Difference between revisions of "1998 AJHSME Problems/Problem 6"
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== Solutions == | == Solutions == | ||
===Solution 1=== | ===Solution 1=== | ||
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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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Bottom-right: Square with area <math>1</math> | Bottom-right: Square with area <math>1</math> | ||
| − | Adding all of these together, we get <math> | + | 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=== | ||
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> | ||
| + | ===Solution 3=== | ||
| − | + | 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> | |
| + | ~sakshamsethi | ||
== See Also == | == See Also == | ||
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]](http://latex.artofproblemsolving.com/e/5/1/e512d80ea585bdc760432f8f8d7ee955400e1d71.png)
Solutions
Solution 1
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 
which is the same as 
Solution 2
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 3
Notice that the extra triangle on the top with area can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area . This creates a 
rectangle, with a area of . The answer is
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
