Difference between revisions of "2004 AMC 12A Problems/Problem 24"
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Claudeaops (talk | contribs) (Solution 2) |
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<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> | ||
+ | |||
+ | ==Alternate Solution== | ||
+ | Credit to the Math Jam for the solution. | ||
+ | (Diagram needed) | ||
+ | |||
+ | Let <math>\triangle ABX</math> be an equilateral triangle "above" segment <math>AB</math>. Now imagine having a "moving" circle of radius 1 pegged through point <math>B</math> and sliding it around as much as possible. The center of the circle must trace an arc of a circle <math>\omega_1</math> with radius 1 centered at <math>B</math>. Note that <math>\omega_1</math> passes through <math>X</math>. In addition, | ||
+ | the point on the "moving" circle farthest from <math>B</math> must always be 2 units away from <math>B</math>, so it traces out an arc of a circle <math>\omega_2</math> with radius 2 centered at <math>B</math>. Now as we move around the circle the center will move around arc <math>AX</math> of <math>\omega_1</math>. Let <math>W</math> be the intersection of <math>AB</math> with <math>\omega_2</math> such that <math>WA<WB</math>. Let <math>M</math> be the midpoint of <math>AB</math>, and <math>Z</math> be the intersection of <math>BX</math> with <math>\omega_2</math> such that <math>ZX<ZB</math>. | ||
+ | Now the point on the "moving" circle farthest from <math>B</math> traces out arc <math>WZ</math> on <math>\omega_2</math>. Finally, one of the extreme positions of the moving circle is when the circle is centered at <math>X</math>. Let the circle centered at <math>X</math> with radius <math>1</math> be called <math>\omega_3</math>. Let <math>Y</math> be the intersection of <math>MX</math> and <math>\omega_3</math> such that <math>YX<YM</math>. Once the center of the moving circle has moved from <math>A</math> to <math>X</math>, the moving circle will have finished one-fourth of its total motion. Thus, <math>S</math> is made up of 4 quadrants (consult the diagram in solution 1). Focus on the upper left quadrant of <math>S</math>. The upper left quadrant is bounded by arc <math>WZ</math> of <math>\omega_2</math>, arc <math>ZY</math> of <math>\omega_3</math>, segments <math>YM</math> and <math>WM</math>. The area of this region is equal to | ||
+ | the area of sector <math>WBZ</math> of <math>\omega_2</math> plus the area of sector <math>ZXY</math> of <math>\omega_3</math> minus the area of <math>\triangle BMX</math>. <math>\angle WBZ=60, \angle ZXY=\angle BXM=30</math>, and <math>\triangle BXM</math> is a 30-60-90 triangle. Routine calculations yield answer choice <math>\textbf {(C)}</math>. | ||
==See also== | ==See also== |
Revision as of 23:34, 6 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
Alternate Solution
Credit to the Math Jam for the solution. (Diagram needed)
Let be an equilateral triangle "above" segment . Now imagine having a "moving" circle of radius 1 pegged through point and sliding it around as much as possible. The center of the circle must trace an arc of a circle with radius 1 centered at . Note that passes through . In addition, the point on the "moving" circle farthest from must always be 2 units away from , so it traces out an arc of a circle with radius 2 centered at . Now as we move around the circle the center will move around arc of . Let be the intersection of with such that . Let be the midpoint of , and be the intersection of with such that . Now the point on the "moving" circle farthest from traces out arc on . Finally, one of the extreme positions of the moving circle is when the circle is centered at . Let the circle centered at with radius be called . Let be the intersection of and such that . Once the center of the moving circle has moved from to , the moving circle will have finished one-fourth of its total motion. Thus, is made up of 4 quadrants (consult the diagram in solution 1). Focus on the upper left quadrant of . The upper left quadrant is bounded by arc of , arc of , segments and . The area of this region is equal to the area of sector of plus the area of sector of minus the area of . , and is a 30-60-90 triangle. Routine calculations yield answer choice .
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 |
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