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bonanova

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Everything posted by bonanova

  1. Nice work. Bonus solution is the one I had in mind.
  2. Bumping this puzzle and offering a clue - unless someone's actively searching for the final case.
  3. In a previous puzzle @plasmid found that by successively doubling the jelly beans on a plate by transferring from one of the other of three plates, it's possible to empty one of the plates. Suppose the starting number of jelly beans distributed among three plates is a sufficiently nice multiple of 3, namely b = 3x2n. By making successive doubling moves, as in the first puzzle, is it always possible to end up with an equal number ( 2n ) of jelly beans on the three plates?
  4. Hi Cygnet, and welcome to the Den. And nice pic.
  5. What fraction of triangles in a circle are obtuse?
  6. This puzzle is an ancient one that doesn't have a definitive answer so far as I can find. Points given to ThunderCloud and plasmid for proofs of possible answers, but will leave the puzzle open for further comments. I have a criticism of this puzzle, closely related to Pickett's comments, that I haven't seen raised elsewhere: There is no such thing as a random triangle in the plane. How do you pick random points in the plane? We can impose a coordinate system that makes (0, 0) a reference point, and we can add (1, 0) to provides a scale factor and orient the axes, but that's it. What we can't do is pick three arbitrary points in the plane. The origin can be the first point, WOLOG, but the other two, if truly chosen at random, are both points at infinity. It's like asking the average value of the integers. A finite value would, by any measure be disproportionately "close" to the origin. I haven't found any reference to this objection in other discussions of this puzzle. The issue plagues any attempt at a solution. In the analyses plasmid gave us, the two divergent answers are equally correct -- or equally incorrect. They assume one of the sides of the triangle has finite length. But any random line segment in the plane must have infinite length. And if so, then the analysis compares areas that are both of infinite extent. The analysis ThunderCloud gave us includes the premise that "scale doesn't matter" so let b=1, and then let a and c be anything. Same issue: if b is constrained, we can't let a and c be infinite. Or, if b=1, then it's not random over an infinite space, where things cannot be "scaled". Other approaches that I found let the triangle, instead, be randomly chosen in a circle. This preserves angular randomness and permits comparison of side lengths, by eliminating the infinity problem. As a bonus, it gives a unique answer. Follow-on puzzle: What fraction of triangles in a circle are obtuse?
  7. In some cases we can get around this point. We could for example divide the plane into increasingly small squares. If we picked a point at random we could say it lies in all squares with equal probability and then count squares. Since a finite circle includes a finite number of squares but excludes an (uncountably) infinite number, the fraction of squares inside the circle is no longer indeterminate -- it's zero. That would allow us to reach the reasonable conclusion say that the probability of hitting a finite circle embedded within an infinite dartboard is unambiguously zero. So in some cases where we're picking points at random we can start out picking very small areas, getting an answer, then take the limit of that answer as the areas go to zero. In cases where we're dealing with two infinite areas, however, this approach does not work. (Unless perhaps if the infinities are of different cardinalities.) Using these "geometric probabilities" is something like saying that points have equal "density" everywhere. It's kind of a reasonable approach, but it's contradicted by the point that you make, namely that there is a surjection between the interior and exterior of a circle.
  8. You can fit all the answers on a sheet of paper. Nice thinking tho.
  9. Sorry, I should have pointed out this was a thought experiment and all three hands are distinguishable.
  10. Ten minutes for the bunch ... Quickie 1 Quickie 2 Quickie 3
  11. Having second thoughts about your answer, or difficulty seeing it.
  12. The number abc has a decimal representation of ax102 + bx101 + cx100. But let's change things around a bit, and drop the base ten. We could then write abc = a2 + b1 + c0 and ask what values of a b c satisfy the equation. There actually may not be a solution. But if we play with the exponents a bit, we might come up with some numbers that do work. Try these (The numbers are in order, smallest to largest.) ab = a2 + b3 cd = c2 + d3 efg = e1 + f2 + g3 hij = h1 + i2 + j3 klm = k1 + l2 + m3 nop = n1 + o2 + p3 qrst = q1 + r2 + s3 + t4 uvwx = u1 + v2 + w3 + x4 yz@$ = y1 + z2 + @3 + $4 There is a shorthand notation. These can also be written ab (2,3), cd (2,3), efg (1,3,5), ... yz@$ (1,2,3,4). So here's a bonus challenge: abcdefgcc (4, 3, 8, 5, 7, 9, 0, 8, 8)
  13. What fraction of triangles are obtuse? Provide a proof.
  14. In case it helps, here's a clarification and restatement of the question. There are moments when it is not possible to identify which hand is the minute hand. An "ambiguous moment" is when that fact makes the time uncertain. How many ambiguous moments occur in any 24-hour period? (Assume we always know the angles of the hands as accurately as needed.)
  15. I have seen neat puzzles where the answer is flat out obvious but (1) wrong or (2) has a simple proof that hides in the shadows. This puzzle, and the one about tiling a hexagon with diamonds, are type (2).
  16. Closer but not all have been found.
  17. @Molly Mae I gave your description a point and will mark it as solved in a couple days to allow some time to offer a proof.
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