13th USA Mathematical Olympiad 1984 Problems
1. Two roots of the real quartic x4 - 18x3 + ax2 + 200x - 1984 = 0 have product -32. Find a.
2. Can one find a set of n distinct positive integers such that the geometric mean of any (non-empty, finite) subset is an integer? Can one find an infinite set with this property?
3. A, B, C, D, X are five points in space, such that AB, BC, CD, DA all subtend the acute angle θ at X. Find the maximum and minimum possible values of ∠AXC + ∠BXD (for all such configurations) in terms of θ.
4. A maths exam has two papers, each with at least one question and 28 questions in total. Each pupil attempted 7 questions. Each pair of questions was attempted by just two pupils. Show that one pupil attempted either nil or at least 4 questions in the first paper.
5. A polynomial of degree 3n has the value 2 at 0, 3, 6, ... , 3n, the value 1 at 1, 4, 7, ... , 3n-2 and the value 0 at 2, 5, 8, ... , 3n-1. Its value at 3n+1 is 730. What is n?
Solution
13th USA Mathematical Olympiad 1984
Problem 1
Two roots of the real quartic x4 - 18x3 + ax2 + 200x - 1984 = 0 have product -32. Find a.
Solution
Let the two roots satisfy the quadratic x2 + hx -32 = 0 (we have not yet shown that h is real). The other two roots have product -1984/-32 = 62. Let them satisfy the quadratic x2 + kx + 62. So x4 - 18x3 + ax2 + 200x - 1984 = (x2 + hx -32)(x2 + kx + 62) = x4 + (h+k) x3 + (hk + 30)x2 + (62h - 32k)x - 1984 = 0. Equating coefficients: h + k = -18, 62h - 32k = 200. Solving, h = -4, k = -14. Hence a = 86. Problem 2
Can one find a set of n distinct positive integers such that the geometric mean of any (non-empty, finite) subset is an integer? Can one find an infinite set with this property?
Solution
Answer: yes, no.
Take each member to be an n! power (for example, 1n!, 2n!, ... , nn!).
Suppose we could find an infinite set. Take any two members m and n. Then for sufficiently large k, (m/n)1/k must be irrational. But now if we take any other a1, a2, ... , ak-1 in the set, (m a1 ... ak)1/k and (n a1 ... ak)1/k cannot both be integers. Contradiction.
A, B, C, D, X are five points in space, such that AB, BC, CD, DA all subtend the acute angle θ at X. Find the maximum and minimum possible values of ∠AXC + ∠BXD (for all such configurations) in terms of θ.
Solution
Answer: minimum 0 (see below); maximum 2 cos-1(2 cos θ - 1), achieved by a square pyramid.
If we take A, B, C, D to lie in a plane so that AC and BD meet at X with the angle AXC = θ, then AB, BC, CD, DA all subtend the angle θ at X, but AC and BD subtend the angle zero. So the minimum is 0.
A little playing around suggests that we should take ABCD to be a square, with X on the normal through the center, so that XABCD is a square pyramid. We calculate the result for this case. Suppose XA = XB = XC = XD = 1, ∠AXB = ∠BXC = ∠CXD = ∠DXA = θ. Then AB = 2 sin θ/2, so AC = 2√2 sin θ/2. So sin AXC/2 = √2 sin θ/2. Hence cos AXC = 1 - 2 sin2AXC/2 = 1 - 4 sin2θ/2 = 2 cos θ - 1. Hence ∠AXC + ∠BXD = 2 cos-1(2 cos θ - 1). We note that this increases monotonically from 0 (at θ = 0) to 2π (at θ = π/2). For θ > 0, we have cos θ < 1, hence 2 cos θ - 1< cos θ and hence 2 cos-1(2 cos θ - 1) > 2θ. In other words, except for θ = 0, we can certainly do better than 2θ.
Take vectors origin X. Write XA = A etc. Since we are only interested in the angles, it is convenient to take a to be the unit vector in the direction A. Then we have a.b = b.c = c.d = d.a = cos q. So (a - c).(b - d) = 0. So either a = c, or b = d or a - c is perpendicular to b - d. Suppose a = c. Then X lies on the line AC. In this case we certainly have AD and CD subtending the same angle at X, and AB and BC subtending the same angle at X. But ∠AXC = 0. XABD is a tetrahedron with ∠AXD = ∠AXB = θ. ∠BXD ≤ ∠AXB + ∠AXD (with equality only if A, B, D, X lie in a plane). So ∠BXD + ∠AXC ≤ 2θ and we have just seen that this is not maximal. Similarly if b = d. So we may assume that a - c is perpendicular to b - d and both are non-zero.
