## 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

1. moneybird Group Title

|dw:1322004729461:dw|

2. moneybird Group Title

Determine the general equation of ink length for Figure n.

3. asnaseer Group Title

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

4. moneybird Group Title

This question is from today's Canadian Intermediate Mathematics Contest

5. Tomas.A Group Title

all sides are equal of that pentagon or not?

6. moneybird Group Title

yes it's a regular pentagon

7. Tomas.A Group Title

sorry I am not familiar with terminology in english

8. asnaseer Group Title

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

9. asnaseer Group Title

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

10. asnaseer Group Title

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

11. moneybird Group Title

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

12. asnaseer Group Title

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

13. asnaseer Group Title

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

14. asnaseer Group Title

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

15. moneybird Group Title

Why is it 5n?

16. asnaseer Group Title

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

17. asnaseer Group Title

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

18. Tomas.A Group Title

|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 Group Title

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

20. Tomas.A Group Title

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 Group Title

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 Group Title

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

23. asnaseer Group Title

24. Tomas.A Group Title

yeah all resulst are equivalent :D

25. moneybird Group Title

yeah so i got it correct on the contest!

26. asnaseer Group Title

we're ALL geniuses! :=)

27. Tomas.A Group Title

28. moneybird Group Title

29. moneybird Group Title

I am still in Grade 10?

30. asnaseer Group Title