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ThunderCloud

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Posts posted by ThunderCloud

  1. 2 hours ago, CaptainEd said:

    Ok I believe Al is empty...

     

      Reveal hidden contents

     

    ok I see that Al’s box will have the property that for any finite number n, n is not in his box. Thanks, Bonanova. Infinity is even more confusing than I knew.

    It’s clear (I thought it was clear) that Bert will have every odd coin in his box.

    So I guess your claim is that the random removal will remove ALL of Charlie’s coins! I have to keep ruminating...

     

     

    Spoiler

    If you were to choose a coin that is inside Charlie's box after a pair has been added at some step n, the probability that your chosen coin will not be removed at that step is n / (n + 1), because there are n + 1 coins in the box at that moment. The probability that it will still not be removed by the next step is ( n / (n + 1) ) * ( (n + 1) / (n + 2) )… and so, the probability that it remains after k additional steps is n / (n + k + 1). Clearly, as k →∞, the probability of any chosen coin remaining in the box approaches zero.

     

    This was a fun one. ^.^

     

  2. Spoiler

    I think plasmid's second analysis and Cygnet's answer will hold if we define random triangles by this process: Consider a straight stick of finite length. Break it into three pieces, choosing the two break points at random with uniform distribution (which we can define over a finite interval). So long as no stick fragment is larger than (or equal to) the sum of the other two, form them into a triangle. The probability that a triangle thus formed is obtuse should be as given by Cygnet and plasmid.

     

  3. I think...

    Spoiler

    Let a diagonal road connect each of the four towns to the endpoints of a vertical road that runs through the center of the square. Call the length of the vertical road x. Then the total length of all of the roads, geometrically, is

     

    T = x + 2√(x2 -2x +2)

     

    Then, dT/dx = 1 + (2x -2) / √(x2 -2x +2), which has a zero at x = 1 - (1/√3). The total road length, by the above formula, is then ~2.73205.

    Special thanks to Cygnet for suggesting there might be a minimum hidden in such a configuration. ^_^

  4. Hmmm… I'd guess…

    Spoiler

    Every triangle has three sides; let's call them a, b, and c, and say that 0 < abc. In order to form a triangle, it must be the case that c < a + b, so then b2 ≤ c2 < (a + b)2 = a2 + b2 + 2ab. The triangle is obtuse if c2 > a2 + b2.

    Since the scale of the triangle doesn't matter, let's assume that b = 1. So then, c2 can fall anywhere between 1 and a2 + 2a + 1, but only yields an obtuse triangle if it is larger than a2 + 1.

    The fraction of obtuse triangles is then ( (a2 + 2a + 1) - (a2 + 1) ) / ( (a2 + 2a + 1) - 1 ) = 2a / (a2 + 2a) = 2 / (a + 2), for a triangle whose shortest side is length a (and whose next shortest side is length 1). Integrating over a from 0 to 1, this gives 2( ln(3) - ln(2)) = ~81%.

     

  5. I think…

    Spoiler

    99 can be saved. Let's assign numeric values to the colors: Red = 0, White = 1, Blue = 2. The person in the back of the line guesses the color corresponding to the sum of the 99 hat colors in front of him modulo 3. Then each of the other 99 prisoners can deduce their hat color from that total and from the other 98 hats (which they see ahead of them or listen for behind them).

     

  6. 1 hour ago, rocdocmac said:
      Reveal hidden contents

     

     

    Colored cube.xlsx

    Spoiler

    I think the answer is really 57. There are two cubes missing from the spreadsheet... it is possible for there to be exactly two sides of each color such that each side is adjacent to another side of the same color. There are two distinct ways to do this, poorly drawn here:5a363df037101_MissingCubes.thumb.jpg.f273abcd4c4b3a029e528436d5bfaeb6.jpg

     

  7. 20 hours ago, ThunderCloud said:

    I think...

      Hide contents

    1 color: 3 possibilities

    2 colors:

    a. one side a different color from the others: 6 possibilities

    b. two sides a different color: 6 choices of colors and 2 configurations adds 12 possibilities

    c. three sides a different color than the other three: 3 choices of colors, 2 configurations, 6 more possibilities

     

    3 colors:

    a. four of one color, one each of other colors: 3 ways to choose colors, 2 configurations, 6 possibilities

    b. three of one color, two of another, one of another: 6 ways to choose colors, 3 configurations, 18 possibilities

    c. two of each color: 1 color choice, 4 configurations, 4 more possibilities

     

    ...so, in total, there should be 3 + (6 + 12 + 6) + (6 + 18 + 4) = 55 possible cubes

    Oops, I missed a couple... a slight correction:

    Spoiler

    ...

    3 colors:

    c. two of each color, 6 configurations, 6 more possibilities...

     

    In total, 3 + (6 + 12 + 6) + (6 + 18 + 6) = 57 possible cubes. I think Pickett got it.

     

  8. 5 hours ago, bonanova said:

    I'll just point out the interesting fact that with a tetrahedron, saying "two sides white" exhausts all permutations, since all pairs of sides share an edge. But with the cube this is not the case. So, I'm wondering what "patterns" is intended to mean: combinations? or permutations?

    This is a good point.  I should clarify that for my answer, I assumed that if one configuration of colors can be transformed into another by rotating the cube, then they should not be counted as two distinct configurations.

