Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AMC 10B Problems/Problem 19: Difference between revisions

← Older editNewer edit →
Xantos C. Guin (talk | contribs)
editor
580 edits
m added link to previous and next problem
Xantos C. Guin (talk | contribs)
editor
580 edits
m added category
Line 46: Line 46:


*[[2006 AMC 10B Problems/Problem 20|Next Problem]]
*[[2006 AMC 10B Problems/Problem 20|Next Problem]]
[[Category:Introductory Geometry Problems]]

Revision as of 20:33, 3 August 2006

Problem

A circle of radius $2$ is centered at $O$. Square $OABC$ has side length $1$. Sides $AB$ and $CB$ are extended past $B$ to meet the circle at $D$ and $E$, respectively. What is the area of the shaded region in the figure, which is bounded by $BD$, $BE$, and the minor arc connecting $D$ and $E$?

$\mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3}) \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3}$

Solution

The shaded area is equivilant to the area of sector $DOE$, minus the area of triangle $DOE$ plus the area of triangle $DBE$.

Using the Pythagorean Theorem:

$(DA)^2=(CE)^2=2^2-1^2=3$

$DA=CE=\sqrt{3}$

Clearly, $DOA$ and $EOC$ are $30-60-90$ triangles with $\angle EOC = \angle DOA = 60^\circ$.

Since $OABC$ is a square, $\angle COA = 90^\circ$.

$\angle DOE$ can be found by doing some subtraction of angles.

$\angle COA - \angle DOA = \angle EOA$

$90^\circ - 60^\circ = \angle EOA = 30^\circ$

$\angle DOA - \angle EOA = \angle DOE$

$60^\circ - 30^\circ = \angle DOE = 30^\circ$

So, the area of sector $DOE$ is $\frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3}$.

The area of triangle $DOE$ is $\frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1$.

Since $AB=CB=1$ , $DB=ED=(\sqrt{3}-1)$.

So, the area of triangle $DBE$ is $\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}$.

Therefore, the shaded area is $(\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Rightarrow A$

See Also

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_19&oldid=9195"