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  3. MPOTD - Math Problem Of The Day

MPOTD - Math Problem Of The Day

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  • M MikeBeard

    Yes. It's the (r squared) in the area formula. Therefore it would take four times as many circles because 2 squared is 4 and 1 squared is still 1. Therefore you'd need four times as many circles. That's my answer and I'm sticking with it! :-D BTW, is there a way to do super/sub scripts in the forums? Mike

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    David Stone
    wrote on last edited by
    #8

    How about the <sup> and <sub> tags? :)


    When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek

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    • M MikeBeard

      Yes. It's the (r squared) in the area formula. Therefore it would take four times as many circles because 2 squared is 4 and 1 squared is still 1. Therefore you'd need four times as many circles. That's my answer and I'm sticking with it! :-D BTW, is there a way to do super/sub scripts in the forums? Mike

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      David Crow
      wrote on last edited by
      #9

      MikeBeard wrote: BTW, is there a way to do super/sub scripts in the forums? Yes. 22 is 4 11 is still 1


      "When I was born I was so surprised that I didn't talk for a year and a half." - Gracie Allen

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      • X Xiangyang Liu

        Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? * No advanced knowledge is required to solve this problem. P.S. The answer to the last problem I posted can be found here[^].[

        My articles and software tools

        ](http://mysite.verizon.net/XiangYangL/index.htm)

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        Jorgen Sigvardsson
        wrote on last edited by
        #10

        Assuming that the circles are lieing on the triangle in a grid layout, each circle is lieing on an imaginary square. If a circle with diameter 2 fits on such a square, then you can fit 4 circles with diameter 1 on it. It's simple really - a circle with diameter 2, is as high as two circles with diameter 1, and wide as two circles with diameter 1. 2 by 2 is 4. Hence, there will be four times as many circles with diameter 1. 100/25 = 4, thus the first question can be answered: Yes. (That's the simplest explanation I could come up with without fumbling with pi and radiuses. A drawing would convince anyone I believe) -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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        • J Jorgen Sigvardsson

          Assuming that the circles are lieing on the triangle in a grid layout, each circle is lieing on an imaginary square. If a circle with diameter 2 fits on such a square, then you can fit 4 circles with diameter 1 on it. It's simple really - a circle with diameter 2, is as high as two circles with diameter 1, and wide as two circles with diameter 1. 2 by 2 is 4. Hence, there will be four times as many circles with diameter 1. 100/25 = 4, thus the first question can be answered: Yes. (That's the simplest explanation I could come up with without fumbling with pi and radiuses. A drawing would convince anyone I believe) -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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          Steve Mayfield
          wrote on last edited by
          #11

          The problem says that the triangle must be completly covered. That does not mean that the circles must fit inside the triangle - the triangle fits inside a group of 25 overlaped circles. With the pattern you describe, there would be gaps in the center of the larger circle. Steve

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          • S Steve Mayfield

            The problem says that the triangle must be completly covered. That does not mean that the circles must fit inside the triangle - the triangle fits inside a group of 25 overlaped circles. With the pattern you describe, there would be gaps in the center of the larger circle. Steve

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            Jorgen Sigvardsson
            wrote on last edited by
            #12

            Steve Mayfield wrote: With the pattern you describe, there would be gaps in the center of the larger circle. Think about it. No triangle can ever be fully covered with circles with finite sizes. There will always be gaps! Hence the original layout had gaps, and it's ok to assume that the new layout may also have gaps. I admit that the problem description is a bit vague. -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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            • X Xiangyang Liu

              Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? * No advanced knowledge is required to solve this problem. P.S. The answer to the last problem I posted can be found here[^].[

              My articles and software tools

              ](http://mysite.verizon.net/XiangYangL/index.htm)

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              Jason De Arte
              wrote on last edited by
              #13

              the size of the triangle is undefined. the layout of the circles are undefined - we do not know if they stacked, everlapping or even touching? I say that they are stacked verticaly (like coins) over a triangle smaller than "diameter 1". [ Jason De Arte | Toy Maker | 1001010.com ]

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              • J Jorgen Sigvardsson

                Steve Mayfield wrote: With the pattern you describe, there would be gaps in the center of the larger circle. Think about it. No triangle can ever be fully covered with circles with finite sizes. There will always be gaps! Hence the original layout had gaps, and it's ok to assume that the new layout may also have gaps. I admit that the problem description is a bit vague. -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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                Xiangyang Liu
                wrote on last edited by
                #14

                Jörgen Sigvardsson wrote: No triangle can ever be fully covered with circles with finite sizes. What do you mean by that? One large pizza could completely cover your mouse pad, which is of rectangular shape, and a rectangle can be seen as two triangles. Since one circle can cover two triangles, there is certainly no problem for 25 circles to cover one, right? ;P Yes, the problem does not say whether the 25 circles are overlaping or not, which means there is no restriction that they must not overlap.[

                My articles and software tools

                ](http://mysite.verizon.net/XiangYangL/index.htm)

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                • X Xiangyang Liu

