Difference between revisions of "2006 AMC 12B Problems/Problem 24"
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== Solution == | == Solution == | ||
+ | We start out by solving the equality first. | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </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. | ||
+ | |||
+ | <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 <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. | ||
== 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 22:55, 16 April 2009
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Problem
Let be the set of all point
in the coordinate plane such that
and
. What is the area of the subset of
for which
Solution
We start out by solving the equality first.
We end up with three lines that matter:
,
, and
. We plot these lines below.
Note that by testing the point
, 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.
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 |