321

THE ACADEMY CORNER

No. 13
Bruce Shawyer
All communications about this column should be sent to Bruce
Shawyer, Department of Mathematics and Statistics, Memorial University
of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Mathematics Competitions
There are many mathematical competitions at many di erent levels.
There is in fact a journal devoted to this. It is called Mathematics Compe-
titions, and is the Journal of the World Federation of National Mathematics
Competitions. It is published by the Australian Mathematics Trust, and in-
formation on it can be obtained by contacting them at PO Box 1, Belconnen,
ACT 2616, Australia.
The journal covers all sorts of mathematical competitions, and articles
not only give problems and solutions, but also cover the philosophy of com-
petition and alternative ways of stimulating talented students.
The World Federation also makes two awards, the David Hilbert In-
ternational Award (which recognises contributions of mathematicians who
have played a signi cant role over a number of years in the development
of mathematical challenges at the international level and which have been
a stimulus for mathematical learning), and the Paul Erdos National Award
(which recognises contributions of mathematicians who have played a sig-
ni cant role over a number of years in the development of mathematical
challenges at the national level and which have been a stimulus for the en-
richment of mathematics learning).
322

THE OLYMPIAD CORNER

No. 184
R.E. Woodrow
All communications about this column should be sent to Professor R.E.
Woodrow, Department of Mathematics and Statistics, University of Calgary,
Calgary, Alberta, Canada. T2N 1N4.
We begin this number with the problems of the Mock Test for the In-
ternational Mathematical Olympiad team of Hong Kong. My thanks go to
Richard Nowakowski, Canadian Team Leader, for collecting them at the 35 th

IMO in Hong Kong.

INTERNATIONAL MATHEMATICAL OLYMPIAD

1994
Hong Kong Committee | Mock Test, Part I
Time: 4.5 hours

1. In 4ABC , we have \C = 2\B. P is a point in the interior of

4ABC satisfying AP = AC and PB = PC . Show that AP trisects the
angle \A.
2. In a table-tennis tournament of 10 contestants, any two contestants
meet only once. We say that there is a winning triangle if the following
situation occurs: i contestant defeated j contestant, j contestant defeated
th th th

k contestant, and k contestant defeated i contestant. Let Wi and Li be

th th th

respectively the number of games won and lost by the i contestant. Suppose th

Li + Wj  8 whenever the i contestant beats the j contestant. Prove that

th th

there are exactly 40 winning triangles in this tournament.

3. Find all the non-negative integers x, y, and z satisfying that
7x + 1 = 3y + 5z :

Mock Test, Part II

Time: 4.5 hours

1. Suppose that yz + zx + xy = 1 and x, y, and z  0. Prove that

p
x(1 , y 2)(1 ,z 2) + y(1 , z 2 )(1 ,x
2) + z(1 , x 2 )(1 , y  4 9 3:
2)
 323

2. A function f (n), de ned on the natural numbers, satis es:

f (n) = n , 12 if n > 2000; and f (n) = f (f (n + 16)) if n  2000:
(a) Find f (n).
(b) Find all solutions to f (n) = n.
3. Let m and n be positive integers where m has d digits in base
ten and d  n. Find the sum of all the digits (in base ten) of the product
(10n , 1)m.

As a second Olympiad set we give the problems of the Final Round of

the 45 Mathematical Olympiad written in April, 1994. My thanks go to
th

Marcin E. Kuczma, Warszawa, Poland; and Richard Nowakowski, Canadian

Team leader to the 35 IMO in Hong Kong, for collecting them.
th

45th MATHEMATICAL OLYMPIAD IN POLAND

Problems of the Final Round | April 10{11, 1994
First Day | Time: 5 hours
1. Determine all triples of positive rational numbers (x; y; z) such that
x + y + z, x,1 + y,1 + z,1 and xyz are integers.
2. In the plane there are given two parallel lines k and l, and a cir-
cle disjoint from k. From a point A on k draw the two tangents to the given
circle; they cut l at points B and C . Let m be the line through A and the mid-
point of BC . Show that all the resultant lines m (corresponding to various
points A on k) have a point in common.
3. Let c  1 be a xed integer. To each subset A of the set f1, 2, : :
: : , ng we assign a number w(A) from the set f1, 2, : : : , cg in such a way
that
w(A \ B) = min(w(A);w(B)) for A; B  f1; 2; : : : ; ng:
p
Suppose there are a(n) such assignments. Compute limn!1 n a(n).
Second Day | Time: 5 hours
4. We have three bowls at our disposal, of capacities m litres, n litres
and m + n litres, respectively; m and n are mutually coprime natural num-
bers. The two smaller bowls are empty, the largest bowl is lled with water.
Let k be any integer with 1  k  m + n , 1. Show that by pouring
water (from any one of those bowls into any other one, repeatedly, in an
unrestricted manner) we are able to measure out exactly k litres in the third
bowl.
324

5. Let A1; A2; : : : ; A8 be the vertices of a parallelepiped and let O be

its centre. Show that
4(OA21 + OA22 +


+ OA28)  (OA1 + OA2 +


+ OA8)2:
6. Suppose that n distinct real numbers x1; x2; : : : ; xn (n  4) satisfy
the conditions x1 + x2 +


+ xn = 0 and x21 + x22 +


+ x2n = 1. Prove
that one can choose four distinct numbers a, b, c, d from among the xi 's in
such a way that
a + b + c + nabc  x31 + x32 +


+ x3n  a + b + d + nabd:

We now give three solutions to problems given in the March 1996 Cor-
ner as the Telecom 1993 Australian Mathematical Olympiad [1996: 58].

TELECOM 1993 AUSTRALIAN

MATHEMATICAL OLYMPIAD
Paper 1
Tuesday, 9th February, 1993
(Time: 4 hours)
6. In the acute-angled triangle ABC , let D, E, F be the feet of alti-
tudes through A, B , C , respectively, and H the orthocentre. Prove that
AH + BH + CH = 2:
AD BE CF
Solution by Mansur Boase, student, St. Paul's School, London, Eng-
land.
AH + BH + CH = 3 ,  HD + HE + HF 
AD BE CF AD BE CF

[BHC ] [CHA ] [ AHB
= 3 , [ABC ] + [ABC ] + [ABC ] ] 

= 3 , [[ABC ] = 2:
ABC ]
7. Let n be a positive integer, a1; a2; : : : ; an positive real numbers
and s = a1 + a2 +


+ an . Prove that
n
X ai  n n
X s , ai  n(n , 1):
and
i=1 s , ai n , 1 i=1 ai
 325

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.
We give the solution by Boase.
n
X s , ai = n
X s , 1 = X
 n s
,n
i=1 ai i=1 ai i=1 ai
n
X ai = 1 Xn s X n ai X
and  ( 1)2 = n2;
i=1 s a i
i=1 i=1 s
by the Cauchy{Schwarz inequality.
Thus n X s  n2:
i=1 ai
n
X s,ai  n2 , n = n(n , 1).
Hence
i=1 a i
To prove the rst inequality, rst note that
n
X n
X n
X
!2
1 a2i  ai = s2 :
i=1 i=1 i=1
n
X s2
Hence a2i  .
i=1 n
By the Cauchy{Schwarz inequality,
n n n !2
ai(s , ai) s ,aia 
X X X
ai = s2 :
i=1 i=1 i i=1
Therefore
n
X ai  P s2 s2 P
=
n ai (s , ai ) s n ai , n a2
i=1 s , ai
P
i=1 i=1 i=1 2
 s2 s, s2 = 1 ,1 1 = n ,n 1 :
2

n n
So, both inequalities are proved.
8. [1996: 58] Telecom 1993 Australian Mathematical Olympiad.
The vertices of triangle ABC in the xy {plane have integer coordinates,
and its sides do not contain any other points having integer coordinates. The
interior of ABC contains only one point, G, that has integer coordinates.
Prove that G is the centroid of ABC .
326

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.
A

C0 B0
r
G
B A0 C
By Pick's Theorem
[ABC ] = 1 + 3 12 , 1 = 32 ;
 

[ABG] = 0 + 3 12 , 1 = 12 ;
 

[BCG] = 12 and
[CAG] = 12 :

Therefore
[ABG] = [BCG] = [CAG] = 1 :
[ABC ] [ABC ] [ABC ] 3
Hence
GA0 = GB0 = GC 0 = 1 :
AA0 BB0 CC 0 3
The unique point satisfying this above is well-known to be the centroid.

Next we give one solution from the Japan Mathematical Olympiad 1993
given in the March 1996 Corner.
2. [1996: 58] Japan Mathematical Olympiad 1993.
Let d(n) be the largest odd number which divides a given number n.
Suppose that D(n) and T (n) are de ned by
D(n) = d(1) + d(2) +


+ d(n);
T (n) = 1 + 2 +


+ n:
Prove that there exist in nitely many positive numbers n such that
3D(n) = 2T (n).
Solutions by Mansur Boase, student, St. Paul's School, London, Eng-
land, and Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Wa-
terloo, Ontario. We give the solution by Boase.
 327

T (n) = n(n2+ 1) :
Thus we need to prove that there are in nitely many n for which
D(n) = n(n3+ 1) so that 3D(n) = 2T (n) holds:
Consider
D(2n) = d(1) + d(3) +


+ d(2n , 1) + d(2) + d(4) +


+ d(2n)
= 1 + 3 +


+ (2n , 1) + d(1) + d(2) +


+ d(2n,1 )
= 1 + 3 +


+ (2n , 1) + D(2n,1 ):
Now
n n n,1 n,1
1 + 3 +


+ (2n , 1) = 2 (2 2 + 1) , 2 2 (2 2 + 1)
= 2n,1 (2n , 2n,1 )
= 22n,2:
Thus D(2n ) = D(2n,1 ) + 22n,2.
Now, D(21 ) = 2 and we shall prove by induction that D(2n ) =
22n + 2 for
n  0. 3
This holds for n = 0 and for n = 1. Suppose it holds for n = k.
Thus D(2k ) =
22k + 2 .
3
Then
2k
D(2k+1) = D(2k) + 22k = 2 3+ 2 + 22k
2k
= 4(2 3) + 2
2k+2
= 2 3+2
and the result follows by induction. Now consider D(2n , 2).
D(2n , 2) = D(2n) , d(2n , 1) , d(2n)
2n
= 2 3+ 2 , (2n , 1) , 1
2n
= 2 3+ 2 , 2n
= 2 , 3(2
2n n) + 2 (2n , 1)(2n , 2)
3 = 3 :
328

Thus D(x) =
x(x + 1) for x = 2n , 2, and there are in nitely many such x.
3
Next we turn to comments and solutions from the readers to problems
from the April 1996 number of the Corner where we gave the selection test
for the Romanian Team to the 34 IMO as well as three contests for the
th

Romanian IMO Team [1996: 107{109].

SELECTION TESTS FOR THE ROMANIAN TEAM,
34th IMO.
Part II | First Contest for IMO Team
1st June, 1993
1. Find the greatest real number a such that
p
x + pz 2y+ x2 + p 2z 2 > a
y 2+ 2 z x +y
is true for all positive real numbers x, y , z .
Solutions by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-
sity, Waterloo, Ontario.
We claim that a = 2. Let
f (x;y;z) = p
x + pz 2y+ x2 + p 2z 2 :
y z 2+ 2 x +y
We show that f (x;y;z ) > 2. Since f (x; y;z ) ! 2 as x ! y and z ! 0, the
lower bound 2 is sharp. Without loss of generality, assume that x  y  z .
Since by the arithmetic{harmonic{mean inequality, we have
pz2 + x2 py2 + z2
y2 + z2 + pz2 + x2  2;
p

it suces to show that

pz2 + x2 py2 + z2
f (x;y;z) > py2 + z2 + pz2 + x2
or equivalently,
z pz2 + x2 , x py2 + z2 , y
x2 + y2 > y2 + z2 + pz2 + x2 :
p p

By simple algebra, this is easily seen to be equivalent to

p pz
y2 + z2( z2 + x2 + x)
+ p 2 2 pz 2 2 1
z + x ( y + z + y) < y: (1)
p
x2+ 2
 329

p p p p p
Since y 2 + z 2  2z 2 = 2 z , z 2 + x2 > x and 2 x  x2 + y 2,
p

we have
p pz < p z = p1  p 21 2 :
y z z + x + x)
2 + 2( 2 2 2 z ( x + x) 2 2x 2 x +y
Thus to establish (1), it remains to show that
pz2 + x2(pzy2 + z2 + y) < 2px21+ y2
or equivalently s
2z z2 + x2 :
y2 + z2 + y < x2 + y2
p

