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2006 AIME I Problems/Problem 7: Difference between revisions

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== Problem ==
== Problem ==
Find the number of [[ordered pair]]s of [[positive]] [[integer]]s <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a zero digit.
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>
 
[[Image:2006AimeA7.PNG]]


== Solution ==
== Solution ==
There are <math>\left\lfloor\frac{999}{10}\right\rfloor = 99</math> numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when <math>a</math> or <math>b</math> have a 0 in the tens digit, and since the equation is [[symmetry|symmetric]], we will just count when <math>a</math> has a 0 in the tens digit and multiply by 2 (notice that the only time both <math>a</math> and <math>b</math> can have a 0 in the tens digit is when they are divisible 100, which falls into the above category, so we do not have to worry about [[Principle of Inclusion-Exclusion|overcounting]]).  
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 <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>.
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.
 
Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
 
<cmath>
\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}
</cmath>
 
Solve this to find that <math>s = \frac{5}{6}</math>.


Excluding the numbers divisible by 100, which were counted already, there are <math>9</math> numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling <math>9 \cdot 9 = 81</math> such numbers; considering <math>b</math> also and we have <math>81 \cdot 2 = 162</math>. Therefore, there are <math>999 - (99 + 162) = 738</math> such ordered pairs.
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 ==
{{AIME box|year=2006|n=I|num-b=6|num-a=8}}
{{AIME box|year=2006|n=I|num-b=6|num-a=8}}


[[Category:Intermediate Combinatorics Problems]]
[[Category:Intermediate Geometry Problems]]

Revision as of 18:30, 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 $\mathcal{A}.$

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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