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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>
  
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==Solution 1==
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== Solutions ==
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===Solution 1===
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We could count the area contributed by each square on the <math>3 \times 3</math> grid:
  
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By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
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Top-left: <math>0</math>
  
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This creates a <math>2\times3</math> box, which has area <math>2\times3=\boxed{6}</math>
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Top: Triangle with area <math>\frac{1}{2}</math>
  
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==Solution 2==
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Top-right: <math>0</math>
  
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We could take the area of each square.
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Left: Square with area <math>1</math>
  
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Top-left: <math>0</math>
 
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Top: Triangle with area <math>\frac{1}{2}</math>
 
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Top-right: <math>0</math>
 
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Left: Square with area <math>1</math>
 
 
Center: Square with area <math>1</math>
 
Center: Square with area <math>1</math>
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Right: Square with area <math>1</math>
 
Right: Square with area <math>1</math>
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Bottom-left: Square with area <math>1</math>
 
Bottom-left: Square with area <math>1</math>
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Bottom: Triangle with area <math>\frac{1}{2}</math>
 
Bottom: Triangle with area <math>\frac{1}{2}</math>
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Bottom-right: Square with area <math>1</math>
 
Bottom-right: Square with area <math>1</math>
  
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Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math>
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Adding all of these together, we get <math>6</math> which is the same as <math>\boxed{B}</math>
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===Solution 2===
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By Pick's 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>
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===Solution 3===
  
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Notice that the extra triangle on the top with area <math>1</math> can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area <math>1</math>. This creates a <math>2*3</math> rectangle, with a area of <math>6</math>. The answer is <math>\boxed{\text{(B) 6}}</math>
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~sakshamsethi
  
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== 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]]
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{{MAA Notice}}

Latest revision as of 16:52, 26 October 2020

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]

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

Solutions

Solution 1

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 $6$ which is the same as $\boxed{B}$

Solution 2

By Pick's 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 $\boxed{\text{(B) 6}}$

Solution 3

Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$. This creates a $2*3$ rectangle, with a area of $6$. The answer is $\boxed{\text{(B) 6}}$ ~sakshamsethi

See Also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 AJHSME/AMC 8 Problems and Solutions

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