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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.
 3 years ago
 3 years ago
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.
 3 years ago
 3 years ago

This Question is Closed

moneybird Group TitleBest ResponseYou've already chosen the best response.4
dw:1322004729461:dw
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
Determine the general equation of ink length for Figure n.
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
thinking... looks like at each step you add 5n and remove 2(n1)
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
This question is from today's Canadian Intermediate Mathematics Contest
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
all sides are equal of that pentagon or not?
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
yes it's a regular pentagon
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
sorry I am not familiar with terminology in english
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
ok, I think it is:\[\frac{5n(n+1)}{2}n(1n)\]
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
sorry  I think it should be "+" after the fraction
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{5n(n+1)}{2}+n(1n)\]
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
My answer is \[5+2(n1)+\frac{3(n+2)(n1)}{2}\]
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
mine simplifies to:\[\frac{n(3n+7)}{2}\]
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
it is basically the sum of two series: 1) 5, 5+10, 5+10+15, ... 2) 0, 02, 024, ...
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
it matches my initial thoughts on adding 5n and removing 2(n1) after each term. interesting problem.
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
Why is it 5n?
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
because at each step you are adding a new regular pentagon where each side has length n. so 5 sides makes 5n.
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
and every time you add a new pentagon, you cover up 2 of the previous pentagons sides  hence 2(n1)
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
dw:1322005697310:dw \[a_1=3\] \[d=3\] \[S_n=\frac{2a_1+(n1)d}{2}n=\frac{6+3n3}{2}n=\frac{3n+3n^2}{2}\] \[P=2n+S_n=3n+\frac{2n+3n^2}{2}=\frac{4n+3n+3n^2}{2}=\frac{7n+3n^2}{2}\] Let's test if n=2 and answer is 13 and it's correct
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
My approach is f(n) = f(n1) + 3n + 2
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
I would like to edit my answer but can't so to make it clearer we can see that \[a_1=1+1+1=3\] \[a_2=2+2+2=6\] \[d=a_2a_1=3\]
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
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(n1) + 3 (2+3+4+5+6...+n)
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
5+2(n1) + 3/2 (n+2)(n1)
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
@moneybird  your answer also simplifies to the same result :)
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
yeah all resulst are equivalent :D
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
yeah so i got it correct on the contest!
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
we're ALL geniuses! :=)
 3 years ago

Tomas.A Group TitleBest ResponseYou've already chosen the best response.0
what grade contest is it?
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
Grade 8,9, and 10
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
I am still in Grade 10?
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I guess even in mathematics  "all roads lead to Rome"!
 3 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.4
LOL I like that quote
 3 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
thanks for posing the question @moneybird  I needed some food for my brain before going to bed :)
 3 years ago
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