Soluzione del problema n°3370

(1)

SOLUTIONS

Au unprobleme n'estimmuable.L'editeur esttoujoursheureux

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

portantsurdesproblemes anterieurs.

3363. [2008:362,364℄ ProposedbyToshioSeimiya,Kawasaki,Japan.

Let

ABC

beatrianglewith

∠ACB = 90

◦

₊

1

2 ∠ABC

. Let

M

bethe

midpointof

BC

. Provethat

∠AM C < 60

◦

.

Composite of similar solutions by Ri ardo Barroso Campos, University of

Seville,Seville,Spain;Ri hardI.Hess,Ran hoPalosVerdes,CA,USA;Matti

Lehtinen,National Defen eCollege, Helsinki,Finland;MadhavR.Modak,

formerly of Sir Parashurambhau College, Pune, India; and Peter Y. Woo,

BiolaUniversity,LaMirada,CA,USA.

In

△ABC

let

4β = ∠B

andlet

W

bethefootofthebise torof

∠A

. Wearegiventhat

∠C = 90

◦

_{+ 2β}

,when e

∠A = 90

◦

_{− 6β}

. Wededu ethat

6β < 90

◦

,sothat

β <

15 ◦

. Moreover,be ause

∠AW C

isanexternalangle of

△ABW

,itsatises

∠AW C = 4β + 45

◦

− 3β

= 45

◦

_{+ β < 60}

◦

.

Itremainstoshowthat

∠AM C < ∠AW C

. Be ausewearegiventhat

∠C

is obtuse, we have

AB > AC

, when e the foot

W

of the angle bise tor satises

BW > W C

. Thatis,

W

liesbetween

M

and

C

;thus

∠AW C

isan externalangleof

△AM W

andthereforesatises

∠AM C < ∠AW C < 60

◦

,

asdesired.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e;

SEFKET

ARSLANAGI

C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA,

Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; CHIP CURTIS,

Missouri Southern State University, Joplin, MO, USA; GERALD EDGECOMB and JULIE

STEELE,students,CaliforniaStateUniversity,Fresno,CA,USA;OLIVERGEUPEL,Bruhl, NRW,

Germany;JOHNG.HEUVER,GrandePrairie,AB;WALTHERJANOUS,Ursulinengymnasium,

Innsbru k, Austria; V ACLAV KONE CN

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

China;TAICHIMAEKAWA,TakatsukiCity,Osaka,Japan;THANOSMAGKOS,3 rd

HighS hool

ofKozani,Kozani,Gree e;SALEMMALIKI

C,student,SarajevoCollege,Sarajevo,Bosniaand

Herzegovina;SKIDMORECOLLEGEPROBLEMSOLVINGGROUP,SkidmoreCollege,Saratoga

Springs, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e; TITU ZVONARU, Comanesti,

Romania;andtheproposer.

Let

ABC

beatrianglewith

∠BAC = 120

◦

and

AB > AC

. Let

M

be themidpointof

BC

. Provethat

∠M AC > 2 ∠ACB

.

(2)

I.SolutionbyPeterY.Woo,BiolaUniversity,LaMirada,CA,USA.

Let

β

= ∠CBA

and

γ

= ∠BCA

; then (be ause

∠BAC = 120

◦

)

β

+ γ = 60

◦

. Let

D

bethepoint ofsegment

BC

forwhi h

∠BAD = 2β

; then

∠DAC = 2γ

. TheLawofSinestellsusthat

BD

=

AD

sin 2β

sin β

= 2AD cos β

;

also,

CD

=

AD

sin 2 60

◦

_{− β}

sin 60

◦

_{− β}

= 2AD cos 60

◦

− β

. But

β+γ = 60

◦

and

γ > β

;hen e,

60 ◦

_{−β > β}

sothat

cos 60

◦

_−β

_{< cos β}

.

Consequently,

CD < BD

,when e

M

liesbetween

B

and

D

. Itfollowsthat

∠M AC > ∠DAC = 2∠ACB

,asdesired.

II.SolutionbyMattiLehtinen,NationalDefen eCollege,Helsinki,Finland.

Let

O

bethe entreofthe ir um ir le

Γ

of

△ABC

,and

D

bethe mid-point of thear

CAB

of

Γ

. Note that

A

is onthe smallerar

DC

. Sin e

120 ◦

_{= ∠BAC = ∠BDC}

,

DB

= DC = DO

. Drawthe ir le

Γ

′

with

en-tre

D

andradius

DB

(passingthrough

B

,

O

,and

C

).Be ause

M

isthe om-monmidpointof

BC

and

DO

, ir les

Γ

and

Γ

′

aresymmetri withrespe tto

M

. Let

AM

meet

Γ

′

at

E

(sothat

M

isalsothemidpointof

AE

);notethat

∠EDC = ∠AOB = 2∠ACB

. Extend

AE

tomeet

Γ

againat

F

andextend

DE

to meet

Γ

again at

G

. Be ausethe line

F A

separates

D

from

G

and

C

,

G

isonthear

F C

opposite

D

and

A

; onsequently,

∠F AC > ∠GDC

. But,

∠F AC = ∠M AC

,and

∠GDC = ∠EDC = ∠AOB = 2∠ACB

,sowe aredone.

