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>AE = 6</math>, <math>EC = 8\ \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>AE = 6</math>, <math>EC = 8\ \textbf{(E)}</math>. | ||
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+ | ===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. | ||
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+ | Then, by similar triangles, the ratio of <math>AE</math> to <math>EC</math> is <math>3:4</math>, so <math>AE=6</math> | ||
== See also == | == See also == |
Revision as of 23:20, 5 February 2012
- 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 has and . Diagonals and intersect at , , and and have equal areas. What is ?
Contents
Solution
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 . .
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 , .
Solution 3 (which won't work when justification is required)
Consider an isosceles trapezoid with opposite bases of and and a diagonal of length . 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.
Then, by similar triangles, the ratio of to is , so
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 |