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

Revision as of 12:42, 16 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 $3: 4$ . $AE = \frac {3}{7}AC = 6\ \boxed{\textbf{(E)}}$ .

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

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 $AE = 6$ , $EC = 8\ \textbf{(E)}$ .

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

@@ Line 9: / Line 9: @@
 == Solution ==
 === Solution 1 ===
-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}AD = 6\ \boxed{\textbf{(E)}}</math>.
+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}AC = 6\ \boxed{\textbf{(E)}}</math>.
 <center><asy>pathpen = linewidth(0.7);pointpen = black;

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