Difference between revisions of "2013 AMC 12A Problems/Problem 13"

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==Solution==
 
==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.
 
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.
  
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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>.
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==Solution 2 (Shoelace)==
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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>.
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Since the areas are equal, we can use [[Shoelace Theorem]] to find the area. This gives <math>3 + 3x - 3y = 4y</math>.
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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>.
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==Video Solution==
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https://www.youtube.com/watch?v=XQpQaomC2tA
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~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}}
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[[Category:Introductory Geometry Problems]]
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{{AMC10 box|year=2013|ab=A|num-b=17|num-a=19}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:00, 31 December 2021

Problem

Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$, where these fractions are in lowest terms. What is $p + q + r + s$?

$\textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad  \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75$

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

$A$ = $I$ $+$ $\frac{B}{2}$ - $1$, where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.

In this case,

$A$ = $5$ $+$ $\frac{7}{2}$ - $1$ = $7.5$

so

$\frac{A}{2}$ = $3.75$

The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$. This height is the y coordinate of our desired intersection point.


Note that segment CD lies on the line $y = -3x + 12$. Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$.

Therefore the point of intersection is ($\frac{27}{8}$, $\frac{15}{8}$), and our desired result is $27+8+15+8=58$, which is $B$.

Solution 2 (Shoelace)

Let the point of intersection be $E$, with coordinates $(x, y)$. Then, $ABCD$ is cut into $ABCE$ and $AED$.

Since the areas are equal, we can use Shoelace Theorem to find the area. This gives $3 + 3x - 3y = 4y$.

The line going through $CD$ is $y = -3x + 12$. Since $E$ is on $CD$, we can substitute this in, giving $3 + 3x = -21x + 84$. Solving for $x$ gives $\frac{27}{8}$. Plugging this back into the line equation gives $y = \frac{15}{8}$, for a final answer of $\boxed{58}$.

Video Solution

https://www.youtube.com/watch?v=XQpQaomC2tA

~sugar_rush

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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