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Difference between revisions of "2009 AMC 12A Problems/Problem 11"

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</asy></center><math>\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761</math>
 
</asy></center><math>\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761</math>
  
−
== Solution ==
+
== Solution 1==
  
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===Solution 1===
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Split <math>F_n</math> into <math>4</math> congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to <math>4</math> times the <math>(n-2)</math>th triangular number (i.e. the diamonds within the triangles or between the diagonals) and <math>4(n-1)+1</math> (the diamonds on sides of the triangles or on the diagonals). The <math>n</math>th triangular number is <math>\frac{n(n+1)}{2}</math>. Putting this together for <math>F_{20}</math> this gives:
−
 
 
−
Split <math>F_n</math> into <math>4</math> congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to <math>4</math> times the <math>(n-2)</math>th triangular number (i.e. the diamonds within the triangles or between the diagonals) and <math>4(n-1)+1</math> (the diamonds on sides of the triangles or on the diagonals). The <math>n</math>th triangular number is <math>\frac{n(n+1)}{2}</math>. Putting this together for <math>F_2_0</math> this gives:
 
  
 
<math>\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}</math>
 
<math>\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}</math>
  
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=== Solution 2 ===
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== Solution 2 ==
  
 
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
 
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
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Hence the total number of diamonds in <math>F_{20}</math> is <math>20^2 + 19^2 = \boxed{761}</math>.
 
Hence the total number of diamonds in <math>F_{20}</math> is <math>20^2 + 19^2 = \boxed{761}</math>.
  
−
=== Solution 3 ===
+
== Solution 3 ==
  
 
When constructing <math>F_n</math> from <math>F_{n-1}</math>, we add <math>4(n-1)</math> new diamonds. Let <math>d_n</math> be the number of diamonds in <math>F_n</math>. We now know that <math>d_1=1</math> and <math>\forall n>1:~ d_n=d_{n-1} + 4(n-1)</math>.
 
When constructing <math>F_n</math> from <math>F_{n-1}</math>, we add <math>4(n-1)</math> new diamonds. Let <math>d_n</math> be the number of diamonds in <math>F_n</math>. We now know that <math>d_1=1</math> and <math>\forall n>1:~ d_n=d_{n-1} + 4(n-1)</math>.
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
 +
 +
== Solution 4 ==
 +
 +
The sequence <math>\{ d_n\}</math> goes <math>1, 5, 13, 25, 41,\dots</math>. The first finite differences go <math>4, 8, 12, 16, \dots</math>. The second finite differences go <math>4, 4, 4, \dots</math>, so we see that the second finite difference is constant. Thus, <math>d_n</math> can be represented as a quadratic, <math>d_n = an^2 + bn + c</math>. However, we already know <math>d_1 = 1</math>, <math>d_2 = 3</math>, and <math>d_3 = 13</math>. Thus,
 +
<cmath> a + b + c = d_1 = 1</cmath>
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<cmath> 4a + 2b + c = d_2 = 3</cmath>
 +
<cmath> 9a + 3b + c = d_3 = 13</cmath>
 +
Solving this system for <math>a</math>, <math>b</math>, and <math>c</math> gives <math>a = 2</math>, <math>b = -2</math>, <math>c = 1</math>. Finally, <math>d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{(E)761}</math>
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2009|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2009|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2009|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2009|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 15:48, 19 April 2022

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, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{(E)761}$

See Also

2009 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions
2009 AMC 10A (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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