eytanz
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Are the pencils sharpened, or not (i.e., are they cylinders, or do they taper to a point)?

[Edited by eytanz on 01-23-2004 at 08:30 PM GMT: Typos galore]

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Schik
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Eesh, thanks - they're not sharpened. Brand new cylinders.

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Oneiromancer
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Dammit, them being sharpend was the crux of my entire solution. Thanks.

Game on,

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Oneiromancer
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I can get 4 easily...make the base 3, like this:

____
 __

where the top line is two pencils and the bottom one is touching the top line, so all 3 base pencils are touching one another. Then place another pencil on top of them, so it touches all 3 in the base.

I'll keep thinking if there's a way to get more, but I thought I'd throw this out first.

Game on,

[Edited by Oneiromancer on 01-23-2004 at 08:40 PM GMT]

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Schik
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That's a very nice solution of four, however you can get more than four.


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DiMono
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I can get five easily. Arrange the first four like this:

  1
  1
  1222
3334
   4
   4

and then stand one up in the middle

[Edited by DiMono on 01-23-2004 at 08:43 PM GMT]

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Schik
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That's a very nice solution of five, however you can get more than five.

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DiMono
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Alright, if you raise pencil 1 on the z axis, which will require tipping the vertical pencil and adjusting 2, 3 and 4 accordingly, and then slip another pencil on the other side (bottom) at an angle, with the tip in the jumbled mess in the middle, it should be able to touch all 5 of the other pencils without disconnecting any of them. How about that?

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Schik
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I'm having trouble visualizing your description.

If you lift #1, then it's no longer touching #4.

In fact, I think I'm being too lenient in allowing 4 cylinders to touch at one point like that. These are real world objects after all, not mathematical abstractions.

To that end, I will make an addendum to the puzzle (sorry...): only two pencils may intersect at any given point.

This puts Onei back in the lead with four pencils.

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DiMono
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Schik wrote:
I'm having trouble visualizing your description.

If you lift #1, then it's no longer touching #4.

In fact, I think I'm being too lenient in allowing 4 cylinders to touch at one point like that. These are real world objects after all, not mathematical abstractions.

To that end, I will make an addendum to the puzzle (sorry...): only two pencils may intersect at any given point.

This puts Onei back in the lead with four pencils.

In that case, you lay 1, 2, 3 and 4 on top of each other so that 1 is on 2, 2 is on 3, 3 is on 4, and 4 is on 1, and then slip the fifth one in the middle. And I'm back in the game!

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eytanz
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DiMono wrote:
In that case, you lay 1, 2, 3 and 4 on top of each other so that 1 is on 2, 2 is on 3, 3 is on 4, and 4 is on 1, and then slip the fifth one in the middle. And I'm back in the game!

I don't see how 1 and 3 (and 2 and 4) touch each other this way.

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DiMono
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eytanz wrote:

DiMono wrote:
In that case, you lay 1, 2, 3 and 4 on top of each other so that 1 is on 2, 2 is on 3, 3 is on 4, and 4 is on 1, and then slip the fifth one in the middle. And I'm back in the game!

I don't see how 1 and 3 (and 2 and 4) touch each other this way.

Well darn, you're right. How about this then: Take three of them and make a triangle. Let's call the pencils A, B, and C, and the pencil intersections opposite the sides a, b, and c, respectively. Now put a fourth pencil so that it intersects 'A' and 'a' both on one side of the triangle. Now take a fifth pencil and put it on the other side of the triangle, also touching 'A' and 'a', but now touching the fourth pencil as well. That's five.

If you then rotate pencil 'A' such that one end is on the top side of 'b' and the other is on the bottom of 'c', you should be able to place another pencil such that one end is on the bottom side of 'b' and the other is on the top of 'c', and still be touching the fourth and fifth pencils. This gives six. Alternatively, if this paragraph is crack, the previous paragraph gives five, but that's still less than you can get, apparently, and so we'll need to keep working on it.

