2013 AMC 12A Problems/Problem 13
Let points and . Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. What is ?
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,
= - =
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 .
Solution 2 (Shoelace)
Let the point of intersection be , with coordinates . Then, is cut into and .
Since the areas are equal, we can use Shoelace Theorem to find the area. This gives .
The line going through is . Since is on , we can substitute this in, giving . Solving for gives . Plugging this back into the line equation gives , for a final answer of .
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