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

m (→‎See also)
(some cleanup (haven't verified my work), + box)
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== Solution ==
 
== Solution ==
−
Apex of the angle is not on the parallel lines.
+
Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].
  
−
Let...
+
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>. The x-axis [[bisect]]s the angle; the top side will be <math>y = x - h</math>.
−
*The set of parallel lines...
+
 
−
:be perpendicular to x-axis
+
Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
−
:& cross x-axis at 0, 1, 2...
+
 
−
*Base of region <math>\mathcal{A}</math> be at <math>x = 1</math>; bigger base of region <math>\mathcal{D}</math> at <math>x = 7</math>
+
:<math>
−
*One side of the angle be x-axis.
 
−
*The other side be <math>y = x - h</math>
 
−
<br>
 
−
Then...
 
−
<br><br>
 
−
As area of triangle =.5 base x height...
 
−
<br><br>
 
−
<math>
 
 
\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5}
 
\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5}
−
= \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}
+
= \frac{\frac 12(5-h)^2 - \frac 12(4-h)^2}{\frac 12(3-h)^2 - \frac12(2-h)^2}
 
</math>
 
</math>
−
<br><br>
 
−
<math>h = \frac{5}{6}</math>
 
  
−
By similar method, <math>\frac{Region \mathcal{D}}{Region \mathcal{A}}</math> seems to be 408.
+
Solve this to find that <math>h = \frac{5}{6}</math>.
 +
 
 +
By a similar method, <math>\frac{Region \mathcal{D}}{Region \mathcal{A}} = \frac{\frac 12(7-h)^2 - \frac 12(6-h)^2}{\frac 12(1-h)^2}</math> is <math>408</math>.
  
 
== See also ==
 
== See also ==
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* [[2006 AIME I Problems/Problem 6 | Previous problem]]
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{{AIME box|year=2006|n=I|num-b=6|num-a=8}}
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* [[2006 AIME I Problems/Problem 8 | Next problem]]
 
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* [[2006 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 20:14, 11 March 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 $x = 7$. The x-axis bisects the angle; the top side will be $y = x - h$.

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

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

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

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