131

THE OLYMPIAD CORNER

No. 181

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,
Calgary, Alberta, Canada. T2N 1N4.
We begin this number with two contests. Thanks go to Richard
Nowakowski, Canadian Team Leader to the 35th IMO in Hong Kong, for
collecting them and forwarding them to us.
SELECTED PROBLEMS FROM THE ISRAEL
MATHEMATICAL OLYMPIADS, 1994
1. p and q are positive integers. f is a function dened for positive
numbers and attains only positive values, such that f (xf (y )) = x y . Prove
p q

that q = p2 .
2. The sides of a polygon with 1994 sides are a = p4 + i2, i =
1; 2; : : : ; 1994. Prove that its vertices are not all on integer mesh points.
i

3. A \standard triangle" in the plane is a (lled) isosceles right triangle

whose sides are parallel to the x and y axes. A nite family of standard
triangles, containing at least three, is given. Every three of this family have
a common point. Prove that there is a point common to all triangles in that
family.
4. A shape c0 is called \a copy of the planar shape c" if the following
conditions hold:
(i) There are two planes  and  0 and a point P that does not belong
to either of them.
(ii) c 2  and c0 2  0 .
(iii) A point X 0 satises X 0 2 c0 i X 0 is the intersection of  0 with the
line passing through X and P .
Given a planar trapezoid, prove that there is a square which is a copy
of this trapezoid.
5. Find all polynomials p(x), with real coecients, satisfying
(x , 1)2p(x) = (x , 3)2p(x + 2)
for all x.
132

PROBLEMS FROM THE BI-NATIONAL

ISRAEL-HUNGARY COMPETITION, 1994
1. a1; : : : ; a ; a +1 ; : : : ; a are positive numbers (k < n). Suppose
that the values of a +1 ; : : : ; a are P
k k n
xed. How should one choose the values
of a1 ; : : : ; a in order to minimize
k n

6= j ?
k
a i
i;j;i j a

2. Three given circles pass through a common point P and have the
same radius. Their other points of pairwise intersections are A, B , C . The
3 circles are contained in the triangle A0 B 0 C 0 in such a way that each side of
4A0 B0 C 0 is tangent to two of the circles. Prove that the area of 4A0 B0 C 0
is at least 9 times the area of 4ABC .
3. m, n are two dierent natural numbers. Show that there exists a
real number x, such that 13  fxng  23 and 13  fxmg  23 , where fag is
the fractional part of a.
4. An \n-m society" is a group of n girls and m boys. Show that there
exist numbers n0 and m0 such that every n0 -m0 society contains a subgroup
of ve boys and ve girls in which all of the boys know all of the girls or none
of the boys knows none of the girls.

Last issue we gave ve more Klamkin Quickies. Next we give his

\Quicky" solutions to these problems. Many thanks to Murray S. Klamkin,
the University of Alberta, for creating the problems and solutions.
ANOTHER FIVE KLAMKIN QUICKIES
October 21, 1996

6. Determine the four roots of the equation x4 + 16x , 12 = 0.

Solution. Since
x4 + 16x , 12 = (x2 + 2)2 , 4(x , 2)2 = (x2 + 2x , 2)(x2 , 2x + 6) = 0;
p
the four roots are ,1  3 and 1  i 5.
p
7. Prove that the smallest regular n-gon which can be inscribed in a
given regular n-gon is one whose vertices are the midpoints of the sides of
the given regular n-gon.
Solution. The circumcircle of the inscribed regular n-gon must intersect
each side of the given regular n-gon. The smallest that such a circle can be
is the inscribed circle of the given n-gon, and it touches each of its sides at
its midpoints.
 133

8. If 311995 divides a2 + b2, prove that 311996 divides ab.

Solution. If one calculates 12 ; 22 ; : : : ; 302 mod 31 one nds that the
sum of no two of these equals 0 mod 31. Hence, a = 31a1 and b = 31b1 so
that 311993 divides a21 + b21. Then, a1 = 31a2 and b1 = 31b2. Continuing in
this fashion (with p = 31), we must have a = p998m and b = p998 n so that
ab is divisible by p1996.
More generally, if a prime p = 4k + 3 divides a2 + b2 , then both a
and b must be divisible by p. This follows from the result that \a natural n
is the sum of squares of two relatively prime natural numbers if and only if
n is divisible neither by 4 nor by a natural number of the form 4k + 3" (see
J.W. Sierpinski, Elementary Theory of Numbers, Hafner, NY, 1964, p. 170).
9. Determine the minimum value of
p
S = p(a + 1)2 + 2(b , 2)2 + (c + 3)2 + p
p
(b + 1)2 + 2(c , 2)2 + (d + 3)2 )
+ (c + 1)2 + 2(d , 2)2 + (a + 3)2 + (d + 1)2 + 2(a , 2)2 + (b + 3)2

where a, b, c, d are any real numbers.

