12th Irish 1999

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A1.  Find all real solutions to x2/(x+1-√(x+1))2 < (x2+3x+18)/(x+1)2.
A2.  The Fibonacci sequence is defined by F0 = 1, F1 = 1, Fn+2 = Fn+1 + Fn. Show that some Fibonacci number is divisible by 1000.
A3.  In the triangle ABC, AD is an altitude, BE is an angle bisector and CF is a median. Show that they are concurrent iff a2(a-c) = (b2-c2)(a+c).
A4.  Show that a 10000 x 10000 board can be tiled by 1 x 3 tiles and a 2 x 2 tile placed centrally, but not if the 2 x 2 tile is placed in a corner.
A5.  The sequence u0, u1, u2, ... is defined as follows. u0 = 0, u1 = 1, and un+1 is the smallest integer > un such that there is no arithmetic progression ui, uj, un+1 with i < j < n+1. Find u100.
B1.  Solve: y2 = (x+8)(x2+2) and y2 - (8+4x)y + (16+16x-5x2) = 0.
B2.  f(n) is a function defined on the positive integers with positive integer values such that f(ab) = f(a)f(b) when a, b are relatively prime and f(p+q) = f(p)+f(q) for all primes p, q. Show that f(2) = 2, f(3) = 3 and f(1999) = 1999.
B3.  Show that a2/(a+b) + b2/(b+c) + c2/(c+d) + d2/(d+a) ≥ 1/2 for positive reals a, b, c, d such that a + b + c + d = 1, and that we have equality iff a = b = c = d.
B4.  Let d(n) be the number of positive divisors of n. Find all n such that n = d(n)4.
B5.  ABCDEF is a convex hexagon such that AB = BC, CD = DE, EF = FA and ∠B + ∠D + ∠F = 360o. Show that the perpendiculars from A to FB, C to BD, and E to DF are concurrent.

To avoid possible copyright problems, I have changed the wording, but not the substance, of the problems.

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© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 2003
Last updated/corrected 12 Oct 03