19981999 Olympiad Correspondence Problems
Set 4
 19.

The following statics problems involving vectors was
given and two separate solutions provided. Determine whether
either of them is correct and explain the discrepancy between
the answers.


Problem. A force R of magnitude 200 N (Newtons)
is the resultant of two forces F and G for which
2 F  = 3 G  and the angle between
the resultant and G is twice the angle between the resultant
and F. Determine the magnitudes of F and G.


Both solutions use
the parallelogram representation of the vectors as
illustrated, where 3u = F , 2u = G and v = R  = 200. From the Law of Sines, we
have that
whence cos
q = 3/4, cos2
q = 1/8 and
cos3
q = 4cos
^{3} q 3cos
q = 9/16.
>From here, there are two ways to proceed:


(i) From the Law of Cosines, we find that
v^{2} = 4u^{2} + 9u^{2}  12u^{2} cos(180^{°}  3q) = 13u^{2} + 12u^{2} cos3q = 
25u^{2} 4



so that 200 =
^{5}/
_{2}u, u = 80,
F  = 240
and
G  = 160.


(ii) From the Law of Cosines, we find that
4u^{2} = 9u^{2} + v^{2}  6uv cosq = 9u^{2} + 200^{2}  900u 

so that
0 = 5(u^{2}  180u + 8000) = 5(u  80)(u  100) . 

Hence, u = 80,
F  = 240,
G = 160 or u = 100,
F  = 300,
G  = 200.


Why does method (i) lead to one solution while
method (ii) yields two? QED
View solution
 20.

Using a pair of compasses with a fixed radius
exceeding half the length of a line segment AB, it is possible
to determine a point C for which the triangle ABC is
equilateral. Here is how it is done.


With centres A and B construct circles using the
compasses and let P be one of the two points of intersection
of these circles. Draw the circle with centre P, and let
it intersect the circle of centre A in Q and the circle of
centre B in R. There are different possible configurations,
but we will select one so that QR is parallel to AB.
Now construct, using the compasses again, circles of centres
Q and R. These will intersect in P and a second point
C. Prove that triangle ABC is equilateral.
View solution
 21.

In the following problem, we will begin by making some
empirical observations. Your task will be to formulate some
general results exemplified by them and provide proofs. You may
observe some other general results not really pointed to below;
if so, formulate and justify these.


The numbers 1, 2, 3, ¼ are placed in a triangular
array and certain observations concerning row sums are
made as indicated below:
The odd row sums of this array are 1, 35, 189, 559, ¼ and
their running totals are 1 = 1^{2}, 36 = 2^{2} ·3^{2} = (1+2+3)^{2}, 225 = 3^{2} ·5^{2} = (1+2+3+4+5)^{2}, 784 = 4^{2} ·7^{2} = (1+2+3+4+5+6+7)^{2}, ¼,
You will notice that the square roots are the sum of odd sums of
consecutive integers. Even sums are not to be left out. The even
row sums in the array are 9, 91, 341, ¼ and their running
totals are 9 = 1^{2} ·3^{2} = (1+2)^{2}, 100 = 2^{2} ·5^{2} = (1+2+3+4)^{2}, 441 = 3^{2} ·7^{2} = (1+2+3+4+5+6)^{2}, ¼.
We can continue in this vein. Writing the triangular array with
1, 4, 7, 10, ¼ numbers in the consecutive rows, we find the
running totals of the odd sums to be 1, 64 = 8^{2} = 2^{2} ·4^{2},
441 = 21^{2} = 3^{2} ·7^{2} and so on. With 1, 5, 9, 13, ¼
numbers in the consecutive rows, the running totals of the odd
sums are 1, 100 = 10^{2} = 2^{2} ·5^{2}, 729 = 27^{2} = 3^{2}·7^{2}.
View solution
22.
The diagonals of a concyclic quadrilateral ABCD
intersect in a point O. Establish the inequality

AB CD

+ 
CD AB

+ 
BC AD

+ 
AD BC

£ 
OA OC

+ 
OC OA

+ 
OB OD

+ 
OD OB

. 

View solution
23.
Let A_{1}, A_{2}, ¼, A_{r} be subsets of
{ 1, 2, ¼, n } such that no A_{i} contains another.
Suppose that A_{i} has a_{i} elements (1 £ i £ r). Prove
that

å
 
æ ç
è

n
a_{i}

ö ÷
ø

1

£ 1 . 

