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

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== Problem ==
 
== Problem ==
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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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Find the number of ordered pairs of positive integers <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a zero digit.
  
 
== Solution ==
 
== Solution ==

Revision as of 14:29, 25 September 2007

Problem

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

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 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
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