202
M397.
Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
Determine all pairs (x, y) of integers suh that
x4 − x + 1 = y 2 .
M398.
Proposed by the Mayhem Sta.
(a) The ubi equation w3 − bw2 + cw − d = 0 has roots r, s, and t.
Determine b, c, and d in terms of r, s, and t.
(b) Suppose that a is a real number. Determine all solutions to the system
of equations
x+y+z
xy + yz + zx
xyz
=
=
=
a,
−1 ,
−a .
M399. Proposed by Neulai Staniu, Saint Mueni Sava Tehnologial
High Shool, Bera, Romania.
Determine all triples (a, b, c) of positive integers for whih
a positive integer.
M400.
3ab − 1
abc + 1
is
Proposed by Mihaly Benze, Brasov, Romania.
Suppose that a, b, and c are positive real numbers. In addition, suppose
that an + bn = cn for some positive integer n with n ≥ 2. Prove that if k is
a positive integer with 1 ≤ k < n, then ak , bk , and ck are the side lengths
of a triangle.
Mayhem Solutions
M357.
Proposed by the Mayhem Sta.
Determine all real numbers x that satisfy 32x+2 + 3 = 3x + 3x+3 .
Solution by Shamil Asgarli, student, Burnaby South Seondary Shool,
Burnaby, BC.
The equation 32x+2 + 3 = 3x + 3x+3 an be written in the form
3 (3 ) + 3 = 3x + 33 3x . Substituting k = 3x , we obtain the equivalent
quadrati equations 9k2 + 3 = k + 27k and 9k2 − 28k + 3 = 0.
Continuing to solve for k by fatoring, we obtain (k − 3)(9k − 1) = 0.
Hene, k = 3 or k = 19 .
Substituting k = 3x , we have 3x = 3 or 3x = 19 , so x = 1 or x = −2.
2
x 2
203
Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosnia
and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Greee; CAO MINH QUANG,
Nguyen Binh Khiem High Shool, Vinh Long, Vietnam; JACLYN CHANG, student, Western
Canada High Shool, Calgary, AB; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; JOSE
HERNANDEZ
SANTIAGO, student, Universidad Tenologia
de la Mixtea, Oaxaa, Mexio;
IES \Abastos", Valenia,
GIANNIS G. KALOGERAKIS, Canea, Crete, Greee; RICARD PEIRO,
Spain; KUNAL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India; MRIDUL SINGH,
student, Kendriya Vidyalaya Shool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler
Publi Shool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Greee; EDWARD T.H. WANG,
Wilfrid Laurier University, Waterloo, ON; JIXUAN WANG, student, Don Mills Collegiate
Institute, Toronto, ON; and CARLIN WHITE, student, California State University, Fresno, CA,
USA. There were two inorret solutions submitted.
M358.
Proposed by Neulai Staniu, Saint Mueni Sava Tehnologial
High Shool, Bera, Romania.
How many integers in the list 1, 2008, 20082 , . . . , 20082009 are simultaneously perfet squares and perfet ubes?
Solution by Luis De Sousa, student, IST-UTL, Lisbon, Portugal, modied by
the editor.
The integer 2008 has prime
fatorization 23 · 251. Thus, 2008k has
k
3
prime fatorization 2 · 251 = 23k · 251k . For a positive integer greater
than 1 to be a perfet square, its prime fatorization must inlude only even
exponents; for a positive integer greater than 1 to be a perfet ube, its prime
fatorization must inlude only exponents divisible by 3.
The exponents 3k and k are both even if and only if k is even. The
exponents 3k and k are both multiples of 3 if and only if k is a multiple of 3.
Therefore, the powers of 2008 whih are perfet squares are the ones
with even exponents. Also, the powers of 2008 whih are perfet ubes are
those with exponents that are multiples of 3. Hene, the powers of 2008
whih are perfet squares and perfet ubes are those whose exponents are
both multiples of 2 and of 3. In other words, they are those whose exponents
are multiples of 6.
