Let the enemy ships' course be t degrees east of North. Let the current distance between the enemy ship and the Caine be d.

Maryk plots a course that has an intersection point that is d * cos(t) units away from the Caine and d * sin(t) units away from the enemy ship. That means that the Caine needs to be cot(t) faster than the enemy ship to reach the intersection point first for a successful interception.

Queeg plots a course that has an intersection point that is d * cot(t) units away from the Caine and d / sin(t) units away from the enemy ship. That means that the Caine needs to be cos(t) faster than the enemy ship to reach the intersection point first for a successful interception.

We assume, in both cases, that the Caine can see when the enemy ship is on a direct course to the Caine's position and can stop accordingly.

Based on the parameters for the problem, we know that 0 <= t <= 90. So, cot(t) can potentially be > 1, while cos(t) <= 1. In other words, Queeg's plan will work for a greater range of speeds for the Caine, while Maryk's plan will work ONLY under certain conditions (where the Caine is sufficiently faster than the enemy ship.)

Thus, Queeg's plan guarantees interception (if possible), though not necessarily in the least time.

Actually, the worst case is if the enemy ship is just barely traveling east. In that situation the Caine must be nearly as fast as the enemy ship for successful interception. The same condition applies to a spherical Earth (with no land masses). In that case, we can assume that once the enemy ship gets to the North Pole that it will leave on some unknown vector, which provides a further limitation for interception. (In the flat Earth scenario, the interception point might be quite a large distance away, while in the spherical Earth scenario, obviously, the interception point cannot be more than the distance from the Caine to the north pole.)

Let the current distance between the two ships be d, the enemy ship is traveling x degrees north of east, at a speed of v:

Perpendicular path:
Caine's distance: d*sin(x)
Enemy's distance: d*cos(x)
Caine's speed for intercept: v*tan(x)


North path:
Caine's distance: d*tan(x)
Enemy's distance: d*sec(x)
Caine's speed for intercept: v*sin(x)

Since sin(x) < tan(x) for all x in (0 deg , 90 deg), heading straight north would be easier. Thus, Queeg is right.