CRUXv 23 N 1

1
Welcome!
Welcome to CRUX MATHEMATICORUM with MATHEMATICAL
MAYHEM. We hope that those of you new to any part of our journal will
enjoy all the material. We see this merger as bene cial to all our readers.
We now o er a broader range of problems. High School students and teach-
ers will nd more material at that level. Please note that we will accept
solutions to problems in the MAYHEM section only from students. These
are the people that we wish to stimulate. When they nd they are successful
with the MAYHEM and SKOLIAD problems, we hope they will then nd a
further challenge with the problems in the OLYMPIAD corner, and, eventu-
ally, with the PROBLEMS section. Good luck, and we look forward to hearing
from you.
You will note that the majority of the members of the Crux Editorial
Board are continuing in their previous jobs. Sadly, Colin Bartholomew has
taken early retirement from Memorial University, and I shall miss him. How-
ever, we are pleased to welcome Clayton Halfyard as Associate Editor. There
will be a short pro le of him elsewhere in this issue. We are also delighted
to welcome Naoki Sato as Mayhem Editor and Cyrus Hsia as Assistant May-
hem Editor. These two Canadian IMO medallists bring with them a wealth
of experience of problem solving, and many years of experience of writing for
MAYHEM.
There are some minor administrative changes: the table of contents
(which is larger) is now on the back outside cover; and the ordering of the
various sections has been changed. There is no change in the quantity of
material that has been recently published in CRUX. Whereas previously
MAYHEM appeared ve times per year, it now appears eight times per year.
The annual quantity of material will be approximately the same.
Here are some details that will help you in cross-referencing previous
material:
1. Please note that the volume number is consecutive for CRUX.
2. Material occurring in previous volumes of CRUX and material occuring
in CRUX with MAYHEM will be referenced as [year: page no].
For example, the last page of the last issue of CRUX is [1996: 384], and
the rst page of CRUX with MAYHEM is [1997: 1].
3. Material occurring in previous volumes of MAYHEM will be referenced
as [MAYHEM volume: issue, page no: year].
For example, [MAYHEM 8: 5, 28: 1996] is the last page of the last issue
of MAYHEM.
Bruce Shawyer
Editor-in-Chief
2
Bienvenue!
Bienvenue au CRUX MATHEMATICORUM with MATHEMATICAL
MAYHEM. Nous esperons que ceux qui ne connaissent pas l'une des
nouvelles parties de notre journal en apprecieront tout le contenu. A notre
avis, la fusion de ces deux magazines sera tout a l'avantage de nos lecteurs.
La gamme des problemes publiesy est plus vaste qu'auparavant, et les e leves
et les professeurs du secondaire y trouveront davantage de problemes a leur
niveau. Nous vous signalons au passage que nous accepterons uniquement
les solutions aux problemes de la section MAYHEM provenant des e tudiants;
apres tout, c'est eux que nous voulons stimuler! Lorsqu'ils pourront reussir
les problemes des sections MAYHEM et SKOLIAD, nous esperons qu'ils
s'attaqueront a ceux de la partie OLYMPIAD, et même, un peu plus tard,
a ceux de la section PROBLEMS. Bonne chance et au plaisir de recevoir de
vos nouvelles!
Vous aurez remarque que la majorite des membres de l'equipe e ditoriale
du Crux sont demeures en poste. Malheureusement, Colin Bartholomew a
pris une retraite anticipee de l'Universite Memorial; il me manquera. Nous
sommes toutefois heureux d'accueillir Clayton Halfyard au poste de redacteur
adjoint. Vous trouverez son pro l ailleurs dans ce numero. Nous sommes
e galement enchantes d'avoir parmi nous Naoki Sato et Cyrus Hsia, qui
occupent respectivement les postes de redacteur et de redacteur adjoint du
Mayhem. Ces deux medailles canadiens de l'OIM apportent avec eux tout un
bagage d'expertise en resolution de problemes, et des annees d'experience a
la redaction du MAYHEM.
Vous constaterez en outre quelques petits changements d'ordre admin-
istratif : la table des matieres (elargie) est desormais en quatrieme de couver-
ture et les sections ne sont plus dans le même ordre. Le volume d'information
du CRUX sera comparable a la quantite d'information qu'il contenait avant
la fusion. Le MAYHEM, qui paraissait cinq fois l'an, sera publie huit fois par
annee; ainsi, le volume annuel de contenu demeurera approximativement le
même.
Voici quelques details sur la facon de noter les renvois :
1. Les numeros de volume du CRUX continuent dans la même sequence.
2. On notera ainsi les renvois a des volumes pre-fusion du CRUX et aux
volumes du CRUX with MAYHEM : [anne: no de page].
Par exemple, la derniere page du dernier numero du CRUX sera [1996:
384], et la premiere page du CRUX with MAYHEM sera [1997: 1].
3. On notera ainsi les renvois a des volumes pre-fusion du MAYHEM :
[volume du MAYHEM: no, no de page: annee].
Par exemple, [MAYHEM 8: 5, 28: 1996] est la derniere page du dernier
numero du MAYHEM.
Le redacteur en chef
Bruce Shawyer
3
THE ACADEMY CORNER

No. 8
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
In this issue, courtesy of Waldemar Pompe, student, University of War-
saw, Poland, we print an international contest paper for university students.
Please send me your nice solutions.
INTERNATIONAL COMPETITION
FOR UNIVERSITY STUDENTS IN MATHEMATICS
July 31 { August 1996, Plovdiv, Bulgaria
First day | August 2, 1996
1. Let for j = 0; 1; : : : ; n, aj = a + jd, where a , d are xed real
0 0
numbers. Put
0a a a ::: an 1
BB a a0 1
a :::
2
an, CC
A=B B@ : : : : : : : : : :: :: ::
a a a an, CC :
1 0 1 1
2 1 0
:::
2
A
an an, an, : : :
1 2 a
0
Calculate det A | the determinant of A.

2. Evaluate the integral
Z sin nx dx ;
, (1 + 2x ) sin x
where n is a natural number.
3. A linear operator A on a vector space V is called an involution if
A = E, where E is the identity operator on V .
2
Let dim V = n < 1.

(i) Prove that for every involution A on V there exists a basis of V
consisting of eigenvectors of A.
(ii) Find the maximal number of distinct pairwise commuting involu-
tions on V .
4
nP,
4. Let a = 1, an = n akan,k for n 2.
1
1
1
k =1
Show that
(i) limsup jan j =n < 2, = ;
1 1 2
n!1
(ii) limsup jan j =n 2=3.
1
n!1
5. (i) Let a, b be real numbers such that b 0 and 1 + ax + bx 2
0 for
every x 2 [0; 1].
Prove that
Z
n dx = ,1=a if a < 0;
1
lim
n!1 n (1 + ax + bx ) 2
+ 1 if a 0:
0
(ii) Let f : [0; 1] ! [0; 1) be a function with continuous second deriva-

tive and f 00 (x) 0 for every x 2 [0; 1]. Suppose that
Z
(f (x))n dx
1
L = nlim
!1 n
0
exists and 0 < L < +1.

, 1
0
Prove that f 0 has a constant sign and L = 2[0;1] jf (x)j
xmin :
6. Upper content of a subset E of the plane R R is de ned as
(X
n )
C(E) = inf diam (Ei)
i=1
where the in mum is takenn over all nite families of sets E ; E ; : : : ; En
S
in R R such that E Ei.
1 2
i=1
Lower content of E is de ned as
K(E) = supflength (L)g
such that L is a closed line segment onto which E can be contracted.
Show that
(i) C (L) = length (L), if L is a closed line segment.
(ii) C (E ) K(E ).
(iii) equality in (ii) need not hold even if E is compact.
Hint: If E = T [ T 0 where T is the triangle with vertices (,2; 2), (2; 2)
and (0; 4), and T 0 is its re ection about the x-axis, then C (E ) = 8 >
K(E).
5
Remarks:
All distances used in this problem are Euclidean.
Diameter of a set E is diam (E ) = supfdist (x; y ) j x; y 2 E g.
Contraction of a set E to a set F is a mapping f : E ! F such that
dist (f (x);f (y )) dist (x; y ) for all x; y 2 E .
A set E can be contracted onto a set F if there is a contraction f of E
to F which is onto, that is such that f (E ) = F .
Triangle is de ned as the union of the three line segments joining its
vertices, so that it does not contain the interior.
Problems 1 and 2 are worth 10 points, problems 3 and 4 are worth 15 points,
problems 5 and 6 are worth 25 points.
You have 5 hours.
Please write the solutions on separate sheets of paper. Good luck!
6
THE OLYMPIAD CORNER

No. 179
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.
Another year has passed, and the rst with Bruce Shawyer as Editor-
in-Chief. He has made the transition pleasant and easy. Special thanks go
to Joanne Longworth whose TEX skills have made changed fonts and formats
easy to incorporate. Thanks also go to the many contributors to the two
Corners including:
Miguel Amengual Covas Cyrus C. Hsia Dieter Ruo
Sefket Arslanagic Murray Klamkin Toshio Seimiya
Mansur Boase Derek Kisman Michael Selby
Seung-Jin Bang Ted Lewis D.J. Smeenk
Christopher Bradley Joseph Ling Daryl Tingley
Francisco Bellot Rosado Beatriz Margolis Panos E. Tsaoussoglou
Paul Colucci Stewart Metchette Ravi Vakil
Hans Engelhaupt Richard Nowakowski Edward T.H. Wang
Tony Gardiner Michael Nutt Hoe Teck Wee
Solomon Golomb Siu Taur Pang Chris Wildhagen
Gareth Grith Bob Prielipp Siming Zhan
Georg Gunther Chandan Reddy
Let me open with a major apology. In the November 1996 number of
the corner we gave twelve of the problems proposed to the jury but not used
at the 36th International Mathematical Olympiad held in Canada. Those
familiar with the process of selection will know that the problems do not
initiate with the host country. They come from proposers in other countries,
and the responsibility of the host country selection committee is to re ne
and select from these submissions the ocial list of problems proposed to
the jury. Over the years I have loosely referred to the problems proposed
to the jury at the Xth International Olympiad in Y as the \Y-problems" for
short. However, when that became part of a longer more ocial sounding
sub-title \Canadian Problems for consideration by the International Jury," I
should have seen that it read as if the original proposers are from Canada,
thus insulting the creators. In retrospect I do not understand why that in-
terpretation did not jump o the page. To the many creative non-Canadians
who submitted problems for possible use at the 36th IMO my sincere apolo-
gies.
7
As an Olympiad Contest this issue we give the problems of the 44th

Mathematical Olympiad from Latvia. My thanks go to Richard Nowakowski,
Canadian Team Leader to the IMO in Hong Kong for collecting the problems
for me.
LATVIAN 44 MATHEMATICAL OLYMPIAD
Final Grade, 3rd Round
Riga, 1994
1. It is given that cos x = cos y and sin x = , sin y. Prove that
sin 1994x + sin 1994y = 0.
2. The plane is divided into unit squares in the standard way. Consider
a pentagon with all its vertices at grid points.
(a) Prove that its area is not less than 3=2.
(b) Prove that its area is not less than 5=2, if it is given that the pentagon
is convex.
3. It is given that a > 0, b > 0, c > 0, a + b + c = abc. Prove that at
least one of the numbers a, b, c exceeds 17=10.
4. Solve the equation 1!+2!+3!+ + n! = m in natural numbers.
3
5. There are 1994 employees in the oce. Each of them knows 1600
of the others. Prove that we can nd 6 employees, each of them knowing all
5 others.
1st SELECTION ROUND
1. It is given that x and y are positive integers and 3x + x = 4y + y.
2 2
Prove that x , y , 3x + 3y + 1 and 4x + 4y + 1 are squares of integers.

2. Is it possible to nd 2 di erent pairs of natural numbers (ai; bi)
1994
such that the following 2 properties hold simultaneously:

1 + 1 + + 1
(1)
ab ab a 1994 b 1994 = 1,
(2) (a + a + + a 1994 ) + (b + b + + b 1994 ) = 3 ?
1 1 2 2 2 2
1995
1 2 2 1 2 2
3. A circle with unit radius is given. A system of line segments is

called a cover i each line with a common point with the circle also has some
common point with some of the segments of the system.
(a) Prove that the sum of the lengths of the segments of a cover is more
than 3,
(b) Does there exist a cover with this sum less than 5?
4. A natural number is written on the blackboard. Two players move
alternatively. The rst player's move consists of replacing the number n on
the blackboard by n=2, by n=4 or by 3n ( rst two choices are allowed only
if they are natural numbers). The second player's move consists of replacing
the number n on the blackboard by n + 1 or by n , 1. The rst player wants
the number 3 to appear on the blackboard (no matter who writes it down).
Can he always achieve his aim?
8
5. Three equal circles intersect at the point O and also two by two at
the points A, B , C . Let T be the triangle whose sides are common tangents
of the circles; T contains all the circles inside itself. Prove that the area of T
is not less than 9 times the area of ABC .
2nd SELECTION ROUND
1. It is given that 0 xi 1, i = 1; 2; : : : ; n. Find the maximum of
the expression
x x xn
+ + + x x : : : xn, + 1 :
1 2
x x : : : xn + 1 x x x : : : xn + 1
2 3 1 3 4 1 2 1
2. There are 2n points on the circle dividing it into 2n equal arcs. We

must draw n chords having these points as endpoints so that the lengths of
all chords are di erent. Is it possible if:
(a) n = 24,
(b) n = 1994?
3. A triangle ABC is given. From the vertex B, n rays are constructed
intersecting the side AC . For each of the n +1 triangles obtained, an incircle
with radius ri and excircle (which touches the side AC ) with radius Ri is
constructed. Prove that the expression
r r : : : rn
1 2 +1
R R : : : Rn
1 2 +1
depends on neither n nor on which rays are constructed.

