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Can You Solve This

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  • R Raj Lal

    For all the Mathematicians OUT THERE


    Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?


    Don't worry, even from here I can clearly hear you say, "Huh?"
    :)


    "The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005

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    Rodney Sullivan
    wrote on last edited by
    #34

    My guess is 27. Here's why: Start with an empty list of dots. For each dot added to the list, the number of new intersection lines can be found using the following product: number of planes present before adding new dot * number of new planes created by adding the new dot Here's how it goes with 5 dots: dot number: previous planes * new planes 1: 0 * 0 = 0 2: 0 * 0 = 0 3: 0 * 1 = 0 (Our first plane) 4: 1 * 3 = 3 (Think of the 4 dots as a pyramid: we get three intersection lines along the base) 5: 4 * 6 = 24 (This is just a guess) Adding the intersection lines from dots 4 and 5 gives us 27 intersection lines in all. How does that sound? Rodney

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    • R Rui A Rebelo

      I don't know. It makes sense to me, but I suspect there might be something I didn't notice.:confused::~ Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

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      Raj Lal
      wrote on last edited by
      #35

      Check Clue for the Clueless Below ! "Not everything that counts can be counted..." -Albert Einstein

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      • R Raj Lal

        check " Rui A. Rebelo " comment above , is there a conflict "Not everything that counts can be counted..." -Albert Einstein

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        David Stone
        wrote on last edited by
        #36

        Ah, I misunderstood. I thought that only three of them could have lines going through them. If this were a question on a midterm or final, I would have thrown a fit. IMO, the question was very poorly worded and needed clarification.


        Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson

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        • R Rodney Sullivan

          My guess is 27. Here's why: Start with an empty list of dots. For each dot added to the list, the number of new intersection lines can be found using the following product: number of planes present before adding new dot * number of new planes created by adding the new dot Here's how it goes with 5 dots: dot number: previous planes * new planes 1: 0 * 0 = 0 2: 0 * 0 = 0 3: 0 * 1 = 0 (Our first plane) 4: 1 * 3 = 3 (Think of the 4 dots as a pyramid: we get three intersection lines along the base) 5: 4 * 6 = 24 (This is just a guess) Adding the intersection lines from dots 4 and 5 gives us 27 intersection lines in all. How does that sound? Rodney

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          Raj Lal
          wrote on last edited by
          #37

          Sounds good logic, you are on the right track ! but i am afraid the problem is a bit deeper but simpler :) I will post the answer in the evening "Not everything that counts can be counted..." -Albert Einstein

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          • R Raj Lal

            For all the Mathematicians OUT THERE


            Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?


            Don't worry, even from here I can clearly hear you say, "Huh?"
            :)


            "The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005

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            Rodney Sullivan
            wrote on last edited by
            #38

            Well, my previous answer of 27 might be an upper bound, because the 3 intersection lines from the first 4 points will be repeated in my contribution for the 5th point, so 24 is probably a better answer. Rodney

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            • R Rui A Rebelo

              Huh? Ok, let me try it. So if you have 5 dots to be arranged by 3 to define planes you'd have: combination of 5 by 3= (5!) / ( (5-3)! * 3!) = 10 planes Now, for 1 plane you'd have 0 lines. For each extra plane you add you would have the existing lines plus a number of lines equal to the previously existing number of planes (since the new plane will intercept the existing just once). So:

              planes lines
              1 0
              2 1
              3 3
              4 6
              5 10
              6 15
              7 21
              8 28
              9 36
              10 45

              or, as a formula:

              lines( 1 )=0 if planes=1
              lines( planes )= lines( planes-1)+ planes-1 if planes > 1

              Makes sense? :confused: Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 13:46 Monday 24th October, 2005

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              El Corazon
              wrote on last edited by
              #39

              sounds smarter than me, and follows along the number of shapes and planes I was counting as I add each additional combination of 3. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)

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              • R Raj Lal

                no more than three at a time can have a single flat surface pass through them :suss: ???? "Not everything that counts can be counted..." -Albert Einstein

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                El Corazon
                wrote on last edited by
                #40

                Quartz... wrote:

                no more than three at a time can have a single flat surface pass through them

                In other words no 4th lies in the same plane as any other 3. If you laid down four pens on a table in the shape of a W or M you have 5 points, but all 5 lie on the same plane (your table providing the plane). The object was to position those points such that no more than 3 share a plane. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)

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                • R Raj Lal

                  10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight


                  You can use the weighing machine EXACTLY ONCE


                  How will you fnd which bag has coins of 9 gms


                  You have time till the evening Ask me if any doubt about the question


                  [DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005

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                  longsnet
                  wrote on last edited by
                  #41

                  Wild guess here. Well, technically one weighing, using the scale once, would mean to have all 10 bags on the scale. So by putting one bag on the scale at a time,you would be able to see which bag does not keep the scale balance. Or if the incremental weight is not linear from one bag would also do the trick. i.e., 9 bags are 1000gms, 1 bag is 900gms. SO if I put 1 1000gms on a scale and then the next bag is the 900gms, it will make the scale off balance or the incremental weight will be 100gms less. Does this make any sense?

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                  • E El Corazon

                    The five dots are not a red herring... however my headache means I am not the best to predict the results this evening. Consider this: Arrange three pens or pencils on a table, or tooth picks if you prefer in a triangle shape:

                    /\
                    / \

                    This represents three of your points and one plane. Now hold a single pen/pencil or toothpic over the top of the triangle suspended in mid air. Imagine building up a connecting object by connecting all points to all points. The result is an object with six sides, but a very odd shape.

