2009 AMC 12A Problems/Problem 11

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The following problem is from both the 2009 AMC 12A #11 and 2009 AMC 10A #15, so both problems redirect to this page.

Problem

The figures $F_1$ , $F_2$ , $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ , $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$ ?

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$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

Solution

Solution 1

Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:

In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square. Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$ .

Solution 2

When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$ .

Hence we get: $\begin{align*}d_{20} & = d_{19} + 4\cdot 19 \\& = d_{18} + 4\cdot 18 + 4\cdot 19 \\& = \cdots \\& = 1 + 4(1+2...$

2009 AMC 12A (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

2009 AMC 10A (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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