51
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.
2299. [1997: 503] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.
Let x, y , z > 0 be real numbers such that x + y + z = 1. Show that
Y
cyclic
(1 , y )(1 , z ) (1,y)(1,z)=x 256 .
x
81
Determine the cases of equality.
Composite solution by Theodore Chronis, Athens, Greece and Kee-Wai
Lau, Hong Kong.
The given inequality is clearly equivalent to
A=
xy 4 ln 4 .
1 + xy
ln
1
+
z
z
3
X cyclic
(1)
Since the function x ln(x) is convex on (0; 1), we have, by Jensen's
Inequality, that
A 1 3 + xy + yz + zx ln 1 3 + xy + yz + zx . (2)
3
3
z x y
3
z x y
Since (xy )2 +(yz )2 +(zx)2 (xy )(yz )+(yz )(zx)+(zx)(xy ) = xyz ,
we get
xy + yz + zx 1 .
z x y
(3)
From (2) and (3), we obtain (1) immediately.
It is clear that equality holds if and only if x = y = z = 13 .
Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY
S. KLAMKIN, University of Alberta, Edmonton, Alberta; MICHAEL LAMBROU, University of
Crete, Crete, Greece; VEDULA N. MURTY, Visakhapatnam, India; HEINZ-JURGEN
SEIFFERT,
Berlin, Germany; and the proposer.
2300. [1997: 503] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK.
Suppose that ABC is a triangle with circumradius R. The circle passing
through A and touching BC at its mid-point has radius R1 . Dene R2 and
R3 similarly. Prove that
27 R2 .
R21 + R22 + R23 16
52
Solution by G. Tsintsifas, Thessaloniki, Greece.
We use mi for the medians and hi for the altitudes. Let M be the midpoint of BC and O1 the centre of the circle through A touching BC at M .
From triangle O1 MA we have: 12 m1 = 12 AM = R1 cos ! = R1 mh11 , where
! = O1MA or 12 m21 = R1h1. Similarly, 21 m22 = R2h2; 12 m23 = R3h3.
Therefore,
\
3 (a2 + b2 + c2 ) = m21 + m22 + m23 = R h + R h + R h .
1 1
2 2
3 3
8
2
(1)
From the Cauchy-Schwarz inequality, we have
R1h1 + R2h2 + R3h3 ,R21 + R22 + R23 ,h21 + h22 + h231=2 . (2)
But, it is well known that:
2 2
2 2
2 2
h21 + h22 + h23 = a b + b4Rc2 + c a .
(3)
9R2 ,a2 + b2 + c22 R2 + R2 + R2 .
1
2
3
16 a2b2 + b2c2 + c2 a2
(4)
Using (1), (2), (3) we obtain
Obviously we have:
,
2
2
2 2
3 2 ba2 ++bb2c2++c c2 a2 .
So, from (4), (5), it follows that 27
R2 R21 + R22 + R23.
16
(5)
Also solved by NIELS BEJLEGAARD, Stavanger, Norway; THEODORE
CHRONIS, Athens, Greece; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG,
student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
Y,
Ferris
MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV
KONECN
State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,
Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England;
D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
2301. [1998: 45] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK.
Suppose that ABC is a triangle with sides a, b, c, that P is a point
in the interior of 4ABC , and that AP meets the circle BPC again at A0 .
Dene B 0 and C 0 similarly.
Prove that the perimeter P of the hexagon AB 0 CA0 BC 0 satises
P 2 pab + pbc + pca .
Solution by Florian Herzig, student, Cambridge, UK.
53
Let x = \BPC , y = \CPA, and z = \APB .
Then, as A0 CPB is cyclic, the angles in 4A0 CB are , x, , z , , y
and so by the Sine Law:
sin z and A0 C = a sin y .
A0B = asin
x
sin x
Thus the perimeter of 4A0 CB equals
a(sin x + sin y + sin z) .
sin x
Analogously we obtain the perimeters of triangles B 0 AC and C 0 BA. Summing these yields
(sin x + sin y + sin z ) sina x + sinb y + sinc z .
