Hard | Strategy |

An aircraft hovers above sea, trying to catch a submarine moving with a constant velocity under the sea. The submarine is completely invisible, but using a human radar only once, the aircraft knows the exact location of submarine under the sea. The direction of submarine is unknown, but constant. The aircraft can move at twice the speed of submarine. As soon as the aircraft is just vertically above the submarine, Aircrafet can magnetically pick it up. How does the aircraft catch the submarine? How much time ill it take?

PS: This scene is from X-men: First Class. Good x-men are in the aircraft called blackbird, human radar is Banshee, Magnet is Magneto. Bad x-men are in submarine, with Sebastian Shaw is about to cause a war, better catch him soon!

PS: This scene is from X-men: First Class. Good x-men are in the aircraft called blackbird, human radar is Banshee, Magnet is Magneto. Bad x-men are in submarine, with Sebastian Shaw is about to cause a war, better catch him soon!

Hint

The locus of submarine from its spotted location is a circle of radius (speed of sub)*(time passed). We only need to move around this locus for some time.

Solution

Let speed of plane be 2s, submarine: s, original distance d

Locus of submarine after time t is circle of radius s*t centered at original location of submarine. Thus the plane moves 2/3 distance towards submarine, and then spirals out by increasing radius with speed 's'. The submarine is caught after one round.

For General case, with speeds p > s, see Palak's Answer:

Let Op, Os be the initial positions of the plane and the submarine, resp. For time t0 = d/(s+p), the plane will move towards Os in the OpOs direction. After that, with Os as the origin, the plane will maintain a constant velocity of s along the radial direction. Thus, at an angle theta from OpOs, tangential displacement in time dt is sqrt(p^2-s^2)*dt = r*d(theta) = s*t*d(theta) => dt/t = s/sqrt(p^2-s^2)*d(theta). Integrating t from t0 to tf, and theta from 0 to 2*pi, we get tf = t0*exp[2*pi*s/sqrt(p^2-s^2)] = [d/(s+p)]*exp[2*pi*s/sqrt(p^2-s^2)].

Note that the plane can be vertically above the submarine any time between t0 and tf, depending on the direction theta of the velocity of the submarine wrt OpOs, thus making tf the worst case time.

Locus of submarine after time t is circle of radius s*t centered at original location of submarine. Thus the plane moves 2/3 distance towards submarine, and then spirals out by increasing radius with speed 's'. The submarine is caught after one round.

For General case, with speeds p > s, see Palak's Answer:

Let Op, Os be the initial positions of the plane and the submarine, resp. For time t0 = d/(s+p), the plane will move towards Os in the OpOs direction. After that, with Os as the origin, the plane will maintain a constant velocity of s along the radial direction. Thus, at an angle theta from OpOs, tangential displacement in time dt is sqrt(p^2-s^2)*dt = r*d(theta) = s*t*d(theta) => dt/t = s/sqrt(p^2-s^2)*d(theta). Integrating t from t0 to tf, and theta from 0 to 2*pi, we get tf = t0*exp[2*pi*s/sqrt(p^2-s^2)] = [d/(s+p)]*exp[2*pi*s/sqrt(p^2-s^2)].

Note that the plane can be vertically above the submarine any time between t0 and tf, depending on the direction theta of the velocity of the submarine wrt OpOs, thus making tf the worst case time.

Source: Inspired from Rustan Leino's puzzle

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