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

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== Solution ==
 
== Solution ==
−
Let:
+
Apex of the angle is not on the parallel lines.
−
The set of parallel lines...perpendicular to x-axis & cross x-axis at 0, 1, 2...
+
 
−
The base of area A is at x = 1.
+
Let...
−
One side of the angle be the x-axis.
+
*The set of parallel lines be
−
The other side be y = x-h...
+
:perpendicular to x-axis
−
as point of the angle isn't on parallel lines
+
:& cross x-axis at 0, 1, 2...
 +
*Base of area A be at x = 1; Lower base of Area D at x = 7.
 +
*One side of the angle be x-axis.
 +
*The other side be <math>y = x - h</math>
 +
<br>
 
Then...
 
Then...
 +
<br><br>
 +
As area of triangle = .5 base x height...
 +
<br><br>
 
<math>
 
<math>
−
Area C / Area B = 11 / 5
+
\frac{Area C}{Area B} = \frac{11}{5}
−
= [.5(5-h)^2 - .5(4-h)^2] / [.5(3-h)^2 - .5(2-h)^2]
+
= \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}
−
Thus h = 5/6
 
 
</math>
 
</math>
−
By similar method, D/A seems to be 408.
+
<br><br>
 +
h = <math>\frac{5}{6}</math>
 +
 
 +
By similar method, <math>\frac{Area D}{Area A}</math> seems to be 408.
  
 
== See also ==
 
== See also ==

Revision as of 19:37, 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 $\mathcal{A}.$



Solution

Apex of the angle is not on the parallel lines.

Let...

  • The set of parallel lines be
perpendicular to x-axis
& cross x-axis at 0, 1, 2...
  • Base of area A be at x = 1; Lower base of Area D at x = 7.
  • One side of the angle be x-axis.
  • The other side be $y = x - h$


Then...

As area of triangle = .5 base x height...

$\frac{Area C}{Area B} = \frac{11}{5} = \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}$

h = $\frac{5}{6}$

By similar method, $\frac{Area D}{Area A}$ seems to be 408.

See also

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