Since
z2 + x2 = 1 , y2 , z2 ;
x2 + y2 x2 + y2
which is an non-decreasing function of x, we have
z2 + x2  z2 + y2 ;
x2 + y2 2y 2
and thus it suces to show that
p
zp2 + y2 > 2z
y2 + z2 + y ;
p
2y
or equivalently
p p
y2 + z2 + y y2 + z2 > 2 2 yz: (2)
Since y 2 + z 2  2yz , we have
p p
y2 + z2 + y y2 + z2  2yz +py 2z2 p
= (2 + 2)yz > 2 2 yz
and thus (2) holds. This completes the proof.
2. Show that if x, y, z are positive integers such that x2 + y2 + z2 =
1993, then x + y + z is not a perfect square.
Solutions by Mansur Boase, student, St. Paul's School, London, Eng-
land; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-
tario. We give Wang's solution.
We show that the result holds for nonnegative integers x, y , and z .
Without loss of generality, we may assume that 0  x  y  z . Then
3z2  x2 + y 2 + z 2 = 1993
330

implies that
z2  665; z  26:
On the other hand z 2  1993 implies that z  44 and thus 26  z  44.
Suppose that x + y + z = k2 for some nonnegative integer k. By the
Cauchy{Schwarz Inequality we have
k4 = (x + y + z)2  (12 + 12 + 12)(x2 + y2 + z2) = 5979
p
and so k  b 4 5979c = 8. Since k2  z  26, k  6. Furthermore, since
x2 + y2 + z2 is odd, it is easily seen that x + y + z must be odd, which implies
that k is odd. Thus k = 7 and we have x + y + z = 49.
Let z = 26 + d, where 0  d  18. Then
x + y = 23 , d ) y  23 , d ) x2 + y2  2y2
= 2(23 , d)2 = 1058 , 92d + 2d2 : (1)
On the other hand, from x2 + y 2 + z 2 = 1993 we get
x2 + y2 = 1993 , z2 = 1993 , (26 + d)2 = 1317 , 52d , d2: (2)
From (1) and (2), we get
1317 , 52d , d2  1058 , 92d + 2d2
or
3d2 , 40d  259
which is clearly impossible since 3d2 , 40d = d(3d , 40)  18  14 = 252.
This completes the proof.
Remark: It is a well-known (though by no means easy) result in clas-
sical number theory that a natural number n is the sum of three squares (of
nonnegative integers) if and only if n 6= 4l (8k + 7) where l and k are non-
negative integers. Since 1993  1(mod 8) it can be so expressed and thus
the condition given in the problem is not \vacuously" true. In fact, 1993 can
be so expressed in more than one way; for example,
1993 = 02 + 122 + 432
= 22 + 152+  422
= 22 + 302 + 332
= 112 + 242 + 362:
These representations also show that the conclusion of the problem is false
if we allow x, y , and z to be negative integers; e.g. if x = ,2, y = ,30,
z = 33 then x2 + y2 +2 z2 2= 1993 and x + y + z = 12 ; and if x = ,11,
y = 24, z = 36 then x + y + z = 1993 and x + y + z = 49 = 72.
2
 331

4. Show that for any function f : P (f1; 2; : : : ; ng) ! f1; 2; : : : ; ng

there exist two subsets, A and B , of the set f1; 2; : : : ; ng, such that A 6= B
and f (A) = f (B ) = maxfi j i 2 A \ B g.
Comment by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,
Ontario.
The problem, as stated, is clearly incorrect since for maxfi : i 2 A \ B g
to make sense, we must have A \ B 6= ;. For n = 1 clearly there are no
subsets A and B with A 6= B and A \ B 6= ;. A counterexample when n = 2
is provided by setting f (;) = f (f1g) = f (f2g) = 1 and f (f1; 2g) = 2.
This counterexample stands if max is changed to min. The conclusion is still
incorrect if A \ B is changed to A [ B . A counterexample would be f (;) = 2
and f (f1g) = f (f2g) = f f(1; 2)g = 1.
Part III | Second Contest for IMO Team
2nd June, 1993
3. Prove that for all integer numbers n, with n  6, there exists an
n{point set M in the plane such that every point P in M has at least three
other points in M at unit distance to P .
Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-
sity, Waterloo, Ontario.
The three diagrams displayed below illustrate the existence of such a
set. M1 is for all n = 3k, M2 is for all n = 3k+1 and M3 is for all n = 3k+2,
where k = 2; 3; 4; : : : . In each diagram, the solid lines connecting two points
all have unit length and the dotted lines, also of unit length, indicate how to
construct an (n + 3){point set with the described property from one with
n points.
c
c c c
c c

c c s c
s s s
s s

s s s s s
s s s
s s

s s s s s

M1 M2 M3

(n = 6; 9; 12; : : : ) (n = 7; 10; 13; : : : ) (n = 8; 11; 14; : : : )

332

Part IV | Third Contest for IMO Team

3rd June, 1993

1. The sequence of positive integers fxngn1 is de ned as follows:

x1 = 1, the next two terms are the even numbers 2 and 4, the next three
terms are the three odd numbers 5, 7, 9, the next four terms are the even
numbers 10, 12, 14, 16 and so on. Find a formula for xn .
Solutions by Mansur Boase, student, St. Paul's School, London, Eng-
land; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University,
Waterloo, Ontario. We give the solution of Shan and Wang.
We arrange the terms of the sequence fxn g in a triangular array ac-
cording to the given rule so that the 1 level consists of a single 1 and for
st

all k  2, the k level consists of the k consecutive even (odd) integers that
th

follow the last odd (even) integer in the (k , 1) level: st

1
2 4
5 7 9
10 12 14 16
17 19 21 23 25

For any given n 2 N , let ln denote the unique integer such that
ln  < n  ln + 1;


2 2
that is,
(ln , 1)ln < n  ln(ln + 1) : (1)
2 2
Note that ln is simply the number of the level which xn is on. We claim that
xn = 2n , ln : (2)
When n = 1, clearly l1 = 1 and thus 2n , ln = 1 = x1 . Suppose (2) holds
for some n  1. There are two possible cases:
Case (i): If n + 1  ln2+1 ; that is, if xn+1 and xn are on the same
,

level, then ln+1 = ln and hence

xn+1 = xn + 2 = 2n , ln + 2 = 2(n + 1) , ln+1 :
 333

Case (ii): If n + 1 > ln2+1 then xn is the last number on the ln level
,
th

and xn+1 is the rst number on the ln+1 level. Thus

th

ln+1 = ln + 1
and
xn+1 = xn + 1 = 2n , ln + 1 = 2(n + 1) , ln+1 :
Hence, by induction, (2) is established. From (1) we get
l2n , ln < 2n  (ln + 1)2 , (ln + 1):
Solving the inequalities, we easily obtain
1 + p1 + 8n
ln < 2  ln + 1:
Hence
& p '
ln = 1 + 21 + 8n , 1 (3)

where dxe denotes the least integer greater than or equal to x (that is, the
ceiling function). From (2) and (3) we conclude that
1 + p1 + 8n
& '
xn = 2n + 1 , 2 :

That completes this number of the Corner. Send me your nice solutions
as well as Olympiad contests.
334

BOOK REVIEWS
Edited by ANDY LIU
A Mathematical Mosaic, by Ravi Vakil,
published by Brendan Kelly Publishing Inc., 1996,
2122 Highview Drive, Burlington, ON L7R3X4,
ISBN# 1-895997-04-6, softcover, 253+ pages, US$16.95 plus handling.
Reviewed by Jim Totten, University College of the Cariboo.
So, you have a group of students who have decided they want the extra
challenge of doing some mathematics competitions. You want a source of
problems which will pique the students' interest, and which also lead to fur-
ther exploration. The problem source should lend itself well to independent
work. The question is: where do you nd the appropriate level enrichment
material? Many of us have already tried to answer this question and have a
collection of such problem books. Well, here is a book to be added to your
collection!
It is certainly a problem book, but it is much more than that. The au-
thor at one moment guides the reader through some very nice mathematical
developments, and throws out problems as they crop up in the development,
and in the next moment uses a problem as a starting point for some inter-
esting mathematical development.
With a few exceptions the problems in this book are not new, nor are the
solutions. They are, however, well organized, both by topic and by level of
mathematical maturity needed. Answers are NOT always provided; instead
there is often simply a solution strategy or hint given, and occasionally there
is simply a reference to some other source for a full-blown treatment. Even
when answers are provided, they are not tucked away at the end of the book,
but rather they are worked into another topic (usually later in the book, but
not always), where they become part of the development of another topic or
problem.
The author is a PhD candidate in pure mathematics at Harvard Univer-
sity (at the time the book was written). Being still very young, he knows how
to speak to today's teenagers. His sense of humour and general puckishness
is present throughout: just when you are lulled into some serious compu-
tation in probability, he deviously throws a trick question at you, that has
a totally non-obvious answer (non-obvious, that is, until you CAREFULLY
re-read the question).
Many mathematics books published today include short biographies on
famous mathematicians through history, especially those whose names come
up in the theory developed in the book. This book is no exception. But what
is unique about this book is the inclusion of Personal Pro les of young math-
ematicians from several countries that he has met at International Mathe-
matical Olympiads (IMOs) in the past. The pro les are quite diverse, which
 335

means that most bright students could nd one to identify with and to use as
a role model. The author and those he pro les have taken a risk in doing this:
they have tried to predict some of the important mathematicians in the early
part of the next century. It should be interesting to follow their careers and
see if those predictions can come true, or if by placing them in the spotlight,
they nd too much pressure to deal with.
The problems range from puzzles that elementary school children can do
to problems that provide training for Putnam candidates (toward the end of
the book). There are many cross-references and connections between seem-
ingly unrelated problems from di erent areas of mathematics, connections
that most students would be unable to make. Many of these connections
are new to this reviewer. However, once made these connections are quite
clear.
As for his credentials, Ravi Vakil placed among the top ve in the Put-
nam competition in all four of his undergraduate years at the University of
Toronto. Before that he won two gold medals and a silver medal in IMOs
and coached the Canadian team to the IMO from 1989 to 1995.

Tests for Divisibility

Some tests for divisibility are well known, such as the test for divisibil-
ity by 3: nd the sum of the digits of the whole number n | if that sum is
divisible by 3, then the original whole number n is divisible by 3.
Many people know this test, but do not know why it works. So, why
does it work?
The basic principle is that the remainder obtained when 10k is divided
by 3, is 1. This is easy to check, since 10k = 3  3  10k,1 + 10k,1 .
This enables one to step down one power at a time. So, if d is a digit, the
\remainder" obtained when d  10k is divided by 3 is d. (NB: this is not the
true remainder one gets when dividing by 3!)
If n = abc : : : def is a whole number, then we write it as
n = a  10k + b  10k,1 + c  10k,2 + : : : + d  100 + e  10 + f:
When we divide by 3, we get a \remainder" of a + b + c + : : : + d + e + f .
If this is divisible by 3, the the true remainder is 0, and so n is divisible
by 3.
Now, do you know, or can you construct, a test for divisibility by 7?
336

Packing Boxes with N {tetracubes

Andris Cibulis
Riga, Latvia
Introduction
With the popularity of the video game Tetris, most people are aware of
the ve connected shapes formed by four unit squares joined edge to edge.
They are called the I {, L{, N {, O{ and T {tetrominoes, after the letter of the
alphabet whose shapes they resemble. They form a subclass of the polyomi-
noes, a favourite topic in research and recreational mathematics founded by
Solomon Golomb.
Here is a problem from his classic treatise, Polyominoes. Is it possible
to tile a rectangle with copies of a particular tetromino? Figure 1 shows that
the answer is armative for four of the tetrominoes but negative for the
N {tetromino, which cannot even ll up one side of a rectangle.

Figure 1
Getting o the plane into space, we can join unit cubes face to face to
form polycubes. By adding unit thickness to the tetrominoes, we get ve
tetracubes, but there are three others. They are shown in Figure 2, along
with the N {tetracube.
Is it possible to pack a rectangular block, or box, with copies of a par-
ticular tetracube? The answer is obviously armative for the I {, L{, O{
and T {tetracubes, and it is easy to see that two copies of each of the three
tetracubes not derived from tetrominoes can pack a 2  2  2 box. Will the
N {tetracube be left out once again? Build as many copies of it as possible
and experiment with them.
If the k  m  n box can be packed with the N {tetracube, we call it an
N {box. Are there any such boxes? Certain types may be dismissed immedi-
ately.
 337

PPPP PPPPPP
PPPPPPP
PPP PP PPPPPPPPPPP
PP
PPP PPPPPP
PPP PPPPPPP
PPPPPPP
 PPP
PPP  PPPP P
P
PPPP PPPPP
PPP PPPP 
P
Figure 2
Observation 1.
The k  m  n box cannot be an N {box if it satis es at least one of the
following conditions:
(a) one of k; m and n is equal to 1;
(b) two of k; m and n are equal to 2;
(c) kmn is not divisible by 4.
It follows that the 2  3  4 box is the smallest box which may be an
N {box. Figure 3 shows that this is in fact the case. The box is drawn in two
layers, and two dominoes with identical labels form a single N {tetracube.

0 0
1 2
3 2 1 3
Figure 3
So there is life in this universe after all! The main problem is to nd all
N {boxes.
N {cubes
If k = m = n, the k  m  n box is called the k{cube, and a cube
which can be packed by the N {tetracube is called an N {cube. We can easily
assemble the 12{cube with the 2  3  4 N {box, which makes it an N {cube.
This is a special case of the following result.
338

Observation 2.
Suppose the k  m  n and `  m  n boxes are N {boxes. Let a, b and c be
any positive integers. Then the following are also N {boxes:
(a) (k + `)  m  n;
(b) ak  bm  cn.
The 12{cube is not the smallest N {cube. By Observation 1, the rst
candidate is k = 4. It turns out that this is indeed an N {cube. It can
be assembled from the 2  4  4 N {box, whose construction is shown in
Figure 4.

0 0
1 2
1 3 2 3
Figure 4
The next candidate, the 6{cube, is also an N {cube, but a packing is not
that easy to nd. In Figure 5, we begin with a packing of a 2  6  6 box,
with a 1  2  4 box attached to it. To complete a packing of the 6{cube,
add a 2  3  4 N {box on top of the small box, ank it with two 2  4  4
N {boxes and nally add two more 2  3  4 N {boxes.