View solution
24.
Without recourse to a calculator or a computer, give
an argument that 512^{3} + 675^{3} + 720^{3} is composite.
View solution
Problem 19.
 19.

First solution. Method (i) leads to the equation
0 = (5u  2v)(5u + 2v) while (ii) leads to
4u^{2} = 9u^{2} + v^{2}  9uv/2 or 0 = (5u  2v)(2u  v). Let us
examine closely the second answer that is provided by (ii).
In this case, the triangle formed by the two vectors and their
resultant is isosceles with sides of magnitudes 3u, 2u and
2u with the base angle equal to q and the apex angle
equal to 180^{°}  2q. In this configuration,
cosq is indeed 3/4, but the apex angle is not
2q as specified in the statement of the problem.
Indeed, method (i) made use of the angle between R
and G. However in method (ii) the result of the sine
law remained valid with 180^{°}  2q in place of
2q but the cosine law did not make use of the angle
between R and G. So it is not surprising that
the second method leads to a spurious possibility.
Problem 20.
 20.

There are various configurations,
and we give the solution for one of these. The solutions for the
other are similar.
 20.

First solution. See Figure 20.1.
We have that AP = AQ = QP = QC. Let
ÐQAC = ÐQCA = q, so that ÐAQC = 180^{°} 2q. Then ÐPQC = ÐAQC  60^{°} = 120^{°}  2q, so that ÐQCP = ÐQPC = 30^{°} + q. Thus, ÐACP = 30^{°}.
Since CP ^AB, ÐCAB = 60^{°}. Similarly,
ÐCBA = 60^{°} and so DABC is equilateral.
 20.

Second solution.
See Figure 20.2.Consider the reflection in the
right bisector of AB. It interchanges circles with centres
A and B, and hence fixes their intersection points,
in particular P. This reflection also carries the circle
with centre P to itself. Hence the point C is fixed by the
reflection. It follows that P and C are on the right bisector
of AB.


Suppose that PC intersects AB in S. Let ÐSAP = q. Then ÐAPB = 180^{°}  2q. Also
ÐQPC = 180^{°}  ÐAPQ  ÐAPS = 180^{°}  60^{°}  (90^{°}  q) = 30^{°} + q 

whence
ÐPQC = 180^{°}  2(30^{°} + q) = 120^{°} 2q 

and
ÐAQC = 60^{°} + ÐPQC = 180^{°}  2q . 

Hence DAQC º DAPB (SAS) so that AC = AB and the
result follows.


Third solution. See Figure 20.3.
A 60^{°} clockwise rotation about A followed by a
60^{°} clockwise rotation about B takes
A ® A ® D and Q ® P ®R, where D is the third vertex of an equilateral triangle
with side AB. Hence AQ ® DR so that DR  = AQ , the radius of the circle.


Similarly, the composition of two counterclockwise rotations
with respective centres B and A takes B ® B ®D and R ® P ® Q, so that DQ  = BR , the radius of the circle. Hence D is a point
of intersection of the circles with centres Q and R. When the
radius is not equal to AB , this will be distinct from
P, so that D = C, and the result follows.
Problem 21.
 21.

First solution. Consider the triangular array
in which the consecutive positive integers are written in rows
with only the number 1 in the top row and
with k ³ 0 more elements in each row than the previous one.
For r ³ 1, the rth row has (r1)k + 1 elements
beginning with ((r1)  2)k + r and ending with
(r  2)k + r. The sum of the numbers in the rth row is



é ê
ë


æ ç
è


æ ç
è

r1
2

ö ÷
ø

+ 
æ ç
è

r
2

ö ÷
ø


ö ÷
ø

k + 2r 
ù ú
û


 
= 
1 2

[(r1)k + 1][(r1)^{2}k + 2r] 
 
= 
1 2

[ (r1)^{3} k^{2} + (3r^{2}  4r + 1)k + 2r] . 

 

When r = 2s  1, this sum is
4(s1)^{3} k^{2} + (6s^{2}  10s + 4)k + (2s  1) 

and the sum of the first m oddnumbered rows is the sum of
these terms over 1 £ s £ m, namely
[m(m1)k]^{2} + 2m^{2}(m1)k + m^{2} = [m((m1)k + 1)]^{2} . 



