bonanova

To celebrate the Grand Opening of his new casino, k-man offered the first 1000 patrons the opportunity to win some pocket change by playing a game that carried the catchy phrase Flip while you're a-Head. Each patron paid $100 to flip a coin multiple times, winning $100 for every H (heads) that appeared, and stopping (without penalty) at the first appearnace of a T (tails.) What was k-man's expected cost for this gesture of good will? A coveted bonanova gold star for an answer from the The Book.

Oh, the right- or left- shoulder "rule" only guarantees that you will find the exit of a maze. The path from entrance to exit divides the maze into two disjoint sets of barriers. Your shoulder will follow the inner surface of one of sets until the exit is reached. If you continue, you will trace the outer surface of the barriers (hedges), and return (by another route) to the entrance. I'm not sure what my points was, actually, except that there is a route from start to finish that demonstratively will have (only) odd visits, and the same is true of a maze -- not necessarily because it is true of a maze. I may have overstated the importance of that similarity -- I didn't give it that much thought -- although similar observations do apply to both cases.

It probably does, but I had not thought about that point. If we create all the 5 6 7 8 sides by truncation, would that chew up the original solid? Could any one or two of 5 6 7 8 be a side created by truncation? Upon reflection the OP should also have asked the amount by the surface area was decreased. Since the area created at each vertex is also removed from its three associated faces, The answer is, trivially, 2 x (1+2+3+4) = 20.

[spoiler='Related problem - labyrinth']The classical solution for the garden variety labyrinth, that has high hedges that offers no view from above, is to keep your right shoulder in contact with the hedge. That has the effect of sending you down roughly half of the dead ends, (you do the other dead ends with your left shoulder) and returning from them, thus labeling every dead-end path as having been traversed twice, and avoiding all circular paths. Every labyrinth is isomorphic with a straight path from entrance to exit, with spurious side paths that are either circular or have dead ends. As k-man observes, these can all be deleted without severing the tourist's path to his destination. Thus, when the exit is reached, every spurious path will have been traversed twice while the direct entrance-to-exit path will have been traversed only once. This puzzle translates to having the train station at the entrance to a maze and the restaurant at its exit. The fact that a maze has dead ends that don't exist in Moscow does not change this fact. One can simply create a maze that conforms to the tourist's path, placing dead ends wherever the tourist decided to turn around. Upon return, the tourist simply follows the once-trod straight path.

What happens if you reflect the pairs?

Is the distance center to center or edge to edge? I'm assuming edge to edge, which makes a better puzzle, but not certain.

Only in case the table's dimensions are rational: length/width = p/q. Now for the head scratching.

[spoiler='Nice solution indeed']I had a more complicated approach. Then I looked at plasmid's graph and what it meant to move the squares up and down: If the table makes a full rotation, each patron gets his order exactly once: there are n matches. If any position has 0, some position must have at least 2.

Sharpen your wits and play with the big boys.

@Perhaps check it again I thought of that distinction as well, and it led me back to the OP, where I found that the word "interior" is not mentioned. It has been agreed that a line cannot always pass through the interior of polygons so described. It has been clarified that BMAD asks something different. Not all Denizens share English as their primary language, so discussions like this one are generally helpful if taken in that regard.

Perhaps begin by deciding how to control the area of (at least three of) the resulting faces when the vertices are truncated by planar cuts. It's a nice result, by itself.

Probably there are different interpretation of "crossing." In particular, when the endpoint of one line segment belongs to the other line segment. For example, does the line segment {(0,0), (1,0)} "cross" or "intersect" the y-axis? I looked at a number of sites by Googling "line segments that intersect." The consensus seems to be: yes, they do intersect. Given that interpretation my post, at least, does not provide a counterexample, and BMAD's question remains unanswered.

I think what BMAD means by intersect is overlap. So, "do not intersect" probably should be taken to mean "are mutually disjoint."

A B C D and E played each other once in a series of games of tiddly winks. An individual win loss and tie earned 1, 0 and 1/2 points, respectively. They finished in alphabetical order, all with different total scores. Afterwards the following comments were heard. B: Only I finished without a loss. E: Only I finished without a win. What were the results of the individual games?

Yes, the sequence matters. 11 is not next. Clue: the sequence never ends, suggesting that the answer, perhaps, is irrational.

The solid is illuminated at every point on its surface where a straight line can be drawn uninterrupted between it and a light source.

Great job. Three sets of letters that work. Your second set was the set I had in mind. I mentioned Scrabble just to suggest the idea of moving tiles around to make anagrams.

@harey, all good points. I totally agree. The answers given previous to your post apply only if we interpret the OP to ask How many lights are necessary to completely illuminate any/a general convex solid in N dimensions? Otherwise, we can enumerate special cases, as you have. And a complete answer would be a complete enumeration. These thoughts suggest what I think is an interesting question: for each N, find an example of a convex solid that can be illuminated with the fewest lights. So far we have, N=0 point. zero. (a point may not qualify as a solid) N=1 line segment. zero (a line or line segment may not qualify as a solid) N=2 triangle. two. (a triangle may not qualify as a solid.) N=3 two (truncated cone, cube lighted outside two opposite corners) N=4 three? N=5 four? . N=Q Q-1? We could eliminate 0, 1 and 2 if we require the object to have non-zero volume. Then we wouldn't have to debate whether dots, lines and circles are solids.

That's the idea. The sequence is important, but not until later.

Not 13. Actually, the numbers are less meaningful than the digits themselves. And how might the title be of use?