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

m (Solution 1)
(Solution 4)
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==Solution 3 (which won't work when justification is required)==
 
==Solution 3 (which won't work when justification is required)==
 
The easiest way for the areas of the triangles to be equal would be if they were congruent. A way for that to work would be if <math>ABCD</math> were simply an isosceles trapezoid! Since <math>AC = 14</math> and <math>AE:EC = 3:4</math> (look at the side lengths and you'll know why!), <math>\boxed{AE = 6}</math>
 
The easiest way for the areas of the triangles to be equal would be if they were congruent. A way for that to work would be if <math>ABCD</math> were simply an isosceles trapezoid! Since <math>AC = 14</math> and <math>AE:EC = 3:4</math> (look at the side lengths and you'll know why!), <math>\boxed{AE = 6}</math>
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==Solution 4 (Easiest Way)==
 +
Using the fact that <math>\triangle AED = \triangle BEC</math> and the fact that <math>\triangle AEB \sim \triangle EDC</math> (which should be trivial given the two equal triangles) we have that
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<cmath>\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}</cmath>
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 +
We know that <math>DC=EC,</math> so we have
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<cmath>\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}</cmath>
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Thus
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<cmath>\frac{AE}{EC} = \frac{3}{4}</cmath>
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But <math>EC = 14 - AE</math> so we have
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<cmath>\frac{AE}{14 - AE} = \frac{3}{4}</cmath>
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Simplifying gives <math>AE = \boxed{6}.</math>
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~mathboy282
  
 
== See also ==
 
== See also ==

Revision as of 00:23, 20 October 2021

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 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 $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$.

[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("12",(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 $EC = 8$, so $AE = 6\ \textbf{(E)}$.

Solution 3 (which won't work when justification is required)

The easiest way for the areas of the triangles to be equal would be if they were congruent. A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\boxed{AE = 6}$

Solution 4 (Easiest Way)

Using the fact that $\triangle AED = \triangle BEC$ and the fact that $\triangle AEB \sim \triangle EDC$ (which should be trivial given the two equal triangles) we have that

\[\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}\]

We know that $DC=EC,$ so we have

\[\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}\]

Thus

\[\frac{AE}{EC} = \frac{3}{4}\]

But $EC = 14 - AE$ so we have

\[\frac{AE}{14 - AE} = \frac{3}{4}\]

Simplifying gives $AE = \boxed{6}.$

~mathboy282

See also

2009 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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