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Posts posted by bonanova

  1. 10 hours ago, CynPyn said:

    Here goes...

    Rmax must be a circle of diameter one. To prove this, assume it isn’t. Then all other possible diameters of Rmax must be one or shorter than one. If this is true, then Rmax can fit inside a circle of diameter one, which has greater area than Rmax. Therefore Rmax doesn’t enclose the maximum area. By contradiction then, Rmax must be a circle of diameter one.

    Not my cleanest proof, but I’m thrilled to be the first response.

    Hi @CynPyn and welcome to the Den.

    Let's accept from this that  Rmax is the unit diameter circle.

    Now imagine a rectangle with unit diameter (diagonal has length 1.)
    That can be made to fit into
    Rmax .

    The question is does every unit-diameter region into Rmax ?


  2. The diameter of a closed, topologically bounded region of the plane is the greatest distance between two points in the region. Example: the diameter of a rectangle is the length of its diagonal. Of all the regions whose diameter equals 1, one of them, call it Rmax,  encloses the largest area.

    Can you prove, or disprove, that Rmax also encloses all other regions of diameter 1? That is, that all other regions of diameter 1 can be made to fit inside Rmax?

  3. Previously, Maiden’s boat could change its heading instantaneously. Ogre’s heading could change only by virtue of following a circular path along the shore at his current speed. His rotational speed was thus far from infinite, and perhaps that disadvantage was unfair.

    So in this final puzzle iteration we’ll limit the boat’s linear speed to be
    f times that of Ogre, as before, but now we’ll also limit the boat’s angular speed to be never greater than g times Ogre’s top angular speed.

    A moment’s thought tells us that unless
    g is greater than unity the boat’s best strategy is to run at full speed from the center to the shore, keeping its initial bearing, no matter where on the shore Ogre initially stands. That is, never to turn the boat. That sucks for Maiden (e.g., she loses if Ogre initially stands at the boat's initial heading) and it sucks as a puzzle. So we’ll say the boat can change heading faster than Ogre can. For clarity we’ll set g = 2.

    We’ll implement that limit by giving the boat’s motor three discrete settings that can be switched instantly an unlimited number of times: clockwise (CW), full speed ahead (FSA), and counterclockwise (CCW.) In the two turning modes the boat turns but maintains its position; in FSA mode it moves forward but does not turn. Boat’s path is thus a succession of arbitrarily short line segments joined at angles of Maiden's choosing, with the time cost of the angle depending on its size.

    If the boat starts in the middle of the lake, how large must
    f now be for Maiden to escape?

    Edit: Extra credit (tough):
    If Ogre's top speed is 1 lake-radius per minute, and Maiden chooses the boat's initial heading at the center, what's her shortest time safely to shore?

  4. On 8/28/2019 at 8:27 PM, mtngoat said:


      Hide contents

    Isn't it possible that a bridge could split, connecting 3 islands instead of 2? Just a thought.


    Yes it is possible. The Triborough Bridge in New York connects Manhattan, the Bronx and' Queens.

  5. On 8/28/2019 at 9:05 PM, mtngoat said:

    But unless the problem has been edited, it is stated:  "along with the fatal gun that belonged to one of them ... the murderer." But not absolutely necessary for the solve.


    Hmmm. I hadn't considered that. Is there a different solution if the murderer did not own the murder weapon?

  6. On 8/27/2019 at 1:38 AM, EventHorizon said:
      Reveal hidden contents


    Nicely done. 

    As for adding some wrinkle to the problem, how about this?

    Suppose we have to add some time for her to get out of the boat before she starts to run. We could say the ogre must have an angular separation of s radians from the boat when it lands, and then minimize f. Hmmm. I'm guessing the same path just minimizes f to a slightly larger value.

    On 8/28/2019 at 6:23 AM, The Lonewolf Brand said:

    The minimum value of f  is 1.0.

    Hi @The Lonewolf Brand and welcome to the Den.

    That would mean the maiden's boat and the ogre have the same speed. She can escape from a faster ogre.

  7. 15 hours ago, EventHorizon said:

    Another interesting addition might be, once the minimum  f  is found, to find the minimum travel distance (e.g., amount of gas) needed for some  f  a little higher than the minimum.

    Interesting idea. How would we pose that question exactly?


    Say the lake has radius of 1.
    The boat's starting point
    P0 is on a shared diameter, 1 + f  from the ogre and 1 - f  from shore.
    The shortest travel distance is the length of a straight line from
    P0 to the landing point P1.
    Among the landing points that make sense, one minimizes
    f , as asked by OP; another minimizes travel distance, namely 1 - f .
    What if, between those landing points, both
    f  and the distance | P0 - P1 | change monotonically?


  8. If you think you've heard this one before, read it carefully. It's not the standard puzzle.

    A beautiful maiden sits in a boat at the center of circular lake. On shore waits an ogre anxious to have his way with her. Being an excellent sprinter she knows she can outrun and therefore escape the lumbering ogre if only she can land her boat safely. But should the ogre reach her landing point first, alas, all will be lost.

    The boat is propelled by a motor capable of only a fraction f of the ogre's speed.

    What is the minimum value of f that will permit the Maiden to escape?




  9. @EventHorizon said,

    But once Plato says that "one die is a 4," it doesn't merely prune out the ones without 4's

    That’s the crux of it. 

    EH is correct that this reads on Monty Hall. But more directly on the long-running Teanchy-Beanchy post (one of his two kids is a girl, what’s the probability he has two girls.) First people said has to be 1/2. Then others (including me) said (all that matters is that) it can’t be BB so it’s 1/3. Both wrong. 

    so I made this one up to show that the informant’s algorithm has to be known. 

    Nicely explained. 

