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2006 AIME I Problems/Problem 7: Difference between revisions

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Line 13: Line 13:
:perpendicular to x-axis
:perpendicular to x-axis
:& cross x-axis at 0, 1, 2...
:& cross x-axis at 0, 1, 2...
*Base of area A be at x = 1; Lower base of Area D at x = 7.
*Base of Region <math>\mathcal{A}</math> be at <math>x = 1</math>; Lower base of Region <math>\mathcal{B}</math> at <math>x = 7</math>
*One side of the angle be x-axis.
*One side of the angle be x-axis.
*The other side be <math>y = x - h</math>
*The other side be <math>y = x - h</math>
Line 22: Line 22:
<br><br>
<br><br>
<math>
<math>
\frac{Area C}{Area 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{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}
</math>
</math>
<br><br>
<br><br>
h = <math>\frac{5}{6}</math>
<math>h = \frac{5}{6}</math>


By similar method, <math>\frac{Area D}{Area A}</math> seems to be 408.
By similar method, <math>\frac{Region \mathcal{D}}{Region \mathcal{A}}</math> seems to be 408.


== See also ==
== See also ==

Revision as of 18:41, 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 Region $\mathcal{A}$ be at $x = 1$; Lower base of Region $\mathcal{B}$ 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...

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

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

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

See also

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