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: | ||
Line 43: | Line 38: | ||
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 17:52, 26 October 2020
Problem
Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
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.