SEFKET

ARSLANAGI

Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; CHIP CURTIS,

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

Germany; V ACLAV KONE CN

Y, Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, China;

THANOS MAGKOS,3 rd

High S hoolofKozani, Kozani, Gree e; SALEMMALIKI

C,student,

SarajevoCollege,Sarajevo,BosniaandHerzegovina; GEORGETSAPAKIDIS,Agrinio,Gree e;

TITUZVONARU,Comanesti,Romania;andtheproposer.

A square

ABCD

is ins ribed in a ir le

Γ

. Let

P

be a point onthe minor ar

AD

of

Γ

,and let

E

and

F

betheinterse tions of

AD

with

P B

and

P C

,respe tively. Provethat

AE

· DF = 2 [P AE] + [P DF ]

,

(3)

SolutionbyJohnG.Heuver,GrandePrairie,AB.

Thealtitudesfrompoint

P

tosegments

AB

and

CD

leadusto on lude that

[P AB] + [P DC] =

1

2 AB

2

, Furthermore,

1

2 AB

2 _{= [P AE] + [AEB] + [P DF ] + [DF C]}

, Sin e

[AEB] + [DF C] =

1

2 AB(AE + DF )

, wehave

.

A

B

C

D

E

F

P

q

b

[P AE] + [P DF ] =

1

2 AB

2 − [AEB] + [DF C]

=

1

2 AB

2 ₋

1

2 AB(AE + DF )

=

1

2 AB

(AB − AE − DF ) =

1

2 AB

· EF

. Now,

∠BP C = ∠CP D = ∠45

◦

asanglessubtendedbyar sequal to

aquarterofthe ir le. Hen e

P F

isthebise torof

∠EP D

,andtherefore,

EF

F D

=

P E

P D

.

Also,

∠ABP = ∠P DA

asanglessubtendedbythear

AP

,sothattriangles

P DE

and

ABE

aresimilar,andtherefore,

P E

P D

=

AE

AB

. Consequently,

EF

F D

=

AE

AB

, or

EF

· AB = AE · DF

,andthen

[P AE] + [P DF ] =

1

2 AB

· EF =

1

2 AE

· DF

, whi h ompletestheproof.

SEFKET

ARSLANAGI

Lebanese University, Fanar,Lebanon; RICARDOBARROSO CAMPOS,UniversityofSeville,

Seville, Spain; MICHELBATAILLE,Rouen, Fran e; CHIPCURTIS, MissouriSouthern State

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

Ursulinengymnasium,Innsbru k,Austria;V

ACLAVKONE

CN

(4)

MALIKI

C,student,SarajevoCollege,Sarajevo,BosniaandHerzegovina;MADHAVR.MODAK,

formerlyofSirParashurambhauCollege,Pune,India;JOELSCHLOSBERG,Bayside,NY,USA;

D.J. SMEENK,Zaltbommel,theNetherlands; PETER Y.WOO,BiolaUniversity,La Mirada,

CA,USA;TITUZVONARU,Comanesti,Romania;andtheproposer. Therewasonein orre t

ommentsubmitted.

Kone nyandtheproposerea hbeganbyletting

Z

bethefootoftheperpendi ularfrom

P

to

AD

,thendedu ingthat

△DP Z ∼ △BEA

and

△AP Z ∼ △CF D

,whi hthenenables the al ulationof

AE · DF

.

3366. [2008:362, 365℄ Proposed byOvidiuFurdui,CampiaTurzii,Cluj,

Romania.

Let

{x}

denote the fra tional part of the real number

x

; that is,

{x} = x − ⌊x⌋

, where

⌊x⌋

is the greatest integer not ex eeding

x

. Evaluate

Z

1

0 ₁

x

4 dx

.

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

USA.

Let

I

betheintegraltobeevaluated. Then

I

=

lim

N →∞

N

X

n=1

Z

_n

1

1 n+1

₁

x

4 dx

=

lim

N →∞

N

X

n=1

Z

n+1

n

{y}

4 y

2 dy

=

lim

N →∞

N

X

n=1

Z

n+1

n

(y − n)

4 y

2 dy

=

_{N →∞}

lim

N

X

n=1

I

n

, where

I

n

=

Z

n+1

n

(y − n)

4 y

2 dy

. Wehave

I

n

=

Z

n+1

n

y

2 _{− 4ny + 6n}

2 ₋

4n

3 y

+

n

4 y

2 dy

=

y

3

3 − 2ny

2 _{+ 6n}

2 _y

− 4n

3 ln y −

n

4 y

n+1

n

= 3n

2 _{− n +}

1

3 −

n

4 n

+ 1

+ n

3 _{− 4n}

3 _ln

1 +

1 n

= 3n

2 − n +

1

3 +

n

2 − n −

1 n

+ 1

− 4n

3 ln

1 +

1 n

= 4n

2 − 2n +

4

3 −

1 n

+ 1

− 4n

3 _ln

1 +

1 n

.