3rd SELECTION ROUND
1. A square is divided into n cells. Into some cells \1" or \2" is
2
written so that there is exactly one \1" and exactly one \2" in each row and
in each column. We are allowed to interchange two rows or two columns;
this is called a move. Prove that there is a sequence of moves such that after
performing it \1"-s and \2"-s have interchanged their positions.
2. Let aij be integers, jaij j < 100. We know that the equation
a x + a y + a z + a xy + a xz + a yz = 0
11
2
22
2
33
2
12 13 23
has a solution (1234; 3456; 5678). Prove that this equation also has a solu-
tion with x, y , z pairwise relatively prime which is not proportional to the
given one.
3. Let ABCD be an inscribed quadrilateral. Its diagonals intersect
at O. Let the midpoints of AB and CD be U and V . Prove that the lines
through O, U and V , perpendicular to AD, BD and AC respectively, are
concurrent.
9
Next we give ve Klamkin Quickies. My thanks go to Murray Klamkin,

the University of Alberta, for supplying them to us. Next issue we will give
the \quick" solutions along with another ve of his special teasers.
FIVE KLAMKIN QUICKIES
October 21, 1996
1. For x; y;z> 0, prove thatx
1 1
(i) 1 +
(x + 1) 1 + x(x + 2) ,
(ii) [(x + y )(x + z )]x [(y + z )(y + x)]y [(z + x)(z + y )]z [4xy ]x [4yz ]y [4zx]z .
2. If ABCD is a quadrilateral inscribed in a circle, prove that the four
lines joining each vertex to the nine point centre of the triangle formed by
the other three vertices are concurrent.
3. How many six digit perfect squares are there each having the prop-
erty that if each digit is increased by one, the resulting number is also a
perfect square?
4. Let ViWi, i = 1; 2; 3; 4, denote four cevians of a tetrahedron
V V V V which are concurrent at an interior point P of the tetrahedron.
1 2 3 4
Prove that
PW + PW + PW + PW max ViWi longest edge:
1 2 3 4
5. Determine the radius r of a circle inscribed in a given quadrilateral

if the lengths of successive tangents from the vertices of the quadrilateral to
the circle are a; a; b; b; c; c; d; d, respectively.
We now turn to solutions from the readers to problems posed in the

May 1995 number of the corner on the Sixth Irish Mathematical Olympiad,
May 8, 1993 [1995: 151{152].
SIXTH IRISH MATHEMATICAL OLYMPIAD
May 8, 1993 | First Paper
(Time: 3 hours)
1. The real numbers , satisfy the equations
3
, 3 + 5 , 17 = 0;
2 3
, 3 + 5 + 11 = 0:
2
Find + .

Solutions by Sefket Arslanagic, Berlin, Germany; by Beatriz Margolis,
Paris, France; by Vedula N. Murty, Andhra University, Visakhapatnam, In-
dia; by D.J. Smeenk, Zaltbommel, the Netherlands; by Panos E. Tsaous-
soglou, Athens, Greece; and comment by Edward T.H. Wang, Wilfrid Laurier
University, Waterloo, Ontario. We give Margolis's solution.
10
De ne f (x) = x , 3x + 5x. We show that if f ( ) + f ( ) = 6, then

3 2
+ = 2. Since f (x) = (x , 1) + 2(x , 1) + 3, we have

3
f ( ) , 3 = ( , 1) + 2( , 1) 3
f ( ) , 3 = ( , 1) + 2( , 1) 3
Adding gives
0 = ( , 1) + ( , 1) + 2( + , 2)
3 3
= ( + , 2)[( , 1) + ( , 1)( , 1) + ( , 1) + 2]
2 2
and, since the second factor is positive, we obtain the result. (See Olympiad
Corner 142 and its solution.)
[Wang's Comment:] The problem is strikingly similar to problem 11.2
of the XXV Soviet Mathematical Olympiad, 11th Form [1993: 37]. The
method used by Bradley given in the published solution [1994: 99] works for
the present problem as well and, in fact, yields the same answer: + = 2.
3. The line l is tangent to the circle S at the point A; B and C are
points on l on opposite sides of A and the other tangents from B , C to S
intersect at a point P . If B , C vary along l in such a way that the product
jABj jAC j is constant, nd the locus of P .
Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;
and by D.J. Smeenk, Zaltbommel, the Netherlands. We use Smeenk's solu-
tion.
P
t t
E
D r
r I
p S q
=2 r =2
=2 =2
B p
H A q C
Let S be the incircle s(I; 2) of 4BCP . We denote \PBA = ,
\PCA =
AB = pAC = q
11
with pq = k , a constant.
2
Let S touch BP and CP at D and E respectively. For 4PEI we have

\EIP = ( + ). Thus
1
2
t = r tan 12 ( + ) = (ppq+,qr)r :
2
The semiperimeter of 4BCP is

p + q + t = p + q + (ppq+,qr)r = pqpq(p,+rq) :
2
2 2
The area, F , of 4BCP is

r pqpq(p,+rq) = 12 (p + q)PH;
2
where PH is the altitude to BC . It follows immediately that

PH = pq2pqr 2k r : 2
,r k ,r= 2 2
So the locus of P is a line parallel to BC .

4. Let a ; a ; : : : ; an, be real numbers, where n 1, and let
f (x) = xn + an, xn, + + a be such that jf (0)j = f (1) and each
0 1 1
1
root of f is real and satis es 0 < < 1. Prove that the product of the
1 0
roots does not exceed 1=2n.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,
Ontario.
Let f (x) = (x , )(x , ) : : : (x , n ) where the i denote Q(1 , the) =n
f i ; ; : : : ; n j f j f
1 2
real roots of , = 1 2 . Then from (0) = (1) we get
Q . (All products are over i = 1; 2; : : : ; n.) Using the Arithmetic Mean{ i
i
Geometric Mean Inequality we then get
Y Y Y i + (1 , i) 1 2
= i (1 , i ) =
2
i 2 2n 2
from which
Q follows. Equality holds if and only if i = for all
i 1
2n
1
i = 1; 2; : : : ; n. 2
May 8, 1993 | Second Paper

(Time: 3 hours)
3. For non-negative integers n, r the binomial coecient ,nr denotes
, of n objects chosen r at a time, with the con-
the number of,ncombinations
vention that = 1 and nr = 0 if n < r. Prove the identity
0
X1 n , r + 1r , 1 n
d=1
d d,1 = r
12
for all integers n, r with 1 r n.

Ontario.
We use a combinatorial argument to establish the obviously equivalent
identity
X k n,r+1 r,1 n
d r,d = r ()
=1d
where k = minfr;n , r + 1g. It clearly suces to demonstrate that the
left hand side of () counts the number of ways of selecting r objects from
n distinct objects (without replacements). Let jS j = r , 1. For each xed
d = 1; 2; : : : ; k, any selection of d objects from S (SnS ) together with any
2
selection of r , d objections from S would yield a selection of r objects from

1 2
S. The total number of such selections is ,n,dr ,rr,,d. Conversely, each

2
+1 1
selection of r objects from S clearly must arise in this manner. Summing

over d = 1; 2; : : : , () follows.
4. Let x be a real number with 0 < x < . Prove that, for all natural
numbers n, the sum
sin x + sin33x + sin55x + + sin(2 n , 1)x
2n , 1
is positive.

Solutions by Sefket Arslanagic, Berlin, Germany; and by Vedula N.
Murty, Andhra University, Visakhapatnam, India. We give Murty's solu-
tion.
We use mathematical induction. Let
X n sin(2k , 1)x
Sn(x) = (2k , 1) :
k =1
S (x) = sin x > 0 for x 2 (0; ). Thus the proposed inequality is true
for n = 1. Let Sr (x) > 0 for r = 1; 2; : : : ; n , 1. We will deduce that
1
Sn(x) > 0 for x 2 (0;). Suppose that Sn(x ) 0 ford some x 2 (0; ),
and that Sn (x) attains its minimum at x = x . Hence dx [Sn (x)]x x0 = 0.
0 0
0 =
That is
0 Xn
Sn(x ) = cos((2k , 1)x ) = 0;
0 0
k=1
so that
X
n
2 sin x Sn0 (x ) =
0 0 2 cos((2k , 1)x ) sin x 0 0
k=1
Xn
= [sin(2kx ) , sin((2k , 2)x )]
0 0
k =1
= sin 2nx : 0
13
Thus Sn0 (x ) = sin 2nx0 = 0 implying sin2nx0 = 0. Hence

0 2 sinx0
2 3 (2 n , 1)
x 2 2n ; 2n ; 2n ; : : : ; 2n :
0
It is easily veri ed that at each of these values Sn (x ) > 0, a contradiction.

Hence Sn (x) > 0 for x 2 (0; ).
0
Editor's Note: Both solutions used the calculus. Does anyone have a
more elementary solution?
We complete this number of the Corner with solutions by our readers

to problems of the 1992 Dutch Mathematical Olympiad, Second Round given
in the June 1995 number of the Corner [1995; 192{193].
1992 DUTCH MATHEMATICAL OLYMPIAD
Second Round
September 18, 1992
1. Four dice are thrown. What is the chance that the product of the
numbers equals 36?
Ontario.
There are four di erent kinds of outcomes in which the product is 36:
each of f1; 1; 6; 6g and f2; 2; 3; 3g can occur in = 6 ways; f1; 4; 3; 3g can
4!
occur in = 12 ways; and f1; 2; 3; 6g can occur in 4! = 24 ways. Hence the

2!2!
4!
probability that the product equals 36 is 4 = .

2!
48 1
2. In the fraction and its decimal notation (with period of length 4)

6 27
every letter represents a digit. Di erent letters denote di erent digits. The
numerator and denominator are mutually prime. Determine the value of the
fraction:
ADA = :SNELSNELSNELSNEL : : :
KOK
[Note. ADA KOK is a famous Dutch swimmer. She won gold in the
1968 Olympic Games in Mexico. SNEL is Dutch for FAST.]
Solution by the Editor.
ADA = :SNEL. Then 10 x = SNEL:SNELSNEL : : :
Let x = KOK 4
and 10 x , x = SNEL. So
4
x = SNEL SNEL SNEL SNEL

9999 = 11 909 = 33 303 = 99 101 :
Taking KOK = 909 we obtain SNEL = 11 ADA, and A = L, which is
impossible.
Taking KOK = 303, we obtain SNEL = 33 ADA, so 3A < 10, as
the product has four digits and 3A = L. Because S 6= L 6= 0, 3 3A < 9,
14
giving A = 1 or A = 2. Now A = 1 gives L = 3 = K , which is impossible,

so A = 2. This gives L = 6, and D 2, so there is a carry. This gives
D 4, as A = 2, K = 3.
ADA = :SNEL is
For D = 4, KOK = :7986, a solution.
242
For D = 5, ADA = 252 is not coprime to KOK = 303.

303
For D = 6, SNEL = 8646 and N = L.

For D = 7, SNEL = 8976 and D = E .
For D = 8, SNEL = 9036 and O = N .
For D = 9, SNEL = 9636 and N = L.
Taking KOK = 101 gives SNEL = 99 ADA forcing A = 1 for
SNEL to have four digits, but then A = K .
Thus the only solution is = :7986.
242
303
3. The vertices of six squares coincide in such a way that they enclose
triangles; see the picture. Prove that the sum of the areas of the three outer
squares (I, II and III) equals three times the sum of the areas of the three
inner squares (IV, V and VI).
aaa " "T
aaa "
"T II TT
I VI T T
"
aaa P "T "
aaa,,,V@@@P IVPPT"
@@ ,,PPB
@,B B
BB III BBB
BB
Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;
and by Vedula N. Murty, Andhra University, Visakhapatnam, India. We
give Murty's solution.
Let the gure on the next page be labelled as shown:
Let MN = x , LU = x , UY = x , AC = x , AB = x , BC = x ,
\MBN = , \LAU = , \V CY = , \BAC = A, \ACB = C and
1 3 2 4 5 6
\ABC = B .
Then we have + = , + A = , + C =
x = x + x , 2x x cos 9
2 2 2
>
> 5 6
=
1 6 5
x = x + x , 2x x cos > : : :
2 2 2
4 6 (1)
>
2 4 6
x = x + x , 2x x cos ;
2
3
2
4
2
5 4 5
15
Pa X
aaa " "
" TT
aaa N Y"" " TT
a
T II TT
I VI TT "W
T " "
Q aaa B
, @ PCPPTPT"""
aaa ,, @@ IV V
aa, V @PP
M@@ , ,APPU
@@,, BB
LB B
BB III BB
BB BB
BB T
S
x = x + x , 2x x cos B 9
2 2
> 2
4 5
>
=
6 5 6
x = x + x , 2x x cos C > : : :
2 2 2
4 6 (2)
>
4 4 6
x = x + x , 2x x cos A ;
2
6
2
4
2
5 4 6
From (2), we have

x + x + x = 2x x cos A + 2x x cos B + 2x x cos C
2
4
2
5
2
6 4 5 5 6 4 6
= ,2x x cos , 2x x cos , 2x x cos : : : (3)

4 5 5 6 4 6
From (1), we have

x + x + x = 2(x + x + x ) , 2x x cos
2
1
2
2
2
3
2
4
2
5
2
6 5 6
,2x x cos , 2x x cos 4 5 4 6
using (3)
= 2(x + x + x ) + x + x + x
2
4
2
5
2
6
2
4
2
5
2
6
= 3(x + x + x ): 2
4
2
5
2
6
That is, Area of (I + II + III ) = 3 Area of (IV + V + V I ).

4. For every positive integer n, n? is de ned as follows:
(
n? = 1 n for
for
n=1
n2
n, ? ( 1)
p p
Prove 1992 < 1992? < 1992. 4
3
16
Solutions by Vedula N. Murty, Andhra University, Visakhapatnam, In-

dia; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-
tario. We give Wang's solution and his remark.
Using the more convenient notation f (n) for n?, we show that in gen-
eral p p
n + 1 < f (n) < 43 n ()
p
for all even n 6. In particular, for n = 1992, we would get 1993 <
f (1992) < p1992.
4
First note that f (n) = f nn, = nn, f (n , 2) for all n 3. If N = 2k

3
where k 2, then multiplying f (2q ) = q,q f (2q , 2) for q = 2; 3; : : : ; k,

( 1) 1
2
we get
2 1
f (2k) = 34 65 2k2,k 1 f (2)

= 12 43 65 2k2,k 1 > 32 54 67 2k2+k 1 :
Hence
(f (2k)) > 1 32 54 6 (2 k2,k 1) 3 52 74 6 (2 k2+k 1) = 2k + 1;
2
from which it follows that

p
f (n) = f (2k) > 2k + 1 = pn + 1: (1)
On the other hand, for k 3 we have
2 4 6 2k , 2
2(2k) = 3 5 7 2k , 1 2k
2 5 7 2k , 1
< 3 6 8 2k 2k:
Hence
2 4 6 (2k , 2) 5 7 (2k , 1)
2
(f (2k)) < 3 5 7 (2k , 1) 6 8 2k (2k)

2 2

= 32 4 2k;
2
from which it follows that

p
f (n) = f (2k) < 43 2k = 43 pn: (2)
The result follows from (1) and (2).
17
Remark: Using similar arguments, upper and lower bounds for f (n)
when n is odd can also be easily derived. In fact, if we set P = k,k 1 3 5 (2 1)
(usually denoted by k,k ) then various upper and lower bounds for P
2 4 6 2
(2 1)!!
(2 )!!
abound in the literature; for example, it is known that
s s
1 5 P 1 3
2 4k + 1 2 2k + 1
and
q 1 < P p1n :
(n + ) 1
2
(Compare, for example, x3.1.16 on p. 192 of Analytic Inequalities by D.S.

