PROBLEMS

174
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, Department
of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,
Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,
together with references and other insights which are likely to be of help to the editor.
When a proposal is submitted without a solution, the proposer must include sucient
information on why a solution is likely. An asterisk (?) after a number indicates that
a problem was proposed without a solution.
In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can
be located, it should not be submitted without the originator's permission.
To facilitate their consideration, please send your proposals and solutions
on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be
typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief,
to arrive no later than 1 November 2002. They may also be sent by email to
[email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated
postscript. Solutions received after the above date will also be considered if there
is sucient time before the date of publication. Please note that we do not accept
submissions sent by FAX.
Each problem is given in English and French, the ocial languages of Canada.
In issues 1, 3, 5 and 7, English will precede French, and in issues 2, 4, 6 and
8, French will precede English. In the solutions section, the problem will be
given in the language of the primary featured solution.
2704. Proposed by Mihaly Bencze, Brasov, Romania. Correction.
Prove that 0
1
2 + r 2 + 4Rr
1 @ X p
s
A 0,
R ; 2r 2(b2 + c2 ) ; a2 ;
12
R
cyclic
where a, b and c are the sides of a triangle, and R, r and s are the circumradius, the inradius and the semi-perimeter of the triangle, respectively.
.................................................................
Montrer que
R ; 2r
1
12
0
1
2 + r 2 + 4Rr
X
p
s
@
A
2(b2 + c2 ) ; a2 ;
R
cyclique
0,
ou a, b et c sont les c^otes d'un triangle, et R, r et s sont respectivement
le rayon du cercle circonscrit, le rayon du cercle inscrit et le demi-perimetre
du triangle.
175
2724?. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.
Let a, b, c be the sides of a triangle and ha , hb , hc , respectively, the
corresponding altitudes. Prove that the maximum range of validity of the
inequality
p3 at + bt + ct 1=t
ht + ht + ht 1=t
a
b
c
,
3
2
3
ln 4
where t 6= 0 is ; ln 4 < t <
ln 4 ; ln 3
ln 4 ; ln 3 .
.................................................................
Soit a, b, c les c^otes d'un triangle et ha , hb , respectivement, les hauteurs
correspondantes. Montrer que le domaine de validite maximal de l'inegalite
p3 at + bt + ct 1=t
ht + ht + ht 1=t
a
b
c
,
3
2
3
ln 4 .
ou t 6= 0 est ; ln 4 < t <
ln 4 ; ln 3
ln 4 ; ln 3
2725. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck,
Austria.
n
P
For k 1, let Sk (n) = (2j ; 1)k be the sum of the kth powers of
j =1
the rst n odd numbers.
1. Show that the sequence fS3 (n), n 1g contains innitely many squares.
2.? Prove that this sequence contains only nitely many squares of other
exponents k.
.................................................................
n
P
Pour k 1, soit Sk (n) = (2j ; 1)k la somme des k-iemes puisj =1
sances des n premiers nombres impairs.
1. Montrer que la suite fS3 (n), n 1g contient une innite de carres.
2.? Montrer que cette suite ne contient qu'un nombre ni de carres d'autres
exposants k.
2726. Proposed by Armend Shabini, University of Prishtina, Prishtina, Kosovo, Serbia.
Given the nite sequence of real numbers, fak g, 1 k 2n, where
the terms satisfy
a2k ; a2k;1 = d , 1 k n , and a2k+1 = q , 1 k n ; 1 ,
prove that, when q 6= 1,
a2k
176
2n
X
ak = 2qa2n ; 2qa1;;1nd(1 + q) ,
k=1
n!
2
n;2
1
;
q
.
(b) a2n = a1 q n + d
(a)
and
1;q
.................................................................
On donne la suite nie de nombres reels fak g, 1 k 2n, satisfaisant
a2k ; a2k;1 = d , 1 k n ,
et
Montrer que si q 6= 1,
2n
X
(a)
ak = 2qa2n ; 2qa1;;1nd(1 + q) ,
k=1
n!
2
n;2
1
;
q
n
(b) a2n = a1 q + d
.
a2k+1 = q , 1 k n ; 1:
a2k
et
1;q
2727. Proposed by Armend Shabini, University of Prishtina, Prishtina, Kosovo, Serbia.
Given the nite sequence of real numbers, fak g, 1 k n, where the
terms satisfy
ak ; ak;1 = ak;1 ; ak;2 + d , k > 2 , d 2 R ,
n
nd a closed form expression for P ak .
k=1
nX
;1 2k + 2
Use this to nd the value of
.
k=0
2k
.................................................................