(5)

Substitutingtheseriesexpansions

x

+ 1

=

∞

X

k=0

(−1)

k

x

k+1

and

ln(1 + x) =

∞

X

k=1

(−1)

k+1

_x

k

, with

x

=

1 n

(n ≥ 2)

intothelastexpressionfor

I

n

yields

I

n

= 4n

2 − 2n +

4

3 −

∞

X

k=0

(−1)

k

n

k+1

− 4n

3 ∞

X

k=1

(−1)

k+1

kn

k

= 4n

2 − 2n +

4

3 − 4n

3 ₁

n

−

1 2n

2 +

1 3n

3 −

∞

X

k=0

(−1)

k

n

k+1

− 4n

3 ∞

X

k=4

(−1)

k+1

kn

k

= −

∞

X

k=0

(−1)

k

n

k+1

− 4n

3 ∞

X

k=4

(−1)

k+1

kn

k

=

∞

X

k=1

(−1)

k

n

k

− 4

∞

X

k=4

(−1)

k+1

kn

k−3

=

∞

X

k=1

(−1)

k

n

k

− 4

∞

X

k=1

(−1)

k

(k + 3)n

k

=

∞

X

k=2

−

1 n

k

− 4

∞

X

k=2

(−1)

k

(k + 3)n

k

=

1 n(n + 1)

− 4

∞

X

k=2

(−1)

k

(k + 3)n

k

,

andthelastformulaisvalidfor

I

1

aswell. Thus,

I

=

∞

X

n=1

1 n(n + 1)

− 4

∞

X

n=1

∞

X

k=2

1 k

+ 3

−

1 n

k

=

1 − 4

∞

X

k=2

(−1)

k

+ 3

∞

X

n=1

1 n

k

=

1 − 4

∞

X

k=2

(−1)

k

+ 3

ζ(k)

.

AlsosolvedbyWALTHERJANOUS,Ursulinengymnasium,Innsbru k,Austria;RICHARD

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

R. MODAK, formerly of Sir Parashurambhau College, Pune, India; PAOLO PERFETTI,

DipartimentodiMatemati a, Universita deglistudidiTor VergataRoma,Rome, Italy; and

theproposer.Thereweretwoin orre tsolutionssubmitted.

Theproposerobtainedtheanswer

−

1

3 − γ + 2 ln 2π − 12 ln A + 12 ln B

,where

γ

is Euler's onstantandwhere

A = lim

n→∞

1

1 ₂

2 _{· · · n}

n

n2

2 +

n

2 +

1

12 e

−n2

4

;

B = lim

n→∞

1

2 ₂

2

2 _{· · · n}

n

2 n

n3

3 +

n2

2 +

n

6 e

−n3

9 +

n

12

(6)

aretheGlaisher{Kinkelin onstantsoforder

1

andorder

2

,respe tively. Geupelobtained

I = 24

∞

P

k=2

ζ

_{(k) − 1}

k(k + 1)(k + 2)(k + 3)

= 0.14553289 . . .

,whi hhe remarked onvergesmorerapidlythantheseries

1 −

∞

P

k=2

(−1)

k

ζ(k)

k

+ 3

andisthusbettersuited fornumeri alapproximation.

Janousreportsthegeneralization

Z

1

0

1 x

N

dx =





N −2

X

j =0

(−1)

j

N

j

1 N − j − 1

2 N −j −1

− 1





− (−1)

N

N ln 2 −

1

2 +

∞

X

p=2

(−1)

p

_{(p − 1)}

p + N − 1

ζ(p) − 1

.

KeithEkblaw,WallaWalla,WA,USA,obtainedtheestimate

I ≈ 0.146

byaMonteCarlo approa h,whi his orre tto

3

de imalpla esafterrounding.

3367. [2008:362, 365℄ Proposedby LiZhou, PolkCommunity College,

WinterHaven,FL,USA.

Let

p(x) = a

n

x

n

_{+ a}

n−1

x

n−1

+ · · · + a

1 x

beapolynomialwithinteger oeÆ ients,where

a

n

>

0

and

n

P

k=1

a

k

= 1

. Proveordisprovethatthereare innitelymanypairsofpositiveintegers

(k, ℓ)

su hthat

p

(k + 1) − p(k)

and

p

(ℓ + 1) − p(ℓ)

arerelativelyprime.

Solution by Cristinel Morti i, Valahia University of Targoviste, Romania,

modiedslightlybytheeditor.