Mitronovic.)
Clearly, each pair of these double inequalities would yield correspond-
ing upper and lower bounds for the function f (n) considered in the given
problem.
5. We consider regular n-gons with a xed circumference 4. We call
the distance from the centre of such a n-gon to a vertex rn and the distance
from the centre to an edge an .
a) Determine a ; r ; a ; r .
4 4 8 8
b) Give an appropriate interpretation forpa and r . 2 2
c) Prove: a n = (an + rn ) and r n = a n rn .
2
1
2 2 2
an rn
Let u ; u ; u ; u ; : : : be de ned as follows:

0 1 2 3
u = 0; u = 1; un = 21 (un, + un, ) for n even and

0 1 2 1
un = pun, un, for n odd: 2 1
d) Determine: limn!1 un .
Solution by Vedula N. Murty, Andhra University, Visakhapatnam, In-
dia.
Let O be the centre of the regular n-gon. Let A A denote one side of
the regular n-gon
1 2
18
O
rn
an
A 1 A 2
Then we have \A OA = n , \OA A = \OA A = , n . Thus

1 2
2
1 2 2 1 2
r
jA A j = rn + rn , 2rn cos 2n
,,,! 1 2
2 2 2
r
= 2rn(1 , cos 2n ) 2
r
= 4rn sin n = 2rn sin n : 2 2
The circumference of the regular n-gon is 2nrn sin n = 4 whence

2 ;
rn = n sin
n

an = rn sin 2 , n = rn cos n = n2 cot n :
In particular
p
r = 12 sin1 = 22 ;
4 a = 24 cot 4 = 12 ;
4
4
2 = 1 :
r = 8 sin
8 4 sin
8 8
Now, cos = p = 1 , 2 sin gives

4
1
2
2
8
q p
sin 8 = 12 2 , 2;
so
r = 14 p 2 p = 12 p 1 p ;
2, 2 2, 2
8
and s p

a = r cos 8 = 4 22 +
1 p 2 = 1 1p p2;
8 8
, 2 42, 2
since cos = 2 cos , 1.
4
2
8
For (b), r = 1, a = 0 as the 2-gon is a straight line with O lying at

the middle of A and A .
2 2
1 2
19
For (c), we have

an + rn = rn 1 + cos n = 2rn cos 2n 2
= n sin4 cos

2
n 2n
= 2n sin cos cos 2n = n2 cot 2n :
4 2
n n 2 2
Thus (an + rn ) = n cot( n ) = a n , and

1
2
1
2 2

a nrn = n1 cos
2
n 2 = 1

cos n =
sin n n sin n n sin n cos n n sin n
2

1 ;
2 2
2
2 2
2 2 2 2
so pa n rn = n 2n = r n .
2
1
sin 2
For (d), note u = 0, u = 1, and u = . For n 2 we have 1
that un is either the arithmetic or geometric mean of un, and un, and
0 1 2 2
1 2
in either case lies between them. It is also easy to show by induction that
u ; u ; u ; : : : form an increasing sequence, and u ; u ; u ; : : : form a de-
creasing sequence with u l u s for all l; s 0. Let limk!1 u k = P
0 2 4 1 3 5
and limk!1 u k = I . Then P I . We also have from u n = (u n, +

2 2 +1 2
1
2 +1 2 2 1
u n, ) that P = (I + P ) so that I = P and limn!1 un exists. Let

2
1
limn!1 un = L.
2 2 2
With a = 0 and r = 1, let u k = a k+1 and u k = r k+1 , for

k = 0; 1; 2; : : : . From (c), u = a 1 = a = 0 and u = r 1 = r = 1.
2 2 2 2 2 +1 2
Also for n = 2k + 2, u k = a k+1+1 = a k+1 = (a k+1 + b k+1 ) =

0 2 2 1 2 2
1
2 +2 2 2 2 2 2
(u k + u k ); that is un = (un, + un, ) and for n = 2k + 3
2
1 1
2 2 2 +1 2 2 1
u k = up k = r k+1+1 = r k+1
= a k+1 r k+1 = pa k+1+1 r k+1
2 +3 2( +1)+1 2 2(2 )
= pu k u k
2(2 ) 2 2 2
2( +1) 2 +1
p
so un = un, un, . Thus un and un satisfy the same recurrence and it
follows that L = limk!1 a k+1 = limk!1 r k+1 . Now, from the solution
1 2
2 2
to (c),
2 =2 n ;
rn = n sin sin
n n

so limn!1 rn = since n ! 0. Therefore limn!1 un = .
2 2
That completes the column for this issue. Olympiad season is approach-
ing. Send me your contests and nice solutions.
20
BOOK REVIEWS
Edited by ANDY LIU
Shaking Hands in Corner Brook and other Math Problems,
edited by Peter Booth, Bruce Shawyer and John Grant McLoughlin,
published by the Waterloo Mathematics Foundation, Waterloo, 1995,
153 pages, paperback, ISBN 0-921418-31-0.
Reviewed by Robert Geretschlager and Gottfried Perz.
One representative from each of six regions met in Corner Brook, New-
foundland, to discuss math problems. Each of these delegates shook
hands with each other delegate. How many handshakes were there?
This is the rst of many problems that can be found in this collec-
tion of problems from the Newfoundland and Labrador Teachers' Association
(NLTA) Senior Mathematics League, which conveniently o ers an explanation
as to the whereabouts of Corner Brook.
The NLTA Math League was started in 1987, and has since developed
into a very interesting competition at the regional level. A number of aspects
make this competition di erent from most math competitions. First of all,
it is purely a team competition, with four students from each participating
school comprising a team. There is no individual ranking, and so students
are motivated to work together at nding solutions. For each of ten ques-
tions posed, a team can receive ve points for a correct team answer. If
the members of a team cannot agree on the correct answer, they can submit
individual answers, for which their team can get one point each, if correct.
Finally, there is a relay question, made up of four parts. In the relay, each
part yields an answer, which is necessary to be able to solve the next part
(much as in the American Regions Mathematics League (ARML), which may
be better known to many readers). The relay section can yield a maximum of
15 points (made up of ve points for the solution and extra points for solving
the problems in a short time), for a possible total of 65 points. If teams end
up with the same point sum, a tie breaker question is posed.
The concept behind this competition is geared to fostering cooperative
problem solving, something that is generally ignored in olympiad-style com-
petitions. The level of diculty of the problems posed is adequate to the
time allowed (usually from 3 to 10 minutes per question) and the intentions
of the competition, and ranges from fairly easy to pre-olympiad level. The
book is divided into sections covering regular questions, relay questions, tie
breakers and solutions. The problems are in a random order, and no indica-
tion is given of which questions were posed at which competition. Perhaps
at least one example of ten speci c questions posed at one competition, and
the order they were posed in, might have been of interest.
21
Here are a few problems to whet your appetite:

How many three digit numbers include at least one seven but have no
zeros?
Al, Betty, Charles, Darlene and Elaine play a game in which each is
either a frog or a moose. A frog's statement is always false while a
moose's statement is always true.
Al says that Betty is a moose.
Charles says that Darlene is a frog.
Elaine says that Al is not a frog.
Betty says that Charles is not a moose.
Darlene says that Elaine and Al are di erent kinds of animals.
How many frogs are there?
Triangle ABC is isosceles, with \ABC = \ACB. There are points
D; E and F on BC; CA and AB, respectively, that form an equilateral
triangle. Given that \AFE = x and \CED = y , calculate \BDF
in terms of x and y .
The book has a very pleasing layout, with the cover showing the densest
packing of seven circles in an equilateral triangle. The solutions are nicely
presented, and in several cases, alternate solutions are given, occasionally
labeled the \routine way" and the \subtle" or \smart way".
Shaking Hands in Corner Brook should be of interest to anyone involved
with high school mathematics, either in competitions, or simply seeking en-
richment material for the interested student.
Copies of the above reviewed book may be obtained from:

Canadian Mathematics Competition
Faculty of Mathematics, University of Waterloo
Waterloo, Ontario, Canada. N2L 3G1
The cost is $12 in Canadian funds (plus 7% GST for shipping to Canadian
addresses). Cheques or money orders in Canadian funds should be made
payable to: Canadian Mathematics Competition. All pro ts from the sale of
this book are for the Newfoundland Mathematics Prizes Fund.
22
Teachers interested in providing a lively and stimulating high school

mathematics competition for their students may be interested in participat-
ing in a NLTA Senior Mathematics League in their own area. Shawn Godin,
St. Joseph Scollard Hall, North Bay, Ontario, a regular contributor to CRUX,
is already a participant. Meetings can be held within individual schools, or
between teams from more than one school.
Sample games and other information on how the NLTA Senior Mathe-
matics League is organised may be obtained free from:
Dr. Peter Booth
Department of Mathematics and Statistics
Memorial University of Newfoundland
St. John's, Newfoundland, Canada. A1C 5S7
Tel: int+ 709{737{8786
Fax: int+ 709{737{3010
email: pbooth@fermat.math.mun.ca
Schools that participate on a regular basis will be sent questions and
detailed solutions ve times per year (October, November, February, March
and May). There is an annual fee of $50 (Canadian funds) for each group of
schools participating. Cheques or money orders should be made payable to
Newfoundland Mathematics Prizes Fund.
23
A Probabilistic Approach
to Determinants
with Integer Entries
Theodore Chronis
It is well known that the probability for an integer number to be odd

is equal to the probability for the number to be even. What about determi-
nants? What is the probability for a rectangular matrix with integer entries
to have odd determinant? More generally, if m is a natural number, what is
the probability for which det A mi mod m, where mi is chosen from the
set f0; 1; 2; : : : ; m , 1g?
I have the following problem to propose; I hope you will nd it inter-
esting.
Let A be an n n matrix whose elements are integers. What is the
probability the determinant of A is an odd number?
Solution:
Let A = [aij ] ; i; j = 1; : : : ; n. It is obvious that
det A det ([aij mod 2]) mod 2:
So the problem is to nd the probability that the determinant of an n n
matrix with elements from the set f0; 1g is an odd number. Let An be an
n n matrix with elements from the set f0; 1g. Let also N (det An) be the
number of odd n n determinants, and P (det An ) be the corresponding
probability. Let
0 k :: :: kn 1
k
BB a
1
:: :: a n C
a
2
C;
f (K = fk ; k ; : : : ; kng) = N B : C
21 22 2
B@ : : C
: A
1 2
: :
an an :: :: ann
1 2
where k ; k ; : : : ; kn 2 f0; 1g are xed.

1 2
P
Then N (det An ) = (f (K )), where the sum is calculated for all
the 2n , 1 possible permutations K = fk ; k : : : ; kn g with ki 2 f0; 1g,
i = 1; : : : ; n and k ; k ; : : : ; kn not all zero. (Note that f (0; 0; : : : ; 0) = 0:)
1 2
1 2
24
Lemma: f (K ) = 2n, N (det An, ), where k + k + : : : + kn 6= 0.

1
1 1 2
Proof. It is well known that a determinant remains unchanged if from the

elements of one of its columns we substract the corresponding elements of
another column. It is also obvious that the same is true for N (det An ).
Additionally, a determinant just changes sign if we interchange two columns,
while N (det An ) remains unchanged. So
0 k k:: :: kn 1 0 1 0 :: :: 0 1
BB a
1
a:: :: a n C
2
C BB a a :: :: a n CC
NB : C=NB
C CC
21 22 2 21 22 2
B@ : : B@ :: :: :
: : : A : A
an an :: :: ann
1 2 an an :: :: ann 1 2
() f (K ) = 2n, N (det An, ): 1

1
Hence N (det An ) = 2n, N (det An, )(2n , 1).

1
Of course N (det A ) = 1 and so

1
Y
n
N (det An) = 2n n, = (2i , 1):( 1) 2
i=1
Finally, P (det An ) = N n2An () P (det An ) = ni (1 , 2,i ).
(det ) Q
Q 2
Note: The in nite sequence ni (1 , 2,i ); n = 1; 2; : : : is decreasing and

=1
bounded below by 0, so limn!1 P (det An ) exists. Using Mathematica, we

=1
found that

!1 P (det An ) = 0:288788095086602421278899721929 : : : :
nlim
Ayras 15, Ki sia

Athens 14562, GREECE
e-mail: tchronis@egnatia.ee.auth.gr
25
THE SKOLIAD CORNER

No. 19
R.E. Woodrow
The problem set we give in this issue comes to us with our thanks from
Tony Gardiner of the UK Mathematics Foundation, School of Mathematics,
University of Birmingham. The Nat West Junior Mathematical Challenge was
written Tuesday, April 26, 1994 by about 105,000 students. Students from
England and Wales must be in school year 8 or below. The use of calculators,
calendars, rulers, and measuring instruments was forbidden.
1994 NAT WEST UK JUNIOR MATHEMATICAL
CHALLENGE
Tuesday, April 26, 1994 | Time: 1 hour
1. 38 + 47 + 56 + 65 + 74 + 83 + 92 equals
A. 425 B. 435 C. 445 D. 456 E. 465.
2. What is the largest possible number of people at a party if no two of

them have birthdays in the same month?
A. 11 B. 12 C. 13 D. 23 E. 334.
3. I have $500 in 5p coins. How many 5p coins is that?

