)

Theresultin(a)followssin e

△BAC

and

△F AE

arehomotheti . Forpart (b) notethatthehomothetythattakesthe ir le

BAC

(namely

Γ

)to

F AE

(namely

Γ

′

) will take

OB

to the radius of

Γ

′

(8)

tangentat

F

,itspreimageistangentto

Γ

at

B

,andisthereforeperpendi ular to

OB

. We on ludethat

OB

⊥ DF

. Similarly,

OC

⊥ DE

. Let

H

′

denote

theortho entreof

△EDF

. Thehomothetywith entre

P

thattakes

B

to

E

willtake

C

to

F

. Sin e

OB

kH

′

_E

and

OC

kH

′

_F

,thishomothetymusttake

O

to

H

′

, when e

H

′

_{, P}

, and

O

are ollinear,whi h ompletestheproofof part(b).

Also solvedbyCHIPCURTIS,MissouriSouthernStateUniversity,Joplin, MO,USA;

OLIVERGEUPEL,Bruhl, NRW,Germany;andPETERY.WOO,BiolaUniversity, LaMirada,

CA,USA.

Allthesubmittedsolutionsex eptBataille'swerebasedonthepropertythat

AD

isa symmedianof

△ABC

ifandonlyif

D

dividestheside

BC

intheratio

c

2 _{: b}

2

(Court,page

248,Theorem561).

Showthat,foranyrealnumber

x

,

lim

n→∞

1 n

2 n−1

X

k=1

k

_·

x

+

n

− k − 1

n

=

⌊x⌋ + {x}

2

,

where

⌊a⌋

istheintegerpartoftherealnumber

a

and

{a} = a − ⌊a⌋

. SolutionbyRi hardI.Hess,Ran hoPalosVerdes,CA,USA.

Let

T

n

=

n−1

P

k=1

k

_·

j

x

+

n

− k − 1

n

k

. Givenan integer

n

≥ 2

, let

m

be theuniqueintegersu hthat

m

n

≤ {x} <

m

+ 1

n

. Notethat

m

= m(n)

isa fun tionof

n

,andthat

m

n

→ {x}

as

n

→ ∞

sin e

{x} −

m

n

<

1 n

. Now,the rst

m

−1

termsof

T

n

are

⌊x⌋+1

1+2+· · ·+(m−1)

,andthenal

n

−m

termsof

T

n

are

⌊x⌋

m

_{+ (m + 1) + · · · + (n− 1)}

. Colle talltermsinvolving

⌊x⌋

and losethesumstoobtain

1 n

2 T

n

=

1

2 ⌊x⌋

n

− 1

n

+

1

2 m

n

m

− 1

n

;

itthenfollowsthat

1 n

2 T

n

→

1

2 ⌊x⌋ + {x}

2

as

n

→ ∞

.

AlsosolvedbyARKADYALT,SanJose,CA,USA;CHIPCURTIS,MissouriSouthernState

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

Ursulinengymnasium,Innsbru k,Austria;MISSOURISTATEUNIVERSITYPROBLEMSOLVING

GROUP,Springeld, MO,USA;JOEL SCHLOSBERG, Bayside,NY,USA; ALBERTSTADLER,

Herrliberg,Switzerland;andtheproposer.

TheMissouriStateUniversityProblemSolvingGroupredu edtheproblemto al ulating

R

1

0 y⌊x + 1 − y⌋ dy

bynotingthattherequiredlimitisthelimitofaRiemannsumforthe integrandon

[0, 1]

plusatermthatvanishesas

n → ∞

.

(9)

Let

p

1

,

p

2

,

. . .

,

p

6

beprimeswith

p

k+1

= 2p

k

+ 1

for

k

= 1

,

2

,

. . .

,

5

. Showthat

X

1≤i<j≤6

p

i

p

j

isdivisibleby

15

.

A ompositeofsimilarsolutionsby

SefketArslanagi ,UniversityofSarajevo,

Sarajevo,BosniaandHerzegovina;ChipCurtis,MissouriSouthernState

Uni-versity,Joplin,MO,USA;CristinelMorti i,ValahiaUniversityofTargovi^ ste,

Romania;andTituZvonaru,Comane sti,Romania.