By the Cauchy-Schwarz inequality,
2
p
P + a + b + c pa + b + pc ;
or equivalently,
Equality holds if
P 2
pbc + pca + pab .
a
b = c .
=
sin x
sin2 y
sin2 z
2
Also solved by MANSUR BOASE, student, Cambridge, England; GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; JOEL
SCHLOSBERG, student, Bayside, NY, USA; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK,
Zaltbommel, the Netherlands; G. TSINTSIFAS, Thessaloniki, Greece; and the proposer.
Most of the submitted solutions are similar to the one given above.
2302. [1998: 45] Proposed by Toshio Seimiya, Kawasaki, Japan.
Suppose that the bisector of angle A of triangle ABC intersects BC
at D. Suppose that AB + AD = CD and AC + AD = BC .
Determine the angles B and C .
I. Solution by Vaclav Konecny, Ferris State University, Big Rapids,
Michigan, USA (somewhat modied by the editor).
54
C
C
E
X
B
B
C
B0
C
D
B
A
B
B
The gure shows the circle with centre A and radius AB intersecting
CB at X , CA at E and AB (extended) again at B0 . Subtract the rst of
the given equalities from the second to get AC , AB = BC , CD, which,
because AB = AE , implies that CE = BD (as the gure shows). Using the
symmetry of AB and AE about the angle bisector AD, we see that BD also
equals ED, and that B equals the angle at E (\AED). This latter angle is
an exterior angle of the isosceles triangle EDC , so that B = \AED = 2C .
The exterior angle at A of triangle ABC (namely \B 0 AE ) is therefore 3C ,
while \B 0 AX = 4C (since the angle at the centre is twice the angle at the
circumference). This implies that
\XAC = C ,
(1)
so that triangle XAC is isosceles, with XC equal to the radius AX . The
gure now recalls Archimedes' trisection of \B 0 AE by compass and marked
ruler (as in David R. Davis, Modern College Geometry, Addison-Wesley,
1958, section 10{8): using CX = AB , we see that the given condition
AB + AD = CD yields XD = AD, so that triangle DXA is isosceles,
and (since \DXA is an exterior angle of triangle AXC ) \DAX = 2C .
This, together with (1) says that half of angle A, namely \DAC , equals 3C ,
so that A = 6C . Thus 180 = A + B + C = 6C + 2C + C = 9C . We
conclude that A = 120, B = 40 and C = 20 .
II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.
Let A, B , C be the angles and a, b, c be the sides of 4ABC . Then,
from c + AD = CD and b + AD = a, we get a , CD = b , c, so that
BD = b , c and AD = a , b.
If E is the point between A and C with AE = AB , then
EC = AC , AE = b , c = BD = DE, so that 4DEC is isosceles,
and B = \DEA = 2C .
[Editor's comment. Note that the argument up to here has repeated the
rst step of solution I, avoiding reference to the gure. Several other solvers
derived B = 2C by rst showing that ca + c2 = b2 , noting that these
equivalent conditions are familiar to CRUX with MAYHEM readers, having appeared in [1976: 74], [1984: 278], [1996: 265], etc.] [It follows that
55
sin A = sin(C +B ) = sin(3C ), and sin , A2 = sin 2 , (B+2 C) = cos , 32C .]
The Sine Law applied to 4ABD, and then to 4ABC , gives
AD = BD =) a , b = b , c
sin B sin(A=2)
sin B sin(A=2)
sin
A , sin B = sin B , sin C
=)
sin B
sin(A=2)
) , sin(2C ) = sin(2C ) , sin C
=) sin(3Csin(2
C)
cos(3C=2)
2)cos(5C=2) = 2 sin(C=2)cos(3C=2)
=) 2 sin(C=
sin(2C )
cos(3C=2)
5
C
=) cos(5C=2) = sin(2C ) =) + 2C = 90 .
Thus, C = 20 , B = 40 and A = 120.
