50th Polish 1999

------
A1.  D is a point on the side BC of the triangle ABC such that AD > BC. E is a point on the side AC such that AE/EC = BD/(AD-BC). Show that AD > BE.
A2.  Given 101 distinct non-negative integers less than 5050 show that one choose four a, b, c, d such that a + b - c - d is a multiple of 5050.
A3.  Show that one can find 50 distinct positive integers such that the sum of each number and its digits is the same.
B1.  For which n do the equations have a solution in integers:
x12 + x22 + 50 = 16x1 + 12x2
x22 + x32 + 50 = 16x2 + 12x3
...
xn-12 + xn2 + 50 = 16xn-1 + 12xn
xn2 + x12 + 50 = 16xn + 12x1
B2.  Show that ∑1≤i<j≤n (|ai-aj| + |bi-bj|) ≤ ∑1≤i<j≤n |ai-bj| for all integers ai, bi.
B3.  The convex hexagon ABCDEF satisfies ∠A + ∠C + ∠E = 360o and AB·CD·EF = BC·DE·FA. Show that AB·FD·EC = BF·DE·CA.

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

Polish home
 
© John Scholes
jscholes@kalva.demon.co.uk
1 March 2004
Last corrected/updated 1 Mar 04