Difference between revisions of "1998 AJHSME Problems/Problem 6"
Sinhaarunabh (talk | contribs) (→Solution 2) |
Sinhaarunabh (talk | contribs) (→Solution 3) |
||
Line 48: | Line 48: | ||
==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. | + | 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 also == | == See also == |
Revision as of 09:48, 3 September 2012
Problem 6
Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
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 |