We also have (a + c).(a - c) = a2 - c2 = 0 and (a + c).(b - d) = 0. So a + c is normal to the plane containing a - c and b - d. Similarly, b + d. Hence a + c and b + d are multiples of each other and (a + c).(b + d) = |a + c| |b + d|. But lhs = 4 cos θ and rhs = (2 cos AXC/2)(2 cos BXD/2). So cos θ = cos AXC/2 cos BXD/2 = 1/2 cos(AXC/2 + BXD/2) + 1/2 cos(AXC/2 - BXD/2). We want to maximise AXC/2 + BXD/2 and hence to minimise cos(AXC/2 + BXD/2), so we must maximise cos(AXC/2 - BXD/2) and hence take ∠AXC = ∠BXD. That then gives cos(AXC/2 + BXD/2) = 2 cos θ - 1, so AXC + BXD = 2 cos-1(2 cos θ - 1). So the square pyramid is indeed maximal.
A maths exam has two papers, each with at least one question and 28 questions in total. Each pupil attempted 7 questions. Each pair of questions was attempted by just two pupils. Show that one pupil attempted either nil or at least 4 questions in the first paper.
Solution
Each pupil attempts 7 questions and hence 21 pairs of questions. There are 28·27/2 = 378 pairs of questions in total and each is attempted by 2 pupils. So there must be 378·2/21 = 36 pupils. Suppose n pupils solved question 1. Each solved 6 pairs involving question 1, so there must be 3n pairs involving question 1. But there are 27 pairs involving question 1, so n = 9. The same applies to any other question. So each question was solved by 9 pupils.
Suppose the result is false. Suppose there are m questions in the first paper, that the number of pupils solving 1, 2, 3 questions in the first paper is a, b, c respectively. So a + b + c = 36, a + 2b + 3c = 9m. Now consider pairs of problems in the first paper. There are m(m-1)/2 such pairs. Pupils solving just 1 solve no pairs, those solving 2 solve 1 pair and those solving 3 solve 3 pairs, so we have b + 3c = m(m-1). Solving for b we get b = - 2m2 + 29m - 108 = -2(m - 29/4)2 - 23/8 < 0. Contradiction. So the result must be true.
A polynomial of degree 3n has the value 2 at 0, 3, 6, ... , 3n, the value 1 at 1, 4, 7, ... , 3n-2 and the value 0 at 2, 5, 8, ... , 3n-1. Its value at 3n+1 is 730. What is n?
Solution
Answer: n = 4.
The (3n+1)th differences of the polynomial are zero. Call it p(x), so we have p(3n+1) - (3n+1)C1 p(3n) + (3n+1)C2 p(3n-1) - ... + (-1)3n+1 p(0) = 0, where rCs is the binomial coefficient. Hence p(3n+1) = 2( (3n+1)C1 - (3n+1)C4 + ... ) + ( (3n+1)C3 - (3n+1)C6 + ... ). Putting n = 1, we get: p(4) = 2( 4C1 - 4C4) + 4C3 = 6 + 4 = 10. So n is not 1. Putting n = 2, we get: p(7) = 2( 7C1 - 7C4 + 7C7) + ( 7C3 - 7C6) = 2( 7 - 35 + 1) + (35 - 7) = -26. So n is not 2. Putting n = 3, we get: p(10) = 2( 10C1 - 10C4 + 10C7 - 10C10) + ( 10C3 - 10C6 + 10C9) = 2(10 - 210 + 120 - 1) + (120 - 210 + 10) = -162 -100 = -262. So n is not 3. Putting n = 4, we get: p(13) = 2( 13C1 - 13C4 + 13C7 - 13C10 + 13C13) + ( 13C3 - 13C6 + 13C9 - 13C12) = 2(13 - 715 + 1716 - 286 + 1) + (286 - 1716 + 715 - 13) = 1458 - 728 = 730. So n = 3 works.
Comment. That looks kludgy and indeed it is. But it is also fast. Less than 5 minutes. Of course, the elegant approach is to find a general expression for those binomial sums. But there is no need, so why take time (unless of course one has finished all the other questions and has time to chase prizes for elegance!)
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