**Problem A1**Construct equilateral triangles ABK, BCL, CDM, DAN on the inside of the square ABCD. Show that the midpoints of KL, LM, MN, NK and the midpoints of AK, BK, BL, CL, CM, DM, DN, AN form a regular dodecahedron.

**Solution**

The most straightforward approach is to use coordinates. Take A, B, C, D to be (1,1), (-1,1), (-1,-1), (1,-1). Then K, L, M, N are (0, -2k), (2k, 0), (0, 2k), (-2k, 0), where k = (√3 - 1)/2. The midpoints of KL, LM, MN, NK are (k, -k), (k, k), (-k, k), (-k, -k). These are all a distance k√2 from the origin, at angles 315, 45, 135, 225 respectively. The midpoints of AK, BK, BL, CL, CM, DM, DN, AN are (h, j), (-h, j), (-j, h), (-j, -h), (-h, -j), (h, -j), (j, -h), (j, h), where h = 1/2, j = (1 - 1/2 √3). These are also at a distance k√2 from the origin, at angles 15, 165, 105, 255, 195, 345, 285, 75 respectively. For this we need to consider the right-angled triangle sides k, h, j. The angle x between h and k has sin x = j/k and cos x = h/k. So sin 2x = 2 sin x cos x = 2hj/k^{2} = 1/2. Hence x = 15.

So the 12 points are all at the same distance from the origin and at angles 15 + 30n, for n = 0, 1, 2, ... , 11. Hence they form a regular dodecagon.

**Problem A2**In a finite sequence of real numbers the sum of any seven successive terms is negative, and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

**Solution**

Answer: 16.

x_{1} + ... + x_{7} < 0, x_{8} + ... + x_{14} < 0, so x_{1} + ... + x_{14} < 0. But x_{4} + ... + x_{14} > 0, so x_{1} + x_{2} + x_{3} < 0. Also x_{5} + ... + x_{11} < 0 and x_{1} + ... + x_{11} > 0, so x_{4} > 0. If there are 17 or more elements then the same argument shows that x_{5}, x_{6}, x_{7} > 0. But x_{1} + ... + x_{7} < 0, and x_{5} + ... + x_{11} < 0, whereas x_{1} + ... + x_{11} > 0, so x_{5} + x_{6} + x_{7} < 0. Contradiction.

If we assume that there is a solution for n = 16 and that the sum of 7 consecutive terms is -1 and that the sum of 11 consecutive terms is 1, then we can easily solve the equations to get: 5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5 and we can check that this works for 16.

**Problem A3**Given an integer n > 2, let V_{n} be the set of integers 1 + kn for k a positive integer. A number m in V_{n} is called indecomposable if it cannot be expressed as the product of two members of V_{n}. Prove that there is a number in V_{n} which can be expressed as the product of indecomposable members of V_{n} in more than one way (decompositions which differ solely in the order of factors are not regarded as different).

**Solution**

Take a, b, c, d = -1 (mod n). The idea is to take abcd which factorizes as ab.cd or ac.bd. The hope is that ab, cd, ac, bd will not factorize in V_{n}. But a little care is needed, since this is not necessarily true.

Try taking a = b = n - 1, c = d = 2n -1. a^{2} must be indecomposable because it is less than the square of the smallest element in V_{n}. If ac = 2n^{2} - 3n + 1 is decomposable, then we have kk'n + k + k' = 2n - 3 for some k, k' >= 1. But neither of k or k' can be 2 or more, because then the lhs is too big, and k = k' = 1 does not work unless n = 5. Similarly, if c^{2} is decomposable, then we have kk'n + k + k' = 4n - 4. k = k' = 1 only works for n = 2, but we are told n > 2. k = 1, k' = 2 does not work (it would require n = 7/2). k = 1, k' = 3 only works for n = 8. Other possibilities make the lhs too big.

So if n is not 5 or 8, then we can take the number to be (n - 1)^{2}(2n - 1)^{2}, which factors as (n - 1)^{2} x (2n - 1)^{2} or as (n - 1)(2n - 1) x (n - 1)(2n - 1). This does not work for 5 or 8: 16·81 = 36·36, but 36 decomposes as 6·6; 49·225 = 105·105, but 225 decomposes as 9·25.

For n = 5, we can use 3136 = 16·196 = 56·56. For n = 8, we can use 25921 = 49·529 = 161·161.

**Problem B1**Define f(x) = 1 - a cos x - b sin x - A cos 2x - B sin 2x, where a, b, A, B are real constants. Suppose that f(x) ≥ 0 for all real x. Prove that a^{2} + b^{2} ≤ 2 and A^{2} + B^{2} ≤ 1.

**Solution**

Take y so that cos y = a/√(a^{2} + b^{2}), sin y = b/√(a^{2} + b^{2}), and z so that cos 2z = A/√(A^{2} + B^{2}), sin 2z = B/√(A^{2} + B^{2}). Then f(x) = 1 - c cos(x - y) - C cos2(x - z), where c = √(a^{2} + b^{2}), C = √(A^{2} + B^{2}).

f(z) + f(π + z) ≥ 0 gives C ≤ 1. f(y + π/4) + f(y - π/4) ≥ 0 gives c ≤ √2.

**Problem B2**Let a and b be positive integers. When a^{2} + b^{2} is divided by a + b, the quotient is q and the remainder is r. Find all pairs a, b such that q^{2} + r = 1977.

**Solution**

a^{2} + b^{2} >= (a + b)^{2}/2, so q ≥ (a + b)/2. Hence r < 2q. The largest square less than 1977 is 1936 = 44^{2}. 1977 = 44^{2} + 41. The next largest gives 1977 = 43^{2} + 128. But 128 > 2.43. So we must have q = 44, r = 41. Hence a^{2} + b^{2} = 44(a + b) + 41, so (a - 22)^{2} + (b - 22)^{2} = 1009. By trial, we find that the only squares with sum 1009 are 28^{2} and 15^{2}. This gives two solutions 50, 37 or 50, 7.

**Problem B3**The function f is defined on the set of positive integers and its values are positive integers. Given that f(n+1) > f(f(n)) for all n, prove that f(n) = n for all n.

**Solution**

The first step is to show that f(1) < f(2) < f(3) < ... . We do this by induction on n. We take S_{n} to be the statement that f(n) is the unique smallest element of { f(n), f(n+1), f(n+2), ... }.

For m > 1, f(m) > f(s) where s = f(m-1), so f(m) is not the smallest member of the set {f(1), f(2), f(3), ... }. But the set is bounded below by zero, so it must have a smallest member. Hence the unique smallest member is f(1). So S_{1} is true.

Suppose S_{n} is true. Take m > n+1. Then m-1 > n, so by S_{n}, f(m-1) > f(n). But S_{n} also tells us that f(n) > f(n-1) > ... > f(1), so f(n) ≥ n - 1 + f(1) ≥ n. Hence f(m-1) ≥ n+1. So f(m-1) belongs to { n+1, n+2, n+3, .. }. But we are given that f(m) > f(f(m-1)), so f(m) is not the smallest element of { f(n+1), f(n+2), f(n+3), ... }. But there must be a smallest element, so f(n+1) must be the unique smallest member, which establishes S_{n+1}. So, S_{n} is true for all n.

So n ≤ m implies f(n) <= f(m). Suppose for some m, f(m) ≥ m+1, then f(f(m)) ≥ f(m+1). Contradiction. Hence f(m) ≤ m for all m. But since f(1) ≥1 and f(m) > f(m-1) > ... > f(1), we also have f(m) ≥ m. Hence f(m) = m for all m.