BMAD

nice.

Alas, the problem seemed more challenging in my head

Can BMAD Polygons () ever be regular? Thanks for the idea Bonanova

Does this hold for shapes with sides > 3? >4?

For example: BMAD Quadrilateral Perimeter = 4 units Area = 4 units^2 (assuming the above is possible) Could I find a BMAD Triangle where Perimeter = 4 units Area = 4 units^2 assuming the same size unit in both cases

wouldn't the area be infinitesimally off

no not necessarily.

Let's define BMAD polygons as polygons where the perimeter and area values are identical. Does there exist two such polygons (p1, p2) where the number of sides of p1 > the number of sides of p2 but the perimeters are identical?

....never mind

Yes. your edit is correct. The edge of the paper are the limits of the finite total area and the points split this area up into 8 distinct smaller areas. Each area is closer to its point than any other point.

Ahh. I stand corrected

You and an opponent are sharing a regular sheet of paper. You will play an area game. Your opponent goes first and marks a single point in the center of the paper. You will make a mark in a different location and then it is their turn, your turn and so on. The game continues until you each have four points on the paper. As this is the "Voronoi Game", the paper is then divided into 8 areas. An area is created by drawing perpendicular lines between the nearest points (Inspired by Phil's puzzle) until enough perpendicular lines define the area (ensuring that the border defines area that is closest to that receptive point). Is there a strategy that you can utilize to ensure you have the largest total area at the end of the game? example game:

Shouldn't your average intervals be of a size more like 16 1/3?

If you give me $10, I give you three cards (from an ordinary playing deck). I then deal me 1 card. You get to look at your hand and then must claim whether my card is lesser in value than your cards, greater in value of your cards, or between your least & greatest valued cards. If you guess correctly you will receive $20 (your 10 plus my 10), otherwise i keep your $10. Then we play again (if you like). Who is advantaged in this game?

Let's add one more caveat then.... After the first person choises, you have the option to redistribute the marbles again.

Each jar feels identical (since you and your contestant are blindfolded the look shouldn't matter) No matter how many marbles they weigh an imperceptible difference The other person only knows the amount of jars and the starting count and colors in each jar and of course knows the value of each marble. I did leave this part out, you are only blindfolded after the first person picks their jar and then the jars are shuffled again while you are blindfolded. The jars are not transparent

you are a contestant on a game show competing against another person. There are four jars each containing 20 marbles. At the start, one jar has all blue marbles, another has only green marbles, a third has white marbles, and the last has black marbles. You see which jar has which at the beginning (the other contestant does not). You may redistribute the marbles however you like except three conditions must persist: every jar must contain at least 1 marble, every marble must be in a jar, no jar may contain more than 50 marbles. Once the redistribution occurs, you and the other person will be blindfolded and jars shuffled and randomly placed on the table. They will pick one jar first then you will pick one from the remaining three. The marbles in each jar are then counted and scored according to their values: blue = 1 green = 3 white = 5 black = 10 Two questions: What would be the best configuration of marbles that would most help assist you in winning? What is the probability you would win this game?

Alas, that is part of the discovery

Somewhere there exists another world where the time-space fabric is discrete. We don't know the space fabric it might be 1D, 2D, 3D, or the space might be curved. Two lover breaks at zero point at zero time then start random walk. What is the probability they will they run into each other again?

There are N balls in the bag. Each one has a distinct color. You draw one ball randomly from the bag, record its color and put it back. Calculate the probability that after T draws you've seen M distinct colors. (T > 0, M > 0, M <= min(T, N) )

What is the expected distance of two points in a unit square? What is the expected distance of two points in a regular unit pentagon? Does the expected distance between two points in a regular unit n-gon converge as n goes to infinity?

A number is picked, and five guesses are made to find this number. The guesses are written on a paper, and the differences between the guesses and the picked number (by subtracting the smaller from the bigger) are written on another paper. Later one number is randomly erased from each paper. As a result there are 4 numbers on each paper. Guesses: 27, 32, 44, 45 Differences: 5, 7, 10, 13 If this is all the information needed, what is the number?

Darn. I couldn't get less until 17 moves.

In rethinking...if black can only move 15 times I would venture to guess that means white can move 16 meaning that the state shown in the op is the position before black makes its 16th move.