Tag Archives: Futility Closet

Counterfeit Coin in Base 3

Futility Closet presented a nifty method of solving the “counterfeit coin in 12 coins” problem in a way I had not seen before by mapping the problem into numbers in base 3. It wasn’t immediately clear to me how their solution worked, so I decided to write up my own explanation.

Futility Closet: “You have 12 coins that appear identical. Eleven have the same weight, but one is either heavier or lighter than the others. How can you identify it, and determine whether it’s heavy or light, in just three weighings in a balance scale? This is a classic puzzle, but in 1992 Washington State University mathematician Calvin T. Long found a solution ‘that appears little short of magic.’ ”

See Counterfeit Coin in Base 3.

Pool Party

Futility Closet offers another interesting puzzle:

“A billiard ball is resting on a table that measures 10 feet by 5 feet. A player hits it with no ‘English’ and it strikes four different cushions and returns to its starting point. University of Alberta mathematician Murray Klamkin asks: How far did it travel?”

After solving the problem myself, I verified that Futility Closet provides an answer, but without real justification. So I thought I would write up my solution.

Answer.

See Pool Party for a solution.

Containing an Arc

This problem from Futility Closet proved quite challenging.

“University of Illinois mathematician John Wetzel called this one of his favorite problems in geometry. Call a plane arc special if it has length 1 and lies on one side of the line through its end points. Prove that any special arc can be contained in an isosceles right triangle of hypotenuse 1.”

My attempts were futile (maybe that is where the title of the website comes from). Maybe this qualifies for another Coffin Problem. But I did have one little comment about the Futility Closet solution. See Containing an Arc.

Four of a Kind

From Futility Closet we have another intriguing problem with what turns out to be a simple and elegant solution.

“If squares are drawn on the sides of a triangle and external to it, then the areas of the triangles formed between the squares each equal the area of the triangle itself.”

I originally assumed that the center triangle was a right triangle as suggested by the picture. But then I realized there was a solution that did not depend on that. See Four of a Kind.

Polygon Areas Problem

This is another problem from Futility Closet, though Futility Closet provides a “solution” of sorts. They provide a set of steps without explaining where they came from. So I thought I would fill in the gap. The problem is to find the area of an irregular polygon, none of whose sides cross one another, if we are given the coordinates of the vertices of the polygon.

Answer.

See Polygon Areas Problem for a solution.

The Four Travelers Problem

This is another Futility Closet puzzle.

“Four straight roads cross a plain. No two are parallel, and no three meet in a point. On each road is a traveler who moves at some constant speed. If Blue and Red meet each other at their crossroad, and each of them meets Yellow and Green at their respective crossroads, will Yellow and Green necessarily meet at their own crossroad?”

Answer.

I was not able to understand the solution given at first, so I tried to solve the problem on my own. Once I did, I was able to see what the Futility Closet solution was getting at. Certainly diagrams were needed to make sense of it all, and that is what I provided. See the Four Travelers Problem.

Corner Reflectors

I came across the following entry in the Futility Closet website that cried out for justification. “An arrangement of three mutually perpendicular planes, like those in the corner of a cube, have a pleasing property: They’ll reflect a ray of light back in the direction that it came from.” So the question is, why is this reflection property true? See Corner Reflectors.

The Two Errand Boys

This is another puzzle from the Futility Closet that was originally from Henry Dudeney’s Canterbury Puzzles.

“A country baker sent off his boy with a message to the butcher in the next village, and at the same time the butcher sent his boy to the baker. One ran faster than the other, and they were seen to pass at a spot 720 yards from the baker’s shop. Each stopped ten minutes at his destination and then started on the return journey, when it was found that they passed each other at a spot 400 yards from the butcher’s. How far apart are the two tradesmen’s shops? Of course each boy went at a uniform pace throughout.”

Answer.

See the Two Errand Boys for solutions.

Keyhole Problem

This is another problem from the Futility Closet website. It turned out to be pretty simple. The idea is to show the length of BC remains the same no matter where A is chosen on its arc of C1. 

(Update 7/1/2020) There is more to this problem than I realized, thanks to a revisit prompted by a question from Deb Jyoti Mitra.  See the revised Keyhole Problem.

  1. Pages:
  3. 1
  4. 2
  5. 3
  6. 4