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Difference between revisions of "2006 AIME A Problems/Problem 7"

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== Solution ==
 
== Solution ==
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There are <math>\frac{1000}{10} = 100</math> numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities for <math>a</math> are when <math>a</math> or <math>b</math> have a 0 in the tens digit, or if the tens digit of <math>a</math> is 0 or 9. Excluding the hundreds, which were counted above already, there are <math>9 \cdot 2 = 18</math> numbers in every hundred numbers that have a tens digit of 0 or 9, totaling <math>10 \cdot 18 = 180</math> such numbers. However, the numbers from 1 to 9 and 991 to 999 do not have 0s, so we must subtract <math>18</math> from that to get <math>162</math>. Therefore, there are <math>1000 - (100 + 162) = 738</math> such ordered pairs.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 14:30, 25 September 2007

Problem

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is 11/5. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}.$

2006AimeA7.PNG

Solution

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 $0, 1, 2 \ldots$. The base of region $\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\mathcal{D}$ is on the line $x = 7$. Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as halve the angle by folding doesn't change the problem.

Since the area of the triangle is equal to $\frac{1}{2}bh$,

$\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}$

Solve this to find that $s = \frac{5}{6}$.

By a similar method, $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$ is $408$.

See also

2006 AIME II (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
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