Difference between revisions of "2009 AMC 12A Problems/Problem 20"
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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>. | ||
− | ==Solution 3 | + | ==Solution 3== |
− | 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 [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> |
+ | |||
+ | ==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 | ||
+ | |||
+ | <cmath>\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}</cmath> | ||
+ | |||
+ | We know that <math>DC=EC,</math> so we have | ||
+ | |||
+ | <cmath>\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}</cmath> | ||
+ | |||
+ | Thus | ||
+ | |||
+ | <cmath>\frac{AE}{EC} = \frac{3}{4}</cmath> | ||
+ | |||
+ | But <math>EC = 14 - AE</math> so we have | ||
+ | |||
+ | <cmath>\frac{AE}{14 - AE} = \frac{3}{4}</cmath> | ||
+ | |||
+ | Simplifying gives <math>AE = \boxed{6}.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ===Note=== | ||
+ | |||
+ | They are similar triangles because common sense. The two triangles that are equal in area implies 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> (which is obvious why). By AAA similarity, <math>\triangle AEB \sim \triangle EDC.</math> | ||
+ | |||
+ | ~mathboy282 | ||
== See also == | == See also == |
Latest revision as of 22:11, 2 November 2021
- The following problem is from both the 2009 AMC 12A #20 and 2009 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
Convex quadrilateral has and . Diagonals and intersect at , , and and have equal areas. What is ?
Solution 1
Let denote the area of triangle . , so . Since triangles and share a base, they also have the same height and thus and with a ratio of . , so .
Solution 2
Using the sine area formula on triangles and , as , we see that
Since , triangles and are similar. Their ratio is . Since , we must have , so .
Solution 3
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 were simply an isosceles trapezoid! Since and (look at the side lengths and you'll know why!),
Solution 4 (Easiest Way)
Using the fact that and the fact that (which should be trivial given the two equal triangles) we have that
We know that so we have
Thus
But so we have
Simplifying gives
~mathboy282
Note
They are similar triangles because common sense. The two triangles that are equal in area implies that is parallel to which implies that and Furthermore, since (which is obvious why). By AAA similarity,
~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.