On donne la suite nie de nombres reels fak g, 1 k n, satisfaisant
ak ; ak;1 = ak;1 ; ak;2 + d , k > 2 , d 2 R:
Trouver une expression explicite pour
n
P
ak .
k=1
X 2k + 2
Utiliser le resultat pour trouver la valeur de
2k .
k=0
n;1
177
2728. Proposed by Eckard Specht, Otto-von-Guericke University,
Magdeburg, Germany.
The distance between two well-known points in 4ABC is
bc p
a + b + c 2(cos A + 1):
What are the points ?
.................................................................
La distance de deux points bien connus dans un triangle ABC est
bc p
a + b + c 2(cos A + 1):
Quels sont ces points ?
2729. Proposed by Vaclav Kone c ny, Ferris State University, Big Rapids, MI, USA.
Let Z (n) denote the number of trailing zeros of n!, where n 2 N.
(a) Prove that Z (n) < 1 .
n
4
1
Z (n)
(b) ? Prove or disprove that nlim
!1 n = 4 .
.................................................................
Soit Z (n) le nombre de zeros apparaissant en queue de n!, ou n 2 N.
(a) Montrer que Z (n) < 1 .
n
4
Z (n)
1
(b) ? Montrer si oui on non, nlim
!1 n = 4 .
USA.
2730. Proposed by Peter Y. Woo, Biola University, La Mirada, CA,
Let AM(x1 ; x2 ; : : : ; xn ) and GM(x1 ; x2 ; : : : ; xn ) denote the arithmetic mean and the geometric mean of the real numbers x1 , x2 , : : : , xn , respectively.
Given positive real numbers a1 , a2 , : : : , an , b1 , b2 , : : : , bn , prove that
(a) GM(a1 + b1 ; a2 + b2 ; : : : ; an + bn )
GM(a1; a2 ; : : : ; an) + GM(b1 ; b2 ; : : : ; bn ).
For each real number t 0, dene f (t) = GM(t + b1 ; t + b2 ; : : : ; t + bn ) ; t.
(b) Prove that f (t) is a monotonic increasing function of t, and that
lim f (t) = AM(b1 ; b2 ; : : : ; bn ) .
t!1
178
Soit AM(x1 ; x2 ; : : : ; xn ) et GM(x1 ; x2 ; : : : ; xn ) la moyenne arithmetique, respectivement la moyenne geometrique des nombres reels x1 , x2 ,
: : : , xn . Etant donne des nombres reels positifs a1, a2, : : : , an, b1 , b2 , : : : ,
bn , montrer que
(a) GM(a1 + b1 ; a2 + b2 ; : : : ; an + bn )
GM(a1; a2 ; : : : ; an) + GM(b1 ; b2 ; : : : ; bn ).
Pour tout nombre reel t 0, soit f (t) = GM(t + b1 ; t + b2 ; : : : ; t + bn ) ; t.
(b) Montrer que f (t) est une fonction monotone croissante de t, et que
lim f (t) = AM(b1 ; b2 ; : : : ; bn ) .
t!1
2731. Proposed by Juan-Bosco Romero Marquez, Universidad de
Valladolid, Valladolid, Spain.
Let C be a conic with foci F1 , F2 , and directrices D1 , D2 , respectively.
Given any point M on the conic, draw the line passing through M , perpendicular to the directrices, intersecting D1 , D2 , at M1 , M2 , respectively.
Let R be the point of intersection of the lines M1 F1 and M2 F2 . Prove that
(a) F1 R is independent of the choice of M ;
M1 R
(b) the normal to the conic at M passes through R.
.................................................................
Soit C une conique de foyers F1 , F2 et de directrices D1 , D2 , respectivement. Par un point donne quelconque M sur la conique, on dessine la
perpendiculaire aux directrices, et soit M1 , M2 les points d'intersection respectifs. Si R est le point d'intersection des droites M1 F1 et M2 F2 , montrer
que
(a) F1 R est independant du choix de M ;
M1 R
(b) la normale a la conique en M passe par R.
2732. Proposed by Mihaly Bencze, Brasov, Romania.
Let ABC be a triangle with sides a, b, c, medians ma , mb , mc , altitudes
ha , hb , hc, and area . Prove that
m m m p
2
2
2
a + b + c 4 3 max h a , h b , h c .
a
b
c
.................................................................