2
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia; DAVID DOSTER, Choate Rosemary Hall, Wallingford,
Connecticut, USA; IAN JUNE L. GARCES, Ateneo de Manila University, The Philippines and
GIOVANNI MAZZARELLO, Ferrovie dello Stato, Florence, Italy; FLORIAN HERZIG, student,
Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,
London, England; ECKARD SPECHT, Otto-von-Guericke-Universitat, Magdeburg, Germany;
D.J. SMEENK, Zaltbommel, the Netherlands; PARAYIOU THEOKLITOS, Limassol, Cyprus;
PANOS E. TSAOUSSOGLOU, Athens, Greece; JOHN VLACHAKIS, Athens, Greece; and the proposer. There was one incomplete solution.
2303. [1998: 45] Proposed by Toshio Seimiya, Kawasaki, Japan.
Suppose that ABC is a triangle with angles B and C satisfying
C = 90 + 12 B, that the exterior bisector of angle A intersects BC at D, and
that the side AB touches the incircle of 4ABC at E .
Prove that CD = 2AE .
Solution by Nikolaos Dergiades, Thessaloniki, Greece.
We know that 2AE = b + c , a (see, for example, 290, Roger A.
Johnson, Modern Geometry, Houghton-Miin, 1929). In the extension of
BA we take AF = b = AC .
Since AD is the external bisector of the angle A of 4ABC , F is the
symmetric point of C with axis of symmetry AD. Then
\B
\AFD = \ACD = 180 , \C = 90 ,
,
2
so 4BFD is isosceles; that is, BD = BF or
a + CD = b + c or CD = b + c , a = 2AE .
56
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain (2 solutions);
SAM BAETHGE, Nordheim, Texas, USA; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANSUR BOASE, student, Cambridge, England; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia;
DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG,
student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
Y,
Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU,
VACLAV
KONECN
University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;
JOEL SCHLOSBERG, student, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; ECKARD SPECHT, Otto-von-Guericke Universitat, Magdeburg, Germany; PARAYIOU
THEOKLITOS, Limassol, Cyprus; JOHN VLACHAKIS, Athens, Greece; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer.
2304. [1998: 46] Proposed by Toshio Seimiya, Kawasaki, Japan.
An acute angled triangle ABC is given, and equilateral triangles ABD
and ACE are drawn outwardly on the sides AB and AC . Suppose that CD
and BE meet AB and AC at F and G respectively, and that CD and BE
intersect at P .
Suppose that the area of the quadrilateral AFPG is equal to the area
of the triangle PBC . Determine angle BAC .
Solution by Michael Lambrou, University of Crete, Crete, Greece.
If [AFPG] = [PBC ], then [ABG] = [BFC ] and so 21 AGc sin A =
1
2 BFa sin B ; hence, by the Sine Rule,
AG = a sin B = b sin A = b .
BF
c sin A c sin A c
AF = c , so that AG = BF .
Similarly,
CG b
CG AF
Writing \AEG = ', \BDF = ! , we have from the equilateral triangles
sin ' = sin \AEG = AG = BF = sin \BDF = sin ! .
sin(60 , ') sin \GEC CG AF sin \FDA sin(60 , !)
!
!
p3
p3
1
1
Hence, sin '
2 cos ! , 2 sin ! = 2 cos ' , 2 sin ' sin ! and
sin(' , !) = 0. This shows that ' = ! (as both are acute).
Recall that P (Fermat or Napoleon point) satises \FPG = 120 .
Now summing the angles of quadrilateral DAEP , we have
,
360 = \ADP + \DPE + \PEA + 60 + \BAC + 60
= 60 , ! + 120 + ' + 120 + \BAC ,
and hence \BAC = 60 .
57
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR I A
LOPEZ
ASCENSION
CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER
J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Varazdin,
Varazdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student,
Cambridge, UK; GERRY LEVERSHA, St. Paul's School, London, England; D.J. SMEENK, Zaltbommel, the Netherlands; ECKARD SPECHT, Otto-von-Guericke Universitat, Magdeburg, Germany; PARAYIOU THEOKLITOS, Limassol, Cyprus; and the proposer.
