Difference between revisions of "2004 AMC 12A Problems/Problem 24"
Claudeaops (talk | contribs) (Solution 2) |
Claudeaops (talk | contribs) (See Math Jam for the 2nd solution) |
||
Line 78: | Line 78: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ | A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ | ||
− | A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi(1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ | + | A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ |
A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ | A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ | ||
A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}</cmath> | A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}</cmath> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==See also== | ==See also== |
Revision as of 19:20, 10 December 2015
Problem 24
A plane contains points and with . Let be the union of all disks of radius in the plane that cover . What is the area of ?
Solution
As the red circles move about segment , they cover the area we are looking for. On the left side, the circle must move around pivoted on . On the right side, the circle must move pivoted on However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.
This egg-like shape is .
The area of the region can be found by dividing it into several sectors, namely
See also
2004 AMC 12A (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.