12 Cutting Edge

The Competition scene, as it has evolved over recent decades, originating in Eastern Europe, has been invaluable in developing the careers of young mathematicians. Many of the best young mathematicians in Australia, many of the best in the world, had their starts in the competition and Olympiad environments, and many will acknowledge that they would not be in the same career positions had it not been for this experience. Let me briefly just note a few really outstanding examples.

One of the world's most famous living mathematicians is Australian Terry Tao, now a Professor at UCLA. He was known by his parents to have been of prodigious talent, but it was not until he won a medal in his first attempt at the AMC, as a Senior at the age of 10, that there was formal external evidence. His career is well documented. He is still the only student who had completed his exam for an IMO Gold Medal before his 13th birthday. He set records through his studies and was a Professor at UCLA at the age of 24. He conducts research in an unusually wide range of mathematics disciplines, has a capacity for coming to grips with complicated issues quickly, and for explaining complex results in a straightforward manner. He holds a Fields Medal.

Maryam Mirzakhani (Iran) won Gold Medals at IMO and became the first, and to this stage only, woman to win a Fields Medal. Sir Timothy Gowers (UK), a Fellow at Trinity College, Cambridge, and another Fields Medalist, famously won an IMO Gold Medal.

The reclusive Russian mathematician Grigorij Perelman, who not only proved the Poincare Conjecture but generalised it in a proof he published on the internet, and who declined a Fields Medal and USD 1 million dollar Clay Trust prize, won an IMO Gold Medal for the USSR. The list goes on.

The IMO is the world's premier competition. Obviously its problems are original and highly challenging each year, but they do fall under a fairly tight, if not formally prescribed syllabus under the headings of geometry, algebra, number theory and combinatorics. There are also many other competitions,, some not related to the IMO, at international, regional and national level, some of which go well outside IMO bounds and have innovative ideas. I find the Russian-based International Mathematics Tournament of Towns to provide some of the most beautiful and creative ideas, often pushing new grounds in developing problem solving new techniques.

Mathematicians setting problems for competitions often discover new ideas. I used to enjoy, for example, working on the Problems Committee of the Mathematics Challenge for Young Australians. There were occasions when we discovered new results, sometimes resulting in published refereed papers.

Sometimes old techniques, long forgotten by most, can be rediscovered with the ability to solve problems which can be easily described but not conducive to conventional techniques. One, which I will illustrate below, is the use of barycentric coordinates, developed by Mobius in the 19th century. My Bulgarian colleague Jordan Tabov and I have a chapter on these in our book Methods of Problem Solving, Book 1, published by AMT Publishing. There was a British IMO student in recent years who discovered he could solve many geometrical problems using them, I think much to the chagrin of his leader, who had to mark his work.

Barycentric coordinates can be particularly conducive to solving collinearity problems. The 1997 IMO was the last one where I saw a serious three dimensional geometry problem, involving collinearity, posed and considered. One by one team leaders produced Euclidean solutions to this difficult problem and it started to shape as a likely problem 6. Then a Team Leader produced a barycentric coordinates proof which demolished the problem in a few lines. Unnerved, the Jury dropped this problem and I didn't see another of this type in later years.

Consider the following problem which appeared in the International Mathematics Tournament of Towns.

Example 12.1

We are given 100 points. N of these are vertices of a convex N-gon and the other 100-N of these are inside this N-gon.

The labels of these points make it impossible to tell whether or not they are vertices of the N-gon. It is known that no three points are collinear and that no 4 points belong to two parallel lines. It has been decided to ask questions of the following type:

What is the area of the triangle XYZ, where X, Y and Z are labels representing three of the 100 given points? Prove that 300 such questions are sufficient in order to clarify which points are vertices and to determine the area of the N-gon.

Comment

I tried very hard to solve this problem, but no method available to me at the time got close. Exasperated, I gave up and asked Jordan Tabov for help. It was his elegant solution given below which introduced me to barycentric coordinates. The solution will look technical (I apologise) to most lay mathematicians as barycentric coordinates are not widely known. But for those familiar, and they are described in the book referred to above, it is not a difficult proof. Without trying to define barycentric coordinates here, I note that in two dimensions, a triangle has three vertices with coordinates (1,0,0), (0,1,0) and (0,0,1), the coordinates of the points inside are positive, with centroid at (¹/₃,¹/₃,¹/₃) and points outside will have some coordinates negative, depending on which sides of the triangle they lie.

