gavinksong

Yes. Everything you have said is correct, although it does not matter what is printed on each side of the die. All that matters is that they are distinguishable.

Well done, DejMar! That was quite good. However, I claim that you can do better.

Bonanova! You just blew my mind. That simple thought never even occurred to me. Just because of that, I might consider it an alternative answer. However, I meant that the 2x2 tiles should be in line with the grid. I will not mark it solved until someone comes up with an elegant solution to this (clarified) problem.

Oh whoops. Sorry about that. :/

Can you place eight 2x2 tiles on a standard 8x8 checkerboard such that there is no room for a ninth? There is an elegant solution to this problem.

Two players take turns naming numbers from 1 to 9. Players may not name numbers that have already been named. If at any point, three of the numbers named by one of the players adds up to exactly 15, that player wins. If neither player has won after all 9 numbers have been named, the game is declared a tie. Who has the advantage in this game? There is a very clever way of answering this.

Credit for this puzzle goes to the author of the blog, By Way of Contradiction. Imagine that you have a collection of very weird dice. For every prime number between 1 and 1000, you have a fair die with that many sides. Your goal is to generate a uniform random integer from 1 to 1001, inclusive. What is the fewest expected number of die rolls needed to complete this task?

I finally got around to reading the old geometry solution by CaptainEd. It was quite confusing, but it seems our first two steps were exactly the same: solving for the red and blue areas. However, while I calculate the result in just one more step, the old solution appears to involve at least half a dozen more constructions. So it looks like I've found a much more trimmed-down proof. That said, BMAD, why don't you mark this thread as solved? The answer is clearly that the radius is √(π / (1 - √3 - π/3)), is it not?

I hope I've made this puzzle reasonable to solve.. Here is a hint. You will need it after you've solved the first part. Of course, Monaco will do as well. I hope that's of some consolation.

I mean, obviously the answer is yes. Like bonanova said, everything just scales. Anyways, I hope you don't mind if I try my hand at a geometry solution as well. I haven't read the previous solution, so apologies if mine is redundant.

You might need a text editor to see it.

Back in the day, huh?

Whaaaaaaat? Edit -- typo

Edit -- decided to use subscripts and the actual symbol for π (which is alt + p on macs btw)

I have stared at this dress for hours.