                  Jörgen Sigvardsson wrote: No triangle can ever be fully covered with circles with finite sizes. What do you mean by that? One large pizza could completely cover your mouse pad, which is of rectangular shape, and a rectangle can be seen as two triangles. Since one circle can cover two triangles, there is certainly no problem for 25 circles to cover one, right? ;P Yes, the problem does not say whether the 25 circles are overlaping or not, which means there is no restriction that they must not overlap.[

                  My articles and software tools

                  ](http://mysite.verizon.net/XiangYangL/index.htm)

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                  Jorgen Sigvardsson
                  wrote on last edited by
                  #15

                  Well, I made assumptions, so bite me. ;P :-D -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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                  • J Jason De Arte

                    the size of the triangle is undefined. the layout of the circles are undefined - we do not know if they stacked, everlapping or even touching? I say that they are stacked verticaly (like coins) over a triangle smaller than "diameter 1". [ Jason De Arte | Toy Maker | 1001010.com ]

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                    Jorgen Sigvardsson
                    wrote on last edited by
                    #16

                    Xiangyang does this periodically to screw with our brains. I think he enjoys it. :-D -- Ich bin Joachim von Hassel, und ich bin Pilot der Bundeswehr. Welle: Erdball - F104-G Starfighter

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                    • X Xiangyang Liu

                      Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? * No advanced knowledge is required to solve this problem. P.S. The answer to the last problem I posted can be found here[^].[

                      My articles and software tools

                      ](http://mysite.verizon.net/XiangYangL/index.htm)

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                      Andy Brummer
                      wrote on last edited by
                      #17

                      Ah, I didn't get a chance to think about it until I got home from work. The circles can cover a similar triangle of the same proportion and 1/4 the area. Just assemble 4 of those triangles to cover the original and you have the entire original triangle covered, hence your 75 circle comment. Just a tesselation problem.:doh:


                      I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon

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                      • X Xiangyang Liu

                        Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? * No advanced knowledge is required to solve this problem. P.S. The answer to the last problem I posted can be found here[^].[

                        My articles and software tools

                        ](http://mysite.verizon.net/XiangYangL/index.htm)

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                        Ray Gregory
                        wrote on last edited by
                        #18

                        Xiangyang Liu wrote: Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? Yes. Assume the given triangle has vertices A, B, and C. Now define points D, E, and F as the midpoints between each of the vertices. These six points can be used as the vertices of 4 smaller triangles (ADF, BDE, CFE, and DEF) that are congruent to and 1/4 the area of the original triangle ABC. These 4 triangles represent a partition of the original triangle. The idea is to cover each of the smaller triangles with 25 circles of diameter 1, and this is possible because this is just applying the known covering at a 1/2 length scale. Thus, we have covered the entire triangle with 100 circles of diameter 1.

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                        • A Andy Brummer

                          Ah, I didn't get a chance to think about it until I got home from work. The circles can cover a similar triangle of the same proportion and 1/4 the area. Just assemble 4 of those triangles to cover the original and you have the entire original triangle covered, hence your 75 circle comment. Just a tesselation problem.:doh:


                          I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon

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                          John Carson
                          wrote on last edited by
                          #19

                          andy brummer wrote: Ah, I didn't get a chance to think about it until I got home from work. The circles can cover a similar triangle of the same proportion and 1/4 the area. Just assemble 4 of those triangles to cover the original and you have the entire original triangle covered Beautiful! :rose: John Carson "I believe in an America where the separation of church and state is absolute--where no Catholic prelate would tell the President (should he be Catholic) how to act, and no Protestant minister would tell his parishoners for whom to vote ... and where no man is denied public office merely because his religion differs from the President who might appoint him or the people who might elect him. - John F. Kennedy

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                          • A Andy Brummer

                            Ah, I didn't get a chance to think about it until I got home from work. The circles can cover a similar triangle of the same proportion and 1/4 the area. Just assemble 4 of those triangles to cover the original and you have the entire original triangle covered, hence your 75 circle comment. Just a tesselation problem.:doh:


                            I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon

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                            Xiangyang Liu
                            wrote on last edited by
                            #20

                            You are absolutely correct! andy brummer wrote: Just a tesselation problem. If only software development is this simple: Any big project can be divided into smaller ones of equal complexity and all developers have the same capability. All management has to do is hire more developers for bigger projects. :-D[

                            My articles and software tools

                            ](http://mysite.verizon.net/XiangYangL/index.htm)

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                            • R Ray Gregory

                              Xiangyang Liu wrote: Suppose a triangle is completely covered by 25 circles of diameter 2. Can the same triangle be completely covered by 100 circles of diameter 1? why? Yes. Assume the given triangle has vertices A, B, and C. Now define points D, E, and F as the midpoints between each of the vertices. These six points can be used as the vertices of 4 smaller triangles (ADF, BDE, CFE, and DEF) that are congruent to and 1/4 the area of the original triangle ABC. These 4 triangles represent a partition of the original triangle. The idea is to cover each of the smaller triangles with 25 circles of diameter 1, and this is possible because this is just applying the known covering at a 1/2 length scale. Thus, we have covered the entire triangle with 100 circles of diameter 1.

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                              Xiangyang Liu
                              wrote on last edited by
                              #21

                              You are right, Ray. Your proof is even mathematically complete. :)[

                              My articles and software tools

                              ](http://mysite.verizon.net/XiangYangL/index.htm)

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