385

THE ACADEMY CORNER

No. 36
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, we present the problems of the 7th International Mathemat-

ics Competition for University Students, held at University College, London,
UK, on 26 July and 31 July 2000. Our thanks to Moubinool Omarjee, Paris,
France, for sending us the problems. We invite our readers to send in their
nice solutions.
The International Mathematics Competition for
University Students
First Day Problems, 26 July 2000
1. Is it true that if f : [0; 1] ! [0; 1] is
(a) monotone increasing
(b) monotone decreasing
then there exists an x 2 [0; 1] for which f (x) = x?
2. Let p(x) = x + x and q (x) = x + x . Find all pairs (w;z ) of complex
5 5 2

numbers with w 6= z for which p(w) = p(z ) and q (w) = q (z ).

3. Suppose that A and B are square matrices of the same size with
rank (AB ; BA) = 1 .
Show that (AB ; BA) = 0.
2

4. (a) Show that, if fxk g is a decreasing sequence of positive numbers,

then
Xn ! 1
2
X n xk
2
xk  p . k
k=1 k=1
(b) Show that there is a constant C such that, if fxk g is a decreasing
sequence of positive numbers, then
X1 1 1 !
X X1 1
2

pm xk  C xk . 2

m =1 k m = k =1
386

5. Let R be a ring of characteristic zero (not necessarily commutative). Let

e, f and g be idempotent elements of R satisfying e + f + g = 0. Show
that e = f = g = 0.
(R is of characteristic zero means that, if a 2 R and n is a positive
integer, then na 6= 0 unless a = 0. An idempotent x is an element
satisfying x = x .) 2

6. Let f : R ! (0; 1) be an increasing di erentiable function for which

!1 f (x) = 1 and f is bounded.
0
xlim
Zx
Let F (x) = f (t) dt. De ne the sequence fang inductively by
0

a = 1 , an = a + n + f (1a ) ,
0 +1
n
and the sequence fbng simply by bn = F ; (n). 1

!1(an ; bn ) = 1.
Prove that nlim
Second Day Problems, 31 July 2000

1. (a) Show that the unit square can be partitioned into n smaller squares
if n is large enough.
(b) Let d  2. Show that there is a constant N (d) such that, whenever
n  N (d), a d{dimensional unit cube can be partitioned into n
smaller cubes.
2. Let f be continuous and nowhere monotone on [0; 1]. Show that the
set of points on which f attains local minima is dense in [0; 1].
(A function is nowhere monotone if there exists no interval where the
function is monotone. A set is dense if each non-empty open interval
contains at least one element of the set.)
3. Let p(z ) be a polynomial of degree n with complex coecients. Prove
that there exist at least n + 1 complex numbers z for which p(z ) is 0
or 1.
4. Suppose that the graph of a polynomial of degree 6 is tangent to a
straight line at the 3 points A , A , A , where A lies between A
and A .
1 2 3 2 1

(a) Prove that if the lengths of the segments A A and A A are

1 2 2 3
equal, then the areas of the gures bounded by these segments
and the graph of the polynomial are also equal.
 387

(b) Let k =
AA
A A , and let K be the ratio of the areas of the appro-
2 3

priate gures. Prove that

1 2

2k < K < 7k .
5 5

7 2
5. Let R be the set of positive real numbers.
+

Find all functions f : R ! R such that, for all x, y 2 R ,

+ + +

f (x) f (yf (x)) = f (x + y) .

6. For an m  m real matrix A, de ne eA to be
P
1
An .1

n n=0
!

(The sum is convergent for all matrices.)

; m 
m real ma-
Prove or disprove,; that
for all real polynomials p and
trices A and B , p eAB is nilpotent if and only if p eBA is nilpotent.
(A matrix A is nilpotent if Ak = 0 for some positive integer k.)
388

THE OLYMPIAD CORNER

No. 209
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.
As a rst set of problems, we give the 11th Grade of the XXIII All Russian
Olympiad of the Secondary Schools. My thanks go to Richard Nowakowski,
Canadian Team Leader to the IMO in Argentina, for collecting them.
XXIII ALL RUSSIAN OLYMPIAD
OF THE SECONDARY
th
SCHOOLS
11 Grade
First Day
1. Solve, in integers, the equation (x ; y ) = 1 + 16y .
2 2 2

2. The Council of Wizards is tested in the following way: The King lines
the wizards up in a row and places on the head of each of them either a white
hat or a blue hat or a red hat. Each wizard sees the colours of hats of the
people standing in front of him, but he neither sees the colour of his hat nor
the colours of hats of the people standing behind. Every minute some of the
wizards must announce one of the three colours (it is allowed to speak out just
once). After completion of this procedure the King executes all the wizards
who failed to guess the right colour of their hats. Prior to this ceremony all
100 members have agreed to minimize the number of executions. How many
of them are de nitely secure against the punishment?
3. Two circles intersect at the points A and B. A line is drawn through
the point A. This line crosses the rst circle again at the point C and it
crosses the second circle again at the point D. Let M and N be the mid-
points of the arcs BC and BD, respectively (these arcs do not contain A).
Let K be the mid-point of the segment CD. Prove that the angle MKN is
a right angle. (It may be assumed that A lies between C and D).
4. An n  n  n cube is constituted of unit cubes. You are given a
closed broken loop without self-crossings such that each link of it joins the
centres of the two neighbouring cubes (the neighbours have common faces).
We say the faces of the cubes that are crossed by this loop are marked. Prove
that the edges of the cubes can be painted in two colours in such a way that
marked faces would have an odd number of edges of both colours, and any
unmarked faces would have an even number of edges of both colours.
 389

Second Day
5. Given all possible quadratic trinomials of the type x + px + q, with
2

integer coecients p and q , 1  p  1997, 1  q  1997, consider the sets

of the trinomials:
(a) having integer zeros,
(b) not having real zeros.
Which of those sets is larger?
6. Suppose a polygon, a line l and a point P on the line l are in general
position (that is, all the lines which are extensions of the sides of the polygon
intersect l in di erent points that are di erent from P ). We mark those of
the vertices of the polygon for which the extensions of the sides leaving the
vertex cross l at points with P between them. Prove that P lies inside this
polygon if and only if there are odd numbers of marked vertices lying on both
sides of l.
7. A sphere is inscribed in a tetrahedron. It touches the rst of the
pyramid's faces at the incentre, the second face at the orthocentre, and the
third face at the point of intersection of the medians. Prove that this is a
right tetrahedron.
8. Dominoes of size 2  1 are arranged in a rectangular box sized m  n,
where m and n are odd numbers. They cover almost the whole box except
for a corner where there is a 1  1 hole. If a domino has a short common
edge with this hole it is allowed to shift along itself for one unit and cover this
hole (while doing this another hole opens). Prove that using some sequence
of those shifts it is possible to move the hole into any corner of the box.

As a second Olympiad set for this number, we give the problems of the
Fourth National Mathematical Olympiad of Turkey. My thanks go to Richard
Nowakowski, Canadian Team Leader at the IMO in Argentina, for collecting
them for us.
FOURTH NATIONAL MATHEMATICAL OLYMPIAD
OF TURKEY
Second Round, First Day
December 6, 1996 | Time: 4.5 hours
1. Let fAng1n and f ng1n be sequences of positive integers. As-
sume that, for each positive integer x there exist a unique positive integer
=1 =1

N and a unique N -tuple (x ; x ; : : : ; xN ) of integers such that

1 2

X
N
x= xn An , 0  xn  n (n = 1, 2, : : : , N ) and xN 6= 0 .
n=1
390

Prove that
(i) An = 1 for some n ;
0 0

(ii) if k 6= j , then Ak 6= Aj ,
(iii) if Ak  Aj , then Ak divides Aj .
2. Given a square ABCD of side length 2, let M and N be points on
the edges [AB ] and [CD], respectively. The lines CM and BN meet at P ,
while the lines AN and MD meet at Q. Show that jPQj  1.
3. n integers on the real axis are coloured. Determine for which posi-
tive integral values of k there exists a family K of closed intervals satisfying
the following conditions:
(i) The union of all closed intervals in K contains all the coloured integers.
(ii) Any two distinct closed intervals in K are disjoint.
(iii) For each I 2 K , abII = k , where aI is the number of integers in I , and bI
1

the number of coloured integers in I .

Second Round, Second Day
December 7, 1996 | Time: 4.5 hours
4. Given a quadrangle ABCD, the circle which is tangential to [AD],
[DC ] and [CB ] touches these edges at K , L and M , respectively. Denote the
point at which the line which passes through L and is parallel to AD meets
[KM ] by N , and the point at which [LN ] and [KC ] meet by P . Prove that
jPLj = jPN j .
5. Show that nkQ; (2n ; 2k) is divisible by n! for each positive integer n.
1

=0

6. Let R stand for the set of all real numbers. Show that there is no
function f : R ! R such that f (x + y ) > f (x)(1 + yf (x)) for all positive
real x, y .

Next, some catching up. Somehow, we misplaced Edward T.H. Wang's

solution to problem #1 of the XI Italian Mathematical Olympiad. Here is his
solution.
1. [1998 : 323] XI Italian Mathematical Olympiad 1995.
Determine for which values of the integer n it is
possible to cover up, without overlapping, a square
of side n with tiles of the type shown in the picture
where each small square of the tile has side 1.
 391

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario.
Call a tile of the given shape a \nonatile". We show that a perfect
cover of an n  n board by nonatiles is possible if and only if n is a mul-
tiple of 6. First note that a nonatile consists of 9 unit squares. Hence, if a
perfect cover exists, then 9 j n and thus, 3 j n. We claim that n must be
2

even. Suppose that an n  n board has been covered up by non-overlapping

nonatiles where n is odd. We colour all the n unit squares of the board
2

alternatingly black and white, as on an ordinary chessboard. Then clearly:

jnumber of black squares ; number of white squaresj = 1. We now classify
all the nonatiles into two kinds: black or white, depending on whether the
top (protruded) square is coloured black or white. Note that a black nonatile
has 6 black squares and 3 white squares, while a white nonatile has 6 white
squares and 3 black squares.
Hence, if there are b black nonatiles and w white nonatiles, then
jnumber of black squares ; number of white squaresj = 3jb ; wj 6= 1 and
this is a contradiction. Hence, n is even and 6 j n follows. To nish the
proof, it suces to demonstrate that a 6  6 board admits a perfect cover
(using 4 nonatiles) as shown by the diagram below.
1 1 1 1 1 2
3 1 1 1 2 2
3 3 1 2 2 2
3 3 3 4 2 2
3 3 4 4 4 2
3 4 4 4 4 4

Next we give a di erent generalization and solution of a problem of the

Dutch Mathematical Olympiad 1993 [1997 : 197; 1998 : 389].
5. [1997 : 197; 1998 : 389{390] Dutch Mathematical Olympiad, 1993.
P , P , : : : , P are eleven distinct points on a line, PiPj  1 for every
pair Pi, Pj . Prove that the sum of all (55) distances PiPj , 1  i, j  11 is
1 2 11

smaller than 30.

Solution by Achilleas Sinefakopoulos, student, University of Athens,
Greece.
More generally, consider n distinct adjacent points P , P , : : : , Pn on
a line (n  2) such that PiPj  1 for every pair P Pi, Pj . We shall prove by
1 2

induction on m that if n = 2m + 1 > 3, then i<j n Pi Pj < m(m + 1).

1
392

First, note that, if n = 3, then P P + P P + P P P= 2P P  2, with

equality if and only if P P = 1. Now, if n = 5, then <i<j< PiPj < 2
1 2 2 3 1 3 1 3

1 3
(since P P 3 < P P  1). Therefore,
1 5

2 4 1 5

X X X
PiPj = PiPj + P P +
1 5 (P Pk + Pk P )
1 5
i<j 5 <i<j<5 <k<5
1 1
X 1
X
= PiPj + P P +
1 5 PP1 5
<i<j<5 <k<5
1
X 1

= PiPj + 4P P < 2 + 4 = 6 .
1 5
1 <i<j<5
Suppose that the result is true for some n = 2m + 1 > 3. We show
that it is also true for n + 2. Indeed, by the inductive hypothesis, we have
P
<i<j<n Pi Pj < m(m + 1). Hence,
1 +1

X X X
PiPj = PiPj + P Pn +
1 +2 (P Pk + PkPn )
1 +2
i<j n+2 <i<j<n+2 <k<n+2
1 1
X 1
X
= PiPj + P Pn +
1 +2 P Pn1 +2
<i<j<n+2 <k<n+2
1
X 1

= PiPj + (n + 1)P1Pn+2
1 <i<j<n+2
< m(m + 1) + (n + 1) = (m + 2)(m + 1) .
The proof is complete.

To round things out with respect to the February 1999 number here is
an additional solution.
4. [1999 : 6{7] Vietnamese Mathematical Olympiad 3 / 1996,
Category B.
Determine all functions f : Z ! Z satisfying simultaneously two
conditions:
(i) f (1995) = 1996
(ii) for every n 2 Z, if f (n) = m, then f (m) = n and f (m + 3) = n ; 3,
(Z is the set of integers).
Solution by Pierre Bornsztein, Courdimanche, France.
Let f : Z ! Z satisfy the two conditions. From (ii), we deduce:
f (f (n)) = n for all n 2 Z; that is, f is involutory.
The equality f (m+3) = n;3 then gives m+3 = f (n;3) or f (n;3) =
f (n)+ 3. From this, we easily obtain, by induction, f (n ; 3k) = f (n)+ 3k
for all positive integers k.
 393

Let m be any integer; m may be written m = f (n) for an n (for

n = f (m), actually). Condition (ii) gives f (m + 3) = f (m) ; 3, and by
induction: f (m + 3k) = f (m) ; 3k for all positive integers k. All this can
be summed up by f (n + 3k) = f (n) ; 3k for all n; k 2 Z.
It follows that for every k 2 Z,
f (3k) = f (0) ; 3k ,
f (3k + 1) = f (1) ; 3k ,
f (3k + 2) = f (2) ; 3k .
Let k = 665. We have f (1995) = f (0) ; 1995 = 1996 and f (1996) =
1995 = f (1) ; 1995. Then f (0) = 3991 and f (1) = 3990.
Thus, for every k 2 Z,
f (3k) = 3991 ; 3k ,
f (3k + 1) = 3990 ; 3k ,
f (3k + 2) = f (2) ; 3k .
We remark that
f (3Z) = 3Z + 1 ,
f (3Z + 1) = 3Z ,
and so, f (3Z + 2)  3Z + 2 .
Moreover f is bijective, so f (3Z + 2) = 3Z + 2. Denote f (2) = 3a + 2,
where a 2 Z. Then:
 3991 ; m if m 6 2 (mod 3) 
if f is a solution then f (m) = 3a + 4 ; m if m  2 (mod 3) .
Conversely, a straightforward calculation shows that any such f satis es the
conditions.

Now, we turn to solutions from our readers to problems given in the

March 1999 number of the Corner with problems of the 19th Austrian-Polish
Mathematics Competition 1996 [1999 : 70{71].
1. Let k  1 be an integer. Show that there are exactly 3k; positive
1

integers n with the following properties:

(a) The decimal representation of n consists of exactly k digits.
(b) All digits of n are odd.
(c) The number n is divisible by 5.
(d) The number m = n has k odd (decimal) digits.
5
394

Solution by Jacqueline Freeman and Edward T.H. Wang, Wilfrid Laurier

University, Waterloo, Ontario.
Call a natural number \ideal" if it satis es conditions (a){(d). We show
that n = ak ak;


a 2 N is ideal if and only if ai 2 f5, 7, 9g for all i = 1,
2, : : : , k and a = 5. This is clearly true when k = 1, and thus, we assume
1 1

that k  2.
1

If n is ideal, then (c) implies that a = 5. We show that none of the

ai's can equal 1 or 3. Let n=5 = bkbk;


b and consider the process of
1

long division of n by 5. Clearly, ak 6= 1, 3 for otherwise n=5 would have only

1 1

k ; 1 digits. If ai = 1 or 3 for any i, 2  i  k ; 1, then clearly bi = 0, 2, 4,

6, or 8, depending on whether the remainder \carried over" when dividing
ai by 5 is 0, 1, 2, 3, or 4, respectively. This is a contradiction to (d).
+1

Conversely, suppose n = ak ak;


a , where ai 2 f5, 7, 9g for

all i = 1, 2, : : : , k and a = 5. To show that n is ideal, it clearly suf-
1 1

ces to verify (d). Since ak  5, m = n=5 has k digits, and we can write
1

m = bkbk;


b . Clearly, b is odd. If bi is even for some i, 2  i  k,

then from n = 5m we see that ai = 0, 1, 2, 3, or 4 depending on whether
1 1 1

the \carry" (when bi; is multiplied by 5) is 0, 1, 2, 3, or 4, respectively.

This is a contradiction. Hence, all the bi's are odd. This completes the proof
1

of our claim.
Finally, since a = 5 and each of the ai 's (i = 2, 3, : : : , k) can take on
any one of the 3 values 5, 7, or 9, the total number of ideal integers is 3k; .
1
1

Remark. In fact, it is not dicult to show that if n is ideal then all the
digits of n=5 must be 1, 5, or 9.
2. A convex hexagon ABCDEF satis es the following conditions:
(a) The opposite sides are parallel (that is, AB k DE , BC k EF , CD k FA).
(b) The distances between the opposite sides are equal (that is, d(AB; DE ) =
d(BC; EF ) = d(CD; FA), where d(g;h) denotes the distance between
lines g and h).
(c) \FAB and \CDE are right angles.
Show that diagonals BE and CF intersect at an angle of 45 .
Solution by Toshio Seimiya, Kawasaki, Japan.
Let X , Y be the feet of the perpendiculars from C to AF , EF , respec-
tively. (See gure.)
Since AF k CD, it follows that CX = d(AF; CD).
Since BC k EF , it follows that CY = d(BC; EF ).
Since d(AF; CD) = d(BC; EF ), we have CX = CY . Thus, CF
bisects \AFE . Similarly, BE bisects \DEF . Let T be the intersection of
AF and DE. Since AB k DE and \A = 90, we get
\FTE = 90 .
 395

A X F T

Y
 E


C D

Then,
\AFE + \DEF
= (\T + \TEF ) + (\T + \TFE )
= 2\T + (\TEF + \TFE )
= 90  2 + 90 = 270 .
Thus, we have
\PFE + \PEF = (\AFE + \DEF ) = 135 .
1
2

Hence, we get
\FPE = 180 ; (\PFE + \PEF )
= 180 ; 135 = 45 .
3. The polynomials Pn(x) are de ned recursively by P (x) = 0,
P (x) = x and
0

Pn(x) = xPn; (x) + (1 ; x)Pn; (x) for n  2 .

1 2

For every natural number n  1, nd all real numbers x satisfying the equa-
tion Pn (x) = 0.
Solutions by Mohammed Aassila, CRM, Universite de Montreal,
Montreal, Quebec; and by Pierre Bornsztein, Courdimanche, France. We
give Bornsztein's write-up.
We will prove that for n  1, the only real solution of Pn (x) = 0 is
x = 0.
For n  2,
Pn(x) ; Pn; (x) = (x ; 1)(Pn; (x) ; Pn; (x)) .
1 1 2
396

Then an easy induction leads to

Pn(x) ; Pn; (x) = (x ; 1)n; (P (x) ; P (x)) = x(x ; 1)n; .
1
1
1 0
1

That is, Pn (x) = Pn; (x) + x(x ; 1)n; for n  2, and we note that it
1

remains true for n = 1.

1

We deduce that, for n  1,

Pn(x) = x(x ; 1)n; + x(x ; 1)n; +


+ x + P (x)
1 2
0

= x((x ; 1)n; + (x ; 1)n; +


+ 1) .

1 2

Then, if x = 2, Pn (2) = 2n =6 0 and if x =6 2, Pn(x) = x
x;x;n; . Thus,

(( 1) 1)

Pn(x) = 0 if and only if x = 0 or (x ; 1)n = 1 for x = 6 2.

2

If n is even, (x ; 1) = 1 if and only if x = 0 or x = 2. Then Pn (x) = 0

n
if and only if x = 0.
If n is odd, (x ; 1)n = 1 if and only if x = 2. Then Pn (x) = 0 if and
only if x = 0.
Thus, for n  1, Pn (x) = 0 if and only if x = 0.
4. The real numbers x, y, z, t satisfy the equalities x + y + z + t = 0
and x + y + z + t = 1. Prove that ;1  xy + yz + zt + tx  0.
2 2 2 2

Solutions by Mohammed Aassila, CRM, Universite de Montreal,

Montreal, Quebec; by Michel Bataille, Rouen, France; by Pierre Bornsztein,
Courdimanche, France; and by Edward T.H. Wang, Wilfrid Laurier Univer-
sity, Waterloo, Ontario. We give Bataille's solution.
First, we have:
xy + yz + zt + tx = (x + z)(y + t) = ;(x + z)  0 2

(since y + t = ;(x + z )). Then,

jxy + yz + zt + txj  (x + y + z + t ) = (y + z + t + x ) = = 1
2 2 2 2 1 2 2 2 2 2 1 2

(by the Cauchy-Schwarz inequality). The conclusion follows.

Remark. Equality xy + yz + zt + tx = 0 holds if and only if x + z =
y + t = 0. Therefore, inequality xy + yz + zt + tx  0 becomes an equality
for the quadruplets (x; y;z; t) = (a; b; ;a; ;b) where a, b are real numbers
such that a + b = .
2 2 1
2

If equality xy + yz + zt + tx = ;1 holds, then we have equality in

the Cauchy-Schwarz inequality so that (x; y;z; t) and (y;z; t; x) are propor-
tional. Since at least one of the numbers x, y , z , t must be non-zero, this
leads to x = y = z = t or y = ;x, z = x, t = ;x. The rst case is
incompatible with the hypotheses, and the second case provides only the
quadruplets:
 397
   
1
2
; ; 21 ; 12 ; ; 21
and ; 12 ; 21 ; ; 12 ; 21 .
Conversely, these two quadruplets satisfy xy + yz + zt + tx = ;1 and we
may
; ; ;conclude: xy + yz + zt + tx = ;1 holds if and only if (x; y; z; t) =
1
2
; ; ;
or ;; ; ; ; ;
.
1
2
1
2
1
2
1
2
1
2
1
2
1
2

6. Natural numbers k, n are given such that 1 < k < n. Solve the
system of n equations
xi
(xi + xi +


+ xi k; ) = xi; for 1  i  n ,
3 2 2
+1
2
+ 1
2
1

with n real unknowns x , x , : : : , xn . Note: x = xn , xn = x ,

xn = x , and so on.
1 2 0 +1 1

+2 2

Solution by Pierre Bornsztein, Courdimanche, France.

We prove that there are two solutions
x = x =


= xn = 0
1 2

and
x = x =


= xn = p1k .
1 2 3

First, suppose that (x ; : : : ; xn ) is a solution.

1

First Case. There is i 2 f1, : : : , ng such that xi = 0.

From the cyclic symmetry, we suppose, without loss of generality, that
x = 0. Then
x (x +


+ xk ) = 0 = xn .
1
3 2 2 2
1 1

Thus, xn = 0.
Continuing, we deduce that xi = 0 for all i.
Second Case. xi 6= 0 for all i.
Since xi (xi +


+ xi k; ) = xi; , we deduce that xi > 0 for all i.
3 2 2
+ 1
2
1
From cyclic symmetry, without loss of generality, we have
x = min fxi : i = 1, 2, : : : , ng .
1

Then x  x and x  xk . Thus,

3
1
3
2
2
1
2
+1

x + x +


+ xk  x + x +


+ xk .
2
1
2
2
2 2
2
2
3
2
+1

We deduce that
xn = x (x +


+ xk )
2 3
1
2
1
2

 x (x +


+ xk ) = x .
3
2
2
2
2
+1
2
1

Thus, xn  x . 1
398

What we have actually shown is that if xi = minfx , : : : , xn g, then

xi; = xi (= minfx , : : : , xn g). An easy induction leads to x = x =
1

= xn.
1 1 1 2

Let a denote the common value. Then a (ka ) = a ; that is, a = pk .

3 2 2 1

 
3

Conversely, it is easy to verify that (0; 0; : : : ; 0) and pk ; : : : ; pk 3

1
3
1

are solutions.
7. Show that there do not exist non-negative integers k and m such
that k! + 48 = 48(k + 1)m .
Solution by Pierre Bornsztein, Courdimanche, France; and by Edward
T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's
solution.
Suppose the given equation holds for some non-negative integers k and
m. Then 48 j k!. Since 48 =m2  3, we must 4
have k  6. If k = 6 or 7,
the equation becomes 16 = 7 or 106 = 8m , respectively. Clearly, neither
is possible. Hence, k  8 and the given equation can be rewritten as
3  5  7  8 


 (k ; 1)  k + 1 = (k + 1)m . (1)
Suppose k + 1 is a composite. Then it has a prime divisor q . Since q  k,
we have q j k!, which implies q j 48. Since k  8, the left side of (1) is odd
and thus, q must be odd. Hence, q = 3, which is clearly impossible in view
of (1), since 3 6 j 1.
Therefore, k + 1 = p is a prime. Then k! = (p ; 1)!  ;1 (mod p) by
Wilson's Theorem; that is, p j k! + 1.
Rewriting the given equation as k! + 1 + 47 = 48pm , we infer that
p j 47 andmso p = 47. Then we have 46! + 48 = 48  47m or
46! = 48(47 ; 1). Since the prime divisors of 46! include 5, 7, 11 which
are all coprime with 48, we have 47m  1 (mod 5; 7; 11). Now, straightfor-
ward checking reveals that ord (47) = 4 (that is, the least positive integer n
such that 47n  1 (mod 5) is n = 4), ord (47) = 6 and ord (47) = 5.
5

Hence, m is divisible by lcmf4; 6; 5g; that is, 60 j m. So m  60 and we

7 11

have 48  47m  48  47 . Clearly, 48  47 > 46! + 48 and we have a

60 60

contradiction.

Next we turn to solutions to problems of the 3rd Turkish Mathematical

Olympiad [1999 : 72].
2. For an acute triangle ABC , k , k , k are the circles with diameters
[BC ], [CA], [AB ], respectively. If K is the radical centre of these
1 2 3

circles, [AK ] \ k = fDg, [BK ] \ k = fE g, [CK ] \ k = fF g and

area(ABC ) = u, area(DBC ) = x, area(ECA) = y , and area(FAB) = z ,
1 2 3

show that u = x + y + z .
2 2 2 2
 399

Solutions by Michel Bataille, Rouen, France; and by Toshio Seimiya,

Kawasaki, Japan. We give the solution by Seimiya.
A
k3 k2

D
R Q
K
E F
B P C
k1

Let P , Q, and R be the feet of the perpendiculars from A, B , and C ,

to BC , CA, and AB , respectively.
AP , BQ, and CR are common chords of k , k ; k , k ; and k , k ,
respectively. Thus, K is the orthocentre of 4ABC . Since \BDC = 90
2 3 3 1 1 2

and DP ? BC , we get
DP = BP
CP . 2
(1)
Since BK ? AC and AP ? BC , we have \BKP = \ACP . Further, we
have 4BKP  4ACP . It follows that BP : AP = KP : CP ; that is
AP
KP = BP
CP . (2)
From (1) and (2) we have
DP = AP
KP .
2

Hence, we have
DP
BC = (AP
BC )  (KP
BC ) .
2 2

From this we get

x = u  area (KBC ) .
2
(3)
Similarly, we have
y = u  area (KCA) ,
2
(4)
400

and
z = u  area (KAB) .
2
(5)
Therefore, we obtain from (3), (4), and (5),
x + y + z = u  farea (KBC ) + area (KCA) + area (KAB)g
2 2 2

= u  area (ABC )
= u . 2

3. Let N denote the set of positive integers. Let A be a real number

and (an)1
n be a sequence of real numbers such that a = 1 and
=1 1

1 < aan +1
 A for all n 2 N .
n
(a) Show that there is a unique non-decreasing surjective function k : N ! N
such that 1 < Aaknn  A for all n 2 N.
( )

(b) If k takes every value at most m times, show that there exists a real
number C > 1 such that C n  Aan for all n 2 N.
Solutions by Michel Bataille, Rouen, France; and by Pierre Bornsztein,
Courdimanche, France. We give Bataille's solution.
(a) The condition 1 < Aaknn  A is equivalent to Ak n ;  an < Ak n
( )
( ) 1 ( )

or k(n) ; 1j aAkn < k(n). Hence, k is necessarily the function given by
ln(
ln(
)
)

k(n) = 1 + aAn = 1 + blogA (an)c for all n 2 N. This shows the unicity
ln( )

of k. Since fan g is increasing and a = 1, we have an  1. We also note

ln( )

A > 1. Hence, aAn is a non-negative real number and 1 + blogA (an)c is

1
ln( )

a positive integer. Thus, we can de ne a function k : N ! N by the formula

ln( )

k(n) = 1 + blogA (an)c. For all n 2 N : an < an , so that

logA (an ) < logA (an ) and k(n)  k(n + 1). Thus, k is non-decreasing.
+1

Now, let us remark that, for s 2 N, k(n) = s is equivalent to

+1

As;  an < As. Therefore,

1
if fan g is bounded above, say an  M
for all n, an s satisfying As; > M (such an s exists since A > 1) cannot be
1

an image under k. Thus, ; if f
ang satis es the hypothesis and is convergent

( for instance an = 1 + n n , the surjectivity of k cannot be obtained.
1 1

Consequently, we will henceforth assume that fan g is not bounded above.

