Difference between revisions of "1998 AJHSME Problems/Problem 6"
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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> | 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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== See Also == | == See Also == | ||
Revision as of 12:16, 9 July 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]](http://latex.artofproblemsolving.com/e/5/1/e512d80ea585bdc760432f8f8d7ee955400e1d71.png)
Solutions
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 Pick's 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 
Solution 4
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
