1.A gardener has 5 trees with the instructions that all 5 trees must be

1.A gardener has 5 trees with the instructions that all 5 trees must be planted so that they are equidistant from each other. That is, no 2 trees can be farther apart and any other 2 trees. How does the gardener solve this problem?

2.You have a grapefruit, several toothpicks and string. You first place some of the string around the middle of the grapefruit making sure it fits snugly. You cut the string to the same size as the grapefruit. You then take some toothpicks and line them up around the grapefruit in the same place the string was placed. You then take some more string and place it atop the toothpicks so that it looks something like a telephone pole with cable running across the top. You measure the string so that it fits the tops of the toothpicks and cut it. You then measure the difference between the first string you cut and the second one.

Now, imagine that you have a whole lot of string and you are able to place the string around the earth's equator. You measure carefully and cut it to the same size as the earth. You then take many more toothpicks and place them along the earth's equator. You now take a second string and place it atop the toothpicks so that it looks like miniature telephone pole lines. You measure carefully and cut the string to the correct length.

Question: Is the difference between the two strings placed around the earth larger than, smaller than, or the same size as the difference between the two strings placed around the grapefruit?

3.There are two groups of beings on a mythical planet in a far away galaxy: they are the Hobbits and the Orks. One day three Hobbits got lost while exploring the homeland of the Orks. The Hobbits could get home safely if they could cross the river that separated their two homelands. The Orks agreed to help the Hobbits cross the river, but the only boat they had could hold only two beings at a time, and the Hobbits could not let themselves ever be outnumbered by the Orks or the Orks would eat them.

Your problem is to figure out a sequence of moves that will carry all three Hobbits to the other side of the river and return all three Orks on their own side. The constraints are that only two beings can fit in the boat at the same time, and if at anytime the Orks on one shore outnumber the Hobbits, you will have to start over. An empty boat cannot travel across the river since it needs someone to row.

He doesn't. Three trees equidistant from one another must be in an equilateral triangle. A fourth can be placed equidistant from the other three in 3-dimensional space, so he could do this if he plants four trees somewhere around the earth. But there is nowhere the fifth could go.

Increase in length = 2*pi*(r+t) - 2* pi*r = 2*pi*t. The increase in length remains the same whatever the value of r

Out: (HO) Back: (O) Out: (OO) Back: (H) Out: (HH) Back: (OO) Out: (H) Back: (O)

(NB: the Orks outnumber the Hobbits occasionally at the points when people are getting in and out of boats, but I don't think that's avoidable)

I'm not a horticulturist, I just remember that my parents had an apple tree in their back garden which had three branches - each of which produced a different variety of apple.

The Hubble telescope observes 4 different galaxies that are all equidistant from each other. These 4 are also all the very same distance away from a fifth galaxy - our own Milky Way. What is the red shift (i.e., z value) of these galaxies.

Art Neuendorffer

This sequence aviods having Orks ever outnumbering Hobbits:

Start: HHHOOO Out: (OO) HHHO OO Back: (O) HHHOO O Out: (00) HHH OOO Back: (O) HHHO OO Out: (HH) HO HHOO Back: (HO) HHOO HO Out: (HH) OO HHHO Back: (O) OOO HHH

The distance from vertex to vertex of a tetrahedron is 2/3 sqrt(6) times the length from centroid to vertex. Thus, the velocity of the galaxies away from the observer must be going sqrt(10)/4 (0.791) times the speed of light to achieve the necessary length extension (2/3

Clearly, the galaxies at d lightyears would form a tetrahedron with the Milky Way at the center of mass:

( Z, Y, X) Milky Way: ( 0, 0, 0) Galaxy A: ( d, 0, 0) Galaxy B: (-d/3, 0, sqrt(8)*d/3) Galaxy C: (-d/3, . . . ) Galaxy D: (-d/3, . . . )

The Hubble redshift distance [z] is some function of the actual distance [d] such that wavelengths are lengthened (redshifted) by [1+z]

But then essentually *all characteristic lengths* (at distance) are only 1/[1+z] what they appear to be from earth. (E.g., while the Cosmic background plasma radiation is very far away it is a shell only 300,000 years in diameter.)

Therefore, at some point the tetrahedral sides connecting distant Galaxies A, B, C & D are actually smaller than the distance (d) that these galaxies are distant.

Note, however, that light geodesics connecting A, B, C & D are no longer straight lines but are closer to great circles on the 2-D spheric surface on which these galaxy points lie. (As if there was a variable index of refraction: n = [1+z])

Therefore: [1+z] ~ great circle distance from A to B divided by the spherical radius (d).

Approximate [1+z] using this assumption (or a better one).

Is the great circle light geodesic assumption a valid one for this particular [1+z]?

Art Neuendorffer

Okay, I see. The Gardener must plant a tree at the center of each of these galaxies!! Why that wasn't clear from the original puzzle I don't know.

Obviously you have no practical gardening experience.