2

Given s 2 N, we prove that the equation k(n) = s has at least one

solution. For s = 1, n = 1 is obviously a solution, so now we suppose that
s  2. From the supplementary hypothesis, there exists n 2 N such that
an  As; . Let r be the
1
least of these n's, so that ar; < As;  ar . 1

Then ar  Aar; < A so that As;  ar < As and r is a solution. The

s
1
1

function k ful ls all the demands and (a) follows.

1
 401

(b) For each s 2 N, denote by N (s) the number of n such that k(n) = s;
that is, such that As;  an < As . By hypothesis, we have N (s)  m for
1

each s.
We will show that (b) holds with C = A =m > 1. Let n be an arbitrary
1

integer. If k(n) = s, then

n = N (1) + N (2) +


+ N (s ; 1) + j where j 2 f1, 2, : : : , N (s)g
(because there are N (1) + N (2) +


+ N (s ; 1) terms of the sequence
fang which are < As; ). On the one hand, anA  As; A = As = C ms,
1 1

and, on the other hand,

n = N (1) + N (2) +


+ N (s ; 1) + j
 N (1) + N (2) +


+ N (s ; 1) + N (s)  sm .
Hence, C n  C ms and we obtain C n  an A. So (b) is proved.
4. In a triangle ABC with jABj 6= jAC j, the internal and external
bisectors of the angle A intersect the line BC at D and E , respectively. If the
feet of the perpendiculars from a point F on the circle with diameter [DE ] to
the lines BC , CA, AB are K , L, M , respectively, show that jKLj = jKM j.
Solutions by Michel Bataille, Rouen, France; and by Toshio Seimiya,
Kawasaki, Japan. We give the solution by Seimiya.
M
A


L
F

B D C K E
Since the circle with diameter DE is the Apollonius circle, we have
FB = AB . (1)
FC AC
By the Law of Sines, we have
AB = sin C . (2)
AC sin B
402

From (1) and (2), we get

FB = sin C ;
FC sin B
that is,
FB sin B = FC sin C . (3)
Since the circle with diameter BF passes through K and M , we have
KM = FB sin B .
Similarly, we have
KL = FC sin C .
Therefore, we obtain from (3)
KL = KM .
5. Let t(A) denote the sum of elements of A for a non-empty subset
A of integers, and de ne t() = 0. Find a subset X of the set of positive
integers such that for every integer k there is a unique ordered pair of subsets
(Ak; Bk) of X with Ak \ Bk =  and t(Ak ) ; t(Bk) = k.
Solution by Pierre Bornsztein, Courdimanche, France.
We will prove that X = f3i : i 2 Ng has the desired property.
LEMMA. Every integer n can be written n =
P 3i where for all i,
i
2 f;1, 0, 1g and the sum has only a nite number of non-zero terms.
Moreover, such a decomposition is unique.
Proof
P of the Lemma. P First we prove the unicity:
If i i3i = j j 3j with i, j 2 f;1, 0, 1g, we recognize
=0 =0
two possibilities.
Case 1. For all i, j , i = j = 0, we are done.
Case 2. Otherwise, let i be the smallest subscript such that i 6= 0 or
i 6= 0. Then
0 0

X X
( i ; i )3i = ;
0 0
0
i3i + j 3j .
i>i0 j>i0
The right-hand side is divisible by 3i . We deduce that i ; i is divisible
0 +1

by 3. But ;2  i ;Pi  2. Thus, i = i . It follows that, for

0 0

0  i  i , i = i and i>i i3i = Pj>i j 3j .

0 0 0 0
0 0 0

If necessary, we denote by i the smallest subscript such that i > i

and i 6= 0 or i 6= 0. Then if 0  i < i , i = i , and with the same
1 1 0

1
reasoning as above, we get i = i , and so on. Unicity follows.
1 1

1 1
 403

Now, let p be a xed positive integer. The numbers of the form

N = Ppi i3i with i 2 f;1, 0, 1g are all distinct, from above. For each
i p2 f0, : : : , pg, there are 3 possible choices for i. Thus, we obtain exactly
=0

3 +1
such integers. Moreover
X
p X
p X
p
; 3i  i 3i  3i ;
i=0 i=0 i=0
that is, ! !
3p ;1 X p
i  3 ;1 .
p
; 
+1 +1

2 i 3 2
i =0
h p ; p+1 ;1 i
But there are exactly 3p integers in the interval ; ; . It
+1
+1 3 1 3
2 2
follows that each integer can be expressed in this form, proving the lemma.
P
Now, let k be an integer. From the lemma, k = i i3i , where
i 2 f;1, 0, 1g, and i 6= 0 for at most nitely many values of i. Denote
0

Ak = f3i : i = 1g and Bk = f3i : i = ;1g. Then Ak, B0 k 0 X ,

Ak \ Bk = ; and t(Ak) ; t(Bk) = k. Moreover, suppose that Ak, Bk are
subsets of X with A0k \ Bk0 = ; and t(A0k ) ; t(Bk0 ) = k. Note that A0k and
Bk0 must be nite. Then,
X X X X
3i ; 3i = k = 3i ; 3i .
3i 2Ak 3 i 2Bk 3i 2A0k 0
3i 2Bk

The unicity of the form

P 3i gives A = A0 and B = B0 .
i k k k k
6. Let N denote the set of positive integers. Find all surjective func-
tions f : N ! N satisfying the condition
m j n (=) f (m) j f (n)
for all m; n 2 N.
Solutions by Michel Bataille, Rouen, France; and by Pierre Bornsztein,
Courdimanche, France. We give Bataille's solution.
For such a function f , we have:
(a) f (1) = 1: Since 1 j n, we have f (1) j f (n) for all n 2 N. But f is
surjective, so that f (n) may be any positive integer; hence, f (1) divides all
positive integers and f (1) = 1.
(b) f is bijective: f is already surjective; moreover, if f (m) = f (n), then
f (m) j f (n) and f (n) j f (m) so that m j n and n j m; that is, m = n.
Hence, f is also injective.
Obviously, the condition on f also reads: m j n () f ; (m) j f ; (n) 1 1

for all m; n 2 N.
(c) f (p) is prime whenever p is prime: If k divides f (p), f ; (k) divides p, 1

so that f ; (k) = 1 or p and k = 1 or f (p).

1
404

More generally, f (pa) = (f (p))a for all positive integers a: Let q be a

prime integer dividing f (pa). Then f ; (q ) is a prime integer dividing pa . It
1

follows that f ; (q ) = p and q = f (p). Therefore, f (pa) is a power of f (p),

1

say f (pa) = (f (p))b. Now, the a + 1 distinct divisors of pa , namely 1, p,

: : : , pa, provide a + 1bdistinct divisors of f (pa) = (f (p))b, namely 1, f (p),
: :;: , f (p ). As (f (p)) has b + 1adivisors, wea have a  b. Similarly, using
a
f , b  a. Thus, a = b and f (p ) = (f (p)) .
1

(d) If u, v are coprime positive integers, then f (u), f (v ) are also coprime
and f (uv ) = f (u)f (v ). That f (u), f (v ) are coprime is immediate since
any prime integer p dividing f (u) and f (v ) would provide a prime integer,
namely f ; (p), dividing u and v .
1

Since u j uv and v j uv , f (u) j f (uv ) and f (v ) j f (uv ). Taking into

account the result we have just shown, we get f (u)
f (v ) j f (uv ). With
f ; , f (u), f (v) instead of f , u, v, we obtain u
v j f ; (f (u)
f (v)) and
1 1

consequently f (uv ) j f (u)
f (v ). Hence, f (uv ) = f (u)f (v). Let P be

the set of prime integers. We have obtained the following concerning f : The
restriction of f to P Qis a bijection from P onto PQ and, combining the last
two results, if n = p2P pvp n , then f (n) = p2P (f (p))vp n . [Here,
( ) ( )

vp(n) denotes the exponent of p in the standard factorization of n into prime

powers: of course, we have vp (n) = 0 for all but a nite number of p and
the sequence (vp(n))p2P is uniquely determined by n.]
Conversely, let  : P ! P be any bijection from P onto P and let
f : N ! N be de ned by
Y Y
f (n) = ((p))vp n ( )
when n = p vp n .
( )
(1)
p2P p2P
Clearly, f is surjective. Moreover,
m j n (=) vp(m)  vp(n) for all p 2 P
(=) v p (f (m))  v p (f (n)) for all p 2 P
(=) vq(f (m))  vq(f (n)) for all q 2 P
( ) ( )

(=) f (m) j f (n) .

In conclusion, the solutions are the functions extending to N the bijections
 : P ! P by means of the formula (1).

That completes the Olympiad Corner for this issue. Send me your
Olympiad contest materials and your nice solutions!
 405

BOOK REVIEWS
ALAN LAW
Calculus Mysteries and Thrillers, by R. Grant Woods
published by the Mathematical Association of America, 1998,
ISBN 0-88385-711-1, softcover, 131+ pages, $24.95 (US).
Reviewed by G.J. Grith, University of Saskatchewan, Saskatoon, Saskat-
chewan.
This book consists of eleven single variable calculus problems together
with their solutions. The problems are posed in story form. In each case,
the student is supposedly a client who is to provide a complete solution to a
judge, a government agency or some other \math freak", which implies that
slip-shod solutions are unacceptable.
My favourite problem is the rst, which involves the mathematics of
banking pool balls o a parabolic rail. I like it because it is relatively easy and
should be accessible to some of my better freshmen students (and perhaps it
re ects my mis-spent youth). I also like the Sunken Treasure problem, which
involves lowering a parabolic hull onto the ocean oor (classi ed as dicult)
and the somewhat standard Designing Dipsticks problem, which requires the
student to calculate volumes of solids of revolution.
I dislike The Case of the Swivelling Spotlight, Saving Lunar Station
Alpha and The Case of the Alien Agent, since I consider these to be far too
complicated even for top rate freshmen.
Dr. Woods has assigned seven of the eleven problems to groups of
students in his calculus classes and claims that \this experience has convinced
[him] that these projects are feasible for reasonably bright freshmen working
in groups." I can only comment that from my experience, this implies that
Manitoba freshmen are signi cantly ahead of their counterparts who reside
in the neighbouring province to the West.

The Math Chat Book, by Frank Morgan,

published by the Mathematical Association of America
(Spectrum Series), 2000.
ISBN 0-88385-530-5, softcover, 113 pages, $19.95 (U.S.).
Reviewed by Sandy Graham, University of Waterloo, Waterloo, Ontario.
This book is a compilation of questions and answers inspired by prob-
lems posed to the author during a live call-in TV show also called Math Chat.
The book/show does for mathematics, what the CBC program Quirks and
Quarks does for science. The author tries to pose interesting problems that
will make everyone enjoy math regardless of their past experiences. He has
406

divided the book into ve sections called stories: Time, Probabilities and
Possibilities, Prime Numbers and Computing, Geometry, and Physics and
the World. Each story is divided into several short episodes of related ques-
tions. There is even a puzzle at the beginning of the book with a $1000 prize
for the best solution submitted to the Mathematical Association of America's
website.
The problems range from modular arithmetic relating to leap years, to
some game theory analysis of tic-tac-toe, to classic statistics fallacies when
ipping a coin, to discussing the largest prime number computed to date.
Some of the problems posed, however, seem to have little to do with math-
ematics, such as \Which makes the water level in a bucket rise more, adding
a pound of salt or a pound of sand?" In a few cases, the author poses a prob-
lem but does not give a solution. In most cases, the solutions go into very
little detail of the underlying mathematics.
The book is quick to read for anyone interested in mathematics, and it
may spark some interest for those who have shied away from this discipline
in the past. The problems could provide interesting subject matter for dis-
cussion in a high school math club. With its wide variety of topics, teachers
could use some of the Math Chat scenarios as starting points for more de-
tailed analysis in almost any mathematics class, from high school to univer-
sity. Teachers may even be inspired to generate more quirky mathematical
problems relating to the topics in their courses.
Although the mathematical content is sometimes questionable,
overall the book is worth reading. My favourite excerpt comes from
Episode 7 | Predicting the Random. It is typical of many of the \problems"
of the book | a little math mixed with a little fun.
Mathematical Trick
1. What is 2 + 2?
2. What is 4 + 4?
3. What is 8 + 8?
4. What is 16 + 16?
5. Quick, pick a number between 12 and 5.
Math Chat predicts that your answer is 7. Try it out on your friends; it works.
Why not start your next discussion about it?
 407

A Nice COMC Problem

Daryl Tingley
Question B12 of the 1999 Canadian Open Mathematics Challenge was:
Q: Triangle ABC is any one of the set of trianglesphaving base BC equal
to a and height from A to BC equal to h, with h < a. P is a point inside
3

the triangle such that the value of

2

]PAB = ]PBA = ]PCB = .

