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

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{{empty}}
 
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== Problem ==
 
== Problem ==
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Let <math>S</math> be the set of all point <math>(x,y)</math> in the coordinate plane such that <math>0 \le x \le \frac{\pi}{2}</math> and <math>0 \le y \le \frac{\pi}{2}</math>.  What is the area of the subset of <math>S</math> for which
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Let <math>S</math> be the set of all point <math>(x,y)</math> in the [[coordinate plane]] such that <math>0 \le x \le \frac{\pi}{2}</math> and <math>0 \le y \le \frac{\pi}{2}</math>.  What is the area of the subset of <math>S</math> for which <cmath>
−
 
 
−
<cmath>
 
 
\sin^2x-\sin x \sin y + \sin^2y \le \frac34?
 
\sin^2x-\sin x \sin y + \sin^2y \le \frac34?
 
</cmath>
 
</cmath>
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\sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\
 
\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{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\
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\sin x &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\
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&= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\
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\sin x &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\
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&= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\
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\sin x &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\
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&= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\
 
\sin x &= \sin (y \pm \frac{\pi}{3})
 
\sin x &= \sin (y \pm \frac{\pi}{3})
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
 
We end up with three lines that matter: <math>x = y + \frac\pi3</math>, <math>x = y - \frac\pi3</math>, and <math>x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y</math>.  We plot these lines below.
 
We end up with three lines that matter: <math>x = y + \frac\pi3</math>, <math>x = y - \frac\pi3</math>, and <math>x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y</math>.  We plot these lines below.
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<asy>
 
<asy>
 
size(5cm);
 
size(5cm);
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MP("\frac{\pi}{2}", (0,3), plain.W);
 
MP("\frac{\pi}{2}", (0,3), plain.W);
 
</asy>
 
</asy>
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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 we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.)
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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 square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.)
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<cmath>
 
<cmath>
 
\begin{align*}
 
\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 \\
 
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 ( \frac14 - \frac1{36} - \frac1{18}\right ) \\
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&= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\frac{\pi^2}{6}}
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&= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\text{(C)}\ \frac{\pi^2}{6}}
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
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==Solution 2==
 +
We can write the given equation as <cmath>\sin^3x + \sin^3y \le \frac{3}{4}(\sin x + \sin y).</cmath> Note that when <math>x = 0</math>, we have <math>\sin y \le \frac{\sqrt{3}}{2}</math> which implies <math>y \le \frac{\pi}{3}</math>. Similary we have <math>x \le \frac{\pi}{3}</math> when <math>y = 0</math>. Then we see what happens at <math>x = \frac{\pi}{2}</math>. Clearly at <math>x = \frac{\pi}{2}</math>, we see that <math>y \le \frac{\pi}{6}</math>. By symmetry we have <math>x \le \frac{\pi}{6}</math> when <math>y = \frac{\pi}{2}</math>. So we get a graph like
 +
<asy>
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size(5cm);
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D((0,0)--(3,0)--(3,3)--(0,3)--cycle);
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D((1,-0.1)--(1,0.1));
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D((2,-0.1)--(2,0.1));
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D((-0.1,1)--(0.1,1));
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D((-0.1,2)--(0.1,2));
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D((2,0)--(3,1)--(1,3)--(0,2));
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MP("\frac{\pi}{6}", (1,0), plain.S);
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MP("\frac{\pi}{3}", (2,0), plain.S);
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MP("\frac{\pi}{2}", (3,0), plain.S);
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MP("\frac{\pi}{6}", (0,1), plain.W);
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MP("\frac{\pi}{3}", (0,2), plain.W);
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MP("\frac{\pi}{2}", (0,3), plain.W);
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</asy>
 +
Thus the area of what we are interested in is <cmath>\frac{\pi^2}{4} - (\frac{\pi^2}{18} + \frac{\pi^2}{36}) = \frac{\pi^2}{6}.</cmath>
 +
~coolmath_2018
 
== 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}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:45, 24 October 2021

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} \\ &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ &= \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 square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.) \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{\text{(C)}\ \frac{\pi^2}{6}} \end{align*}

Solution 2

We can write the given equation as \[\sin^3x + \sin^3y \le \frac{3}{4}(\sin x + \sin y).\] Note that when $x = 0$, we have $\sin y \le \frac{\sqrt{3}}{2}$ which implies $y \le \frac{\pi}{3}$. Similary we have $x \le \frac{\pi}{3}$ when $y = 0$. Then we see what happens at $x = \frac{\pi}{2}$. Clearly at $x = \frac{\pi}{2}$, we see that $y \le \frac{\pi}{6}$. By symmetry we have $x \le \frac{\pi}{6}$ when $y = \frac{\pi}{2}$. So we get a graph like [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] Thus the area of what we are interested in is \[\frac{\pi^2}{4} - (\frac{\pi^2}{18} + \frac{\pi^2}{36}) = \frac{\pi^2}{6}.\] ~coolmath_2018

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 12 Problems and Solutions

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