You may discuss the problems here. Note, C = 3 x 10^8, It's sort of hard to view it from the picture given.
Problem #1 is solved by Aristocrat
Problem #3 is solved by Vrael

Post has been edited 6 time(s), last time on Jun 16 2010, 5:47 am by BeDazed.

Man these problems are hideous and barely even understandable.
To be more exact, I don't get the easy one either. Are you -sure- the difficulty is correct?

I'm 100% positive the difficulty is correct. These should be a piece of cake for most SEN'ers who are too awesome at Math. I had an urge to rate them all very easy, however, that would have been too subjective. The first question refers to taking a cylinder with infinite heights, but with perceivable bottoms. And since you cut it 150 from the bottom, 60 degrees, it shouldn't be hard to draw what the shape looks like. From there, you find volume, surface area, and the circumference of the ellipse that would be created from the cut.

I liked "All their rights belong to me". Light a cross of "All your base are belong to me" and "All your slaves are belong to me". Problems have rights?

Volume of an ellipse? wut?

Problem 3 is trivial: teleport a nuke to the middle of the sphere. Its acceleration is insufficient to allow it to escape the blast radius of 750m.

EDIT> Ah, wait, missed the "as of 2.5 seconds ago" part.

I assume deploying the nukes in the vertices of an octahedral formation and the railguns in a 3-deep nest of hexahedrons in a triangular lattice (with distance between lattice points 150sqrt(3) meters) would suffice?

Wut?

∑ k³ {k=1 to n}

Assuming that in this series 5 is a good approximation of n.

1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125

n=1 -> 1 = 1³
n=2 -> 9 = 1³ + 2³
n=3 -> 36 = 1³ + 2³ + 3³
n=4 -> 100 = 1³ + 2³ + 3³ + 4³
n=5 -> 225 = 1³ + 2³ + 3³ + 4³ + 5³

We can then create an expression that is equal to ∑ k³ {k=1 to n}.

We go step by step from n=1 to n=5:
n=1 -> n⁴
n=2 -> n⁴/2 + n/2
n=3 -> n⁴/3 + n²
n=4 -> n⁴/4 + n³/2 + n
n=5 -> n⁴/4 + n³/2 + n²/4

We can now say that:
∑ k³ {k=1 to n} ≈ n⁴/4 + n³/2 + n²/4
∑ k³ {k=1 to n} ≈ (n²/4)(n² + 2n + 1)
∑ k³ {k=1 to n} ≈ (n²/4)(n + 1)²

On Problem #1: I can't really visualize what you're trying to say. The problem probably needs more information on the cut. The cut could be 60 degrees radial, like in my attachment, it could be 60 degrees straight and it could be a number of different cuts.

On Problem #2: I've never heard of a permutation, probably because I took all my mathematics classes in Spanish and we have some other word for it. I just solved it like I do in my Signals & Systems class.

On Problem #3: Conceptually it's a simple problem, but it's quite long since you've got to take into consideration movement, speed of projectile, radii of explosion and time. With the projectiles you assume that the projectile is infinitely smaller than the target. You place the projected hit on any place & direction where the ship could be. Since the ship has a 300m and is a sphere, you use probability equations to determine the best 19 places. For a nuke you assume the projectile is 750m and the target is 300m. You assume that if the outer exterior of any sphere comes into contact with the other, then the target is automatically destroyed.
Simple problem, but really long.

Attachments:

I believe this is problem #1. Find the volume.

Also, for beer keg, a permutation is a mixing up of ordered numbers. For example,
1264735 is a permutation of 1234567.
If you have n unique objects, n! is the number of permutations you can make with those objects.

Post has been edited 1 time(s), last time on Jun 12 2010, 2:00 am by Vrael.

That makes more sense. Should have said that the cut was from the edge of the cylinder. Since he didn't specify, I assumed it was on the z axis.

Nice 60-degree angle you have there . 10 == 150 as well.

Volume would be 100pi^2 * (150 + 10sqrt(3)). Axes of ellipse are obviously 20/40 so 200pi^2 for surface area and 40pi for perimeter.

For anyone who may wonder where Aristocrat got those numbers from, he used simple geometry and the pythagorean theorem to find the sides of the triangle, and noticed that the volume of Cylinder B + (1/2)Cylinder A = the total volume.

