Difference between revisions of "2006 AMC 10A Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | {{image}} | + | {{image}}<center><asy> |
| + | /begin{figure} | ||
| + | unitsize(1cm); | ||
| + | draw((5,0){up}..{left}(0,5),red); | ||
| + | draw((-5,0){up}..{right}(0,5),red); | ||
| + | draw((5,0){down}..{left}(0,-5),red); | ||
| + | draw((-5,0){down}..{right}(0,-5),red); | ||
| + | draw((6,0){up}..{left}(0,6),blue); | ||
| + | draw((-6,0){up}..{right}(0,6),blue); | ||
| + | draw((6,0){down}..{left}(0,-6),blue); | ||
| + | draw((-6,0){down}..{right}(0,-6),blue); | ||
| + | \end{figure} | ||
| + | </center></asy> | ||
Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \Longrightarrow \mathrm{(D)}</math>. | Since <math>d = rt</math>, we note that Odell runs one lap in <math>\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}</math> minutes, while Kershaw also runs one lap in <math>\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}</math> minutes. They take the same amount of [[time]] to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47</math> meeting points <math> \Longrightarrow \mathrm{(D)}</math>. | ||
Revision as of 13:02, 27 July 2010
Problem
Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
/begin{figure}
unitsize(1cm);
draw((5,0){up}..{left}(0,5),red);
draw((-5,0){up}..{right}(0,5),red);
draw((5,0){down}..{left}(0,-5),red);
draw((-5,0){down}..{right}(0,-5),red);
draw((6,0){up}..{left}(0,6),blue);
draw((-6,0){up}..{right}(0,6),blue);
draw((6,0){down}..{left}(0,-6),blue);
draw((-6,0){down}..{right}(0,-6),blue);
\end{figure}
</center> (Error making remote request. Unknown error_msg)
Since 
, we note that Odell runs one lap in 
minutes, while Kershaw also runs one lap in 
minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are 
laps run by both, or 
meeting points 
.
See Also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 10 Problems and Solutions | ||