0 1 0 1 6 7
2 3 6 7
2 3 8 9
8 9
4 5 4 5
Figure 5
Can we pack the 8{cube, the 10{cube, or others? It would appear that
as size increases, it is more likely that we would have an N {cube. How-
ever, it is time to stop considering one case at a time. We present a recur-
sive construction which expands N {cubes into larger ones by adding certain
N {boxes.
Theorem 1.
The k{cube is an N {cube if and only if k is an even integer greater than 2.
Proof:
That this condition is necessary follows from Observation 1. We now
 339

prove that it is sucient by establishing the fact that if the k{cube is an N {

cube, then so is the (k + 4){cube. We can then start from either the 4{cube
or the 6{cube and assemble all others.
From the 2  3  4 and 2  4  4 N {boxes, we can assemble all 4  m  n
boxes for all even m; n  4, via Observation 2. By attaching appropriate
N {boxes from this collection, we can enlarge the k{cube rst to the
(k + 4)  k  k box, then the (k + 4)  (k + 4)  k box and nally the
(k + 4){cube. This completes the proof of Theorem 1.
Further Necessary Conditions
Observation 1 contains some trivial necessary conditions for a box to
be an N {box. We now prove two stronger results, one of which supersedes
(c) in Observation 1.
Lemma 1.
The k  m  n box is not an N {box if at least two of k; m and n are odd.
Proof:
We may assume that m and n are odd. Place the box so that the hor-
izontal cross-section is an m  n rectangle. Label the layers L1 to Lk from
bottom to top. Colour the unit cubes in checkerboard fashion, so that in any
two which share a common face, one is black and the other is white. We may
assume that the unit cubes at the bottom corners are black. It follows that
Li has one more black unit cube than white if i is odd, and one more white
unit cube than black if i is even.
Suppose to the contrary that we have a packing of the box. We will call
an N {tetracube vertical if it intersects three layers. Note that the intersec-
tion of a layer with any N {tetracube which is not vertical consists of two or
four unit cubes, with an equal number in black and white. The intersection
of a vertical N {tetracube with its middle layer consists of one unit cube of
each colour.
Since L1 has a surplus of one black unit cube, it must intersect `1 ver-
tical N {tetracubes in white and `1 + 1 vertical N {tetracubes in black, for
some non-negative integer `1 . These N {tetracubes intersect L3 in `1 black
unit cubes and `1 + 1 white ones. Hence the remaining part of L3 has a
surplus of two black. They can only be packed with `3 vertical N {tetracubes
intersecting L3 in white, and `3 + 2 vertical N {tetracubes in black, for some
non-negative integer `3. However, the surplus in black unit cubes in L5 is
now three, and this surplus must continue to grow. Thus the k  m  n
box cannot be packed with the N {tetracube. This completes the proof of
Lemma 1.
Lemma 2.
The k  m  n box is not an N {box if kmn is not divisible by 8.
340

Proof:
Suppose a k  m  n box is an N {box. In view of Lemma 1, we may
assume that at least two of k; m and n are even. Place the packed box so
that the horizontal cross-section is an m  n rectangle, and label the layers
L1 to Lk from bottom to top. De ne vertical N {tetracubes as in Lemma 1
and denote by ti the total number of those which intersects Li ; Li+1 and
Li+2; 1  i  k , 2.
Since at least one of m and n is even, each layer has an even number
of unit cubes. It follows easily that each ti must be even, so that the total
number of vertical N {tetracubes is also even. The same conclusion can be
reached if we place the box in either of the other two non-equivalent orien-
tations. Hence the total number of N {tetracubes must be even, and kmn
must be divisible by 8. This completes the proof of Lemma 2.
N {Boxes of Height 2
We now consider 2  m  n boxes. By Observation 1, m  3 and
n  3. By Lemma 2, mn is divisible by 4. We may assume that m is even.
First let m = 4. We already know that the 2  4  3 and 2  4  4 boxes are
N {boxes. Figure 6 shows that so is the 2  4  5 box. If the 2  4  n box
is an N {box, then so is the 2  4  (n + 3) box by Observation 2. It follows
that the 2  4  n box is an N {box for all n  3.

0 0
1 2
3 2 1 3
Figure 6
Now let m = 6. Then n is even. We already know that the 2  6  4 box
is an N {box. However, the 2  6  6 box is not. Our proof consists of a long
case-analysis, and we omit the details. On the other hand, the 2  6  10
box is an N {box. In Figure 7, we begin with the packing of a 2  3  6 box
with a 2  2  3 box attached to it. We then build the mirror image of this
solid and complete the packing of the 2  6  10 box by adding a 2  4  3
N {box.
If the 2  6  n box is an N {box, then so is the 2  6  (n + 4) box by
Observation 2. It follows that the 2  6  n box is an N {box for n = 4 and
all even n  8.
Theorem 2.
The 2  m  n box is an N {box if and only if m  3; n  3 and mn is
divisible by 4, except for the 2  6  6 box.
 341

1 0 0
1
3 2
4 5 2 3 4 5
Figure 7
Proof:
If m is divisible by 4, the result follows immediately from Observa-
tion 2. Let m = 4` + 2 for some positive integer `. We already know that
the 2  10  6 box is an N {box. If the 2  (4` + 2)  n box is an N {box,
then so is the 2  (4` + 6)  n box by Observation 2. This completes the
proof of Theorem 2.
The Main Result
Theorem 3.
The k  m  n box, k  m  n, is an N {box if and only if it satis es all of
the following conditions:
(a) k  2;
(b) m  3;
(c) at least two of k, m and n are even;
(d) kmn is divisible by 8;
(e) (k;m; n) 6= (2; 6; 6).
Proof:
Necessity has already been established, and we deal with suciency.
We may assume that k  3, since we have taken care of N {boxes of height
2. Consider all 3  m  n boxes. By (c), both m and n must be even. By
(d), one of them is divisible by 4. All such boxes can be assembled from the
3  2  4 box.
Consider now the k  m  n box. We may assume that m and n are
even. If k is odd, then one of m and n is divisible by 4. Slice this box into
one 3  m  n box and a number of 2  m  n boxes. Since these are all
N {boxes, so is the k  m  n box.
Suppose k is even. Slice this box into a number of 2  m  n boxes,
each of which is an N {box unless m = n = 6. The 4  6  6 box may be
assembled from the 4  2  3 box, and we already know that the 6{cube is
an N {cube. If the k  6  6 box is an N {box, then so is the (k + 4)  6  6
box. This completes the proof of Theorem 3.
342

Research Projects
Problem 1.
Try to prove that the 2  6  6 box is not an N {box. It is unlikely that any
elegant solution exists.
Problem 2.
An N {box which cannot be assembled from smaller N {boxes is called a prime
N {box. Find all prime N {boxes.
Problem 3.
Prove or disprove that an N {box cannot be packed if we replace one of the
N {tetracubes by an O{tetracube.
Problem 4.
For each of the other seven tetracubes, nd all boxes which it can pack.

Bibliography
Bouwkamp, C. J. & Klarner, David, Packing a Box with Y {pentacubes, Jour-
nal of Recreational Mathematics, 3 (1970), 10-26.
Gobel, F. & Klarner, David, Packing Boxes with Congruent Figures, Indaga-
tiones Mathematicae, 31 (1969), 465-472.
Golomb, Solomon G., Polyominoes | Puzzles, Patterns, Problems & Pack-
ings, Princeton University Press, Princeton, 1994.
Klarner, David, A Search for N{pentacube Prime Boxes, Journal of Recre-
ational Mathematics, 12 (1979), 252-257.

Acknowledgement
This article has been published previously in a special edition of delta-k,
Mathematics for Gifted Students II, Vol. 33, 3 1996, a publication of the
Mathematics Council of the Alberta Teachers' Association and in AGATE,
Vol. 10, 1, 1996, the journal of the Gifted and Talented Education Council
of the Alberta Teachers' Association. It is reprinted with permission of the
author and delta-k.
 343

THE SKOLIAD CORNER

No. 24
R.E. Woodrow
This number we give the problems of the 1995 Concours Mathematique
du Quebec. This contest comes to us from the organizers via the Canadian
Mathematical Society which gives partial support to the contest. My thanks
go to Therese Ouellet, secretary to the contest. The contest was written by
over 2000 students February 2, 1997 over a three hour period. This will also
test your French! We will, of course, accept solutions in either English or
French.

CONCOURS MATHEMATIQUE 
DU QUEBEC 1995
February 2, 1995
Time: 3 hours
1. LA FRACTION SIMPLIFIE E
Simpli er la fraction
1 358 024 701 :
1 851 851 865
2. LA FORMULE MYSTERE

Considerons les e quations suivantes
xy = p; x + y = s; x1993 + y1993 = t; x1994 + y1994 = u:
En faisant appel aux lettres p, s, t, u mais pas aux lettres x, y , donnez une
formule pour la valeur
x1995 + y1995:
3. LA DIFFE RENCE ETONNANTE

Lorsque la circonference d'un cercle est divisee en dix parties e gales,
les cordes qui joignent les points de division successifs forment un decagone
regulier convexe. En joignant chaque point de division au troisieme suivant,
on obtient un decagone regulier e toile. Montrer que la di e rence entre les
c^otes de ces decagones est e gale au rayon du cercle.
344

4. LE TERRAIN EN FORME DE CERF-VOLANT

Abel Belgrillet est membre du Club des aerocervidophiles (amateurs de
cerfs-volants) du Quebec. Il dispose de quatre troncons de cl^oture rectilignes
AB, BC , CD, DA pour delimiter un terrain (en forme de cerf-volant, voir
gure) sur lequel il s'adonnera a son activite favorite cet e te. Sachant que les
troncons AB et DA mesurent 50 m chacun et que les troncons BC et DA
mesurent 120 m chacun, determiner la distance entre les points A et C qui
maximisera l'aire du terrain.
D
50 120
A C
50 120
B
5. L'INEGALIT
 E MODIFIE E D'AMOTH DIEUFUTUR
(a) (2 points) L'inegalite x2 + 2y 2  3xy est-elle vraie pour tous les
entiers?
(b) (8 points) Montrer que l'inegalite x2 + 2y 2  145 xy est valide pour
tous reels x et y .
6. L'ECHIQUIER
 COQUET
Trouver l'unique facon de colorier les 36 cases d'un e chiquier 6  6 en
noir et blanc de sorte que chacune des cases soit voisines d'un nombre impair
de cases noires.
(Note: deux cases sont voisines si elles se touchent par un c^ote ou par
un coin). On ne demande pas de demontrer que la solution est unique.
7. LA FRACTION D'ANNE GRUJOTE
En base 10, 13 = 0:333 : : : . Ecrivons
 0:3 pour ces decimles repetees.
Comment e crit-on 3 dans une base b, ou b est de la forme
1
(i) b = 3t (trois points),
(ii) b = 3t + 1 (trois points),
(iii) b = 3t + 2 (quatre points),
ou t est un entier positif quelconque?
 345

To nish this number of the Skoliad Corner we give the ocial solutions
to the 1995 P.E.I. Mathematics Competition given last issue [1997: 278].
My thanks go to Gordon MacDonald, University of Prince Edward Island for
forwarding the materials.
1995 P.E.I. Mathematics Competition
1. Find the area of the shaded region inside the circle in the following
gure.
61
S
,1 T2 O -1
T1

P
?,1
Solution. First determine the coordinates of the point P . Since P lies
on the line y = x, P = (a; a) for some value of a. Since P lies on the circle
x2 + y2 = 1, a2 + a2 = 1 and so a = p,12 . Hence P = ( p,12 ; p,12 ). Partition
the shaded region as shown. Then the area of the quarter circle S is 4 and
the area of the triangle T1 is 12 (base)(height). The base is 1 and the height
is p12 . (Turn the page upside down.) So the area of T1 is 2p1 2 . The triangle
T2 has the same area so the total area of the shaded region is
 + p1 :
4 2
2. \I will be n years old in the year n2", said Bob in the year 1995.
How old is Bob?
Solution. Assuming Bob's age is an integer, Bob must have a reasonable
chance to be alive in a year that is a perfect square. Since 442 = 1936,
452 = 2025 and 462 = 2116, Bob must be 45 years old in the year 2025.
Hence Bob is now 15 years old (or 14 years old if he hasn't had his birthday
yet this year).
3. Draw the set of points (x; y) in the plane which satisfy the equation
jxj + jx , yj = 4.
346

Solution. Consider the following four possible cases:

Case 1: When x  0 and x  y . Then x + x , y = 4 so 2x , y = 4.
Case 2: When x  0 and x < y . Then x , (x , y ) = 4 so y = 4.
Case 3: When x < 0 and x  y . Then ,x + x , y = 4 so y = ,4.
Case 4: When x < 0 and x < y . Then ,x , (x , y ) = 4 so ,2x + y = 4.
Thus the set of points is made up of four line segments. When these
line segments are drawn we obtain the following parallelogram:
{axis
6r
y

4 r

d - {axis
,4 ,2 2 4
x

,2
r r
,4
4. An autobiographical number is a natural number with ten digits
or less in which the rst digit of the number (reading from left to right)
tells you how many zeros are in the number, the second digit tells you how
many 1's, the third digit tells you how many 2's, and so on. For example,
6; 210; 001; 000 is an autobiographical number. Find the smallest autobio-
graphical number and prove that it is the smallest.
Solution. When you add the digits of an autobiographical number, you
count the total number of digits. (For example the digits of the above 10 digit
autobiographical number must sum to 10.) Using this fact and the process of
elimination we can nd the smallest autobiographical number.
The only possible one-digit number whose digits sum to 1 is 1 and it is
not autobiographical so there are no one digit autobiographical numbers.
The only possible two digit numbers whose digits sum to two are (in
increasing order) 11 and 20 and they are not autobiographical so there are
no two digit autobiographical numbers.
The only possible three digit numbers whose digits sum to three are (in
increasing order) 102; 111; 120; 201; 210 and 300 and they are not autobio-
graphical so there are no three digit autobiographical numbers.
The possible four digit numbers whose digits sum to four are (in in-
creasing order) 1003; 1012; 1021; 1030; 1102; 1111; 1120; 1201; : : : .
 347