As for the ellipse axes, one must be exactly as wide as the original cylinder, since it is cut from the cylinder, so 10 + 10 = 20. The other axis is going to lie along the 40 length edge in the profile view, hence 40.




For problem #3, should we assume that the target is at rest? If both the target and our spaceships are moving at .5c, then the target is moving 0 m/s relative to us, but it makes the nukes completely useless unless we know which way the thing is going. Or is that just part of the problem?

Post has been edited 1 time(s), last time on Jun 12 2010, 2:54 am by Vrael.

I believe so.

Oh and also, you guys are forgetting the rail guns. The rail guns kill everything in the path, which would be a line.

Post has been edited 1 time(s), last time on Jun 12 2010, 3:15 am by BeDazed.

I wasn't forgetting the railguns, I was just noting how useless the nukes will be if the target spaceship has an initial velocity which is sufficiently large enough. It drasticaly alters the weaponry usage. For example, if both ships are travelling at .5c already and we have no idea which direction the enemy ship is going, there is no point in even deploying the nukes unless you want to take a chance, since the .75 km blast radius is so small compared to 1.25light-seconds. However, if the ship has a zero initial velocity and a maximum acceleration of 75m/s^2 over 5 seconds, then the maximum distance moved at time of detonation is .9375km, and the 6 nukes will be extremely valuable, since a single nuke placed at the origin of the enemy ship would likely hit it. .9375km - .15km (due to ship radius) = .787km, only slightly larger than 1 nuke. Then a few of the railguns could simply be shot about .1km away from the outer blast circumference and 1 through it to just ensure the destruction of the ship.



The red circle is the initial nuke at the center of the spaceship 2.5 seconds ago, then space the 19 railgun shots evenly along the blue ring at 937.5m from the origin with 155.012m of arc length between each. Then fire a second nuke at positive distance greater than 37.5m along the z axis and a third at a negative distance of greater than 37.5m along the z axis, and BAM, ship is dead, nowhere to run, nowhere to hide, badda bing badda boom explode. Assuming an initial velocity of 0m/s though, if its nonzero and sufficiently large that may change things.

Post has been edited 4 time(s), last time on Jun 14 2010, 7:35 am by Vrael.

I do believe this is the problem we're talking about. I think we assume the blast will be sufficiently fast enough so that it will reach maximum blast radius within the time frame of unnoticeable difference in traveled distance for the object. So in that sense, you could just assume the conditions are that they are completely in halt, and accelerate from that moment on. Also, the 3rd problem notes that you have one shot at this, or you're dead. So you're probably going to deploy all 6 nukes unless you want to be dead. You want it so that the chance of that target surviving being close to 0 or 0 itself. So I don't think that's a viable solution.

Also, you're forgetting that you are firing the guns from one point, and this is not referenced in a 2 dimensional plane. So your guns will probably be shaped like a cone. There will be plenty of space left without using the other 5 nukes. And if you used one nukes in the center, there will be plenty of space left... somewhere.

Post has been edited 2 time(s), last time on Jun 12 2010, 4:15 am by BeDazed.

Already covered the space you're talking out, I used 3 nukes. They're spaced inside the cylinder of railgun fire. At a distance of 2.5 light seconds the cone will approximate a cylinder without worry.

The angle the railguns are fired at is
arctan ( 937.5 / (2.5 * 3 * 10^8)
Approximately 0, so the cylindrical approximation is valid, then since there is only 37.5m leeway on either side of the nuke along the z axis, I can use a nuke detonated at 937.5m on the positive z axis and another at 937.5m along the negative z axis (z axis is inside the cylinder) and the spaceship has nowhere to go.

Are you certain that your calculation is correct? Last time I checked, a circle's circumference is not radius x pi.

Disregard what I said about the ellipse's perimeter; I forgot that arc length doesn't double with a horizontal stretch, and you need an integral to find it, namely:


EDIT> Devourer PLEASE remove the 429 pixel restriction on image width D:. It screws over a lot of things...

Post has been edited 1 time(s), last time on Jun 12 2010, 2:58 pm by Aristocrat.

That wasn't what I was really talking about. But good job.

I was talking about problem 1...

Oh, then you should check Vrael's calculations. In the mean while, why don't you try and solve number two?