8th Australian Mathematical Olympiad Problems 1987

A1.  ABC is an isosceles triangle with AB = AC. M is the midpoint of AC. D is a point on the arc BC of the circumcircle of BMC not containing M, and the ray BD meets the ray AC at E so that DE = MC. Show that MD² = AC·CE and CE² = BC·MD/2.

A2.  Show that (2p)!/(p! p!) - 2 is a multiple of p if p is prime.

A3.  A graph has 20 points and there is an edge between every two points. Each edge is colored red or green. There are two points which are not joined by a path of red edges. Show that every two points are either joined by a green edge or joined to a third point by green edges.

B1.  ABC is a triangle. P and Q are interior points such that ∠PBA = ∠QBC, and ∠PCA = ∠QCB. Show that ∠PAB = ∠QAC.

B2.  m is a fixed even positive integer and n > 1 is a fixed integer. f is a real-valued function defined on the non-negative reals such that f( (x₁^m + x₂^m + ... + x_n^m)/m) = ( |f(x₁)|^m + ... + |f(x_n)|^m)/m for all x_i, f(1988) is non-zero and f(1986) - 1986 is non-zero. Show that f(1987) = 1.

B3.  Show that 1/√1 + 1/√2 + 1/√3 + ... + 1/√n < √(n+1) + √n - √2 for n > 1. 

Solutions

Problem A1

ABC is an isosceles triangle with AB = AC. M is the midpoint of AC. D is a point on the arc BC of the circumcircle of BMC not containing M, and the ray BD meets the ray AC at E so that DE = MC. Show that MD² = AC·CE and CE² = BC·MD/2.

Solution

Cosine rule on MED gives ME² = DE² + MD² - 2MD·DE cos MDE (*). But -cos MDE = cos MDB = cos C = BC/2AC, so (*) becomes (MC + CE)² = DE² + MD² + MD·DE·BC/AC, or 2MC·CE + CE² = MD² + MD·BC/2 (**), since MC = DE and AC = 2MC = 2DE.

Triangles EDM, ECB are similar, so DE/DM = CE/CB. Substituting from this for BC makes (**) become: AC·CE(1 + CE/AC) = MD²(1 + CE/AC), so MD² = AC·CE. Subtracting from (**) gives CE² = MD·BC/2. 

Problem A2

Show that (2p)!/(p! p!) - 2 is a multiple of p if p is prime.

Solution

(2p)!/(p!p!) = (2p/p) ((2p-1)/(p-1)) ... (p+1)/1. Note that 1, ... , p-1 and (2p-1), (2p-2), ... , (p+1) are both complete sets of non-zero residues mod p. So their product is equal mod p. Hence (2p)!/(p!p!) = 2p/p = 2 mod p. 

Problem A3

A graph has 20 points and there is an edge between every two points. Each edge is colored red or green. There are two points which are not joined by a path of red edges. Show that every two points are either joined by a green edge or joined to a third point by green edges.

Solution

Suppose A and B are the points not joined by a path of red edges. Suppose X and Y are any two other points. If the edge XY is green we are done, so assume it is red. If both XA and YA are green we are done, so wlog XA is red. Now YB must be green or we have the red path AXYB. Now XB cannot be red or we have the red path AXB, so XB is green and B is joined to both X and Y by green edges.

It remains to consider the cases where one or other of X, Y belong to {A, B}. Obviously the edge AB must be green, or we would have a red path A to B. That deals with the case where both X and Y belong to {A, B}. So suppose Y = A. If the edge AX is green we are done, so assume it is red. But now XB must be red, or we would have the red path AXB. But now B is joined to A and X by green edges.

Note that 20 is a red herring.

Problem B1

ABC is a triangle. P and Q are interior points such that ∠PBA = ∠QBC, and ∠PCA = ∠QCB. Show that ∠PAB = ∠QAC.

Solution

Using the sine rule repeatedly and putting ∠PBA = x, ∠PCA = y, ∠PAB = z, ∠QAC = z', we have PA/PB = sin x/sin z, PB/PC = sin(C-y)/sin(B-x), PC/PA = sin(A-z)/sin y. So sin z/sin(A-z) = (sin x/sin(B-x))(sin(C-y)/sin y). Similarly, writing down QA/QB etc we get sin z'/sin(A-z') = (sin x/sin(B-x))(sin(C-y)/sin y). Hence sin z sin(A-z') = sin z' sin(A-z). Expanding, sin z(sin A cos z' - cos A sin z') = sin z'(sin A cos z - cos A sin z), so sin z cos z' = sin z' cos z, so tan z = tan z'. But both z and z' lie in the range 0^o to 180^o, so z = z'.

 

Problem B2

m is a fixed even positive integer and n > 1 is a fixed integer. f is a real-valued function defined on the non-negative reals such that f( (x₁^m + x₂^m + ... + x_n^m)/n) = ( |f(x₁)|^m + ... + |f(x_n)|^m)/n for all x_i, f(1988) is non-zero and f(1986) - 1986 is non-zero. Show that f(1987) = 1.

Solution

The absolute values signs are unnecessary since m is even. Putting all x_i = x, for example, we get f(x^m) = f(x)^m. Putting x = 0, 1 gives f(0) = 0 or 1 and f(1) = 0 or 1. Also given any y we can find x such that x^m = y, so f(y) = f(x)^m ≥ 0. In other words f always takes non-negative values.

Substituting back into the main relation gives f( (x₁^m + x₂^m + ... + x_n^m)/n) = (f(x₁)^m + ... + f(x_n)^m)/n = (f(x₁^m) + ... + f(x_n^m) )/n. But given any y_i ≥ 0 we can find x_i ≥ 0 such that y_i = x_i^m, so for any y₁, y₂, ... , y_n we have f( (y₁ + y₂ + ... + y_n)/n) = (f(y₁) + ... + f(y_n))/n.

Taking y₁ = x-1, y₂ = x+1 and any remaining y_i = x (if n > 2) we get f(x) = f(x-1)/n + f(x+1)/n + (n-2)/n f(x), so 2f(x) = f(x-1) + f(x+1). Hence f(0), f(1), f(2), ... form an arithmetic progression. If f(0) = f(1) = 0, then all f(n) = 0, but we are told f(1988) ≠ 0. If f(0) = 0, f(1) = 1, then all f(n) = n, but we are told f(1986) ≠ 1986. If f(0) = 1, f(1) = 0, then f(n) < 0 for n > 1, but we showed above that f(n) is always non-negative. So we must have f(0) = f(1) = 1 and hence f(n) = 1 for all positive integers. 

Problem B3

Show that 1/√1 + 1/√2 + 1/√3 + ... + 1/√n < √(n+1) + √n - √2.

Solution

Induction on n. For n = 1 we have 1 = √2 + 1 - √2. We show that ≤ for n implies < for n+1. It is sufficient to show that 1/√(n+1) < √(n+2) - √n, or √n + 1/√(n+1) < √(n+2). Squaring, that is equivalent to 2√(n/(n+1)) < 2 - 1/(n+1), or squaring again to n/(n+1) < 1 - 1/(n+1) + 1/(4(n+1)²), which is obviously true. 

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