Difference between revisions of "2006 AIME I Problems/Problem 7"
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[[Image:2006AimeA7.PNG]] | [[Image:2006AimeA7.PNG]] | ||
− | == Solution == | + | == Solution 1 == |
Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]]. | Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]]. | ||
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Using the same reasoning as above, we get <math>\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}</math>, which is <math>\boxed{408}</math>. | Using the same reasoning as above, we get <math>\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}</math>, which is <math>\boxed{408}</math>. | ||
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+ | == Solution 2 == | ||
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+ | Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be <math>x</math> and the area of it be <math>x^2</math>. Also, let all sections of the line on the same side as the side with length <math>x</math> on a trapezoid be equal to <math>1</math>. | ||
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+ | Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is <math>(\frac{x+1}{x})^2</math>. Multiplying, we get <math>(x+1)^2</math> as the area of the triangle, so the area of the trapezoid is <math>2x+1</math>. Repeating this process, we get that the area of B is <math>2x+3</math>, the area of C is <math>2x+7</math>, and the area of D is <math>2x+11</math>. | ||
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+ | We can now use the given condition that the ratio of C and B is <math>\frac{11}{5}</math>. | ||
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+ | <math>\frac{11}{5} = \frac{2x+7}{2x+3}</math> gives us <math>x = \frac{1}{6}</math> | ||
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+ | So now we compute the ratio of D and A, which is <math>\frac{2(\frac{1}{6} + 11)}{(\frac{1}{6})^2} = \boxed{408.}</math> | ||
== See also == | == See also == |
Revision as of 21:00, 13 January 2016
Contents
[hide]Problem
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region to the area of shaded region is 11/5. Find the ratio of shaded region to the area of shaded region
Solution 1
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.
Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at . The base of region is on the line . The bigger base of region is on the line . Let the top side of the angle be and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to ,
Solve this to find that .
Using the same reasoning as above, we get , which is .
Solution 2
Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be and the area of it be . Also, let all sections of the line on the same side as the side with length on a trapezoid be equal to .
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is . Multiplying, we get as the area of the triangle, so the area of the trapezoid is . Repeating this process, we get that the area of B is , the area of C is , and the area of D is .
We can now use the given condition that the ratio of C and B is .
gives us
So now we compute the ratio of D and A, which is
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.