A. 100 B. 500 C. 1000 D. 2500 E. 10000.
4. What was the precise date exactly sixty days ago today? (No calendars!)
A. Friday 25th February B. Saturday 26th February
C. Friday 26th February D. Saturday 27th February
E. Tuesday 26th February.
5. You have to nd a route from A to B moving hor- 3 9 B

izontally and vertically only, from one square to
an adjacent square. Each time you enter a square 8 5 6
you add the number in that square to your total. 9 11 7
What is the lowest possible total score for a route A 8 10
from A to B ?
A. 28 B. 29 C. 30 D. 31 E. 34.
26
6. On a clock face, how big is the angle between the lines joining the centre
to the 2 O'clock and the 7 O'clock marks?
A. 160 B. 150 C. 140 D. 130 E. 120.
7. Gill is just six and boasts that she can count up to 100. However, she
often mixes up nineteen and ninety, and so jumps straight from nine-
teen to ninety one. How many numbers does she miss out when she
does this?
A. 70 B. 71 C. 72 D. 78 E. 89.
8. In how many ways can you join the two shapes

shown here to make a gure with a line of
symmetry?
A. 0 B. 1 C. 2 D. 3 E. 4.
9. If you divide 98765432 by 8, which non-zero digit does not appear in

your answer?
A. 2 B. 4 C. 6 D. 8 E. 9.
10. How many numbers between 20 and 30 (inclusive) cannot be written

as a multiple of 5, or as a multiple of 7, or as the sum of a multiple of
5 and a multiple of 7?
A. 1 B. 2 C. 3 D. 5 E. 6.
11. Four children are arguing over a broken toy. Alex says Barbara broke
it. Barbara says Claire broke it. Claire and David say they do not know
who broke it. Only the guilty child was lying. Who broke the toy?
A. Alex B. Barbara C. Claire D. David E. can't be sure.
12. The diagram is made up of one circle

and two semicircles. Which of the three
regions | the black region, the white
region and the large circle | has the
longest perimeter?
27
A. the black region B. the circle C. the white region

D. black and white are equal and longest
E. all three perimeters are equal.
13. From my house the church spire is in the direction NNE. If I face in this
direction and then turn anticlockwise through 135 I can see the Town
Hall clock. In which direction am I then facing?
A. WSW B. due West C. SW D. due South E. SSE.
14. Samantha bought seven super strawberry swizzles and ten tongue twist-
ing to ees for $1.43. Sharanpal bought ve super strawberry swizzles
and ten tongue twisting to ees for $1.25. How much is one tongue
twisting to ee?
A. 7p B. 8p C. 9p D. 10p E. 18p.
15. I am forty eight years, forty eight months, forty eight weeks, forty eight
days and forty eight hours old. How old am I?
A. 48 B. 50 C. 51 D. 52 E. 53.
16. The numbers 1 to 12 are to be placed

so that the sum of the four numbers 4
in each of the six rows is the same.
Where must the 7 go? A B
3 1
8 5
C 6 D E
9
A. at A B. at B C. at C D. at D E. at E
17. Three hedgehogs | Roland, Spike and Percival | have a leaf collecting
race. Roland collects twice as many as Percival, who collects one and
a half times as many as Spike. (Spike is moulting, and so has fewer
prickles for her to stick the leaves onto.) Between them they collect
198 leaves. How many did Spike manage to collect?
A. 18 B. 22 C. 36 D. 44 E. 66.
28
18. The four digits 1, 2, 3, 4 are writtin in increasing order. You must insert
one plus sign and one minus sign between the 1 and the 2, or between
the 2 and the 3, or between the 3 and the 4, to produce expressions
with di erent answers. For example,
1 , 23 + 4 gives the answer , 18:
How many di erent positive answers can be obtained in this way?
A. 2 B. 3 C. 4 D. 5 E. 6.
19. Roger Rabbit has twice as many sisters as brothers. His sister Raquel
notices that 2=5 of her brothers and sisters are boys. How many Rabbit
children are there in the family?
A. 2 B. 4 C. 8 D. 16 E. 32.
20. The population of a new town in 1990 was 10; 000. It has since doubled
every year. If it kept on doubling every year for ten years, what would
its population be in the year 2000?
A. 100; 000 B. 200; 000 C. 1; 000; 000 D. 2; 000; 000 E. 10; 000; 000.
21. LMNO is a square. P is a point inside the square such that NOP is
an equilateral triangle. How big is the angle PMN ?
A. 75 B. 70 C. 60 D. 45 E. 30 .
22. If the perimeter of a rectangle is 16x + 18 and its width is 2x + 6, what

is its length?
A. 18x + 24 B. 7x + 6 C. 12x + 6 D. 6x + 3 E. 14x + 12.
23. In a group of fty girls each one is either blonde or brunette and is either
blue-eyed or brown-eyed. Fourteen are blue-eyed blondes, thirty one
are brunettes and eighteen are brown-eyed. How many are brown-eyed
brunettes?
A. 5 B. 7 C. 9 D. 13 E. 18.
24. A bottle of Jungle Monster Crush (JMC) makes enough drink to ll sixty
glasses when it is diluted in the ratio 1 part Crush to 4 parts water. How
many glasses of drink would a bottle of JMC make if it is diluted in the
ratio 1 part Crush to 5 parts water?
A. 48 B. 60 C. 72 D. 75 E. 80.
29
25. A 3 by 3 by 3 cube has three holes, each

with a 1 by 1 cross section running from
PP
P PPP
the centre of each face to the centre of
PP

P
P

PP P
P

the opposite face. What is the total sur- PP
face area of the resulting solid? PP

PP
PP
PP
PP
PP
A. 24 B. 48 C. 72 D. 78 E. 84.
That completes the Skoliad Corner for this issue. Send me your con-
tests, suggestions, and recommendations to improve this feature.
Introducing the new Associate Editor-in-Chief

For those of you who do not know Clayton, here is a short pro le:
Born: Haystack, Newfoundland 1
Educated Haystack School School, Newfoundland

Thornlea School, Newfoundland
Queen's University, Kingston, Ontario
Employment Random South School Board,
Trinity Bay, Newfoundland
Mathematical Mathematical Education
Interests Commutative Algebra
1 You may have diculty nding Haystack on a map of Newfoundland | it is not lost
| it no longer exists!
30
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, University of
Toronto, Toronto, ON Canada M5S 1A1. The electronic address is
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).
Editorial
It gives me great pleasure to unveil the premiere issue of \Crux Mathemati-
corum with Mathematical Mayhem". This merger has been in the works for
quite some time, and it has nally been successfully realized.
For the bene t of those who have not heard of Mayhem, I will provide a
brief description. Mayhem was founded in 1988 by two high school students
Ravi Vakil and Patrick Surry, who wished to establish a journal speci cally
oriented towards students, and totally operated by students. Although the
journal has been passed down through many hands, and though it has not
always been easy, this mandate has always been resolutely upheld; it has
made Mayhem a unique and exceptional journal. And rest assured, we will
still be running our share of \Crux with Mayhem".
Our features include articles, olympiads, and a problems section. The
material is generally focused towards contests and olympiads, and how to
prepare for them, and the topics range from high school mathematics to un-
dergraduate material. We cannot emphasize enough that we are a journal
dedicated to mathematics students. I myself am a fourth-year student at the
University of Toronto, and Cyrus Hsia (the Mayhem Assistant Editor) is a
third-year student, also at the University of Toronto.
Our fearless sta also consists of undergraduate and high school stu-
dents.
This brings me to my next point. After considerable discussion, \May-
hem" has decided to restrict itself to publishingsolutions only from students.
The rationale behind this move is that the Crux problems already draw many
solutions, and if the same people were to respond to our problems, which are
considerably easier, it would simply overwhelm the section. We know that
31
there are many non-students who have contributed to the problems sections
over the years, who have our full gratitude, and hope they understand our
position. We are, however, prepared to make exceptions in, well, exceptional
cases.
However, we warmly welcome submissions for articles and problems
from all people. Back issues are available; the information is inside the back
cover. Any correspondence about Mayhem should be sent to Mathemat-
ical Mayhem, c/o Naoki Sato, Department of Mathematics, University of
Toronto, M5S 1A1, or at the e-mail address <mayhem@math.toronto.edu>.
Subscriptions, however, should be sent to the Canadian Mathematical So-
ciety oces, as mentioned on the inside back cover.
Well, I think that's about it. For people who have subscribed to May-
hem, welcome back, I know it's been a long wait. I look forward to working
with Bruce Shawyer on our new project (I think he will bring a certain \dis-
cipline" to Mayhem, but we will resist it as much as possible. Don't tell him
I said that.) Here's to a new year and a new era for Mayhem.
Naoki Sato
Mayhem Editor
Shreds and Slices

Positive Matrices and Positive Eigenvalues
Theorem. An n n matrix M with positive real entries has at least one
positive eigenvalue.
Proof. Let S = f(x ; x ; : : : ; xn ) j x ; x ; : : : ; xn 0; x + x + + xn =
2 2 2
1g; that is, S is the portion of the unit sphere in Rn with all coordinates non-
1 2 1 2 1 2
negative.
De ne the map f : S ! S by f (~v) = jM~ v
M~vj . Since M has all positive real
entries, for any ~v 2 S , M~v also has all non-negative coordinates, so f is
well-de ned, and does indeed map S into S .
Note that S is a closed, simply connected set. Then by Brouwer's Fixed
Point Theorem, there is a xed point of f ; that is, for some ~v 2 S , f (~v) =
~v = jM~ v
M~vj ) M~v = ~v, for some positive value (since cannot be zero),
namely = jM~v j. This is a positive eigenvalue of M .
32
Newton's Relations
Given n reals a , a , : : : , an , let Sk be the sum of the products of the
ai taken k at a time, and let Pk = ak + ak + + akn . Consider the generating
1 2
1 2
functions
S + S x + S x + + Snxn
0 1 2
2
= 1 + (a + a + + an )x + (a a + a a + + an, an )x
1 2 1 2 1 3 1
2
+ + a a an xn 1 2
= (1 + a x)(1 + a x) (1 + anx)
1 2
and
P , P x +P x ,
0 1 2
2
= (1 , a x + a x , ) + (1 , a x + a x , )
1
2
1
2
2
2
2
2
+ + (1 , an x + an x , ) 2 2
= 1 +1a x + 1 +1a x + + 1 +1a x :

1 2 n
Their product is
(n , P x + P x , )(1 + S x + S x + )
1 2
2
1 2
2
= (1 + a x)(1 + a x) (1 + anx)
1
2
1 +1a x + 1 +1a x + + 1 +1a x ;

1 2 n
since P = n and S = 1.
0 0
We claim the expression is equal to n + (n , 1)S x + (n , 2)S x + + 2
Sn, xn, . To see this, consider the coecient of xk . Since the expression
1 2
1
is symmetric, the coecient is some multiple of Sk . How many times does

1
the term a a ak xk appear? It must have appeared in the product (1 +

a x)(1 + a x) (1 + anx) as a a akal, where k < l n, before
1 2
having the term al divided out. There are n , k choices for l, and hence the
1 2 1 2
coecient is (n , k)Sk .
Hence,
(n , P x + P x , )(1 + S x + S x + )
1 2
2
1 2
2
= n + (n , 1)S x + (n , 2)S x + + Sn, xn, ; 1 2

2
1
1
and equating coecients:

nS , P = (n , 1)S 1 1 1
nS , S P + P = (n , 2)S 2 1 1 2 2
nS , S P + S P , P = (n , 3)S 3 2 1 1 2 3 3

2
n ,
nSn, , Sn, P + Sn, P , + (,1) Pn, = Sn, ;
1 1 3 2
1
1 1
33
or
P ,S 1 1 = 0
P , S P + 2S 2 1 1 2 = 0
P , S P + S P , 3S 3 1 2 2 1 3 = 0

1 1 2
n ,
Pn, , S Pn, + S Pn, , + (,1) (n , 1)Sn,
2 3
1
1 = 0;
and
Pm , S Pm, + S Pm, , + (,1)nPm,n Sn = 0
1 1 2 2
for m n:
The last equation is the well-known recursion sequence for the Pi , and the
previous equations (known as Newton's relations) can help pin down the
values of P , P , : : : , Pn, , or vice-versa.
1 2 1
Problem. If
x + y + z = 1;
x + y + z = 2; 2 2 2
x + y + z = 3; 3 3 3
determine the value of x + y + z . 4 4 4
Solution. Newton's relations become

P , S = 1 , S = 0;
1 1 1
P , S P + 2S = 2 , S + 2S = 0;
2 1 1 2 1 2
P , S P + S P , 3S = 3 , 2S + S , 3S = 0;
3 1 2 2 1 3 1 2 3
which imply that S = 1, S = ,1=2, and S = 1=6. Also, P , 3S +

2S , S = 0 ) P = 25=6.
1 2 3 4 1
2 3 4
Here is the last problem of the 1995 Japan Mathematical Olympiad, Final
Round.
Problem. Let 1 k n be positive integers. Let a , a , : : : , ak be complex 1 2
numbers satisfying
a + a + + ak = n 1 2
a + a + + ak = n 2
1
2
2
2

ak + ak + + akk = n 1 2
, ,
Show that (x + a )(x + a ) (x + an) = xk + n xk, + n xk, + + nk .
, 1 2
1 2 1 2
Solution. Given P = P, n = = Pk = n, we must nd S , S , : ,: n: , Sk . We

will prove that S = by induction. Clearly S = P = n = .
1 2 1 2
m m 1 1 1
34
, ,
Now, for some m, assume S = n , S = n , : : : , Sm, = mn, . Then
,
1 2 1
by the equations above, Pm , S Pm, + S Pm, , +(,1)m, Sm, P +

1 2 1
1
m(,1)mSm = 0, or
1 1 2 2 1 1
n n n
m,
n , n 1 + n 2 , + (,1) n m , 1 + m(,1)mSm = 0
1
so that
m S = n , n + n , = n , 1 ;
n m m,1 m,2 m,3 m,1
and further, n , 1 n
n
Sm = m m , 1 = m :
So by induction, we are done.
Mathematically Correct Sayings