Soit ABC un triangle de c^otes a, b, c, de medianes ma , mb , mc , de
hauteurs ha , hb , hc , et d'aire . Montrer que
m m m p
2
2
2
a + b + c 4 3 max a , b , c .
ha hb hc
179
2733?. Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta.
It is a known result that if O is the circumcentre of 4A1A2 A3 , and if
O1 , O2 , O3, are the circumcentres of 4OA2A3 , 4OA3A1 , 4OA1A2 , respectively, then the lines A1 O1 , A2 O2 and A3 O3 are concurrent.
Does the corresponding result hold for simplexes ? That is, if O is the
circumcentre of a simplex A0 A1 : : : An and Ok is the circumcentre of the
simplex determined by O and the face opposite Ak , are the lines Ok Ak ,
k = 0, 1, : : : , n, concurrent ?
.................................................................
On conna^t le resultat armant que si O est le centre du cercle circonscrit du triangle A1 A2 A3 , et si respectivement O1 , O2 , O3 sont les centres
des cercles circonscrits aux triangles OA2 A3 , OA3 A1 et OA1 A2 , alors les
droites A1 O1 , A2 O2 et A3 O3 sont concourantes.
Le resultat ci-dessus reste-t-il valable pour des simplexes ? En d'autres
termes, si O est le centre de la sphere circonscrite au simplexe A0 A1 : : : An
et si Ok est le centre de la sphere circonscrite au simplexe determine par O
et la face opposee Ak , est-ce que les droites Ok Ak , k = 0, 1, : : : , n sont
concourantes ?
2734. Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta.
Prove that
(bc)2n+3 + (ca)2n+3 + (ab)2n+3 (abc)n+2 (an + bn + cn ) ,
where a, b, c, are non-negative reals, and n is a non-negative integer.
.................................................................
Montrer que
(bc)2n+3 + (ca)2n+3 + (ab)2n+3 (abc)n+2 (an + bn + cn ) ,
ou a, b, c sont des reels non negatifs et n un entier non negatif.
2735?. Proposed by Richard I. Hess, Rancho Palos Verdes, CA,
USA.
Given three Pythagorean triangles with the same hypotenuse, is it possible that the area of one triangle is equal to the sum of the areas of the other
two triangles ?
.................................................................
Etant donne trois triangles pythagoriciens ayant la m^eme hypotenuse,
est-il possible que l'aire de l'un des triangles soit e gale a la somme des aires
des deux autres ?
180
2736. Proposed by Juan-Bosco Romero Marquez, Universidad de
Valladolid, Valladolid, Spain.
Let ABCD be a convex quadrilateral. From points A and B , draw lines
parallel to sides BC and AD, respectively, giving points G and F on CD,
respectively.
Let P and Q be the points of the intersection of the diagonals of the
trapezoids ABFD and ABCG, respectively.
Prove that PQ k CD.
.................................................................
Soit ABCD un quadrilatere convexe. Des sommets A et B , on trace
des paralleles aux c^otes BC et AD, determinant respectivement des points
G et F sur CD.
Soit P et Q les points d'intersection des diagonales des trapezes ABFD
et ABCG, respectivement.
Montrer que PQ et CD sont paralleles.
2737. Proposed by Lyubomir Lyubenov, teacher, and Ivan Slavov,
student, Foreign Language High School \Romain Rolland", Stara Zagora, Bulgaria.
Find all solutions of the equation
xn ; 2nxn;1 + 2n(n ; 1)xn;2 + axn;3 + bxn;4 + + c = 0 ,
given that there are n real roots.
.................................................................
Trouver toutes les solutions de l'equation
n
x ; 2nxn;1 + 2n(n ; 1)xn;2 + axn;3 + bxn;4 + + c = 0 ,
sachant qu'il y a n racines reelles.
2738. Proposed by Sefket
Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina.
Let x, y and z be positive real numbers satisfying x2 + y2 + z2 = 1.
Prove that
x + y + z 3p3 .
1 ; x2 1 ; y2 1 ; z2
2
.................................................................
Soit x, y et z des nombres reels positifs satisfaisant x2 + y2 + z2 = 1.
Montrer que
x + y + z 3p3 .
1 ; x2 1 ; y2 1 ; z2
2