When k = 0, the sum of the elements in the sth
oddnumbered row is 2s  1 and the sum of the elments in the
first m oddnumbered rows is m^{2}, the sum of the first m
odd integers.


When k = 1, the sum of the elements in the sth
oddnumbered row is


+ (6s^{2}  10s + 4) + (2s  1) = 4s^{3}  6s^{2} + 4s  1 
 
= s^{4}  (s1)^{4} = [s^{2}  (s1)^{2}][s^{2} + (s1)^{2}] 
 
= [(s1) + s][(s1)^{2} + s^{2}] 

 

and the sum of all the elements in the first m oddnumbered
rows is m^{4}.


When k = 2, the sum of the elements in the first
m oddnumbered rows is
[m(2m  1)]^{2} = [(1/2)(2m)(2m1)]^{2} = [1 + 2 + ¼+ 
2m1

]^{2} . 



When r = 2s, the sum of the elements in the rth row
is

1 2

[(2s  1)^{3} k^{2} + (12s^{2}  8s + 1)k + 4s] . 

For k = 0, this sum is 2s. When k = 1, the sum is
4s^{3} + s = s[1 + (2s)^{2}] and the sum of all the elements
in the first m evennumbered rows is


+ 
1 2

m(m+1) = 
æ ç
è

m+1
2

ö ÷
ø

[m^{2} + (m+1)^{2}] 
 
= [1 + 2 + ¼+ m][m^{2} + (m+1)^{2}] . 

 

For k = 2, the sum of the numbers in the sth evennumbered
row is
2(2s  1)^{3} + (12s^{2}  8s + 1) + 2s = 16s^{3}  12s^{2} + 6s  1 = (2s  1)^{3} + (2s)^{3} 

and the sum of all the elements in the first m evennumbered
rows is

m å
s = 1

[(2s  1)^{3} + (2s)^{3}] = 
2m å
r = 1

r^{3} = [1 + 2 + ¼+ 2m]^{2} . 

Problem 22.
 22.

First solution. Let the points and lengths be
as labelled in the diagram, and let q = ÐAOB,
a = ÐBAD, b = ÐABC. Then
ÐBCD = 180^{°}  a and ÐCDA = 180^{°}  b. Then, where [ ¼] denotes
area,
2[ABC] = ab sinb = (p + q)rsinq 

2[ACD] = cd sinb = (p + q)s sinq 

so that r/s = (ab)/(cd). Similarly p/q = (ad)/(bc).
Hence

p q

+ 
r s

= 
a c


æ ç
è


d b

+ 
b d


ö ÷
ø

³ 2 
a c




p q

+ 
s r

= 
d b


æ ç
è


a c

+ 
c a


ö ÷
ø

³ 2 
d b




q p

+ 
r s

= 
b d


æ ç
è


c a

+ 
a c


ö ÷
ø

³ 2 
b d




q p

+ 
s r

= 
c a


æ ç
è


b d

+ 
d b


ö ÷
ø

³ 2 
c a



and the result follows.


Second solution. From similar triangles,
we find that a/c = r/q = p/s, b/d = r/p = q/s, so that
pq = rs. Then


+ 
q p

+ 
r s

+ 
s r


ö ÷
ø

 
æ ç
è


a c

+ 
c a

+ 
b d

+ 
d b


ö ÷
ø


 
= 
æ ç
è


p q

+ 
q p

+ 
r s

+ 
s r


ö ÷
ø

 
æ ç
è


r q

+ 
s p

+ 
q s

+ 
p r


ö ÷
ø


 
= 
p^{2} + q^{2}  pr  qs pq

+ 
r^{2} + s^{2}  qr  ps rs

. 

 

Since pq = rs, this is a fraction over a common
denominator with numerator


r^{2} + s^{2}  pr  qs  rq  sp 
 
= 
1 2

[ (p  r)^{2} + (q  s)^{2} + (q  r)^{2} +(s  p)^{2}] , 

 

which is nonnegative. The result follows.


Third solution. Begin as in the second solution. Then


 
 
 
£ 
 ______________ Öp^{2} + q^{2} + r^{2} + s^{2}


 ______________ Ör^{2} + s^{2} + q^{2} + p^{2}

pq


 
= 
p^{2} + q^{2} + r^{2} + s^{2} pq


 
= 
p^{2} + q^{2} pq

+ 
r^{2} + s^{2} rs


 

 

from the CauchySchwarz Inequality.
Problem 23.
 23.

