(1 + a1) ... (1 + an) < 1 + s + s2/2! + ... + sn/n! .
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A1. ai are positive reals. s = a1 + ... + an. Prove that for any integer n > 1 we have
(1 + a1) ... (1 + an) < 1 + s + s2/2! + ... + sn/n! . |
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| A2. Prove that 5n2 = 36a2 + 18b2 + 6c2 has no integer solutions except a = b = c = n = 0. |
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| A3. ABC is a triangle. X lies on the segment AB so that AX/AB = 1/4. CX intersects the median from A at A' and the median from B at B''. Points B', C', A'', C'' are defined similarly. Find the area of the triangle A''B''C'' divided by the area of the triangle A'B'C'. |
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| A4. Show that a graph with n vertices and k edges has at least k(4k - n2)/3n triangles. |
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| A5. f is a strictly increasing real-valued function on the reals. It has inverse f-1. Find all possible f such that f(x) + f-1(x) = 2x for all x. |
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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
5 April 2002