Let

Q

(x) = p(x + 1) − p(x)

. Thenwehave

Q

(0) = p(1) − p(0) =

n

X

k=1

a

k

− 0 = 1

.

Hen e,

Q(x) = xq(x) + 1

forsomepolynomial

q(x)

ofdegree

n

− 1

. We need to nd innitely many positive integers

k

and

l

su h that

gcd Q(k), Q(l) = 1

.

Sin e the leading term in

Q(x)

is

na

n

x

n−1

we have

Q(k) > 0

for suÆ ientlylargepositiveintegers

k

. Forsu h

k

let

l

= Q(k) = kq(k) + 1

, sothat

l

isapositiveinteger. If,onthe ontrary,

gcd Q(k), Q(l)

6= 1

,then there is a prime number

p

su h that

p

| Q(k)

and

p

| Q(l)

. Then, sin e

Q(l) = lq(l) + 1 = Q(k)q(l) + 1

,we on ludethat

p

| 1

,a ontradi tion. AlsosolvedbyROYBARBARA,LebaneseUniversity,Fanar,Lebanon;OLIVERGEUPEL,

Bruhl, NRW,Germany;WALTHERJANOUS,Ursulinengymnasium,Innsbru k,Austria;andthe

proposer.

Morti iremarkedthatthe on lusioninfa tholdsforanypolynomial

Q(k)

whose on-stanttermiseither

1

or

−1

. Thisis learfromtheprooffeaturedabove. Barbaraprovedthe strongerresultsthat(1)if

a

n

6= 0

and

n

P

k=1

(7)

integers

k

su hthat

p(k + 1) − p(k)

are allpairwise relativelyprime, and(2)if

f(x)

isa polynomialwithinteger oeÆ ientsandpositivedegreesu hthat

f(0) = 1

,thenthereare innitelymanyintegers

k

1 < k

2 < k

3 < · · ·

su hthatthe

f(k

i

)

'sareallpairwiserelatively prime.

3368. [2008:363,365℄ ProposedbyNevenJuri ,Zagreb,Croatia.

Let

m

be aninteger,

m

≥ 2

, andlet

A

= [A

ij

]

bea blo kmatrixof dimension

2 m

_{× 2}

m

with

A

ij

∈ M

4,4

(N)

for

1 ≤ i

,

j

≤ 2

m−2

,denedby

A

ij

= 2

m

B

ij

+ C

ij

,where

B

ij

=







2 m

_{− 4i + 4}

_{4i − 4}

₂

m

_{− 4i + 4}

4i − 3

2 m

_{− 4i + 3 2}

m

_{− 4i + 3}

_{4i − 3}

4i − 2

2 m

_{− 4i + 2 2}

m

_{− 4i + 2}

_{4i − 2}

2 m

_{− 4i + 1}

_{4i − 1}

₂

m

_{− 4i + 1}







, and

C

ij

=







4 − 4j 4j − 2 4j − 1 1 − 4j

4j − 3 3 − 4j 2 − 4j

4j

4j − 3 3 − 4j 2 − 4j

4j

4 − 4j 4j − 2 4j − 1 1 − 4j







.

Showthatmatrix

A

isamagi squareoforder

2 m

.

SolutionbyOliverGeupel,Bruhl, NRW,Germany.

For onvenien e,write

B

ij

= B

(i)

₌

_b

(i)

kℓ

and

C

ij

= C

(j)

₌

_c

(j)

kℓ

,

andnotethatthese

4 × 4

matri eshaveintegerentries. Claim1Forea hentry

a

of

A

,

1 ≤ a ≤ 2

2m

. Proof: Wehave

a

= 2

m

_b

(i)

kℓ

+ c

(j)

kℓ

forsome

i

,

j

,

k

, and

ℓ

. Ifwealsohave

(k, ℓ) ∈ {(1, 2)

,

(1, 3)}

,then

b

(i)

kℓ

≥ 0

and

c

(j)

kℓ

≥ 1

;hen e

a

≥ 1

. Otherwise,

b

(i)

_kℓ

≥ 1

and

c

(j)

kℓ

≥ 1−2

m

;hen e

a

≥ 2

m

₊₁₋₂

m

_{= 1}

. Thisprovesthelower

bound. Fortheupperbound,weobservethatfor

(k, ℓ) ∈ {(1, 1)

,

(1, 4)}

we have

b

(i)

kℓ

≤ 2

m

and

c

(j)

kℓ

≤ 0

;hen e

a

≤ 2

m

_{· 2}

m

_{= 2}

2m

. Inanyother ase

b

(i)

_kℓ

_{≤ 2}

m

_{− 1}

and

c

(j)

kℓ

≤ 2

m

;soagain

a

≤ 2

m

₂

m

_{− 1}

_{+ 2}

m

_{= 2}

2m

.

