Difference between revisions of "2009 AMC 12A Problems/Problem 20"
m (→Solution 1) |
Mathboy282 (talk | contribs) (→Solution 4) |
||
Line 24: | Line 24: | ||
==Solution 3 (which won't work when justification is required)== | ==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 <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. 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 | ||
== See also == | == See also == |
Revision as of 00:23, 20 October 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 (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!),
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
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.