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

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{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #20]] and [[2009 AMC 10A Problems|2009 AMC 10A #23]]}}
 
{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #20]] and [[2009 AMC 10A Problems|2009 AMC 10A #23]]}}
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__TOC__
 
== 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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<math>\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</math>
 
<math>\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</math>
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
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__TOC__
 
  
 
== Solution 1 ==
 
== Solution 1 ==
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Let <math>[ABC]</math> denote the area of triangle <math>ABC</math>. <math>[AED] = [BEC]</math>, so <math>[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]</math>. Since triangles <math>ABD</math> and <math>ABC</math> share a base, they also have the same height and thus <math>\overline{AB}||\overline{CD}</math> and <math>\triangle{AEB}\sim\triangle{CED}</math> with a ratio of <math>3: 4</math>. <math>AE = \frac {3}{7}</math>, so <math>AC = 6\ \boxed{\textbf{(E)}}</math>.
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Let <math>[ABC]</math> denote the area of triangle <math>ABC</math>. <math>[AED] = [BEC]</math>, so <math>[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]</math>. Since triangles <math>ABD</math> and <math>ABC</math> share a base, they also have the same height and thus <math>\overline{AB}||\overline{CD}</math> and <math>\triangle{AEB}\sim\triangle{CED}</math> with a ratio of <math>3: 4</math>. <math>AE = \frac {3}{7}\times AC</math>, so <math>AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}</math>.
  
 
<center><asy>pathpen = linewidth(0.7);pointpen = black;
 
<center><asy>pathpen = linewidth(0.7);pointpen = black;
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Since <math>\angle AEB = \angle DEC</math>, triangles <math>AEB</math> and <math>DEC</math> are similar. Their ratio is <math>\frac {AB}{CD} = \frac {3}{4}</math>.  Since <math>AE + EC = 14</math>, we must have <math>EC = 8</math>, so <math>AE = 6\ \textbf{(E)}</math>.
 
Since <math>\angle AEB = \angle DEC</math>, triangles <math>AEB</math> and <math>DEC</math> are similar. Their ratio is <math>\frac {AB}{CD} = \frac {3}{4}</math>.  Since <math>AE + EC = 14</math>, we must have <math>EC = 8</math>, so <math>AE = 6\ \textbf{(E)}</math>.
  
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==Solution 3 (which won't work when justification is required)==
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==Solution 3(Fakesolve)==
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Consider an isosceles trapezoid with opposite bases of <math>9</math> and <math>12</math> and a diagonal of length <math>14</math>. This trapezoid exists and satisfies all the conditions in the problem (the areas are congruent by symmetry). So, we can simply use this easier case to solve this problem, because the problem implies that the answer is invariant for all quadrilaterals satisfying the conditions.
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The easiest way for the areas of the triangles to be equal would be if they were congruent [https://artofproblemsolving.com/wiki/index.php/Constraints_Strategy]. 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)==
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Using the fact that <math>[AED] = [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
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===Note===
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The two triangles that are equal in area imply that <math>AB</math> is parallel to <math>DC</math> which implies that <math>\angle{EAB} = \angle{CDE}</math> and <math>\angle{EBA} = \angle{DCE}.</math> Furthermore, since <math>\angle{AEB} = \angle{DEC}</math> (vertical angles). By AAA similarity, <math>\triangle AEB \sim \triangle EDC.</math>
  
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Then, by similar triangles, the ratio of <math>AE</math> to <math>EC</math> is <math>3:4</math>, so <math>AE=\boxed{\textbf{(E)}6}</math>.
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~mathboy282
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Triangle Area Ratio Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:39, 13 October 2024

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(Fakesolve)

The easiest way for the areas of the triangles to be equal would be if they were congruent [1]. 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 $[AED] = [BEC]$ 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

Note

The two triangles that are equal in area imply that $AB$ is parallel to $DC$ which implies that $\angle{EAB} = \angle{CDE}$ and $\angle{EBA} = \angle{DCE}.$ Furthermore, since $\angle{AEB} = \angle{DEC}$ (vertical angles). By AAA similarity, $\triangle AEB \sim \triangle EDC.$

~mathboy282

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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