Checking these numbers we nd that the rst autobiographical number

in this list, and hence the smallest autobiographical number is 1201.
5. A solid cube of radium is oating in deep space. Each edge of the
cube is exactly 1 kilometre in length. An astronaut is protected from its
radiation if she remains at least 1 kilometre from the nearest speck of radium.
Including the interior of the cube, what is the volume (in cubic kilometres) of
space that is forbidden to the astronaut? (You may assume that the volume
of a sphere of radius r is 43 r3 and the volume of a right circular cylinder of
radius r and height h is r2 h.)
Solution. Along with the cube of radium, on each of the six faces of
the cube there will be a cube with sides of length 1 km of forbidden space.
Along each of the 12 edges of the cube of radium there will be a quarter of
a cylinder of radius 1 km and length 1 km of forbidden space, and at each of
the eight corners of the cube of radium there will be an eighth of a sphere of
radius 1 km of forbidden space.
Putting it all together, there are 7 cubes (with sides of length 1 km),
3 cylinders of radius 1 km and length 1 km, and 1 sphere of radius 1 km of
forbidden space. So the volume of forbidden space is:
7 + 3 + 34  km3 = 7 + 16
3  km3 :

6. Which is greater, 999! or 500999? (Where 999! denotes 999 factorial,

the product of all the natural numbers from 1 to 999 inclusive.) Explain your
reasoning.
Solution. Write
999! = 1
2
3
4


995
996
997
998
999
and successively pair o numbers from the left with numbers from the right
so
999! = (1
999)(2
998)(3
997)(4
996)


(449
501)(500):
(Note that no number is paired with 500 and there are 499 such pairs.)
Now use the fact that (500 , k)(500+ k) = 5002 , k2 < 5002 for each
value of k from 1 to 499 so
999! < (5002)(5002)(5002)(5002)


(5002)(500)
where 5002 is repeated 499 times. Hence
999! < (5002)499(500) = 5002(499)+1 = 500999;
so 500999 is greater.
348

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic
address is still
mayhem@math.toronto.edu
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai
Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Invariants of Inscribed Regular n-gons
There are two unexpected (in their simultaneity) invariants of inscribed
regular n-gons.
Let P1P2


Pn be a regular n-gon, and let P be a point on its circum-
circle. Then both
PP12 + PP22 +


+ PPn2 and PP14 + PP24 +


+ PPn4
are independent of the position of P .
Proof. We will show the result using complex numbers. Without loss
of generality, assume the radius of the circumcircle is 1, and that Pk is at the
complex number ! k,1 in the complex plane, where ! = cis( 2n ). Let z be
the complex number corresponding to P , so jz j = 1, and zz = jz j2 = 1.
Note
PPk2+1 = jz , !kj2 = (z , !k)(z , !k)
= zz , !kz , ! kz + ! k ! k = 2 , ! kz , ! n,k z;
k2
since ! k = j!!kj = !1k = !!nk = ! n,k . Hence,
,1
nX ,1
nX
PP12 + PP22 +


+ PPn2 = jz , !kj2 = (2 , ! kz , ! n,k z )
k=0 k=0
,1
nX ! ,1
nX !
= 2n , !k z, !n,k z
k=0 k=0
= 2n;
 349

since ,1
nX
!k = 11,,!! = 0:
n
k=0
Therefore, the sum is indeed independent of z . Similarly,
PPk4+1 = (2 , !kz , !n,kz)2
= 4 + ! 2kz2 + ! 2(n,k)z 2 , 4!kz , 4! n,k z + 2
= 6 + ! 2kz2 + ! 2(n,k)z 2 , 4!kz , 4! n,k z;
so
PP14 + PP24 +


+ PPn4
,1
nX
= (6 + ! 2k z 2 + ! 2(n,k)z 2 , 4! kz , 4! n,k z )
k=0
,1
nX ! ,1
nX !
= 6n + ! 2k z 2+ ! 2(n,k) z2
k=0 k=0
,1
nX ! ,1
nX !
,4 !k z , 4 !n,k z = 6n:
k=0 k=0

Four Large Spheres and a Small Sphere

Wai Ling Yee
Problem: Given four identical spheres of radius R which are mutually
tangent, nd the radius r of the sphere that will t in the hole at the centre
of the tetrahedron that they form, in terms of R.
When I rst tried this problem, I looked at several triangles and, after
many calculations, produced an answer. I came across this problem again in
chemistry when calculating the size of tetrahedral holes in ion lattices. Try
this:
Construct a cube so that the centres of the spheres are at alternate cor-
ners of the cube. The length of a face diagonalpof the cube is 2R. Therefore,
the length of a body diagonal of the cube is 6R. Note that we can also
express the length of the body diagonal as 2(R + r). Hence,
p p
2(R + r) = 6R implies that r = 62, 2 R:
Rider. This is the radius of the sphere that is externally tangent to all
four. What is the radius of the sphere that is internally tangent to all four?
350

IMO REPORT
Adrian Chan
Two weeks of training at St. Mary's University in Halifax in early July
kicked o the month-long journey the 1997 Canadian IMO team would en-
dure. Four ights later, the team found itself in Mar del Plata, Argentina,
ready to participate in the 37th International Mathematical Olympiad.
This year's team members included: Adrian \Terrible Taco" Birka, Sabin
\Cursed Corner" Cautis, Adrian \Flash Flood" Chan, Jimmy \Fruits and Veg-
gies" Chui, Byung Kyu \Argentinian Polite" Chun, and Mihaela \My Dear-
est" Enachescu. We gave our team leader Dr. Richard \Richard" Nowakowski
and our deputy leader Naoki \Raspberry" Sato all they could handle. Spe-
cial thanks goes out to our coaches Dr. Bruce Shawyer and leader observer
Dr. Chris \Bob's Your Uncle" Small.
This year's contest was not as dicult as last year's killer in India, but
still challenging. It was the largest IMO ever, with a record 82 countries
competing. Canada was well represented, bringing back 2 silver, 2 bronze,
and an honourable mention. Our team's scores were as follows:
CAN1 Adrian Birka 6
CAN2 Sabin Cautis 16 Bronze Medal
CAN3 Adrian Chan 25 Silver Medal
CAN4 Jimmy Chui 10 Honourable Mention
CAN5 Byung Kyu Chun 29 Silver Medal
CAN6 Mihaela Enachescu 21 Bronze Medal
As a team, Canada nished 29 out of the 82 countries. Best of luck
th

to the graduating members of the team, Sabin and Byung, as they take their
math skills to the University of Waterloo next year. The remaining four mem-
bers of the team are eligible to return to Taiwan for next year's IMO. How-
ever, it seems that the team should focus more in the area of geometry!
Special thanks must also go out to Dr. Graham Wright of the Canadian
Mathematical Society for paying all our bills, and Professor Ed Barbeau of
the University of Toronto for conducting a year-long correspondence program
for all IMO hopefuls.
Visiting Argentina was a new experience for many of us, and was def-
initely a memorable one to all. The food was great, especially the beef and
the chocolate alfajores, and the IMO was very well-organized and run. Best
of luck to all IMO hopefuls who are working hard to be on the 1998 IMO
team, which will compete in Taipei, Taiwan.
 351

Unique Forms
Naoki Sato
For the mathematician, there is something re-assuring about being able
to put an object into a unique form, or a canonical form, such as the prime
factorization of a positive integer, or a matrix in Jordan Canonical Form. Such
forms make working with these objects much easier and more tractable, as
well as allowing a way to get a handle on them. Here, we explore the idea
of the unique, canonical form.
1. Let p(x) = an xn + an,1 xn,1 +


+ a1 x + a0 , and let c be a real
number. Show that there exist unique constants bn , bn,1 , : : : , b1, b0,
such that
p(x) = bn(x , c)n + bn,1(x , c)n,1 +


+ b1(x , c) + b0:
Can you nd the bi explicitly?
What may seem like a lot of (linear) algebra to untangle, comes easily
undone with one idea: consider p(x + c). This is a polynomial of degree at
most n, so there exist unique bi such that
p(x + c) = bnxn + bn,1xn,1 +


+ b1x + b0;
which is equivalent to
p(x) = bn(x , c)n + bn,1(x , c)n,1 +


+ b1(x , c) + b0:
The expression may look familiar, because it is the n approximant to a
th

power series around x = c, a central object of calculus. Suppose we take the

k derivative of both sides. The \constant term" is then k!bk, so evaluating
th

both sides at x = c leads to

(k)
p(k)(c) = k!bk so that bk = p k!(c) :
A quick corollary: If p(x) is a polynomial, then (x , c)k j p(x) if and
only if p(c) = p0 (c) = p00 (c) =


= p(k,1) (c) = 0. (Both conditions are
equivalent to b0 = b1 =


= bk,1 = 0.)
2. Let N be a non-zero integer. Show that there exist unique positive
integers k and a1 < a2 <


< ak such that
N = (,2)ak + (,2)ak,1 +


+ (,2)a1 :
(In other words, show that there is a unique representation of N in
base ,2.)
352

We will use strong induction. The statement is true for N = 1 =

(,2)0 and for N = ,1 = (,2)1 + (,2)0. Now, assume there is a positive
integer n such that the statement is true for N = 1; 2; : : : ; n and N =
,1; ,2; : : : ; ,n.
Let N = n + 1. If N is odd, then n is even, and n=(,2) is an inte-
ger between ,1 and ,n, so by the induction hypothesis, for some unique
distinct ai ,
n=(,2) = (,2)ak + (,2)ak,1 +


+ (,2)a1 ;
which is equivalent to
N = n + 1 = (,2)ak+1 + (,2)ak,1+1 +


+ (,2)a1+1 + (,2)0:
Note that the exponents are still distinct. The same trick works if N is even,
and for N = ,(n + 1). Subtracting o 1 = (,2)0 if necessary, and then
dividing by ,2, we obtain a form which is simultaneously seen to exist and
be unique, by the induction hypothesis. Hence, by strong induction, the
statement is proved for all N . Note that this also provides an algorithm for
nding the exponents.
One may be tempted to isolate the higher exponents, but this tends
to get very messy. Instead, one can start with the lower exponents - after
seeing if (,2)0 is in the sum, they can then be picked o inductively.
3. Let r be a positive rational number. Show that there exist unique pos-
itive integers k and a1  a2 


 ak such that
r = a1 + a 1a + a a1 a +


+ a a a1


a :
1 1 2 1 2 3 1 2 3 k
Note that
r = a1 + a 1a + a a1 a +


+ a a a1


a ;
1 1 2 1 2 3 1 2 3 k
which is equivalent to
a1 r , 1 = a1 + a 1a +


+ a a 1


a :
2 2 3 2 3 k
This suggests nding the right a1 , reducing, and applying the same argument
to a1 r , 1.
De ne the sequences fai g and frig as follows: set r1 = r, ai = d1=ri e,
ri+1 = airi , 1, i  1, and we terminate when ri+1 = 0. For example, if
r1 = 3=7, then we obtain a1 = d7=3e = 3, r2 = 3
3=7 , 1 = 2=7, a2 = 4,
r2 = 1=7, a3 = 7, and r3 = 0. And we see
3=1+ 1 + 1 :
7 3 3
4 3
4
7
 353

We will show that in general, this is the only possible sequence which works.
Consider the choice of a1 . Since a1 r  1, a1 must be at least d1=re.
However, if a1 was set to d1=re + 1, then
r = a1 + a 1a + a a1 a +


+ a a a1


a
1 1 2 1 2 3 1 2 3 k
1 1 1
< a + a2 + a3 +



1 1 1
= a , 1 = d11=re  11=r
1
1
= r;
contradiction. Hence, there is only one choice for a1 , the one we have chosen
above. By the above reduction, there is only one choice for a2 , a3 , and so
on. Hence, we have uniqueness.
We also know terms will be generated, but how do we know the se-
quence always terminates? Let ri = pi=qi, as a reduced fraction. Then
ai = dqi=pie, so
l m
p i q
 
i pi pqii , qi
ri+1 = q p , 1 = qi :
i i
However,
q i
  
q i

pi p , qi < pi p + 1 , qi = pi:
i i
Hence, the numerators of the ri are strictly decreasing. But as they are all
positive integers, the sequence must terminate at some point.
Finally, we must show the sequence fai g is non-decreasing. We have
2 3

ai+1 = 66 l
qmi 7  l
qmi > pqi > pqi , 1 = ai , 1:
 

6 pi qi ,q 7
i7 pi qi , qi i i
pi pi
Therefore, ai+1  ai .
Rider. Does an analogous
p result hold for positive irrational numbers?
What is the sequence for 2? For e?
4. Let be a root of x3 , x , 1 = 0, and let p, q be polynomials with
rational coecients, such that q ( ) 6= 0. Show that there exist unique
rationals a, b, and c such that
p( ) = a 2 + b + c:
q( )
354

