Difference between revisions of "2009 AMC 12A Problems/Problem 20"
m (→Solution 3 (which won't work when justification is required)) |
m (→=Solution 3 (which won't work when justification is required)) |
||
Line 23: | Line 23: | ||
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 (which won't work when justification is required)= | + | ==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 <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 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> | ||
Revision as of 10:23, 11 June 2019
- 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
[hide]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 areas of 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.