2013 AMC 12A Problems/Problem 13
Problem
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 
?
Solution
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 .
See also
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 AMC 12 Problems and Solutions | |
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