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Difference between revisions of "2006 AMC 12B Problems/Problem 24"

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</asy>
 
</asy>
 
Note that by testing the point <math>(\pi/6,\pi/6)</math>, we can see that we want the area of the pentagon.  We can calculate that by calculating the area of the sqaure and then subtracting the area of the 3 triangles.
 
Note that by testing the point <math>(\pi/6,\pi/6)</math>, we can see that we want the area of the pentagon.  We can calculate that by calculating the area of the sqaure and then subtracting the area of the 3 triangles.
 +
 +
<cmath>
 +
\begin{align*}
 +
A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\
 +
&= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\
 +
&= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\frac{\pi^2}{6}}
 +
\end{align*}
 +
</cmath>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2006|ab=B|num-b=23|num-a=25}}

Revision as of 23:01, 16 April 2009

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Problem

Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \le x \le \frac{\pi}{2}$ and $0 \le y \le \frac{\pi}{2}$. What is the area of the subset of $S$ for which

\[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\]

$\mathrm{(A)}\ \dfrac{\pi^2}{9} \qquad \mathrm{(B)}\ \dfrac{\pi^2}{8} \qquad \mathrm{(C)}\ \dfrac{\pi^2}{6} \qquad \mathrm{(D)}\ \dfrac{3\pi^2}{16} \qquad \mathrm{(E)}\ \dfrac{2\pi^2}{9}$

Solution

We start out by solving the equality first. \begin{align*} \sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\ \sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\ \sin x &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ \sin x &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ \sin x &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\ \sin x &= \sin (y \pm \frac{\pi}{3}) \end{align*} We end up with three lines that matter: $x = y + \frac\pi3$, $x = y - \frac\pi3$, and $x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y$. We plot these lines below.

[asy] size(5cm); D((0,0)--(3,0)--(3,3)--(0,3)--cycle); D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1)); D((-0.1,1)--(0.1,1)); D((-0.1,2)--(0.1,2)); D((2,0)--(3,1)--(1,3)--(0,2)); MP("\frac{\pi}{6}", (1,0), plain.S); MP("\frac{\pi}{3}", (2,0), plain.S); MP("\frac{\pi}{2}", (3,0), plain.S); MP("\frac{\pi}{6}", (0,1), plain.W); MP("\frac{\pi}{3}", (0,2), plain.W); MP("\frac{\pi}{2}", (0,3), plain.W); [/asy] Note that by testing the point $(\pi/6,\pi/6)$, we can see that we want the area of the pentagon. We can calculate that by calculating the area of the sqaure and then subtracting the area of the 3 triangles.

\begin{align*} A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\ &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\ &= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\frac{\pi^2}{6}} \end{align*}

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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