Solution 12.1

Take 3 arbitrary points A, B and C; let Δ be the area of ΔABC.

Let D be any other point. We shall show that, using 3 questions, we can determine the barycentric coordinates of D with respect to ΔABC.

In order to do this, let us ask for the areas of triangles BCD, CAD and ABD and let these areas be d₁, d₂ and d₃.

Then the barycentric coordinates of D with respect to ΔABC are

ε₁ ^d₁/Δ, ε₂ ^d₂/Δ and ε₃ ^d₃/Δ, where ε_i = ±1.

It remains to show that the triplet (ε₁,ε₂,ε₃) is determined from d₁, d₂ and d₃ and Δ in a unique way.

We shall use

Δ = ε₁d₁ + ε₂d₂ + ε₃d₃ (1)

This equation is clear from a geometric point of view.

Suppose that (ε₁',ε₂',ε₃') and (ε₁'',ε₂'',ε₃'') satisfy (1) and that ε₁' ≠ ε₁'', i.e. that ε₁' = -ε₁''.

Then since Δ ≠ 0, either ε₂' = -ε₂'' and ε₃' = ε₃'', or ε₂' = ε₃'' and ε₃' = -ε₃''.

Without loss of generality we may assume the first.

Then

Δ - ε₃'d₃ = ε₁'d₁ + ε₂'d₂ = -ε₁''d₁ - ε₂''d₂ = -Δ + ε₃''d₃ = -(Δ - ε₃'d₃).

Hence ε₃'d₃ = Δ, which means that CD is parallel to AB, a contradiction to the condition that no 4 points lie on 2 parallel lines.

So using 3 questions for each of the given points we can find its barycentric coordinates with respect to ΔABC.

This information for all given points is sufficient to determine which points are vertices of the given polygon and to find the area of the polygon.

Developing new Methods of Problem Solving

Just as methods from the past can be rediscovered for useful application, as the above example illustrate, but new methods evolve. Consider the problem below, which was set in 1989 in the International Mathematics Tournament of Towns.

Example 12.2

A regular hexagon is cut up into N parallelograms of equal area. Prove that N is divisible by three.

Comments

Like the Chameleon problem and many others this is a typical Russian problem, very easy to describe, with a very surprising result and on the face of it very difficult. However you know there is a way through. But conventional methods do not exist. Towards the latter part of the 20th century there was a lot of published work on dissection problems and the Soviet student maths journal Kvant had been discussing a method of moving parallels. Andy Liu provides the solution below. There is another brilliant problem of the same style in the Tournament's Book 1, which I published through AMT Publishing.

Solution 12.2

We refer to the diagram below, which shows a typical case in which the hexagon ABCDEF is divided into 27 parallelograms of equal area, the parallelograms being those surrounded by solid lines.

We note that each side of each parallelogram must be parallel to a side of a hexagon. This classifies the parallelograms into three types. Type I consists of those with sides parallel to AB and BC. Type II consists of those with sides parallel to CD and DE. Type III consists of those with sides parallel to EF and FA.

For each line that is parallel to AB and containing a side of a parallelogram of Type I or II, draw the parts that are inside all other parallelograms of Types I or II (for example, in the diagram, these are shown as dotted lines). This partitions existing parallelograms of Types I and II into ``smaller'' parallelograms of the same type.

The sum of the bottom edges of these parallelograms on the bottom row is clearly equal to AB. Hence the sum of their top edges is also equal to AB. However this is equal to the sum of the bottom edges of the Type I and II parallelograms in the second row from the bottom (including ``smaller'' ones), and so on. Since there are finitely many rows, we can conclude that the total area of these parallelograms is equal to AB times BD, or two thirds that of ABCDEF. It follows that the total area of the parallelograms of Type III, which we have left out of the picture until now, is one third that of ABCDEF. By symmetry, the same can be said about parallelograms of Types I and II. Since all parallelograms have equal area, it will follow that there is an equal number of each type.

Since there are three types, the total number of parallelograms is divisible by three.