Difference between revisions of "2005 AMC 10A Problems/Problem 25"
(→See also) |
(→See also) |
||
Line 94: | Line 94: | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | <img>https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg</img> |
Revision as of 10:00, 24 May 2020
Contents
[hide]Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution 1
We have that
But , so
Solution 2(no trig)
We can let . Since , . So, . This means that . Thus,
-Conantwiz2023
Solution 3(trig)
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem
Solution 4
Let be on such that then we have Since we have Thus and Finally, after some calculations.
~ Nafer
~ LaTeX changes by dolphin7
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.
<img>https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg</img>