2305. [1998: 46] Proposed by Richard I. Hess, Rancho Palos Verdes,
California, USA.
An integer-sided triangle has angles p and q, where p and q are
relatively prime integers. Prove that cos is rational.
Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria
(slightly modied by the editor).
Let Q denote the set of all rationals. By the Cosine Law, we see that
cos(p) 2 Q and cos(q) 2 Q.
Furthermore, since cos((p + q )) = , cos( , (p + q )), we have as
well that cos((p + q )) 2 Q.
From cos(p) cos(q) = 21 (cos((p + q )) + cos((p , q ))), we infer
that cos((p , q )) 2 Q.
Similarly, from cos(p) cos((p q )) = 12 (cos(2p q ) + cos(q)),
we infer that cos((2p q )) 2 Q.
Analogously, we have cos((p 2q )) 2 Q.
A simple induction then yields that cos((lp + kq )) 2 Q for all integers
l and k. The result now follows, since there exist integers l and k such that
lp + kq = 1.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE
CHRONIS, Athens, Greece; DIANE and RAY DOWLING, University of Manitoba, Winnipeg,
Manitoba; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG,
student, Cambridge, UK; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta;
MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY
LEVERSHA, St. Paul's School, London, England; PAVLOS MARAGOUDAKIS, Pireas, Greece;
JOEL SCHLOSBERG, student, Bayside, NY, USA; and the proposer.
Besides Janous, Godin also obtained the stronger result. Klamkin showed that
cos((lp , kq ) ) 2 Q for all non-negative integers l and k. Their arguments, which are very
similar, all used induction. The Dowlings gave a proof based on de Moivre's Theorem and the
Binomial Theorem. Bradley gave the following example to show that such triangles do exist:
a = 135, b = 5144, c = 31. In this
,case,
the angles of the triangle are 2, 3 and , 5,
where cos = 6 . (Thus, = cos,1 56 0:5857 radians, or 33:5573 | Ed.)
2306. [1998: 46, 175] Proposed by Vedula N. Murty, Visakhapat-
nam, India.
58
(a) Give an elementary proof of the inequality:
2
2x2 ;
>
sin x
2
1 + x2
(0 < x < 1) .
(1)
(b) Hence (or otherwise) show that
8
<
<
tan x :
>
x(1,x) ;
1,2x
x(1,x) ;
1,2x
,
0 < x < 21 ,
(2)
,1
<
x
<
1
.
2
sin(x)
(c) Find the maximum value of f (x) =
x(1 , x) on the interval (0; 1).
Comments on (a) and (c) by the editor.
As pointed out by several solvers (indicated by a dagger y before their
names), part (a) is identical to the rst half of problem 2296 [1997: 503] with
solution on [1998: 533]., Part
(c) follows immediately from the second half
x)
1
of that problem since f 2 = 4, where f (x) = xsin(
(1,x) . This, together with
the (corrected) inequality f (x) 4, shows that max f (x) = 4.
Solution to (b) by yWalther Janous, Ursulinengymnasium, Innsbruck,
Austria.
x(1 , x) , tan(x), 0 < x < 1.
Consider the function f (x) =
1 , 2x
Then f (0) = 0. We claim that f 0(x) > 0 for all x 2 (0; 21 ).
Since f 0(x) = show that
(1 , 2x)2 + 2x(1 , x) , 1 , it suces to
(1 , 2x)2
cos2(x)
2
cos2 (x) > (1 , 2x(1)2,+22xx) (1 , x)
(3)
Letting y = 1 , 2x, we have y 2 (0; 1) and x = 12 (1 , y ). Simple
manipulations show that (3) is equivalent to
2
2
y
y
y
2
sin 2 > y2 + (1 , y )(1 + y )=2 or sin 2 > 1 2+y y 2 ,
which is valid by (a). Hence f (x) > 0 for all x 2 (0; 12 ).
