6th ASU 1966 Problem 11 The distance from A to B is d kilometers. A plane P is flying with constant speed, height and direction from A to B. Over a period of 1 second the angle PAB changes by α degrees and the angle PBA by β degrees. What is the minimal speed of the plane? Solution
: 20πd√(αβ) kilometers per hour. Let the plane be at height h and a (horizontal) distance y from A. Let the angle PAB be θ+α and the angle PBA be φ. After 1 second, the angle PAB is θ and the angle PBA is φ+β. We have immediately that: h/y = tan(θ+α), h/(d-y) = tan φ, h/(y+x) = tan θ, h/(d-y-x) = tan(φ+β). Eliminating θ, we obtain: h/y = (tan α + tan θ)/(1 - tan α tan θ) = (a(y+x) + h)/(y+x-ah), where a = tan α. Hence x = a(h 2 + y 2 )/(h - ay). Similarly, eliminating φ, we obtain x = b(h 2 + (d-y) 2 )/(h + (d-y)b). At this point I do not see how to make further progress without approximating. But approximating seems reasonable, since α and β, are certainly small, at least when expressed in radians. For example, typical values might be 10,000 ft for h and more than 10 miles for y or d-y and 500 mph for the aircraft speed. That gives x = 0.14 miles, so x/y = 0.014 and x/h = 0.07. So, let us neglect a/h, b/h, a/y etc. Then we get the simplified expressions: x = a(h 2 + y 2 )/h = b(h 2 + (d-y) 2 )/h. If a = b, then we quickly obtain y = d/2, h = d/2, x = ad. Assume a > b. Then we can solve for h, substitute back in and obtain an expression for x in terms of y. It is convenient to divide through by d and to write X = x/d, Y = y/d. Note that since we are assuming a > b, we require Y < 1/(1 + √(a/b)). After some manipulation we obtain: X = ab(1 - 2Y)/√((a-b)(b(1-Y) 2 - aY 2 ). Differentiating, we find that there is a minimum at Y = b/(a+b), which is in the allowed range, and that the minimum value of X is √(ab). By symmetry, we obtain the same result for a < b and we notice that it is also true for a = b. So in all cases we have that the minimum value of x is d√(ab). We are assuming α and β are small, so we may take a = α, b = β. However, the question specified that α and β were measured in degrees, so to obtain the final answer we must convert, giving: x = d (π/180) √(αβ), and hence speed = 20πd√(αβ) kilometers per hour. 6th ASU 1966 (C) John Scholes jscholes@kalva.demon.co.uk 6 Nov 1998