Claim2Theentriesof

A

aredistin t. Proof: Iftwoentriesof

A

areequalthen

2 m

_b

(i)

kℓ

+ c

(j)

kℓ

= 2

m

b

(i

′

₎

kℓ

+ c

(j

′

₎

kℓ

for some

j

,

j

′

,

k

,

k

′

,

ℓ

,and

ℓ

′

. Let

b

= b

(i)

kℓ

,

b

′

_{= b}

(i

′

)

k

′

ℓ

′

,

c

= c

(j)

kℓ

,and

c

′

_{= c}

(j

′

)

k

′

ℓ

′

. Wethenhave

1 ≤ |c − c

′

_{| ≤ 4 · 2}

m−2

_{− 1 − 4 · 2}

m−2

_{= 2}

m+1

_{− 1 < 2}

m+1

.

Without loss ofgenerality, suppose that

c < c

′

. Then

c

′

− c = 2

m

and

b

= b

′

+ 1

,hen e

c

≡ c

′

_{(mod 4)}

and

b

≡ b

′

_{+ 1 (mod 4)}

. Therearefour

asesfor

c

and

c

′

: Case1

(k, ℓ) ∈ {(1, 1)

,

(4, 1)}

and

(k

′

_{, ℓ}

′

_{) ∈ {(2, 4)}

,

(3, 4)}

; Case2

(k, ℓ) ∈ {(1, 4)

,

(4, 4)}

and

(k

′

_{, ℓ}

′

_{) ∈ {(2, 1)}

,

(3, 1)}

;

(8)

Case3

(k, ℓ) ∈ {(2, 3)

,

(3, 3)}

and

(k

′

_{, ℓ}

′

_{) ∈ {(1, 2)}

,

(4, 2)}

; Case4

(k, ℓ) ∈ {(2, 2)

,

(3, 2)}

and

(k

′

_{, ℓ}

′

_{) ∈ {(1, 3)}

,

(4, 3)}

; ea hofthemin ompatiblewiththe ondition

b

≡ b

′

_{+ 1 (mod 4)}

.

Claim3Thesumoftheentriesalonganyhorizontal,verti al,ormaindiagonal

lineof

A

is

2 3m−1

_{+ 2}

m−1

.

Proof: Thesumalonganygivenhorizontalorverti allineof

B

(i)

and

C

(j)

is

S

B

= 2

m+1

and

S

C

= 2

,respe tively. Therefore,thesumoftheentriesin ea hlineof

A

is

2 m−2

₂

m

_S

B

+S

C

= 2

m−2

2 m

·2

m+1

+2 = 2

3m−1

+2

m−1

. The maindiagonal sumsof

B

(i)

and

C

(j)

are

b

(i)

_{= 2}

m+2

_{− 16i + 10}

and

c

(j)

_{= 10 − 16j}

. Theentriesinea hmaindiagonal of

A

thenalsohavethe

sum

2 m−2

P

i=1

2 m

_b

(i)

₊

2 m−2

P

j=1

c

(j)

_{= 2}

3m−1

_{+ 2}

m−1

.

Alsosolvedbytheproposer.Onein ompletesolutionwasre eivedthatveriedtherow,

olumn,anddiagonalsums,butdidnotshowthattheentriesofthemagi square onsistedof

1

,

2

,

. . .

,

2 2m

.

3369. [2008 : 363, 365℄ Proposed by George Tsintsifas, Thessaloniki,

Gree e.

Let

A

1 A

2 A

3 A

4

be a tetrahedron whi h ontains the entre

O

of its ir umsphereasaninteriorpoint. Let

ρ

i

bethedistan efrom

O

tothefa e oppositevertex

A

i

. If

R

istheradiusofthe ir umsphere,provethat

4

3 R

≥

4 X

i=1

ρ

i

.

SolutionbyOliverGeupel,Bruhl, NRW,Germany.

The laimisfalse;thefollowing ounterexampleisadaptedfrom[1℄.

First, onsiderthedegeneratetetrahedron

A

′

1 A

′

2 A

′

3 A

′

4

,where

A

′

₁

= A

′

₂

=

(−1, 1, 0)

,

A

′

₃

=

(1, 1, 0)

,

A

′

₄

=

(−1, −1, 0)

,

O

= (0, 0, 0)

,and

R

=

√

2

. Wehave

ρ

1 = ρ

2 = 0

,and

ρ

3 = ρ

4 = 1

, so that

4R

4 P

i=1

ρ

i

=

4 √

2

2 = 2

√

2 < 3

.

(9)

Now,let

A

1 =

−1 + ε, 1 − ε,

p2ε(2 − ε)

,

A

2 =

−1 + ε, 1 − ε, −p2ε(2 − ε)

,

A

3 =

√

2 cos

π

4 − ε

,

√

2 sin

π

4 − ε

,

0

,

A

4 =

√

2 cos

−

3π

₄

+ ε

,

√

2 sin

−

3π

₄

+ ε

,

0

.

