Six missiles on a crash course?

Question

Six missiles are positioned at each of the six vertices of regular hexagon ABCDEF. Each side of ABCDEF is one mile in length. The missiles take off simultaneously, each one always heading directly toward the missile in front of it -- the missile starting at point A always heads toward the missile starting at point B, etc.

The missiles clearly follow a spiral pattern. Do the missiles simultaneously meet in the center, or do they spiral away from each other? If they meet in the center, how far does each missile travel?

Aspx: There will be an infinite number of hexagons for sure, and, in fact, the missiles will circle the center of the hexagon an infinite number of times. BUT, the distance is finite!

Agreed, the answer is 2, and there seems to be a lot of ways to solve it. In my non-calculus approach, I figure the distance is s / (1 - sin(θ - 90)), where s is the length of a side and θ is the interior angle of the regular polygon, which is 120° for a hexagon. Yes, the missiles converge eventually, regardless of how many sides the polygon has. For example, a 360-sided polygon has interior angles of 179°, so the distance is s / (1 - sin(89°) = s / .00015.

Aspx makes a very enthusiastic attempt to argue the missiles can never meet. However, since the distance is finite, if the missiles are moving at a constant velocity, the time before they meet is a simple application of time/rate/distance. Now, it is true that the missiles circle the center an infinite number of times, which is impossible from a Physics standpoint. However, this is mathematics -- no one said the path of the missiles has to be physically possible. :)

Answer 1

dr/dA = -tan 30

dl = ((r dA)^2 + dr^2)^.5
dl = ((-r*dr/tan30)^2 + dr^2)^.5
dl = dr*(1+(1/tan30)^2)^.5
dl = 2 dr
integrating dr from 1 to 0 implies
L = 2 miles

Answer 2

They will never meet and here is the proof:

For simiplicity lets consider that at each instant, A moves toward B with half the distance between them and so on. Note that F will move toward the initial A because all points will move at the same time.

Try to draw to visualize the movement. Now connect the new points, you will find that you have drawn a new hexagon but with smaller size. Again, move the new A toward the new B, half the distance between them and so on. You will find that you are drawing a new hexagon each time but with smaller size. Now this will never end, you will always be drawing a new hexagon.


You can now try 1/4 distance between A and B. You will also find the same thing.

Important: I am not considering that the speed is relative to distance difference. They will always have constant speed. I am just trying to explain that for any movement, a new hexagon with similar shape but with smaller size will be formed.

So, we will always be forming a new hexagon from the previous state of the points (missiles).

The points will tend to meet at the center. I mean, this is what it would look like. But theoretically, they will never meet.

Although the limit length of the spiral formed can be calculated which is radius/cos60 which is 2 * r = 2 *1 = 2 miles, this value is never achieved (it is the limit).

A simple clear example about unreachable limit is when a ball travels towards a wall by 1/2 the instantaneous distance between it and the wall, each second. The ball will never reach the wall. (at t=1sec, Distance covered= 1/2, t=2, Dc=1/2 + (1/2)(1/2), and so on. At t= inf, the limit of distance covered is 1 (but never achieved).

Conclusion:
- At TIME=infinity, distance between A and B tends to zero (but they never meet)
- The LIMIT of the distance covered by a missile is 2 miles (but is never achieved).


A proof that 2 miles cannot be achieved:
Since, we have proved that a missile will continue to circle the center of the hexagon and gets closer to it as time passes with no end of this process, then the distance covered by the missile will remain increasing slightly as time passes, example:

At a certain time, Distance Covered(DC) is 1.999 then later we might have DC= 1.999999
it will remain increasing and might reach 1.999999999999 at a certain time.
It will remain increasing forever.

If you consider that DC will reach 2, then the missile would have stopped, which CONTRADICTS what we have proved.
2 is nothing but an approximation or the limit that the missile will remain under.

Hence, as time goes to infinity, a missile will approach distance covered: 2 miles but will NEVER EVER reach it.

