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
,andc
bepositiverealnumbers. Provethata
2a
2
+ bc
+
b
2b
2
+ ca
+
c
2c
2
+ ab
≤
Êa
−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
)
ands = (s
1
, s
2
, . . . , s
n
)
bearbitraryn
-tuples ofnonnegativerealnumbers. Wewillwritet ≻ s
if(i)
t
1
≥ · · · ≥ t
n
ands
1
≥ · · · ≥ s
n
, (ii)k
Pi=1
t
i
≥
k
Pi=1
s
i
forallk = 1
,2
,. . .
,n
,withequalitywhenk = n
.Let
R
+
denotethesetofpositiverealnumbers,letP
bethesetofall permutationsof{1, 2
,. . .
,n}
,anddene[t] : R
n
+
→ R
by[t](x) =
Xσ∈P
x
t
1
σ(1)
x
t
2
σ(2)
· · · x
t
n
σ(n)
forallx = (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)
forallx ∈ R
n
+
. Now,bysquaring andsimplifying,thegiveninequalityisequivalenttoA ≥ B
,whereA
= 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]
,[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. Provethat1
2
n−1
n
Xk=0
√
k
2n
k
≤
Ên
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
for0 ≤ k ≤ 2n
andk
2n
k
= 2n
2n−1
k−1
for
1 ≤ k ≤ 2n
,andalsotheCau hy{S hwarzInequality, wehavethat "n
Xk=0
√
k
2n
k
#2
=
"n
Xk=0
Ê2n
k
√
k
Ê2n
k
#2
≤
"n
Xk=0
2n
k
#"n
Xk=0
k
2n
k
#=
1
2
"2n
n
+
2n
Xk=0
2n
k
#"2n
n
Xk=1
2n − 1
k − 1
#=
1
2
2n
n
+ 2
2n
"2n
n−1
Xk=0
2n − 1
k
#=
1
2
2n
n
+ 2
2n
"2n ·
1
2
2n−1
Xk=0
2n − 1
k
#=
1
2
2
2n
+
2n
n
(n · 2
2n−1
)
=
n
2
2n
+
2n
n
· 2
2n−2
,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
,andc
benonnegativerealnumberssu hthata + 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, modiedby the
editor.
(a) Fornonnegativeintegers
ℓ
,m
,andn
,let[ℓ, m, n] =
Psymm.
a
ℓ
b
m
c
n
.
Thefollowinginequalityisa onsequen eofMuirhead'sTheorem,
27
Y y li 9(a + b + c)
2
− 32ab
−
11(a + b + c)
2
+ 16(ab + bc + ca)
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
2
≥ 49
,fromwhi htheinequalityin(a)follows.
(b) Let
x = 3a
,y = 3b
,andz = 3c
. Thenx
,y
,andz
arenonnegative realnumberssu hthatx + y + z = 3
,andwearetoshowthatX
p
Noterstthat 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
=
3
8
(z − 1)
2
. (2) Hen e,(3 + z)
2
8
≥ 3 −
(x + y)
2
4
, (3) and(1)isequivalentto X y li p3 − xy −
Ê3 −
(x + y)
2
4
!≥
X y li Ê(3 + z)
2
8
−
Ê3 −
(x + y)
2
4
! . (4)Let
H
andK
denotetheleftandrightsideof(4),respe tively. ThenH =
1
4
X y li(x − y)
2
√
3 − xy +
È3 −
(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
+
È3 −
(x+y)
4
2
=
3
8
X y li(z − 1)
2
È(3+z)
2
8
+
È3 −
(x+y)
4
2
≤
3
8
X y li(z − 1)
2
2
È3 −
(x+y)
4
2
≤
3
8
X y li(z − 1)
2
2
È3 −
9
4
=
√
3
8
X y li(z − 1)
2
. (6) Finally, X y li(x − y)
2
= 3
X y lix
2
!−
X y lix
!2
= 3
X y lix
2
!− 9
= 3
X y lix
2
!− 6
X y lix
!+ 9 = 3
X y li(z − 1)
2
! . (7)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 urswhena = b = 1/2
,c = 0
.3491. [2009:515,518℄ProposedbyDorinMarghidanu, ColegiulNational
\A.I.Cuza",Corabia,Romania.
