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2006 AMC 10A Problems/Problem 15

Revision as of 11:02, 17 December 2021 by Dairyqueenxd (talk | contribs) (→‎Solution 2)

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 $603$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

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

Solution

[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 $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{\textbf{(D) } 47}$ meeting points.

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, we get $a=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. Then, from the first equation, we have $\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi$. 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 $\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=\boxed{\textbf{(D) }47}$.

~apsid

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

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