If 12 < x < 1, then 0 < 1 , x < 12 , and hence, by the result just
established, we have f (1 , x) > 0; that is,
x(1 , x) > tan((1 , x)) , or x(1 , x) > , tan(x) ,
1 , 2(1 , x)
,(1 , 2x)
x(1 , x) .
from which we obtain that tan(x) >
1 , 2x
2
59
Also solved by GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia;
Y,
Ferris State
RICHARD I. HESS, Rancho Palos Verdes, California, USA; yVACLAV
KONECN
University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,
Greece; PHIL McCARTNEY, Northern Kentucky University, Highland Heights, KY, USA;
yHEINZ-JURGEN
SEIFFERT, Berlin, Germany; and the proposer. Part (a) only was solved by
AISSA GUESMIA, Institut de recherche mathematique avancee, Universite Louis Pasteur et
CNRS, Strasbourg, France. There was also one incorrect solution.
Seiert also cited the following known inequalities (V.D. Mascioni und K. Schutte,
Aufgabe 979, Elem. Math. 44, 1989, 20):
8x(1 , x)
x
1
(1 , 2x) < tan(x) < 1 , 4x2 , 0 < x < 2 ,
and noted that the right inequality of this renes the right inequality in (b). Replacing x by
x , 12 , he derived the following inequalities:
(1
(1 , x)
, 2x)(3 , 2x) <
tan(
x) < 8x(1(1,,2xx)) ,
1
2
< x < 1,
and remarked that the left inequality of this is sharper than the left inequality in (b). He also
pointed out that this problem and problem 2296 are both simple consequences of problem 519
[1980: 44; 1981: 65] and problem 144 [Matyc J. 14, 1980, 72 and 15, 1981, 156]. All of these
problems were posed by this proposer.
2307. [1998: 46] Proposed by Joaqun Gomez
Rey, IES Luis Bu~nuel,
Alcorcon, Madrid, Spain.
, It is known that every regular 2n{gon can be dissected into n2 rhombuses with the same side length.
(a) How many dierent classes of rhombuses are there?
(b) How many rhombuses are there in each class?
Solution by Michael Lambrou, University of Crete, Crete, Greece.
It is stated on page 10 of Disections, Plane and Fancy by G.N.
Frederickson, Cambridge University Press, 1977, that a regular 2n{gon can
be dissected into rhombi with the same side length according to the following
rules:
(a) If n is odd with n = 2m + 1, then the 2n{gon can be dissected into n
such rhombi, each of angle j=n, for each j = 1, 2, : : : , m. In other
words, we have m = n,2 1 classes of, rhombi
with n rhombi in each
n
1
class, making a total of 2 n(n , 1) = 2 pieces.
(b) If n is even with n = 2m, then the 2n{gon can be dissected into n such
rhombi, each of angle j=n, for each j = 1, 2, : : : , m , 1, plus m
squares. In other words, we have m = n2 classes of rhombi with a
total of n(m ,,n1)
+ m = 2m(m , 1) + m = m(2m , 1) =
1
n
(
n
,
1)
=
pieces.
2
2
Although the proof is not given in detail in Frederickson, the accompanying gures for the cases n = 2, 3, 4, 5 and 6 make it obvious how to do
the general case, so there is no need to repeat the details here.
Also solved by the proposer.
60
2308. [1998: 46] Proposed by Joaqun Gomez
Rey, IES Luis Bu~nuel,
Alcorcon, Madrid, Spain.
A sequence fvn g has initial value v0 = 1 and, for n 0, satises the
recurrence relation
n
X
vn+1 = 2n+1 , vk vn,k .
k=0
Find a formula for vn in terms of n.
Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.
We use the method of generating functions and show
vn = bn=n2c .
where bxc is the greatest integer not exceeding x. Let
f (x) =
1
X
n=0
vnxn .
