Difference between revisions of "2013 AMC 12A Problems/Problem 13"
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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 | + | == Problem== |
+ | |||
+ | Let points <math> A = (0,0) , \ B = (1,2), \ C = (3,3), </math> and <math> D = (4,0) </math>. Quadrilateral <math> ABCD </math> is cut into equal area pieces by a line passing through <math> A </math>. This line intersects <math> \overline{CD} </math> at point <math> \left (\frac{p}{q}, \frac{r}{s} \right ) </math>, where these fractions are in lowest terms. What is <math> p + q + r + s </math>? | ||
+ | |||
+ | <math> \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 </math> | ||
+ | |||
+ | ==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 | 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 | + | 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. |
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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</math>, which is <math>B</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</math>, which is <math>B</math>. | ||
+ | |||
+ | ==Solution 2 (Shoelace)== | ||
+ | Let the point of intersection be <math>E</math>, with coordinates <math>(x, y)</math>. Then, <math>ABCD</math> is cut into <math>ABCE</math> and <math>AED</math>. | ||
+ | |||
+ | Since the areas are equal, we can use [[Shoelace Theorem]] to find the area. This gives <math>3 + 3x - 3y = 4y</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 == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:13, 24 November 2020
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.