1998 AJHSME Problems/Problem 6
Problem 6
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 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.
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 | ||
