What were the problems of the 29th International Mathematical Olympiad (Canberra, 1988)?
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Problem
What were the problems of the 29th International Mathematical Olympiad
(Canberra, 1988)?
Here are the problems from the International Mathematical Olympiad,
just being wound up in Canberra at the moment. Contestants sat two
exams of four and a half hours each, facing questions 1-3 and 4-6
respectively. Question 6 proved the hardest, with only 9 correct
solutions (from about 260 students). Five contestants (from Romania
(2), Vietnam, USSR and China) obtained perfect scores on all questions.
I've slightly changed the notation in a few places to suit an ascii
terminal. Otherwise [apart from my notes in square brackets] these are
verbatim.
Q1. Consider two coplanar circles of radii R and r (R > r) with the
same centre. Let P be a fixed point on the smaller circle and B a
variable point on the larger circle. The line BP meets the larger
circle again at C. The perpendicular p to BP at P meets the smaller
circle again at A (if p is tangent to the circle at P then A = P).
(i) Find the set of values of (BC)^2 + (CA)^2 + (AB)^2.
[ (BC)^2 means the square of the distance from B to C. ]
(ii) Find the locus of the midpoint of AB.
Q2. Let n be a positive integer and let A[1], A[2], ..., A[2n+1] be
subsets of a set B. Suppose that
(a) each A[i] has exactly 2n elements,
(b) each A[i]_intersect_A[j] (1 <= i < j <= 2n+1) contains exactly one
element,
(c) every element of B belongs to at least two of the A[i].
For which values of n can one assign to every element of B one of the
numbers 0 and 1 in such a way that each A[i] has 0 assigned to exactly
n of its elements?
Q3. A function f is defined on the positive integers by
f(1) = 1, f(3) = 3,
f(2n) = f(n),
f(4n+1) = 2f(2n+1) - f(n),
f(4n+3) = 3f(2n+1) - 2f(n),
for all positive integers n. Determine the number of positive integers
n, less than or equal to 1988, for which f(n) = n.
Q4. Show that the set of real numbers x which satisfy the inequality
SUM(k=1..70) k / (x - k) >= 5/4
is a union of disjoint intervals, the sum of whose lengths is 1988.
Q5. ABC is a triangle right-angled at A, and D is the foot of the
altitude from A. The straight line joining the incentres of the
triangles ADB, ACD intersects the sides AB, AC at the points K, L
respectively. S and T denote the areas of the triangles ABC and AKL
respectively. Show that S >= 2T.
Q6. Let a and b be positive integers such that ab+1 divides a^2+b^2.
Show that (a^2+b^2)/(ab+1) is the square of an integer.
-- Brendan McKay <bdm@anucsd.oz>
Show solution
[ Editor: only question 6 was solved in the collected postings; no
solution to questions 1-5 was found. ]
In article <772@anucsd.oz>, bdm@anucsd.oz (Brendan McKay) posted the
problems from the 1988 International Mathematical Olympiad, held in
Canberra.
I tried question 6 (the hardest one, as was said), and here is what I
found.
> Q6. Let a and b be positive integers such that ab+1 divides
> a^2+b^2. Show that (a^2+b^2)/(ab+1) is the square of an integer.
By the way, did *you* already try to prove this? I would suggest you
first work on it, before reading further on.
In what follows, a, b and m stand for positive integers (eventually
zero).
Call (a,b,m) a ``good'' triple if (ab+1)m = a^2 + b^2.
We have to prove that for a good triple (a,b,m), m is always a square.
Notice that m is never 0, except when both a and b are 0.
Lemma 1: If (a,b,m) is a good triple, then also (b,a,m).
Lemma 2: If (0,b,m) is a good triple, i.e. a=0, then m = b^2.
Lemma 3: If (a,b,m) is a good triple, then also (a , am-b , m).
Moreover, if a > 0, then am-b >= 0.
Proof: Suppose (a,b,m) is a good triple, i.e., (ab+1)m = a^2 + b^2.
1. (ab+1) m = a^2 + b^2 <==> ((am-b)a+1)m = (am-b)^2 + b^2
(this is elementary calculus)
2. am-b is certainly an integer, so it suffices to prove that
am-b > -1. Suppose for a moment that am-b <= -1.
Then (ab+1)(am-b+1) <= 0 (because ab+1 > 0), so
a(a^2+b^2) <= (ab+1)(b-1), or a^3+1 <= b(1-a),
which is impossible because a^3+1 > 1, while b(1-a) <= 0.
Lemma 4: If (a,b,m) is a good triple with 0 < a <= b, and b'=am-b,
then (a,b',m) is a good triple with 0 <= b' < a.
Proof: Suppose 0 < a <= b (so m > 0).
Using lemma 3, we only have to prove that am-b < a.
Notice that (a^2-1)a < (a^2+1)b , so a^3-b < (ab+1)a ,
so a^3 + a b^2 < a^2 b +a+b + a b^2 ,
so am(ab+1) < (a+b)(ab+1) , so am < a+b.
Conclusion: suppose you are given a ``good'' triple (a,b,m), and you
have to prove that m is square. Eventually interchange a and b, such
that a <= b (see lemma 1). If a=0, you are done, because of lemma 2.
If a > 0, apply lemma 4 to get a new ``good'' triple (a,b',m), with b'
smaller than before. Repeat this procedure, until a or b is 0. (This
has to happen, because a and b are both strictly decreasing after two
steps, but they remain positive.) Then, because of lemma 2, m is
square, which proves the whole story.
P.S. Did someone find a (completely) different and/or shorter proof?
Let it know to the interested people!
-- Peter Vanroose <vanroose@kulmath.uucp>, Department of Mathematics,
K.U.Leuven, Belgium