Incidentally, I've edited this post about three hundred times as I get more ideas and realize they don't work, so I'm sorry to anyone who's been checking back and reading an old version of it. I just think it would have been poor form to post half a dozen replies to myself.

[Edited by DiMono on 01-23-2004 at 10:25 PM GMT]

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Schik
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I'm too tired to figure out if your 6 was correct. Here's an easy way to get six:

Lay two pencils with ends touching about 30 degrees apart. Slide a third in the middle of them so it's touching both. There's three all touching. Now add the exact same pattern on *top* of those three, rotated 90 degress. Bingo - 6.

But you can get more than 6.


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Scott
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I tried that with real pencils schik. I can't see how any of the top triangle can be touching all 3 on the bottom.

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File: pic1.gif (10.9 KB)
Downloaded 131 times.
License: Other
From: Unspecified

It's not a triangle - I didn't describe it very well. Instead of sticking that third pencil in sideways, put it the other way, so it looks kinda like you're holding three fingers up. Here, I'll attach a pic of 6 pencils all touching.

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zex20913
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Can we break the pencils? Or is that too easy for the answer?

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Schik
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Hmmm... I'll say no, you can't break them.

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Skylancer64
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I've seen this exact problem before (except it was with cigarettes).

And more than 6 is possible.



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Schik
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I guess it's time for a hint.

The image I posted showed a solution for 6 that you can recreate by placing them on a flat surface. The solution for 7 (the highest solution) is more 3-dimensional - it couldn't sit on a table. If nobody can come up with it by tomorrow, I'll post the solution and pass it along to the best-so-far solution, which is DiMono. Or Skylancer64 can claim it, since he's seen it and we've gone over 24 hours.

[Edited by Schik on 01-25-2004 at 06:06 AM GMT: typo]

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Skylancer64
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File: Solution.jpg (9.2 KB)
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Hmmm. I wonder if we have different solutions. Mine can be made on a simple surface, except for the pencil standing up. Anyways, here is one solution. I've tried it physically with pencils, as you see it in the picture. It works.

So there we go. 7-pencil solution.

[Edited by Skylancer64 on 01-25-2004 at 11:45 PM GMT]

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Schik
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That's it - You're up.


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Skylancer64
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Okay-dokey. Here it is.

A group of airplanes is planning to fly around the world. Their base is on a small island. Each plane flies at constant speed and will go halfway around the world on one tank of fuel. The planes may only refuel at the island, though a plane can transfer any amount of fuel to any other plane next to it in flight.

What is the least number of planes required to allow for at least one of the planes to make a flight all the way around the world, and what is the minimum time it will take to accomplish that flight with that number of planes? Say that a plane can go 10 hours on one tank (and will thus travel 1/2 way around the world on that single tank). Also, just for puzzle purposes, refueling and transfering of fuel is instant.

Edit - All the planes must land safely at the base at the end!

[Edited by Skylancer64 on 01-27-2004 at 01:34 AM GMT: Fixed mistake]

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The_Red_Hawk
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Four. They fly one quarter of the way, then one plane transfers all its fuel, bringing another plane's tank back up to full, then the third plane does it to the fourth, so there are two planes with full tanks. At the halfway point, both planes will be half out of fuel, and one plane transfers the rest of its fuel, giving the last enough to finish the rest.

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Scott
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That is in fact 3 planes.

First plane flies 1/4 of the way around. 1/2 a tank left.
Second plane flies to the same point 1/2 a tank left. Transfer fuel to first plane. First plane full tank, second plane crash (very sad).
First plane flies another 1/2 way around. No fuel left.
Third plane flew 1/4 of the way around to meet the first plane. 1/2 a tank left.
Transfer from third to first 1/2 a tank in first meaning it can finish the trip.

Skylancer64
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LOL! Oops. I forgot to mention that all planes must be intact at the end. That was funny though.

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Oneiromancer
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Scott wrote:
Third plane flew 1/4 of the way around to meet the first plane. 1/2 a tank left.