First, let us express p( )=q ( ) as a polynomial in . Note that x3 ,x,1

is irreducible over the rationals. Hence, there exist polynomials u(x) and
v(x) with rational coecients such that
u(x)q(x) + v(x)(x3 , x , 1) = gcd(q(x);x3 , x , 1) = 1:
Then by substituting x = ,
u( )q( ) = 1 implies that pq(( )) = p( )u( ) = r( );
where r(x) = p(x)u(x), a polynomial with rational coecients.
Now, by the division algorithm for polynomials,
r(x) = (x3 , x , 1)w(x) + ax2 + bx + c;
for some rationals a, b, and c (dividing r(x) by x3 , x , 1 leaves, at most, a
quadratic remainder). Hence, r( ) = a 2 + b + c, which shows existence.
To see uniqueness, suppose r( ) = a0 2 + b0 + c0 implies that
(a , a0 ) 2 + (b , b0 ) + (c , c0 ) = 0; that is, is a root of the quadratic
(a , a0 )x2 + (b , b0 )x + (c , c0 ) = 0. But as indicated above, the cubic
x3 , x , 1 is irreducible; hence, all the coecients of the quadratic must be
zero, and a = a0 , b = b0 , c = c0 , showing uniqueness.
5. Let a and b be distinct positive integers, and let d = gcd(a; b). Con-
sider the Diophantine equation ax + by = d. It is well-known that
all solutions are given by (x; y ) = (u + bt=d; v , at=d), where t is an
integer and (u; v ) is some solution.
(i) Show that there exists a unique solution satisfying jxj  b=2d,
jyj  a=2d.
(ii) Show that the solution in (i) is given by the Euclidean Algorithm.
For example, for a = 4 and b = 7, we have
7 = 1
4 + 3;
4 = 1
3 + 1;
3 = 1
3;
so gcd(4,7) = 1, and
1 = 4 , 3 = 4 , (7 , 4) = 2
4 , 7:
Hence, the Euclidean Algorithm gives the solution (x; y ) = (2; ,1),
which satis es jxj  7=2 and jy j  4=2,
We can assume d = 1, as a way of normalizing the equation, since
ax + by = d is equivalent to (a=d)x + (b=d)y = 1. It is easy to check that
d may be factored back in.
 355

(i) Uniqueness is immediate, since the solutions in x form an arithmetic pro-

gression with common di erence b=d = b, and similarly for y , with di erence
a=d = a.
If b = 1, then (x; y ) = (0; 1) serves as a solution. If b = 2, then a is
odd, and (x; y ) = (1; (1 , a)=2) serves as a solution. Assume that b > 2.
Because the solutions in x form an arithmetic progression with di er-
ence b, there exists a solution (x; y ) with jxj  b=2, or ,b=2  x  b=2.
Then y = (1 , ax)=b, so
b
y  1 , ba
2 = 1b , a2 > , a2 ;
and b
y  1 + ba
2 = 1b + a2 < a +2 1 ;
so y  a=2 implies that jy j  a=2.
(ii) We must qualify the statement somewhat | if either a or b is equal to 1,
then the Euclidean Algorithm does not explicitly provide a solution. We may
stipulate now that the solution is (x; y ) = (1; 0) or (0,1), if a or b is 1,
respectively (they cannot both be 1, since they must be distinct).
We will use strong induction on max(a; b). The statement is easily ver-
i ed for max(a; b)  2. Assume the statement is true for max(a; b)  n,
for some positive integer n. We will prove the statement for
max(a; b) = n + 1. Assume b = n + 1.
We seek solutions of ax + (n + 1)y = 1. By the division algorithm,
there exist unique non-negative integers q and r such that n + 1 = aq + r,
0  r  a , 1. Then re-arranging, the equation becomes ax + (aq + r)y =
a(x + qy) + ry = 1. Since a, r  n, by the induction hypothesis, there exist
solutions satisfying jx + qy j  r=2, jy j  a=2. By the triangle inequality,
jxj  r=2+ jqyj  r=2+ qa=2 = (r + qa)=2 = (n +1)=2. Hence, by strong
induction, the statement is proved for all values of max(a; b).

1996 Balkan Mathematical Olympiad

1. Let O and G be the circumcentre and centroid of a triangle ABC re-
spectively. If R is the circumradius and r is the inradius of ABC , show
that p
OG  R(R , 2r) :
2. Let p > 5 be a prime number and X = fp , n2 jn 2 Z+; n2 < pg.
Prove that X contains two distinct elements x, y , such that x 6= 1 and
x divides y.
356

3. Let ABCDE be a convex pentagon. Denote by M , N , P , Q, R the

midpoints of the segments AB , BC , CD, DE , EA respectively. If
the segments AP , BQ, CR, DN have a common point of intersection,
prove that this point also belongs to the segment EN .
4. Show that there exists a subset A of the set f1; 2; 3; : : : ; 21996 , 1g
having the following properties:
(a) 1 2 A and 21996 , 1 2 A,
(b) Every element of Anf1g is the sum of two (not necessarily distinct)
elements of A, and
(c) The number of elements of A is at most 2012.

Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,
Cyrus Hsia Mayhem Advanced Problems Editor,
Ravi Vakil Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will nd only solutions | the
next issue will feature only problems.
We warmly welcome proposals for problems and solutions. With the
new schedule of eight issues per year, we request that solutions from the pre-
vious issue be submitted by 1 November 1997, for publication in the issue 5
months ahead; that is, issue 4 of 1998. We also request that only students
submit solutions (see editorial [1997: 30]), but we will consider particularly
elegant or insightful solutions from others. Since this rule is only being im-
plemented now, you will see solutions from many people in the next few
months, as we clear out the old problems from Mayhem.
 357

High School Problems | Solutions

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>
H211. Proposed by Richard Hoshino, grade OAC, University of Toronto
Schools
Let p, q , and r be the roots of the cubic equation ax3 , bx2 + cx + d = 0,
where a, b, c, and d are real coecients.
Given that Arctan p + Arctan q + Arctan r = 4 , where Arctan de-
notes the principal value, prove that a = b + c + d.
Solutionby Miguel Carrion 
 Alvarez, Universidad Compluteuse de Mad-
rid, Spain.
Let  = Arctan p,  = Arctan q , and = Arctan r. Then we have
 +  + = 4 and
+tan 
1 = tan( +  + ) = 1tan( ,tan  tan  + tan
 + ) + tan = 1tan
, tan( + ) tan 1 , 1tan +tan 
,tan  tan  tan
= 1tan  + tan  + tan , tan  tan  tan = p + q + r , pqr :
, tan  tan  , tan  tan , tan  tan 1 , pq , pr , qr
Finally, a(x,p)(x,q )(x,r) = a[x3 ,(p+q +r)x2 +(pq +pr +qr)x,pqr] =
ax3 , bx2 + cx + d. From the relationship between the coecients and the
roots of this cubic, we get p + q + r = ab , pq + pr + qr = ac , and pqr = , da .
Using these relations in the above equation, we have
b d
1 = 1a ,+ ca ) 1 , ac = ab + ad ) a = b + c + d :
a
Also solved by BOB PRIELIPP, University of Wisconsin-Oshkosh, WI,
USA. Prielipp also points out that this problem is similar to problem E2299
(pp. 520-521 of the May 1972 issue of The American Mathematical Monthly.)
H212. Given two sequences of real numbers of length n, with the
property that for any pair of integers (i;j ) with 1  i < j  n, the i term
th

is equal to the j term in at least one of the two sequences. Prove that at
th

least one of the sequences has all its values the same.
Solutionby Miguel Carrion 
 Alvarez, Universidad Compluteuse de Mad-
rid, Spain.
Let the two sequences be fai g and fbig. Assume that fai g does not
have all its values the same. Then, there is a pair (i; j ) such that ai 6= aj ,
and hence bi = bj . Now, it is impossible that both ak = ai and ak = aj .
Hence, either ak 6= ai or ak 6= aj . In either case, bk = bi = bj . This is true
for all k, so the sequence of bi 's has all its values the same.
358

H213. Two players begin with one counter each, initially on opposite
corners of an n  n chessboard. They take turns moving their counter to an
adjacent square. A player wins by being the rst to reach the row opposite
from their initial starting row, or landing on the opponent's counter. Who
has the winning strategy? (Generalize to an m  n board.)
Solutionby Miguel Carrion 
 Alvarez, Universidad Compluteuse de Mad-
rid, Spain.
On a square board, the rst player always wins by crossing the board
to the opposite row. In n , 2 moves,
t

?t
d
becomes ?d 
and the rst player wins.
If there are more columns than rows, the same strategy works for the
rst player.
If there are more rows than columns, one of the players wins by \cap-
turing" the opponent. If the parity of both dimensions of the board is the
same, both players start from squares of the same colour. In this case, the
rst player always moves to the opposite colour and cannot capture the sec-
ond player. The second player wins by \closing in" on the rst player and
capturing the rst player. [Ed: Why is this always possible?] Now if the
parity of both dimensions of the board is opposite, the rst player moves to-
wards the colour of the second player, and so can capture the second player
using the same strategy.

Advanced Problems | Solutions

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.
M1G 1C3 <hsia@math.toronto.edu>
A190. Find all positive integers n > 1 such that ab  ,1(mod n)
implies that a + b  0(mod n).
Solution byQ Wai Ling Yee, University of Waterloo, Waterloo, Ontario.
Let n = ki=1 pi i where the pi are distinct primes. By the Chinese
Remainder Theorem, the given property holds for n if and only if it holds for
each pi i ; that is, ab  ,1(mod pi i ), implies that a + b  0(mod pi i ).
Also, both a and b must be relatively prime with n.
 359

Let Si be the set of positive integers both relatively prime with and less
than pi i . Take x, u, and v 2 Si. Note that ux  vx mod pi i is equivalent
to x(u , v )  0(mod pi i ), which, in turn, is equivalent to u = v , since x is
relatively prime with pi i and u and v are less than pi i . Hence, each element
of fx; 2x; : : : ; (pi , 1)x; (pi + 1)x; : : : ; (pi i , 1)xg leaves a unique residue
modulo pi i . Also note that the product of two numbers in Si is itself in Si .
Therefore, the set fx; 2x; : : : ; (pi , 1)x; (pi + 1)x; : : : ; (pi i , 1)xg taken
modulo pi i is a permutation of Si. Since ,1 2 Si, for each x 2 Si, there
exists a unique y 2 Si such that xy  ,1(mod pi i ).
Case I: pi is odd.
Let a = 2. Then there exists a b such that 2b  ,1(mod pi i ).
Thus, 2 + b  0(mod pi i ), so that 4  1(mod pi i ). Therefore
3  0(mod pi i ), from which it follows that pi = 3, i = 1. Therefore,
the only possible odd prime factor of n is 3, and only one such factor
may appear. By a simple check, n = 3 satis es the given property.
Case II: pi = 2.
Let a = 3. Then there exists a b such that 3b  ,1(mod 2 i ). Thus
3 + b  0(mod 2 i ), so that 9  1(mod 2 i ). Therefore 2 i j 8, from
which it follows that i  3. Thus n can have 0, 1, 2, or 3 factors of 2,
and again by a simple check, all satisfy the given property.
From the above two cases, the solutions for n are 2, 3, 4, 6, 8, 12, and 24.
This problem, which generalizes problem B-1 on the 1969 Putnam Com-
petition, is equivalent to nding all n > 1, such that a relatively prime to n
implies that a2  1(mod n).
A191. Taken over all ordered partitions of n, show that
k1k2


km = m2+mn,,1 1 :
 
X

k1+k2 +


+km =n
Solution.
The problem is particularly suitable for generating functions. As is con-
vention, let [xn ]f (x) denote the coecient of xn in f (x). Consider each ki
being contributed from a factor of x + 2x2 + 3x3 +


. Then the number
we seek is:
[xn ] |(x + 2x2 +


)(x + 2x2 {z+


)


(x + 2x2 +


)}
m factors
m
= [xn ] (1 ,x x)2m = [xn,m ] (1 ,1x)2m
n , m

2 m , 1  
2 m  
2
= [x ] 2m , 1 + 2m , 1 x + 2m , 1 x +


m + 1 
2



m + n , 1 
= 2m , 1 :
360

A192. Let ABC be a triangle, such that the Fermat point F lies in the
interior. Let u = AF , v = BF , w = CF . Derive the expression
2 2 2 p
u2 + v2 + w2 , uv , uw , vw = a + b2 + c , 2 3K:
Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.
Using the law of cosines on 4BCF , 4ACF , and 4ABF :
a2 = v2 + w2 , 2vw cos 120 = u2 + v2 + vw; (1)
2 2 2 
b = u + w , 2uw cos 120 = u + v + uw;2 2 (2)
c2 = u2 + v2 , 2uv cos 120 = u2 + v2 + uv: (3)
Adding the areas of 4BCF , 4ACF , and 4ABF :
K = 12 uv sin 120 + 21 uw sin120 + 21 vw sin120
p3
= 4 (uv + vw + uw): (4)
p
Therefore, u2 + v 2 +p w2 , uv , uw , vw = a2 +b22 +c2 , 2 3K , from
1 [(1) + (2) + (3)] , 2 3(4).
2
Since u2 + v 2 + w2 , uv , uw , vwp 0 for all real u, v , w, it follows
from this problem that a2 + b2 + c2  4 3K .

Challenge Board Problems | Solutions

Editor: Ravi Vakil, Department of Mathematics, Princeton Univer-
sity, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA
<vakil@math.princeton.edu>
C70. Prove that the group of automorphisms of the dodecahedron
is S5, the symmetric group on ve letters, and that the rotation group of
the dodecahedron (the subgroup of automorphisms preserving orientation)
is A5 .
Solution.
Assign a number to each of the dodecahedron's edges as shown in Fig-
ure 1. (Alternatively, think of it as painting each of the edges one of ve
di erent colours.) Notice that if an edge on a face F is numbered n, then
the edge from the opposite vertex in F (that isn't an edge of F ) has the same
number. Thus the entire numbering can be recovered from the numbering of
the edges of a simple face. Notice also that the edges around each face are
numbered 1 through 5 (in some order).
 361

1
3 4
2 5
3 4
5 4 3
1 2
1 5 2 1
2
4 3 5
2 5
3 3 4
5 2
4 3
1

Figure 1.
 
By observation, if a1 2b 3c d4 5e is an element of A5 , then there
is a unique face where the edges are numbered
 (clockwise) a-b-c-d-e. In
1 2 3 4 5
this way each element a b c d e of A5 induces a rotation of the
dodecahedron (by sending the face 1-2-3-4-5 to the face a-b-c-d-e, with edge
1 of the rst face sent to edge a of the second). Conversely, each rotation
induces an element of A5 . This set map is clearly a group homomorphism,
so the group of rotations of the dodecahedron is A5 .
Essentially the same argument shows that the group of automorphisms
of the dodecahedron (including re ections) is S5 .
Comments.
1. What are the rotation and automorphism groups of the icosahedron?
How about the other platonic solids?
2. The group A5 comes up often in higher mathematics. For example,
because of a simple property of A5 , there are quintic polynomials with
integer co-ecients whose roots can't be written in terms of radicals.
(This is not true of polynomials of degree less than 5. The proof involves
Galois theory.) Surprisingly, the description of A5 as the automorphism
group of the dodecahedron comes up in a variety of contexts.
C72. A nite group G acts on a nite set X transitively. (In other
words, for any x; y 2 X , there is a g 2 G with g
x = y .) Prove that there
is an element of G whose action on X has no xed points.
362

(The problem as stated is incorrect. We need to assume that X has

more than one element.)
Solution.
Fix any element x0 of X . The number of elements of G sending x0 to
y is independent of y 2 X : if y1 and y2 are any two elements of X , and
Si = fg 2 G j g
x0 = yig; i = 1; 2;
and z is an element of G sending y1 to y2 (z exists by the transitivity of
the group action), then zS1 = S2. In other words, the elements of S2 are
obtained from the elements of S1 by multiplication on the left by z . In par-
ticular, jS1 j = jS2j.
Let f (g;x) = 1 if g
x = x and 0 otherwise. Then by the previous
comment, for any xed x 2 X ,
g2G f (g;x) = 1 :
P

jGj jX j
The average number of xed points (over all elements of the group) is 1:
!
g2G f (g;x)
P
1 X X
f (g;x) =
X


jGj g2G x2X x2X jGj

=
X 1
x2X jX j
= 1:
One element of the group (the identity) has more than the average (as
jX j > 1), so there is an element with less than the average number (and
hence zero) xed points.
Comments.
1. How does this relate to problem 1 on the 1987 IMO?:
Let pn (k) be the number of permutations of the set f1; : : : ; ng, n  1,
which have exactly k xed points. Prove that
n
X
kpn(k) = n!:
k=0
(Hint: Let X be the set f1; : : : ; ng and G the group of permutations on
X , acting on X in the natural way.)
2. Can the result be salvaged if the group action is not transitive?
 363

PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-
partment of Mathematics and Statistics, Memorial University of Newfound-
land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-
companied by a solution, together with references and other insights which
are likely to be of help to the editor. When a submission is submitted with-
out a solution, the proposer must include sucient information on why a
solution is likely. An asterisk (?) after a number indicates that a problem
was submitted without a solution.
In particular, original problems are solicited. However, other inter-
esting 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 so-
lutions 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 April 1998. They may also
be sent by email to crux-editors@cms.math.ca. (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.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX,
either to the Editor-in-Chief's departmental FAX machine in St. John's, New-
foundland, or to the Canadian Mathematical Society's FAX machine in Ot-
tawa, Ontario. While we understand the reasons for solvers wishing to use
this method, we have found many problems with it. The major one is that
hand-written material is frequently transmitted very badly, and at times is
almost impossible to read clearly. We have therefore adopted the policy that
we will no longer accept submissions sent by FAX. We will, however, con-
tinue to accept submissions sent by email or regular mail. We do encourage
email. Thank you for your cooperation.
364

2263. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle, and the internal bisectors of \B, \C , meet AC ,
AB at D, E, respectively. Suppose that \BDE = 30.
Characterize 4ABC .
2264. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a right angled triangle with the right angle at A. Points D and
E are on sides AB and AC respectively, such that DEkBC . Points F and
G are the feet of the perpendiculars from D and E to BC respectively.
Let I , I1 , I2 , I3 be the incentres of 4ABC , 4ADE , 4BDF , 4CEG
respectively. Let P be the point such that I2 P kI1 I3 , and I3 P kI1 I2 .
Prove that the segment IP is bisected by the line BC .
2265. Proposed by Waldemar Pompe, student, University of War-
saw, Poland.
Given triangle ABC , let ABX and ACY be two variable triangles
constructed outwardly on sides AB and AC of 4ABC , such that the angles
\XAB and \Y AC are xed and \XBA + \Y CA = 180. Prove that all
the lines XY pass through a common point.
2266. Proposed by Waldemar Pompe, student, University of War-
saw, Poland.
BCLK is the square constructed outwardly on side BC of an acute
triangle ABC . Let CD be the altitude of 4ABC (with D on AB ), and let
H be the orthocentre of 4ABC . If the lines AK and CD meet at P , show
that
HP = AB :
PD CD
2267. Proposed by Clark Kimberling, University of Evansville, Evans-
ville, IN, USA and Peter Y , Ball State University, Muncie, IN, USA.
In the plane of 4ABC , let F be the Fermat point and F 0 its isogonal
conjugate.
Prove that the circles through F 0 centred at A, B and C meet pairwise
in the vertices of an equilateral triangle having centre F .
2268. Proposed by Juan-Bosco Romero Marquez, Universidad de
Valladolid, Valladolid, Spain.
Let x, y be real. Find all solutions of the equation
s
2xy + x2 + y 2 = pxy + x + y :
x+y 2 2
 365

2269. Proposed by Cristobal  S a nchez{Rubio, I.B. Penyagolosa,

Castellon,
 Spain.
Let OABC be a given parallelogram with \AOB = 2 (0; =2].
A. Prove that there is a square inscribable in OABC if and only if
OA  sin + cos
sin , cos  OB
and
sin , cos  OB
OA  sin + cos :
B. Let the area of the inscribed square be Ss and the area of the given
parallelogram be Sp . Prove that
2Ss = tan2
,
OA2 + OB2 , 2Sp
:
2270. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.
Given 4ABC with sides a, b, c, a circle, centre P and radius  inter-
sects sides BC , CA, AB in A1 and A2 , B1 and B2 , C1 and C2 respectively,
so that
A1A2 = B1B2 = C1C2 =   0:
a b c
Determine the locus of P .
2271. Proposed by F.R. Baudert, Waterkloof Ridge, South Africa.
A municipality charges householders per month for electricity used ac-
cording to the following scale:
rst 400 units | 4.5cj per unit;
next 1100 units | 6.1cj per unit;
thereafter | 5.9cj per unit.
If E is the total amount owing (in dollars) for n units of electricity used,
nd a closed form expression, E (n).
2272?. Proposed by no name on proposal { please identify!
Write r - s if there is an integer k satisfying r < k < s. Find, as a
function of n (n  2), the least positive integer k satisfying
k - k - k - : : : - k - k:
n n,1 n,2 2
[The proposer has not seen a proof, and has veri ed the conjectured
solutions for 2  n  600.]
366

2273. Proposed by Tim Cross, King Edward's School, Birmingham,

England.
Consider the sequence of positive integers: f1, 12, 123, 1 234, 12 345,
: : : g, where the next term is constructed by lengthening the previous term at
its right-hand end by appending the next positive integer. Note that this next
integer occupies only one place, with \carrying" occurring as in addition: thus
the ninth and tenth terms of the sequence are 123 456 789 and 1 234 567 900
respectively.
Determine which terms of the sequence are divisible by 7.
2274. Proposed by Vaclav Konecny, Ferris State University, Big
Rapids, Michigan, USA.
n Y
X m
A. Let m be a non-negative integer. Find a closed form for (k + j ).
k=1 j =0
n Y
X m
B. Let m 2 f1; 2; 3; 4g. Find a closed form for (k + j )2.
k=1 j =0
C?. Let m and j (j = 0; 1; : : : ; m) be non-negative integers. Prove or
n Y
X m mY
+1
disprove that (k + j ) j is divisible by (n + j ).
k=1 j =0 j =0

2275. Proposed by M. Perisastry, Vizianagaram, Andhra Pradesh,

India.
Let b > 0 and ba  ba for all a > 0. Prove that b = e.
 367

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.
1940. [1994: 108; 1995: 107, 205; 1996: 321; 1997: 170] Proposed
by Ji Chen, Ningbo University, China.
Show that if x; y;z > 0,
(xy + yz + zx) (x +1 y )2 + (y +1 z )2 + (z +1 x)2  94 :
 

Comment by Vedula N. Murty, Andhra University, Visakhapatnam, In-

dia.
In response to Marcin Kuczma's comment [1997: 170], I present the
details of my proof of the assertion:
bc(b , c)2(2a2 , bc) + ca(c , a)2(2b2 , ca) > 0
when
2a2 , bc < 0 < 2b2 , ca  2c2 , ab;
where a, b, c are the side lengths of a triangle satisfying 0 < a  b  c.
We have
a + b > c: (1)

So, 2 a2 + b2  (a + b)2 > c(a + b) implies that 2

bc , 2a2
2b , ca < 1; and
,

further that
bc bc , 2a2 < bc : (2)
ac 2b2 , ca ac
Thus
c , a  1 implies that (c , a)2  c , a  b ; and further that
c ,2 b (c ,2 b)2 c,b a
(c , a)  bc > bc bc , 2a : So we have


(c , b)2 ac ac 2b2 , ca
bc(b , c)2(2a2 , bc) + ca(c , a)2(2b2 , ca) > 0:
368

2158. [1996: 218] Proposed by P. Penning, Delft, the Netherlands.

Find the smallest integer in base eight for which the square root (also
in base eight) has 10 immediately following p the `decimal' point.
In base ten, the answer would be 199, with 199 = 14:10673 : : : .
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.
Let n be a positive integer with the given property. Let m be the integer
part of pn and n = m2 + p. The base 8 representation of a real number x
is denoted by (x)8. By hypothesis,
m + 18 = m + (0:10)8  pn < m + (0:11)8 = m + 64
9;

which is equivalent to m2 + 14 m + 641  n < m2 + 329 m + 64922 :

It is easy to see that these inequalities can be replaced by the following
stronger ones by taking into account that the expressions on both sides can-
not be integers:
m2 + m 4+ 1  n  m2 + 932m :
[For example, any integer greater than or equal to m2 + m4 + 641 must be
greater than or equal to m2 + m4+1 , since m is an integer.|Ed.] So
m + 1  p  9m :
4 32
The lower limit has to be less than the upper limit, hence m  8. For
8  m  10 the lower limit is greater than 2 and the upper limit less than
3 [so no integer value of p exists]. If m = 11 then p = 3; therefore the
smallest m is 11 and so the smallest n is m2 + p = 124 = (174)8, where
p
(174)8 = (13:10531 : : : )8:
Also solved by CHARLES ASHBACHER, Hiawatha, Iowa, USA; MIGUEL
ANGEL CABEZON  OCHOA, Logro~no, Spain; THEODORE CHRONIS, student,
Aristotle University of Thessaloniki, Greece; MIHAI CIPU, Instituteof Math-
ematics, Romanian Academy, Bucharest, Romania; GEORGI DEMIREV,
Varna, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse,
Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Ger-
many; ROBERT GERETSCHLAGER,  Bundesrealgymnasium, Graz, Austria;
SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD
I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul-
inengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic
High School, Beaverton, Oregon, USA; HEINZ-JURGEN  SEIFFERT, Berlin,
Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
The curious fact that the smallest solution for base 8 is 124 = 112 + 3
and for base 10 is 199 = 142 + 3 (both written in base 10) is no coincidence!
Geretschlager, Godin, and the proposer consider other bases as well, and
 369

show that the smallest (base 10) integer whose square root when written in
base b has 10 immediately following the \decimal" point is
8 
>
>
<
b , 1 2 + 1 if b is odd
 2
>
>
:
3b , 2 2 + 3 if b is even:
2
Janous lists all (base 10) positive integers < 10000 which satisfy the
condition of the problem, and gets 144 such integers, starting 124, 229, 329,
: : : . From this list one can nd, for example, that: the rst occurrence of
a consecutive pair is 1860; 1861 (also pointed out by Engelhaupt); the last
integer not part of a consecutive pair appears to be 3616; the rst consecutive
triple is 5644; 5645; 5646; and so on. Can anyone nd any general patterns
here for arbitrary bases?
Tsaoussoglou reports that the smallest integer whose square root in
base
p 10 has 100 immediately following the decimal point is 992 +20 = 9821
( 9821 = 99:100958p: : : ), and that the corresponding answer for base 8 is
712 + 18 = 5059 ( 5059 = (107:100657 : : : )8). What if we want 1000
following the decimal point? Or 1 followed by n zeros?

2159. [1996: 218] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.
Let the sequence fxn ; n  1g be given by
xn = n1 (1 + t + : : : + tn,1)
where t > 0 is an arbitrary real number.
Show that for all k, l  1, there exists an index m = m(k;l) such that
xk
xl  xm .
Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton,
Oregon, USA.
Since xn = 1 for all n if t = 1, any index m will suce. We distinguish
two other cases.
First, let t > 1. By the AM{GM inequality,
p
xn = 1 + t +
n

+ t > n 1
t
: : :
tn,1
n,1
p p
= n t1+


+(n,1) = n tn(n,1)=2
= t(n,1)=2:
This last quantity increases without bound as n ! 1, so there must be an
index m for which xk xl  xm for given indices k; l.
370

Second, let 0 < t < 1. Then

xn = 1 + t +
n

+ t < 1 + 1 +n


+ 1 = 1;
n,1

so xk xl < xk and xk xl < xl , and we may choose the index m to equal either
k or l.
Also solved by THEODORE CHRONIS, student, Aristotle University
of Thessaloniki, Greece; MIHAI CIPU, Institute of Mathematics, Romanian
Academy, Bucharest, Romania; GEORGI DEMIREV, Varna, and MITKO
KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; FLORIAN
HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos
Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Ed-
monton, Alberta; THOMAS C. LEONG, City College of City University of New

York, New York, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the
proposer.
Some solvers noted that since x1 = 1 for all t, m = 1 will work for all
k and l whenever t < 1. Cipu and the proposer found that m = k + l , 1
works when t > 1.