Show that the measure of is the same for every triangle in the set.
The problem is one implication of the following locus problem:
L: LetpBC be parallel to the line ` such that the distance from BC to ` is less
than of the length of BC . Let A be a variable point on `. Find the locus
3

of all points P inside 4ABC such that

2

]PAB = ]PBA = ]PCB . ()

p3
The two constraints: \the distance from BC to ` is less than of the
length of BC ", and \P inside 4ABC " restrict the more general question:
2

G: Let BC be parallel to the line `. Let A be a variable point of `. Find the

locus of all points P that satisfy ().
These are nice problems for computer investigation. The locus for L is
hard to guess. When G is considered, the locus consists of several branches,
some of which are unfamiliar and dicult curves. A computer geometry pack-
age, such as Geometer's Sketchpad, easily allows one to nd solutions P
associated with each A. (Note that P must lie on the perpendicular bisector
of AB .) Such an investigation should reveal the answer to L, and indicate
that the answer to G is more dicult.
To use the geometry package to produce a trace of the locus, a compass
and straight edge type construction for P must be found. That is, given A, a
compass and straight edge construction must be found for the various P that
satisfy (). Such constructions are (of course) tied closely to a synthetic proof
of the problem. Finally, a computer package such as MAPLE can be used to
nd an equation for the unfamiliar branches of the trace. (After eliminating
the radicals, an 8th degree polynomial, with highest order term x y , is the
4 4

result.)
In this article, we outline a solution of G, leaving much of the work to
the reader. To avoid tiresome details, special cases are ignored: cases such
as when P lies on the lines that form the edges of the triangle. The reader
can verify (easily) that there is nothing exciting about them.
Copyright c 2000 Canadian Mathematical Society
408
S A `
P
r

B C
Theorem 1. Let BC be parallel to `. Let A be on ` and P inside 4ABC
with ]PAB = ]PBA = ]PCB = . If the line PC meets ` at S , then
the points B , P , A and S lie on a circle. As a consequence, SB = BC and
]ASC = ]BCS = ]BSC .
Proof. ]BCS = ]ASC = (interior opposite angles), showing that
]ASC = ]ASP = ]ABP , and that ASBP is concyclic. Hence,
]BSP = ]BAP = (angles on the same chord), so that SB = BC .
Remarks:
1. If h > a, it follows from Theorem 1 that there are no points P inside
4ABC satisfying ().
2. If h < a (we leave the case h = a to the reader), there are two points
(independent of A), call them S and S 0 , on `, with SB = S 0 B = BC
(we assume that S and S 0 are labelled so that ]SBC > 90 and
]S 0 BC < 90 ). Theorem 1 shows that the locus of P (inside 4ABC )
lies on SC [ S 0 C .
3. The point P is either at the intersection of the line SC , the circle SAB
and the perpendicular bisector of AB , or at the intersection of the line
S0C , the circle S0AB and the perpendicular bisector of AB. Thus, it is
easily constructed with compass and straight edge.
4. With the orientation of the above diagram, both S and S 0 must be to
the left of A, since the line CP must pass though the segment AB if P
is inside 4ABC .
p3
To answer Q, note that if h < a, then P cannot be inside 4ABC
and lie on S 0 C . For then we would have ]S 0 BC < 60 , giving that
2

]BCA > ]BCS 0 = ]BS 0 C > 60 (remember that S 0 must be to the left
of A) and that the sum of the angles of 4ABC is greater than
]PCB + ]PAB + ]PBA = 3]BCS 0 > 180 .
S
r
S0 A `

P r

B C
 409

To complete the discussion of L, we must decide which points on SC

can be the point P associated with some choice of A. As in the proof of
Theorem 1, SAPB is concyclic. Thus, given A on `, we can nd P on SC ,
and vice versa, using that circle.
Let M be the mid-point of SB , and let R be the point on SC such that
RM ? SB. Since ]SBC > 90, the point R exists.
S ` T
r
r R P r

M
B C
Theorem 2. The locus of question L is RC .
Proof. Let T be the point on ` with ]SBC = ]BCT . Suppose that P
is on RC . Let A be the point (other than S ) where circle SBP meets `.
(Consider the case that circle SBP is tangent to ` separately. Then, A = S ,
P = R.) If A is to the left of S, then ]ASP + ]ABP = 180 (opposite
angles of a cyclic quadrilateral). However, ]TSP = ]SBR < ]SBP <
]ABP , so that ]ASP + ]ABP > ]ASP + ]TSP = 180, a contradiction.
Thus, A must be to the right of S . Since ASBP is concyclic, it follows that P
is inside 4ABC .
We have already shown that if A and P satisfy the conditions of prob-
lem L, then P is on SC , SAPB is concyclic, and A is to the right of S . We
leave it to the reader to complete this theorem by showing that P must be
on the segment RC , and also to show that A is on the segment ST .
We now turn to problem G. First, consider the case in which P is (again)
inside 4ABC .
p
If a < h < a, the above ideas can be used to show that there is a
3

segment R0 C on S 0 C such that RC [ R0 C is the locus of P . The point R0

2

is the intersection of the perpendicular bisector of S 0 B with S 0 C . For there

to be a solution P on R0 C , the point A must lie on the segment S 0 T 0 , and
S0 must be to the left of T 0. In this case, there are two solutions for P
inside 4ABC : onep on RC and one on R0 C . Note that S 0 is to the left of T 0
if and only if h > a.
2
3

S S0 T 0 T
R
r

9R0
r

B C
410

For a given A, allowing P to be either inside or outside 4ABC ,

results in more solutions. Indeed, as A moves to the left of S , and P is
the intersection of the circle ASB with the line SC , the point P moves on
line SC past R, then past S . It can be shown that any point
the ;! ;;! P on the
ray CS satis es (). Likewise, P can be any point on the ray CS (even if 0
there are no solutions on CS 0 that lie inside 4ABC ). Thus, the locus of G
;! ;;
includes the rays CS and CS 0 .
!
If A is to the right of T , the circle ASB intersects SC at a point P
below BC . Then, ]PCB is supplementary to the equal angles ]PAB and
]PBA. Calling such a P a supplementary solution, we have that any point P
on lines SC or S 0 C is either a solution or a supplementary solution to G;
that is, there is an A on ` with ]PAB = ]PBA = ]PCB , or ]PAB =
]PBA = 180 ; ]PCB .
We have by no means solved G. We have seenp that if P is required to
be inside 4pABC , then the locus is RC [ R0 C ( a < h < a), or RC
2
3

(0 < h < a). If P is also allowed to be outside 4ABC , then the locus
3

;! ;;!
2

includes CS [ CS 0 .
For each A, the lines AB , BC and CA divide the plane into seven
regions: the inside of 4ABC , the in nite regions adjacent to the vertices
A, B, C (each bounded by two rays), and the in nite regions adjacent to the
sides AB , BC and CA (each bounded by two rays and a line segment).
Suppose that P is in the in nite region adjacent to A. Then ]PAB >
]PCB . (Extend PA to meet BC , and use the result that an exterior angle of
a triangle is greater than either of the two interior opposite angles.) Thus, P
cannot be a solution to G. Similarly, there is no solution P to G inside the
in nite regions adjacent to B or C .
For any A on `, the point P where the perpendicular bisector of AB
meets the circle ABC to the left of AB is a solution to G. (Use the circle
theorems!) Conversely, if P is a solution to G inside the in nite region ad-
jacent to AB , then PACB is concyclic, since ]PCB = ]PAB . Thus, P is
at the intersection of the perpendicular bisector of AB and the circle ABC ,
the intersection to the left of AB .
If P is a solution to G in the in nite region adjacent to BC , then P is
again a point of intersection of the circle ABC and the perpendicular bisector
of AB | this time, the intersection to the right of AB . However, if h > a,
then there is no choice of A that leads to a solution P in this region.
Finally, we must consider P in the in nite region adjacent to AC . We
have already seen some solutions;! in this region: if A is to the left of S (S 0 ),
;;!
then there is a solution P on RS (R0 S 0 ) which will be in the region adjacent
to AC . We leave it to the reader to prove that these are the only solutions
in this region.
 411

For some points A, there are supplementary solutions for G in the re-
gion adjacent to AC , namely, the point of intersection of the perpendicular
bisector of AB and the circle ABC that lies to the right of AB .
Theorem 3. The locus of G is indicated in the gures below for h < a and
h > a. The solutions are indicated with thick dashed lines, while the supple-
mentary solutions are indicated with thin dashed lines. Together, these loci
consist of the lines SC and SC 0 , and of those points where, for each A on `,
the perpendicular bisector of AB intersects the circle through A, B and C .

SP T 0
S
PPPP   `
PPPP 
B C
h=1
a 3

h=2
a B C
Daryl Tingley
Department of Mathematics and Statistics
University of New Brunswick
Fredericton, New Brunswick
Canada E3B 5A3
daryl@math.unb.ca
412

THE SKOLIAD CORNER

No. 49
R.E. Woodrow
This issue we continue the theme with the problems of the Final Round
of the Junior High School Mathematics Contest of the British Columbia Col-
leges. On the basis of the Preliminary Round, students are invited to write
this contest as part of a visit and tour of a college campus. My thanks go to
Jim Totten, The University College of the Cariboo, one of the organizers, for
forwarding the materials to us.
BRITISH COLUMBIA COLLEGES
Junior High School Mathematics Contest
Part A | Final Round | May 5, 2000
1. The last (ones) digit of a perfect square cannot be:
(a) 1 (b) 4 (c) 5 (d) 6 (e) 8
2. Suppose a string art design is con-
structed by connecting nails on a vertical s

axis and on a horizontal axis by line seg- s

ments as follows: The nail furthest from the

origin on the vertical axis is connected to
s

the nail nearest the origin on the horizon- s

tal axis. Then proceed toward the origin on s

the vertical axis and away on the horizontal s

axis as shown in the diagram. If this were

done on a project with 10 nails on each axis,
s

the number of points of intersection of line s s s s s s s s

segments would be:

(a) 45 (b) 46 (c) 47 (d) 48 (e) none of these
3. Assume there is an unlimited supply of pennies, nickels, dimes, and
quarters. An amount (in cents) which cannot be made using exactly 6 of these
coins is:
(a) 91 (b) 87 (c) 78 (d) 51 (e) 49
4. Given x + y = 28 and xy = 14, the value of x ; y equals:
2 2 2 2

(a) ;14 (b) 0 (c) 14 (d) 28 (e) 42

5. Given that 0 < x < y < 20, the number of integer solutions (x; y)
to the equation 2x + 3y = 50 is:
(a) 16 (b) 9 (c) 8 (d) 5 (e) 3
 413

6.
The numbers 1; 3; 6; 10; 15 : : : are known as triangular numbers.
Each triangular number can be expressed as n n where n is a natural num-
( +1)

ber. The largest triangular number less than 500 is:

2

(a) 494 (b) 495 (c) 496 (d) 497 (e) 498
7. An 80 m rope is suspended at its two ends from the tops of two
50 m agpoles. If the lowest point to which the mid-point of the rope can be
pulled is 36 m from the ground, then the distance, in metres, between the
agpoles is:
p
(a) 6 39
p
(b) 36 13
p
(c) 12 39
p
(d) 18 13
p
(e) 12 26
8. At a certain party, the rst time the door bell rang one guest arrived.
On each succeeding ring two more guests arrived than on the previous ring.
After 20 rings, the number of guests at the party was:
(a) 39 (b) 58 (c) 210 (d) 361 (e) 400
9. An operation  is de ned such that
A  B = AB ; BA .
The value of 2  (;1) is:
(a) ;3 (b) ;1 (c) ; 1
2
(d) 0 (e) 3
2

10. Three circles with a common

centre P are drawn as shown with
PQ = QR = RS. The ratio of the
area of the region between the inner and
middle circles (shaded with squares) to the P Q R
r r r r
S
area of the region between the middle and
outer circles (shaded with lines) is:

(a) 1
3
(b) 4
9
(c) 1
2
(d) 3
5
(e) 2
3

Part B | Final Round | May 5, 2000

1. (a) How many three-digit numbers can be formed using only the
digits 1, 2, and 3 if both of the following conditions hold:
(i) repetition is allowed;
(ii) no digit can have a larger digit to its left.
(b) Repeat for a four-digit number using the digits 1, 2, 3, and 4.
414

2. The square ABCD is inscribed in a P C

circle of radius one unit. ABP is a straight
line, PC is tangent to the circle. Find the
length of PD. Make sure you explain thor- B D
oughly how you got all the things you used O
q

to nd your solution!
A
3. If a diagonal is drawn in a 3  6 r

rectangle, it passes through four vertices of r

smaller squares. How many vertices does

the diagonal of a 45  30 rectangle pass r

through? r

4. Let a and b be any real numbers. Then (a ; b) is also a real number,

and consequently (a ; b)  0. Expanding gives a ; 2ab + b  0. If we
2 2 2

add 2ab to both sides of the inequality, we get a + b  2ab. Thus, for any
2 2

real numbers a and b, we have a + b  2ab. 2 2

Prove that for any real numbers a, b, c, d

(a) 2abcd  b c + a d .
2 2 2 2

(b) 6abcd  a b + a c + a d + b c + b d + c d .
2 2 2 2 2 2 2 2 2 2 2 2

5. A circular coin is placed on a table. Then identical coins are placed

around it so that each coin touches the rst coin and its other two neighbours.
(a) If the outer coins have the same radius as the inner coin, show that there
will be exactly 6 coins around the outside.
(b) If the radius of all 7 coins is 1, nd the total area of the spaces between
the inner coin and the 6 outer coins.

Last issue we gave the problems of the Preliminary Round of the British
Columbia Colleges Senior High School Contest for 2000. Here, thanks to Jim
Totten, The University College of the Cariboo, one of the organizers, are some
of the \ocial" solutions. Look for the rest in the next issue.
BRITISH COLUMBIA COLLEGES
Senior High School Mathematics Contest
Preliminary Round | March 8, 2000
1. Antonino sets out on a bike ride of 40 km. After he has covered half
the distance he nds that he has averaged 15 km/hr. He decides to speed
up. The rate at which he must travel the rest of the trip in order to average
20 km/hr for the whole journey is:
(a) 25 km/hr (b) 30 km/hr (c) 35 km/hr (d) 36 km/hr (e) 40 km/hr
 415

Solution. The answer is (b). Antonino averages 15 km/hr for the rst
20 km. This means it takes him 20=15 = 4=3 hours to cover the rst
20 km. In order to average 20 km/hr for a 40 km distance, he must cover
the distance in 2 hours. He only has 2=3 hours remaining in which to cover
the last 20 km. His speed over this last 20 km then must be (on average)
20=(2=3) = 30 km/hr.

2.
A circle with centre at (3; 2) intersects
the x{axis at the origin, O, and at the
r

; point B . The tangents to the circle at

(3 2)
O and B intersect at the point P . The
O B y{coordinate of P is:

(a) ;3 1
2
(b) ;4 (c) ;4 1
2
(d) ;5 (e) none of these

Solution. The answer is (c). Let C be the centre of the circle. Since
the points O and B are equidistant from the centre of the circle and also
equidistant from the point P , and since both O and B lie on the x{axis,
we see that P has coordinates (3; y ) with y < 0. The slope of OC is 2=3.
Since PO is the tangent line to the circle at O, we know that PO ? OC .
Therefore the slope of PO is ;3=2. However, the slope of PO is computed
to be (y ; 0)=(3 ; 0) = y=3. Together these imply that y = ;9=2.

3. From ve students whose ages are 6, 7, 8, 9, and 10, two are ran-
domly chosen. The probability that the di erence in their ages will be at
least 2 years is:

(a) 1
2
(b) 2
5
(c) 3
5
(d) 7
10
(e) 3
4

Solution. The answer is (c). First of all the number;of
possible ways
to choose a pair of distinct students from a set of ve is = = 10.
5 5!

