Try Your Hand At This Surprisingly Difficult Matchstick Puzzle

Matchstick puzzles (aka toothpick puzzles) typically involve adding to, subtracting from, or rearranging an initial configuration of matchsticks to create another, target configuration of matchsticks. Some of these puzzles can get rather complicated. This matchstick puzzle is more straightforward than most, but still quite challenging.

Sunday Puzzle # 19: Six Matches, Four Triangles

Using six matchsticks of equal length, create four identical, equilateral triangles. There’s no need for snapping, burning, or otherwise altering the matchsticks. The solution, I assure you, is elegant, and should be obvious when you find it.

Some of you will no doubt solve this almost immediately, but I suspect many of you will wrestle with it as I did. This one took me longer to solve than I’d care to admit, but it was satisfying when I finally did. If you’re in desperate need of a clue, read the “even bigger hint” that I provided in
the first part of Sunday Puzzle #3: The Bear Hunter.

Think You Know The Solution To This Classic Riddle? Think Again.

We’ll be back next week with the solution – and a new puzzle! Got a great brainteaser, original or otherwise, that you’d like to see featured?
E-mail me with your recommendations. As always, be sure to include “Sunday Puzzle” in the subject line!

Top image via Shutterstock

SOLUTION To Sunday Puzzle #18: A Puzzle For Pirates

Last week, I asked you how ten rational pirates would divide a loot of 100 gold coins.

You’ll Have To Think Like A Pirate To Solve This Week’s Puzzle

Many of you took a handy approach to this puzzle by working backwards from a scenario involving a smaller number of pirates. How would this puzzle play out if there were only 2 pirates involved? Three pirates? Four? Working backwards, a solution emerges that allows the fiercest pirate to keep a surprisingly large portion of the loot. Many of you arrived at the answer, but the first to do so, I believe, was
artzfreak (followed just a few minutes later by whoa):

The fiercest pirate, Pirate 10, proposes that she gets 96 gold, Pirate 9 gets 0 gold, 8(1), 7(0), 6(1), 5(0), 4(1), 3(0), 2(1), 1(0). Pirates 10, 8, 6, 4, and 2 vote in favor, and the proposal passes.

With two pirates, Pirate 2 can propose that she gets 100 gold, because she will vote in favor of the proposal, it doesn’t matter what Pirate 1 votes, the vote will be 50% in favor, so it will automatically pass.

With three pirates, Pirate 3 knows that Pirate 2 will vote against, because P2 can guarantee she will get 100 gold if P3 fails. P1 knows that if P3 fails and it goes to P2, he will get no gold. Therefore, so long as P1 is offered 1gold, he will vote in favor of the solution. So P3 proposes 3(99), 2(0), 1(1) and it passes 2-1.

For N pirates, Pirate N offers 1 gold to all pirates that match her in being odd/even and 0 gold to those pirates that do not match her in being odd/even, and then takes the remainder of the gold for herself. This is because the pirates that match her know they will get no gold if Pirate N fails and Pirate N-1 takes over, so they will vote in favor of Pirate N for only 1 gold. The pirates that don’t match her know that they will get 1 gold if Pirate N-1 took over, so they will vote against the proposal (Pirate N could convince them to vote her way by offering them 2 gold, but she doesn’t need their vote and doing so would lessen how much she gets, so she won’t do it).

artzfreak’s analysis is spelled out more explicitly in the solution provided by Ian Stewart, in
the May 1999 installment of Scientific American’s “Mathematical Recreations” column. As several of you noted, this solution sort of flies off the rails the second you dismiss the stipulation that these are all perfectly logical pirates – this led to some rather interesting discussions on the hypothetical subtleties of pirate psychology, which you will find peppered throughout last week’s comments section.

For a twist on this puzzle, consider this modification, originally suggested by
Steve Omohundro: How does the distribution of coins change in a scenario involving not ten, but 500 pirates divvying up the 100 gold pieces? Once you’ve had a chance to mull it over, head here for the solution.