### 10th Putnam 1950

Problem A4

Do either (1) or (2)

(1)   P is a prism with triangular base. A is a vertex. The total area of the three faces containing A is 3k. Show that if the volume of P is maximized, then each of the three faces has area k and the two lateral faces are perpendicular to each other.

(2)   Let f(x) = x + x3/(1·3) + x5/(1·3·5) + x7/(1·3·5·7) + ... , and g(x) = 1 + x2/2 + x4/(2·4) + x6/(2·4·6) + ... . Show that ∫0x exp( - t2/2) dt = f(x)/g(x).

Solution

(1) Let the sides of the base at A be a, b with angle between them θ. Let the height be h. Then the volume is V = 1/2 abh sin θ and 3k = ah + bh + 1/2 ab sin θ. Now V2 = (1/2 ab sin θ) (ah) (bh) 1/2 sin θ. By the arithmetic geometric mean theorem we have that (1/2 ab sin θ) (ah) (bh) ≤ k3, with equality iff ah = bh = 1/2 ab sin θ. So V2 ≤ k3/2 sin θ. Hence V ≤ √(k3/2) with equality iff θ = π/2 and ah = bh = 1/2 ab.

(2) exp(- t2/2) = 1 - t2/2 + t4/2·4 - t62·4·6 + ... . We notice that this is similar to g(t). In fact g(t) = exp( t2/2). So the required relation is f(x) = exp( x2/2) ∫0x exp( -t2/2) dt. Differentiating gives f '(x) = x f(x) + 1. All that is just motivation.

We start from the series for f(x). By the ratio test it is absolutely convergent, so we may differentiate term by term to get f '(x) = 1 + x2/1 + x4/1·3 + x6/1·3·5 + ... = x f(x) + 1. Multiplying this by exp(-x2/2) we get exp(-x2/2) f '(x) - x exp(-x2/2) f(x) = exp(-x2/2). Integrating and using f(0) = 0, we get exp(-x2/2) f(x) = ∫0x exp( -t2/2) dt or f(x)/g(x) = ∫0x exp( -t2/2) dt.

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002