From this we need only eliminate those whose age di erence is 1. Clearly
2 2!3!

there are exactly 4 such, namely (6; 7), (7; 8), (8; 9), and (9; 10). Thus, our
probability of success is 6 out of 10, or 3=5.
416

4. The centres of three circles of radius 2 units are located at the points
(0; 0), (12; 0) and (0; 5). If the circles represent pulleys, what is the length
of the belt which goes around all 3 pulleys as shown in the diagram?
;
(0 5)

;
(0 0) ;
(12 0)

(a) 30 +  (b) 30 + 4 (c) 36 +  (d) 60 ; 4 (e) none of these

Solution. The answer is (b). The straight sections of the belt are tangent
to all 3 pulleys and thus perpendicular to the radius of each pulley at the
point of contact. Thus the straight sections of the belt are the same lengths
as
p5the+ distances between the centres of the pulleys, which are 5, 12, and
2
12 = 13. Thus, the straight sections of belt add up to 30 units. The
2

curved sections of belt, when taken together, make up one complete pulley,
or a circle of radius 2. Thus the curved sections add up to 2 (2) = 4 . Thus,
the full length of the belt is 30 + 4 units.
5. If Mark gets 71 on his next quiz, his average will be 83. If he gets 99,
his average will be 87. How many quizzes has Mark already taken?
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8
Solution. The answer is (c). Let n be the number of quizzes Mark has
already taken. Let x be his total score on all n quizzes. Then we have the
following:
x + 71 = 83 ==) x = 83(n + 1) ; 71 ,
n+1
x + 99 = 87 ==) x = 87(n + 1) ; 99 .
x+1
Solving this system yields n = 6.

That completes the Skoliad Corner for this issue. The solutions to the
problems of the British Columbia Colleges Senior High School Contest for
2000 will be completed in the next issue.
Send me suitable contest materials and suggestions for the future of the
Corner.
 417

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
Mathematical Mayhem, Department of Mathematics, University of Toronto,
100 St. George St., Toronto, Ontario, Canada. M65 3G3. The electronic
address is
mayhem@math.toronto.edu
The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-
tario). The rest of the sta consists of Adrian Chan (Harvard University),
Jimmy Chui (University of Toronto), Donny Cheung (University of Waterloo),
and David Savitt (Harvard University)

Mayhem Problems
The Mayhem Problems editors are:
Adrian Chan Mayhem High School Problems Editor,
Donny Cheung Mayhem Advanced Problems Editor,
David Savitt 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 warmly welcome proposals for problems and solutions. With the
schedule of eight issues per year, we request that solutions from this issue
be submitted in time for issue 8 of 2001.

High School Problems

Editor: Adrian Chan, 1195 Harvard Yard Mail Center, Cambridge, MA,
USA 02138-7501 <ahchan@fas.harvard.edu>
H277.
(a) Find all right triangles with integer sides with perimeter 60.
(b) Find all right triangles with integer sides with area 600.
418

H278. Consider the time as seen on a digital clock in 24 hour mode.

(24 hour mode is representing the time relative to 12 midnight. For example,
6:25 am is 06:25, but 6:25 pm is 18:25. Also, 12:45 am counts as 00:45.) Let
n be the number we get when we remove the colon from the time T as seen
on a digital clock in 24 hour mode. Find all times T such that:
(i) n is a palindrome, [Ed. reads the same backwards as forwards.]
(ii) m, the number of minutes that T is after midnight, is a palindrome,
and
(iii) n = m.
H279. Proposed by Ho-joo Lee, student, Kwangwoon University,
Kangwon-Do, South Korea.
Let a and b be integers such that a  b (mod 3). Prove that
2
;a + ab + b

2 2
3

can be expressed as a sum of three non-negative squares.

H280. Proposed by Fotifo Casablanca, Bogota, Colombia.
In the spirit of the Olympics: There are 9 regions inside the rings of
the Olympics. Put a di erent positive whole number in each so that the ve
products of the numbers in each ring form a set of ve consecutive integers.

Advanced Problems
Editor: Donny Cheung, c/o Conrad Grebel College, University of
Waterloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>
A253. Proposed by Mohammed Aassila, Strasbourg, France.
Does there exist a polynomial f (x; y;z ) with real coecients, such that
f (x; y;z) > 0 if and only if there exists a non-degenerate triangle with side
lengths jxj, jy j, and jz j?
A254. In the acute triangle ABC , the bisectors of \A, \B, and \C
intersect the circumcircle again at A , B and C , respectively. Let M be the
point of intersection of AB and B C , and let N be the point of intersection
1 1 1

of BC and A B . Prove that MN passes through the incentre of triangle

1 1

ABC .
1 1

(1997 Baltic Way)

 419

A255. Proposed by Ho-joo Lee, student, Kwangwoon University,

Kangwon-Do, South Korea. qQ
P
De ne A = ( ni ai )=n, G = n ni ai , and H = n=( ni 1=ai )
P
for positive real numbers a , a , : : : , an . It is known that A  G  H ,
=1 =1 =1

from which it follows that 0  log(G=A) and 0  1 ; A=H . Prove that

1 2

0  log(G=A)  1 ; A=H , and determine when equality holds.

A256. Proposed by Mohammed Aassila, Strasbourg, France.
Prove that for any positive integer n, there exist n + 1 points
M , M , : : : , Mn in Rn such that for any integers i and j for which
1  i < j  n + 1, the Euclidean distance between Mi and Mj is 1.
1 2 +1

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>
C97. Given a positive integer n, let 0, 1, : : : , n ; 1 denote the inte-
gers modulo n (so that a is the reduction of a modulo n). Find all positive
integers n with the property that the set
fa j 0 < a < n=2 with a and n relatively primeg
is a group under multiplication.
C98. Find all pairs of integers (x; y) which satisfy the equation
x ; 34y = ;1 .
2 2
420

Problem of the Month

Jimmy Chui, student, University of Toronto
Problem. The graph of the function y = f (x) is concave downward over the
interval a  x  b. A and B are the points (a; f (a)), (b; f (b)), respectively.
The tangent to y = f (x) at any point P (x; f (x)), a  x  b, meets the line
x = a at Q and the line x = b at R. If the area bounded by the curve, the
tangent at P , the line x = a, and the line x = b, is minimized, prove that
the x{coordinate of the point P is independent of the function f (x).
(1997 Descartes, Problem 12)
Solution.
R
P
Q
A B

S T- x
a c b

Let P be at (c; f (c)). Let S and T be the points (a; 0) and (b; 0),
respectively. Without loss of generality, we can assume that the points A,
B, P , Q, and R, lie above the x-axis. (If they do not, we can merely translate
the points upward until they do lie above the x-axis, and this translation will
not a ect the area in question.)
Now, the area under the curve between x = a and x = b is xed.
Hence, we want to nd the c such that the area of trapezoid QRTS is a
minimum.
The area of the trapezoid is (b ; a)(Qy + Ry ), where Qy and Ry
1

are the y {coordinates of Q and R, respectively. But since (b ; a) is con-

2
1

stant, that only means we must nd the c that yields the minimum value of
2

Qy + Ry . This is equivalent to nding the c that yields the minimum value

of (Qy + Ry ); that is, the average value of the y {coordinates of Q and R.
1
2

Consider the tangent line y = g (x). This tangent line varies with c.
Because f (x) is concave down, the line g (x) lies entirely above f (x) except
at the contact point. In other words, g (x)  f (x) with equality if and only
if x = c.
 421

Now (Qy + Ry ) = (g (a) + g (b)) = g ( a b ), with the last equality

1 1 +

holding since g (x) is a linear function. But g ( a b )  f ( a b ), with equality

2 2 2
+ +

if and only if c = a b . In turn, this means that the minimum value of

2 2
+

(Qy + Ry ) is f ( a b ), and this occurs only when c = a b .

2
1 + +
2 2 2

Hence, the area in question is minimized when c = a b , which is in- +

dependent of f (x), QED.

2

Remark. This problem can also be solved using calculus.

J.I.R. McKnight Problems Contest 1986 | Solution

4. (b) Prove that in any acute triangle ABC ,
cot A + cot B + cot C = a +4bK + c ,
2 2 2

where K is the area of triangle ABC .

Solution by Vedula N. Murty, Dover, PA, USA.
We use the following formulae:
cos A = b +2cbc; a ,
2 2 2

sin A = a=(2R), where R is the circumradius of the triangle ABC , and

K = abc=(4R).
From these, we have
c ;a ),
cot A = R(b +abc
2 2 2

Using this equation and similar equations for cot B and cot C , we have
cot A + cot B + cot C = a +4bK + c .
2 2 2

This appears to be a little simpler than the one given in [1999 : 296{
297]. Now, why do we need the triangle to be acute?
422

Another Do-It-Yourself Proof of the n = 3

case of Fermat's Last Theorem
Andy Liu

Part Zero.
Everyone knows Fermat's Last Theorem, which states that the Dio-
phantine equation xn + y n = z n has no solution in non-zero integers for
all n  3. This article o ers a proof of the case n = 3 di erent from the
one presented in [2]. The idea of the present proof is taken from [1], with
two changes. First, the concept of quadratic residues, though very useful in
general, would take us too far a eld. It was discovered that the reference to
quadratic residues in the argument in [1] is actually redundant, and is ac-
cordingly excised. Second, the main lemma in the argument in [1] is rather
unmotivated. It is now presented in a manner which makes its discovery
plausible.
Part One.
We wish to prove that the Diophantine equation x + y = z has no
3 3 3

solutions in non-zero integers. Our approach is indirect. Suppose that such

a solution exists.
Problem 1.
Prove that no two of x, y and z are equal.
By the Well Ordering Principle, there exists a solution such that jxyz j
is minimal.
Problem 2.
Prove that x, y and z are pairwise relatively prime to one another, and
that exactly one of them is even.
We will attempt to construct another solution for which jxyz j is smaller.
This will yield the desired contradiction. We may assume that z is even. Then
x + y and x ; y are both even. Hence, there exist integers u and w such that
x + y = 2u and x ; y = 2w. It follows that x = u + w and y = u ; w. We
then have z = (u + w) + (u ; w) = 2u(u + 3w ).
3 3 3 2 2

Problem 3.
Prove that u and w are relatively prime to each other, and that exactly
one of them is odd.
Case 1.
Suppose u is not divisible by 3. Then u and 3w are relatively prime to
each other.
Copyright c 2000 Canadian Mathematical Society
 423

Problem 4.
Prove that 2u and u + 3w are relatively prime to each other.
2 2

It follows that there exist integers r and s such that 2u = r and 3

u + 3w = s . We shall continue the analysis in Part Four.

2 2 3

Case 2.
Suppose u is divisible by 3. Then u = 3v for some integer v , and
z = 18v(3v + w ).
3 2 2

Problem 5.
Prove that 3v and w are relatively prime to each other, as are 18v and
3v + w .
2 2

It follows that there exist integers r and s such that 18v = r and 3

3v + w = s . We shall continue the analysis in Part Four.

2 2 3

Part Two.
Both cases in Part One lead us to consider the Diophantine equation
a + 3b = s where (a; 3b) = 1 and a + b  1 (mod 2). We rst explore
2 2 3

the case where s is a prime.

Problem 6.
Prove that s > 3 and that neither a nor b is divisible by s.
Let b; be the inverse of b modulo s and let g = ab; . From
a + 3b p 0 (mod s), we have g + 3  0 (mod s). Let q = bpsc.
1 1

2 2 2

Then q < s < q + 1.

Problem 7.
Prove that g (i0 ; i00 )  j 0 ; j 00 (mod s) for some i0 , j 0, i00 and j 00 , each
an integer between 0 and q inclusive, such that (i0 ; j 0 ) 6= (i00 ; j 00 ).
De ne i = ji0 ; i00 j and j = jj 0 ; j 00 j. Then either gi + j  0 (mod s)
or gi ; j  0 (mod s). In any case, g i ; j  0 (mod s). 2 2 2

Problem 8.
Prove that 0 < i < ps and 0 < j < ps.
From g + 3  0 (mod s), we have g i + 3i  0 (mod s). Hence,
2 2 2 2

3i + j  0 (mod s), so that 3i + j = hs for some integer h.

2 2 2 2

Problem 9.
Prove that h = 1 or 3.
If h = 1, then we have s = 3i + j . If h = 3, then 3 divides j also,
2 2

so that j = 3k for some integer k. We then have s = i + 3k . In summary, 2 2

we have s = m + 3n for some integers m and n.

2 2

Problem 10.
Prove that (m; 3n) = 1 and m + n  1 (mod 2).
We have a + 3b = s = (m + 3n ) = m + 9m n + 27m n +
2 2 3 2 2 3 6 4 2 2 4

27n . We may take a = m ; Amn and b = Bm n ; 3n , so that

6 3 2 2 3

a + 3b = m + (3B ; 2A)m n + (A ; 18B)m n + 27n .

2 2 6 2 4 2 2 2 4 6
424

Problem 11.
Prove that the system of equations 3B ; 2A = 9 and A ; 18B = 27 2 2

has two solutions, namely, (A; B ) = (;3; ;1) and (9; 3).
Since in Part One we attempt to nd a solution smaller than the minimal
solution, we take a = m ; 9mn and b = 3m n ; 3n . 3 2 2 3

Part Three.
From our exploration in Part Two, we claim that a + 3b = s , with 2 2 3

(a; 3b) = 1 and a + b  1 (mod 2), has a solution given by s = m + 3n , 2 2

a = m ; 9mn and b = 3m n ; 3n for some integers m and n such

3 2 2 3

that (m; 3n) = 1 and m + n  1 (mod 2). We shall justify the claim using
induction on the number ` of prime factors of s.
Problem 12.
Prove the claim for the case ` = 0.
We could also have used the case ` = 1 as the basis of our induction.
Consider now the case where s has ` +1 prime factors. Let p be one of them.
Then s = tp where t has ` prime factors.
We have a + 3b = s , so that a + 3b  0 (mod p). From Part
2 2 3 2 2

Two, there exist integers c and d such that p = c + 3d . Moreover, 3 2 2

p = m + 3n , c = m ; 9m n and d = 3m n ; 3n where m and n

2 2 3 2 2 3

are integers such that (m ; 3n ) = 1 and m + n  1 (mod 2). Now,

1 1 1 1 1 1 1 1 1 1

1 1 1 1

t p = (a + 3b )(c + 3d ) = (ac + 3bd) + 3(ad ; bc)

3 6 2 2 2 2 2 2

= (ac ; 3bd) + 3(ad + bc) . 2 2

Problem 13.
Prove that (ad + bc)(ad ; bc) = p (t d ; b ) and deduce that p divides 3 3 2 2

exactly one of ad + bc and ad ; bc.

