Because of space limitations, the solutions below will often be
abbreviated and you are encouraged to fill in the details
yourself. Part of the art of writing up a solutions is to
know what details to cover and which can be passed over
quickly. Basically, each step that is
not completely standard should justified or at least
provided with a cue that will remind the reader of a
suitable background result. In recording manipulations,
you do not have to put in each step but you should indicate
at key points what the reader must do to reproduce the
calculation. Remember: if the
marker has not got a clue what you are up to, even after
much reflection, you will not get the mark. There are two
stages to solving a problem: (a) figuring out what the
solution is; (b) writing it up. The order in which the
ideas occur in attaining (a) is not always the best order
for implementing (b), and you should regard (a) and (b)
as separate processes. This means that you may have to
refine your ideas and reorganize your material.

- 1.
- Let $M$ be a set of eleven points consisting of the four vertices along with seven interior points of a square of unit area.

- (a) Prove that there are three of these points that are vertices of a triangle whose area is at most 1/16.

- (b) Give an example of a set $M$ for which no four of the interior points are collinear and each nondegenerate triangle formed by three of them has area at least 1/16.

(b) Let the square have vertices at
$(0,0)$,
$(1,0)$,
$(0,1)$,
$(1,1)$ and
let the interior points be at
$(\frac{1}{8},\frac{5}{8})$,
$(\frac{2}{8},\frac{2}{8})$,
$(\frac{3}{8},\frac{7}{8})$,
$(\frac{4}{8},\frac{4}{8})$,
$(\frac{5}{8},\frac{1}{8})$,
$(\frac{6}{8}.\frac{6}{8})$
and
$(\frac{7}{8},\frac{3}{8})$. There is a triangulation for which each
of the triangles has area exactly equal to 1/16. (*Exercise:* produce the
diagram.) Any other triangle determined by three of the points will have
area at least as large as some triangle in the triangulation.

(b) Most people did not produce an example. The ones who did either transgressed
the condition about not having too many collinear points, or else did not ensure
that the area of *every* triangle, even those not involved in the triangulation,
was at least 1/16.

- 2.
- Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Suppose that $u={a}^{2}+{b}^{2}+{c}^{2}$ and $v=(a+b+c){}^{2}$. Prove that

and that the fraction 1/2 on the right cannot be replaced by a smaller number.

By the triangle inequality, $a<b+c$, $b<c+a$ and $c<a+b$, so that the right side is always positive. Since all variables are positive, the right inequality follows.

The numerator of
$\frac{u}{v}-\frac{1}{3}$ is equal to

The right side, being a sum of squares, is nonnegative and it vanishes if and only if $a=b=c$. The left inequality follows.

To see that
$u/v$ can be arbitrarily close to 1/2, let
$(a,b,c)=(\epsilon ,1,1)$ where
$0<\epsilon <4$. Then

This can be made as close to 0 as desired by taking $\epsilon $ sufficiently close to 0.

Some students tried to argue the inequality by
looking at extreme situations, without attempting to draw a
relationship between the extreme and general situations.
Others tried a kind of variation argument, seeming to feel that
since we can change the variables to a situation where the
inequality holds, it will hold on the way there. While it is
often useful to think in terms of extreme cases and variation
of the unknowns in an inequality problem in order to get a handle
on the situation, this approach generally becomes completely
unmanageable in the write-up. It is best to think of the
variables as being fixed at certain values and perhaps try to
compare the value of a function to a purported extreme value.
In many inequalities, including this
one, it is often conceptually cleanest to look at the difference
between two sides; this difference can usually be manipulated into
a form which is clearly positive or negative, in a way in which
the reader can easily follow the steps. If you ever depart from
this format, you should have a good reason for doing so.
Avoid in the write-up
starting with what you have to prove and working backwards; this
makes the logic harder to follow and
could lead to trouble with steps that are not reversible.
*Begin with what you know and proceed
by logical steps to what has to be obtained.*

- 3.
- Suppose that $f(x)$ is a function satisfying

for all positive rational numbers $n$ and $m$. Show that, for all natural numbers $k$,

for each nonnegative integer $i$. Hence

so that

This problem provided a nice illustration of the maxim that
a little knowledge is a dangerous
thing. Students who had seen some calculus thought that somehow a
derivative was involved, and then were lost forever beneath the waves.
The context does not suggest a place for calculus. Calculus
deals with functions defined on the real numbers (not just the
rationals) and possessing derivatives. The most important thing
for you to
learn about calculus is when not to use it. More seriously, turning
to calculus right off the bat inhibits the ability to take the
problem on its own terms and come up with a more elementary solution.
Resist the urge to categorize problems too quickly. Once the key idea
of looking at consecutive powers of 2 is found, the solution then falls
right out.

- 4.
- Is it true that any pair of triangles sharing a common angle, inradius and circumradius must be congruent?

- 5.
- Each point of the plane is coloured with one of 2000 different colours. Prove that there exists a rectangle all of whose vertices have the same colour.

- 6.
- Let $n$ be a positive integer, $P$ be a set of $n$ primes and $M$ a set of at least $n+1$ natural numbers, each of which is divisible by no primes other than those belonging to $P$. Prove that there is a nonvoid subset of $M$, the product of whose elements is a square integer.

where the vector $({a}_{1},{a}_{2},\dots ,{a}_{n})$ has all of its entries equal to either 0 or 1. There are ${2}^{n}$ possibilities for this vector. However, there are at least ${2}^{n+1}-1>{2}^{n}$ nonvoid subsets of $M$ and so that many possible products $x$. Hence there are two such products, $x={y}^{2}z$ and $u={v}^{2}w$ which give rise to the same vector. It may transpire that both $x$ and $u$ have factors from $M$ in common; divide through by the product of these factors to obtain products of disjoint subsets $P$ and $Q$ of $M$. These two products will have the same vector, and so must have the form ${r}^{2}t$ and ${s}^{2}t$, where $t$ is a product of distinct primes from $P$. The product of the numbers in $P\cup Q$ is ${r}^{2}{s}^{2}{t}^{2}$, and this is the desired square.