Then using f 2(x) to mean [f (x)] we get
2
f (x) =
2
1
X
n
X
n=0 k=0
!
vkvn,k xn =
Hence (via ,v0 + 1 = 0) we get
xf 2(x) =
1
X
1
X
(,vn+1 + 2n+1 )xn .
n=0
(,vn+1 + 2n+1)xn+1 =
n=0
1
X
1
X
m=1
(,vm + 2m )xm
(,vm + 2m )xm = ,f (x) + 1 ,1 2x ;
m=0
1 ,
that is, 0 = xf 2 (x) + f (x) ,
1 , 2x
=
from which it follows that
s
!
2x .
f (x) = 21x ,1 11 +
, 2x
Because of f (0) = 1 we have to use the positive root. Thus
f (x) = 21x ,1 + (1 + 2x)(1 , 4x2),1=2 .
It is known that
(1 , 4z),1=2 =
1
X
k=0
2kz k .
k
61
Hence
that is
!
1 2k
X
1
f (x) = 2x ,1 + (1 + 2x)
x2k
k
k=0
1 2k
1 1 2k
X
X
=
x2k + 2 k x2k,1 ;
k
k=0
k=1
2
k
v2k = k
and
showing vn = bn=n2c , as claimed.
,
1
2
k
2
k
,
1
v2k,1 = 2 k = k , 1 ,
Also solved by MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, CRM, Universite de Montreal, Montreal, Quebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID DOSTER,
Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG, student, Cambridge,
UK; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;
HEINZ-JURGEN
SEIFFERT, Berlin, Germany; and the proposer. There was one incomplete
solution submitted.
Dimminie remarks that if fCn g is the sequence of Catalan numbers dened by C0 = 1
and
n
Cn+1
=
XC C
k n,k
k=0
for n 0, then the solution may also be expressed as
v2n
and
v2n,1
= (
we get
f (x)
=
=
=
=
=
n , 1)Cn,1
= (2
Janous also observes that by using
(1
n + 1)Cn
, 4z)1=2
=
for n 0
for n 1 .
1 k
, X 2kk 2kz, 1
k=0
!
(1 , 4x2 )1=2
,
1+
x
1 , 2x
!
1
1
2k x2k X
X
, 21x 1 +
2` x`
k 2k , 1 `=0
=0
2 k1
0
1 3
X @ X 2k 2` A m 5
1 4
, 2x 1 +
x
m=0 2k+`=m k 2k , 1
1 0 X 2k 2`,1 1
X
A xm,1
, @
m=1 2k+`=m k 2k , 1
1 0 X 2k 2`,1 1 n
X
Ax .
, @
n=0 2k+`=n+1 k 2k , 1
1
2
62
Hence we get the curious (and apparently new) identity [compare coecients and replace ` by
n + 1 , 2k]:
b(n+1)
X =2c 2k 2n+1,2k,1
which simplies to
k=0
k
b(n+1)
X =2c 2k
k=0
k,1
2
1
k 4k (2k , 1)
=
=
, bn=n2c ,
, 21n bn=n2c .
2310. [1998: 47] Proposed by K.R.S. Sastry, Dodballapur, India.
Let n 2 N. I call a positive integral divisor of n, say d, a unitary divisor
if gcd(d; n=d) = 1.
Let (n) denote the sum of the unitary divisors of n.
Find a characterization of n so that (n) 2(mod 4).
Solution by Mansur Boase, student, Cambridge, England.
Suppose n = p1 1 p2 2 : : : pr r , where pi are distinct primes and i > 0
for all i, 1 i r. Then
(n) = (1 + p1 1 )(1 + p2 2 ) : : : (1 + pr r ) ,
as each term when this is multiplied out is a unitary divisor of n and every
unitary divisor of n occurs exactly once in the expansion.
If (n) 2 (mod 4), then it must be even, but not divisible by 4. For
odd pi, the term (1 + pi i ) is always even, so at most one odd prime can
divide n and n must be of the form 2m pk .