Let
a
1
,a
2
,. . .
,a
n+1
bepositiverealnumberswherea
n+1
= a
1
. Prove thatn
Xi=1
a
4
i
(a
i
+ a
i+1
)(a
2
i
+ a
2
i+1
)
≥
1
4
n
Xi=1
a
i
.SolutionbyGeorgeApostolopoulos,Messolonghi,Gree e.
Let
A
=
n
Xi=1
a
4
i
(a
i
+ a
i+1
)(a
2
i
+ a
2
i+1
)
,
B
=
n
Xi=1
a
4
i+1
(a
i
+ a
i+1
)(a
2
i
+ a
2
i+1
)
. Then
A − B =
n
Xi=1
a
4
i
− a
4
i+1
(a
i
+ a
i+1
)(a
2
i
+ a
2
i+1
)
=
n
Xi=1
a
i
− a
i+1
= 0
, andhen eA = B
.Wenowshowthatforallpositiverealnumbers
a
andb
wehavea
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
Xi=1
a
4
i
+ a
4
i+1
(a
i
+ a
i+1
)(a
2
i
+ a
2
i+1
)
≥
1
4
n
Xi=1
(a
i
+ a
i+1
) =
1
2
n
Xi=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,
3492
⋆
. [2009 : 515, 518℄ Proposed by Ovidiu Furdui, Campia Turzii, Cluj,Romania.Let
P
beapointintheinterioroftetrahedronABCD
su hthatea h of∠P AB
,∠P BC
,∠P CD
,and∠P DA
isequal toarccos
È
2
3
. ProvethatABCD
isaregulartetrahedronandthatP
isits entroid.Theproblemremainsopen. Theonlysubmission,fromPeter Y.Woo,
BiolaUniversity,LaMirada,CA,USA,gavea ounterexamplewhere
ABCD
isadegeneratetetrahedron. Inparti ular,heprovidedanelegantproofthatif
P
is the entreofaparallelogramABCD
withsidesAD = BC = 3
√
2
andAB = CD =
√
6
,anddiagonalsAC = 2
√
3
andAC = 6
,then∠P AB = ∠P BC = ∠P CD = ∠P DA = arccos
r2
3
.This ertainly addresses the question that was asked, and itsuggests that
there are innitely many tetrahedra with an interiorpoint
P
that satises thegivenanglerequirement,butitfailstoprovideanexpli itnondegenerateexample.
3494. [2009:516,518℄ ProposedbyMi helBataille,Rouen,Fran e.
Let
n > 1
beanintegerandforea hk = 1
,2
,. . .
,n
letσ(n, k) =
X1≤i
1
<···<i
k
≤n
i
1
i
2
· · · i
k
. Provethatn
Xk=1
ln n
n + 1 − k
· σ(n, k) ∼ (n + 1)! ∼
n
Xk=1
n + 1 − k
ln n
· σ(n, k)
,where
f (n) ∼ g(n)
meansthatf (n)
g(n)
→ 1
asn → ∞
.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)
. IfU
n
denotesn
Pk=1
σ(n, k)
n + 1 − k
,thenU
n
=
Z1
0
P
n
(x) dx
−
1
n + 1
.Clearly,
P
′
n
(x)
P
n
(x)
=
1
x + 1
+
1
x + 2
+ · · · +
1
x + n
,sothatforallx ∈ [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
, whereH
n
= 1 +
1
2
+ · · · +
1
n
denotesthen
thharmoni number. Sin ewe
alsohave
P
n
(1) − P
n
(0) = (n + 1)! − n! =
n
n + 1
· (n + 1)!
,weobtainn
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 allingthatH
n
∼ ln(n)
,theSqueezeTheorem forlimitsyieldslim
n→∞
U
n
ln(n)
(n + 1)!
= 1
,thatis,n
Xk=1
ln(n)
n + 1 − k
· σ(n, k) ∼ (n + 1)!
