34th British Mathematical Olympiad 1998 Problems

34th British Mathematical Olympiad 1998 Problems

1.  A station issues 3800 tickets covering 200 destinations. Show that there are at least 6 destinations for which the number of tickets sold is the same. Show that this is not necessarily true for 7.

2.  The triangle ABC has ∠A > ∠C. P lies inside the triangle so that ∠PAC = ∠C. Q is taken outside the triangle so that BQ parallel to AC and PQ is parallel to AB. R is taken on AC (on the same side of the line AP as C) so that ∠PRQ = ∠C. Show that the circles ABC and PQR touch.

3.  a, b, c are positive integers satisfying 1/a - 1/b = 1/c and d is their greatest common divisor. Prove that abcd and d(b - a) are squares.

4.  Show that:     xy + yz + zx = 12

      xyz - x - y - z = 2 have a unique solution in the positive reals. Show that there is a solution with x, y, z distinct reals.

Solutions

Problem 1

A station issues 3800 tickets covering 200 destinations. Show that there are at least 6 destinations for which the number of tickets sold is the same. Show that this is not necessarily true for 7.

Solution

There is some ambiguity as to whether the station can issue 0 tickets for a particular destination. Suppose 5 destinations had 0 tickets, then would it be true that the station issued "3800 tickets covering 200 destinations"? It would be more natural to say 3800 tickets covering 195 destinations. However, the point is not important, because the result is true either way.

If we allow zero, then the smallest possible number of tickets that could be sold if no more than 5 destinations had the same number of tickets would be 5(1 + 2 + ... + 39) = 5·39·40/2 = 3900. The number of tickets would obviously be higher if we did not allow zero. In any case, only 3800 tickets were sold, so at least 6 destinations have the same number of tickets.

However, 6(1 + 2 + ... 33) + 34 + 400 = 6·33·34/2 + 434 = 3800 and 6·33 + 2 = 200, so there could be 6 destinations with just 1 ticket, 6 with just 2 tickets and so on, up to 6 with just 33 tickets, then 1 destination with 34 tickets and 1 with 400 tickets. In this case there is no group of 7 or more destinations with the same number of tickets. 

Problem 2

The triangle ABC has ∠A > ∠C. P lies inside the triangle so that ∠PAC = ∠C. Q is taken outside the triangle so that BQ parallel to AC and PQ is parallel to AB. R is taken on AC (on the same side of the line AP as C) so that ∠PRQ = ∠C. Show that the circles ABC and PQR touch.

Solution

Let the lines BQ and AP meet at X. We show that the circles touch at X.

By construction ∠PAC = ∠C. BQ is parallel to AC, so ∠BXA = ∠C. Also by construction ∠PRQ = ∠C. So ∠PRQ = ∠PXQ and hence X lies on the circle PQR. But since ∠BXA (another name for ∠PXQ) = ∠C, X also lies on the circle ABC.

Finally, since PQ is parallel to AB, ∠BAX = ∠QPX, so the angles made by the tangents at X to the circles ABC and PQR with the chord BQX are the same. Hence the tangents coincide and the circles touch at X. 

Problem 3

a, b, c are positive integers satisfying 1/a - 1/b = 1/c and d is their greatest common divisor. Prove that abcd and d(b - a) are squares.

Solution

Let A = a/d, B = b/d, C = c/d, so that A, B, C have greatest common divisor 1. We have 1/A - 1/B = 1/C, so AB = C(B - A). We show that (B - A) must be a square.

If not, then there is a prime p such that the highest power of p dividing (B - A) is p^2r+1 for some r. Now if p^r+1 divides A, then it also divides B = (B - A) + A. Hence p^2r+2 divides AB. But p cannot divide C (since A, B, C have no common factor), so p^2r+2 divides (B - A). Contradiction. So at most p^r divides A, and similarly B. Hence at most p^2r divides AB and hence (B - A). Contradiction. That establishes that (B - A) is a square. But ABC(B - A) = ABAB, so ABC(B - A) is a square, and hence also ABC(B - A)/(B - A) = ABC.

Hence (b - a)d = d²(B - A) and abcd = d⁴ABC are squares. 

Problem 4

Show that:     xy + yz + zx = 12
      xyz - x - y - z = 2

have a unique solution in the positive reals. Show that there is a solution with x, y, z distinct reals.

Solution

Obviously 2, 2, 2 is a solution (in the positive reals). If we regard z as known, then we have x + y = (11z - 2)/(z² + 1), xy = (z² + 2z + 12)/(z² + 1). So necessary and sufficient conditions for x, y, z to be positive reals are: (11z - 2)² > 4(z² + 2z + 12)(z² + 1) (*) (which ensures that x and y are real) and z > 2/11 (which then ensures that they and z are positive).

(*) becomes 4z⁴ + 8z³ - 69z² + 52z + 44 ≤ 0, or (z - 2)²(2z + 11)(2z + 1) ≤ 0. This is satisfied for -11/2 ≤ z ≤ -1/2 and z = 2. So the only positive solution is z = 2 (and hence x = 2, y = 2 also, since the problem is symmetrical between x, y and z).

For other real solutions we should look at values in the range -11/2 to -1/2. z = -1 does not help (we get the non-distinct -1, -1, -11/2). z = -2 gives x = - (12 + 2√21)/5, y = - (12 - 2√21)/5, which are real and distinct.

A better solution to the first part was provided by Arne Smeets

So assume x, y, z are postive and put w = (xyz)^1/3. Using AM/GM on xy + yz + zx = 12, we get 12 ≥ 3 w², so w ≤ 2. Using AM/GM on the other equation we get w³ ≥ 2 + 3w, or (w - 2)(w + 1)² >= 0, so w ≥ 2. Hence w = 2. But we only get equality in AM/GM if the terms are equal, so x = y = z = 2.

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