  10. 11 hours ago, Thalia said:


      Reveal hidden contents


    And so thought poor Aristotle until today's class.

    He knew the probability of two dice making seven, until his teacher told him the value of one of his dice. Then he reasoned it to be different. Ah, the magic of conditional probability, he thought. But then he reasoned further that it was not the knowledge of which value one of his dice had, for it did not matter whether that value was 1 2 3 4 5 or 6. It was seemingly only that it had a value. But what kind of conditional probability is that? Was he not already aware of that?

    One of his dice has a particular value. Six values engender six cases. In each particular case he reasons the probability of seven changes to a new value. Worse, there are no other cases. Therefore in every case it changes to a new value. How then could it have the value he originally imagined?

    So there's the question that lurks within the flavor text. Beneath the surface perhaps, but now fully revealed for all to ponder.

  11. Now, what was it we were talking about? Oh yes ... wherrzzz Beeel? I'll tell you. You know why I tell you? Because Bill would want me to. But this time the old man was not talking to the impossibly gorgeous Beatrix Michelle Kiddo. It was to a guy. A guy who was on a mission to kill Bill. A guy with the name of Al, Jack, Joe or Tom. Which one? Well, it's your job to determine which.

    Here's what we know from police interviews that followed the discovery of Bill's cold body on a lonely stretch of country road late last night, along with the fatal gun that belonged to one of them ... the murderer. So read on and see where the evidence leads. But fair warning, exactly half of each of the suspects' four statements are lies.

    • Al:
      I didn't do it.
      Tom did it.
      Sure I own a gun.
      Joe and I were playing poker last night when Bill was shot.

    • Jack:
      I didn't do it.
      Al did it.
      Joe and I were at the movies last night when Bill was shot.
      Bill was shot with Joe's gun.

    • Joe:
      I was asleep when Bill was shot.
      Al lied when he said that Tom killed Bill.
      Jack is the only one of us who owns a gun.
      Tom and Bill were pals.

    • Tom:
      I've never fired a gun in my life.
      I don't know who did it.
      Joe doesn't own a gun.
      I never saw Bill until they showed me the body.

    Who was the rat who done poor Bill in?

  12. Plato: Good morning, class, today's lesson is on probability.

    Aristotle: Fantastic.
    I'm headed to Vegas this weekend, and I can use some pointers.

    P: Curb your enthusiasm kid, this is serious stuff.
    Here, roll this pair of dice, but don't look at the the result. OK, good.
    Now without looking, tell me the probability that you rolled a seven.
    If you're going to play craps this is important.
    By the way, I can tell you that one of your dice is a four.

    A: Hmm...
    So I could have rolled 41 42 43 44 45 46 14 24 34 54 or 64,
    all with equal likelihood, with 34 and 43 making seven.
    That's a probability of 2/11.

    P: You're on a roll kid, now let's do it again. Great.
    Again without looking, what's the probability you rolled a seven?
    By the way, I can tell you one of your dice is a one.

    A: So I could have rolled 11 12 13 14 15 16 61 51 41 31 or 21,
    with 16 and 61 making seven. Hmm... it's the same as before - 2/11.

    P: And if I had told you one of your dice was a five?

    A: Well ... I guess it really doesn't matter what number you tell me.
    It will always come out the same. The probability will be 2/11.

    P: So what can we deduce from that?

    A: That the probability of two dice making seven is ... 2/11.
    But wait... Hey, you're not really Professor Plato, are you?

    P: No. I'm an insurance salesman.

    So ... what exactly is the probability of making seven?

  13. 1 hour ago, bonanova said:

    After each man, at random, was given one of the boxes, they were given the following test. Each in turn, and out of sight of the others, was to blindly remove two of the balls from his box, read the label on his box, and then endeavor to tell to the others the color of the third ball, which remained in the box.

    Yes. The restriction is only that they can't read each others' labels.

  14. Four identical boxes, each box containing 3 balls, each ball being either black or white, were placed before four men: Al, Bert, Cal and Don. No two boxes contained the same selection of ball colors. Each box bore a label that correctly identified the contents of one and only one of the boxes, but no individual box bore its correct label.

    After each man, at random, was given one of the boxes, they were given the following test. Each in turn, and out of sight of the others, was to blindly remove two of the balls from his box, read the label on his box, and then endeavor to tell to the others the color of the third ball, which remained in the box.

    It did not seem a difficult task, but the results were a bit surprising:

    1. Al:
      I've drawn two black balls.
      I can tell you the color of the third ball.

    2. Bert:
      I've drawn one white ball and one black ball.
      I can tell you the color of the third ball.

    3. Cal:
      I've drawn two white balls.
      I'm not able to tell you the color of the third ball.

    4. Don: (who was blind, and could not read his box's label)
      I don't need to draw.
      I can tell you the colors of the balls in my box.
      I can tell you the color of the third ball in each of the others' boxes.

    And then he proceeded to do so. How did he tell?


  15. On 6/24/2019 at 1:15 AM, Thalia said:

    A couple assumptions:

    "We didn't all break 100." includes the possibility that no one did. 

    "Bill placed between Ed and Jim" means directly between. 


      Reveal hidden contents

    Does this solution imply that


    Jim's statement 2 and Tom's statement 4 are both false?



  16. You're right. Nice solve.


    Specifying the team of the first player to count would distinguish the teams, if they had been grouped. As it is, it only identified that player's team. I had the grouping in my mind but didn't put the basis into the clues.


  17. 8 hours ago, Ben said:

    I got:

      Hide contents

    2 different solutions. 364 x 27 = 9828 or 312 x 27 = 8424. I don't think there are any others but I may have made a mistake.


    You can eliminate the second case else there would be another X in the puzzle statement. Nicely done. 

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