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( (x1m + x2m + ... + xnm)/m) = ( |f(x1)|m + ... + |f(xn)|m)/m for all xi, 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 0o to 180o, 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( (x1m + x2m + ... + xnm)/n) = ( |f(x1)|m + ... + |f(xn)|m)/n for all xi, 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 xi = x, for example, we get f(xm) = 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 xm = 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( (x1m + x2m + ... + xnm)/n) = (f(x1)m + ... + f(xn)m)/n = (f(x1m) + ... + f(xnm) )/n. But given any yi ≥ 0 we can find xi ≥ 0 such that yi = xim, so for any y1, y2, ... , yn we have f( (y1 + y2 + ... + yn)/n) = (f(y1) + ... + f(yn))/n.
Taking y1 = x-1, y2 = x+1 and any remaining yi = 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.



Fun Math Games for Kids

 
Return to top of page Copyright © 2010 Copyright 2010 (C) CoolMath4Kids - Cool Math 4 Kids - Cool Math Games 4 Kids - Coolmath4kids Bloxorz - Coolmath-4kids - Math games, Fun Math Lessons, Puzzles and Brain Benders, Flash Cards for Addition, Subtration, Multiplication, Fraction, Division - Cool Math 4 Kids - Math Games, Math Puzzles, Math Lessons - Cool Math 4 Kids Math Lessones - 4KidsMathGames - CoolMath Games4Kids coolmath4kids.info. All right reseved.