7th Australian Mathematical Olympiad Problems 1986

A1.  Given a positive integer n and real k > 0, what is the largest possible value for (x₁x₂ + x₂x₃ + x₃x₄ + ... + x_n-1x_n), where x_i are non-negative real numbers with sum k?

A2.  What is the smallest tower of 100s that exceeds a tower of 100 threes? In other words, let a₁ = 3, a₂ = 3³, and aⁿ⁺¹ = 3^a_n. Similarly, b₁ = 100, b₂ = 100¹⁰⁰ etc. What is the smallest n for which b_n > a₁₀₀?

A3.  Three chords of the circumcircle bisect the three angles of a triangle. Show that the sum of their lengths exceeds the perimeter of the triangle.

B1.  C is a circle of unit radius. For any line L define d(L) to be the distance between the two points of intersection if L meets C in two points, and zero otherwise. For any point P define f(P) to be the maximum value of d(L) + d(L') for two perpendicular lines through P. Which P have f(P) > 2?

B2.  Define the sequence a₁, a₂, a₃, ... by a₁ = 1, a₂ = b, a_n+2 =2a_n+1 - a_n + 2, where b is a positive integer. Show that for any n we can find an m such that a_na_n+1 = a_m.

B3.  A real polynomial of degree n > 2 has all roots real and positive. The coefficient of xⁿ is 1, the coefficient of x^n-1 is -1. The coefficient of x¹ is non-zero and -n² times the coefficient of x⁰. Show that all the roots are equal. 

Solutions

 
Problem A1

Given a positive integer n and real k > 0, what is the largest possible value for (x₁x₂ + x₂x₃ + x₃x₄ + ... + x_n-1x_n), where x_i are non-negative real numbers with sum k?

Answer

k²/4

Solution

Let x_h = max(x_i). Then x₁x₂ + x₂x₃ + x₃x₄ + ... + x_h-1x_h + x_hx_h+1 + x_h+1x_h+2 + ... + x_n-1x_n ≤ x₁x_h + x₂x_h + ... + x_h-1x_h + x_hx_h+1 + x_hx_h+2 + ... + x_hx_n = x_h(k - x_h) ≤ k²/4 (last step AM/GM). But k²/4 can be realized by x₁ = x₂ = k/2 and others 0. 

Problem A2

What is the smallest tower of 100s that exceeds a tower of 100 threes? In other words, let a₁ = 3, a₂ = 3³, and aⁿ⁺¹ = 3^a_n. Similarly, b₁ = 100, b₂ = 100¹⁰⁰ etc. What is the smallest n for which b_n > a₁₀₀?

Answer

99

Solution

We have 3³ < 100 and hence by a trivial induction a_n+1 < b_n (a_n+1 = 3^a_n < 100^a_n < 100^b_n-1 = b_n).

We claim that a_n+1 > 6b_n-1. Obviously 3³³ > 600 so it is true for n = 2. Suppose it is true for n. Note that 3⁶ = 729 > 600, so a_n+2 > 3^6b_n-1 > 600^b_n-1 = 6^b_n-1b_n > 6b_n.

Problem A3

Three chords of the circumcircle bisect the three angles of a triangle. Show that the sum of their lengths exceeds the perimeter of the triangle.

Solution

Bizarrely, this is a repeat of 82/A3. 

Problem A3

In the triangle ABC, let the angle bisectors of A, B, C meet the circumcircle again at X, Y, Z. Show that AX + BY + CZ is greater than the perimeter of ABC.

Solution

The sine rule gives BC/sin A = CA/sin B = AB/sin C. Put k = BC/sin A, so perimeter = k(sin A + sin B + sin C). Applying sine rule to BCY we have BC/sin Y = BC/sin A = k = BY/sin BCY. But ∠BCY = ∠C + ∠ACY = ∠C + ∠B/2. So BY = k sin(C+B/2). Similarly for the other chords, so AX + BY + CZ = k(sin(C+B/2) + sin(A+C/2) + sin(B+A/2)).

But sin(C+B/2) = sin(90^o + C/2 - A/2) = cos(A/2 - C/2) > cos(A/2 - C/2) sin(A/2 + C/2) = ½ sin A + ½ sin C. Adding the two similar relations gives the required result. Note that we cannot have equality because A + C < 180^o, so sin(A/2 + C/2) < 1.

Problem B1

C is a circle of unit radius. For any line L define d(L) to be the distance between the two points of intersection if L meets C in two points, and zero otherwise. For any point P define f(P) to be the maximum value of d(L) + d(L') for two perpendicular lines through P. Which P have f(P) > 2?

Answer

points inside (but not on) the circle with the same center and radius √(3/2)

Solution

It is obvious that points inside C qualify (take L to be a diameter), and fairly obvious that points on C qualify (L and L' meet C again at the ends of a diameter). So we consider P outside C.

Let O be the center of C. Obviously if PO ≥ √2, then only one of L, L' meets C, so f(P) = 2. So suppose P is close enough for L and L' to meet C.

Let OP = r. Take the angle x as shown. Then d = d(L) + d(L') = 2√(1 - r²cos²x) + 2√(1 - r²sin²x). So d²/4 = 2 - r² + 2√(1 - r² + r⁴sin2xcos2x) = 2 - r² + √(4 - 4r² + r⁴sin2x). That is obviously maximised by taking x = π/4 giving 2 - r² + √(4 - 4r² + r⁴) = 4 - 2r², since r² < 2. So f(P) > 2 iff 4 - 2r² > 1, or r < √(3/2).

Problem B2

Define the sequence a₁, a₂, a₃, ... by a₁ = 1, a₂ = b, a_n+2 =2a_n+1 - a_n + 2, where b is a positive integer. Show that for any n we can find an m such that a_na_n+1 = a_m.

Solution

A trivial induction shows that a_n+2 = (n+1)b + n². So a_n+1a_n+2 = n(n+1)b² + (2n³-n²-n+1)b + n²(n-1)² = a_N+2, where N = nb+n²-n. 

Problem B3

A real polynomial of degree n > 2 has all roots real and positive. The coefficient of xⁿ is 1, the coefficient of x^n-1 is -1. The coefficient of x¹ is non-zero and -n² times the coefficient of x⁰. Show that all the roots are equal.

Solution

Let the roots be a₁, a₂, ... , a_n. Then a₁ + a₂ + ... + a_n = 1 and 1/a₁ + 1/a₂ + ... + 1/a_n = n², so (a₁ + ... + a_n)(1/a₁ + ... + 1/a_n) = n². But (a₁ + ... + a_n)(1/a₁ + ... + 1/a_n) ≥ n² with equality iff all a_i are equal. (A well-known result = prove for example by Cauchy-Schwartz on √a_i and 1/√a_i). 

Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust