### 20th Putnam 1959

Problem A5

At a particular moment, A, T and B are in a vertical line, with A 50 feet above T, and T 100 feet above B. T flies in a horizontal line at a fixed speed. A flies at a fixed speed directly towards B, B flies at twice T's speed, also directly towards T. A and B reach T simultaneously. Find the distance traveled by each of A, B and T, and A's speed.

Solution

Answer: A travels 25(3 + √73)/3 = 96.2 ft, B travels 400/3 ft = 133.3 ft, T 200/3 ft = 66.7 ft, A's speed is (3 + √73)/8 = 1.443 times T's speed.

Take the x-axis vertical and the y-axis horizontal, so that at t = 0, the target is at (0, 0) and the pursuer is at (a, 0). Assume the target (T) has speed v and the pursuer (A or B) has speed kv with k > 1. At time t, the target is at (0, vt). The equations of motion are: (dx/dt)2 + (dy/dt)2 = kv and y' = (y - vt)/x. The first equation gives kv/(dx/dt) = -√(1 + y'2) (negative because x decreases with time).

Differentiating the second equation gives xy'' = - v/(dx/dt). So kxy'' = √(1 + y'2). Integrating, ln(y' + √(1 + y'2) ) = ln x + const. At x = a, y = 0, so (x/a)1/k = y' + √(1 + y'2). Hence 2y' = (x/a)1/k - (x/a)-1/k. Integrating again, 2y = a/(1+1/k) (x/a)1+1/k - a/(1-1/k) (x/a)1-1/k - a/(1+1/k) + a/(1-1/k), since y = 0 at x = a. When pursuit ends, x = 0 and hence y = ak/(k2 - 1).

For B, k = 2, so y = 2a/3 = 200/3 ft. That is the distance travelled by T. B is travelling at twice the speed, so B travels a distance 400/3 ft. For A reaches T at the same y, hence 200/3 = 50 k/(k2 - 1), so 4k2 - 3k - 4 = 0, so k = (3 + √73)/8.

This looks straightforward, but it is easy to get bogged down. Switching x and y axes, for example, makes it significantly harder.