If

p

1 = 3

, then

p

2 = 7

and

p

3 = 15

, when e

p

3

is notprime and, therefore, our sequen e annot begin with the prime

p

1 = 3

. Nor an it begin with

p

1 ≡ 1 (mod 3)

,otherwisewewouldhave

p

2 ≡ 0 (mod 3)

,so that

p

2

ouldnotbeprime.We on ludethatne essarily,

p

1 ≡ −1 (mod 3)

, andhen ethat

p

i

≡ −1 (mod 3)

forea h

p

i

. Theresultingsumsatises

X

1≤i<j≤6

p

i

p

j

≡

X

1≤i<j≤6

(−1)

2 ≡

6

2 ≡ 0 (mod 3)

.

Similarly,workingmodulo

5

,weseethat

(a) If

p

1 = 5

,then

p

5 ≡ 0 (mod 5)

andisnotprime.

(b) If

p

1 ≡ 1 (mod 5)

,then

p

4 ≡ 0 (mod 5)

andisnotprime. ( ) If

p

1 ≡ 2 (mod 5)

,then

p

6 ≡ 0 (mod 5)

andisnotprime. (d) If

p

1 ≡ 3 (mod 5)

,then

p

3 ≡ 0 (mod 5)

andisnotprime. On eagain,theonlypossiblevaluefor

p

i

is

−1 (mod 5)

,when e

X

1≤i<j≤6

p

i

p

j

≡

X

1≤i<j≤6

(−1)

2 _≡

6

2 ≡ 0 (mod 5)

.

We have seen that the sum is divisible by

3

and by

5

, and thus by

15

as laimed. Finally,wenotethattheresultisnotva uouslytrue:

89

,

179

,

359

,

719

,

1439

,

2879

is anexample ofsu hasequen e(and iseasilyseentobe thesmallestexample).

Also solvedbyOLIVERGEUPEL,Bruhl, NRW, Germany;RICHARD I.HESS, Ran ho

PalosVerdes,CA,USA;WALTHERJANOUS,Ursulinengymnasium,Innsbru k,Austria;DAVID

E.MANES, SUNYat Oneonta, Oneonta, NY,USA;JOELSCHLOSBERG, Bayside,NY,USA;

ROBERTP.SEALY,MountAllisonUniversity,Sa kville, NB;ALBERT STADLER,Herrliberg,

Switzerland;PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;andtheproposer.

Manespointsoutthataprime

p

is alledaSophieGermainprimeif

2p + 1

isalsoa prime;moreover,asequen eof

n − 1

SophieGermainprimes,

p

,

2p + 1

,

2(2p + 1) + 1

,

. . .

, that annotbeextendedineitherdire tion(thatis,therstprimeisnotoftheform

2q + 1

for

q

aprime,whilethenalprimeofthesequen eisnotaSophieGermainprime)is alleda Cunningham hainoftherstkindoflength

n

. AsidefromtheCunningham hainoflength

5

thatbeginswith

2

(namely,

2

,

5

,

11

,

23

,

47

),thenaldigitofanyprimeinaCunningham hain oflengthfourorgreatermustbea

9

(be ausethenaldigit y les

1

,

3

,

7

,

5

,

. . .

). A ording

(10)

totheCunninghamChainRe ordswebpage,thelongestknownCunningham hainoftherst

kindhaslength

17

.

Onarelatednote, ompareProblem10ontheMathemati alCompetitionBalti Way

2004[2008:212;2009:153℄whereoneisaskedtoprovearesultimplyingthataCunningham

hain anneverbeinnite.

3386. [2008:432, 434℄ Proposed byOvidiuFurdui,CampiaTurzii,Cluj,

Romania.

Evaluatetheintegral

Z

∞

0 e

−x

Z

x

0 e

−t

− 1

t

dt

ln x dx

.

Solution by Walther Janous, Ursulinengymnasium, Innsbru k, Austria and

GerhardKir hner,UniversityofInnsbru k,Innsbru k,Austria.

