Difference between revisions of "2010 AIME I Problems/Problem 13"
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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>. | 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]], | + | Finally, denote <math>DT = a, and AD = x.</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> | <math>\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}</math> | ||
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<math>x^2 = (63)(126)(3) = (2)(3^5)(7^2)</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> | + | <math>x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}</math>, so the answer is <math>\boxed{069}.</math> |
== See also == | == See also == |
Revision as of 11:20, 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 ,
, 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |