Difference between revisions of "2010 AIME I Problems/Problem 13"
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Finally, the answer is <math>63 + 6 = \boxed{069}</math>. | Finally, the answer is <math>63 + 6 = \boxed{069}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let <math>O</math> be the center of the semicircle. It follows that <math>AU + UO = AN = NO = 126</math>, so triangle <math>ANO</math> is [[equilateral]]. | ||
+ | |||
+ | Let <math>Y</math> be the foot of the altitude from <math>N</math>, such that <math>NY = 63\sqrt{3}</math> and <math>NU = 21</math>. | ||
+ | |||
+ | Finally, denote <math>DT = a.</math> Extend <math>U</math> to point <math>Z</math> so that <math>Z</math> is on <math>CD</math> and <math>UZ</math> is perpendicular to <math>CD</math>. It then follows that <math>ZT = a-84</math>. Using [[similar triangles]], | ||
+ | |||
+ | <math>\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}</math> | ||
+ | |||
+ | Given that line <math>NT</math> divides <math>R</math> into a ratio of <math>1:2</math>, we can also say that | ||
+ | |||
+ | <math>(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3} = (\frac{1}{3})(252x + \frac{126^2\pi}{2})</math> | ||
+ | |||
+ | where the first term is the area of trapezoid <math>AUTD</math>, the second and third terms denote the areas of <math>\frac{1}{6}</math> a full circle, and the area of <math>NUO</math>, respectively, and the fourth term on the right side of the equation is equal to <math>R</math>. Cancelling out the <math>\frac{126^2\pi}{6}</math> on both sides, we obtain | ||
+ | |||
+ | <math>(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})</math> | ||
+ | |||
+ | By adding and collecting like terms, | ||
+ | <math>\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})</math> | ||
+ | |||
+ | <math>\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})</math>. | ||
+ | |||
+ | Since <math>a - 84 = \frac{x}{3\sqrt{3}}</math>, | ||
+ | |||
+ | <math>\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})</math> | ||
+ | |||
+ | <math>\frac {x^2}{\sqrt{3} = (63)(126)(\sqrt{3})</math> | ||
+ | |||
+ | <math>x^2 = (63)(126)(3) = (2)(3^5)(7^2)</math> | ||
+ | |||
+ | <math>x = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}</math>, so the answer is <math>\boxed{069}.</math> | ||
== See also == | == See also == |
Revision as of 11:17, 18 March 2010
Contents
[hide]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
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 Extend to point so that is on and is perpendicular to . It then follows that . Using similar triangles,
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 ,
$\frac {x^2}{\sqrt{3} = (63)(126)(\sqrt{3})$ (Error compiling LaTeX. Unknown error_msg)
, so the answer is
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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All AIME Problems and Solutions |