Geometry Puzzle: Crush the Rebellion

Source: CMU Puzzle Toad

Problem:

Ten rebel encampments have sprung up on the plane of Usyan. The Martian Federation plans to send flying saucers to deal with them. They are pretty ruthless. They will simply land on top of the encampments. The encampments are small and the saucers are huge. It must be done simultaneously, or the rebels will flee. Also, the saucers must not overlap when they land. Can the Martians prevail?

Mathematically, the encampments are points in the plane and the saucers are non-overlapping disks of equal radius.

So, The problem is 
Given Radius r > 0 of circle, is it possible to arrange 10 points on the plane such that no number of non-overlapping circles of radius r would be such that all 10 points lie in a circle.

 Recent Geometry Puzzles on CSE Blog:



Solution:

Highlight the part between the * symbols for the answer.
* Yes - the Martians can always prevail, no matter how the 10 encampments are arranged. (This is the official CMU Puzzle Toad solution, heard by them via Peter Winkler; making it precise answers the request in the comments for a formal proof.)

Consider the hexagonal packing of the plane by non-overlapping disks of the given radius R: centers on a triangular lattice with nearest-neighbor distance 2R, so the disks touch but never overlap. Each disk sits inside a regular hexagon of area 2.sqrt(3).R^2, so the disks cover a fraction pi.R^2 / (2.sqrt(3).R^2) = pi/(2.sqrt(3)) = 0.9069... of the plane.

Now place this packing on the plane uniformly at random (fix one disk D of the packing, pick a uniformly random point x of D, and translate the whole packing so that the center of D lands at x). Whatever a fixed point P of the plane is, it ends up covered with probability exactly the covered fraction rho = pi/(2.sqrt(3)) - translation by a uniform point of one period cell makes coverage uniform over the cell.

Let Z_i be the indicator that encampment i is crushed and Z = Z_1 + ... + Z_10. By linearity of expectation - no independence needed, which is why the argument works for ARBITRARY point configurations - E[Z] = 10.rho = 9.069... > 9. But Z is an integer, so at least one placement of the packing has Z > 9, i.e. Z = 10: all ten encampments are covered by ten non-overlapping saucers landing simultaneously.

The same argument handles any n <= 10 encampments, since n.(1 - pi/(2.sqrt(3))) < 1 exactly for n <= 10. For n = 11 it proves nothing (E[Z] = 9.98, so all you get is a placement crushing at least 10).

(solution posted by AI) *

Comments

  1. Stephen Jones8:54 PM, November 12, 2012

    I'm not sure you've defined your problem carefully enough. As stated, there's no limit to the radius of the circles (just that the be equal) so one circle is always sufficient to cover all 10 points, assuming that the Martians know the placement of all points. Is there a limit to the number of circles?

    Likewise, you did not constrain the geometry of the plane such as max dimensions or whether circle centers must lie inside the rectangular field, etc.

    OK, so the problem is only interesting if you assume that there's a limit to the size of the circles and that size is smaller than the field. Also, the aliens must know the locations of the points. Otherwise, it becomes a probability problem.

    If all those assumptions are true, then the answer is for the colonists to place points at corners and the centers of isosceles triangles whose sides are 2R where R is the radius of the saucers. That should allow some colonies to hide between adjacent saucers.

    ReplyDelete
    Replies
    1. Pratik Poddar5:30 PM, February 04, 2013

      Thanks for your comment Stephen.

      Regarding the assumptions:
      1) Its a standard problem. I have not made it up.
      2) "Given R", means that the radius is given, and its finite.
      3) If not given, plane is assumed to be 2-dimensional, I have been told.
      4) Since I am arranging the points first, its assumed that aliens would know the locations of the points.

      Regarding the solution:
      Please provide a formal proof that your construction will work. Thanks

      Delete
  2. JDGM10:26 AM, November 28, 2012

    This comment has been removed by the author.

    ReplyDelete
  3. JDGM8:14 PM, December 05, 2012

    It looks like this.

    A formal written proof accompany the picture but it's a lot of words for something intuitively gleaned from staring at the diagram and visualising the circles.

    ReplyDelete
    Replies
    1. Pratik Poddar5:24 PM, February 04, 2013

      can you please give the outline of a formal proof. Thanks a ton

      Delete

Post a Comment

Popular posts from this blog

Polya's Urn Problem: Expected Balls in the Smaller Urn

Puzzle: There are two urns with one ball each. Each of subsequent n-2 balls is placed into one of these urns, with probability proportional to the number of balls already in that urn. What is the expected number of balls in the smaller sized urn? Source: P. Winkler's Puzzles book. (Chapter: Probability). Solution: Highlight the part between the * symbols for the answer. * This problem can be reformulated as the following problem. Suppose I have a stack of black cards and one red card. Initially I take red card in my hand. Now I add black cards randomly between any two cards (so, initially its either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent. No...
Read more

Lion and Man in a Circular Cage: Can the Lion Catch the Tamer?

Source: Asked to me by Pramod Ganapathi (PhD Student at Stony Brook University) Problem: A lion and a lion tamer are enclosed within a circular cage. If they move at the same speed but are both restricted by the cage, can the lion catch the lion tamer? (Represent the cage by a circle, and the lion and lion tamer as two point masses within it.) Solution: Highlight the part between the * symbols for the answer. * Answer: surprisingly, no - the tamer can avoid being caught forever. This is the famous Lion and Man problem, posed in the 1930s and settled by Besicovitch; the proof appears in Littlewood's "A Mathematician's Miscellany" (and opens Bollobas's "The Art of Mathematics"). What the lion CAN do (Rado's strategy): move to the center, then always stay on the radius through the tamer, using any spare speed to move outward. Since the lion runs on a smaller circle, it can match the tamer's angular position while steadily increasing its radiu...
Read more

Expected Tosses for Consecutive Heads: 2 Heads vs 3 Heads Puzzle

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially) ...
Read more