9th USA Mathematical Olympiad 1980 Problems

9th USA Mathematical Olympiad 1980 Problems

1.  A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight A, when placed in the left pan and against a weight a, when placed in the right pan. The corresponding weights for the second object are B and b. The third object balances against a weight C, when placed in the left pan. What is its true weight?

2.  Find the maximum possible number of three term arithmetic progressions in a monotone sequence of n distinct reals.

3.  A + B + C is an integral multiple of π. x, y, z are real numbers. If x sin A + y sin B + z sin C = x² sin 2A + y² sin 2B + z² sin 2C = 0, show that xⁿ sin nA + yⁿ sin nB + zⁿ sin nC = 0 for any positive integer n.

4.  The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.

5.  If x, y, z are reals such that 0 ≤ x, y, z ≤ 1, show that x/(y + z + 1) + y/(z + x + 1) + z/(x + y + 1) ≤ 1 - (1 - x)(1 - y)(1 - z).

Solution

9th USA Mathematical Olympiad 1980

Problem 1

A balance has unequal arms and pans of unequal weight. It is used to weigh two objects of unequal weight. The first object balances against a weight A, when placed in the left pan and against a weight a, when placed in the right pan. The corresponding weights for the second object are B and b. A third object balances against a weight C, when placed in the left pan. What is its true weight?

Solution

The effect of the unequal arms and pans is that if an object of weight x in the left pan balances an object of weight y in the right pan, then x = hy + k for some constants h and k. Thus if the first object has true weight x, then x = hA + k, a = hx + k. So a = h²A + (h+1)k. Similarly, b = h²B + (h+1)k. Subtracting gives h² = (a - b)/(A - B). and hence (h+1)k = a - h²A = (bA - aB)/(A - B).

The true weight of the third object is thus hC + k = √( (a-b)/(A-B) ) C + (bA - aB)/(A - B) 1/(√( (a-b)/(A-B) ) + 1). Problem 2

Find the maximum possible number of three term arithmetic progressions in a monotone sequence of n distinct reals.

Solution

Answer: m² for n = 2m+1, m(m-1) for n = 2m.

Let the reals be a₁, ... , a_n. Suppose n = 2m+1 and the middle term is a_k. If k < m+1, then we are constrained by the shortage of first terms. If k > m+1 we are constrained by the shortage of third terms. Thus if k = 1, a_k cannot be the middle term. If k = 2, there is only one candidate for the first term. If k = 3, there are two candidates for the middle terms and so on. Thus the total number of possible progressions certainly cannot exceed: 1 + 2 + ... + m + m-1 + m-2 + ... + 1 = m(m+1)/2 + m(m-1)/2 = m². But this bound is achieved by the sequence 1, 2, 3, ... , n.

Similarly, if n = 2m, then the upper bound is 1 + 2 + ... + m-1 + m-1 + ... + 1 = m(m-1). Again, this is achieved by the sequence 1, 2, ... , n.

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Problem 3

A + B + C is an integral multiple of π. x, y, z are real numbers. If x sin A + y sin B + z sin C = x² sin 2A + y² sin 2B + z² sin 2C = 0, show that xⁿ sin nA + yⁿ sin nB + zⁿ sin nC = 0 for any positive integer n.

Solution

The juxtaposition of x² and sin 2A strongly suggests considering cos A + i sin A. So put u = x(cos A + i sin A), v = y(cos B + i sin B), w = z(cos C + i sin C). Put a_n = uⁿ + vⁿ + wⁿ. So a₁ and a₂ are real. Hence also uv + vw + wu = (a₁² - a₂)/2 is real. Also uvw = xyz exp( i(A+B+C) ) = ± xyz, since A+B+C is an integral multiple of π. Thus u, v, w are roots of some cubic p³ + ap² + bp + c = 0 with real coefficients. Putting p = u, v, w and adding, we get a₃ + a a₂ + b a₁ + 3c = 0, so a₃ is real. Also multiplying through by pⁿ, then putting p = u, v, w and adding, we get: a_n+3 + a a_n+2 + b a_n+1 + c a_n = 0. So by a trivial induction a_n is real for all positive n. Hence result.

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Problem 4

The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.

Solution

Let the tetrahedron be ABCD. Let G be the centroid of ABC and H the centroid of ACD. Let AM be a median in ABC and AN a median in ACD. Then AG and AH are tangents to the insphere, so they are equal. CG and CH are also tangents and hence equal. So the triangles ACG and ACH are congruent. Hence ∠AGC = ∠AHC and so ∠CGM = ∠CHN. But GM = AG/2 = AH/2 = GN, so the triangles CGM and CHN are also congruent. Hence CM = CN. Hence CB = CD. So every pair of adjacent edges is equal. Hence all the edges are equal and the tetrahedron is regular.

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Problem 5

If x, y, z are reals such that 0 ≤ x, y, z ≤ 1, show that x/(y + z + 1) + y/(z + x + 1) + z/(x + y + 1) ≤ 1 - (1 - x)(1 - y)(1 - z).

Solution

Consider x/(y+z+1) + y/(z+x+1) + z/(x+y+1) + (1-x)(1-y)(1-z) as a function of x, with y and z fixed. Each term is convex, so the whole function is convex. Hence its maximum value occurs at its endpoints. The same is true for x and y, so we need only check the eight possible values x, y, z = 0 or 1. In fact, we easily find the expression has value 1 at all eight points. The result follows.

A function is convex if for any three points a < b < c, the point (b, f(b) ) lies on or below the chord joining the points (a, f(a) ) and (c, f(c) ). Analytically, this means that if b = ha + (1-h)c, where 0 ≤ h ≤ 1, then f(b) ≤ h f(a) + (1-h) f(c). The linear nature of this relation implies immediately that a sum of convex functions is convex and that a positive multiple of a convex function is convex. Linear functions are obviously convex. It is obvious from the graph that the function a/(b+x) is convex. To prove it analytically we must show that a/(b+hx+(1-h)y)) ≤ ha/(b+x) + (1-h)a/(b+y) or a(b² + bx + by + xy) ≤ ha(b+y)(b+hx+(1-h)y) + (1-h)a(b+x)(b+hx+(1-h)y). After cancelling some terms, we have to show that xy <= h(1-h)x² + (1-h)²xy + h²xy + h(1-h)y². This is obviously true for h = 1 or 0. Otherwise we may divide by 1-h, then h to get 2xy ≤ x² + y², which is true.

To see that the maximum value of a convex function must occur at its endpoints just draw a chord between the endpoints. All other points of the curve must lie below the chord.