## moneybird 3 years ago 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.

1. moneybird

|dw:1322004729461:dw|

2. moneybird

Determine the general equation of ink length for Figure n.

3. asnaseer

thinking... looks like at each step you add 5n and remove 2(n-1)

4. moneybird

This question is from today's Canadian Intermediate Mathematics Contest

5. Tomas.A

all sides are equal of that pentagon or not?

6. moneybird

yes it's a regular pentagon

7. Tomas.A

sorry I am not familiar with terminology in english

8. asnaseer

ok, I think it is:$\frac{5n(n+1)}{2}-n(1-n)$

9. asnaseer

sorry - I think it should be "+" after the fraction

10. asnaseer

$\frac{5n(n+1)}{2}+n(1-n)$

11. moneybird

My answer is $5+2(n-1)+\frac{3(n+2)(n-1)}{2}$

12. asnaseer

mine simplifies to:$\frac{n(3n+7)}{2}$

13. asnaseer

it is basically the sum of two series: 1) 5, 5+10, 5+10+15, ... 2) 0, 0-2, 0-2-4, ...

14. asnaseer

it matches my initial thoughts on adding 5n and removing 2(n-1) after each term. interesting problem.

15. moneybird

Why is it 5n?

16. asnaseer

because at each step you are adding a new regular pentagon where each side has length n. so 5 sides makes 5n.

17. asnaseer

and every time you add a new pentagon, you cover up 2 of the previous pentagons sides - hence -2(n-1)

18. Tomas.A

|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

19. moneybird

My approach is f(n) = f(n-1) + 3n + 2

20. Tomas.A

I would like to edit my answer but can't so to make it clearer we can see that $a_1=1+1+1=3$ $a_2=2+2+2=6$ $d=a_2-a_1=3$

21. moneybird

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)

22. moneybird

5+2(n-1) + 3/2 (n+2)(n-1)

23. asnaseer

24. Tomas.A

yeah all resulst are equivalent :D

25. moneybird

yeah so i got it correct on the contest!

26. asnaseer

we're ALL geniuses! :=)

27. Tomas.A

28. moneybird

29. moneybird

I am still in Grade 10?

30. asnaseer

31. moneybird

LOL I like that quote

32. asnaseer

thanks for posing the question @moneybird - I needed some food for my brain before going to bed :-)