We may assume that p divides ad ; bc but not ad + bc, since the other
case can be handled in an analogous manner.
Problem 14.
Prove that p divides ad ; bc as well as ac + 3bd.
3

Let e = (ac + 3bd)=p and f = (ad ; bc)=p . 3 3

Problem 15.
Prove that e + 3f = t , a = ce + 3df and b = de ; cf , and deduce
2 2 3

that (e; 3f ) = 1 and e + f  1 (mod 2).

By the induction hypothesis, there exist integers m and n such that
(m ; 3n ) = 1, m + n  1 (mod 2), t = m + 3n , e = m ; 9m n
2 2
2 2 3 2

and f = 3m n ; 3n . Let m = m m + 3n n and n = m n ; n m .

2 2 2 2 2 2 2 2 2
2 3
2 2 2 1 2 1 2 2 1 2 1

Problem 16.
Prove that (m; 3n) = 1, m + n  1 (mod 2), s = m + 3n , 2 2

a = m ; 9mn and b = 3m n ; 3n .
3 2 2 3
 425

This completes the induction argument and justi es the claim.

Part Four.
We now complete the analysis begun in Part One.
Case 1.
The equation u + 3w = s has a solution in which u = m ; 9mn
2 2 3 3 2

for some integers m and n such that (m; n) = 1 and m + n  1 (mod 2).
Then r = 2u = 2m(m ; 3n)(m + 3n).
3

Problem 17.
Prove that 2m, m ; 3n and m + 3n are pairwise relatively prime to
one another.
It follows that 2m = , m ; 3n = and m + 3n = for some
3 3 3

integers , and .
Problem 18.
Prove that = + and 0 < j j < jxyz j.
3 3 3

This is the desired contradiction.

Case 2.
The equation 3v + w = s has a solution in which v = m n ; 3n
2 2 3 2 3

for some integers m and n such that (m; n) = 1 and m + n  1 (mod 2).
Then r = 18v = 3 (2n)(m ; n)(m + n).
3 3

Problem 19.
Prove that 2n, m ; n and m + n are pairwise relatively prime to one
another.
It follows that 2n = , m ; n = and m + n = for some integers
3 3 3

, and .
Problem 20.
Prove that = + and 0 < j j < jxyz j.
3 3 3

This is the desired contradiction.

Bibliography.
[1] W. Sierpinski, Elementary Theory of Numbers, North-Holland, Amster-
dam (1988) pp. 30 and 415{418.
[2] R. Vakil, A Do-It-Yourself Proof of the n = 3 case of Fermat's Last
Theorem, CRUX with MAYHEM 26 (2000) pp. 36{44.
Andy Liu
Department of Mathematical Sciences
University of Alberta
Edmonton, Alberta T6G 2G1
aliu@vega.math.ualberta.ca
426

An Interesting Application of the

Sophie Germain Identity
Carl Johan Ragnarsson

Abstract
The readers of this article should be familiar with modular arithmetic,
and may also recall Sophie Germain's identity. This article deals with an
application of the identity in solving the equation 3x + 4y = 5z .
A word about the identity
As the name says, Sophie Germain's identity was rst discovered by
Sophie Germain. It reads
a + 4b = a + 4a b + 4b ; 4a b = (a + 2b ) ; 4a b
4 4 4 2 2 4 2 2 2 2 2 2 2

= (a + 2ab + 2b )(a ; 2ab + 2b ) .

2 2 2 2

What is interesting about this identity is that sums of even powers do not
generally factor. Further, such sums factor only when the term we complete
the square with is, in itself, a perfect square. Its main application in contest
problem solving has, so far, often been trivial because of everyone knowing
the identity. When starting to read the chapter about number theory in [1],
I found that the identity was used in solving some simple problems related
to factoring integers, and the result was an immediate consequence of it. At
the time, I was quite sure that whenever I saw another problem involving
the identity, I would solve it immediately. I was totally wrong! We present
here an interesting application of the identity which is by no means obvious!
The problem
We propose the following problem, which was on the IMO Short List
as late as 1991, and which also appears with this source in [2].
Problem. Solve 3x + 4y = 5z in non-negative integers.
It is natural to try to prove that the only solutions are the well-known
triple (2; 2; 2) and the trivial (0; 1; 1). After starting to try to prove this, one
may always change one's course if an obstacle is found.
Counting modulo 4, we have (;1)x  1 (mod 4), so that x is even.
Let x = 2n, so that 9n + 4y = 5z . Now, counting modulo 5, we get
(;1)n + (;1)y  0 (mod 5), since z  1, showing that n and y have
opposite parities. We split into two cases:
Copyright c 2000 Canadian Mathematical Society
 427

Case 1: n even and y odd. Let n = 2m and y = 2t + 1, giving that

9 m + 4 t = 5z , or, by application of Sophie Germain's identity,
2 2 +1

(3m) + 4  (2t )
4 4
 
= (3m) + 2  3m 2t + 2  (2t ) (3m) ; 2  3m 2t + 2  (2t)
2 2 2 2

= 5z .
Noting that the di erence between the two factors is 3m 2t , which is not
+2

divisible by 5, we conclude that both brackets are not multiples of 5. This

implies
(3m ) ; 2  3m 2t + 2  (2t ) = (3m ; 2t) + (2t) = 1 ,
2 2 2 2

and since the two squares sum up to 1, they are 0 and 1 respectively, with
m = 0 and t = 0 as the only solution. Tracing back, we easily see that this
yields the triple (0; 1; 1) in the original equation.
Case 2: n odd and y even. Now, similarly to Case 1, letting y = 2t,
the equation becomes 9n + 16t = 5z . Now, counting modulo 8, we get
1  5z (mod 8), and thus, z is even. Thus, letting z = 2w, we obtain that
9n + 16t = 25w , or equivalently
(5w ) ; (4t) = (5w + 4t )(5w ; 4t) = 3 n ,
2 2 2

and since the di erence between the brackets is 2 t , which is not divisible
2 +1

by 3, we conclude that 5w ; 4t = 1, or 5w = 4t + 1. Again, counting

modulo 5, we have 0  (;1)t + 1 (mod 5), and thus, t is odd. Now, letting
t = 2s + 1, we get, by a second application of Sophie Germain's identity,
1 + 4 s = 1 + 4  (2s )
2 +1 4
 
= 1 + 2  2s + 2  (2s ) 1 ; 2  2s + 2  (2s )
2 2

= 5w .
Finally, since the di erence between the two brackets in the expression above,
2s , is not divisible by 5, we conclude that
+2

1 ; 2  2s + 2  (2s) = (1 ; 2s ) + (2s ) = 1 ,
2 2 2

and further, that s = 0. Tracing back, we easily see that this case leads back
to a unique triple as well, namely (2; 2; 2).
Summing up, we conclude that the equation has in all, only the two
solutions (0; 1; 1) and (2; 2; 2).
It is interesting how much we have learned from solving this problem.
Not least, it uses only the least bit of elementary number theory. It is in-
teresting to note that once again, the triple (3; 4; 5), which has appeared so
many times, appears here again, and allows for some interesting problem
solving methods. This also allows for brushing up some classics using the
same concepts, as seen above.
428

Further investigation
1. Find another solution of the proposed problem!
2. For which positive integers m, n can we factor mak + nbk?
3. In the reals, solve the system
3x + 4y = 5z ,
3y + 4z = 5x ,
3z + 4x = 5y .
4. Prove that for n  2, n + 4n is composite.
4

References
[1] Arthur Engel, Problem Solving Strategies, New York: Springer, 1998.
[2] Naoki Sato, Number Theory (unpublished).
Carl Johan Ragnarsson
Paalsjoevaegen 16 SE{223 63 Lund
Sweden
cjr2000@telia.com
 429

PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, Department
of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,
Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,
together with references and other insights which are likely to be of help to the editor.
When a proposal is submitted without a solution, the proposer must include sucient
information on why a solution is likely. An asterisk (?) after a number indicates that
a problem was proposed without a solution.
In particular, original problems are solicited. However, other interesting prob-
lems may also be acceptable provided that they are not too well known, and refer-
ences are given as to their provenance. Ordinarily, if the originator of a problem can
be located, it should not be submitted without the originator's permission.
To facilitate their consideration, please send your proposals and solutions
on signed and separate standard 8 12 "11" or A4 sheets of paper. These may
be typewritten or neatly hand-written, and should be mailed to the Editor-in-
Chief, to arrive no later than 1 April 2001. They may also be sent by email to
crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-
tions were written in LATEX). Graphics les should be in epic format, or encapsulated
postscript. Solutions received after the above date will also be considered if there
is sucient time before the date of publication. Please note that we do not accept
submissions sent by FAX.

2576. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon,

Spain.
Characterize the numbers n such that n! nishes (in base 2 notation)
with exactly n ; 1 zeros.
2577. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon,
Spain.
Let a , a , : : : , an (n  2) be positive integers. Determine the values
of n and k (2  k  n) for which the following identity holds:
1 2

gcd ;lcmfa ; : : : ; a g
= lcm k n;gcdfai ; : : : ; aik g

i ik i <:::<i
1i <:::<ik n
1
1
1 1
1

2578. Proposed by Joaqun Gomez Rey, IES Luis Bu~nuel, Alcorcon,

Spain.
For each integer n, determine the hundreds and the units digits of the
number
1+5 n . 2 +1

6
430

2579. Proposed by Paul Yiu, Florida Atlantic University, Boca Raton,

FL, USA.
The excircle on the side BC of triangle ABC touches AC and AB ,
respectively at YA and ZA . Likewise, the one on CA touches BC and BA
at XB and ZB , and the one on AB touches CA and CB at YC and XC .
Let A0 be the intersection of ZB XB and XC YC , B 0 be that of XC YC and
YA ZA, and C 0 be that of YA ZA and ZB XB . Show that AA0, BB0 and CC 0
are concurrent. What is the point of intersection of these three lines?
2580. Proposed by Ho-joo Lee, student, Kwangwoon University,
Seoul, South Korea.
Suppose that a, b and c are positive real numbers. Prove that
b+c + c+a + a+b  1 + 1 + 1.
a + bc b + ac c + ab a b c
2 2 2

2581. Proposed by Ho-joo Lee, student, Kwangwoon University,

Seoul, South Korea.
Suppose that a, b and c are positive real numbers. Prove that
ab + c + bc + a + ca + b  a + b + c .
2 2 2

a+b b+c c+a

2583. Proposed by Nairi M. Sedrakyan, Yerevan, Armenia.
C
B Given a point M inside the convex quadrangle
(see diagram), such that \AMB = \MAD +
\MCD, \CMD = \MCB + \MAB and
M
MA = MC .
A Prove that AB
CM = BC
MD.
D
2584. Proposed by Nairi M. Sedrakyan, Yerevan, Armenia.
You are given that X , Y , Z and T are points on the chord AB of the
circle ;. Circles ; and ; pass through the points X and Y , and touch the
circle ; at points P and S , respectively, while the circles ; and ; pass
1 2

through the points Z and T , respectively, and touch the circle ; at points Q
3 4

and R, respectively. Also, Q belongs to the arc APB and the segments XY
and ZT do not have common points. Prove that the segments PR, QS and
AB intersect at the same points.
2585. Proposed by Vedula N. Murty, Visakhapatnam, India.
Prove that, for 0 <  < =2,
tan  + sin  > 2 .
 431

2586. Proposed by Michael Lambrou, University of Crete, Crete,

Greece.
Find all (real or complex) solutions of the system
3x + x = y (1 + 3x ) ,
3 2

3y + y = z (1 + 3y ) ,
3 2

3z + z = w(1 + 3z ) ,
3 2

3w + w = x(1 + 3w ) .
3 2

2587. Proposed by Peter Y. Woo, Biola University, La Mirada, CA,

USA.
In the half plane Z = f(x; y ) : y  0g, let f be the union of the set
of all semicircles lying in Z with diameters on the x{axis, with the set of all
lines in Z perpendicular to the x{axis.
Denote by FXY the unique member of f that goes through any two
points X and Y in Z . For any three points A, B and C in Z , denote by
4ABC the curvilinear triangle formed by the arcs fAB , fBC and fCA.
Let A, B and C be any three points on the x{axis. Let P be any point
in the interior of 4ABC . Let A0 = fAP \ fBC , B 0 = fBP \ fCA and
C 0 = fCP \ f0AB . Let be the angle at A0 , interior to 4CAA0 , let be
the angle at B interior to 4ABB 0 , and let be the angle at C 0 interior to
4BCC 0 .      
Prove that tan
2 tan 2 tan 2 = 1 .
2588. Proposed by Niels Bejlegaard, Stavanger, Norway.
Each positive whole integer ak (1  k  n) is less than a given positive
integer N . The least common multiple of any two of the numbers ak is
greater than N .
Xn 1
(a) Show that < 2.
k ak
=1

X n 1
(b) ? Show that
a < 56 .
k
=1 k
X
n 1
(c) ? Find the smallest real number such that
a < .
k=1 k
432

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.

2464. [1999 : 366] Proposed by Michael Lambrou, University of

Crete, Crete, Greece.
Given triangle ABC with circumcircle ;, the circle ;A touches AB and
AC at D and D , and touches ; internally at L. De ne E , E , M , and
F , F , N in a corresponding way. Prove that
1 2 1 2

1 2

(a) AL, BM , CN are concurrent;

(b) D D , E E , F F are concurrent, and that the point of concurrency is
the incentre of 4ABC .
1 2 1 2 1 2

I. Solution by Toshio Seimiya, Kawasaki, Japan.

A

P Q
;A

; D1 D2
I

C
B

L
T
Let P and Q be the second points of intersection of LD and LD with
;, respectively. As shown in the gure above, let LT be the common tangent
1 2

to ; and ;A . Then
\BD L = \TLD
1 and \TLB = \BAL .
1

Hence, \BLP = \TLD ; \TLB = \BD L ; \BAL = \ALP so that

\ACP = \ALP = \BLP = \BCP . Therefore, CP is the bisector of
1 1

\ACB . Similarly, BQ is the bisector of \ABC .

