Difference between revisions of "2010 AIME I Problems/Problem 13"
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Given that line <math>NT</math> divides <math>R</math> into a ratio of <math>1:2</math>, we can also say that | 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> | + | <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 | 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 |
Revision as of 16:55, 18 March 2010
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 .
Contents
[hide]Solution
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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
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 |