bonanova

n people party at a restaurant, sitting at the vertices of an n-gon-shaped dinner table. Their orders have been mixed up -- in fact, none of them have received the correct entree. Show that the table may be rotated so that at least two people are sitting in front of the correct entree.

The Bertrand paradox asks the probability of a random chord being longer than the side of an inscribed equilateral triangle. At least three (see link) answers are possible depending on how "random" is employed in drawing the chord. Jaynes argues that one of the answers is "best." Here's a cute question, not of my making, that might also lead to a preferred answer from among the three. Or maybe not. The question first notes that the sum of the lengths of all sides and diagonals emanating from a vertex of a regular n-gon inscribed in the unit circle is 2 cot (pi/2n) and then asks us to use this fact to find the average length of a chord of the unit circle. Does the answer to this question give credence to one of the three answers (again see link) over the other two? Is it Jaynes' choice?

The aspect ratio is not given. How can we be sure of covering the table?

This is exactly the right direction, and it will get you the ratio of the answers for [1] and [2]. The answer for [1] is not quite right, and you may be confusing the additional plates needed with the total plates you would need? And, yes, it's hexagons.

Happy New Year to BrainDen, and all its amazing denizens!

I believe plasmid's line of analysis will lead to a solution. To achieve a degree of closure this year, here is the 2nd of the two proofs referenced in post 7. It is much simpler than it first appears. Working an example is the best way to understand it.

I may have posted this delightful puzzle already, but search didn't find it for me. Hopefully it is new to most of our current puzzle solvers: I've just placed 13 non-touching plates on a table in such a way that a 14th non-touching plate cannot be added. [1] If the table is so large that a larger table would permit a 14th non-touching plate, what is the smallest number of additional plates that will ensure that the table is completely covered? [2] If the table is so small that a smaller table cannot contain 13 non-touching plates, additional plates that are fewer in number than in case [1] can cover the table. What is that number? Assume: the plates are identical, thin circular disks the table is rectangular, aspect ratio unknown the center of every plate lies within the table's perimeter (it will not fall) "completely covered" means that looking from above, the table is completely obscured

I was going to guess that, next. (not)

You would have to enumerate them, individually or by classification, to do that. Individually, by inspection or argument, conditions can be shown not to exist. By classification, you could show that conditions of a particular type (appropriate to the statement) do not exist. It could be done, but it could also be difficult. You'd have to be certain that you analyzed them all.

A game of my grandson's uses a plastic icosahedron. To play, pegs numbered 1-12 are first inserted at its 12 vertices. Then twenty (removable) triangles, having 0, 1, 2 or 3 dots on each corner, are fastened to its faces. You win if the peg's number at each vertex equals the sum of the dots on the 5 triangle corners that meet there. The game maker asserts that there is a solution (appropriate placement of triangles) for any arrangement of the pegs, which is surprising. I wonder whether it is provable. The image shows a position for which the vertex 10 is solved, but not the others. The sets of numbers on the triangle corners are (symmetrical, but reading CW if you like): 000 111 222 333 001 002 003 011 022 033 012 012 012 021 021 021 123 123 132 132 Note the pegs and dots both sum to 45, the 12th triangular number, as they must.

k-man (post #2) finds the region between local extrema and states (post 7) that in that region there is no real solution. dejMar (post #6) finds the region where the discriminant is negative. Both approaches give the right answer. Post #6 is what I was looking for. But before I mark it, I'll ask for comments regarding whether post #2 an equivalent condition for real solutions.

Find the real numbers a and b for which, when (and only when) a < q < b, x3 + 1/x3 = q does not hold for any real value of x (has no real solutions in x.)

Do the extrema relate to the OP conditions?

Kudos to k-man. Permitting (real) Queen moves only, this seems optimal.

Find the real numbers a and b for which a < q < b if and only if x 3 + 1/x 3 = q is not satisfied for any real x.