International
Mathematical
Olympiad 1968
N. A. Routledge and David Monk.

The Russians were hosts to the Tenth International Mathematical Olympiad (IMO) in which 12 countries took part. It was held in Moscow in July.

The British team was made up as follows:—Head of the Delegation: Dr. N. A. Routledge (Eton College); Deputy Head: Dr. David Monk (Edinburgh University); Malcolm Williamson, Manchester Grammar School; Simon Norton, Eton College; William Porterfield, Westminster School; John Scholes, Winchester College; Clifford Cocks, Manchester Grammar School; Elwyn Davies, Manchester Grammar School; Noel Leaver, Burnley Grammar School; Michael Proctor, Shrewsbury School.

For details of the organisation of an IMO see SCIENCE TEACHER, 11:1:p.30 (October 1967).

The Papers

There were two four-hour papers of three questions. The maximum individual mark was 40: for a country, 320.

The questions were felt to be too easy, no less than 16 contestants (mainly GDR, USSR, Hungary) got full marks; three British got 39.

Categories were divided up as follows: I class, 39+ marks; II, 34+; III, 22+.

(In all, 64 were classed out of 96, (i.e. two-thirds of the candidates.)

The British team scored as follows:—

First—Malcolm Williamson; Simon Norton; William Porterfield.
Second—John Scholes; Clifford Cocks.
Third—Elwyn Davies; Noel Leaver.

A special prize for elegance went to Williamson (for Question 6), to two Hungarians, and a Yugoslav. When Williamson’s very short solution was shown to the delegates, they received it with acclamation.

A very special prize went to Ouoon, the only girl (a Mongolian) who with 19 marks beat all the Mongolian boys.

Performance by countries was as follows:—

Country	Marks	No. in each Class
		I	II	III
GDR	304	5	3	—
USSR	298	5	1	2
Hungary	291	3	3	2
GB	263	3	2	2
Poland	262	2	3	2
Sweden	252	1	2	5
Czechoslovakia	248	2	4	—
Rumania	208	1	1	2
Bulgaria	204	—	3	1
Yugoslavia	177	—	—	3
Italy	132	—	—	1
Mongolia	74	—	—	—

Dr. Routledge writes:

‘As head of the delegation, I went two days ahead of the main party to join the International Jury (of heads of delegations) who control the Olympiad, and select the question papers. Under the efficient chairmanship of Professor Markushevich all went smoothly.

‘The boys were accommodated in the Internat, the well-known Kolmogorov boarding school for boys of exceptional mathematical and scientific talent, and were whisked at once into a vigorous programme of activities.

‘The seniors completed the selection, translation, and copying of the question papers (a task made tedious by the absence of duplicators in Moscow University). At the start of each paper the boys were allowed a brief word with the seniors to make sure the questions were clearly expressed and understood. The seniors then marked their own boys’ papers, but had to justify the marking by discussing it with a Russian co-ordinator (the Russian boys’ work was co-ordinated by the seniors from the countries from which the questions came.)

‘The co-ordinators imposed very few changes on the British marking, and indeed, reference was later made to the fairness, not to say severity, with which we marked. This year our team lost vital marks by: (i) failing to check that a solution of a problem fulfilled the requirements of the problem; (ii) saying $\sqrt{x^2} = x$ ; (iii) thinking an arbitrary number, a, was necessarily positive. Great credit was given to boys who produced generalisations of the questions set: on any future occasion the British team must try to do this.

‘The last meeting of the Jury (in their room on the 15th floor of Moscow University—a gigantic wedding-cake of a building, the last of the great works of Stalin) was to confirm the marking, and to decide where to draw the first-, second-, and third-class dividing lines.

‘Interspersed with the work the Russians had organised a relentless programme of sightseeing for all of us: the Kremlin, Lenin’s flat, Lenin’s country house, the Moscow State Circus, the panorama of the battle of Borodino, the Tretiakov gallery (of social realist art and icons), the Roubliev museum (of icons), Ostankino (a charming country house with a private theatre, and associations with the freeing of the serfs), and so on.

‘For the last four days we took the night train to Leningrad and were taken round that magnificent city. The climax of our visit was the first night of a new ballet, “Antony and Cleopatra” by Lazareff and Chernyshev, at the Mali theatre. In Leningrad we were all accommodated in a boarding school for partially-sighted children (a great change for the seniors from the splendours of the Hotel Rossyar, opposite the Kremlin in Moscow.) This mixed us all up happily together and led to much fruitful fraternisation among the seniors (in Moscow we were each given about £40 and invited to feed ourselves in the excellent restaurants and buffets.) The British boys found language more of a problem, but even so made many friendships through bridge and chess.

