Difference between revisions of "2006 AIME I Problems/Problem 7"
m (rv) |
(→Solution) |
||
| Line 19: | Line 19: | ||
Solve this to find that <math>s = \frac{5}{6}</math>. | Solve this to find that <math>s = \frac{5}{6}</math>. | ||
| − | By a similar method, <math>\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}</math> is <math>408</math>. | + | By a similar method, <math>\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}</math> is <math>\boxed{408}</math>. |
== See also == | == See also == | ||
Revision as of 20:38, 23 July 2012
Problem
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region 
to the area of shaded region 
is 11/5. Find the ratio of shaded region 
to the area of shaded region 
Solution
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.
Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at 
. The base of region 
is on the line 
. The bigger base of region is on the line 
.
Let the top side of the angle be 
and the bottom side be x-axis, as halve the angle by folding doesn't change the problem.
Since the area of the triangle is equal to 
,
![\[\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}\]](http://latex.artofproblemsolving.com/9/c/4/9c4b4fd73c01793ee229b4148f3eb5c1fbcbbb2f.png)
Solve this to find that 
.
By a similar method, 
is 
.
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
