Difference between revisions of "1998 AJHSME Problems/Problem 6"
Sinhaarunabh (talk | contribs) (→Solution 3) |
m |
||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==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 | ||
| Line 16: | Line 16: | ||
<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== | + | == 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. | ||
| Line 22: | Line 23: | ||
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== | + | ===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: | ||
| Line 46: | Line 47: | ||
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== | + | ===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 | + | == 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 07:24, 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]](http://latex.artofproblemsolving.com/e/5/1/e512d80ea585bdc760432f8f8d7ee955400e1d71.png)
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 
box, which has area 
Solution 2
We could count the area contributed by each square on the 
grid:
Top-left: 
Top: Triangle with area 
Top-right:
Left: Square with area 
Center: Square with area
Right: Square with area
Bottom-left: Square with area
Bottom: Triangle with area
Bottom-right: Square with area
Adding all of these together, we get 
or 
Solution 3
By http://www.artofproblemsolving.com/Wiki/index.php/Pick%27s_Theorem, We get the formula, 
where 
is the number of lattice points in the interior and 
being the number of lattice points on the boundary. In this problem, we can see that 
and 
. Substituting gives us 
Thus, the answer is 
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 | ||