[The following shred/slice appeared in the newsgroup rec.humor.funny.]
After applying some simple algebra to some trite phrases and cliches, a new
understanding can be reached of the secret to wealth and success. Here it
goes.
Knowledge is Power,
Time is Money,
and as everyone knows, Power is Work divided by Time.
So, substituting algebraic equations for these time worn bits of wisdom, we
get:
K = P (1)
T = M (2)
P = W=T (3)
Now, do a few simple substitutions. Put W=T in for P in equation (1), which
yields:
K = W=T (4)
Put M in for T into equation (4), which yields:
K = W=M (5)
Now we've got something. Expanding back into English, we get: Knowledge
equals Work divided by Money.
What this MEANS is that:
35
1. The More You Know, the More Work You Do, and
2. The More You Know, the Less Money You Make.
Solving for Money, we get:
M = W=K (6)
Money equals Work divided by Knowledge.
From equation (6) we see that Money approaches in nity as Knowledge ap-
proaches 0, regardless of the Work done.
What THIS MEANS is: The More you Make, the Less you Know.
Solving for Work, we get
W =M K (7)
Work equals Money times Knowledge
From equation (7) we see that Work approaches 0 as Knowledge approaches
0.
What THIS MEANS is: The stupid rich do little or no work.
Working out the socioeconomic implications of this breakthrough is left as
an exercise for the reader.
Contest Dates
Here are some upcoming (or in some cases, already past) contest dates to mark
on your calendar.
Contest Grade Date
Gauss Grades 7 & 8 Wednesday, May 14, 1997
Pascal Grade 9 Wednesday, February 19, 1997
Cayley Grade 10 Wednesday, February 19, 1997
Fermat Grade 11 Wednesday, February 19, 1997
Euclid Grade 12 Tuesday, April 15, 1997
Descartes Grades 12 & 13 Wednesday, April 16, 1997
CIMC Grades 10 & 11 Wednesday, April 16, 1997
AJHSME Grades 7 & 8 Thursday, November 21, 1996
AHSME High School Thursday, February 13, 1997
AIME High School Thursday, March 20, 1997
USAMO High School Thursday, May 1, 1997
AJHSME Grades 7 & 8 Thursday, November 20, 1997
COMC High School Wednesday, November 27, 1996
CMO High School Wednesday, March 26, 1997
APMO High School March, 1997
IMO High School July 18 { 31, 1997
36
A Journey to the Pole | Part I

Miguel Carrion
Alvarez
student, Universidad Complutense de Madrid
Madrid, Spain
For those of us who can not seem to get a strong grip on synthetic ge-
ometry, analytic geometry comes in handy. Even though polar coordinates
can be superior to rectangular coordinates in some situations, they are sys-
tematically ignored by instructors and students alike. The purpose of this
series is to introduce their uses with the idea that, as is always happening
in mathematics, with a little ingenuity, the concepts central to polar coordi-
nates can be applied elsewhere. This rst article uses polar coordinates in
elementary geometry.
De nition
In polar coordinates, the position of a point P is determined by the
distance r from a point O called the pole and the angle between OP and a
semi-in nite line called the polar axis. By convention, the polar axis is taken
to be the positive x-axis, and the
P transformation from polar to carte-
sian coordinates is given by x =
r cos , y = r sin . The in-
r verse change of coordinates is not
so straightforward; p The obvious ex-
pressions are r = x + y , =
2 2
arctan( yx ), but the equation for

is not single-valued, even if is re-
Polar axis stricted to [0; 2 ), and is unde-
O ned at the origin. Fortunately, we
need not worry about this: when
handling curves in polar coordinates, the change from rectangular to polar co-
ordinates is of little use, and it is convenient to allow r and to take on all
real values. With this provision, a point can be referred to by an in nite set of
coordinate pairs: (r; ) = ((,1)nr; + n ). Unless you want to do multiple
integrals, this is not a problem, but rather something to exploit!
Polar Curves
Polar curves are usually written in the form r = r(), and unlike curves
of the form y = y (x), they can be closed and need not be simple (they
can intersect themselves). Implicit curves of the form f (r;) = 0 can be
even more general. From a cartesian equation, the substitution x = r cos ,
y = r sin yields a polar expression. Throughout this article, I have tried
to avoid this whenever possible, and it turns out that it is always possible.
37
Some symmetries of the curves can be detected by checking the functions r()
or f (r;) above for the following simple properties (this is not a complete
list):
The curve is symmetric about the pole if r() = r( + ) or
f (,r;) = f (r;)
The curve has n-fold symmetry about the pole if r() = r( + n )2
The curve is symmetric about the polar axis if r() = r(,)

The curve is symmetric about a line at an angle to the polar axis if
r() = r(2 , )
The following transformations are also useful:
Any curve can be rotated through by substituting , for
The x- and y-axes can be permuted by substituting , for 2
Example 1. The equation of a circle of radius a centered at the origin is

r = a.
Example 2. The equation of a line passing through the origin at an angle
to the polar axis is = .
Exercise 1. Find the equation of a line at an angle to the polar axis
passing at a distance d to the pole.
Exercise 2. Identify the curve r = 2a cos .
The Cosine Law
More often than not, when working in polar coordinates, one uses
nothing but trigonometry, and the cosine law is the starting point of many
derivations. If you think about it, it comes closest to being a `vector addi-
tion rule' to use if you need to translate a curve, although this is best done
in rectangular coordinates. I will not give a translation rule, because it is
cumbersome and is of little use. Instead, I will use the cosine law to derive
the equation of a circle of radius centered at (R; ) (see gure). Applying
the cosine law to side of 4OCP , we have
2
= R + r , 2Rr cos( , )
2 2
= [r , R cos( , )] + R , R cos ( , )

2 2 2 2
=) [r , R cos( , )] = , R sin ( , ):

2 2 2 2
38
From this equation, it is evident that

P if R , then the curve is de-
r ned for a limited range of given by
,
C R sin( , ) R , as we would
expect when the origin lies outside
R the circle. The squared length of the
tangent from O, when sin( , ) =
O
R2, is P = R cos ( , ) = R (1 ,
2 2 2

R2 ) = R , ; this is called the
2 2
potence of the origin w.r.t. the circle. Note that this formula is correct, even
when R < and sin = R has no solution. Incidentally, the solution to
Exercise 2 can be obtained easily by noting that if the origin is on the circle,
then R = , and
r , R cos( , ) = R cos( , ) =) r = 2R cos( , ):
Example 3. Polar equation of the el-

lipse with one focus at the origin and P 2a , r
the main axis at an angle to the F 0
polar axis. The cosine law applied r

to side F 0P of triangle FF 0P gives
2c

F
(2a , r) 2
= r + 4c , 4rc cos( , )
2 2
=) a , ar = c , rc cos( , )
2 2
=) r[a , c cos( , )] = a , c 2 2
=) b =a
r = 1 , e cos(
2
, ) :
A Catalogue of Important Curves

The following curves are all important in their own right, but since their
polar expressions are particularly simple, they make good examples of the use
of polar coordinates.
39
Conic sections
We already have the equation for
the ellipse. The polar equation of d + r cos P
the parabola is even easier to derive. &
In the gure, we have the focus at r
the origin, the axis at an angle to
the polar axis and a distance d from
the focus to the directrix. From the V F
q
q
gure on the previous page, we

have
r = d + r cos( , ) =) r = 1 , cos(d , ) :
Exercise 3. In a similar way, derive the equation for the hyperbola, not-
ing how both branches are handled. Hence, deduce that the general equation
of the conic is r = ,e de , , where e is the eccentricity. This equation
1 cos( )
can be obtained immediately from the de nition of a conic as the locus of the
points whose distances to a line (called the directrix) and a point (called the
focus) are at a constant ratio e.
The Cardioid
The cardioid is the trajectory of a P
point on a circle that rolls on another r R
circle of the same radius. So de- 2# P C

0
ned, it is a special case of the epicy- 0
cloid, which is the curve described O " C "

by a point on a circle rolling on an- 2
other circle with no restriction on
the radii; the general equation of the
epicycloid is best expressed in para-
metric form.
In the gure, the two circles have radius R. The condition that the one cen-
tered at C 0 rolls on the one centered at C implies that triangles OCP 0 and
PC 0P 0 must be congruent.0 In triangles OCP 0 and PC 0P 0, = 2R cos(=2).
Similarly, in triangle OP P , r = 2 cos(=2). Putting all together, we have
r = 4R cos (=2) =) r = 2R(1 + cos ):
2
The cardioid is also a special case of Pascal's Limacon, of the equation

r = b + a cos . The cusp at O becomes a loop if b < a, and a smooth
indentation if b > a. The limacon can be de ned as the locus of the feet
40
of perpendiculars dropped from the origin to tangent lines to a circle. The

radius of the circle is b and the distance from O to its centre is a.
The Lemniscate
The lemniscate is the locus of the P
points such that the product of their at r a=t
distances to two points 2a apart
a
is a . In the gure, the cosine law
2 O
gives
at 2 2
= r
2
+ a + 2ar cos ; a =t = r + a , 2ar cos
2 2 2 2 2
=) a4
= (r + a ) , 4a r cos
2 2 2 2 2 2
=) r
4
= 2a r (2 cos , 1)
2 2 2
=) r
2
= 2a cos(2):
2
The lemniscate is a special case of Cassini's ovals, which are de ned in the
same way, but with no restriction on the product of the distances.
The Rose
Loosely related to the lemniscate are
the roses, of equation
a P
O r = a cos[n( , )]
with integer n. For odd n, the curve
has n `leaves' and it is traced com-
pletely when varies from 0 to .
For even n, the curve has 2n
`leaves', and it is traced completely only when varies from 0 to 2 (see the
gure).
More general curves can be obtained if n is rational or irrational. In
the rst case there is an integer number of overlapping lobes and the curve
is closed, but in the latter case the curve never closes, and in fact it is dense
in the disc r a.
The Spirals
Polar coordinates are particularly suited to spirals. The two most fa-
mous are the Archimedean Spiral r = a, which is the trajectory of a point
whose angular and radial velocities are proportional, and the Logarithmic
Spiral r = ea . In fact, any continuous monotonic function that goes to in-
nity as goes to in nity de ned in a semi-in nite interval will give rise to
a spiral, like r = a ln . A related feature of polar curves is the limit cycle,
occurring when r() has a nite limit r at in nity. In that case, the curve
winds around the origin in nitely many times, approaching the circle r = r .
0
0
Recognizing a limit cycle makes it easier to sketch a polar curve.
41
Straight Lines
We nish by deriving the equation
of a straight line not passing through
the origin. From the gure, we have
d = r cos( , ), where d is the r
distance from the line to the origin
and is the direction of the clos- d
est point. An alternative form is
d = r sin( , ), where is the di- O
rection of the line and 0 > , >
(see the gure).
Additional Problems
Problem 1. Considering r( + ) and r( , ), derive an expression
for a secant line to a conic. Passing to the limit ! 0, write an equation for
0 0
the tangent line at .

0
Problem 2. PQ is a chord through the focus F of a conic, and the

tangents at P , Q meet at T . Prove that T lies on the directrix corresponding
to F and that FT ? PQ.
Problem 3. Let s be the tangent line at the vertex of a parabola and t
be the tangent at P . If r and s meet at Q, prove that FQ bisects the angle
between FP and the axis of the parabola, and that FQ ? s.
IMO Report
Richard Hoshino
student, University of Waterloo
Waterloo, Ontario.
After ten days of intensive training at the Fields Institute in Toronto,

the 1996 Canadian IMO team travelled to Mumbai, India to participate in the
37th International Mathematical Olympiad. For the rst time in our team's
history, every team member brought home a medal, with three silver and
three bronze.
This year's team members were: Sabin \Get me a donut" Cautis, Adrian
\Da Chef" Chan, Byung Kyu \Spring Roll" Chun, Richard \YES! WE'VE GOT
BAGELS!" Hoshino, Derek \Leggo my Eggo" Kisman, and Soroosh \Mr. Bean"
Yazdani. Our team leaders were J.P. \Radishes" Grossman and Ravi \Oli"
Vakil (no, he's not Italian). Special thanks go out to our coaches, Naoki
\Dr. Cow" Sato and Georg \Where's my Ethanol" Gunther.
42
This year's paper was one of the most dicult ever, and thus, the cuto s
for medals were among the lowest in history, 28 for gold, 20 for silver and
12 for bronze. Only one student, a Romanian, received a perfect score of 42.
Our team's scores were as follows:
CAN 1 Sabin Cautis 13 Bronze Medal
CAN 2 Adrian Chan 14 Bronze Medal
CAN 3 Byung Kyu Chun 18 Bronze Medal
CAN 4 Richard Hoshino 22 Silver Medal
CAN 5 Derek Kisman 22 Silver Medal
CAN 6 Soroosh Yazdani 22 Silver Medal
Some weird coincidences: all the silver medallists got the exact same
score, are graduating and are headed to the University of Waterloo in Septem-
ber, and all the bronze medallists are eligible to return to Argentina for next
year's IMO. Overall, Canada nished sixteenth out of seventy- ve countries,
one of our highest rankings ever. A main reason for our success was our
combined team score of 36 out of 42 on question #6, a problem that many
countries answered very poorly (in fact, only two countries had more points
on that problem than we did, and they both got 37 out of 42). Unfortu-
nately, question #2 was answered very poorly by Canada, with only one 7,
even though the problem was created by our own team leader, J.P. Gross-
man.
We all owe special thanks to Dr. Graham Wright of the Canadian Math-
ematical Society, Dr. Bruce Shawyer of Memorial University and Dr. Richard
Nowakowski of Dalhousie University for their hard work and organization in
making our trip possible and Dr. Ed Barbeau of the University of Toronto
for all his commitment and dedication to training all the IMO team hopefuls
with his year-long correspondence program.
Overall, the experience was memorable for all of us, although we could
have done without the cockroaches in our rooms. Best of luck to all the
students who will be working hard to make the 1997 IMO team, which will
be held in Chapadmalal, Argentina near Mar del Plata.
Mayhem Problems
A new year brings new changes and new problem editors. Cyrus Hsia now
takes over the helm as Mayhem Advanced Problems Editor, with Richard
Hoshino lling his spot as the Mayhem High School Problems Editor, and
veteran Ravi Vakil maintains his post as 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 problems | the next issue will feature
only solutions. We intend to have problems and solutions in alternate issues.
43
We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions be submitted
by 1 June 1997, for publication in the issue 5 months ahead; that is, issue 6.
We also request that only students submit solutions (see editorial), but we
will consider particularly elegant or insightful solutions for others. Since this
rule is only being implemented now, you will see solutions from many people
in the next few months, as we clear out the old problems from Mayhem.
High School Problems

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>
H217. Let a , a , a , a , a be a ve-term geometric sequence satis-
fying the inequality 0 < a < a < a < a < a < 100, where each term
1 2 3 4 5
1 2 3 4 5
is an integer. How many of these ve-term geometric sequences are there?
(For example, the sequence 3, 6, 12, 24, 48 is a sequence of this type).
H218. A Star Trek logo is inscribed inside a A
circle with centre O and radius 1, as shown. Points
A, B, and C are selected on the circle so that AB = B
AC and arc BC is minor (that is ABOC is not a BB
convex quadrilateral). The area of gure ABOC is O B
equal to sin m , where 0 < m < 90 and m is an in- SSBB

teger. Furthermore, the length of arc AB (shaded as SB
shown) is equal to a=b, where a and b are relatively SB
prime integers. Let p = a + b + m. B C
(i) If p = 360, and m is composite, determine all possible values for m.