. LetV
n
=
n
Pk=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,forn > 1
,V
n
=
n
Xk=1
(n − k)σ(n, k) +
n
Xk=1
σ(n, k)
=
P
n
′
(1) − n + (n + 1)! − 1
=
P
n
′
(1) + (n + 1)! − (n + 1)
, sothatH
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
3495. [2009:516, 518℄ Proposedby CosminPohoata, TudorVianu
Na-tionalCollege,Bu harest,Romania.
Let
a
,b
,c
bepositiverealnumberswitha + b + c = 2
. Provethat1
2
+
X y lia
b + c
≤
X y li a
2
+ bc
b + c
≤
1
2
+
X y lia
2
b
2
+ c
2
.A ombinationofsolutionsbyGeorgeApostolopoulos,Messolonghi,Gree e
and Paolo Perfetti, Dipartimento diMatemati a, Universita degli studi di
TorVergataRoma,Rome,Italy,modiedbytheeditor.
Forve tors
a = (a
1
, a
2
, . . . , a
n
)
and(b
1
, b
2
, . . . , b
n
)
withrealentries, thenotationa ≺ b
meansthata
1
+ a
2
+ · · · + a
n
= b
1
+ b
2
+ · · · + b
n
anda
1
+ a
2
+ · · · + a
j
≤ b
1
+ b
2
+ · · · + b
j
holdsforea hj = 1
,2
,. . .
,n − 1
. Sin ea + b + c = 2
,theinequalityontheleftisequivalentto
1
2
+
X y lia
b + c
a + b + c
2
≤
X y lia
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 lia
2
b
2
+ c
2
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)
.UsingMuirhead'sinequalityrepeatedlyweobtain:
(6, 4, 0) ≺ (9, 1, 0) =⇒
X symmetri2a
9
b ≥
X symmetri2a
6
b
4
(6, 2, 2) ≺ (7, 2, 1) =⇒
X symmetri2a
7
b
2
c ≥
X symmetri2a
6
b
2
c
2
(6, 3, 1) ≺ (8, 1, 1) =⇒
X symmetri2a
8
bc ≥
X symmetri2a
6
b
3
c
(5, 4, 1) ≺ (8, 1, 1) =⇒
X symmetri2a
8
bc ≥
X symmetri2a
5
b
4
c
(5, 5, 0) ≺ (7, 3, 0) =⇒
X symmetria
7
b
3
≥
X symmetria
5
b
5
(5, 3, 2) ≺ (7, 2, 1) =⇒
X symmetria
7
b
2
c ≥
X symmetria
5
b
3
c
2
(4, 3, 3) ≺ (7, 2, 1) =⇒
X symmetria
7
b
2
c ≥
X symmetria
4
b
3
c
3
(5, 4, 1) ≺ (7, 2, 1) =⇒
X symmetria
7
b
2
c ≥
X symmetria
5
b
4
c
Also,bytheAM-GMInequality,wehave
X symmetri
(2a
6
b
2
c
2
+ 2a
4
b
4
c
2
) ≥
X symmetri4a
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
CompositeofsolutionsbyKee-WaiLau,HongKong,ChinaandD.J.Smeenk,
Zaltbommel,theNetherlands.
With thehelp ofappropriatetrigonometri identities, bothequations
anberedu edtopropertiesofthegoldense tion
τ =
1 +
√
5
2
,whi histhe positiverootofthequadrati equationτ
2
= τ + 1
. (1)Be ause
τ
istheratioofadiagonaltoasideofaregularpentagon,itsatisescos 36
◦
=
τ
2
andcos 72
◦
=
1
2τ
. (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
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 onrmthatthereareseven
4
-tuples(a, b, c, d)
ofdistin tintegersbetween
0
and90
(otherthanthepairfeaturedinourproblem)thatsatisfytherelationtan 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)
.Therstfourare learlyrelatedtothegoldense tionasinourfeaturedpair,whilethenal
threeseemtoberelatedtotheregularenneagon(ornonagon,ifyouprefer)asdis ussedin
\TrigonometryandtheNonagon"byAndrewJobbings(seewww.arbelos. o.uk/papers.html).