Solution. Applying Minkowski's inequality,
p p
S  (4 + s) + 2(s , 8) + (s + 12) = 4s2 + 288
2 2 2
p
where s = a + b + c + d. Consequently, min S = 12 2 and is taken on for
a = b = c = d = 0.
10. A set of 500 real numbers is such that any number in the set is
greater than one-fth the sum of all the other numbers in the set. Determine
the least number of negative numbers in the set.
Solution. Letting a1 ; a2 ; a3 ; : : : denote the numbers of the set and S
the sum of all the numbers in the set, we have
a1 > S ,5 a1 ; a2 > S ,5 a2 ; : : : ; a6 > S ,5 a6 :
Adding, we get 0 > S , a1 , a2 ,


, a6 so that if there were six or less
negative numbers in the set, the right hand side of the inequality could be
positive. Hence, there must be at least seven negative numbers.
Comment. This problem where the \5" is replaced by \1" is due to
Mark Kantrowitz, Carnegie{Mellon University.
134

First a solution to one of the 36th IMO problems:

2. [1995: 269] 36th IMO
Let a, b, and c be positive real numbers such that abc = 1. Prove that
1 1 1 3
a3(b + c) + b3(c + a) + c3(a + b)  2 :
Solution by Panos E. Tsaoussoglou, Athens, Greece.
By the Cauchy{Schwartz inequality
 1 1 1 
[a(b + c) + b(c + a) + c(a + b)] a3 (b + c) + b3 (c + a) + c3(a + b)

1 1 1 2
 a b c + + ;
or
 
2(ab + ac + bc) a3(b1+ c) + b3 (c1+ a) + c3(a1+ b)
2
 (ab +(abc
ac + bc) ;
)2
or
1 1 1 ab + ac + bc ;
a3(b + c) + b3(c + a) + c3(a + b)  2
because abc = 1.
Also
ab + ac + bc  p3 a2b2c2 = 1:
3
Therefore
1 1 1 3
a3(b + c) + b3(c + a) + c3(a + b)  2
holds.
 135

Now we turn to some of the readers' solutions to problems proposed

to the jury but not used at the 35th IMO in Hong Kong [1995: 299{300].
PROBLEMS PROPOSED BUT NOT USED
AT THE 35th IMO IN HONG KONG
Selected Problems
3. A semicircle , is drawn on one side of a straight line `. C and D
are points on ,. The tangents to , at C and D meet ` at B and A respec-
tively, with the center of the semicircle between them. Let E be the point
of intersection of AC and BD, and F be the point on ` such that EF is
perpendicular to `. Prove that EF bisects \CFD.
Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk,
Zaltbommel, the Netherlands. We give Seimiya's write-up.
P

D
C E

 
B T FQ O ` A
Let P be the intersection of AD and BC . Then \PCO = \PDO =
90, \CPO = \DPO and PC = PD. Let Q be the intersection of PE
with AB . Then by Ceva's Theorem, we get
BQ
AD
PC = BQ
AD = 1:
QA DP CB QA CB
Thus we get
BQ = BC : (1)
QA AD
Since \BPO = \APO we get
PB = BO : (2)
PA AO
We put \PAB = , \PBA = .
136

Let T be the foot of the perpendicular from P to AB . Then from (1)

and (2) we have

BC = BO cos  = PB cos  = PT : (3)

AD AO cos  PA cos  TA
From (1) and (3) we have
BQ = PT :
QA TA
Hence Q coincides with T so that P , E , F are collinear. [See page 136.]
Because \PCO = \PDO = \PFO = 90 , P , C , F , O, D are con-
cyclic. Hence \CFE = \CFP = \CDP = \DCP = \DFP = \DFE .
Thus EF bisects \CFD.
P

