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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 7]] |
| − | An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}. </math>
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| − | [[Image:2006AimeA7.PNG]]
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| − | == Solution ==
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| − | Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].
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| − | Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at <math>0, 1, 2 \ldots</math>. The base of region <math>\mathcal{A}</math> is on the line <math>x = 1</math>. The bigger base of region <math>\mathcal{D}</math> is on the line <math>x = 7</math>.
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| − | Let the top side of the angle be <math>y = x - s</math> and the bottom side be x-axis, as halve the angle by folding doesn't change the problem.
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| − | Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
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| − | :<math>
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| − | \frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5}
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| − | = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}
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| − | </math>
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| − | Solve this to find that <math>s = \frac{5}{6}</math>.
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| − | By a similar method, <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> is <math>408</math>.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=6|num-a=8}}
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| − | [[Category:Intermediate Combinatorics Problems]]
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