(ii) If m and p are both prime, determine the value of p.
H219. Consider the in nite sum
a + a + a + a + ;
S = 10 0 1 2 3
10 10 10
0 2 4 6
where the sequence fan g is de ned by a = a = 1, and the recurrence

0
p 1
relation an = 20an, + 12an, for all positive integers n 2. If S can

1
a
be expressed in the form pb , where a and b are relatively prime positive
2
integers, determine the ordered pair (a; b).

44
H220. Let S be the sum of the elements of the set

f1; 2; 3; : : : ; (2p)n , 1g:
Let T be the sum of the elements of this set whose representation in base 2p
consists only of digits from 0 to p , 1.
Prove that 2n TS = (p , 1)=(2p , 1).
Advanced Problems
Editor: Cyrus Hsia, 21 Van Allen Road, Scarborough, Ontario, Canada.
M1G 1C3 <hsia@math.toronto.edu>
A193. If f (x;y) is a convex function in x for each xed y, and a convex
function in y for each xed x, is f (x;y ) necessarily a convex function in x
and y ?
A194. Let H be the orthocentre (point where the altitudes meet) of a
triangle ABC . Show that if AH : BH : CH = BC : CA : AB , then the
triangle is equilateral.
A195. Compute tan 20 tan 40 tan 60 tan 80.
A196. Show that r + ra + rb + rc 4K , where r, ra, rb, rc, and K
2 2 2 2
are the inradius, exradii, and area respectively of a triangle ABC .
Challenge Board Problems

Editor: Ravi Vakil, Department of Mathematics, One Oxford Street,
Cambridge, MA, USA. 02138-2901 <ravi@math.harvard.edu>
C70. Prove that the group of automorphisms of the dodecahedron
is S , the symmetric group on ve letters, and that the rotation group of
5
the dodecahedron (the subgroup of automorphisms preserving orientation)
is A .
5
C71. Let L , L , L , L be four general lines in the plane. Let pij be

the intersection of lines Li and Lj . Prove that the circumcircles of the four
1 2 3 4
triangles p p p , p p p , p p p , p p p are concurrent.

12 23 31 23 34 42 34 41 13 41 12 24
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.
45
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 "11" or A4 sheets of paper.
1
These may be typewritten or neatly hand-written, and should be mailed to

2
the Editor-in-Chief, to arrive no later than 1 September 1997. They may

also be sent by email to cruxeditor@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.
2201. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABCD is a convex quadrilateral, and O is the intersection of its diag-
onals. Let L; M; N be the midpoints of DB; BC; CA respectively. Suppose
that AL; OM; DN are concurrent. Show that
either AD k BC or [ABCD] = 2[OBC ];
where [F ] denotes the area of gure F .
2202. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck,
Austria.
Suppose than n 3. Let A ; : : : ; An be a convex n-gon (as usual with
interior angles A ; : : : ; An ).
1
1
Determine the greatest constant Cn such that
X n 1 X n 1 :
A Cn , Ak
k=1
k k =1
Determine when equality occurs.

46
2203. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck,

Austria.
Let ABCD be a quadrilateral with incircle I . Denote by P , Q, R
and S , the points of tangency of sides AB , BC , CD and DA, respectively
with I .
Determine all possible values of \(PR;QS ) such that ABCD is cyclic.
2204. Proposed by Sefket
Arslanagic, Berlin,
p Germany.
For triangle ABC such that R(a + b) = c ab, prove that
3 a:
r < 10
Here, a, b, c, R, and r are the three sides, the circumradius and the
inradius of 4ABC .
2205. Proposed by Vaclav Konecny, Ferris State University, Big Rapids,
Michigan, USA.
Find the least positive integer n such that the expression
sinn A sinn B sinn C
+2 +1
has a maximum which is a rational number (where A, B , C are the angles of

a variable triangle).
2206. Proposed by Heinz-Jurgen Sei ert, Berlin, Germany.
Let a and b denote distinct positive real numbers.
(a) Show that if 0 < p < 1, p 6= , then
1
2
1 ,apb ,p + a ,p bp < 4p(1 , p)pab + (1 , 4p(1 , p)) a + b :

1 1
2 2
(b) Use (a) to deduce Polya's
Inequality:
a , b < 1 2pab + a + b :
log a , log b 3 2
Note: \log" is, of course, the natural logarithm.
2207. Proposed by Bill Sands, University of Calgary, Calgary, Al-
berta.
Let p be a prime. Find all solutions in positive integers of the equation:
2 + 3 = 5:
a b p
47
2208. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.
1. Find a set of positive integers fx; y;z; a; b; c; kg such that
y z = a +k
2 2 2 2
z x = b +k
2 2 2 2
x y = c +k
2 2 2 2
2. Show how to obtain an in nite number of distinct sets of positive inte-

gers satisfying these equations.
2209. Proposed by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain.
Let ABCD be a cyclic quadrilateral having perpendicular diagonals crossing
at P . Project P onto the sides of the quadrilateral.
1. Prove that the quadrilateral obtained by joining these four projections
is inscribable and circumscribable.
2. Prove that the circle which passes through these four projections also
passes through the mid-points of the sides of the given quadrilateral.
2210?. Proposed by Joaqun Gomez
Rey, IES Luis Bu~nuel, Alcorcon,
Madrid, Spain.
Given a = 1, the sequence fan g (n = 1; 2; : : : ) is given recursively by
0
n n n n
n a n , n,1 a n , + 1
n,2 a n , , : : :2 b n c ab n2 c = 0:
2
Which terms have value 0?

2211. Proposed by Bill Sands, University of Calgary, Calgary, Al-
berta.
Several people go to a pizza restaurant. Each person who is \hungry" wants
to eat either 6 or 7 slices of pizza. Everyone else wants to eat only 2 or 3
slices of pizza each. Each pizza in the restaurant has 12 slices.
It turns out that four pizzas are not sucient to satisfy everyone, but that
with ve pizzas, there would be some pizza left over.
How many people went to the restaurant, and how many of these were \hun-
gry"?
48
2212. Proposed by Edward T.H. Wang, Wilfrid Laurier University,

Waterloo, Ontario.
Let S = f1; 2; : : : ; ng where n 3.
(a) In how many ways can three integers x, y , z (not necessarily distinct)
be chosen from S such that x + y = z ? (Note that x + y = z and
y + x = z are considered to be the same solution.)
(b) What is the answer to (a) if x, y , z must be distinct?
2213. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.
A generalization of problem 2095 [1995: 344, 1996: 373].
Suppose that the function f (u) has a second derivative in the interval (a; b),
and that f (u) 0 for all u 2 (a; b). Prove that
1. (y , z )f (x) + (z , x)f (y ) + (x , y )f (z ) > 0 for all x; y;z 2 (a; b),
z<y<x
if and only if f 00 (u) > 0 for all u 2 (a; b);
2. (y , z )f (x) + (z , x)f (y ) + (x , y )f (z ) = 0 for all x; y;z 2 (a; b),
z<y<x
if and only if f (u) is a linear function on (a; b).
Correction
2137. [1996: 124, 317]Proposed by Aram A. Yagubyants, Rostov na
Donu, Russia.
Three circles of (equal) radius t pass through a point T , and are each
inside triangle ABC and tangent to two of its sides. Prove that:
(i) t =
rR , (ii) T lies on the line segment joining the
R+r
[NB: r instead of 2] centres of the circumcircle and the incircle
of 4ABC .
49
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.
2101. [1996: 33] Proposed by Ji Chen, Ningbo University, China.

Let a; b; c be the sides and A; B; C the angles of a triangle. Prove that
for any k 1,
X ak 3 X k
A a;
where the sums are cyclic. [The case k = 1 is known; see item 4.11, page 170
of Mitrinovic, Pecaric, Volenec, Recent Advances in Geometric Inequalities.]
I. Solution by Hoe Teck Wee, Singapore.
Let f (x) =
sink x for 0 < x < . Then, for 0 < k 1, we have
x
k,
f 0(x) = (kx cos x , xsin x) sin x :
1
For x =2, we have cos x 0, so that kx cos x , sin x 0.

For x < =2, we have cos x > 0 and tan x > x kx, so that kx cos x ,
sin x 0.
Therefore f 0(x) =
(kx cos x , sin x) sink, x 0.
1
x 2
Without loss of generality, we may assume that A B C . For

0 < k 1, we have that f (x) is a non-increasing function, so that f (A)
f (B) f (C ). Thus, by Tchebyshev's inequality, we have
!
sink A + sink B + sink C (A+B +C ) 3 sink A + sink B + sink C :
A B C
By the Sine Rule, we have a = 2R sin A, b = 2R sin B and c = 2R sin C ,
where R is the circumradius of 4ABC . Multiply the inequality by (2R)k
and substitute A + B + C = to get
X ak 3 X k
A a: (1)
For k 0, we have
k k k
a b c =) ak bk ck =) aA bB cC :
Thus, by using Tchebyshev's inequality again, (??) holds for k 0.
50
In conclusion, (??) holds for any k 1.

II. Solution by Kee-Wai Lau, Hong Kong. (Edited)
, loss of generality, a, b c, so that A B C:
We let, without
Next, put f (k) = ak =A + bk =B + ck =C ak + bk + ck , , then
1
X ak bk(A , B)(log a , log b) X k,

f 0(k) = , a 0;
2
AB
where the sums are cyclic. Since we are given that f (1) 3=n, it follows
that f (k) 3=n for k 1.
Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV

KONECNY, Ferris State University, Big Rapids, Michigan, USA; HEINZ-

JURGEN SEIFFERT, Berlin, Germany; and the proposer. One incorrect solu-
tion was received.
Janous establishes the following generalization.
Let k and ` be real numbers. Then:
1. k `
Xa 3 X ak ;

A`
for the cases
0 k ` 1,
k 0 `,
k 0 and ` ,1:
2.
X ak 3 ` X k
A` a;
where
k 0 and , 1 ` 0:
2102. [1996: 33] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with incentre I . Let P and Q be the feet of the
perpendiculars from A to BI and CI respectively. Prove that
AP + AQ = cot A :
BI CI 2
Solution by Panos E. Tsaoussoglou, Athens, Greece, and by eight oth-
ers!
51
In 4APB , we have sin(B=2) =

AP
AB ,
in 4AQC , we have sin(C=2) =
AQ ,
AC
in 4ABI , we have
BI = AB ,
sin(A=2) cos C=2
in 4ACI . we have
CI = AC ,
sin(A=2) cos B=2
so that
AP + AQ = sin(B=2) cos(C=2) + sin C=2 cos B=2
BI CI sin(A=2) sin A=2
= sin(B= 2 + C=2) = cos(A=2) = cot(A=2):
sin(A=2) sin(A=2)
Editor's comment. Our featured solution is based on the property:
\AIB = , A=2 , B=2 = =2 + C=2, so that sin(\AIB ) = cos(C=2).
Other solvers used the equally ecient relations: BI = r= sin(B=2), or
BI = 4R sin(A=2) sin(C=2).
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,
Spain (two solutions); FEDERICO ARDILA, student, Massachusetts Insti-
tute of Technology, Cambridge, Massachusetts, USA; FRANCISCO BELLOT
ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student,
Washburn Rural High School, Topeka, Kansas, USA; CARL BOSLEY, stu-
dent, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER
J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL CABEZON OCHOA,
Logro~no, Spain; THEODORE CHRONIS, student, Aristotle University of Thes-
saloniki, Greece; HAN PING DAVIN CHOR, Student, Cambridge, MA, USA;
DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;
HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLO-
RIAN HERZIG, student, Perchtoldsdorf, Austria; JOHN G. HEUVER, Grande
Prairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, stu-
dent, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursul-

inengymnasium, Innsbruck, Austria; VACLAV Y,
KONECN Ferris State Uni-
versity, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba Tonka School
of Mathematics, Rousse, Bulgaria; P. PENNING, Delft, the Netherlands;
GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR

POMPE, student, University of Warsaw, Poland; HEINZ-JURGEN SEIFFERT,
Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; GEORGE
TSAPAKIDIS, Agrinio, Greece; MELITIS D. VASILIOU. Elefsis, Greece;
CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer.
52
2103. [1996: 33] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle. Let D be the point on side BC produced beyond
B such that BD = BA, and let M be the mid-point of AC . The bisector of
\ABC meets DM at P . Prove that \BAP = \ACB .
Solution by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg,
Germany.
Let PX be parallel to AC with X lying on the line BC . Let Y be the
intersection of PX with AD. P is the midpoint of XY because M is the
mid-point of AC .
Then B is the mid-point of DX [PB is parallel to AD since 2\DAB =
\DAB + \BDA = \ABC = 2\PBA]:
Hence BX = BD = AB: Triangle BPA is congruent to triangle BPX
[PB = PB ; AB = XB ; \ABP = \XBP .]
Therefore, \BAP = \BXP = \BCA [PX kAC ].
X
C
P
Y M =2
=2 B
A =2
D
Also solved by FEDERICO ARDILA, student, MassachusettsInstitute of

Technology, Cambridge, Massachusetts, USA; SEFKET Berlin,
ARSLANAGIC,
Germany; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MARIA
ASCENSION LOPEZ
CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; HAN PING DAVIN
CHOR, Student, Cambridge, MA, USA; TIM CROSS, King Edward's School,
Birmingham, England; DAVID DOSTER, Choate Rosemary Hall, Wallingford,
Connecticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MITKO
53
CHRISTOV KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bul-

garia; KEE-WAI LAU, Hong Kong; VASILIOU MELETIS, Elelsis, Greece;
P. PENNING, Delft, the Netherlands; WALDEMAR POMPE, student, Uni-
versity of Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands;
HOE TECK WEE, Singapore; and the proposer.
2104. [1996: 34] Proposed by K.R.S. Sastry, Dodballapur, India.

