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

Revision as of 18:50, 19 August 2017

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 areas of 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!), $\boxed{AE = 6}$

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.

@@ Line 24: / Line 24: @@
 ==Solution 3 (which won't work when justification is required)==
-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.
+The easiest way for areas of 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>
-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>.
 == See also ==

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