Difference between revisions of "2013 AMC 12A Problems/Problem 13"
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Revision as of 08:56, 7 April 2013
Contents
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
Solution 1
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 | |