Answer 3

SOLUTION SUMMARY: The missiles follow an equiangular spiral with "opening angle" (in this case a closing angle!) of 60 deg., and meet simultaneously at the centre of the original hexagon. The distance each missile travels is 2 miles, independently of the speed v. The time taken, tf, is of course finite at

tf = 2 r0/v,

but the angle θ through which each missile will have rotated is given by

θ = 3^(1/2) ln [r0 / (r0 - vt/2)],

and thus θ ---> ∞ logarithmically as t --> 2 r0 / v.

DETAILS: Because of the hexagonal symmetry and the fact that there is a constant negative component of speed with respect to the radial vector, the hexagons will shrink, producing equiangular spiral paths.

Work in polar coordinates with initial r = r0, and constant speed v angling inwards relative to the radial direction at an angle of 60 deg.

Then dr / dt = - v cos 60 = - v/2; hence r = r0 - vt/2.

Thus the final, finite time taken is

tf = 2 r0/v.

Since ds/dt = v, the total distance travelled is v tf = 2 r0, giving

distance travelled = 2 r0 = 2 miles, for r0 = 1 mile.

(As we would say in solving Tripos problems, "It is amusing to note that the distance traveled is independent of the speed or time taken.")

Meanwhile, r dθ/dt = v sin 60 = v 3^(1/2) / 2.

Since we already know r = r(t), this gives

dθ/dt = [v 3^(1/2) / 2]/[r0 - vt/2], which integrates to

θ = 3^(1/2) ln [r0 / (r0 - vt/2)].


Hence θ ---> ∞ logarithmically as t --> 2 r0 / v.

Live long and prosper.

Answer 4

Let the position of a given missile be defined in polar coordinates by the position vector (r,θ). Because of the symmetry of the problem, the positions of the missiles always form a regular (rotating) hexagon with varying “radius”.

Let v = velocity of each missile

The velocity vector always points to θ + 2π/3 (ie 2π/3 from the radial line).
Thus dr/dt = v*cos(2π/3) = -v / 2

Since dr/dt = -ve, the missiles spiral toward the center of the hexagon. Because of the constant angular displacements, they can only collide at r = 0.

r = r0 – v*t/2
where r0 = 1 (from the geometry of the hexagon)

Time taken for collision, T = r0*2/v
Distance covered by each missile = T*v = 2*r0
ANSWER = 2 miles (independent of v)

* * * *
EXTRAS

We can also find that
r*dθ/dt = v*sqrt(3) / 2
dθ/dt = v*sqrt(3) / (2*r0 – vt)
θ = θo - sqrt(3) * ln (1 – v*t / 2)
and r = exp [(θo - θ) / sqrt(3) ]
This is the equation of the spiral path for each missile.

If you utilize the relation
ds/dr = sqrt [1 + (r*dθ/dr)^2 ]
and integrate from r = 1 to 0, you will also get 2 miles.

* * * *
Now suppose v = v(t) is not constant. But each missile has the same velocity function.
Let s(t) = antiderivative of v(t), such that s(0) = 0
dr/dt = -v/2
r(t) = r0 – ½ * s(t)
r = 0 when t = T
so s(T) = 2*r0

Distance = integral of v(t)*dt from t = 0 to t = T
= s(T) – s(0)
= 2*r0
= 2 miles (same as above)
So this result is independent of the velocity function.

Answer 5

I know that if it were a square instead of a hexagon, the paths would be circular arcs that intersect at the center.
I would think that they'd have to intersect at the center no matter what n-gon you have. I have no idea as to the length (or the shape) of the path.
Maybe they fall into an attractor and follow eachother in a circular path orbiting the center? I'm gonna have to model this now.

Answer 6

They should, undoubtedly, simultaneously meet in the center. But I dont know how to compute distance of the spiral curve

Answer 7

Spoiler: Look up here for the answer.