‘The final ceremony was at Moscow University, where the distinguished grand old man of topology, Alexandrov, presented the certificates (everyone got one, if only a certificate of participation) and the magnificent picture books which all the boys received.

‘Was our performance satisfactory? We occupied the same place on the list as last year in Yugoslavia, beaten by Russia, E. Germany, and Hungary, the three countries which have special mathematical schools and train hard for these competitions. Should we do the same? If our team is chosen earlier we can certainly circulate old papers, and give some tips based on past experience.

‘Also, each year the National Mathematical Contest, which is the qualifying round for the British Mathematical Olympiad, from the winners of which our team is drawn, covers more schools, so we should do better for this reason. Although the team might be selected from the winners of the current year as well as of the previous year, instead of, as at present, just for the previous year.

‘What value does the competition have? The Russians made it extremely clear that they were gratified to have Great Britain represented at what is basically an Eastern bloc affair. The expedition means that the boys who will be among our leading mathematicians in the next generation have a chance to get to know one another at an early age, and are also exposed (vigorously) to foreign cultures. To the participants the mathematical value is not particularly high (the syllabus on which it is based is not advanced), but the competitions in the home countries (including Great Britain) which act as qualifying rounds, do stimulate an interest in mathematics.’

Dr. Monk writes:

‘I feel sure that the team enjoyed their visit to Russia and found it interesting and stimulating. Their performance was most gratifying. For our part, Dr. Routledge and I were entirely satisfied with the fairness of the marking.

‘The fact that in each session of four hours only three problems are set, far from making the competition straightforward, adds to its vigour. For candidates are expected to explain their solutions fully, and deal with exceptional cases. The time also allows for the finding of polished solutions and generalizations; a candidate from Hungary obtained a special mention this year for indicating how one problem could be extended. The working is strict and small points are not overlooked. I consider that the high standards of performance set by the Olympiad make it well worthwhile to continue its support.

‘On this occasion, there was some support among the Jury for the view that the competition should have included at least one problem more testing than those actually set. We agreed with this opinion which was, indeed, shared by some of our boys; we were confident that they could have tackled a harder problem successfully. Such a policy might have avoided the situation in which 16 boys obtained maximum marks, and enabled the very best contestants to be identified other than by the “special elegance” awards. Generally, however, the arrangements for selection and translation of questions worked well, being facilitated by a very efficient interpreter.

‘To a British observer, a noteworthy feature of the collection of problems submitted by participating countries, from which the Olympiad questions were selected, was the amount of geometry involved. This contrasts with the decline in geometry in school syllabuses in Britain, a trend which is evidently not so pronounced in Eastern Europe. However, only two of the final six questions were geometrical in nature. We were impressed also, by the keenness and rivalry of the leading countries, and by the public interest in the competition, as indicated by the extent of press and broadcasting coverage.’

In 1969, the Olympiad is to be held in Rumania.

The Questions

Prove that there exists a unique triangle whose sides are consecutive real numbers and one angle of which is twice as large as another.

Solution

In the similar triangles ABD and ACB

$\therefore$ c² = bx, = x² + xy

But BD = y ( $\triangle BDC$ is isosceles). Thus c² = x² + xy is an equation about $\triangle ABD$ . Therefore in the similar $\triangle ACB$ , b² = c² + ca. (1)

Now a, b, c are n - 1, n, n + 1 in some order, but b > c [from (1)].

Thus a = n - 1, b = n + 1, c = n

or a = n, b = n + 1, c = n - 1

or a = n + 1, b = n, c = n - 1

So from (1) n² - 3n - 1 = 0 or n² - 5n = 0 or n² - 2n = 0. The only integral solutions of these are n = 0, 2 or 5, of which the first two are clearly useless. So 5, 6, 4 is the only possibility.

In this triangle cos C = ¾, and (cosine rule). So , from which B = 2C, as required.
If p(x) is the product of the digits in the decimal expansion of a positive number x, solve

p(x) = x² - 10x - 22

Solution

We first prove . For if x = abc... in its decimal expansion, and has n + 1 figures,

Now x² - 10x - 22 = (x - 5)² - 47 < 0 if and so if x cannot satisfy our equation (since x² - 10x - 22 < 0 and ).

Again x² - 11x - 22 = (x - 5½)² - 52¼ > 0, if (since (7½)² = 56¼)

Thus if , , and so x is not a solution.

