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Difference between revisions of "2000 AMC 12 Problems/Problem 8"

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== Problem ==
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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}
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Figures <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math> consist of <math>1</math>, <math>5</math>, <math>13</math>, and <math>25</math> non-overlapping squares. If the pattern continued, how many non-overlapping squares would there be in figure <math>100</math>?
+
==Problem==
  
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<math>\text {(A)}\ 10401 \qquad \text {(B)}\ 19801 \qquad \text {(C)}\ 20201 \qquad \text {(D)}\ 39801 \qquad \text {(E)}\ 40801</math>
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Figures <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math> consist of <math>1</math>, <math>5</math>, <math>13</math>, and <math>25</math> nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
  
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[[Image:2000_12_AMC-8.png]]
+
<asy>
 +
unitsize(8);
 +
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
 +
draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
 +
draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);
 +
draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);
 +
draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
 +
draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);
 +
draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);
 +
draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
 +
draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);
 +
draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);
 +
label("Figure",(0.5,-1),S);
 +
label("$0$",(0.5,-2.5),S);
 +
label("Figure",(9.5,-1),S);
 +
label("$1$",(9.5,-2.5),S);
 +
label("Figure",(19.5,-1),S);
 +
label("$2$",(19.5,-2.5),S);
 +
label("Figure",(32.5,-1),S);
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label("$3$",(32.5,-2.5),S);
 +
</asy>
  
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== Solution 1==
 
−
By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has <math>1 + 4(1)</math> squares, figure 2 has <math>1+4(1+2)</math> squares, and so on. Figure 100 would have <math>1 + 4(1 + 2 + \cdots + 100) = 1 + 4 \frac{100(101)}{2} = 20201 \Rightarrow \mathrm{(C)}</math>.
 
  
 +
<math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math>
  
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== Solution 2==
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==Solution==
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Note that figure 0 has 1 square, figure 1 has 5 squares, figure 2 has 13 squares, and so on. If we let the number of the figure = <math>N</math>, note that <math>N^2 + (N+1)^2</math> represents the number of squares in the figure. For example, figure 4 has <math>4^2+5^2 = 41</math> squares. Therefore, the number of squares in figure 100 has <math>100^2 + 101^2 = 20201 \Rightarrow\mathrm{(C)}</math>.
 
  
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<math>2^{\text{nd}}</math> alternate solution:
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===Solution 1===
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For the <math>n^{\text{th}}</math> figure, note that it could be constructed by making a <math>(2n+1)\times (2n+1)</math> square, and then removing the <math>n^{\text{th}}</math> triangular number from each of its corners. So, if <math>a_n</math> represents the amount of squares in figure <math>n</math>, <math>a_{n} = (2n+1)^2-4\frac{(n)(n+1)}{2}= 2n^2+2n+1</math>. Therefore, <math>a_{100} = 20201</math>, which gives <math>\mathrm{C}</math>.
+
 
 +
We have a recursion:
 +
 
 +
<math>A_n=A_{n-1}+4n</math>.
 +
 
 +
I.E. we add increasing multiples of <math>4</math> each time we go up a figure.
 +
 
 +
So, to go from Figure 0 to 100, we add
 +
 
 +
<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>.
 +
 
 +
<math>20201</math>
 +
 
 +
<math>\boxed{\text{C}}</math>
 +
 
 +
===Solution 2===
 +
 
 +
We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>:
 +
 
 +
<asy>
 +
draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);
 +
draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);
 +
draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);
 +
draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);
 +
draw((3,0)--(3,7),linewidth(1.5));
 +
</asy>
 +
 
 +
The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math>
 +
 
 +
 
 +
===Solution 3===
 +
 
 +
Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>.  Thus, the original sequence can be generated from a quadratic function.
 +
 
 +
If <math>f(n) = an^2 + bn + c</math>, and <math>f(0) = 1</math>, <math>f(1) = 5</math>, and <math>f(2) = 13</math>, we get a system of three equations in three variables:
 +
 
 +
<math>f(0) = 0</math> gives <math>c = 1</math>
 +
 
 +
<math>f(1) = 5</math> gives <math>a + b + c = 5</math>
 +
 
 +
<math>f(2) = 13</math> gives <math>4a + 2b + c = 13</math>
 +
 
 +
Plugging in <math>c=1</math> into the last two equations gives
 +
 
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<math>a + b = 4</math>
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 +
<math>4a + 2b = 12</math>
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 +
Dividing the second equation by 2 gives the system:
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<math>a + b = 4</math>
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<math>2a + b = 6</math>
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 +
Subtracting the first equation from the second gives <math>a = 2</math>, and hence <math>b = 2</math>.  Thus, our quadratic function is:
 +
 
 +
<math>f(n) = 2n^2 + 2n + 1</math>
 +
 
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Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\text{C}}</math>
 +
 
 +
==See Also==
  
−
== See also ==
 
 
{{AMC12 box|year=2000|num-b=7|num-a=9}}
 
{{AMC12 box|year=2000|num-b=7|num-a=9}}
 +
{{AMC10 box|year=2000|num-b=11|num-a=13}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 22:47, 26 November 2011

The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.

Problem

Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]


$\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801$

Solution

Solution 1

We have a recursion:

$A_n=A_{n-1}+4n$.

I.E. we add increasing multiples of $4$ each time we go up a figure.

So, to go from Figure 0 to 100, we add

$4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200$.

$20201$

$\boxed{\text{C}}$

Solution 2

We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$:

[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7),linewidth(1.5)); [/asy]

The sum of the first $n$ odd numbers is $n^2$, so for figure $n$, there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $20201$, which is choice $\boxed{\text{C}}$


Solution 3

Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can be generated from a quadratic function.

If $f(n) = an^2 + bn + c$, and $f(0) = 1$, $f(1) = 5$, and $f(2) = 13$, we get a system of three equations in three variables:

$f(0) = 0$ gives $c = 1$

$f(1) = 5$ gives $a + b + c = 5$

$f(2) = 13$ gives $4a + 2b + c = 13$

Plugging in $c=1$ into the last two equations gives

$a + b = 4$

$4a + 2b = 12$

Dividing the second equation by 2 gives the system:

$a + b = 4$

$2a + b = 6$

Subtracting the first equation from the second gives $a = 2$, and hence $b = 2$. Thus, our quadratic function is:

$f(n) = 2n^2 + 2n + 1$

Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$, which is choice $\boxed{\text{C}}$

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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