SOLUTION TO PROBLEM 1546 PROPOSED BY B. G. KLEIN, ET AL Summary: SOLUTION TO PROBLEM 1546 PROPOSED BY B. G. KLEIN, ET AL Tewodros Amdeberhan DeVry Institute, Mathematics 630 US Highway One, North Brunswick, NJ 08902 amdberhan@admin.nj.devry.edu Problem 1546: P Given y 1, let P be the set of all real polynomials px with nonnegative coe cients that satisfy p1 = 1 and p3 = y. Prove there exists p0x 2 P such that i fp2 : px 2 Pg = 1;p02 ; ii if px 2 P and p2 = p02, then px = p0x. Solution: We show that p0x = y,1 2 x , 1 + 1 is the unique polynomial satisfying the above conditions with p02 = y + 1=2. Note that since coe cients are nonnegative, all functions in P are strictly increasing as well as concave upwards for 1 x 3. Thus certainly for each px 2 P, we have px p0x;x 2 1;3 since the line p0x joins the end points 1;1 and 3;y. Furthermore, p0x is unique. Else assume that p2 = p02, for some p0x 6= px 2 P. Then by the Intermediate-Value-Theorem for derivatives, there exist two distinct points one in 1;2 and another in 2;3 where tangents to px hence derivatives have same slope, that is, y , 1=2. This cannot be true of the nonlinear px as its derivative is one-to-one because p00x 0 there. Collections: Mathematics