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2006 AMC 10A Problems/Problem 15

Revision as of 16:12, 4 September 2021 by Apsid (talk | contribs) (Added a solution using the angle between the starting point and first meeting point)


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?

$\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad$


[asy] 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); [/asy]

Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ 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 $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47$ meeting points $\Longrightarrow \mathrm{(D)}$.

Solution 2

We first find the amount of minutes, k, until Odell and Kershaw's next meeting. Let $a$ be the angle in radians between their starting point and the point where they first meet, measured counterclockwise. Since Kershaw has traveled 300k meters at this point and the circumference of his track is $120\pi$, $a=\frac{300k}{120\pi}\cdot 2\pi$. Similarly, $2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}$ since Odell has traveled 250k meters in the opposite direction and the circumference of his track is $100\pi$. Solving for a in the second equation, $a=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. Then, from the first equation, $\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. Solving for k, we get $k=\frac{\pi}{5}$. After k minutes, they are back at the same position, except rotated, so they will meet again in k minutes. So the total amount of meetings is $\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=47$. So, the answer is $\boxed{(D)}$.


See Also

2006 AMC 10A (ProblemsAnswer KeyResources)
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

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