Itiseasytoverifythatpoint

O

isintheinteriorofthetetrahedron

A

1 A

2 A

3 A

4

, andthat

A

i

→ A

′

i

forea h

i

as

ε

→ 0

,aswellas

4R

4 P

i=1

ρ

i

−→ 2

√

2

. The bound

2 √

2

isthebest possible. This isa orollaryfrom the fol-lowing:

Theorem[1℄.Let

A

1 A

2 A

3 A

4

beanondegeneratetetrahedronwhose ir um- entre

O

isnotanexteriorpoint. Let

P

beapointnotexteriorto

A

1 A

2 A

3 A

4

. Letthe distan esfrom

P

totheverti esandtothefa es of

A

1 A

2 A

3 A

4

be denotedby

R

i

and

ρ

i

,respe tively. Then

4 P

i=1

R

i

4 P

i=1

ρ

i

>

2 √

2

, and

2 √

2

isthegreatestlowerbound.

Referen es

[1℄ Ni holas D. Kazarino, \D.K. Kazarino's inequality for tetrahedra",

Mi higanMathemati alJournal,Vol.4,No.2(1957),pp.99-104.

CounterexamplealsogivenbyPeterY.Woo,BiolaUniversity,LaMirada,CA,USA.

GeorgeApostolopoulos,Messolonghi,Gree e, itedthebookbyKazarino,Geometri

Inequalities,YaleUniversityPress,1961,p.116.

No ompletesolutionsorother ommentsweresubmitted.

Gree e.

Let

a

i

and

b

i

bepositive realnumbersfor

1 ≤ i ≤ k

, andlet

n

bea positiveinteger. Provethat

k

X

i=1

a

n

1 i

!

n

≤

k

X

i=1

a

i

b

i

!

k

X

i=1

b

1 n−1

i

!

n−1

.

(10)

Compositeofsimilaroridenti alsolutionssubmittedbyallthesolverswhose

namesappearbelow(ex eptthetwosolversidentiedbya\*"beforetheir

names). Let

p

= n

and

q

=

n

− 1

. Then

p >

1

,

q >

1

and

1 p

+

1 q

= 1

. [Ed: learly,

n >

1

forthegiveninequalitytomakesense.℄ Let

x

i

=

_a

_i

b

i

n

1

and

y

i

= b

1 n

i

. Then

x

i

and

y

i

arepositiveforea h

i

.

ByHolder'sInequality,wehave

k

P

i=1

x

i

y

i

≤

_k

P

i=1

x

p

_i

1 p

k

P

i=1

y

q

_i

1 q

whi h be omes

k

P

i=1

a

1 n

i

≤

k

P

i=1

a

i

b

i

n

1 k

P

i=1

b

n−1

_i

n−1

n

.

Theresultfollowsbyraisingbothsidestothe

n

th

power.

SEFKET

ARSLANAGI

C,UniversityofSarajevo,Sarajevo,BosniaandHerzegovina;MICHELBATAILLE,

Rouen,Fran e;MANUELBENITO,

OSCARCIAURRI,EMILIOFERNANDEZ,andLUZRONCAL,

Logro~no,Spain;CAOMINHQUANG,NguyenBinhKhiemHighS hool,VinhLong,Vietnam;

CHIPCURTIS,MissouriSouthernStateUniversity,Joplin,MO,USA;OLIVERGEUPEL,Bruhl,

NRW, Germany;JOEHOWARD,Portales,NM,USA;WALTHERJANOUS,

Ursulinengymnas-ium, Innsbru k,Austria; SALEMMALIKI

C,student, SarajevoCollege, Sarajevo,Bosniaand

Herzegovina; DUNG NGUYENMANH, HighS hool ofHUS, Hanoi, Vietnam; *CRISTINEL

MORTICI, Valahia Universityof Targoviste, Romania; PAOLO PERFETTI, Dipartimento di

Matemati a, Universita deglistudi diTorVergata Roma, Rome, Italy; JOEL SCHLOSBERG,

Bayside,NY,USA;*PETERY.WOO,BiolaUniversity,LaMirada,CA,USA;TITUZVONARU,

Comane sti,Romania;andtheproposer.

Gree e.

Let

ABC

beatrianglewith

a

,

b

,and

c

thelengthsofthesidesopposite the verti es

A

,

B

, and

C

, respe tively, and let

M

be aninterior point of

△ABC

. Thelines

AM

,

BM

,and

CM

interse ttheoppositesidesatthe points

A

1

,

B

1

,and

C

1

,respe tively. Linesthrough

M

perpendi ulartothe sidesof

△ABC

interse t

BC

,

CA

,and

AB

at

A

2

,

B

2

,and

C

2

,respe tively. Let

p

1

,

p

2

,and

p

3

bethedistan esfrom

M

tothesides

BC

,

CA

,and

AB

, respe tively. Provethat

[A

1 B

1 C

1 ]

[A

2 B

2 C

2 ]

=

(ap

1 + bp

2 )(bp

2 + cp

3 )(cp

3 + ap

1 )

8a

2 _b

2 _c

2 a

p

1 +

b

p

2 +

c

p

3

,

where

[KLM ]

denotestheareaoftriangle

KLM

. SolutionbyJoelS hlosberg,Bayside, NY,USA.

