Difference between revisions of "2010 AIME I Problems/Problem 13"
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Now just solve for <math>h</math>. | Now just solve for <math>h</math>. | ||
− | < | + | <cmath>\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ |
0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ | 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ | ||
h^2 & = \frac {9}{4}(6) \\ | h^2 & = \frac {9}{4}(6) \\ | ||
− | h & = \frac {3}{2}\sqrt {6} \end{align*}</ | + | h & = \frac {3}{2}\sqrt {6} \end{align*}</cmath> |
Don't forget to un-rescale at the end to get <math>AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}</math>. | Don't forget to un-rescale at the end to get <math>AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}</math>. |
Revision as of 17:36, 13 March 2015
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,
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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