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

Revision as of 00:06, 2 November 2025

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 (Trigonometry)

Using the sine area formula AREA= absinc/2 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 </asy></center>
-== Solution 2(Trigonometry) ==
+== Solution 2 (Trigonometry) ==
 Using the sine area formula AREA= absinc/2 on triangles <math>AED</math> and <math>BEC</math>, as <math>\angle AED = \angle BEC</math>, we see that

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