bonanova

Is there a general way to produce n outcomes, each with 1/n probability, using two coins and finite number of flips? You are allowed to use a different pair of coins for each n. The answer is yes. The needed coins are [a] fair coin and 1/n coin. But how?

No. Since the upper ball ends up closer to the hook than when it started, the lower ball is moving downward.

This problem is adapted from a problem given in the 1976 USA Mathematics Olympiad. Prove or disprove: Any 2-colored plane contains an m x n monochromatic rectangle (4 vertices of same color) such that m = 1 or 2, and n <= 6.

A little more fun: Let the lower mass be a fraction f of the upper mass. What are the values of f for which the upper ball should be pulled distances of [a] 0.5 and 1.0 from the hook in order for the balls to collide?

If I modeled it correctly, Nice puzzle.

Can you ensure, for every coloring, that a monochromatic triangle with sides of unit length can be found? Note that a simple scaling changes the coloring.

Oh. I was fairly certain of the answer. I'll work on a proof.

I have found recent puzzles here to be as delightfully counter-intuitive as others that deal with probability and infinity. They are the puzzles where each point on a line, a figure, and here the entire plane, receives a specific color. The notion of an isolated point (described by a pair of real numbers) is beyond common experience and sometimes requires proficiency in real number theory to work with. We may wonder what an isolated point actually is, and whether it has "adjacent" points. Nevertheless intuition can sometimes lead us to say, without rigor, that something must be or can't be true. Example For every 2-coloring of the plane (arbitrarily assign to every point (x, y) a red or blue color,) will there always be a unit equilateral triangle whose vertices are monochromatic?

Is what is needed [a] a better answer, a better proof, or [c] both?

Suppose we add a constraint similar to the one I made in answering BMAD's Four colors must be used in equal measure on each side of a unit equilateral triangle. What is the greatest distance between two points of the same color that is unavoidable? Without having given much thought, I'm wondering how this answer relates to 1/2: with added constraint will it be greater?

Agree totally with your point.

@gavinksong, I get W for your example.

@ k-man: You are exactly right, and all I can say at this point is, Well at least I tried. I conveniently ignored the crucial fact that the vertices all must be reached along the closed path. If that really, really unreasonable constraint were lifted, and all the plane had to do was to start and finish at point A, then ignoring the crosswind would be permissable. However, in any event, since the OP only asked for it to be shown (or not) that a wind adds to the travel time, and does not require that we say precisely how much, the fact that I swept some of the time penalty under the rug (and a very transparent rug it turned out to be) leaves my flawed analysis no worse that an understatement of the time penalty incurred by the wind. To wit: the main point, namely the presence of a time penalty, has, however ineptly, been established.

Previous post is incorrect.

OK, here is an accurate description of the simplified analysis. Suppose a plane flies from point A to point B that is due west of point A. A cross (north or south) wind does not impede his westerly progress. If there is a constant north wind, say, the plane will be blown off course and finish its westerly travel at a point that is south of point B. But if for an equal time there is also a constant south wind, the plane will arrive at B, and its travel time will be the same as if there were no cross wind. This assumes the plane maintains a westerly bearing during its flight. In the OP, the wind is constant, and the course is closed. Choose points A and B to be extremal on the path in the direction for which the A to B and B to A travel times are equal. This can always be done. Then collapse the path to the line connecting A and B. Now the cross wind component can be ignored, and the analysis of the head/tail wind applies: the component of the wind along AB is the effective magnitude of the wind.

Inspired by the recent 14th of March observance, we have, simply QED

@harey You're right. I need to rethink the effect of flattening the polygon.

That's a solution for one case.

I'm of two minds whether an infinite number of bents will cover it. You'd have to look at the area between the 180+ angle and a straight line. Can you take that limit and still have a bent?

This is one of these counter-intuitive questions that asks us to take an average over some particular quantity, when there is a choice.