Thus no number or can be a solution. But 12 is, by direct verification.
Consider the system of equations
ax_i² + bx_i + c = x_i+1 (),
and x_n+1 = x₁,
when are real.

If $\Delta = (b - 1)^2 - 4ac$ , prove there are no real solutions if $\Delta < 0$ , just one if $\Delta = 0$ , and more than one if $\Delta > 0$ .
Solution

If $\Delta < 0$ , by adding the equations, we have

$\sum_1^n (ax_i^2 + (b - 1)x_i + c) = 0.$ (1)

But $ax^2 + (b - 1)x + c \equiv a [(x - p)^2 + q^2]$
for some real , p, in this case.

Thus $a\sum_1^n [(x_i - p)^2 + q^2] = 0$ ,
whereas obviously

$\sum_1^n [(x_i - p)^2 + q^2] > 0.$

So there is no solution.

If $\Delta = 0$ , (1) still holds, but

$ax^2 + (b - 1)x + c \equiv a(x - p)^2$

for some real p.

So $a\sum(x_i - p)^2 = 0$ . This can only be so if all x_i = p. It is easy to see that this is a solution.

If $\Delta > 0$ , ax² + (b - 1)x + c = 0 has two real solutions, $\alpha$ , $\beta$ say. It is easy to see that $x_i = \alpha$ , (all i), and $x_i = \beta$ (all i) are two real solutions (as required).
Prove that every tetrahedron has a vertex such that one can construct a triangle whose sides are equal to the three edges meeting at that vertex.

Solution

Assume this is false. We may assume that no edge exceeds AB, and that none of AD, BC, BD exceed AC.

Since AB, AC and AD cannot form a triangle, the longest of them, AB, must be .

But , we have assumed,

$\therefore$

But ABD is a triangle. Therefore AB < BD + AD.

This is absurd.

Hence the required result is true.
Let f(x) be a real-valued function such that for all x

$f(x + \alpha) = ½ + \sqrt{f(x) - (f(x))^2}$

(for a fixed $\alpha > 0$ ).

Prove that f(x) is periodic (i.e. $f(x + \beta) = f(x)$ for all x, for some $\beta > 0$ ).

Give an example of such a function (not a constant) when $\alpha = 1$ .

Solution

Clearly . Also since , .

Now, from the given equation,

$f^2(x + \alpha) - f(x + \alpha) + ¼ = f(x) - f^2(x)$

Put $x + \alpha$ for x:

$f^2(x + 2\alpha) - f(x + 2\alpha) + ¼ = f(x + \alpha) - f^2(x + \alpha)$

So $f^2(x + 2\alpha) - f(x + 2\alpha) = f^2(x) - f(x)$
or $(f(x + 2\alpha) - f(x))(f(x + 2\alpha) + f(x) - 1) = 0$ .

$\therefore$ for each x, $f(x + 2\alpha) = f(x)$
or $f(x + 2\alpha) + f(x) = 1$ .

In the latter case, since f(x) (and $f(x + 2\alpha)$ ) ,

$f(x) = f(x + 2\alpha) = ½$ .

$\therefore$ In all cases $f(x) = f(x + 2\alpha)$ , as required.

Now $f(x) = (2 + \sqrt{2}) / 4$ is the constant solution they do not want.

But the step function

f(x) = ½ ()

= 1 ()

= ½ ()

= 1 ()

etc.

satisfies the required equation, as does

$f(x) = ½(1+|sin(\pi x/2)|),$

as may be immediately verified (if $\alpha = 1$ ).
Find the sum

$[(n + 1)/2] + [(n + 2)/4] + ... + [(n + 2^k)/2^{k+1}] + ...$

where n is a positive integer. ([x] means the integral part of x, i.e. the greatest integer not exceeding x).

Solution

For any real , then $x = 2m + \alpha$ or $x = 2m + 1 + \alpha$ where m is an integer, and $0 <= \alpha < 1$ . In either case it may be verified that

(1)

Thus if n < 2ⁱ,

n =

=

$... + ([n/2^{i-1}] - [n/2^i])$

= $[(n + 1)/2] + [(n + 2)/4] + ... + [(n + 2^{i-1})/2^i]$ (2)

But we can, for any n, always find a positive integer i such that n < 2ⁱ. Also, for ,

$n + 2^k < 2^i + 2^k <= 2^k + 2^k = 2^{k+1},$

So $0 <= (n + 2^k)/2^{k+1} < 1$ , and thus $[(n + 2^k)/2^{k+1}] = 0$ .

So, from (2),

(adding an infinite string of terms equal to 0 to the series there).