superprismatic

Oops, you'r right! Stupid slip up on my part. Thanks!

Matou's got it!

If I ask you to place N equally space points on a circle, you would have no particular trouble doing so. The same is not true for equally spaced points on a sphere, as is evidenced by the irregular-looking pattern of dimples on a golf ball. But what does equally spaced points on a sphere mean? I have come up with a working definition which is probably not original to me: The points are equally spaced if and only if the minimum distance from a point to any other point is maximized. That is, find the closest neighbor to every point, then find the smallest such distance, and make sure that it is as large as possible. So, for a 2-point example, the points would be antipodal -- that's as far away from each other than they can get. For three points, they will all lie on a great circle. For four points, they form the vertices of a regular tetrahedron (a tetrahedron of which all edges have the same length). Suppose one places 8 equally-spaces points (according to the definition above) on a unit sphere (a sphere of radius 1). What would be the shortest euclidean distance between any pair of points? Euclidean distance is the straight-line distance (not the distance over the surface of the sphere) between the points. How would you describe the figure whose vertices are those points?

A small mistake leads to a big error in the final answer, as you know. So far, we have essentially 5 different answers to this problem in this forum. Lots of people are making lots of different mistakes!

The length of the ladder is the hypotenuse (193 inches), but the base of the triangle is a bit smaller than 192 inches because it ends where it touches the bottom edge of the ladder, at the vertical line segment on the right side of your graphic. So the base is 192-y inches for some y.

Wow! Three essentially different calculated answers! Did anyone try to make a scale model of the problem out of paper to get a ballpark on this? I'll say one thing: The off-the-cuff guesses some of you made were all closer to the truth than my original guess. Thanks for your interest in this problem.

This problem actually came up in real life for me. This is definitely not a trick question. So, if there's something you don't understand about the problem statement, I'll be happy to clear it up. I was trying to store a ladder in a shed.....well, you'll see: I would like to store a ladder inside a shed. I want to store it flat against the wall using hooks. The shed wall is 16 feet long (192 inches). The ladder is 16 feet and 1 inch long (193 inches) and it is 18 inches wide. What is the minimum height that the shed wall must be to allow me to mount the ladder flat against this wall without poking through the floor, ceiling, or side walls? My shoot-from-the-hip initial guess was way off. So, I was surprised at the height needed after I worked it out. I would ask the solver to give 2 answers: (1) An initial guess and (2) The actual value (to the nearest inch or so). I would be particularly interested in an "AHA!" solution.

Oops! I even searched to see if I could find this puzzle here. I guess that shows how poor a searcher I really am! Sorry to have duplicated a puzzle.

Suppose we have 2N points in the plane. N of the points are Red and N of them are Blue. Furthermore, no subset of 3 of these 2N points lie along a line. Prove or disprove: It is always possible to pair up the Red and Blue points into N pairs, one Red and one Blue in each pair connected by a line segment, in such a way that no line segment crosses another line segment.

I think this is optimal for the square:

An intriguing problem, indeed. I have some thoughts which may be of some help in approaching it:

Ok, I wasn't rigorous enough in my reply, Mea Culpa!

Yes! But, in practice you don't really have to agree on an n, after all the, likelihood that n gets big enough to be a pain if you let it grow is very very small. So, just keep flipping until you either get smaller or larger than 1/pi. But, what about the second part of the question? If you keep flipping until the matter is decided, what is the expected number of flips?

I should have added that the method should work for any probability -- not just 1/pi.

AH! But to be exact, the lines would have to have no thickness. Ya can't do that!

This is not a trick question. It is indeed possible to use the results of ordinary coin flips to decide something with probability exactly 1/pi. "any required accuracy" means exactly what it says, so Fred and Mike could determine any decimal place, no matter how far out it is, if they required it.

No "approximately", we're talking "exactly" here!

Fred and Mike, two mathematicians, are in a bar. Fred suddenly feels magnanimous and offers to buy Mike a beer with probability 1/pi. As good mathematicians, they would have no trouble calculating 1/pi to any required accuracy. They agree to decide the matter using only the results of flipping a fair coin some number of times. How can they do this? And, on average, how many coin flips would be required to decide the matter? Of course, minimizing the number of average coin flips would be important to them.

tpaxatb asked why I chose the particular numbers in the puzzle: I picked 123456789 because it was easy to remember. I chose the other numbers using a little theory about linear congruential generators and a bit of fiddling in order to get a relatively small cycle with a rather large tail leading into it. By the way, none of the algorithms posted here are general enough to work in all cases; cycles on the order of trillions would be missed by all of them. There are ways to detect cycles which use very little space, are quite fast (i.e., the running time is linear in the length of the cycle), and are completely general. You can find them out there on the web.

Oops, I gave the program incorrect data.

Something's wrong with your numbers: They don't add up to 100%.