C E

B F O ` A

4. A circle ! is tangent to two parallel lines `1 and `2. A second circle

!1 is tangent to `1 at A and to ! externally at C . A third circle !2 is tangent
to `2 at B , to ! externally at D and to !1 externally at E . AD intersects
BC at Q. Prove that Q is the circumcentre of triangle CDE.
Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk,
Zaltbommel, the Netherlands. We give Smeenk's solution.
We denote the three circles as ! (O;R), !1 (O1; R1 ), !2 (O2; R2 ). Now
let ! touch `1 at F and `2 at F 0 . Let the line through O2 parallel to `1
intersect FF 0 at G and the production of AO1 at H .
 137

`1 F L A

R1
R

C O1
O R1
Q
E
K D R
2
G N O 2
R2
`2 F 0
M B

Let the line through D parallel to `1 intersect FF 0 at K .

Let the line through D parallel to FF 0 intersect `1 at L, `2 at M and
GO2 at N . Now AF is a common tangent of ! and !1, so
p
AF = 2 RR1 (1)
and
p
BF 0 = 2 RR2 = GO2: (2)
It follows that p p
HO2 = j2 RR2 , 2 RR1 j;
HO1 = 2R , R1 , R2:
In right triangle O1 HO2 ,
p p
(2 RR2 , 2 RR1 )2 + (2R , R1 , R2 )2 = (R1 + R2)2 :
p
After some reduction R = 2 R1 R2 .
Next consider triangle GOO2 .
p
GO = R , R2; GO2 = 2 RR2; DO2 = R2; DO = R; KDkGO2:
R 2 R pRR
We nd that GN = FL =
GO = 2.
R + R2 2 R + R2
With (1) we have
p
2 R pRR
AL = 2 RR1 , R + R 2 : (3)
2
138

R2 R2(R , R2)
Furthermore DN =
R + R2
GO = R + R2 and

2
DL = 2R , R2 , R2R(R+,RR2) = R2+RR : (4)
2 2

Now AD2 = AL2 + DL2 . With (3) and (4),

p 2 R pRR !2  2R2 2
2 RR1 , R + R 2 + R + R = 4RR1 = AE 2 :
1 2 1 2

So AD = AF .
That means that AD touches ! at D and AD is a common tangent and
the radical axis of ! and !2.
In the same way BC is the radical axis of ! and !1 and Q is the radical
point of ! , !1 and !2.
So QC = QD = QE , as required.
5. A line ` does not meet a circle ! with center O. E is the point on
` such that OE is perpendicular to `. M is any point on ` other than E.
The tangents from M to ! touch it at A and B . C is the point on MA such
that EC is perpendicular to MA. D is the point on MB such that ED is
perpendicular to MB . The line CD cuts OE at F . Prove that the location
of F is independent of that of M .
Solution by Toshio Seimiya, Kawasaki, Japan.
As MA, MB are tangent to ! at A, B respectively, we get \OAM =
\OBM = 90 and OM ? AB . Let N , P be the intersections of AB with
OM and OE respectively.
Since M , E , P , N lie on the circle with diameter MP we get ON

OM = OB2 = r2 where r is the radius of !. Hence P is a xed point. (P
is the pole of `.)
Let G be the foot of the perpendicular from E to AB . As \OBM =
\OAM = \OEM = 90 , O, B , M , E , A are concyclic, so that by Simson's
Theorem C , D, G are collinear.
 139

!
B
r
O
N P
A

D G
C F

M E `
Since A, C , E , G lie on the circle with diameter AE we get
\EGF = \EGC = \EAC = \EAM: (1)
As O, M , E , A are concyclic and OM is parallel to EG we have
\EAM = \EDM = \DEG = \FEG: (2)
From (1) and (2) we get
\EGF = \FEG: (3)
Since \EGP = 90 we get
\FGP = \FPG: (4)
From (3) and (4) we have EF = FG = FP . Thus F is the midpoint of
EP . Hence F is a xed point.
140

