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| − | ==Problem==
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| − | A straight one-mile stretch of highway, <math>40</math> feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at <math>5</math> miles per hour, how many hours will it take to cover the one-mile stretch?
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| − | Note: <math>1</math> mile = <math>5280</math> feet
| + | ==Solution== |
| | + | IF YOU READ THIS YOU ARE GAY |
| | + | ===Solution 1=== |
| | + | GYATTTTTT |
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| − | <asy> size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,1)--(5.5,1)); D((-1.5,0)--(5.5,0),dashed); D((-1.5,-1)--(5.5,-1)); </asy>
| + | ==See Also== |
| − | <math> \textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3} </math>
| + | {{AMC8 box|year=2014|num-b=24|after=Last Problem}} |
| − | | + | {{MAA Notice}} |
| − | ==Solution==
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| − | There are two possible interpretations of the problem: that the road as a whole is <math>40</math> feet wide, or that each lane is <math>40</math> feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be <math>20</math> feet wide, so Robert must be riding his bike in semicircles with radius <math>20</math> feet and diameter <math>40</math> feet. Since the road is <math>5280</math> feet long, over the whole mile, Robert rides <math>\frac{5280}{40} =132</math> semicircles in total. Were the semicircles full circles, their circumference woul be <math>2\pi\cdot 20=40\pi</math> feet; as it is, the circumference of each is half that, or <math>20\pi</math> feet. Therefore, over the stretch of highway, Robert rides a total of <math>132\cdot 20\pi =2640\pi</math> feet, equivalent to <math>\frac{\pi}{2}</math> miles. Robert rides at 5 miles per hour, so divide the <math>\frac{\pi}{2}</math> miles by <math>5</math> mph (because t = d/r time = distance/rate) to arrive at <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
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