6th USA Mathematical Olympiad 1977 Problems

6th USA Mathematical Olympiad 1977 Problems

1.  For which positive integers a, b does (x^a + ... + x + 1) divide (x^ab + x^ab-b + ... + x^2b + x^b + 1)?

2.  The triangles ABC and DEF have AD, BE and CF parallel. Show that [AEF] + [DBF] + [DEC] + [DBC] + [AEC] + [ABF] = 3 [ABC] + 3 [DEF], where [XYZ] denotes the signed area of the triangle XYZ. Thus [XYZ] is + area XYZ if the order X, Y, Z is anti-clockwise and - area XYZ if the order X, Y, Z is clockwise. So, in particular, [XYZ] = [YZX] = -[YXZ].

3.  Prove that the product of the two real roots of x⁴ + x³ - 1 = 0 is a root of x⁶ + x⁴ + x³ - x² - 1 = 0.

4.  ABCD is a tetrahedron. The midpoint of AB is M and the midpoint of CD is N. Show that MN is perpendicular to AB and CD iff AC = BD and AD = BC.

5.  The positive reals v, w, x, y, z satisfy 0 < h ≤ v, w, x, y, z ≤ k. Show that (v + w + x + y + z)(1/v + 1/w + 1/x + 1/y + 1/z) ≤ 25 + 6( √(h/k) - √(k/h) )². When do we have equality?

Solution

6th USA Mathematical Olympiad 1977

Problem 1

For which positive integers a, b does (x^a + ... + x + 1) divide (x^ab + x^ab-b + ... + x^2b + x^b + 1)?

Solution

Answer: a+1 and b relatively prime.

The question is when (x^a+1 - 1)/(x - 1) divides (x^b(a+1) - 1)/(x^b - 1) or when (x^a+1 - 1)(x^b - 1) divides (x^b(a+1) - 1)(x - 1). Now both (x^a+1 - 1) and (x^b - 1) divide (x^b(a+1) - 1). They both have a factor (x - 1), so if that is their only common factor, then their product divides (x^b(a+1) - 1)(x - 1). That is true if a+1 and b are relatively prime, for the roots of x^k - 1 are the kth roots of 1. Thus if a+1 and b are relatively prime, then the only (complex) number which is an (a+1)th root of 1 and a bth root of 1 is 1.

But suppose d is a common factor of a+1 and b, then exp(2πi/d) is a root of both x^a+1 - 1 and x^b - 1. It is a root of x^b(a+1) - 1, but only with multiplicity 1, so (x^a+1 - 1)(x^b - 1) does not divide (x^b(a+1) - 1)(x - 1). Problem 2

The triangles ABC and DEF have AD, BE and CF parallel. Show that [AEF] + [DBF] + [DEC] + [DBC] + [AEC] + [ABF] = 3 [ABC] + 3 [DEF], where [XYZ] denotes the signed area of the triangle XYZ. Thus [XYZ] is + area XYZ if the order X, Y, Z is anti-clockwise and - area XYZ if the order X, Y, Z is clockwise. So, in particular, [XYZ] = [YZX] = -[YXZ].

Solution

The starting point is that [ABC] = [XBC] + [AXC] + [ABX] (*) for any point X. If X is inside the triangle, then all the rotations have the same sense. If X is outside, then they do not. But it is easy to check that (*) always holds.

So [ABC] = [DBC] + [ADC] + [ABD], [DEF] = [AEF] + [DAF] + [DEA]. Now, ignoring sign, the triangles ABD and DEA have equal area, because they have a common base AD and the same height (since AD is parallel to BE). But the sign is opposite, so [ABD] + [DEA] = 0. Similarly, [ADC] + [DAF] = 0, so [ABC] + [DEF] = [DBC] + [AEF]. Adding the two similar equations (obtained from E with [ABC] and B with [DEF], and from F with [ABC] and C with [DEF]) gives the required result.

Alternative solution

Use vectors. Take origin A. Put b = AB, c = AC, d = AD. Then AE = b + h d, AF = c + k d, for some real h, k. Then 2 [ABC] = b x c etc and the rest is purely mechanical. Start with the lhs, expand and use d x d = 0.

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Problem 3

Prove that the product of the two real roots of x⁴ + x³ - 1 = 0 is a root of x⁶ + x⁴ + x³ - x² - 1 = 0.

Solution

Let the roots of the quartic be a, b, c, d. We show that ab, ac, ad, bc, bd, cd are the roots of the sextic. We have a + b + c + d = -1, ab + ac + ad + bc + bd + cd = 0, 1/a + 1/b + 1/c + 1/d = 0, abcd = -1.

Let x⁶ + a₅x⁵ + a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀ be the sextic with roots ab, ac, ad, bc, bd, cd. Since their sum is zero, we have a₅ = 0. Their product is (abcd)³, so a₀ = -1.

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Problem 4

ABCD is a tetrahedron. The midpoint of AB is M and the midpoint of CD is N. Show that MN is perpendicular to AB and CD iff AC = BD and AD = BC.

Solution

Use vectors. Take any origin O and write the vector OX as X. Then MN perpendicular to AB and CD is equivalent to:

(A + B - C - D).(A - B) = 0 and

(A + B - C - D).(C - D) = 0.

Expanding and adding the two equations gives (A - D)² = (B - C)² or AD = BC. Subtracting gives AC = BD.

Conversely, AD = BC and AC = BD gives:

(A - D)² = (B - C)² and

(A - C)² = (B - D)².

Adding gives (A + B - C - D).(A - B) = 0, so MN is perpendicular to AB. Subtracting gives MN perpendicular to CD.

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Problem 5

The positive reals v, w, x, y, z satisfy 0 < h ≤ v, w, x, y, z ≤ k. Show that (v + w + x + y + z)(1/v + 1/w + 1/x + 1/y + 1/z) ≤ 25 + 6( √(h/k) - √(k/h) )². When do we have equality?

Solution

Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but x. Then the expression on the lhs has the form (r + x)(s + 1/x) = (rs + 1) + sx + r/x, where r and s are fixed. But this is convex. That is to say, as x increases if first decreases, then increases. So its maximum must occur at x = h or x = k. This is true for each variable.

Suppose all five are h or all five are k, then the lhs is 25, so the inequality is true and strict unless h = k. If four are h and one is k, then the lhs is 17 + 4(h/k + k/h). Similarly if four are k and one is h. If three are h and two are k, then the lhs is 13 + 6(h/k + k/h). Similarly if three are k and two are h.

h/k + k/h ≥ 2 with equality iff h = k, so if h < k, then three of one and two of the other gives a larger lhs than four of one and one of the other. Finally, we note that the rhs is in fact 13 + 6(h/k + k/h), so the inequality is true with equality iff either (1) h = k or (2) three of v, w, x, y z are h and two are k or vice versa.