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Difference between revisions of "2009 AMC 12A Problems/Problem 20"

(solutions by (1) worthawholebean, (2) e^pi*i = -1)
 
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{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #20]] and [[2009 AMC 10A Problems|2009 AMC 10A #23]]}}
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== Problem ==
 
== Problem ==
 
Convex quadrilateral <math>ABCD</math> has <math>AB = 9</math> and <math>CD = 12</math>.  Diagonals <math>AC</math> and <math>BD</math> intersect at <math>E</math>, <math>AC = 14</math>, and <math>\triangle AED</math> and <math>\triangle BEC</math> have equal areas.  What is <math>AE</math>?
 
Convex quadrilateral <math>ABCD</math> has <math>AB = 9</math> and <math>CD = 12</math>.  Diagonals <math>AC</math> and <math>BD</math> intersect at <math>E</math>, <math>AC = 14</math>, and <math>\triangle AED</math> and <math>\triangle BEC</math> have equal areas.  What is <math>AE</math>?
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2009|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2009|ab=A|num-b=19|num-a=21}}
 +
{{AMC10 box|year=2009|ab=A|num-b=22|num-a=24}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 14:20, 14 February 2009

The following problem is from both the 2009 AMC 12A #20 and 2009 AMC 10A #23, so both problems redirect to this page.

Problem

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?

$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$

Solution

Solution 1

Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}AD = 6\ \boxed{\textbf{(E)}}$.

[asy]pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("14",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); [/asy]

Solution 2

Using the sine area formula on triangles $AED$ and $BEC$, as $\angle AED = \angle BEC$, we see that

\[(AE)(ED) = (BE)(EC)\quad \Longrightarrow\quad \frac {AE}{EC} = \frac {BE}{ED}.\]

Since $\angle AEB = \angle DEC$, triangles $AEB$ and $DEC$ are similar. Their ratio is $\frac {AB}{CD} = \frac {3}{4}$. Since $AE + EC = 14$, we must have $AE = 6$, $EC = 8\ \textbf{(E)}$.

See also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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