By the basi formula forthe area ofa triangle,

[ACM ] =

1

2 bp

2

and

[CM B] =

1 ₂

ap

1

. Itiswellknownthatifsegments

Y Z

and

Y

′

_Z

′

(11)

samelineand

X

isanypoint,then

[XY Z] : [XY

′

_Z

′

_{] = Y Z : Y}

′

_Z

′

,sothat

[ACC

1 ]

[ACM ]

=

[CC

1 B]

[CM B]

=

CC

1 CM

;

[ACC

1 ]

[CC

1 B]

=

[ACM ]

[CM B]

=

AC

1 C

1 B

=

bp

2 ap

1

. By similarreasoning,

BA

1 A

1 C

=

cp

3 bp

2

and

CB

1 B

1 A

=

ap

1 cp

3

. A knownformula(see

Eri W. Weisstein, \Routh's Theorem," at

http://mathworld.wolfram.com/

RouthsTheorem.html

) states that if

A

′

,

B

′

, and

C

′

are points on the sides

BC

,

CA

,and

AB

of

△ABC

,respe tively,then

[A

′

B

′

C

′

] =

AC

′

C

′

_B

· BA

′

A

′

_C

· CB

′

B

′

_A

+ 1

AC

′

C

′

_B

+ 1

BA

′

A

′

_C

+ 1

CB

′

B

′

_A

+ 1

[AB C ]

. Therefore,

[A

1 B

1 C

1 ] =

bp

2 ap

1

· cp

3 bp

2

· ap

1 cp

3 + 1

bp

2 ap

1 + 1

cp

3 bp

2 + 1

ap

1 cp

3 + 1

[AB C ]

=

2abcp

1 p

2 p

3 (ap

1 + bp

2 )(bp

2 + cp

3 )(cp

3 + ap

1 )

[ABC]

.

Sin e

∠M A

2 C

and

∠M B

2 C

arerightangles,

∠A

2 M B

2 = 360

◦

− ∠M A

2 C

− ∠M B

2 C

− ∠A

2 CB

2 = 180

◦

− ∠ACB

,

so

sin ∠A

2 M B

2 = sin C

,andsin eallfouranglesofquadrilateral

A

2 M B

2 C

arelessthan

180 ◦

,

A

2 M B

2 C

is onvex. Thewell-knownareaformulafora triangle,

[XY Z] =

1

2 XY

· XZ sin ∠Y XZ

,yields

[M A

2 B

2 ] =

1

2 M A

2 · M B

2 sin ∠A

2 M B

2 =

p

1 p

2 ab

₁

2 ab

sin C

=

p

1 p

2 ab

[ABC]

. Bysimilarreasoning,

[M B

2 C

2 ] =

p

2 p

3 bc

[ABC]

and

[M C

2 A

2 ] =

p

3 p

1 ca

[ABC]

, and

B

2 M C

2 A

and

C

2 M A

2 B

are onvex. Sin e

A

2 M B

2 C

,

B

2 M C

2 A

,and

C

2 M A

2 B

are onvex,

M

isoutsideof

△A

2 B

2 C

,

△B

2 C

2 A

,and

△C

2 A

2 B

, and sin e

M

is in the interior of

△ABC

,

M

must be in the interior of

△A

2 B

2 C

2

. Therefore,

(12)

andso

[A

2 B

2 C

2 ]

=

p

1 p

2 ab

+

p

2 p

3 bc

+

p

3 p

1 ca

[ABC]

=

p

1 p

2 p

3 abc

a

p

1 +

b

p

2 +

c

p

3 [ABC]

. Finally,

[A

2 B

2 C

2 ]

[A

1 B

1 C

1 ]

=

p

1 p

2 p

3 abc

a

p

1 +

b

p

2 +

c

p

3 [ABC]

2abcp

1 p

2 p

3 (ap

1 + bp

2 )(bp

2 + cp

3 )(cp

3 + ap

1 )

[ABC]

=

(ap

1 + bp

2 )(bp

2 + cp

3 )(cp

3 + ap

1 )

2a

2 _b

2 _c

2 a

p

1 +

b

p

2 +

c

p

3

.

AlsosolvedbyGEORGEAPOSTOLOPOULOS,Messolonghi,Gree e;ARKADYALT,San

Jose, CA,USA;MICHELBATAILLE,Rouen,Fran e; CHIPCURTIS,MissouriSouthern State

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

Ursulinengymnasium, Innsbru k,Austria; PETERY.WOO,BiolaUniversity,La Mirada,CA,

USA;andTITUZVONARU,Comane sti,Romania.Therewasonein orre tsolutionsubmitted.

