Difference between revisions of "1998 AJHSME Problems/Problem 6"

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==Problem 6==
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==Problem==
  
 
Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
 
Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
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<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
 
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
  
==Solution 1==
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== Solution ==
 +
===Solution 1===
  
 
By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
 
By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
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This creates a <math>2\times3</math> box, which has area <math>2\times3=\boxed{6}</math>
 
This creates a <math>2\times3</math> box, which has area <math>2\times3=\boxed{6}</math>
  
==Solution 2==
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===Solution 2===
  
 
We could count the area contributed by each square on the <math>3 \times 3</math> grid:
 
We could count the area contributed by each square on the <math>3 \times 3</math> grid:
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Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math>
 
Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math>
  
==Solution 3==
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===Solution 3===
  
 
By http://www.artofproblemsolving.com/Wiki/index.php/Pick%27s_Theorem, We get the formula, <math>A=I+\frac{b}{2}-1</math> where <math>I</math> is the number of lattice points in the interior and <math>b</math> being the number of lattice points on the boundary. In this problem, we can see that <math>I=1</math> and <math>B=12</math>. Substituting gives us <math>A=1+\frac{12}{2}-1=6</math> Thus, the answer is <math>\boxed{\text{(B) 6}}</math>
 
By http://www.artofproblemsolving.com/Wiki/index.php/Pick%27s_Theorem, We get the formula, <math>A=I+\frac{b}{2}-1</math> where <math>I</math> is the number of lattice points in the interior and <math>b</math> being the number of lattice points on the boundary. In this problem, we can see that <math>I=1</math> and <math>B=12</math>. Substituting gives us <math>A=1+\frac{12}{2}-1=6</math> Thus, the answer is <math>\boxed{\text{(B) 6}}</math>
  
== See also ==
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== See Also ==
 
{{AJHSME box|year=1998|num-b=5|num-a=7}}
 
{{AJHSME box|year=1998|num-b=5|num-a=7}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]

Revision as of 08:24, 6 September 2012

Problem

Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is

[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution

Solution 1

By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.

This creates a $2\times3$ box, which has area $2\times3=\boxed{6}$

Solution 2

We could count the area contributed by each square on the $3 \times 3$ grid:

Top-left: $0$

Top: Triangle with area $\frac{1}{2}$

Top-right: $0$

Left: Square with area $1$

Center: Square with area $1$

Right: Square with area $1$

Bottom-left: Square with area $1$

Bottom: Triangle with area $\frac{1}{2}$

Bottom-right: Square with area $1$

Adding all of these together, we get $\boxed{6}$ or $\boxed{B}$

Solution 3

By http://www.artofproblemsolving.com/Wiki/index.php/Pick%27s_Theorem, We get the formula, $A=I+\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$. Substituting gives us $A=1+\frac{12}{2}-1=6$ Thus, the answer is $\boxed{\text{(B) 6}}$

See Also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions