15th CanMO 1983 Problem 2 Find all real-valued functions f on the reals whose graphs remain unchanged under all transformations (x, y) → (2 k x, 2 k (kx + y) ), where k is real. Solution
: for any fixed real a, b we can take f(x) = x ln 2 |x| + ax for x positive and x ln 2 |x| + bx for x negative. Suppose f(0) = h. So (0, h) is a point on the graph. Hence so is (0, 2 k h) for all k. But f is single-valued, so h must be 0. Suppose f(1) = h. Then (1, h) is a point on the graph. Hence so is (2 k , 2 k k + 2 k h) for all real k. Put k = ln 2 x. Then (x, x ln 2 x + xh) is a point on the graph. In other words, f(x) = x ln 2 x + hx. We need to check that this works. So we have a point (x, x ln 2 x + hx). Under the transformation this goes to the point P (2 k x, 2 k (kx + x ln 2 x + hx) ). Put x' = 2 k x, so ln 2 x' = k + ln 2 x. So we can write P as (x', x'(ln 2 x' - ln 2 x) + x' ln 2 x + hx') = (x', x' ln 2 x' + hx'), which is indeed another point on the graph. We now have to consider negative x. So suppose f(-1) = h. So (-1, h) is a point on the graph. Hence so is (-2 k , -2 k k - 2 k h) for all real k. Put k = ln 2 x. Then (-x, -x ln 2 x - hx) is a point on the graph. In other words, f(x) = x ln 2 |x| + hx for x negative. Again we can easily check that this works. Note that h does not have to be the same for positive and negative x. 15th CanMO 1983 © John Scholes jscholes@kalva.demon.co.uk 21 June 2002