                    /\
                    / \

                    Each of those planes connects once at your object, with I believe 9 lines where the planes cross. However the planes are infinately long. So the question is, does any of those planes cross anywhere else other than at the shape in the center. That is where my headache cannot go beyond today. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)

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                    David Stone
                    wrote on last edited by
                    #42

                    Jeffry J. Brickley wrote:

                    The five dots are not a red herring...

                    Yeah yeah. I misunderstood the original question. Like I said to Quartz, if this question had been on a test, I would have complained loudly to the prof that his question sucked and that he needed to clarify on the board. ;P And I think he probably meant that that planes are perfectly "straight", so they probably won't intersect anywhere but in the center.


                    Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson

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                    • E El Corazon

                      sounds smarter than me, and follows along the number of shapes and planes I was counting as I add each additional combination of 3. _________________________ Asu no koto o ieba, tenjo de nezumi ga warau. Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)

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                      Rui A Rebelo
                      wrote on last edited by
                      #43

                      Yea, but it is wrong!:( Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

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                      • R Raj Lal

                        no more than three at a time can have a single flat surface pass through them :suss: ???? "Not everything that counts can be counted..." -Albert Einstein

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                        Rui A Rebelo
                        wrote on last edited by
                        #44

                        18 [edit]Ooops it's 19 (forgot the vertical line)[/edit] The explanation is left as an exercise for the reader...:laugh: Ok: imagine 2 tetrahedrons (piramids) joined by one face (the whole thing looks like a diamond). Thats 9 lines. Now overlap it with 3 vertical planes sharing only one intersecting line which crosses the top and bottom vertices of the diamond above... Oh well, you'd need a CAD system... Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 16:31 Monday 24th October, 2005

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                        • R Raj Lal

                          For all the Mathematicians OUT THERE


                          Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?


                          Don't worry, even from here I can clearly hear you say, "Huh?"
                          :)


                          "The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005

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                          TheGreatAndPowerfulOz
                          wrote on last edited by
                          #45

                          Place 3 points in a plane in a equilateral triangluar arrangement. This is triangle ABC. Now place the other two points, D and E, exactly in the middle of triangle/plane ABC and move them outward in opposing directions from that plane. This describes a non-rectangular volume with six sides. Now, no more than three of the five points in this volume occupy the same plane. Count the lines formed by each of the triangles joined by points A, B, C, D, E.

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                          • L longsnet

                            Wild guess here. Well, technically one weighing, using the scale once, would mean to have all 10 bags on the scale. So by putting one bag on the scale at a time,you would be able to see which bag does not keep the scale balance. Or if the incremental weight is not linear from one bag would also do the trick. i.e., 9 bags are 1000gms, 1 bag is 900gms. SO if I put 1 1000gms on a scale and then the next bag is the 900gms, it will make the scale off balance or the incremental weight will be 100gms less. Does this make any sense?

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                            Raj Lal
                            wrote on last edited by
                            #46

                            nah ! that means you are using the weighing machine more than once logically also and therotically also try some thing more ( deeper simple mathematics :) ) "Not everything that counts can be counted..." -Albert Einstein

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                            • T TheGreatAndPowerfulOz

                              Place 3 points in a plane in a equilateral triangluar arrangement. This is triangle ABC. Now place the other two points, D and E, exactly in the middle of triangle/plane ABC and move them outward in opposing directions from that plane. This describes a non-rectangular volume with six sides. Now, no more than three of the five points in this volume occupy the same plane. Count the lines formed by each of the triangles joined by points A, B, C, D, E.

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                              Raj Lal
                              wrote on last edited by
                              #47

                              So what is the answer ?????? "Not everything that counts can be counted..." -Albert Einstein

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                              • R Raj Lal

                                So what is the answer ?????? "Not everything that counts can be counted..." -Albert Einstein

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                                TheGreatAndPowerfulOz
                                wrote on last edited by
                                #48

                                The answer is 10 Lines AB, AD, AE, AC and BD, BE, BC and CD, CE and DE

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                                • T TheGreatAndPowerfulOz

                                  The answer is 10 Lines AB, AD, AE, AC and BD, BE, BC and CD, CE and DE

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                                  Raj Lal
                                  wrote on last edited by
                                  #49

                                  And how many planes ??? "Not everything that counts can be counted..." -Albert Einstein

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                                  • R Raj Lal

                                    Check Clue for the Clueless Below ! "Not everything that counts can be counted..." -Albert Einstein

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                                    Rui A Rebelo
                                    wrote on last edited by
                                    #50

                                    Ok. It's 19, check my answer in Clue for the Clueless.:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

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                                    • R Raj Lal

                                      And how many planes ??? "Not everything that counts can be counted..." -Albert Einstein

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                                      TheGreatAndPowerfulOz
                                      wrote on last edited by
                                      #51

                                      Thats the same as the question "How many unique combinations of 5 items in groups of three", so the number of planes: 5!/(5-3)!/3! = (5*4*3*2*1)/(2*1)/3*2*1 = 5*4/2 = 20/2 = 10 -- modified at 17:19 Monday 24th October, 2005

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                                      • T TheGreatAndPowerfulOz

                                        Thats the same as the question "How many unique combinations of 5 items in groups of three", so the number of planes: 5!/(5-3)!/3! = (5*4*3*2*1)/(2*1)/3*2*1 = 5*4/2 = 20/2 = 10 -- modified at 17:19 Monday 24th October, 2005

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                                        TheGreatAndPowerfulOz
                                        wrote on last edited by
                                        #52

                                        [Message Deleted]

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                                        • R Rui A Rebelo

                                          Ok. It's 19, check my answer in Clue for the Clueless.:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)

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                                          Raj Lal
                                          wrote on last edited by
                                          #53

                                          i think you are trying to solve the question in a very complicated way you are quite near though "Not everything that counts can be counted..." -Albert Einstein

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