A1. Find all real solutions to (1 + 4^{2x-y})(5^{y-2x+1}) = 2^{2x-y+1} + 1, y^{3} + 4x + ln(y^{2} + 2x) + 1 = 0. | |
A2. ABC is a triangle. A' is the midpoint of the arc BC of the circumcircle not containing A. B' and C' are defined similarly. The segments A'B', B'C', C'A' intersect the sides of the triangle in six points, two on each side. These points divide each side of the triangle into three parts. Show that the three middle parts are equal iff ABC is equilateral. | |
A3. The sequence a_{1}, a_{2}, a_{3}, ... is defined by a_{1} = 1, a_{2} = 2, a_{n+2} = 3a_{n+1} - a_{n}. The sequence b_{1}, b_{2}, b_{3}, ... is defined by b_{1} = 1, b_{2} = 4, b_{n+2} = 3b_{n+1} - b_{n}. Show that the positive integers a, b satisfy 5a^{2} - b^{2} = 4 iff a = a_{n}, b = b_{n} for some n. | |
B1. Find the maximum value of 2/(x^{2} + 1) - 2/(y^{2} + 1) + 3/(z^{2} + 1) for positive reals x, y, z which satisfy xyz + x + z = y. | |
B2. OA, OB, OC, OD are 4 rays in space such that the angle between any two is the same. Show that for a variable ray OX, the sum of the cosines of the angles XOA, XOB, XOC, XOD is constant and the sum of the squares of the cosines is also constant. | |
B3. Find all functions f(n) defined on the non-negative integers with values in the set {0, 1, 2, ... , 2000} such that: (1) f(n) = n for 0 ≤ n ≤ 2000; and (2) f( f(m) + f(n) ) = f(m + n) for all m, n. |
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
22 July 2002