Wewillusethefa t that

∞

P

n=1

(−1)

n

= − ln 2

and

∞

P

n=1

(−1)

n

2 = −

π

2

12

. Also,fromtheprodu trepresentationoftheGammafun tion,wehave

1 Γ(x)

= xe

γx

Y

∞

k=1

1 +

x

k

e

−x/k

=⇒ −

Γ

′

_(x)

Γ(x)

=

1 x

+ γ +

∞

X

k=1

1 x

+ k

−

1 k

=⇒ −

Γ

′

_{(n + 1)}

Γ(n + 1)

=

1 n

+ 1

+ γ +

∞

X

k=1

1 n

+ 1 + k

−

1 k

= γ −

1 +

1

2 +

1

3 + · · · +

1 n

,

where

γ

isEuler's onstant. [Ed. ForpropertiesoftheGammafun tionsee

http://en.wikipedia.org/wiki/Gamma_function

.℄ Wenow ompute Z

∞

0 e

−x

Z

x

0 e

−t

_{− 1}

t

dt

ln x dx

=

Z

∞

0 e

−x

Z

x

0 ∞

X

n=1

(−1)

n

n!

t

n−1

_dt

!

ln x dx

=

Z

∞

0 e

−x

ln x

∞

X

n=1

(−1)

n

n!

x

n

dx

=

∞

X

n=1

(−1)

n

· n!

Z

∞

0 e

−x

x

n

ln x dx

=

∞

X

n=1

(−1)

n

· Γ

′

_{(n + 1)}

Γ(n + 1)

=

∞

X

n=1

(−1)

n

−γ +

n

X

k=1

1 k

!

=

γ

ln 2 +

∞

X

n=1

(−1)

n

X

k=1

1 k

(11)

= γ ln 2 +

Z

1

0

Z

1

0 ∞

X

n=1

(−1)

n

x

n−1

n

X

k=1

y

k−1

dxdy

= γ ln 2 +

Z

1

0

Z

1

0 ∞

X

n=1

(−1)

n

x

n−1

y

n

_{− 1}

y

_{− 1}

dxdy

= γ ln 2 +

Z

1

0

Z

1

0 −y

1 + xy

+

1 1 + x

y

_{− 1}

dxdy

= γ ln 2 −

Z

1

0

Z

1

0

1 (1 + xy)(1 + x)

dy

dx

= γ ln 2 −

Z

1

0 ln(1 + x)

x(1 + x)

dx

= γ ln 2 −

Z

1

0 ln(1 + x)

1 x

−

1 1 + x

dx

= γ ln 2 −

Z

1

0 ln(1 + x)

x

dx

+

1

2 ln

2 ₂

= γ ln 2 +

1

2 ln

2 2 +

∞

X

n=1

Z

1

0 (−1)

n

x

n−1

_dx

= γ ln 2 +

1

2 ln

2 _{2 +}

X

∞

n=1

(−1)

n

2 = γ ln 2 +

1

2 ln

2 2 −

π

2

12

.

AlsosolvedbyKEE-WAILAU,HongKong,China;DUNGNGUYENMANH,HighS hool

of HUS, Hanoi, Vietnam; MADHAV R. MODAK, formerlyof Sir Parashurambhau College,

Pune, India; MOUBINOOL OMARJEE, Paris, Fran e; PAOLO PERFETTI, Dipartimento di

Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER,

Herrliberg, Switzerland; NGUYENVAN VINH,Belarusian State University, Minsk, Belarus;

andtheproposer.Thereweretwoin ompletesolutionssubmitted.

3387. [2008:432, 434℄ Proposed byOvidiuFurdui,CampiaTurzii,Cluj,

Romania.

Let

k > l

≥ 0

bexedintegers. Find

lim

x→∞

2 x

ζ(x + k)

ζ(x+k)

− ζ(x + l)

ζ(x+l)

,

where

ζ

istheRiemannzetafun tion.

SimilarsolutionsbyPaoloPerfetti,DipartimentodiMatemati a,Universita

deglistudidiTorVergataRoma,Rome,ItalyandAlbertStadler,Herrliberg,

Switzerland.

Forreal-valuedfun tions

f

(x)

and

g(x)

denedontheinterval

(1, ∞)

the notation

f

(x) = O

g(x)

willmeanthatthereexistsan

x

0 >

1

anda positive onstant

C

su hthat

|f(x)| ≤ C|g(x)|

whenever

x

≥ x

0

.