2160. [1996: 218] Proposed by Toshio Seimiya, Kawasaki, Japan.

4ABC is a triangle with \A < 90 . Let P be an interior point of ABC

such that \BAP = \ACP and \CAP = \ABP . Let M and N be the
incentres of 4ABP and 4ACP respectively, and let R1 be the circumradius
of 4AMN . Prove that
1 = 1 + 1 + 1 :
R1 AB AC AP
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.
Let a; b; c denote the sides BC; CA; AB and ; ; the angles of the
triangle. Then
\APB = 180 , \ABP , \BAP = 180 , \ABP , + \CAP = 180 ,
and by the cosine law in 4APB
c2 = AP 2 + BP 2 + 2AP
BP cos :
4ABP and 4CAP are by hypothesis similar, whence BP = c
AP
b and
c2 = AP
2
b2 (b + c + 2bc cos ):
2 2

By setting x2 = b2 + c2 + 2bc cos

AP = bcx and BP = c
bAP = cx :
2
 371

Let X be the point where the incircle of 4APB touches AP . Then

PX = AP + BP , AB = c
(b + c , x)
2 2x
c
(( b + c
= 2x
(b + c + x) ) 2 , x2 )

= x 2
bc
2 sin2
2
(b + c + x)
and since \XPM = 12 \APB = 90 , 2 we get
PM = sin PX = 2bc2 sin 2 :
2 x
(b + c + x)
Since 4NPM and 4CPA are similar, MN = b
AP PM , and by using the fact
that \MAN = 2 , we have
1 2 sin 2
R1 = MN
bc
= 2
2xbc2sin 2
b
x
(b+sin 2
c+x)
= b + bc
c+x
1 + 1 + 1 :
= AB AC AP
Also solved by MIGUEL ANGEL CABEZON  OCHOA, Logro~no, Spain;
RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; MARI A ASCENSION  LOPEZ

CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; D.J. SMEENK, Zalt-
bommel, the Netherlands; and the proposer.

2161. [1996: 219] Proposed by Juan-Bosco Romero Marquez, Uni-

versidad de Valladolid, Valladolid, Spain.
Evaluate 1
X 1
n=1 (2 n , 1)(3 n , 1) :
Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA
(modi ed slightly be the editor).
Let S denote the given summation. We show that
p
S = 2 ln2 , 32 ln 3 + 63  0:64527561:
372

Using partial fractions, we have

1
X

2 3 
S =
n=1 2n , 1 , 3n , 1
X1  2 2 
X1 1 3 
= , + ,
n=1 2n , 1 2n n=1 n 3n , 1
1 1  1
= 2 (,1)n,1 n1 , 3 1
X X
,
n=1 n=1 3n , 1 3n
X1 
1 , 1 :
= 2 ln 2 , 3 (1)
n=1 3n , 1 3n
Since the sequence of functions ffn(x)g, where fn(x) = x3n+1 , x3n+2 , is
uniformly convergent on [0; 1],
X1  1 1  = 1 , 1 + 1 , 1 + 1 , 1 +



,
n=1 3n , 1 3n 2 3 5 6 8 9
Z 1
= (x , x2 + x4 , x5 + x7 , x8 +


) dx
0
Z 1
= x(1 , x)(1 + x3 + x6 +


) dx
0
=
Z 1
x(1 , x) dx = Z 1 x dx
0 1 , x3 0 1 + x + x2
Z 1
x + 2 dx , 1 Z 1
1 1
=
0 1+x+x 2 2 0 x + 2
2 + 43 dx
, 1

=

1 ln(1 + x + x2 ) , 1
p2 tan,1  2xp+ 1  1
2 p 2 3 3 0
= 1 ln3 , 3 tan,1 p3 , tan,1 p1 


2 p3 3
= 1 ln3 , 3
 : (2)
2 3 6
p
Substituting (2) into (1), we nd S = 2 ln 2 , 32 ln 3 + 63 , as claimed.
Also solved by THEODORE CHRONIS, student, Aristotle University
of Thessaloniki, Greece; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO
KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria (jointly);
RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic
High School, Beaverton, OR, USA; MURRAY S. KLAMKIN, University of Al-

berta, Edmonton, Alberta; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and
the proposer. Two incorrect solutions were also received.
 373

About half of the solvers used the 0non{elementary approach of consid-

ering the Euler psi-function (x) = ,,((xx)) where ,(x) denotes the gamma
function. Both Hess and Sei ert pointed out that the evaluation of the more
general sum 1 n=1 (n+p)(n+f ) (p > ,1, f > ,1, p 6= f ) can be found in
P 1
Tables of Integrals, Series, and Products by I.S. Gradsteyn and I.M. Ryzhik
(Academic Press, 1994, 5th edition). Speci cally, the following formulas can
be found on p. 952 and p. 954, respectively:
X1  1 1  ; x > 0;
(x) = , , ,
n=1 n , 1 + x n
1 = , , 2 ln 2;
 

2 p
2 = , + 3 , 3 ln 3:
 

3 6 2
(Here, denotes Euler's constant.)
 
Since S = 1 n=1 n, 12 , n, 13 , one obtains easily
P 1 1 from the above
p3
formulas that S = 3 , 2 = 2 ln2 , 2 ln3 + 6 as in the solution
,2
,1
3
above.

2162. [1996: 219] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.
In 4ABC , the Cevian lines AD, BE , and CF concur at P . [XY Z ] is
the area of 4XY Z . Show that
[DEF ] = PD
PE
PF
2[ABC ] PA PB PC
I. Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain
(slightly shortened by the editor).
It is a standard exercise (for example, #13.5.6 in H.S.M. Coxeter, In-
troduction to Geometry) to show that
[DEF ] =  + 1
[ABC ] ( + 1)( + 1)( + 1)
with  =
AF BD CE
FB ,  = DC ,  = EA .
Since in our case, according to Ceva's Theorem,  = 1, all we have
to prove is that
PA
PB
PC = ( + 1)( + 1)( + 1):
PD PE PF
374

But it is true because

PA = [PCA] ; PB = [PAB] ;
PD [PDC ] PE [PEA]
PC = [PBC ] ;
PF [PFB ]
AF + 1 = [PAF ] + 1 = [PAB] ;
 + 1 = FB [PFB ] [PFB ]
 + 1 = BD [PBD] [PBC ]
DC + 1 = [PDC ] + 1 = [PDC ] ;
and
 + 1 = CE
EA + 1 = [PCE ] + 1 = [PCA] :
[PEA] [PEA]
II. Comment by Murray S. Klamkin, University of Alberta, Edmonton,
Alberta.
The corresponding more general problem for a simplex is given in [2]
and also with a slightly modi ed proof in [3]. It is shown that if V0 , V1 , : : : ,
Vn denote0 the0 n +1 vertices of a simplex S in n{dimensional Euclidean space
and if V0 , V1 , : : : , Vn0 denote the n + 1 vertices of an inscribed simplex S 0
such that the cevians Vi Vi0 are concurrent at a point P within S , then
VOL S 0 n01 : : : n
=
VOL S (1 , 0 )(1 , 1 ) : : : (1 , n )
:
Here the barycentric representation for P is P = 0V0 + 1 V1 + : : : + n Vn
where 0 + 1 + : : : + n = 1 and i  0. It is also shown that the volume
of S 0 is a maximum if P is the centroid.
Also PV 0 =PV = i =(1 , i ). For this and other metric properties of
concurrent cevians of a simplex see [1987: 274{275].
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-
ladolid, Spain; MIGUEL ANGEL CABEZON  OCHOA, Logro~no, Spain; GEORGI
DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of
Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gym-
nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,
Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER
JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN,
University of Alberta, Edmonton, Alberta; TOSHIO SEIMIYA, Kawasaki,
Japan; and the proposer.
Janous refers to [4, p. 342, item 1.3] for a closely related identity.
Bellot quotes Victor Thebault who reported an 1881 reference for the
exercise that began our featured solution. This exercise is often attributed
 375

to Routh (1891). For those who read Bulgarian, he also provides a reference
concerning interesting properties of a cevian triangle DEF : Hristo Lesov,
Projections of noteworthy points of the triangle following the respective ce-
vians (in Bulgarian). Matematyka & Informatyka, 5 (1994) 42{49. He fur-
ther mentions [1, p. 78] where a problem of Langendonck is solved: Given
a 4ABC , nd the probability of selecting a point P inside the triangle such
that it is possible to form a triangle whose sides equal the respective dis-
tances from P to the sides of the 4ABC . The probability turns out to be
[DEF ]=[ABC ].
[1] Heinrich Dorrie, Mathematische Miniaturen, F. Hirt, Breslau, 1943.
[2] M.S. Klamkin, A volume inequality for simplexes, Univ. Beograd Publ.
Elektrotehn. Fak. Ser. Mat. Fiz. No. 357{380 (1971) 3{4.
[3] M.S. Klamkin, International Mathematical Olympiads 1978{1985,
Math. Assoc. Amer., Washington, D.C., 1986, pp. 87{88.
[4] D.S. Mitrinovic et. al., Recent Advances in Geometric Inequalities,
Kluwer Academic Publishers, 1989.

2163. [1996: 219] Proposed by Theodore Chronis, student, Aristotle

University of Thessaloniki, Greece.
Prove that if n;m 2 N and n  m2  16, then 2n  nm.
I. Solution by Florian Herzig, student, Perchtoldsdorf, Austria.
Let x = pn  m  4. Then we rst prove that x2  2x . This inequality is
equivalent to
x  2 :
ln x ln 2
De ne f (x) as x=(ln x). Then
f 0(x) = ln(lnx x,)21
which is positive for all x  4. Moreover f (x) is continuous and di er-
entiable in the interval [4; 1) and f (4) = f (2). Hence x2  2x , and so,
nm  (x2)x  (2x)x = 2n.
II. Solution by D. Kipp Johnson, Valley Catholic High School, Beaver-
ton, Oregon.
We rst show that the inequality is true for any m  4 with the smallest
allowable n, namely n = m2, and then we induct on n.
For the rst step we let m  4 and n = m2 and consider the following
equivalent statements:
2m2  (m2)m () 2m2  22m log2 m
() m2  2m log2 m
() m=2  log2 m:
376

We establish this last statement by noticing that, for m  4, we have

1 1=2
m,1  2 ,1
1 + m 1, 1  21=2
log2 1 + m 1, 1  21
 

log2 m , log2 (m , 1)  12 :
This permits us to write the following m , 4 inequalities:
log2 5 , log2 4  12
log2 6 , log2 5  12
..
.
log2(m) , log2 (m , 1)  12 ;
which, when added, telescope to produce the desired log2 m  m=2.
For the second step we induct on n. We shall assume that for some
 m2  16m we have 2n  nm. Multiplying each
nn+1 side by 2 we obtain
2  2
n . It will suce to show that 2
nm  (n + 1)m , which is
equivalent to 2  (1 + 1=n)m. The following sequence of inequalities does
the trick:
s
m2 m m
2  pe  1 + m12  1 + m12  1 + n1 ;
  

and the proof is nished.


Also solved by SEFKET ARSLANAGIC,  University of Sarajevo, Sarajevo,
Bosnia and Herzegovina; MIHAI CIPU, Romanian Academy, Bucharest, Ro-
mania; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba
Tonka School of Mathematics, Rousse, Bulgaria; CHARLES R. DIMINNIE,
Angelo State University, San Angelo, TX, USA; RUSSELL EULER and JAWAD
SADEK, NW Missouri State University, Maryville, Missouri; SHAWN GODIN,
St. Joseph Scollard Hall, North Bay, Ontario; FABIAN MARTIN HERCE, stu-
dent, Universidad de La Rioja, Logro~no, Spain; RICHARD I. HESS, Ran-
cho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands
University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengymnasium,
Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmon-
ton, Alberta; ROBERT P. SEALY, Mount Allison University, Sackville, New

Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH,
 377

Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Lau-
rier University, Waterloo, Ontario; and the proposer. There was 1 incom-
plete solution.
Hess actually proves a somewhat stronger result by replacing the con-
dition n  m2  16 by the 2 conditions n  16 and n  m2.

2164. [1996: 273] Proposed by Toshio Seimiya, Kawasaki, Japan.

