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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 of the quadrilateral by other methods.
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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 part of the quadrilateral makes a triangle with base <math>4</math>, so we can deduce that the height of the triangle must be <math>\frac{15}{8}</math> in order for the area to be <math>3.75</math>. This height is the y coordinate of our desired intersection point.
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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

$A$ = $I$ $+$ $\frac{B}{2}$ - $1$, where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.

In this case,

$A$ = $5$ $+$ $\frac{7}{2}$ - $1$ = $7.5$

so

$\frac{A}{2}$ = $3.75$

The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$. This height is the y coordinate of our desired intersection point.


Note that segment CD lies on the line $y = -3x + 12$. Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$.

Therefore the point of intersection is ($\frac{27}{8}$, $\frac{15}{8}$), and our desired result is $27+8+15+8=58$, which is $B$.

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