# 2009 AMC 12A Problems/Problem 11

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 $F_{20}$? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker m=marker(scale(5)*d,Fill); path f1=(0,0); path f2=(0,0)--(-1,1)--(1,1)--(1,-1)--(-1,-1); path[] g2=(-1,1)--(-1,-1)--(0,0)^^(1,-1)--(0,0)--(1,1); path f3=f2--(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2); path[] g3=g2^^(-2,-2)--(0,-2)^^(2,-2)--(1,-1)^^(1,1)--(2,2)^^(-1,1)--(-2,2); path[] f4=f3^^(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)-- (3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3); path[] g4=g3^^(-2,-2)--(-3,-3)--(-1,-3)^^(3,-3)--(2,-2)^^(2,2)--(3,3)^^ (-2,2)--(-3,3); draw(f1,m); draw(shift(5,0)*f2,m); draw(shift(5,0)*g2); draw(shift(12,0)*f3,m); draw(shift(12,0)*g3); draw(shift(21,0)*f4,m); draw(shift(21,0)*g4); label("F_1",(0,-4)); label("F_2",(5,-4)); label("F_3",(12,-4)); label("F_4",(21,-4)); [/asy]$ $\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

## Solution 1

Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$th triangular number is $\frac{n(n+1)}{2}$. Putting this together for $F_{20}$ this gives: $\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

## Solution 2

Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern: $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle; draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); [/asy]$

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 3

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+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761} \end{align*}

## Solution 4

The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$. The first finite differences go $4, 8, 12, 16, \dots$. The second finite differences go $4, 4, 4, \dots$, so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$. However, we already know $d_1 = 1$, $d_2 = 3$, and $d_3 = 13$. Thus, $$a + b + c = d_1 = 1$$ $$4a + 2b + c = d_2 = 3$$ $$9a + 3b + c = d_3 = 13$$ Solving this system for $a$, $b$, and $c$ gives $a = 2$, $b = -2$, $c = 1$. Finally, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{(E)761}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 