The key point being that the third plane flies 1/4 of the way around in the reverse direction in order to get to the original plane, which is 3/4 of the way around in the direction it went. Just thought it was worth clarifying in case someone complains.

Game on,

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Skylancer64
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First plane flies 1/4 of the way around. 1/2 a tank left.
Second plane flies to the same point 1/2 a tank left.

Also let me clarify that the planes cannot simply stop in mid-flight. So that would be "First and second planes fly 1/4 of the way around."

The only place a plane can stop is at the island.

[Edited by Skylancer64 on 01-27-2004 at 01:44 AM GMT]

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The_Red_Hawk
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Plane #1 is the one that will get around.

They fly one eighth around, then Plane #2 fills up #1's tanks again. So it will have half a tank left. It transfers a quarter to Plane #3, which will also have a full tank, then returns with just enough fuel to spare.

Planes #1 and #3 fly another eighth and #3 refills #1's tanks. It will have 3/4 of a tank and gives 1/4 to #1, with 1/2 left returns with just enough to spare.

Planes 2 and 3 refuel and join with 4, and do the same thing on the other side. They meet at the 3/4 point (1 and 2). 2 transfers half of its fuel to #1, and they both fly home safely.

So four planes are needed.

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Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

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Scott
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Boo I liked crashing 2 of the planes.

eytanz
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The_Red_Hawk wrote:
Plane #1 is the one that will get around.

They fly one eighth around, then Plane #2 fills up #1's tanks again. So it will have half a tank left. It transfers a quarter to Plane #3, which will also have a full tank, then returns with just enough fuel to spare.

Planes #1 and #3 fly another eighth and #3 refills #1's tanks. It will have 3/4 of a tank and gives 1/4 to #1, with 1/2 left returns with just enough to spare.

Planes 2 and 3 refuel and join with 4, and do the same thing on the other side. They meet at the 3/4 point (1 and 2). 2 transfers half of its fuel to #1, and they both fly home safely.

So four planes are needed.

This is almost right - the very last step is wrong. I'll try to show why below:

For simplicity's sake, lets measure progress in hemispheres (so 1 tank = 1 hemisphere). At least one plane needs to fly 2 hemispheres. We'll use the notation H to stand for hemispheres, and T to stand for tank.

---

1. Planes 1, 2, 3 start at the island.
2. Planes 1, 2, and 3 all fly 1/4H east.
3. Plane 3 gives 1/4T fuel to plane 1, and 1/4T to plane 2.
4. Plane 3 flies 1/4H west, reaches the island, and refuels. At the same time, Planes 1 and 2 fly another 1/4H east.
5. Plane 2 gives 1/4T fuel to plane 1.
6. Plane 2 flies 1/2H west, reaches the island, and refuels. At the same time, Plane 1 flies another 1/2H east.

(half-point recap - Plane 1 has 1/2 a tank left; it's halfway around the world (1H). Planes 2,3 and 4 are at the island, with full tanks.)

7. Planes 2,3,4 fly 1/4H west. Plane 1 flies 1/4H east.
8. Planes 4 transfers 1/4T of it's fuel to Plane 2, and 1/4T of it's fuel to plane 3.
9. Plane 4 flies 1/4H east, and refuels. Planes 1 flies 1/4H east. Planes 2 and 3 fly 1/4 H west.

Note the situation now. Plane 4 is at the island with it's tank full. Plane 1 is at 1/2H west, running on fumes. Planes 2 and 3 are both at 1/2H west, with 3/4T fuel each.

This is where Red_Hawk gets confused; he says that the planes I call 2 and 3 (his 3 and 4) repeat stages 5 and 6 here. But this is obviously wrong; what needs to happen is that planes 2 and 3 give 1/4T each to plane 1. This means that they each have 1/2T of fuel, just enough to fly the final 1/2H east.

----

Anyway, I'd say that Red_Hawk got it - I just posted a minor correction, he actually came up with the right solution.

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