Difference between revisions of "2009 AMC 12A Problems/Problem 11"
(New page: == Problem == The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> ...) |
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+ | {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #11]] and [[2009 AMC 10A Problems|2009 AMC 10A #15]]}} | ||
+ | |||
== Problem == | == Problem == | ||
The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> is constructed from <math>F_{n - 1}</math> by surrounding it with a square and placing one more diamond on each side of the new square than <math>F_{n - 1}</math> had on each side of its outside square. For example, figure <math>F_3</math> has <math>13</math> diamonds. How many diamonds are there in figure <math>F_{20}</math>? | The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> is constructed from <math>F_{n - 1}</math> by surrounding it with a square and placing one more diamond on each side of the new square than <math>F_{n - 1}</math> had on each side of its outside square. For example, figure <math>F_3</math> has <math>13</math> diamonds. How many diamonds are there in figure <math>F_{20}</math>? | ||
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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== |
− | === Solution | + | 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: |
+ | |||
+ | <math>\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}</math> | ||
+ | |||
+ | == 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 == | |
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>. | ||
− | Hence < | + | Hence we get: |
+ | <cmath> | ||
+ | \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*} | ||
+ | </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> | ||
+ | <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 == | ||
{{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}} | ||
+ | {{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 , , , and shown are the first in a sequence of figures. For , is constructed from by surrounding it with a square and placing one more diamond on each side of the new square than had on each side of its outside square. For example, figure has diamonds. How many diamonds are there in figure ?
Solution 1
Split into congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to times the th triangular number (i.e. the diamonds within the triangles or between the diagonals) and (the diamonds on sides of the triangles or on the diagonals). The th triangular number is . Putting this together for this gives:
Solution 2
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure , the blue diamonds form a square, and the red diamonds form a square. Hence the total number of diamonds in is .
Solution 3
When constructing from , we add new diamonds. Let be the number of diamonds in . We now know that and .
Hence we get:
Solution 4
The sequence goes . The first finite differences go . The second finite differences go , so we see that the second finite difference is constant. Thus, can be represented as a quadratic, . However, we already know , , and . Thus, Solving this system for , , and gives , , . Finally,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.