The largest multiple of 6 less than 2009 is 2004 = 6 · 334, so there are
334 multiples of 6 in the list 1, 2, 3, . . . , 2008, 2009. This tells us that 334 of
the integers 20081 , 20082 , . . . , 20082009 are simultaneously perfet squares
and perfet ubes. The integer 1 is also in the list and is both a perfet square
and a perfet ube. Therefore, there are 334 + 1 = 335 integers in the list
that are simultaneously perfet squares and perfet ubes.
Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosnia
and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Greee; JACLYN CHANG,
student, Western Canada High Shool, Calgary, AB; CHANTHOEUN CHAP and JUSTIN
HENDERSHOTT, students, California State University, Fresno, CA; USA; JOSE HERNANDEZ
SANTIAGO, student, Universidad Tenologia
de la Mixtea, Oaxaa, Mexio; RICHARD
IES \Abastos", Valenia, Spain;
I. HESS, Ranho Palos Verdes, CA, USA; RICARD PEIRO,
KUNAL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India; MRIDUL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi
Shool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Greee; and JIXUAN WANG, student,
Don Mills Collegiate Institute, Toronto, ON.
204
M359.
Proposed by the Mayhem Sta.
A trapezoid DEF G is irumsribed about a irle of radius 2, as
shown in the diagram. The side DE
has length 3 and there are right angles
at E and F . Determine the area of the
trapezoid.
D................................................................... E
..........
...
....
......
.............
..... ....
........
.... ....
....
... ...
...
.
.
.
.
.
.
.
.......
......
......
......
...
.... ....
.
.
.
..
... .....
.
...
.
..
..
.....
...
....
......
.
.
.
.
.
.
...
. ..
..
....
... ...
....
.....
.... ...
....
.......
..... .....
....
..
.......
.................................................................................................................................................
G
F
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,
ON.
Let O be the entre of the irle. Let A be the point of intersetion of
the perpendiular from D to GF . Let the points of tangeny of EF , DE ,
GD , and GF to the irle be W , X , Y , and Z , respetively.
We twie use the fat that tanD .................X
E
.. ..... .......q......................................................
......
................
.... .....
gents to the same irle from the same
....... .....
.
.
.
.
.
.
... ...
Y ............q .......
... ...
.......
...
point have equal lengths. Let GZ = x.
....
.....
...
........
O
.
...
.
.
.
...
q
q
... ....
.
.
.
...
...
Then GY = x by equal tangents.
.
...
.
.. W
.
.
.
..
..
..
......
.
.
.
.
.
.
.
...
...
..
. ....
.
.
.
.
.
.
.
.
Sine OX and OW are perpen.... ....
.. ...
....
.... ..
... ...
........
...
...... ....
......
....
.......
diular to DE and EF , respetively,
................................................................................................q..................................................
G
F
and the trapezoid has a right angle at
A Z
E , then OXEW is a square.
Sine the radius of the irle is 2, then OW = 2, so XE = 2. Sine
DE = 3, then DX = DE − XE = 1. By equal tangents, DY = 1.
Now DEF A is also a retangle sine it has three right angles (at E , F ,
and A) hene the fourth angle is also a right angle. Also, XZ is parallel to
EF . Thus, AZ = DX = 1.
We know that DA = XO+OZ = 2+2 = 4, GD = GY +Y D = x+1,
and GA = GZ − AZ = x − 1. By the Pythagorean Theorem,
GD 2 =
(x + 1)2 =
2
x + 2x + 1 =
GA2 + AD 2 ;
(x − 1)2 + 42 ;
x2 − 2x + 1 + 16 ;
hene 4x = 16 and x = 4. Therefore, GF = GZ + ZF = 4 + 2 = 6, and so
the area of the trapezoid is 12 (GF + DE)(DA) or 12 (6 + 3)(4) = 18.
Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosnia
and Herzegovina; SERDAR ALTUNTAS, student, University of Karlsruhe, Karlsruhe, Germany;
MIGUEL AMENGUAL COVAS, Cala Figuera, Mallora, Spain; SCOTT BROWN, Auburn University, Montgomery, AL, USA; CAO MINH QUANG, Nguyen Binh Khiem High Shool,
Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High Shool, Calgary, AB;
COURTIS G. CHRYSSOSTOMOS, Larissa, Greee; JORDAN CRIST, student, Auburn University
at Montgomery, Montgomery, AL, USA; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal;
JOSH GUZMAN and AMANDA PERCHES, students, California State University, Fresno, CA;
RICHARD I. HESS, Ranho Palos Verdes, CA, USA; GIANNIS G. KALOGERAKIS, Canea, Crete,
IES \Abastos", Valenia, Spain; KUNAL SINGH, student, Kendriya
Greee; RICARD PEIRO,
Vidyalaya Shool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi Shool,
Waterloo, ON; and GEORGE TSAPAKIDIS, Agrinio, Greee. There was one inorret and one
inomplete solution submitted.
205
M360.
Proposed by Neulai Staniu, Saint Mueni Sava Tehnologial
High Shool, Bera, Romania.
Determine all positive integers x that satisfy
3x = x3 + 3x2 + 2x + 1 .
Solution by Shamil Asgarli, student, Burnaby South Seondary Shool,
Burnaby, BC.
We will show that there are no suh positive integers.
Suppose that there was a positive integer solution x. We rewrite the
equation in the form
3x = x x2 + 3x + 2 + 1 = x(x + 1)(x + 2) + 1 .
Sine x > 0, the left side is divisible by 3. Therefore, the right side should
be divisible by 3 as well.
But x(x + 1)(x + 2) is divisible by 3, sine it is the produt of three
onseutive integers. Thus x(x + 1)(x + 2) + 1 gives a remainder of 1 when
divided by 3, so is not divisible by 3. This is a ontradition, so there are no
positive integer solutions.
Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosnia
and Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long, Vietnam;
IES \Abastos",
LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; RICARD PEIRO,
Valenia, Spain; ALEX SONG, student, Elizabeth Ziegler Publi Shool, Waterloo, ON; GEORGE
TSAPAKIDIS, Agrinio, Greee; and JIXUAN WANG, student, Don Mills Collegiate Institute,
Toronto, ON. There were two inomplete solutions submitted.
M361.
Proposed by George Tsapakidis, Agrinio, Greee.
Let a, b, and c be positive real numbers. Prove that
ab(a + b − c) + bc(b + c − a) + ca(c + a − b) ≥ 3abc .
Solution by Shamil Asgarli, student, Burnaby South Seondary Shool,
Burnaby, BC.
First, by the Arithmeti Mean{Geometri Mean Inequality, we nd that
a2 b + ab2 + b2 c + bc2 + c2 a + ca2
≥
=
6
q
6
a2 b ab2 b2 c bc2 c2 a ca2
√
6
a6 b6 c6 = abc .
It follows that
a2 b + ab2 + b2 c + bc2 + c2 a + ca2 ≥ 6abc .
206
Rearranging the last expression, we nd that
a2 b + ab2 − abc + b2 c + bc2 − abc + c2 a + ca2 − abc ≥ 3abc ,
whih is equivalent to
ab(a + b − c) + bc(b + c − a) + ca(c + a − b) ≥ 3abc ,
as required.
Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosnia
and Herzegovina; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallora, Spain; GEORGE
APOSTOLOPOULOS, Messolonghi, Greee; CAO MINH QUANG, Nguyen Binh Khiem High
Shool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa, Greee; LUIS DE SOUSA,
student, IST-UTL, Lisbon, Portugal; JOSE LUIS D I AZ-BARRERO, Universitat Politenia
de
Catalunya, Barelona, Spain; ANA GUTIERREZ and MARYLOV INSISIENGMAY, students,
California State University, Fresno, CA, USA; RICHARD I. HESS, Ranho Palos Verdes, CA,
USA; HUGO LUYO SANCHEZ,
Pontiia Universidad Catolia
del Peru, Lima, Peru; MISSOURI
IES
STATE UNIVERSITY PROBLEM SOLVING GROUP, Springeld, MO, USA; RICARD PEIRO,
\Abastos", Valenia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya Shool, Shillong,
India; ALEX SONG, student, Elizabeth Ziegler Publi Shool, Waterloo, ON; EDWARD
T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don Mills
Collegiate Institute, Toronto, ON.
