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moneybird Group Title

A pattern of Figures is shown below. Figure 1 is a regular pentagon with side length 1. Figure 2 is a regular pentagon of side length 2 drawn around Figure 1 so that the two shapes share the top vertex, T, and the sides on either side of T overlap. The pattern continues so that each n>1, Figure n is a regular pentagon of side length n drawn around the previous Figure so that the two shapes share the top vertex, T, and the sides on either side of T overlap. The ink length of each Figure is the sum of the lengths of all of the line segments in the Figure.

  • 2 years ago
  • 2 years ago

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  1. moneybird Group Title
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    |dw:1322004729461:dw|

    • 2 years ago
  2. moneybird Group Title
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    Determine the general equation of ink length for Figure n.

    • 2 years ago
  3. asnaseer Group Title
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    thinking... looks like at each step you add 5n and remove 2(n-1)

    • 2 years ago
  4. moneybird Group Title
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    This question is from today's Canadian Intermediate Mathematics Contest

    • 2 years ago
  5. Tomas.A Group Title
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    all sides are equal of that pentagon or not?

    • 2 years ago
  6. moneybird Group Title
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    yes it's a regular pentagon

    • 2 years ago
  7. Tomas.A Group Title
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    sorry I am not familiar with terminology in english

    • 2 years ago
  8. asnaseer Group Title
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    ok, I think it is:\[\frac{5n(n+1)}{2}-n(1-n)\]

    • 2 years ago
  9. asnaseer Group Title
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    sorry - I think it should be "+" after the fraction

    • 2 years ago
  10. asnaseer Group Title
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    \[\frac{5n(n+1)}{2}+n(1-n)\]

    • 2 years ago
  11. moneybird Group Title
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    My answer is \[5+2(n-1)+\frac{3(n+2)(n-1)}{2}\]

    • 2 years ago
  12. asnaseer Group Title
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    mine simplifies to:\[\frac{n(3n+7)}{2}\]

    • 2 years ago
  13. asnaseer Group Title
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    it is basically the sum of two series: 1) 5, 5+10, 5+10+15, ... 2) 0, 0-2, 0-2-4, ...

    • 2 years ago
  14. asnaseer Group Title
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    it matches my initial thoughts on adding 5n and removing 2(n-1) after each term. interesting problem.

    • 2 years ago
  15. moneybird Group Title
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    Why is it 5n?

    • 2 years ago
  16. asnaseer Group Title
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    because at each step you are adding a new regular pentagon where each side has length n. so 5 sides makes 5n.

    • 2 years ago
  17. asnaseer Group Title
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    and every time you add a new pentagon, you cover up 2 of the previous pentagons sides - hence -2(n-1)

    • 2 years ago
  18. Tomas.A Group Title
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    |dw:1322005697310:dw| \[a_1=3\] \[d=3\] \[S_n=\frac{2a_1+(n-1)d}{2}n=\frac{6+3n-3}{2}n=\frac{3n+3n^2}{2}\] \[P=2n+S_n=3n+\frac{2n+3n^2}{2}=\frac{4n+3n+3n^2}{2}=\frac{7n+3n^2}{2}\] Let's test if n=2 and answer is 13 and it's correct

    • 2 years ago
  19. moneybird Group Title
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    My approach is f(n) = f(n-1) + 3n + 2

    • 2 years ago
  20. Tomas.A Group Title
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    I would like to edit my answer but can't so to make it clearer we can see that \[a_1=1+1+1=3\] \[a_2=2+2+2=6\] \[d=a_2-a_1=3\]

    • 2 years ago
  21. moneybird Group Title
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    5 + 1x2 +2x3 = f(2) = 13 f(3) = 5+ 1x2 + 2x3 + 1x2 + 3x3 = 24 f(n) = 5 + 2 + 2x3 + 2 + 3x3 + 2 + 3n = 5+ 2(n-1) + 3 (2+3+4+5+6...+n)

    • 2 years ago
  22. moneybird Group Title
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    5+2(n-1) + 3/2 (n+2)(n-1)

    • 2 years ago
  23. asnaseer Group Title
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    @moneybird - your answer also simplifies to the same result :-)

    • 2 years ago
  24. Tomas.A Group Title
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    yeah all resulst are equivalent :D

    • 2 years ago
  25. moneybird Group Title
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    yeah so i got it correct on the contest!

    • 2 years ago
  26. asnaseer Group Title
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    we're ALL geniuses! :=)

    • 2 years ago
  27. Tomas.A Group Title
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    what grade contest is it?

    • 2 years ago
  28. moneybird Group Title
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    Grade 8,9, and 10

    • 2 years ago
  29. moneybird Group Title
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    I am still in Grade 10?

    • 2 years ago
  30. asnaseer Group Title
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    I guess even in mathematics - "all roads lead to Rome"!

    • 2 years ago
  31. moneybird Group Title
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    LOL I like that quote

    • 2 years ago
  32. asnaseer Group Title
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    thanks for posing the question @moneybird - I needed some food for my brain before going to bed :-)

    • 2 years ago
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