Now (n) 2 (mod 4) if and only if (1 + pk ) 2 (mod 4). If
p 1 (mod 4), this will be the case for any k; if p 3 (mod 4), then k
must be even.
Therefore, n = 2m pk with m 0, p an odd prime, with k an arbitrary positive integer if p 1 (mod 4) and k an arbitrary even integer if
p 3 (mod 4).
Also solved by GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia;
NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf,
Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;
KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England; MICHAEL
PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; ROBERT
P. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ-JURGEN
SEIFFERT,
Berlin, Germany; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;
KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There were two incorrect solutions submitted.
63
2311. [1998: 47] Proposed by K.R.S. Sastry, Dodballapur, India.
Let e (n) denote the sum of the even unitary divisors, and o (n), the
sum of the odd unitary divisors, of n. Assume that e (n) , o (n) = n.
(a) If n is composed of powers of exactly two distinct primes, show that n
must be the product of two consecutive integers, one of which is a Mersenne
prime.
(b) Give an example of a natural number n that is composed of powers of
more than two distinct primes.
Solution by Kenneth M. Wilke, Topeka, Kansas, USA.
For n = 2dpe11 pe22 : : : pekk where d 1 and each pi denotes an odd
prime which divides n, we have
o (n) =
so
k
Y
(pei i + 1)
i=1
and
e (n) = 2d
k
Y
(pei i + 1) ,
i=1
k
Y
n = e (n) , o(n) = (2d , 1) (pei i + 1) .
i=1
(1)
(a) Here i = 1 so that [writing p for p1 and e for e1]
n = 2dpe = (2d , 1)(pe + 1)
and hence
2d = pe + 1
(2)
where d and e are positive integers. Since p 3, we must have e 1 and
d 2. Suppose e is even, say e = 2l. Then p2l 1 modd4 regardless
of the choice of the odd prime p and the integer l. Then since 2 0 mod 4
for all integers d 2, we have that
pe + 1 = p2l + 1 2 6 0 2d mod 4 ,
which is impossible from (2). Hence e must be odd, say e = 2l + 1. Then
from (2) we have
2d = (p + 1)(p2l , p2l,1 + , p + 1) .
Since the second factor on the right hand side is odd, equality can be maintained only if l = 0 and p = 2d , 1, a Mersenne prime.
(b) From (1) and our choice of n,
d
1 = 2 2,d 1
!
pe11 + 1 pe22 + 1 : : : pekk + 1 .
pe11
pe22
pekk
64
We can use the values d = 3, (p1; e1 ) = (5; 2), (p2; e2) = (7; 2) and
(p3; e3 ) = (13; 1) to nd
n = 23 52 72 13 = 127400
as a solution for part (b).
Editorial remark. The above solution was also found by solver Hess
and by the proposer. The smallest solution to part (b), and the one most
solvers gave, is n = 180. Other solutions found are
n = 441000 (found only by the proposer)
and
n = 2646000 (found by Hess and Leversha).
Are there any others?
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORAN
CONAR, student, Gymnasium Varazdin, Varazdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho
Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's
School, London, England; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JOEL
SCHLOSBERG, student, Bayside, NY, USA; HEINZ-JURGEN
SEIFFERT, Berlin, Germany;
DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposer.
In part (a), one solver obtained equation (2) and then concluded rather hastily that e = 1
and so p is a Mersenne prime. Another solver believed (2) to imply that e = 1 but admitted he
had no proof. Both have been given the benet of the doubt (in the spirit of the season | this
is being written during the Christmas holidays). Other solvers merely said that this is a known
result, or proved it themselves (as Wilke did), or gave a reference. For example, Lambrou
mentioned the article \A note on Mersenne numbers", by Ligh and Neal, on pp. 231{233 of
Math. Magazine 47 (1974), and Manes quoted problem E1221 of the Amer. Math. Monthly,
solution in Volume 64 (1957), p. 110. The solution of the Monthly problem also contains some
much earlier references to this result.
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