Difference between revisions of "2013 AMC 12A Problems/Problem 13"
(Created page with "If you have graph paper, use Pick's Theorem to quickly and efficiently find the area. If not, just find the area of the quadrilateral by other methods. Pick's Theorem states tha...") |
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| − | If you have graph paper, use Pick's Theorem to quickly and efficiently find the area. If not, just find the area | + | If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods. |
Pick's Theorem states that | Pick's Theorem states that | ||
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<math>\frac{A}{2}</math> = <math>3.75</math> | <math>\frac{A}{2}</math> = <math>3.75</math> | ||
| − | The bottom | + | The bottom half of the quadrilateral makes a triangle with base <math>4</math> and half the total area, so we can deduce that the height of the triangle must be <math>\frac{15}{8}</math> in order for its area to be <math>3.75</math>. This height is the y coordinate of our desired intersection point. |
Revision as of 22:43, 6 February 2013
If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
= 
- 
, where is the number of lattice points in the interior of the polygon, and 
is the number of lattice points on the boundary of the polygon.
In this case,
= 
- = 
so
= 
The bottom half of the quadrilateral makes a triangle with base 
and half the total area, so we can deduce that the height of the triangle must be 
in order for its area to be . This height is the y coordinate of our desired intersection point.
Note that segment CD lies on the line 
. Substituting in for y, we can find that the x coordinate of our intersection point is 
.
Therefore the point of intersection is (, ), and our desired result is 
, which is .
