bonanova

I don't think that works. We are saying that the green curve below, connecting two green dots, by virtue of sharing a single point (touching but not crossing) isolates the lower red dot from the upper red dot. (The green characters do not actually touch, they can't, but they represent touching lines.) aO dO|fsO a\| fs/ sa\e| / asd\|/ asd/|\ as/d|f\ s/ad|fa\ /cxv|cxz\ \cxvO|cz/ a\s/|f\/ s/\a|f/ /cx\_/ Touching and crossing cannot be distinguished, once the line has been drawn. It's exactly the same locus of points. But let's say it's all about how the lines were drawn, not how they end up. a line cannot be drawn in a manner by which an existing line is crossed by the pencil. The "touching but not crossing" line still does not isolate the red dots. The red dots can still be connected by a red line that also shares the common touching point without crossing either the left part or the right part of the green line. Thus: aO dO|fsO a\ds|fs/ sa\e| / asd\|/ asd/|\ as/d|f\ s/ad|fa\ /cxv|cxz\ \cxvO|cz/ a\s/|f\/ s/\a|f/ /cx\_/ So I still contend you can isolate a portion of the plane, for the purposes of this game, but only by using a line segment of infinite extent or by crossing an existing line.

To divde the plane requires a line segment of infinite length or a closed loop. A closed loop would require starting and ending on the same point, or a line segment that crosses itself. Both cases seem prohibited by OP. So, somewhere in one of those statements is what I am missing?

It was clear to me what you were asking. I just included, for fun, something outside the "box". Your English is fantastically good, even for a first-language writer, No need to apologize. Did I hear somewhere that Russian is your first language?

Exactly. You obviously have the general formula. Care to share it?

On top, the joined pairs are AF BD and CE The bottom pairs are AE DF and BC. There is not room for a vertical stroke between the adjacent lines B and C, it's implied. Sorry, a limitation of the typed characters.

A simple game to determine who picks up the tab involves drawing six vertical lines on a sheet of paper. We can call the lines A B C D E and F. P 11.11 / 2/3.4/5=.53+ 40 Holding the paper out of sight of Player 2, the first player joins pairs of upper ends of the lines. He then folds the paper in half, to expose only the lower ends and passes the paper to Player 2. Player 2 joins pairs of lower ends, hoping to make the six lines form a single closed loop. Here is a configuration that wins for the second player. Top ends Fold Bottom ends A +----------------------------|--------------------------------+ B | +-----------------------|----------------------------+ | C | +-|-----------------------|----------------------------+ | D | | +-----------------------|--------------------------+ | E | +-+-----------------------|--------------------------|-----+ F +----------------------------|--------------------------+ Is this a fair game? If not, which player has the advantage? As the number of lines (always an even number) increases, the chances of a closed loop decrease. How many lines must be drawn for Player 2's probability of winning to be less than 0.2?

Yes, it's a different result then. I took the "directly behind" interpretation. It also took a second to see "patter" meant "pattern" I'm thinking there is further richness (or simplicity) in the solution, my answer was in the manner of first thoughts.

Nice puzzle. I figured the number of points should be odd, to make other points become candidate pairs once a vertex was chosen. I was able to confirm 9 points were not enough. I fell asleep exploring 11. Nice solve as well!

The probability can reasonably vary by a factor of two, depending on the description of "randomness" for the line. There is a strong argument for "best" implementation of randomness. It leads to one of the answers given above. an answer not given above. The reason it's best is that it gives the same result with circles of any radius, and arbitrary center locations. It also has a physical implementation that is satisfyingly random in nature. What is it?

The problem sounds familiar, but I don't think it has been posted. Here is my approach. A very fine puzzle indeed.

Make a 4x4 grid and fill the major diagonal with L D A E. Make eight four-letter words on the grid by inserting these letters in the blank spaces: AAEEIIMMPPTT L _ _ _ _ D _ _ _ _ A _ _ _ _ E

It seems I've inspired your enigmasticity beyond my powers of discernment. Still thinking.

I saw a clock the other day with only 9's on its face. Thirty-six 9's in all, three each at the twelve hour-markers. Can you design such a clock face? That is, express each of the numbers 1-12 using three 9's Along with standard math symbols. Extra credit for three different expressions for each hour.

oops. I meant to also state that the business is 1000 miles away. How does that qualify as local? The OP might mean the store sells only to nearby customers. As opposed to a national chain or a wholesale distributor.

Assume the business is 3000 miles away. The driver will consume all the apples. Assume the business is one inch away. Almost any scheme delivers all the apples. So, don't we need to know how far away the business is? Or should we take the distance to be x<3000 and optimize for each x?

Does he offer 85% of the median value? Below is ambiguous.