BothBatailleandGeupel al ulatedwithbary entri oordinates.Geupelalsousedthe

formulafortheareaofthepedaltriangleof

M

,towit

[A

2 B

2 C

2 ] = |R

2 _−OM

2 _|/4R

2

,where

O

and

R

arethe ir um entreand ir umradiusoftriangle

ABC

,respe tively.

3372. [2008:364,366℄ ProposedbyVoQuo BaCan,CanThoUniversity

ofMedi ineandPharma y,CanTho,Vietnam.

If

x

,

y

,

z

≥ 0

and

xy

+ yz + zx = 1

,provethat (a)

1 p2x

2 _{+ 3yz}

+

1 p2y

2 _{+ 3zx}

+

1 p2z

2 _{+ 3xy}

≥

2 √

6

3

; (b)

⋆

1 px

2 _{+ yz}

+

1 py

2 _{+ zx}

+

1 pz

2 _{+ xy}

≥ 2

√

2

.

SolutionbyGeorgeApostolopoulos,Messolonghi,Gree e.

(a) Sin e

2x

2 _{≤ 3x}

2

wehave

2x

2 _{+ 3yz ≤ 3x}

2 _{+ 3yz}

,thus

1 p2x

2 _{+ 3yz}

≥

1 p3x

2 _{+ 3yz}

=

1 √

_3px

₂

+ yz

.

Similarinequalitiesholdfortheothertwotermsontheleftsideofthedesired

inequality,andwenowhave

1 p2x

2 _{+ 3yz}

+

1 p2y

2 _{+ 3zx}

+

1 p2z

2 _{+ 3xy}

≥

√

1

3

1 px

2 _{+ yz}

+

1 py

2 _{+ zx}

+

1 pz

2 _{+ xy}

!

.

(13)

Thedesiredinequalityfollowsnowfrom(b),whi hisprovenbelow.

(b)

⋆

Moregenerally,wewillprovethatif

x

,

y

,

z

≥ 0

and

xy+yz+zx > 0

, then

1 px

2 _{+ yz}

+

1 py

2 _{+ zx}

+

1 pz

2 _{+ xy}

≥

2 √

xy

+ yz + zx

. (1)

Sin e thisissymmetri in

x

,

y

,

z

,wemayassumethat

x

≥ y ≥ z

. Noti e that

1 py

2 _{+ zx}

+

1 pz

2 _{+ xy}

≥

2 √

2 py

2 _{+ z}

2 _{+ xy + zx}

. [Ed: we have

1 py

2 _{+ zx}

+

1 pz

2 _{+ xy}

≥ 2

r

₁

py

2 _{+ zx}

1 pz

2 _{+ xy}

from the

AM{GMInequality,andalso

1 py

2 _{+ zxpz}

2 _{+ xy}

≥

2 y

2 _{+ zx + z}

2 _{+ xy}

;the inequalityabovenowfollowsfromthesetwo.℄

SoitsuÆ estoprovethat

1 px

2 _{+ yz}

+

2 √

2 py

2 _{+ z}

2 _{+ xy + zx}

≥

2 √

xy

+ yz + zx

. Let

K

= xy + yz + zx

and

L

= y

2 _{+ z}

2 _{+ xy + zx}

. Then

2 √

K

−

2 √

L

=

2 √

L

−

√

K

√

KL

=

2 √

2 y

2 _{− yz + z}

2 √

KL

√

L

+

√

K

. Itis learthat

L

≥ K

,

L

≥ 2 y

2 _{− yz + z}

2

,and

K

≥ y

px

2 _{+ yz}

.

[Ed: Sin e

x

≥ y ≥ z

, we have

L

= y

2 _{+ z}

2 _{+ xy + zx ≥ y}

2 _{+ z}

2 + y

2 _{+ z}

2 _{≥ 2(y}

2 _{− yz + z}

2 ₎

and

K

= xy + yz + zx ≥ xy + yz + yz

= y

√

x

2 _{+ 4xz + 4z}

2 _{≥ y}

_px

2 _{+ 4yz + 4z}

2 _{≥ y}

_px

2 _{+ yz}

.℄ Thus,

2 √

2 y

2 _{− yz + z}

2 √

KL(

√

L

+

√

K)

≤

2 √

2 y

2 _{− yz + z}

2 √

KL(2

√

K)

=

√

2 y

2 _{− yz + z}

2 K

√

L

≤

√

2 y

2 _{− yz + z}

2 ypx

2 _{+ yz}

q

_{2 y}

2 _{− yz + z}

2 =

py

2 _{− yz + z}

2 ypx

2 _{+ yz}

≤

py

2 ypx

2 _{+ yz}

=

1 px

2 _{+ yz}

,