Next, we give a counterexample to the rst problem of the set of prob-

lems proposed to the jury, but not used at the 35th IMO in Hong Kong given
in the December 1995 number of the corner.
1. [1995: 334] Problems proposed but not used at the 35th IMO in
Hong Kong.
ABCD is a quadrilateral with BC parallel to AD. M is the midpoint
of CD, P that of MA and Q that of MB . The lines DP and CQ meet at
N . Prove that N is not outside triangle ABM .
Counterexamples by Joanna Jaszunska,
 student, Warsaw, Poland; and
by Toshio Seimiya, Kawasaki, Japan. We give Jaszunska's
 example.
B C

X M
P
t
N
A D
We draw a triangle ADM and denote the midpoint of MA by P . Let
C be a point on the half-line DM such that M is the midpoint of CD.
Let N be any point of the segment PD, inside triangle ADM .
We construct a parallelogram MCBX such that MX and BC are par-
allel to AD and X lies on the segment CN .
Let us denote the point where the diagonal MB of this quadrilateral
meets CN by Q. Q is then the midpoint of MB .
Connect points A and B . We have thus constructed a quadrilateral
ABCD with BC parallel to AD, M is the midpoint of CD, P that of MA
and Q that of MB . Lines DP and CQ meet at N .
N is inside triangle ADM ; hence it is outside triangle ABM .
 141

Next we look back to some further solutions to problems of the Sixth

Irish Mathematical Olympiad given in [1995: 151{152] and for which some
solutions were given in [1997: 9{13]. An envelope from Michael Selby ar-
rived which I misled. It contains solutions to problems 1, 2, and 4 of Day 1,
and problems 1, 2, 3 and 4 of Day 2.
1. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.
Given ve points P1 , P2 , P3 , P4 , P5 in the plane having integer coor-
dinates, prove that there is at least one pair (P ; P ) with i 6= j such that
the line P P contains a point Q having integer coordinates and lying strictly
i j

between P and P .
i j

i j

Solution by Michael Selby, University of Windsor, Windsor, Ontario.

The points can be characterized according to the parity of their x and
y coordinates. There are only four such classes: (even, even), (even, odd),
(odd, even), (odd, odd).
Since we are given ve such points, at least two must have the same
parity of coordinates by the Pigeonhole Principle. Suppose they are P and
P , P = (x ; y ), P = (x ; y ). Then x + x is even and y + y is even,
i

since the x , x have the same parity and y , y have the same parity. Hence
j i i i j j j i j i j

i j i j
the midpoint x + x y + y 
Q= i

2 ;
j i

2
j

has integral coordinates.

2. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.
Let a1 ; a2 ; : : : a , b1; b2 ; : : : b be 2n real numbers, where
a1 ; a2; : : : ; a are distinct, and suppose that there exists a real number 
n n

such that the product

(a + b1 )(a + b2 ) : : : (a + b )
i i i n

has the value  for all i (i = 1; 2; : : : ; n). Prove that there exists a real
number  such that the product
(a1 + b )(a2 + b ) : : : (a + b )
j j n j

has the value  for all j (j = 1; 2; : : : ; n).

Solution by Michael Selby, University of Windsor, Windsor, Ontario.
Dene
P (x) = (x + b1)(x + b2)


(x + b ) , :
n (1) n

Then P (a ) = 0 for i = 1; 2; : : : ; n.
n i

Therefore P (x) = (x , a1 )(x , a2 )


(x , a ) by the Factor Theorem.

n n
142

Now (,1) P (,x) = (x + a1 )(x + a2 )


(x + a ). So

n
n n

(,1) P (,b ) = (b + a1 )(b + a2 )


(b + a )

n
n i i i i n

= (,1) +1 by (1):
n

Hence (b + a1 )(b + a2 )


(b + a ) = (,1) +1 for i = 1; 2; : : : ; n.

i i i n
n

Thus, the result is true with  = (,1) +1. n

That completes the Corner for this number. We are in high Olympiad
season. Send me your nice solutions and contests.

Do you believe what occurs in print?

The last sentence of the quoted passage, taken from The Daughters of
Cain by Colin Dexter (Macmillan, 1994), contains two factual erors. What are
they?
`Have you heard of \Pythagorean Triplets"?'
`We did Pythagoras Theorem at school.'
`Exactly. The most famous of all the triplets, that is |
\3, 4, 5" 32 + 42 = 52. Agreed?'
`Agreed.'
`But there are more spectacular examples than that.
The Egyptians, for example, knew all about \5961, 6480, 8161".'
Contributed by J.A. McCallum, Medicine Hat, Alberta.