Itisamusingtonotethattheproposerthoughtthathehadfoundonethatfailstoteitherof
thetwopatterns,butitturnsoutthat
tan 62
◦
diersfrom
tan 48
◦
+ tan 24
◦
+ tan 18
◦
by
about
10
−5
.Wagonfurtherprodu edalistof
49
su hequationsallowingrepeatedangles,anddeterminedthattherewerenosu h
3
-termequationsandnosu h5
-termequations.3497. [2009 :516, 519℄ Proposed by Salem Maliki , student, Sarajevo
College, Sarajevo,BosniaandHerzegovina.
Let
P
be a point in the interior of triangleABC
, and letr
be the inradiusofABC
. Provethatmax{AP
,BP
,CP } ≥ 2r
.I.SolutionbyRoyBarbara,LebaneseUniversity,Fanar,Lebanon.
Re allthatthe onvexhullofatriangle
T
istheunionofitsinteriorand boundary.IfC
isa ir lewithradiusr
inthe onvexhullofatriangleT
1
with inradiusr
1
,thenr ≤ r
1
. (Hereisaproofofthissimplefa t: Considerthe threetangentstoC
thatareparalleltothesidesofT
1
andseparatethe entre ofC
fromthe orrespondingsides;theyformatrianglethatissimilartoT
1
forwhi hC
isthein ir le.Sin eallpointsofC
areinsideoronT
1
,theratioof thesidesofthenewtriangletothesidesofT
1
|whi hisalsotheratioofthe inradii| ouldbeatmost1
;thatis,r ≤ r
1
.)LetT = △ABC
beanarbitrary triangle with in ir leC
and inradiusr
, and letP
bea point in the onvex hull ofT
. Withoutloss ofgenerality, we mayassume thatmax{AP
,BP
,CP } = AP
andshow thatAP ≥ 2r
. Extend (ifne essary) thesegmentsP B
toP B
1
andP C
toP C
1
su h thatP B
1
= P C
1
= P A
. ThenP
is the ir um entreoftriangleT
1
= △AB
1
C
1
,andP A
its ir umradius;letr
1
denoteitsinradius. Notethatbe auseP
isassumedtolieinthe onvexhull ofT
,T
mustlieinthe onvexhullofT
1
; onsequentlythein ir leofT
also liesinthat onvexhull,sothat(fromoursimplefa t)r
1
≥ r
.II.SolutionbyMi helBataille,Rouen,Fran e.
Generalization: Thefollowingresultholdsforanypoint
P
intheplane of△ABC
. LetR
,O
,a
,b
, andc
be the ir umradius, ir um entre, and sidesof△ABC
,andletM = max{AP
,BP
,CP }
;then(a) if
△ABC
is a ute,M ≥ R ≥ 2r
,withM = 2r
ifandonlyifP = O
andthetriangleisequilateral;(b) if
△ABC
isnota ute,M ≥
max{a, b, c}
2
≥ 2r
, withM = 2r
ifand onlyifP
isthemidpointofthelongestside.Let
A
′
,B
′
, andC
′
bethemidpointsofthesidesoppositeverti es
A
,B
,andC
,respe tively. Forpart(a)wexpointsD
,E
,andF
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 thatP
is in or on thesides of angle∠EOF
(boundedbytherays[OE)
and[OF )
)sothatM = P A
. LetE
0
onAB
andF
0
onAC
besu hthatOE
0
||AC
andOF
0
||AB
. Notethatbe auseO
isintheinteriorof△ABC
,E
0
andF
0
belongtotherays[AB)
and[AC)
, whileOE
0
⊥ OE
andOF
0
⊥ OF
. Sin eA
isintheinteriorof∠E
0
OF
0
,the angle∠P OA
isobtuse,hen eM = P A ≥ OA = R
,withequalityexa tly whenP = O
. The inequalityR ≥ 2r
is Euler's inequality, withR = 2r
exa tlywhen△ABC
isequilateral,sotheproofofpart(a)is omplete.For part (b) we rstsuppose that
∠BAC
, say, is obtuse. ThenO
is exteriorto△ABC
withlineBC
separatingO
fromA
,andtheplaneisthe union ofthe three angles∠EOF
,∠EOA
′
, and∠F OA
′
. IfP
is in∠EOF
thenM = P A ≥ R >
a
2
