Difference between revisions of "2004 AMC 10B Problems/Problem 18"
Mathboy282 (talk | contribs) (→Solution 5) |
(added solution 6) |
||
Line 110: | Line 110: | ||
~mathboy282 | ~mathboy282 | ||
+ | |||
+ | ==Solution 6 (Wooga Looga Theorem)== | ||
+ | |||
+ | We know that <math>\frac{CB}{BA}=\frac{AF}{FE}=\frac{ED}{DC}=3</math>, so by the [[Wooga Looga Theorem]] we have <math>\frac{[DBF]}{[ACE]}=\frac{3^2-3+1}{(3+1)^2}=\boxed{\frac7{16}}</math>. | ||
== See also == | == See also == |
Revision as of 17:15, 2 January 2021
Contents
Problem
In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?
Solution 1
Let . Because is divided into four triangles, .
Because of triangle area, .
and , so .
, so .
Solution 2
First of all, note that , and therefore .
Draw the height from onto as in the picture below:
Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is , hence . Now the area of can be computed as = .
Similarly we can find that as well.
Hence , and the answer is .
Solution 3 (Coordinate Geometry)
We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42.
Solution 4
You can also place a point on such that is , creating trapezoid . Then, you can find the area of the trapezoid, subtract the area of the two right triangles and , divide by the area of , and get the ratio of .
Solution 5
It is well known that for when two triangles share an angle, the two sides around the shared angle is proportional to the areas of each of the two triangles.
We can find all the ratios of the triangles except for and then subtract from
In this case, we have sharing with .
Therefore, we have
Also note that shares with .
Therefore, we have
Lastly, note that shares with .
Therefore, we have
Thus, the ratio of to is
~mathboy282
Solution 6 (Wooga Looga Theorem)
We know that , so by the Wooga Looga Theorem we have .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.