First solution. There are n! arrangements
of the first n natural numbers. Suppose that a_{i} = k;
consider the arrangements { x_{1}, x_{2}, ¼, x_{n} }
where A_{i} = { x_{1}, x_{2}, ¼, x_{k} } (i.e.,
the first a_{i} numbers constitute the set A_{i}). There
are k! possible ways of ordering the first k numbers
and (nk)! ways of ordering the remaining numbers
so that there are a_{i}! (n  a_{i})! arrangements of this
type. Let i ¹ j. No arrangement { x_{1}, x_{2}, ¼, x_{n} } can at the same time have its
first a_{i} elements coincide with A_{i} and its first
a_{j} elements coincide with a_{j}, since neither of
A_{i} and A_{j} is contained in the other. Hence
a_{i}! (n  a_{i})! arrangements corresponding to A_{i}
are distinct from the a_{j}! (n  a_{j})! arrangements
corresponding to A_{j}. It follows that

n å
i = 1

a_{i}! (n  a_{i})! £ n! 

and the result obtains.
Problem 24.
 24.

First solution. Observe that
512 = 2^{9}, 675 = 3^{3}·5^{2} and 720 = 2^{4} ·3^{2} ·5, so that
2 ·720^{2} = 3 ·512 ·675.
We now use the identity
x^{3} + y^{3} + z^{3} = x^{3} + y^{3}  z^{3} + 3xyz = (x + y  z)(x^{2} + y^{2} + z^{2}  xy + xz + yz) 

which is valid when 2z^{2} = 3xy to obtain that
512^{3} + 675^{3} + 720^{3} = (512 + 675  720)(512^{2} + 675^{2} + 720^{2}  512·675+ 512 ·720 + 675 ·720) . 

Thus, 467 = 512 + 675  720 is a factor of the sum of cubes.
 24.

Second solution. [P. LeVan]


+ 675^{3} + 720^{3} = 512^{3} + 45^{3} [15^{3} + 16^{3}] 
 
= 512^{3} + 45^{3} [(16  1)^{3} + 16^{3} ] 
 
= 512^{3}  45^{3} + 45^{3} ·16 [2 ·16^{2}  3 ·16 + 3] 
 
= (512  45)(512^{2} + 512 ·45 + 45^{2}) + 45^{3} ·16[512  3 ·15] 
 
= 467[512^{2} + 512 ·45 + 45^{2}] + 45^{3} ·16 [467] 
 
= 467 [512^{2} + 512 ·45 + 45^{2} + 45^{3} ·16] 

 

so that 467 is a factor of the sum of the cubes.
 24.

Third solution. [D. Pritchard] We have the
identity
(a^{9})^{3} + (b^{3} c^{2})^{3} + (a^{4} b^{2} c^{3})^{3} = a^{27} + b^{9} c^{6} + a^{12} b^{6} c^{3} 

= (a^{9} + b^{3} c^{2}  a^{4} b^{2} c)(a^{18}  a^{9} b^{3} c^{2} + b^{6} c^{4}+ a^{13} b^{2} c + a^{4} b^{5} c^{3} + a^{8} b^{4} c^{2}) + (2b  3a)a^{12} b^{5} c^{3} . 

Now take (a, b, c) = (2, 3, 5) to yield the desired result, one
factor being 467.
 24.

Fourth solution. [D. Arthur] Writing
512 = ^{1}/_{2}(x + y), 675 = ^{1}/_{2}(x + z) and
720 = ^{1}/_{2}(y + z) yields (x, y, z) = (467, 557, 883). Modulo 467, we find that

512^{3} + 675^{3} + 720^{3} 

º 8^{1}(2y^{3} + 2z^{3} + 3y^{2}z + 3yz^{2}) 
 
= 8^{1}(y + z)(2y^{2} + 2z^{2} + yz) . 

 

[Note that 8^{1} represents a number a for which
8a º 1 (mod 467).] Now y º 90 and
z º 51, so that


º 2 ×90^{2}+ 2 ×51^{2}  90 ×51 
 
= 9 [4 ×450 + 2 ×289  510] 
 
º 9 [ 4 ×17 + 578  510] = 9 [68 + 68] = 0 

 

and so the sum of the cubes is divisible by 467.
 24.

Comment. In fact,
512^{3} + 675^{3} + 720^{3} = 229 ×467 ×7621 . 
