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Brain Teasers

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  • #71
    Originally posted by Seapahn Ok here's a hint for problem #2 (line drawing) ... highlight to see:

    Think in terms of how many sides each region has. What happens with even and odd number of edges in terms of where your line starts and stops.
    The box in the top middle is the culprit, eh? 5 sides

    See, I was getting to that, but couldn't point my finger as to what it was exactly.

    Am I closer to the truth? *cue X Files theme*

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    • #72
      Originally posted by sSsflamesSs The box in the top middle is the culprit, eh? 5 sides
      Close but no cigar .. hint #2 is that there are actually 3 boxes with 5-sides.
      this post = teh win.

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      • #73
        Originally posted by Seapahn Close but no cigar .. hint #2 is that there are actually 3 boxes with 5-sides.
        Haha, you're right, but I was just being impatient - hence I missed them. The mistakes I always made on my math tests were stupid mistakes like that. Is that it? Any more pointers?

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        • #74
          Seap did you check my revised answers? You haven't pointed out the flaws in my reasoning.

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          • #75
            At first, I did look at the sides, and counted them. I got 16 since I counted the "shared" sides only once. I thought that I was supposed to get an odd number, so I quit with the sides after that. But then, when you take into account that the sides are indeed shared (4 + 5 + 4 +5 + 5), you get an odd number. I'm probably way off with that, huh?

            Hahaha, I crack myself up. Look at what I'm doing instead of studying for my O Chem midterm. My priorities are indeed in check.

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            • #76
              Originally posted by sSsflamesSs At first, I did look at the sides, and counted them. I got 16 since I counted the "shared" sides only once. I thought that I was supposed to get an odd number, so I quit with the sides after that. But then, when you take into account that the sides are indeed shared (4 + 5 + 4 +5 + 5), you get an odd number. I'm probably way off with that, huh?

              Ok my final hint ... "separate" those boxes in your mind. Think of them as separate entities. Now with the line, everytime it crosses and edge it enters or exits the box. Now think what happens with even and odd edges. This is my last hint ... next post on this will be the solution.
              this post = teh win.

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              • #77
                I don't really agree with the idea of the problems not being solvable. The wording of the problems indicate the methods of solution, if one can find loopholes in instructions it is possible to find a solution without braking the rules. Parellel lines DO NOT INTERSECT. If the wording was more detailed and percise then it would indeed make the problem unsolvable. Touching is not acceptable since if the line can be extended it would surely intersect at some point, where parallel lines by the rules of math can never intersect each other.

                I think my solution is valid, especially due to the fact that Seap was unable to prove it otherwise.

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                • #78
                  Eve, parallel lines don't intersect by definition. But what you are drawing, can't be parallel ... and if they are parallel, then they must lay on "top" of eachother ... in a plane, that means they intersect at an infinite number of points. The lines CANNOT touch at all.

                  And in this case, I can PROVE that problem #1 cannot be done. So try as much as you would like ... you won't be able to draw the lines without breaking the laws of Euclidean space
                  this post = teh win.

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                  • #79
                    Originally posted by Seapahn Ok my final hint ... "separate" those boxes in your mind. Think of them as separate entities. Now with the line, everytime it crosses and edge it enters or exits the box. Now think what happens with even and odd edges. This is my last hint ... next post on this will be the solution.
                    Well, as far as I can see, for each 5-sided box, you cannot intersect each side without intersecting one of the sides twice, thus breaking the rules.

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                    • #80
                      Ok here is the solution to problem #1.

                      This is not a rigorous proof but should be good enough

                      Step I. Connect points 'a' and 'b' to points 1 and 2. Note that the placement is arbitrary but however you do it, it will partition the 2D plane into an "inside" and an "outside". Also note that throughout the explanation, a,b,c and 1,2,3 are interchangable.

                      Step II. Now we have 2 possibilities for where to put 3. Either "inside" or "outside". I have drawn both cases and connected 3 to both 'a' and 'b'.

                      Step III. We now have 3 regions as I label as X,Y,Z for the "inside" case and X',Y',Z' for the "outside" case. Now I list from what regions each point 1,2, and 3 are accessible. For example, 1 in the "inside" case is accessible from X and Y but NOT from Z without crossing a line.

                      Step IV. We need to find a region in which to put 'c' such that it is accessible to all 1,2, and 3. But we see in both cases, such a region cannot be found and thus, 'c' cannot be placed.
                      Attached Files
                      this post = teh win.

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