2010 AIME I Problems/Problem 13
Contents
Problem
Rectangle and a semicircle with diameter are coplanar and have nonoverlapping interiors. Let denote the region enclosed by the semicircle and the rectangle. Line meets the semicircle, segment , and segment at distinct points , , and , respectively. Line divides region into two regions with areas in the ratio . Suppose that , , and . Then can be represented as , where and are positive integers and is not divisible by the square of any prime. Find .
Solution
Solution 1
The center of the semicircle is also the midpoint of . Let this point be O. Let be the length of .
Rescale everything by 42, so . Then so .
Since is a radius of the semicircle, . Thus is an equilateral triangle.
Let , , and be the areas of triangle , sector , and trapezoid respectively.
To find we have to find the length of . Project and onto to get points and . Notice that and are similar. Thus:
.
Then . So:
Let be the area of the side of line containing regions . Then
Obviously, the is greater than the area on the other side of line . This other area is equal to the total area minus . Thus:
.
Now just solve for .
Don't forget to un-rescale at the end to get .
Finally, the answer is .
Solution 2
Let be the center of the semicircle. It follows that , so triangle is equilateral.
Let be the foot of the altitude from , such that and .
Finally, denote , and . Extend to point so that is on and is perpendicular to . It then follows that . Since and are similar,
Given that line divides into a ratio of , we can also say that
where the first term is the area of trapezoid , the second and third terms denote the areas of a full circle, and the area of , respectively, and the fourth term on the right side of the equation is equal to . Cancelling out the on both sides, we obtain
By adding and collecting like terms,
.
Since ,
, so the answer is
Solution 3
Note that the total area of is and thus one of the regions has area
As in the above solutions we discover that , thus sector of the semicircle has of the semicircle's area.
Similarly, dropping the perpendicular we observe that , which is of the total rectangle.
Denoting the region to the left of as and to the right as , it becomes clear that if then the regions will have the desired ratio.
Using the 30-60-90 triangle, the slope of , is , and thus .
is most easily found by :
Equating,
Solving,
Solution 4 (Coordinates)
Like above solutions, note that is equilateral with side length where is the midpoint of Then, if we let and set origin at we get Line is then so it intersects the -axis, at giving us point Now the area of region is so one third of that is
The area of the smaller piece of is Setting this equal to and canceling the yields so and the anser is ~ rzlng
See Also
- <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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