Let D be a point on the side BC of triangle ABC , and let E and F be
the incentres of triangles ABD and ACD respectively. Suppose that B , C ,
E, F are concyclic. Prove that
AD + BD = AB :
AD + CD AC
Solution by the Con Amore Problem Group, the Royal Danish School
of Educational Studies and Informatics, Copenhagen, Denmark.
The intersection points of the line EF with AB; AC , and AD will be
called G; H and K , respectively, and we put \B = 2y , and \C = 2z , so
\GBE = \EBD = y , and \DCF = \FCH = z . Since B; C; E; F are
concyclic, \GEB = \FCD = z , and \CFH = \EBD = y ; and so
\AGH = y + z = \AHG
giving
AG = AH: (1)
Since AE bisects \A in 4GAK , and AF bisects \A in 4KAH , we have
[using (1)]: KE KA KA KF KE KF
EG = AG = AH = FH , whence EG + 1 = FH + 1, or
KG = KH : (2)
EG FH
In 4ABD let the inradius, the altitude from D, and the area be r1 , h1 and
T1 , respectively; and in 4ACD let the corresponding quantities be r2; h2
and T2 . In 4AKG, and 4AKH let the altitude from K be k1 and k2 ,
respectively. Then hh12 = kk12 , whence
h1 = h2 (3)
k1 k2
and [using (2)]
k1 = KG = KH = k2 ; (4)
r1 EG FH r2
and further, (3) - (4) imply
h1 = h2 : (5)
r1 r2
378

Now
h1
AB = 2T1 = r1
(AB + BD + DA);
so
h1 = AB + BD + DA = 1 + BD + DA ; (6)
r1 AB AB
and similarly,
h2 = AC + CD + DA = 1 + CD + DA ; (7)
r2 AC AC
From (5) - (7) it easily follows that
AD + BD = AB :
AD + CD AC
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-
ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
RICHARD I. HESS, Rancho Palos Verdes, California, USA; CRISTOBAL 

SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon,  Spain; D.J. SMEENK, Zalt-
bommel, the Netherlands; and the proposer.
Bellot Rosado notes that this problem is a special case of problem 1206
[H. Demir, Mathematics Magazine, 58, no. 1, January 1985; solution by
V. D. Mascioni, Mathematics Magazine, 59, no. 1, February 1986]

2165. [1996: 273] Proposed by Hoe Teck Wee, student, Hwa Chong
Junior College, Singapore.
Given a triangle ABC , prove that there exists a unique pair of points
P and Q such that the triangles ABC , PQC and PBQ are directly similar;
that is, \ABC = \PQC = \PBQ and \BAC = \QPC = \BPQ,
and the three similar triangles have the same orientation. Find a Euclidean
construction for the points P and Q.
Solution by the Con Amore Problem Group, the Royal Danish School
of Educational Studies and Informatics, Copenhagen, Denmark.
We identify the Euclidean plane with the complex plane, and points
with complex numbers. We may assume without loss of generality that
B = 0: (1)
Suppose such a P and Q exist. Since 4ABC  4PQC , there is a
point (complex number) X such that
P = XA; (2)
and
Q = XC; (3)
 379

and besides:
P , Q = A: (4)
C ,Q C
Moreover, if (2) - (4) hold, then P; Q are as desired. Now (2) - (4) imply that
XA , XC = A ;
C , XC C
or
XA , XC = A , XA;
or
X = 2AA, C ; (5)
and then, by (2) - (3):
P = 2AA, C
2
(6)
and
Q = 2AAC
, C: (7)
On the other hand, if X; P; Q satisfy (5) - (7) then (2) - (4) hold. This proves
the existence and uniqueness of P and Q with the desired properties. To nd
a Euclidean construction for P and Q, de ne
R = 2A , C;
and note that
A = R +2 C ;
that is, R is the image of C by re ection in A. Also note that
Q = A;
C R
that is, triangles ABR and QBC are similar, and similarly oriented. Finally
A + Q = 1  AC + A = A2 = P
2 2 2A , C 2A , C
so P is the midpoint of AQ. All this gives us the following construction:
(A) Re ect C in A to get R.
(B) Construct Q such that A and Q are on opposite sides of the line
BC , \CBQ = \RBA, and \QCB = \ARB.
380

(C) Construct P as the midpoint of AQ.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,
Rancho Palos Verdes, California, USA; GOTTFRIED PERZ, Pestalozzigymna-
sium, Graz, Austria (two solutions); TOSHIO SEIMIYA, Kawasaki, Japan;
and the proposer.

2166. [1996: 273] Proposed by K.R.S. Sastry, Dodballapur, India.

In a right-angled triangle, establish the existence of a unique interior
point with the property that the line through the point perpendicular to any
side cuts o a triangle of the same area.
Solution by Toshio Seimiya, Kawasaki, Japan.
Let the right-angled triangle be ABC , with right angle at A. We may
assume without loss of generality that AB  AC . P is an interior point of
4ABC . The line through P perpendicular to BC meets BC; AC at D; E
respectively (if AB = AC , then E = A); the line through P perpendicular to
CA meets BC; CA at F; G respectively, and the line through P perpendicu-
lar to AB meets BC; AB at H; K respectively. We assume that the triangles
BHK; CDE; CFG have the same area. Since these triangles are similar,
they must be congruent. Since 4CDE  4CGF , we have CD = CG, so
that 4CDP  4CGP . Thus we have \PCD = \PCG. Therefore, CP is
the bisector of \C .
Let M be the intersection of AP with BC (if AB = AC; M = D).
Since 4BHK  4FCG, we have BH = FC , so that
BF = HC: (1)
Since PF kAB and PH kAC , we get
BF : BM = AP : AM = CH : CM (2)
From (1) and (2) we have BM = CM . Therefore, AM is the median of
4ABC . Thus P must be the intersection of the median AM with the bi-
sector of \C . It is interesting to note that the bisector of \B does not pass
through P if AB 6= AC .
Conversely, let P be the intersection of the median AM with the bi-
sector of \C . Draw the lines DE; FG;HK through P perpendicular to BC ,
CA, AB respectively, as shown in the gure. Since \PCD = \PCG, we
have 4PCD  4PCG, so that CD = CG. As 4CFG is 4CED we
get 4CFG  CED. Since PF kAB and PH kAC , we have BF : BM =
AP : AM = CH : CM . As BM = CM , we get BF = CH , so that
BH = FC . Because 4KBH is 4GFC , we have 4KBH  4GFC .
Since 4CDE; 4CGF; 4HKB are congruent, they have the same area.
A number of solvers used a coordinate approach. If A = (0; 0),
B = (0; c), C = (b; 0) and BC = a, then P = 2bb+2 a ; 2bbc+a .
 381

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; SAM BAETHGE, Nordheim, Texas, USA; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; CON AMORE PROBLEM GROUP, Royal Dan-
ish School of Educational Studies, Copenhagen, Denmark; C. DIXON, Royal
Grammar School, Newcastle upon Tyne, England; DAVID DOSTER, Choate
Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{
Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Per-
chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,
USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV 
KONECN  Y, Ferris State University, Big Rapids, Michigan, USA; D.J. SMEENK,
Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece;
and the proposer. There were two incorrect solutions.
The proposer comments: \I am unable to establish the existence of such
a point in a general triangle. If we replace \perpendicular" by \parallel" then
the problem is easy and yields the unique point centroid of a triangle."

2167. [1996: 274] Proposed by Sefket

 Arslanagi c , Berlin, Germany.
Prove, without the aid of the di erential calculus, the inequality, that
in a right triangle
a2 (b + c) + b2(a + c)  2 + p2;
abc
where a and b are the legs and c the hypotenuse of the triangle.
Solution by Mihai Cipu, Institute of Mathematics, Romania Academy,
Bucharest, Romania, and CICMA, Concordia University, Montreal Quebec.
Since a2 (b + c)+ b2(a + c) = c(a2 + b2 )+ ab(a + b) the given inequality
is equivalent to
p
c(a , b)2  ( 2c , a , b)ab = pab2c(a+,ab+) b :
2

In an isosceles triangle
p this relation holds trivially, while in the general
case it is equivalent to ( 2c + a + b)c  ab, a simple consequence of c2  ab.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; CON AMORE PROBLEM GROUP, Royal Danish School of Educational
Studies, Copenhagen, Denmark; HANS ENGELHAUPT, Franz{Ludwig{Gym-
nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,
Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN
G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV 
KONECN  Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI
LAU, Hong Kong; VEDULA N. MURTY, Andhra University, Visakhapatnam,
India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB
382

PRIELIPP and JOHN OMAN, University of Wisconsin{Oshkosh, Wisconsin,

USA; JUAN-BOSCO ROMERO MARQUEZ,  Universidad de Valladolid, Val-

ladolid, Spain; HEINZ-JURGEN SEIFFERT, Berlin, Germany; TOSHIO
SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands;
PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid
Laurier University, Waterloo, Ontario; and the proposer. There were four
incorrect solutions.
From the proof above it is obvious that equality holds if and only if
a = b. Both Konecpny and Wang pointed out that a weaker version of this
problem with 2 + 2 replaced by  appeared as problem No. 238 in the
January, 1983 issue of the College Mathematics Journal. Wang remarked
that in the published solution (CMJ 15 (4), 1984; 352{353) Walter Blumberg
and Marshall Fraser independently proved the more general result that
a2(b + c) + b2(a + c)  2 + csc(C=2)
abc
where a, b, and c are the sides of an arbitrary triangle with C being the
largest angle. The present problem is the special case when C = =2.
Klamkin gave a proof for this more general result. Romero Marquez asked
whether the stronger inequality
a2(b + c) + b2(a + c)  p2c
abc a+b,c
holds. The triangle with a = 3, b = 4, and c = 5 provides a simple coun-
terexample. Janous remarked that the given inequality could be considered
as the special case (when n = 1) of the problem of nding the minimum
value of 2n n n + b2n(an + cn )
Tn = a (b + c ()abc )n
where a, b, c are the legs and hypotenuse of a right triangle and n > 0 is
a real number. He proposed the conjecture that Tn  2 + 2(2,n)=2 for all
n  n0 where n0 is the solution of the equation 2(n+4)=2 + n = 2.

2168?. [1996: 274] Proposed by Jan Ciach, Ostrowiec Swi  etokrzyski,

Poland.
Let P be a point inside a regular tetrahedron ABCD, with circumra-
dius R and let R1 ; R2 ; R3 ; R4 denote the distances of P from vertices of the
tetrahedron. Prove or disprove that
R1R2 R3R4  34 R4;
and that the maximum value of R1 R2 R3 R4 is attained.
 383

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-

berta. [Ed: First, we give Klamkin's proof for a triangle, which was not
published as a solution to 2073? [1995: 277; 1996: 282]]
Let P be given by the vector P~ = x1 A~1 + x2 A~2 + x3 A~3 , where
x1 + x2 + x3 = 1 and x1, x2, x3  0 (barycentric coordinates). Then
2
PA21 = P~ , A~1
   2
= x2 A~2 , A~1 + x3 A~3 , A~1
,

= 3 x22 + x2 x3 + x23 ;
p
etc. We have taken R = 1, so that the side of the triangle A~2 , A~1 = 3.
The given inequality now takes the form
3;
x2 + x2x3 + x23
,x23 + x3x1 + x21
,x21 + x1 x2 + x22
 64
, 2

or in homogeneous form
64 ,x22 + x2 x3 + x23
,x23 + x3 x1 + x21
,x21 + x1 x2 + x22

 3 (x1 + x2 + x3)6 (1)
where (only) x1 , x2 , x3  0.
We assume without loss of generality that x1  x2  x3  0 so that
we can let x3 = c, x2 = b + c and x1 = a + b + c with a, b, c  0 and
a + b + c > 0. On substituting back into (1), we get
3a6 + 36a5b + 54a5c + 116a4b2 + 348a4bc + 213a4c2 + 160a3b3
+816a3b2c + 1128a3bc2 + 468a3c3 + 80a2b4 + 864a2b3 c
+2232a2b2c2 + 2232a2bc3 + 765a2 c4 + 480ab4 c + 2208ab3c2
+3888ab2c3 + 3060abc4 + 918ac5 + 192b5c + 1104b4c2
+2592b3c3 + 3060b2c4 + 1836bc5 + 459c6  0
since all terms on the left side are non-negative (this expansion was con-
rmed using Mathematica [Ed: and using DERIVE]). There is equality only if
a = c = 0, so that if x1 + x2 + x3 = 1, x1 = x2 = 0:5 and x3 = 0, then the
point P must be a mid-point of a side.
For a regular tetrahedron, we proceed as before: let P be given by the
vector P~ = x1 A~1 + x2 A~2 + x3 A~3 + x4 A~4 , where x1 + x2 + x3 + x4 = 1
and x1 , x2 , x3 , x4  0. Then
PA1 = P~ , A~1
     
= x2 A~2 , A~1 + x3 A~3 , A~1 + x4 A~4 , A~1 ;
384

so that
PA21 = a2 ,x22 + x23 + x24 + x3x4 + x4x2 + x2x3
;
etc., where a is an edge of the tetrahedron. So if we take R = 1, we have
a2 = 83 . the desired inequality then becomes (in homogeneous form)
9 (x1 + x2 + x3 + x4 )8  28F1 F2F3 F4; (2)
where
F1 = x22 + x23 + x24 + x3x4 + x4 x2 + x2x3 ;
and the other Fi's are obtained by cyclic interchange of the indices 1, 2, 3, 4.
Assuming without loss of generality that x1  x2  x3  x4 , we
take x4 = d, x3 = d + c, x2 = d + c + b, x1 = d + c + b + a, where
a, b, c, d  0 and x1 > 0. On substituting back into (2) and expanding
out (using Mathematica), we get a polynomial [Ed: we have omitted this
polynomial, since it would cover two whole pages of CRUX with MAYHEM]
which is non-negative since all the terms are non-negative. There is equality
only if a = c = d = 0. Hence x1 = x2 = 0:5 and so P must be the
mid-point of an edge.
I also conjecture for the correponding result for a regular n{dimensional
simplex:
the product of the distances from a point within or on a regular simplex
to the vertices is a maximum only when the point is a mid-point of an edge.
Analogously to inequality (2), the conjecture is that
3n,1 (x0 + x1 + : : : + xn)2n+2  22n+2 F0F1 : : : Fn
(x0 , x1 , : : : , xn  0), where Fi is the complete symmetric homogeneous
polynomial of degree 2 with unit coecients of all the variables x0 , x1 , : : : ,
xn , except xi, and there is equality only if two of the variables are equal and
the rest are zero.
Also solved by CON AMORE PROBLEM GROUP, Royal Danish School
of Educational Studies, Copenhagen, Denmark; and RICHARD I. HESS, Ran-
cho Palos Verdes, California, USA.

Crux Mathematicorum
Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell
Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical Mayhem
Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia