2010 AIME I Problems/Problem 13
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 .
$\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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 2
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 1/3 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 $\frac {-3\sqrt{3}}$ (Error compiling LaTeX. Unknown error_msg), and thus .
is most easily found by :
Equating, = \frac {DA^2}{6\sqrt{3}} 63 * 21 * 3 * 6 = DA^2$$ (Error compiling LaTeX. Unknown error_msg)DA = 63 \sqrt{6} \longrightarrow \boxed {069}$
See also
- <url>viewtopic.php?t=338915 Discussion</url>, with a Geogebra diagram.
2010 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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