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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== 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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== Solution 1 == | == Solution 1 == |
Revision as of 17:46, 1 January 2020
- 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 (which won't work when justification is required)
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 were simply an isosceles trapezoid! Since and (look at the side lengths and you'll know why!),
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.