M362.
Proposed by Mihaly Benze, Brasov, Romania.
Suppose that x1 , x2 , . . . , xn is a sequene of integers suh that
|x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn | = 2 .
Prove that at least one of x1 , x2 , . . . , xn equals 1 or −1.
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,
ON.
Without loss of generality, assume that x1 , x2 , . . . , xk are positive and
xk+1 , xk+2 , . . . , xn are not positive (if this is not the ase, then we an
rearrange the numbers to make it so). There are now two ases to onsider.
Case 1 We have x1 + x2 + x3 + · · · + xn ≥ 0. Then
|x1 + x2 + x3 + · · · + xn | = x1 + x2 + · · · + xn
= x1 + x2 + · · · + xk − (−xk+1 ) − (−xk+2 ) − · · · − (−xn )
= |x1 | + |x2 | + · · · + |xk | − |xk+1 | + |xk+2 | + · · · + |xn | .
From this, we have
2
= |x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn |
= |x1 | + |x2 | + · · · + |xn | − |x1 | + |x2 | + · · · + |xk |
− |xk+1 | + |xk+2 | + · · · + |xn | = 2 |xk+1 | + |xk+2 | + · · · + |xn | = 2 |xk+1 | + |xk+2 | + · · · + |xn | .
207
It follows that |xk+1 | + |xk+2 | + · · · + |xn | = 1. Sine the sum onsists of
nonnegative integers, one term in the sum is equal to 1 (and the rest are 0),
hene one of the (nonpositive) numbers xk+1 , xk+2 , . . . , xn equals −1.
Case 2 We have x1 + x2 + x3 + · · · + xn ≤ 0. In a similar way to Case 1, we
see that
|x1 + x2 + x3 + · · · + xn |
= −(x1 + x2 + x3 + · · · + xn )
= −x1 − x2 − · · · − xk + (−xk+1 ) + (−xk+2 ) + · · · + (−xn )
= |xk+1 | + |xk+2 | + · · · + |xn | − |x1 | + |x2 | + · · · + |xk | .
Therefore,
2
= |x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn |
= |x1 | + |x2 | + · · · + |xn | − |xk+1 | + |xk+2 | + · · · + |xn |
− |x1 | + |x2 | + · · · + |xk | = 2 |x1 | + |x2 | + · · · + |xk | = 2 |x1 | + |x2 | + · · · + |xk | .
Therefore, |x1 | + |x2 | + · · · + |xk | = 1. The sum onsists of positive integers,
so it has exatly one term, whih is equal to 1. Hene, k = 1 and x1 = 1.
By the two ases, one of x1 , x2 , . . . , xn is equal to 1 or −1.
Also solved by LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; MISSOURI STATE
UNIVERSITY PROBLEM SOLVING GROUP, Springeld, MO, USA; ALEX SONG, student,
Elizabeth Ziegler Publi Shool, Waterloo, ON; and GEORGE TSAPAKIDIS, Agrinio, Greee.
There was one inomplete solution submitted.
Case 2 an be avoided by replaing x1 , x2 , . . . , xn with −x1 , −x2 , . . . , −xn .
Problem of the Month
Ian VanderBurgh
Contest problems involving omplex numbers don't appear very often...
Problem (2007 Amerian Invitational Mathematis Challenge A) The omplex
number z is equal to 9 + bi, where b is a positive real number and i2 = −1.
Given that the imaginary parts of z 2 and z 3 are equal, nd b.
...and when they do, they're often easier than they look. The one piee
of information that we need to remember is that the imaginary part of a
omplex number is the oeÆient of i. As the famous slogan (almost) says,
just alulate it!