Difference between revisions of "2013 AMC 12A Problems/Problem 13"
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Note that segment CD lies on the line <math>y = -3x + 12</math>. Substituting in <math>\frac{15}{8}</math> for y, we can find that the x coordinate of our intersection point is <math>\frac{27}{8}</math>. | Note that segment CD lies on the line <math>y = -3x + 12</math>. Substituting in <math>\frac{15}{8}</math> for y, we can find that the x coordinate of our intersection point is <math>\frac{27}{8}</math>. | ||
| − | Therefore the point of intersection is (<math>\frac{27}{8}</math>, <math>\frac{15}{8}</math>), and our desired result is <math>27+8+15+8=58 | + | Therefore the point of intersection is (<math>\frac{27}{8}</math>, <math>\frac{15}{8}</math>), and our desired result is <math>27+8+15+8= \boxed{ \text{B}) \ 58}</math>. |
==Solution 2 (Shoelace)== | ==Solution 2 (Shoelace)== | ||
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The line going through <math>CD</math> is <math>y = -3x + 12</math>. Since <math>E</math> is on <math>CD</math>, we can substitute this in, giving <math>3 + 3x = -21x + 84</math>. Solving for <math>x</math> gives <math>\frac{27}{8}</math>. Plugging this back into the line equation gives <math>y = \frac{15}{8}</math>, for a final answer of <math>\boxed{58}</math>. | The line going through <math>CD</math> is <math>y = -3x + 12</math>. Since <math>E</math> is on <math>CD</math>, we can substitute this in, giving <math>3 + 3x = -21x + 84</math>. Solving for <math>x</math> gives <math>\frac{27}{8}</math>. Plugging this back into the line equation gives <math>y = \frac{15}{8}</math>, for a final answer of <math>\boxed{58}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=XQpQaomC2tA | ||
| + | |||
| + | ~sugar_rush | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=12|num-a=14}} | {{AMC12 box|year=2013|ab=A|num-b=12|num-a=14}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | |||
| + | {{AMC10 box|year=2013|ab=A|num-b=17|num-a=19}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:29, 4 September 2022
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 
.
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 
.
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
~sugar_rush
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 | |
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