In how many ways can 111 be written as a sum of three integers in geometric
progression?
Solution by Zun Shan, Nanjing Normal University, Nanjing, China.
The answer is seventeen or sixteen depending on whether we allow the
common ratio of the G.P. (geometric progression) to be zero or not.
Suppose 111 = a + ar + ar where a is an integer and r is a rational number.
2
If r = 0, then we get the trivial solution

111 = 111 + 0 + 0: (1)
n
Suppose r = m 6= 0 where m and n are nonzero integers. Without loss
of generality, we may also assume that m > 0 and (m; n) = 1. Since the
reverse of the G.P. a; ar; ar is another G.P. ar ; ar; a, we may also assume
2 2
that jrj 1 and so 0 < m jnj.

From a(1 + r + r ) = 111 we get a(m + mn + n ) = 111m . Since clearly
2 2 2 2
(m + mn + n ; m ) = 1 we have m ja. Letting a = km where k is an

2 2 2 2 2
integer we then get k(m + mn + n ) = 111 which implies kj111. Since

2 2
m + mn + n > 0 and 111 = 3 37, we have k = 1; 3; 37, or 111. Note

2 2
that m + mn + n = m + jnj(m + jnj) m .

2 2 2 2
Case [1] If k = 1, then m + mn + n = 111 =) m 111 =) m 10.

2 2 2
When m = 1, a = 1 and from n + n = 110 we get n = 10; ,11. Thus

2
r = 10, ,11 and we obtain the solutions:

111 = 1 + 10 + 100 = 1 , 11 + 121: (2)
For 2 m 9 it is easily checked that the resulting quadratic equation
in n has no integer solutions.
When m = 10, a = 100 and from n + 10n = 11 we get n = 1, ,11.
2
Since m jnj, n = ,11 and r = , yielding the solution:

11
10
111 = 100 , 110 + 121: (3)

Case [2] If k = 3, then m + mn + n = 37 =) m 37 =) m 6.
2 2 2
Quick checkings reveal that there are no solutions for m = 1, 2, 5, 6.

When m = 3, a = 27 and from n + 3n = 28 we get n = 4, ,7. Thus
2
r = ; , yielding the solutions:

4
3
7
3
111 = 27 + 36 + 48 = 27 , 63 + 147: (4)

54
When m = 4, a = 48 and from n + 4n = 21 we get n = 3, ,7.

2
Since m jnj, n = ,7 and r = , yielding the solution:

7
4
111 = 48 , 84 + 147: (5)

Case [3] If k = 37, then m + mn + n = 3 =) m 3 =) m = 1 =)
2 2 2
a = 37 and from n + n = 2 we get n = 1, ,2. Thus r = 1 or ,2

2
yielding the solutions:

111 = 37 + 37 + 37 = 37 , 74 + 148: (6)
Case [4] If k = 111, then m + mn + n = 1 =) m 1 =) m = 1 =)
2 2 2
a = 111, and from n + n = 0 we get n = ,1 as n 6= 0. Thus r = ,1

2
and we get the solution

111 = 111 , 111 + 111: (7)
Reversing the summand in (2) , (7) and noting that two of them are
\symmetric", we obtain seventeen solutions in all:
111 = 111 + 0 + 0 = 1 + 10 + 100 = 100 + 10 + 1
= 1 , 11 + 121 = 121 , 11 + 1 = 100 , 110 + 121
= 121 , 110 + 100 = 27 + 36 + 48 = 48 + 36 + 27
= 27 , 63 + 147 = 147 , 63 + 27 = 48 , 84 + 147
= 147 , 84 + 48 = 37 + 37 + 37 = 37 , 74 + 148
= 148 , 74 + 37 = 111 , 111 + 111:
Spain, (who assumed r 6= 0 and found sixteen solutions); F.J. FLANIGAN,
San Jose State University, San Jose, California, USA; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; (both of them found all seventeen
solutions). There were also two incomplete and twenty-three incorrect so-
lutions submitted!
Among these twenty-three, thirteen submissions claimed six solutions; six
submissions claimed ve solutions; two submissions claimed six or nine so-
lutions; one submission claimed two solutions, and one submission claimed
one solution only. Most of the errors were the result of assuming by mistake
that a(1 + r + r ) = 111 =) 1 + r + r must be an integer.
2 2
2105. [1996: 34] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK
Find all values of for which the inequality
2(x + y + z ) + 3(1 + 3)xyz (1 + )(x + y + z )(yz + zx + xy)
3 3 3
holds for all positive real numbers x; y;z .

55
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-

berta.
On setting x = y = 1 and z = 0, we obtain 4 (1 + )2 and thus nd
that must be 1. We now show that the inequality holds for all 1.
First, if = 1, the inequality reduces to
x + y + z + 6xyz (x + y + z)(yz + zx + xy);
3 3 3
which is equivalent to the special case n = 1 of the known Schur inequality

xn(x , y)(x , z) + yn(y , z)(y , x) + zn(z , x)(z , y) 0;
true for all real n, and which has come up many times in this journal. The
rest will follow by showing that for all < 1,
(1 + )(x + y + z)(yz + zx + xy ) , 3(1 + 3)xyz
(1 + 1)(x + y + z)(yz + zx + xy) , 3(1 + 3)xyz: (1)
[Editor's note: rewrite the original inequality as
2(x + y + z ) (1 + )(x + y + z )(yz + zx + xy) , 3(1 + 3)xyz;
3 3 3
then (1) says that the right hand side is largest when = 1, so doing the
case = 1 would be enough.] But (1) can be written
(1 , )[(x + y + z )(yz + zx + xy) , 9xyz] 0
which [after cancelling the positive factor 1 , ] is a known elementary in-
equality, equivalent to Cauchy's inequality

(x + y + z ) x1 + y1 + z1 9:
Also solved by CARL BOSLEY, student, Washburn Rural High School,
Topeka, Kansas, USA; THEODORE CHRONIS, student, Aristotle University
of Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf, Aus-
tria; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; HEINZ-

JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens,
Greece; and HOE TECK WEE, Singapore. There were three incorrect solu-
tions sent in.
2106. [1996: 34] Proposed by Yang Kechang, Yueyang University,

Hunan, China.
A quadrilateral has sides a; b;c; d (in that order) and area F . Prove that
2a + 5b + 8c , d 4F:
2 2 2 2
When does equality hold?

56
Solution by Federico Ardila, student, Massachusetts Institute of Tech-

nology, Cambridge, Massachusetts, USA.
Let ABCD be the quadrilateral with AB = a, BC = b, CD = c, and
DA = d. We can assume, without loss of generality, that AC = 1. There-
fore, we can locate the quadrilateral in a system of Cartesian coordinates
where
A = (0; 0); B = (p; q); C = (1; 0); D = (r; s):
We assume that ABCD is simple so that its area is well-de ned. If ABCD
is not convex we can make it convex and keep the side lengths the same
while increasing the area. This means that we will be done if we can show
that the result is true for convex quadrilaterals. It's also clear from this that
if the result is true for convex quadrilaterals, then equality cannot hold for
non-convex quadrilaterals. Therefore, assume q < 0 and s > 0. Now note
that
2a + 5b =
2 2
2(p + q ) + 5((p , 1) + q )
2 2 2 2
= 7p , 10p + 5 + 7q
2 2
= 7 ,p , , + 5 + 7q
5 2
7
25
7
2
2a + 5b
2 2
7q + ;
2 10
7
(2)
and
8c , d = 8((r , 1) + s ) , (r + s )
2 2 2 2 2 2
= 7r , 16r + 8 + 7s 2 2
= 7 ,r , , + 8 + 7s 8
7
2 64
7
2
8c , d 7s , :
2 2 2 8
7
(3)
Combining (??) and (??), we get
2a + 5b + 8c , d
2 2 2 2
7, q + 7s + ,
2

2 2
7
= 7q + + 7s +2 1 2 1
, ,
7
7
= 7 jqj , + 2jq j + 7 jsj , + 2jsj

1 2 1 2
2,jqj + jsj
7 7
(4)
= 4(ABC ) + 4(CDA):
2a + 5b + 8c , d 4F;
2 2 2 2
as we wished to prove (where ABC and CDA refer to the areas of the
two triangles ABC and CDA respectively). For equality to hold (when
A = (0; 0) and C = (1; 0)), it must hold in steps (??), (??), and (??). There-
fore p = , r = , q = , , and s = . Thus, in general, equality holds if
5 8 1 1
and only if ABCD is directly similar to quadrilateral A B C D , where

7 7 7 7
0 0 0 0
A = (0; 0); B = ( ; , ); C = (1; 0); D = ( ; ):

0 0
5
7
1
7 0 0
8
7
1
7
57
Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,

Austria; VACLAV KONECN Y, Ferris State University, Big Rapids, Michigan,
USA; KEE-WAI LAU, Hong Kong; and the proposer.
KONECN Y makes the observation that the quadrilateral is cyclic and
that \ABD = 135 . The proposer makes the early observation that the
maximum area for a quadrilateral with xed sides occurs when it is cyclic and
uses properties of cyclic quadrilaterals in the proof. He also generalizes the
result to an inequality which JANOUS uses in his proof and which appears in
the Addenda to the Monograph \Recent Advances in Geometric Inequalities"
by D. S. Mitrinovic et al. in I. Journal of Ningbo University 4, No. 2 (Dec.
1991), 79{145.
2107. [1996: 34] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.
Triangle ABC is not isosceles nor equilateral, and has sides a; b; c. D
and E are points of BA and CA or their productions so that
1
BD = CE = a. D and E are points of CB and AB or their productions

1
so that CD = AE = b. Show that D E k D E .

1 1 2 2
2 2 1 1 2 2
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let S be the intersection of AB and D E . 2 1
[Editor's comment by Chris Fisher. Even though there seem to be two choices
for each Di and Ei , no solver had any trouble choosing the positions that
make the result correct; furthermore, it must have been \obvious" to evey-
one but me that AB is not parallel to D E , so that S exists. Alas, perhaps
2 1
I need stronger glasses.]
Then CS is the bisector of \ACB , since CE = CB and CA = CD . 1 2
Therefore
D S = BD , BS = a = BSb , AS = a ;
1 1
SE 2 AE , AS 2 b
since
BS = a . It then follows that D E kD E , since
AS b 1 1 2 2
E S = CE = a = D S :
1 1 1
SD CD 2 b SE 2 2

Technology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnas-
ium, Bamberg, Germany; JOHN G. HEUVER, Grande Prairie Composite High
School, Grande Prairie, Alberta; CYRUS HSIA, student, University of Toronto,
Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; VACLAV KONECN Y,
Ferris State University, Big Rapids, Michigan, USA;
P. PENNING, Delft, the Netherlands; JOEL SCHLOSBERG, student, Hunter
58
College High School, New York NY, USA; GEORGE TSAPAKIDIS, Agrinio,
Greece; MELITIS D. VASILIOU. Elefsis, Greece; and the proposer.
Janous adds the observation that D E and D E are not only par-
allel, but their lengths are in the ratios a : b (as is clear from the featured
1 1 2 2
solution).
2108. [1996: 34] Proposed by Vedula N. Murty, Andhra University,

Visakhapatnam, India.
Prove that
s
a + b + c 1 3 (b + c) (c + a) (a + b) ;
2 2 2
3 4 abc
where a; b; c > 0. Equality holds if a = b = c.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria, (modi ed
slightly by the editor).
By the arithmetic-geometric mean inequality we have
p
a b + ab + b c + bc + c a + ca 6 6 a b c = 6abc;
2 2 2 2 2 2 6 6 6
which implies
9(a b + ab + b c + bc + c a + ca + 2abc)
2 2 2 2 2 2
8(a b + ab + b c + bc + c a + ca + 3abc);
2 2 2 2 2 2
or
9(a + b)(b + c)(c + a) 8(a + b + c)(ab + bc + ca)
= 4(a + b + c)(a(b + c) + b(c + a) + c(a + b)):
Using the arithmetic-geometric mean inequality again, we then have
3 (a + b)(b + c)(c + a)
4
(a + b + c) a(b + c) + b(c 3+ a) + c(a + b)
(a + b + c)pabc(a + b)(b + c)(c + a) (1)
From (1) it follows immediately that
s
1 3 (a + b) (b + c) (c + a) a + b + c :
2 2 2
4 abc 3
Clearly, equality holds if a = b = c. [Ed. In fact, if equality holds, then
from (1) we have a(b + c) = b(c + a) = c(a + b). The rst equality implies
59
a = b and the second one implies b = c. Thus, equality holds in the given
inequality if and only if a = b = c. This was observed by about half of the
solvers.]
Technology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; HAN PING DAVIN CHOR, Student, Cambridge,
MA, USA; THEODORE CHRONIS, student, Aristotle University of Thessa-
loniki, Greece; TIM CROSS, King Edward's School, Birmingham, England;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.
KLAMKIN, University of Alberta, Edmonton, Alberta; KEE-WAI LAU, Hong
Kong; JUAN-BOSCO ROMERO MARQUEZ, Universidad de Valladolid, Val-
ladolid, Spain; JOEL SCHLOSBERG, student, Hunter College High School,
New York NY, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS
E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
Janous commented that upon the transformation
a ! a1 ; b ! 1b ; c ! 1c ;
the given inequality can be shown to be equivalent to
r ab + bc + ca s
3 3 (a + b)(b +8 c)(c + a)
which is known in the literature as Carlson's inequality (cf. eg. P.S. Bullen,
D.S. Mitrinovic and P.M. Vasic, \Means and Their Inequalities", Dordrechf,
1988. An anonymous reader commented that in this form, the inequality
was problem 3 of the 1992 Austrian-Polish Mathematics Competition and
has appeared in Crux before (see [1994:97; 1995:336-337]). Several solvers
showed that the given inequality is equivalent to various other trigonometric
inequalities involving a triangle XY Z , for example
X Y Z 3p3
cos 2 cos 2 cos 2 8
or
3
p3
sin X + sin Y + sin Z 2 ;
etc. These inequalities can be found in \Geometric Inequalities" by
O. Bottema et al.
2109. [1996: 34] Proposed by Victor Oxman, Haifa, Israel.

In the plane are given a triangle and a circle passing through two of the
vertices of the triangle and also through the incentre of the triangle. (The
incentre and the centre of the circle are not given.) Construct, using only an
unmarked ruler, the incentre.
60
Solution by P. Penning, Delft, the Netherlands.