(12)

Wehave

ζ(x) = 1 + 2

−x

_{+ 3}

−x

₊

P

∞

k=4

k

−x

and

1 (x − 1)4

x−1

=

Z

∞

4 ds

s

x

≤

∞

X

k=4

k

−x

≤

Z

∞

3 ds

s

x

=

1 (x − 1)3

x−1

, hen e

∞

P

k=4

k

−x

= O

3 −x

and

ζ(x) = 1 + 2

−x

_{+ O}

3 −x

. Moreover,

ζ(x)

= exp

1 + 2

−x

+ O

3 −x

· ln

1 + 2

−x

+ O

3 −x

= exp

1 + 2

−x

+ O

3 −x

·

2 −x

+ O

3 −x

= exp

2 −x

_{+ O}

3 −x

= 1 + 2

−x

_{+ O}

3 −x

. Finally,wehave

lim

x→∞

2 x

ζ(x + k)

ζ(x+k)

− ζ(x + l)

ζ(x+l)

=

lim

x→∞

2 x

2 −x−k

_{− 2}

−x−l

_{+ O}

3 −x

= 2

−k

_{− 2}

−l

.

AlsosolvedbyOLIVERGEUPEL,Bruhl,NRW,Germany;WALTHERJANOUS,

Ursulinen-gymnasium,Innsbru k,Austria;andtheproposer.

3388. [2008:432, 434℄ Proposed byPaulBra ken,University ofTexas,

Edinburg,TX,USA,inmemoryofMurrayS.Klamkin.

Forallreal

x

≥ 1

,showthat

1

2 √

x

_{− 1 +}

(x − 1)

2 √

x

− 1 +

√

x

+ 1

<

x

2 √

x

+

√

x

+ 2

.

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

studidiTorVergataRoma,Rome,Italy,modiedbytheeditor.

Theinequalityholdsfor

x

= 1

,solet

x >

1

. Sin e

√

x

+ 1 >

√

x

_{− 1}

and

√

x

+

√

x

+ 2 < 2

√

x

+ 1

bythe on avityof

√

x

,itsuÆ estoprove

1

2 √

x

− 1 +

1

2 (x − 1)

3/2

_<

x

2

2 √

x

+ 1

, (1)

as the right side of (1) is less thanthe right side of the desired inequality

andtheleftsideof(1) isgreaterthantheleftsideofthedesiredinequality.

Inequality(1)issu essivelyequivalentto

p

x

2 _{− 1 +}

p

x

2 _{− 1(x − 1) < x}

2

,

x

p

x

2 _{− 1 < x}

2

, p

x

2 _{− 1 < x}

,

(13)

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS,

MissouriSouthernStateUniversity,Joplin,MO,USA;OLIVERGEUPEL,Bruhl,NRW,Germany;

WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; RICHARD I. HESS, Ran ho

PalosVerdes,CA,USA;PAOLOPERFETTI,DipartimentodiMatemati a,Universitadeglistudi

diTorVergataRoma,Rome,Italy(twosolutions);ALBERTSTADLER,Herrliberg,Switzerland;

PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;andtheproposer.Onein orre tsolution

wassubmitted.

STANWAGON,Ma alesterCollege,St.Paul,MN,USAveriedtheinequalityusinga

omputeralgorithm.

In this small spa e that remains, we again put out the all for more

problemproposalsfromourreadersintheareasofGeometry,Algebra,Logi ,

andCombinatori s. Wehaveavaststoreofinterestinginequalitiesandwe

will ontinuetopro essthem anda eptnewones,buttheotherareasare

wanting!

Regarding arti les, wenow have a small ba klog. This is due to the

diligen e oftheArti les Editor, JamesCurrie, butalso duetothefa t that

spa eforarti lesin Cruxis extremelylimited. Forthisreasonweaskthat,

ifpossible,authorswitharti lesappearinginCruxwaitabout18-20months

before submitting another manus ript to Crux. In the meantime, we will

gear upandattempt to lear theba klogin 2010by runningtheo asional

96pageissue.

Va lav Linek

Crux Mathemati orum

with Mathemati al Mayhem

FormerEditors/An iensReda teurs: Bru eL.R.Shawyer,JamesE.Totten

Crux Mathemati orum

FoundingEditors/Reda teurs-fondateurs: Leopold Sauve&Frederi kG.B.Maskell

FormerEditors/An iensReda teurs: G.W.Sands,R.E.Woodrow,Bru eL.R.Shawyer

Mathemati al Mayhem

FoundingEditors/Reda teurs-fondateurs: Patri kSurry&RaviVakil

FormerEditors/An iensReda teurs: PhilipJong,JeHigham,J.P.Grossman,