Let I be the intersection of CP and BQ. Then I is the incentre of
4ABC . Since hexagon ABQLPC is inscribed in ;, by Pascal's Theorem
D , I and D are collinear. Thus, D D passes through the incentre I of
4ABC . Similarly, E E and F F pass through I . Therefore, D D , E E
1 2 1 2

and F F are concurrent at the incentre of 4ABC , and part (b) is proved.
1 2 1 2 1 2 1 2

1 2
 433

Since I is the incentre of 4ABC , we have \D AI = \D AI . Since

AD and AD are tangent to ;A, we have AD = AD . Thus, D I = D I
1 2

and AI ? D D . Since LD and LD are bisectors of \ALB and \ALC ,

1 2 1 2 1 2

1 2 1 2
respectively, we have
BL AL AL CL
BD = AD = AD = CD .
1 1 2 2

Thus, we get
BL = BD . (1) 1

CL CD 2

Since \BAC + \ABC + \ACB = 90 , we have \D AI + \D BI +

1 1 1
1 1

\D CI = 90 , so that
2 2 2
2

\D IB = \AD I ; \D BI = (90 ; \D AI ) ; \D BI = \D CI .
1 1 1 1 1 2

Similarly, we have \D BI = \D IC . Hence, we have 4BD I  4ID C .

1 2 1 2
Thus,
BD = ID = BI . 1 1

ID CD CI 2 2

Thus, we have
BD = BD
ID = BI .
1 1 1
2
(2)
CD ID CD
2 CI 2 2
2

From (1) and (2) we have

BL = BI . 2
(3)
CL CI 2

N A
M
Z
Y

X C
B
L
434

Now let X be the intersection of AL with BC (see diagram above).

Since \ABL + \ACL = 180 , we have from (3)
BX = [ABL] = AB
BL
sin \ABL
1
2
XC [ACL] AC
CL
sin \ACL
1
2

= AB BL AB BI
AC
CL = AC
CI ,
2

where [PQR] denotes the area of triangle PQR. Let Y and Z be the points
of intersection of BM and CN with AC and AB , respectively. Then we can
similarly show:
CY = BC
CI 2
and
AZ = CA
AI . 2

Y A BA AI 2 ZB CB BI 2

Therefore,
BX
CY
AZ = AB
BC
CA
BI
CI
AI = 1 . 2 2 2

XC Y A ZB AC BA CB CI AI BI 2 2 2

By Ceva's Theorem AX , BY and CZ are concurrent. This implies that AL,

BM and CN are concurrent.
II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.
Let AB 0 C 0 be the symmetric triangle of triangle ABC relative to the
bisector of angle A. Then both triangles have the same incircle and excircle
in the angle A. Let Ia be the centre of the excircle which touches AB at U ,
BC at L00 , CA at T , and B0 C 0 at L0 .
We apply inversion with pole A and power AB
AC . The inverse point
of B is C 0 and the inverse point of C is B 0 . The inverse of the circle ; is
therefore the line B 0 C 0 and hence, since the circle ;A touches AB , AC and
;, then its inverse is a circle that touches the sides of triangle AB 0 C 0 . (See
gure on page 435.)
If ;A were outside of ;, then its inverse would be the incircle of 4ABC ,
but now the inverse of ;A is the excircle in the angle A. Thus, the inverse of
D is the point U , the inverse of D is the point T , and the inverse of L is
the point L0 . Since L0 is the symmetric point of L00 , it follows that AL is the
1 2

isogonal line of AL00 . But it is known (F.G.-M. 1242, 1242a Gergonne-like

theorems) that AL00 (and BM 00 and CN 00 , which correspond to the other
angles of 4ABC ) passes through Nagel's point. Thus, the isogonals AL,
BM and CN are also concurrent.
Now, AIa is the bisector of angle A and intersects D D at J . The
inverse of the line D D is a circle which passes through A, U and T , and
1 2

since the points A, U , Ia , T are concyclic, then the inverse of J is the point Ia .
1 2
 435

U
B

L00 C0
I 2 ;A D 1

;
J
L L0
A
C D 2

T B0

Hence,
AJ
AIa = AB
AC =) AC AJ = AB
AIa
=) 4AJC  4ABIa
=) \ACJ = \AIaB = C2 .
Thus, the point J is the incentre of 4ABC , and D D , E E , F F 1 2 1 2 1 2
are concurrent.
Other consequences:
1. It is obvious that if ;A were outside of ;, then the results are the same,
but the point of concurrency would be the excentre instead of the
incentre.
2. The inverse of the circumcircles of 4ABD and 4ACD are the lines
C 0T and B0U , respectively (which are concurrent with AL0, F.G.-M.
2 1

1242), and hence, these circumcircles intersect AL at a point that is the

inverse of the symmetric of the adjoint of Gergonne's point of
4ABC (F.G.-M. 1242).
3. If r, r , r , r are the radii of the incircle and the excircles of 4ABC ,
s is the semiperimeter of 4ABC and R , R , R are the radii of
1 2 3

;A , ;B , ;C , respectively, then since the power of A to the excircle is

1 2 3

AT = s , then
2 2

R = sbc
r = sbc
s sr
1
2
1 =
2
r ,
; a cos (A=2) 2
436

and hence,
 
R + R + R = r cos (1A=2) + cos (1B=2) + cos (1C=2)
1 2 3
2 2 2

 cos ;(A +3Br + C )=6
= 4r ,

2

by Jensen's inequality, since the function f (x) = 1= cos x is convex.

2

Also solved by MICHEL BATAILLE, Rouen, France; CHRISTOPHER J. BRADLEY, Clifton

College, Bristol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; PETER
Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
Janous observes that Banko (Mixtilinear Adventure, CRUX 9 (1983), pp. 2-7) has shown
that the incentre I is even the mid-point of all three line segments; he further mentions a
forthcoming note by Paul Yiu (to appear in the American Mathematical Monthly), entitled
Mixtilinear Incircles, in which he shows that AL, BM and CN are concurrent at the external
centre of similitude of the circumcircle and the incircle of 4ABC .

2467. [1999 : 367] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.
r s
Given is a line segment UV and two
rays, r and s, emanating from V such h
that \(UV;r) = \(r;s) = 60 ,
and two lines, g and h, on U such 
g
that \(UV;g ) = \(g;h) = , 60
where 0 < < 60.
60

U V
The quadrilateral ABCD is determined by g , h, r and s. Let P be the
point of intersection of AB and CD.
Determine the locus of P as varies in (0; 60 ).
(Editor's note: As was pointed out by several solvers, P should be the
intersection of AC and BD.)
Solutionby Nikolaos Dergiades, Thessaloniki, Greece and Michael Lam-
brou, University of Crete, Crete, Greece (amalgamated and adapted by the
editors)
We solve a more general problem by replacing the condition:
\\(UV; r) = \(r; s) = 60 "
by
\UV is the external bisector of \(r;s)",
and replacing the condition
\\(UV; g ) = \(g;h) = "
by
\the lines g and h each meet the rays r and s".
 437

In this case the locus is (part of) the internal bisector of \(r; s). Thus,
for the problem as given originally, the locus is (part of) the perpendicular to
UV at V .
Let A = h \ r, B = g \ r, C = g \ s, D = h \ s, and let P = AC \ BD.
Let V P meet the line h at E .
r s
D h
E
A P g
B C
U V
In triangle V AD we have from Ceva's Theorem:
EA
BV
CD = ;1 , (1)
ED BA CV
and from Menelaus' Theorem (relative to U , B , C ):
UA
BV
CD = 1 . (2)
UD BA CV
From (1) and (2) we have
UA = ; EA ,
UD ED
which means that the points U , E are harmonic conjugates to A, D. Since
V U is the external bisector of \AV D, (that is, \(r; s)), we conclude that
V E is the internal bisector.
Finally, as g turns so that \(UV; g ) runs from 0 until g is parallel to s
and as h turns so that \(UV;h) runs from 0 until h is parallel to r (as in
the original problem), excepting the case when h is parallel to s, the point P
sweeps the said bisector.

Also solved by SEFKET  University of Sarajevo, Sarajevo, Bosnia and
ARSLANAGI C,
Herzegovina; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GERRY LEVERSHA,
St. Paul's School, London, England; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zalt-
bommel, the Netherlands; and PETER Y. WOO, Biola University, La Mirada, CA, USA.

2469. [1999 : 367] Proposed by Paul Yiu, Florida Atlantic University,

Boca Raton, FL, USA.
Given a triangle ABC , consider the altitude and the angle bisector at
each vertex. Let PA be the intersection of the altitude from B and the bisec-
tor at C , and QA the intersection of the bisector at B and the altitude at C .
438

These determine a line PA QA . The lines PB QB and PC QC are analogously

de ned. Show that these three lines are concurrent at a point on the line
joining the circumcentre and the incentre of triangle ABC . Characterize this
point more precisely.
[Ed. the solution is combined with that of the next problem.]
2470. [1999 : 368] Proposed by Paul Yiu, Florida Atlantic University,
Boca Raton, FL, USA.
Given a triangle ABC , consider the median and the angle bisector at
each vertex. Let PA be the intersection of the median from B and the bisector
at C , and QA the intersection of the bisector at B and the median at C . These
determine a line PA QA . The lines PB QB and PC QC are analogously de ned.
Show that these three lines are concurrent. Characterize this intersection
more precisely.
Combined generalization of 2469 and 2470 devised independently by
Nikolaos Dergiades, Thessaloniki, Greece and by Peter Y. Woo, Biola Uni-
versity, La Mirada, CA, USA.
Given 4ABC , its incentre I , and a point X not on the sides of the
triangle, de ne
PA = BXTT CI PB = CXTT AI PC = AX TT BI
Qa = BI CX QB = CI AX QC = AI BX ,
and prove that
(i) PA QA , PB QB , PC QC are concurrent at a point S , and
(ii) S lies on the line joining I to the isogonal conjugate of X .
Solution to (i) by Peter Y. Woo, Biola University, La Mirada, CA, USA.
The concurrency is an immediate consequence of Pappus' theorem: QA ,
PB , X are three points on the line CX whileTQB , PA, I are three points on
CI . The cross-joinsTintersect at S = PAQA QB PB , PC = IQA T QB X ,
and QC = PA X IPB , whose collinearity implies that S is on the axis
PC QC , which proves (i).
Editor's comment. Note the appropriateness of an alternative state-
ment of Pappus' theorem | \If two triangles are doubly perspective, they
are triply perspective." Here we are given that the triangles of P -points and
of Q-points are perspective from both I and X , so our conclusion is that they
are perspective from a third point S .
Solution to (ii) by Michel Bataille, Rouen, France (whose separate so-
lutions to 2469 and 2470 have been combined by the editor).
We shall use trilinear coordinates relative to 4ABC with A(1; 0; 0),
B(0; 1; 0), C (0; 0; 1) and I (1; 1; 1). Let the coordinates of the given point
X be (1=u; 1=v; 1=w) so that its isogonal conjugate is the point X 0 (u; v; w).
Let PA have the unknown coordinates (x; y; z ) that we want to evaluate in
terms of the given u, v and w.
 439

0x 0 1 1
u
PA is on BX so that det B@ y 1 v CA = 0, which gives wx = uz .
1

z 0 w 1

0x 0 11
PA is on CI so that det @ y 0 1 A = 0, which gives y = x.
z 1 1
From this we get PA (w; w; u) and, by suitable permutations, QA (v;u; v ),
PB (v;u; u), QB (w; w;v), PC (v;w; v) and QC (w;u; u). By adding PA to
QA , and so forth, we see that PAQA , PB QB , PC QC concur at the point S
that satis es
S = (v + w; w + u; u + v) .
By adding the coordinates of S to X 0 (u; v;w) we see that S , X 0 and the
incentre I (1; 1; 1) are collinear, as desired.
In 2469 we are given that X is the orthocentre H (sec A; sec B; sec C ); its
isogonal conjugate is the circumcentre X 0 (u; v;w) = O(cos A; cos B; cos C ).
Here S is the point on IO with coordinates (cos B + cos C; cos C + cos A;
cos A + cos B ); this point is X in Kimberling's list [Ed. Clark Kimberling,
65
Central Points and Central Lines in the Plane of a Triangle, Math. Mag. 67:3
(June 1994) 163-187. Alternatively, one can consult his web page:
http://cedar.evansville.edu/ck6/encyclopedia/].
Editor's comments. Janous and Yiu both show that I lies between O
and S and divides that segment in the ratio R : r. Most solvers noted that
S is the isogonal conjugate of the Schier point (X in Kimberling's list),
21
which Kimberling named for the proposer of problem 1018 [1986: 150-152].
In 2470 we are given X = G (1=a; 1=b; 1=c) whose isogonal conjugate
is the Lemoine point L(a; b; c). Here S = (b + c; c + a; a + b), which is
\the simplest unnamed centre" X in Kimberling's list.
37

[Comment. Janous and Yiu show instead that the centroid G lies be-
tween S and the isotomic conjugate of the incentre, and it divides that seg-
ment in the ratio 1 : 2. This fact also appears in Kimberling's table 3.]
Both problems were also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium,
Innsbruck, Austria; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer.
440

2471. [1999 : 368] Proposed by Vedula N. Murty, Dover,PA, USA.


n k; X (;1) k n + 1
For all integers n  1, determine the value of
1

k
=1
k+1 k .
Solution by Heinz-Jurgen Sei ert, Berlin, Germany.
Let Sn , n  1, be the sum in question. It is straightforward to check
that the equality
k n + 1 = n + 1 ; 1 n + 2
k+1 k k n+2 k+1
holds for k = 1, 2, : : : , n. Applying the above equality and the Binomial
Theorem, we nd
X
n n + 1 1 X n n + 2
Sn = (;1)k;1 k ;
k ; n + 2 k (;1) k + 1
1

k =1 =1

= ;(1 ; 1)n + 1 ; (;1)n

+1

; n +1 2 ;(1 ; 1)n ; 1 + (n + 2) ; (;1)n

+2

= 1 ; (;n1)+ (2n + 1) .
n


Also solved by SEFKET ARSLANAGI C, University of Sarajevo, Sarajevo, Bosnia and
Herzegovina; MICHEL BATAILLE, Rouen, France; CHRISTOPHER J. BRADLEY, Clifton College,
Bristol, UK; JAMES T. BRUENING, Southeast Missouri State University, Cape Girardeau, MO,
USA; NIKOLAOS DERGIADES, Thessaloniki, Greece; JOSE LUIS DIAZ, Universitat Politecnica
de Catalunya, Terrassa, Spain; CHARLES R. DIMINNIE, Angelo State University, San Angelo,
TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, CT, USA; RUSSELL EULER and
JAWAD SADEK, NW Missouri State University, Maryville, MO, USA; RICHARD I. HESS, Ran-
cho Palos Verdes, CA, USA; PETER HURTHIG, Columbia College, Vancouver, BC; WALTHER
JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Al-
berta, Edmonton, Alberta; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI
LAU, Hong Kong; HO-JOO LEE, student, Kwangwoon University, Kangwon-Do, South Korea;
DIGBY SMITH, Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Laurier
University, Waterloo, Ontario; and the proposer.