(mu hasin part(a)). Otherwise, withoutlossof generality,we ansupposethatP
is in∠EOA
′
,in whi h ase
M = P C ≥
A
′
C =
a
2
. To he kthattheminimumvalueofM
,namelya
2
,o urswhenP = A
′
, note thatA
andA
′
are on the same side of the perpendi ular
bise torofthesegment
AC
,sothatA
′
A < A
′
C
;thatis, ifP = A
′
,thenM = A
′
C = A
′
B =
a
2
. If∠BAC = 90
◦
, this argument an easily be
adaptedtoshowthat
M ≥
a
2
= R
. To ompletetheproofweshowthatin thepresent asewehavea
2
≥ 2r
. Leth = AH
bethealtitudefromA
,and letA
0
bethepointontheray[HA)
su hthat∠BA
0
C = 90
◦
. Wewantto
showthat
ah ≥ 4rh
;thatis,thata + b + c ≥ 4h
(sin eah
2
=
r(a + b + c)
2
=
area
(△ABC)
). Buth ≤ HA
0
=
√
HB · HC ≤
HB + HC
2
=
a
2
,when e
a ≥ 2h
;moreover,b
,c ≥ h
,sothata + b + c ≥ 4h
,asdesired. Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e (2solutions);
SEFKETARSLANAGI
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: ForapointP
intheinteriorofthesimplex
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-NoordhoPubl.,Groningen,1969.3498. [2009:517, 519℄ ProposedbyJose LuisDaz-Barrero,Universitat
Polite ni a deCatalunya,Bar elona,Spain.
Let
F
n
be then
thFibona i number, that is,
F
0
= 0
,F
1
= 1
, andF
n
= F
n−1
+ F
n−2
forn ≥ 2
. Forea hpositiveintegern
,provethat ÊF
n+3
F
n
+
ÊF
n
+ F
n+2
F
n+1
> 1 + 2
ÊF
n
F
n+3
+
ÊF
n+1
F
n
+ F
n+2
.Solution by Chip Curtis, Missouri Southern State University, Joplin, MO,
USA. Let
x =
ÉF
n+3
F
n
andy =
ÉF
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) Setλ =
F
n+1
F
n
. Thenxy
=
ÊF
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
.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
sF
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 hisequivalentto32λ
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
oorsnumbered1
ton
andanumberofelevatorsall ofwhi hstopatboth oors1
andn
,andpossiblyother oors. Forea hn
, ndtheleastnumberofelevatorsneededinthisbuildingifbetweenanytwooorsthereisatleastoneelevatorthat 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
.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 ardinalityofP
is
n
2
4
,soatleastthismanyelevatorsareneeded.
Toshowthat
n
2
4
elevatorssuÆ e,wegivea onstru tionintwo ases.
Case 1:
n = 2k
. Herek
2
elevators are needed. Let integers
i
andj
be restri tedsothat1 ≤ i ≤ k
andk + 1 ≤ j ≤ 2k
,anddes ribeea helevator bythetupleof oorsitstopsat. Theelevatorsarethen8 < :
(1, i, j, 2k)
, ifi + j = 2k + 1
,(1, 2k + 1 − j, i, j, 2k)
, ifi + j > 2k + 1
,(1, i, j, 2k + 1 − j, 2k)
, ifi + j < 2k + 1
. Case2:n = 2k + 1
. Herek
2
+ k
elevatorsareneeded. Letintegers
i
andj
berestri tedsothat1 ≤ i ≤ k
andk + 1 ≤ j ≤ 2k + 1
,anddes ribeea h elevatorbythetupleof oorsitstopsat. Theelevatorsarethen8 < :
(1, i, j, 2k + 1)
, ifi + j = 2k + 2
,(1, 2k + 2 − j, i, j, 2k + 1)
, ifi + j > 2k + 2
,(1, i, j, 2k + 2 − j, 2k + 1)
, ifi + 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,Springeld,MO,USA;MORTEN
H.NIELSEN,UniversityofWinnipeg,Winnipeg,MB;andPETERY.WOO,BiolaUniversity,La