### 31st Putnam 1970

Problem B2

A weather station measures the temperature T continuously. It is found that on any given day T = p(t), where p is a polynomial of degree ≤ 3, and t is the time. Show that we can find times t1 < t2, which are independent of p, such that the average temperature over the period 9am to 3pm is ( p(t1) + p(t2) / 2. Show that t1 = 10:16am, t2 = 1:44pm.

Solution

Let t be time after 9am and T be 6 hrs after 9am, so that t = 0 represents 9am and t = T represents 3pm. We may write the polynomial p(t) as at3 + bt2 + ct + d for some a, b, c, d. The average temperature is 1/T ∫0T(at3 + bt2 + ct + d) dt = T3 a/4 + T2 b/3 + T c/2 + d. We wish to find t1, t2 so that this equals (t13 + t23) a/2 + (t12 + t22) b/2 + (t1 + t2) c/2 + d for all a, b, c, d.

The terms in d match. The term in c matches provided we take t1 + t2 = T. The term in b matches provided we take t12 + t22 = 2T2/3. These two equations already determine t1 and t2. In fact, solving the quadratic, we get t1 = T(1 - 1/√3)/2 and t2 = T(1 + 1/√3)/2.

For the terms in a to match we need t13 + t23 = T3/2. But we can easily check that the values above satisfy this relation also.

Finally note t1 = 6(1/2 - (√3)/6 ) hrs = 3 - √3 = 3 - 1.732 = 1.268 hrs = 1 hr 16 min (approx), so t1 represents about 10:16am and t2 about 1 hr 16 min before 3pm or 1:44pm.