Let the triangle be ABC , and , the circle passing through B; C and
the incentre. The angles of the triangle are denoted by the symbol for the
corresponding vertex A; B or C . A
B
..
X
B0 C0
S C
,
ANALYSIS:
Let point S be the intersection of the circumcircle and the angular bisec-
tor through A. The arcs SB and SC of the circumcircle are now equal and so
are the chords SB and SC . Introduce X on AS such that SX = SB = SC:
\BSA = \BCA = C ;
4XBS is isosceles with SX = SB, so \XBS = 90 , C=2:
\CBS = \CAS = A=2; \XBC = 90 , C=2 , A=2 = B=2:
So BX is the angular bisector at vertex B , and X must be the incentre. The
circle , apparently has the point S as centre. [It does, see Roger A. Johnson,
Modern Geometry, 292].
[Editor's note: If either AB or AC is tangent to ,, then they both
are and AB = AC . Suppose AB is tangent to ,. Then \SBA = , so
\BSA + \SAB = . Since SC = SB; \BCS = \SBC = \SAC =
2
\SAB . Therefore, \ACS = \ACB + \BCS = \BSA + \SAB = and

2
AC is tangent to ,. In addition, tangents to a circle from an exterior point

2
are equal, so AB = AC:]

So if AB 6= AC , neither line is tangent to ,. Let B 0 and C 0 be the other
intersections of AB , respectively AC , with ,. There is mirror-symmetry
with respect to the line AS : , remains ,; AB re ects into AC and AC
re ects into AB ; B and C 0 are mirror-images and so are C and B 0 . The side
BC becomes B0C 0; as a consequence they must intersect on the mirror-line
AS.
CONSTRUCTION:
Find the other two intersections, B 0 and C 0 , of AB and AC with the
circle ,. The intersection of BC and B 0 C 0 is M . The incentre is the inter-
61
section of AM and ,.
COMMENT:
The construction fails if ABC is isosceles, with AB = AC . In that
case , touches both AB and AC in B and C respectively. M is now the
midpoint of BC , but that cannot be found with unmarked ruler.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; VACLAV

KONECNY, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI
LAU, Hong Kong; and the proposer.
2110. [1996: 35] Proposed by Jordi Dou, Barcelona, Spain.

Let S be the curved Reuleaux triangle whose sides AB , BC and CA
are arcs of unit circles centred at C , A and B respectively. Choose at random
(and uniformly) a point M in the interior and let C (M ) be a chord of S for
which M is the midpoint. Find the length ` such that the probability that
C (M ) > ` is 1=2.
Solution by the proposer.
Let be the locus of the mid-point M of segments of constant length ,
whose ends S and S move on the boundary of S . For the points M inside
the chords bisected by M are greater than .
1 2
(?) The area contained between S and is . It is sucient to show that 4

2
` = [S] :
2
4 2
p3
Since [S ] = , 2
2
4
, we will have
, p3 ! 12
`= ' 0:67:
Brief proof of the assertion (?) (Special Case of Holditch's Theorem)
The ends S , S of the chord of constant length move along the contour
of the closed curve S . The mid-point M describes the curve .
1 2
Let S = (x ; y ); S = (x ; y ), and M = ( x1 x2 ; y1 y2 ). Suppose + +
that x , y , x , y are functions of t such that for t t t , we have S ,

1 1 1 2 2 2 2 2
S describing S.
1 1 2 2 0 1 1
Z t1 Z t1 Z t1
2
[S] = y dx = y dx = 12 (y dx + y dx )
Zt0 t1 Zt0t1 1
1 1 2 2 1 1 2 2
t0
[] = y dx = 4 (y + y )(dx + dx ):
t0 t0
1 2 1 2
Then Z t1
1
[S] , [] = 4 (y , y )(dx , dx ):
t0
2 1 2 1
62
Substitute X = x , x , Y = y , y , giving
2 1 2 1
[S] , [] = 4 1 Z t1 Y dX = 4 ;
2
t0
since (X;Y ) describes a circle of radius .
2111. [1996: 35] Proposed by Hoe Teck Wee, student, Hwa Chong
Junior College, Singapore.
Does there exist a function f : N ,! N (where N is the set of positive
integers) satisfying the three conditions:
(i) f (1996) = 1;
(ii) for all primes p, every prime occurs in the sequence
f (p), f (2p), f (3p); : : : ; f (kp); : : : in nitely often; and
(iii) f (f (n)) = 1 for all n 2 N ?
I. Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, On-
tario.
Yes, a function does exist that satis es the three conditions. It is:
condition holds;
f (x) = p1i ifotherwise ;
Q
where condition is: if the prime factorization of x is x = pei i , there exists
a power ei such that ei > 2 and ei > ej for all j 6= i.
For example, 109850 has condition , since 109850 = 2 5 13 and 2 3
the power of 13 is bigger than 2 and bigger than all other powers in the
factorization; thus f (109850) = 13.
Now
f satis es condition (i) since f (1996) = f (2 499) = 1;
2
f satis es condition (ii) because for any two primes p and q , f (xi) = q
for every xi = q ip, i = 3; 4; : : : ;
f satis es condition (iii) since for all n either f (n) = 1 or f (n) = pi
for some prime pi, and in either case f (f (n)) = 1.
II. Solution by Chris Wildhagen, Rotterdam, the Netherlands.
For each n 2 N let qn be the nth prime and b(n) be the number of 1's
in the binary representation of n. De ne f : N ! N as follows:
q m = pn with p prime and n 2;
f (m) = 1b n ifelse
( )
:
63
Clearly f satis es the three given conditions.

Also solved by FEDERICO ARDILA, student, Massachusetts Institute
of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student,
St. Paul's School, London, England; CARL BOSLEY, student, Washburn Rural
High School, Topeka, Kansas, USA; CHARLES R. DIMINNIE, Angelo State
University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washing-
ton, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS
HSIA, student, University of Toronto, Toronto, Ontario; DAVID E. MANES,
State University of New York, Oneonta, NY, USA; ROBERT P. SEALY, Mount
Allison University, Sackville, New Brunswick; and the proposer. There were
three incorrect solutions sent in.
Most solvers gave a variation of Solution I.
2112. [1996: 35] Proposed by Shawn Godin, St. Joseph Scollard Hall,
North Bay, Ontario.
Find a four-digit base-ten number abcd (with a 6= 0) which is equal to
aa + bb + cc + dd.
Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-
tario (modi ed slightly by the editor).
We rst stipulate that 0 = 1. Let m = abcd, s = aa + bb + cc + dd
0
and assume that m = s. Clearly, 10 m < 10 . If x 6 for any

3 4
x 2 fa; b; c; dg then s 6 > 10 which is a contradiction. So a; b;c; d 5.

6 4
If x < 5 for all x 2 fa; b; c; dg, then s 4 4 = 1024 and furthermore

4
a = b = c = d = 4 is the only combination for which s 10 . However, in

3
this case, s = 1024 6= 4444 = m. Hence x = 5 for some x 2 fa; b; c; dg.

We cannot have more than one 5 or else s 2 5 = 6250 would imply
5
that some digit of m is at least 6. Hence, we have exactly one 5.

Since s > 5 = 3125 and s 5 + 3 4 = 3893 < 4000, we must have
5 5 4
a = 3. Thus, s = 5 +3 +xx +yy = 3152+xx+yy where x; y 2 fa; b; c; dg.

5 3
Without loss of generality, we may assume that 0 y x 4.

If x = 0, then y = 0 and s = 3154 which has no 0 among its digits.
If x = 1, then y = 0; 1 and s = 3154 while m has no 4 among its digits.
If x = 2, then s = 3156 + y y and it is easily veri ed that s has no 2 among
its digits for y = 0; 1; 2.
If x = 3, then s = 3179 + y y and it is easily veri ed that s has no 5 among
its digits for y = 0; 1; 2; 3.
If x = 4, then s = 3408 + y y and, again, it is readily checked that s has no 5
among its digits when y = 0; 1; 2; 4. However, when y = 3, s = 3435 which
is clearly a solution.
To summarize, m = 3435 is the only solution.
64

Spain; FEDERICO ARDILA, student, Massachusetts Institute of Technol-
ogy, Cambridge, Massachusetts, USA; MANSUR BOASE, student, St. Paul's
School, London, England; CHRISTOPHER J. BRADLEY, Clifton College,
Bristol, UK; MIGUEL ANGEL CABEZON OCHOA, Logro~no, Spain;
THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;
TIM CROSS, King Edward's School, Birmingham, England; JEFFREY K.
FLOYD, Newnan, Georgia, USA; FLORIAN HERZIG, student, Perchtolds-
dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.
KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECN Y,
Ferris State University, Big Rapids, Michigan, USA; LUKE LAMOTHE, stu-
dent, St. Joseph Scollard Hall S.S., North Bay, Ontario, KATHLEEN E.
LEWIS, SUNY Oswego. Oswego, NY, USA; DAVID E. MANES, State Univer-
sity of New York, Oneonta, NY, USA; JOHN GRANT McLOUGHLIN, Okana-
gan University College, Kelowna British Columbia; P. PENNING, Delft, the
Netherlands; CORY PYE, student, Memorial University of Newfoundland,
St. John's, Newfoundland; JOEL SCHLOSBERG, student, Hunter College High
School, New York NY, USA; ROBERT P. SEALY, Mount Allison University,
Sackville, New Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany;
DIGBY SMITH, Mount Royal College, Calgary, Alberta; CHRIS
WILDHAGEN, Rotterdam, the Netherlands; and the proposer.
Of the twenty-six solvers (including the proposer), nine of them only gave
the answer 3435. About half of all the solvers claimed, with or without
proof, that 3435 is the only solution. Chronis, Hess, and Janous found the
answer by computer search. Hess remarked that no other solutions were
found for the present problem and the corresponding problem on 5{digit
integers. Janous investigated the corresponding n{digit problem of nding
nX, 1
all n{digit integers an, an, : : :a a which equal aakk . He showed that
1 2 1 0
k
a necessary condition is n 10. For n > 1, he conducted an extensive, but
=0
not exhaustive, computer search, which revealed no solutions other than the
one found by all the solvers! He made a guess that 1 and 3435 are the only
integers with the desired property. Can any reader prove or disprove this?
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

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1

THE ACADEMY CORNER

Calculate det A | the determinant of A.

Let dim V = n < 1.

(ii) Let f : [0; 1] ! [0; 1) be a function with continuous second deriva-

exists and 0 < L < +1.

THE OLYMPIAD CORNER

As an Olympiad Contest this issue we give the problems of the 44th

Prove that x , y , 3x + 3y + 1 and 4x + 4y + 1 are squares of integers.

such that the following 2 properties hold simultaneously:

3. A circle with unit radius is given. A system of line segments is

2. There are 2n points on the circle dividing it into 2n equal arcs. We

depends on neither n nor on which rays are constructed.

Next we give ve Klamkin Quickies. My thanks go to Murray Klamkin,

5. Determine the radius r of a circle inscribed in a given quadrilateral

We now turn to solutions from the readers to problems posed in the

De ne f (x) = x , 3x + 5x. We show that if f ( ) + f ( ) = 6, then

+ = 2. Since f (x) = (x , 1) + 2(x , 1) + 3, we have

Let S touch BP and CP at D and E respectively. For 4PEI we have

The semiperimeter of 4BCP is

The area, F , of 4BCP is

where PH is the altitude to BC . It follows immediately that

So the locus of P is a line parallel to BC .

roots does not exceed 1=2n.

May 8, 1993 | Second Paper

for all integers n, r with 1  r  n.

selection of r , d objections from S would yield a selection of r objects from

S. The total number of such selections is ,n,dr ,rr,,d. Conversely, each

selection of r objects from S clearly must arise in this manner. Summing

Thus Sn0 (x ) = sin 2nx0 = 0 implying sin2nx0 = 0. Hence

It is easily veri ed that at each of these values Sn (x ) > 0, a contradiction.

We complete this number of the Corner with solutions by our readers

occur in = 12 ways; and f1; 2; 3; 6g can occur in 4! = 24 ways. Hence the

probability that the product equals 36 is 4 = .

2. In the fraction and its decimal notation (with period of length 4)

x = SNEL SNEL SNEL SNEL

giving A = 1 or A = 2. Now A = 1 gives L = 3 = K , which is impossible,

For D = 5, ADA = 252 is not coprime to KOK = 303.

For D = 6, SNEL = 8646 and N = L.

From (2), we have

= ,2x x cos , 2x x cos , 2x x cos : : : (3)

From (1), we have

,2x x cos , 2x x cos 4 5 4 6

That is, Area of (I + II + III ) = 3 Area of (IV + V + V I ).

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

First note that f (n) = f nn, = nn, f (n , 2) for all n  3. If N = 2k

where k  2, then multiplying f (2q ) = q,q f (2q , 2) for q = 2; 3; : : : ; k,

f (2k) = 34 65 2k2,k 1 f (2)

from which it follows that

(f (2k)) < 3 5 7 (2k , 1) 6 8 2k (2k)

from which it follows that

(Compare, for example, x3.1.16 on p. 192 of Analytic Inequalities by D.S.

Let u ; u ; u ; u ; : : : be de ned as follows:

u = 0; u = 1; un = 21 (un, + un, ) for n even and

un = pun, un, for n odd: 2 1

Then we have \A OA = n , \OA A = \OA A =  , n . Thus

The circumference of the regular n-gon is 2nrn sin n = 4 whence

Now, cos  = p = 1 , 2 sin  gives

For (b), r = 1, a = 0 as the 2-gon is a straight line with O lying at

for all integers n, r with 1 r n.

First note that f (n) = f nn, = nn, f (n , 2) for all n 3. If N = 2k

where k 2, then multiplying f (2q ) = q,q f (2q , 2) for q = 2; 3; : : : ; k,

Then we have \A OA = n , \OA A = \OA A = , n . Thus

The circumference of the regular n-gon is 2nrn sin n = 4 whence

Now, cos = p = 1 , 2 sin gives

Thus (an + rn ) = n cot( n ) = a n , and

For (d), note u = 0, u = 1, and u = . For n 2 we have 1

and limk!1 u k = I . Then P I . We also have from u n = (u n, +

the opposite face. What is the total sur- PP

face area of the resulting solid? PP

is symmetric, the coecient is some multiple of Sk . How many times does

and equating coecients:

arctan( yx ), but the equation for

The curve is symmetric about the polar axis if r() = r(,)

= [r , R cos( , )] + R , R cos ( , )

=) [r , R cos( , )] = , R sin ( , ):

r , R cos( , ) = R cos( , ) =) r = 2R cos( , ):

=) r[a , c cos( , )] = a , c 2 2

cloid, which is the curve described O " C "

The cardioid is also a special case of Pascal's Limacon, of the equation

the tangent line at .