AMC10 2009 10A, Problem 15

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]

Solution: 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: $\frac{4(18)(19)}{2}+4(19)+1=\fbox{761}$

Answer: 761