  9. I think...

    Spoiler

    1 color: 3 possibilities

    2 colors:

    a. one side a different color from the others: 6 possibilities

    b. two sides a different color: 6 choices of colors and 2 configurations adds 12 possibilities

    c. three sides a different color than the other three: 3 choices of colors, 2 configurations, 6 more possibilities

     

    3 colors:

    a. four of one color, one each of other colors: 3 ways to choose colors, 2 configurations, 6 possibilities

    b. three of one color, two of another, one of another: 6 ways to choose colors, 3 configurations, 18 possibilities

    c. two of each color: 1 color choice, 4 configurations, 4 more possibilities

     

    ...so, in total, there should be 3 + (6 + 12 + 6) + (6 + 18 + 4) = 55 possible cubes

     

  10. Spoiler

    This looks like a Sierpinski triangle. It can be defined iteratively by removing the "middle triangle" from the figure, then the "middle triangles" of the remaining 3 triangles, then the middles of those...

     

    At each step, the shaded portion is 3/4 of what it was before. So the area of the remaining shaded portion after N steps is (3/4)N. If continued ad infinitum, the limit is zero. Therefore the area of the shaded portion is zero, and the white portion must then be 1.

     

  11. 25 minutes ago, plasmid said:

    Sort of combining the two solutions already given

      Hide contents

    f(x) = C 3log2x

    where C is any constant

     

    f(2x) = C 3log2(2x) = C 31+log2x = C (3 3log2x) = 3f(x)

     

    Spoiler

    I suspect the issue with this solution is that f(x) is not defined for all x, and therefore not continuous for all x.

    f(x) = 3log2(|x|)almost works, but still has a discontinuity at x = 0.

     

  12. Spoiler

    I should think the tamer would not be safe.

     

    Assuming he runs around the perimeter of the cage, he is traversing a circle with larger radius than is the lion. Since they run at the same speed, the lion will always be able to match the tamer's angular movement, and with time to spare to start moving outward from the center.

     

  13. It appears that I would need 5 weights to discriminate the number of marbles, which would put me at a $2 loss due to postage.

    I need a weight equal to 2 marbles in order to distinguish the cases of 1 marble from 2 marbles from more than two marbles.

    To distinguish 3 from 4 from more, however, I'd need a 2nd weight equal to 4 marbles.

    Then to distinguish 5 from 6 from more I can combine previous weights, but to discriminate 7 from 8 from more I need a 3rd weight equal to 8 marbles.

    The binary pattern continues: I need a 4th weight equal to 16 marbles, and a 5th equal to 32.

    I do. I can do it with 4 weights, of values 2, 6, 18, and 54.

    I can use the 2w to discriminate cases of 1 marble, 2 marbles, or more. If I then add the 2w to the same side of the scale as the box, I can use the 6w to discriminate 3 marbles from 4 from more. Removing the 2w, I can use the 6w to discriminate 5 from 6 from more. Then, adding the 2w and 6w together I can tell if there are 7 marbles, 8 or more.

    But now I need a third weight that can give me a net weight of 10 marbles, to distinguish cases of 9 from 10 from more. So I choose 18w, since I can counterbalance it with the 2w and 6w to yield a net of 10.

    By similar combinations, these three weights allow me to discriminate up to 2+6+18 = 26 marbles. If there are still more than this, I need a 4th weight that can yield an equivalent of 28 marbles. So I choose 54, since I can counterbalance it with the 2w, 6w, and 18w to yield the desired weight.

    By combining these weights I can measure precisely how many marbles are in the bag, at an $8 profit.

  14. It appears that I would need 5 weights to discriminate the number of marbles, which would put me at a $2 loss due to postage.



    I need a weight equal to 2 marbles in order to distinguish the cases of 1 marble from 2 marbles from more than two marbles.

    To distinguish 3 from 4 from more, however, I'd need a 2nd weight equal to 4 marbles.

    Then to distinguish 5 from 6 from more I can combine previous weights, but to discriminate 7 from 8 from more I need a 3rd weight equal to 8 marbles.

    The binary pattern continues: I need a 4th weight equal to 16 marbles, and a 5th equal to 32.
  15. I'm not sure I understand Picard's Theorem. e^x misses all of the negative points as well as 0; even-degree polynomials also miss infinitely many points.

    e^(e^x) simply misses all of the points below, and including, 1.

    It's a theorem in complex analysis, not in real analysis. Do you know about complex numbers?

    Ah, that was the part I missed. Thanks. :blush:

  16. If I have more than $80, I must have the larger amount and therefore do not want to trade.



    If I have more than $40 but less than $81, I know that if I have the smaller amount then my friend will know this and not want to trade. Therefore I can only lose, so I still don't want to trade.

    If I have more than $20 but less than $41, I know that if I have the smaller sum then my friend will not want to trade, by the above reasoning. So I can still only lose. I still won't trade.

    If I have more than $10 but less than $21, I know that if my friend has the higher sum he will not want to trade, by the above reasoning. Therefore I can only lose, and still do not want to trade.

    If I had exactly $10, I also can only lose because my friend will not trade with me if he has the higher value, by above reasoning.

    I'd be willing to trade if I had less than $10, even though my friend wouldn't take me up on it.

  17. At a Halloween party, three boxes of candy were set out. Each contained 100 candies, individually wrapped in plain black wrapper. One box contained only chocolate candies, another contained only licorice candies, and the remaining box contained a blend (of unknown proportion) of chocolate and licorice candies. All three boxes were labeled, however, some prankster came by and swapped the labels on two of the boxes.

    How many candies would you have to sample in order to set the labels right again?

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