2472. [1999 : 368] Proposed by Vaclav Konecny, Ferris State Uni-

versity, Big Rapids, MI, USA.
If A, B , C are the angles of a triangle, prove that
     
cos A ;2 B cos B ;2 C cos C ;2 A
2 2 2

 A B   C  3

 8 sin 2 sin 2 sin 2 .

 441

I. Editorial comment.
As some solvers pointed out, this follows immediately from
problem 2382 [1999 : 440]. In fact, in [1999 : 441], it is given that
 
cos B ;2 C  8 sin A2 sin B2 sin C2 ( = 2Rr ) ,
2

and, by symmetry, the same is true for cos [(A ; B )=2] and cos [(C ; A)=2],
2 2

and so we are done.

II. Solution by Heinz-Jurgen Sei ert, Berlin, Germany.
In CRUX 585 [1981 : 303] it was shown that
     
cos A ;2 B cos B ;2 C cos C ;2 A  8 sin A2 sin B2 sin C2 . (1)
By squaring (1) and multiplying the right hand side of the resulting inequal-
ity by 8 sin(A=2) sin(B=2) sin(C=2), which is  1 (see O. Bottema et al,
Geometric Inequalities, item 2.12), the desired inequality follows.

Also solved by SEFKET ARSLANAGI C,  University of Sarajevo, Sarajevo, Bosnia and
Herzegovina (two solutions); MICHEL BATAILLE, Rouen, France; NIKOLAOS DERGIADES,
Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete,
Greece; KEE-WAI LAU, Hong Kong; HO-JOO LEE, student, Kwangwoon University, Kangwon-
Do, South Korea; D.J. SMEENK, Zaltbommel, the Netherlands; ECKARD SPECHT, Otto-von-
Guericke University, Magdeburg, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and
the proposer.
Dergiades and Lambrou also derive the stronger inequality (1), though without referring
to CRUX 585 to do it. Janous nds the even stronger inequality
   Y
cos B ; C  9 ; 8 sin A sin A  8 sin A .
Y Y Y
2 2 2 2
In the other direction, Janous also proves
  Y 2
cos B ; C  1 + 3 sin A + 8 sin A .
Y Y
2 2 2 2
Lambrou further connected this problem to another earlier CRUX problem, as follows.
Put A =  ; 2A1 , etc.; then A1 , B1 , C1 are the angles of an acute triangle, and (1) becomes
Y Y
cos(B1 ; A1 )  8 cos A1 .
Rewriting A1 as A, etc., we get that (1), over all triangles, is equivalent to the inequality
Y Y
cos(B ; A)  8 cos A (2)
over all acute triangles. Now let O be the circumcentre of triangle ABC , let AO meet the
circle BOC again at A0 , and de ne B 0 and C 0 similarly. Then the radius R1 of circle BOC
satis es
R1 = 2 cos R :
A
Letting O1 be the centre of circle BOC , we nd that \O1 OA0 = B ; C , so
OA0 = 2R1 cos(B ; C ) = R cos( B ; C) ;
cos A
and cyclically for OB 0 and OC 0 . Thus (2) is equivalent to
OA0
OB 0
OC 0  8R3 ;
which is unused problem 13 from the 1996 IMO; see [1999 : 8] for a solution.
442

2473. [1999 : 368] Proposed by Miguel Amengual Covas, Cala

Figuera, Mallorca, Spain.
Given a point S on the side AC of triangle ABC , construct a line
through S which cuts lines BC and AB at P and Q, respectively, such that
PQ = PA.
Solution by D.J. Smeenk, Zaltbommel, the Netherlands.
Let D be the re ection of A in BC , so that 4DBC is congruent to
4ABC .
Let the line through S parallel to AB intersect BD at T . Let the cir-
cumcircle of 4STD intersect BC at P and P . Let SP and SP intersect
AB and Q and Q , respectively.
1 2 1 2

1 2

Then P Q = P A and P Q = P A.
1 1 1 2 2 2

P 2

D
C
r
O


S  T
P 1

A Q 2 B Q 1

Proof. Note that BC is an axis of symmetry of quadrilaterals

ABCD , ABDP , and ABDP . (1) 1 2

Since quadrilateral SP TD is cyclic, we have

1

\P ST = \P DT =  , say.
1 (2) 1

Now, (1) and (2) imply that \P AB = \P DB = . Since ST kAB , we

1 1
have
\P Q A = \P ST =  .1 1 (3) 1
 443

Now, (2) and (3) imply that P Q = P A.

1 1 1

From (1), we obtain that

\P AB = \P DB .
2 2 (4)
Since quadrilateral P STD is cyclic, we have
2

\P ST = \P Q B = 180 ; \P AB .
2 2 2 2 (5)
Now, (4) and (5) imply that \P AQ = \P Q A, so that P Q = P A.
2 2 2 2 2 2 2

Also solved by MICHEL BATAILLE, Rouen, France; NIELS BEJLEGAARD, Stavanger,

Norway; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengym-
nasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; TOSHIO
SEIMIYA, Kawasaki, Japan; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the
proposer.
Smeenk commented that he found it to be a very interesting problem on which he spent
much time, but found only a quadratic equation with unknown tan , which was quite unsat-
isfactory. He then asked a friend, who asked another friend, who found this elegant solution in
an old issue of Euclides, a Dutch mathematical periodical.
Bataille, Bejlegaard and Lambrou all made reference to part of this problem being a
well-known classical problem. Bataille refered to H. Dorrie, 100 Great Problems of Elemen-
tary Geometry, Dover, 1965, and Lambrou, to B. Bold, Famous Problems of Geometry, also
published by Dover.

2474?. [1999 : 368] Proposed by Mohammed Aassila, CRM, Uni-

versite de Montreal, Montreal, Quebec.
Let f : R ! R be a decreasing continuous function satisfying, for
+ +

all x, y 2 R :+

 
f (x + y) + f ;f (x) + f (y)
= f f ;x + f (y)
+ f ;y + f (x)
.
Obviously f (x) = c=x (c > 0) is a solution. Determine all other solutions.
Editor's remark: This problem should have been starred when posed,
since it was proposed without a solution.
Comment by proposer: One can prove that if f satis es the functional
equation, then f (f (x)) = x for all x 2 R (this was problem 5 of the 1997
+

Iranian Mathematical Olympiad), but determining all the solutions seems to

be a very challenging problem.
This problem remains open.
444

2475. [1999 : 368] Proposed by JoaqunGomez

 Rey, IES Luis Bu~nuel,
Alcorcon, Spain.
Prove that
X
n X
n   n 
(;1)j k 22nj 2k2; 1 = 0.
+

j k =0 =1

I. Solution by David Doster, Choate Rosemary Hall, Wallingford, CT,

USA.
Let An =
Pn (;1)j ; n
and B = Pn (;1)k; n
.
j n 2
k;
2

j =0
2
k =1
2 1

Then
Pn Pn (;1)j k; n
; n
= A B and (1 + i) n = A ; iB .
j k;
+ 2
n n 2
n n 2

j k 2 2 1

;

=0 =1

Thus, (;4)n = (2i) n = (1 + i) n = (1 + i) n = (An ; Bn ) ; 2iAnBn ,

2 2 2 4 2 2

from which we infer that An ; Bn = (;4)n and An Bn = 0.

2 2

II. Solution by Gerry Leversha, St. Paul's School, London, England.

It is sucient to note that the given expression factorizes:
X
n X
n 2n 2n  0 Xn 2n1 X
n  2n !
(;1) j k @ j A k
2j 2k ; 1 = j (;1) 2j k (;1) 2k ; 1
+

j =0 k=1 =0 =1

and then to note that, by the symmetry of Pascal's triangle, the rst bracket
on the right side is zero for odd values of n and the second bracket is zero for
even values of n.

Also solved by SEFKET ARSLANAGI C,  University of Sarajevo, Sarajevo, Bosnia and
Herzegovina; MICHEL BATAILLE, Rouen, France; CHRISTOPHER J. BRADLEY, Clifton College,
Bristol, UK; JAMES T. BRUENING, Southeast Missouri State University, Cape Girardeau, MO,
USA; NIKOLAOS DERGIADES, Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo State
University, San Angelo, TX, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER
JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Al-
berta, Edmonton, Alberta; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI
LAU, Hong Kong; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal
College, Calgary, Alberta; CHOONGYUP SUNG, Pusan Science High School, Pusan, Korea;
EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; PETER Y. WOO, Biola
University, La Mirada, CA, USA; JEREMY YOUNG, student, Nottingham High School, Notting-
ham, UK; and the proposer.
All the submitted solutions are minor variations of either I or II above, with the only
exception of that by the proposer, which uses De Moivre's formula.
 445

2476?. [1999 : 429] Proposed by Mohammed Aassila, CRM, Uni-

versite de Montreal, Montreal, Quebec.
Let n be a positive integer and consider the set f1, 2, 3, : : : , 2ng. Give
a combinatorial proof that the number of subsets A such that
1. A has exactly n elements, and
2. the sum of all elements in A is divisible by n,
is equal to
1 X(;1)n d2d ; n
,
+
n d jn d d
where  is the Euler function. [Ed. Note the correction in the line above.]
Note: When n is prime, proving the formula is problem 6 of the 1995
IMO. A non-combinatorial proof of the formula is due to Roberto Dvornicich
and Nikolay Nikolov.
Editor's remark: This problem should have been starred when posed,
since it was proposed without a solution.
This problem remains open.

2478. [1999 : 429] Proposed by JoaqunGomez

 Rey, IES Luis Bu~nuel,
Alcorcon, Spain.
n n ; k 2k2n ; 2k
X
For n 2 N, evaluate .
k=0 k + 1 k n;k
Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria
(adapted by the editor).
[Ed: This solution assumes familiarity with Newton's generalized
binomial coecients together with some identities thereof.]
Let Sn denote the summation to be evaluated.
The following formulas regarding generalized binomial coecients are
known and easy to verify:
;  1 2k
(;4)k 2
= k , (1)
;k1
= (;1)n , (2)
 mn  m m ; 1 ,
k + 1 = (3)

X l m
n  kl++1m k
k n;k = n . (4)
k=0
446

[Ed: In all formulas, l and m denote arbitrary integers and n and k,

non-negative integers. Formula (4) is usually referred to as the Vander-
monde's Convolution Formula.]
Using (3) with m = and ; , respectively, we obtain
1
2
1
2

  1
1 ;  1

=
2 2
(5)
 k;+ 1 2(k + 1) k
1
; 1  ;  3

and
n ; k = 2(n ; k) n ; k ; 1 . (6)
2 2

Therefore, we have
;1 n ; k
nX ;  ; 
(;4)k ;
1
n k
1

Sn = k (;4) n ; k (by (1))

2 2

k k+1
=0

nX;    
(;4)n 2 k + 1 (; 12 ) n ;;k ; 1
1 1 3

= 2 2
(by (5), (6))
k =0

nX;  1  ; 
;(;4)n
1 3

= k+1 n;k;1
2 2

k
;1  ; !
=0

1 3

;(;4) n
= n ; 0 n (by (4))
2 2

 ; ! 1

= n n
;(;4) (;1) + 2(n + 1) n + 1 2
(by (2), (6))
n + 1 2n + 2
= n
;4 + 2 n + 1 (by (1))
2n
= (2n + 1) n ; 4n .

Also solved by MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, CRM, Universite de
Montre al, Montre al, Quebec; G.P. HENDERSON, Garden Hill, Campbellcroft, Ontario; HEINZ-

JURGEN SEIFFERT, Berlin, Germany; SOUTHWEST MISSOURI STATE UNIVERSITY PROBLEM
SOLVING GROUP, Spring eld, MO, USA; and the proposer.
Most of the other submitted solutions involve di erentiation and integration of the
power series expansion of the function f (x) = (1 ; 4x); 12 for j x j< 41 .
 447

2479. [1999 : 430] Proposed by JoaqunGomez  Rey, IES Luis Bu~nuel,

Alcorcon, Spain.
Writing  (n) for the number of divisors of n, and ! (n) for the number
of distinct prime factors of n, prove that
X
n X
n Xc  bn=kc 
bn=k
( (k)) = 2
2!(k) j .
k=1 k=1 j =1
Solution by Heinz-Jurgen Sei ert, Berlin, Germany. If n; k 2 N, then
there exist m 2 N and r 2 f0; 1; : : : ; k ; 1g such that n ; 1 = km + r.
0
Then n r + 1 n ; 1
k = m+ and
k = m. k
Hence,
 n   n ; 1  ( 1 if kjn.
k ; k = 0 otherwise.
Let Sn , n 2 N , denote the sum on the right hand side of the desired
equation (empty sums are understood to be 0). If n 2 N, then from the
0

above result we have

X Xc  bn=kc 
bn=k  
Sn ; Sn; = 1 2! k
( )

j ; b(n ;j1)=kc ,
kjn j =1
or, by applying the result once again,
X n
Sn ; Sn; = 1 2!(k) k .
kjn
With the notation of Dirichlet products (see, for example, T.M. Apostol, In-
troduction to Analytic Number Theory, 2nd Ed., Springer, 1984, 29-39), the
latter equation may be written as Sn ; Sn; = (f   )(n), n 2 N, where the
arithmetical function f is de ned by f (n) = 2! n , n 2 N. We claim that
1
( )

f   =  (where  means ordinary pointwise multiplication).

2 2
Since f ,  ,
and  are all multiplicative, it suces to show that (f   )(pe) = ( (pe))
2 2

when p is a prime and e 2 N. We have

X
e
(f   )(pe) = (  f )(pe) =  (pj )f (pe;j )
j =0
;1
eX ;1
eX
=  (pe) + 2  (pj ) = (e + 1) + 2 (j + 1)
j =0 j =0
= (e + 1) = ( (pe)) .
2 2
448

It follows that Sn ; Sn; = ( (n)) for all n 2 N. Replacing n by k and

2

summing over k = 1, 2, : : : , n, n 2 N, gives the requested equation.

1

Also solved by MICHEL BATAILLE, Rouen, France; WALTHER JANOUS, Ursulinengym-

nasium, Innsbruck, Austria; KEE-WAI LAU, Hong Kong; and the proposer.
Janous observes that the identity
X !(k)  n 
( (n))2 = 2 
kjn k
(implied in the above proof, but not explicitly stated) may be new and is worth noting in its
own right, and that applying the Mobius inversion formula to it yields another identity which
may be new:
(k) 2 nk
X !(k) X  
2 =
kjn kjn

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