anonymous
  • anonymous
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.
Mathematics
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anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
|dw:1322004729461:dw|
anonymous
  • anonymous
Determine the general equation of ink length for Figure n.
asnaseer
  • asnaseer
thinking... looks like at each step you add 5n and remove 2(n-1)

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anonymous
  • anonymous
This question is from today's Canadian Intermediate Mathematics Contest
anonymous
  • anonymous
all sides are equal of that pentagon or not?
anonymous
  • anonymous
yes it's a regular pentagon
anonymous
  • anonymous
sorry I am not familiar with terminology in english
asnaseer
  • asnaseer
ok, I think it is:\[\frac{5n(n+1)}{2}-n(1-n)\]
asnaseer
  • asnaseer
sorry - I think it should be "+" after the fraction
asnaseer
  • asnaseer
\[\frac{5n(n+1)}{2}+n(1-n)\]
anonymous
  • anonymous
My answer is \[5+2(n-1)+\frac{3(n+2)(n-1)}{2}\]
asnaseer
  • asnaseer
mine simplifies to:\[\frac{n(3n+7)}{2}\]
asnaseer
  • asnaseer
it is basically the sum of two series: 1) 5, 5+10, 5+10+15, ... 2) 0, 0-2, 0-2-4, ...
asnaseer
  • asnaseer
it matches my initial thoughts on adding 5n and removing 2(n-1) after each term. interesting problem.
anonymous
  • anonymous
Why is it 5n?
asnaseer
  • asnaseer
because at each step you are adding a new regular pentagon where each side has length n. so 5 sides makes 5n.
asnaseer
  • asnaseer
and every time you add a new pentagon, you cover up 2 of the previous pentagons sides - hence -2(n-1)
anonymous
  • anonymous
|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
anonymous
  • anonymous
My approach is f(n) = f(n-1) + 3n + 2
anonymous
  • anonymous
I would like to edit my answer but can't so to make it clearer we can see that \[a_1=1+1+1=3\] \[a_2=2+2+2=6\] \[d=a_2-a_1=3\]
anonymous
  • anonymous
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)
anonymous
  • anonymous
5+2(n-1) + 3/2 (n+2)(n-1)
asnaseer
  • asnaseer
@moneybird - your answer also simplifies to the same result :-)
anonymous
  • anonymous
yeah all resulst are equivalent :D
anonymous
  • anonymous
yeah so i got it correct on the contest!
asnaseer
  • asnaseer
we're ALL geniuses! :=)
anonymous
  • anonymous
what grade contest is it?
anonymous
  • anonymous
Grade 8,9, and 10
anonymous
  • anonymous
I am still in Grade 10?
asnaseer
  • asnaseer
I guess even in mathematics - "all roads lead to Rome"!
anonymous
  • anonymous
LOL I like that quote
asnaseer
  • asnaseer
thanks for posing the question @moneybird - I needed some food for my brain before going to bed :-)

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