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Points A $(0,1)$ and B $(1,0)$ start moving along the circumference of a unit circle with center $(0,0)$ in the same, positive (that is, counterclockwise) direction. Every minute, points A and B traverse arcs respectively of $60$° and $42$°. Visually:

                                                     enter image description here

Determine moments $t_1, \ldots, t_k,\ldots,$ such that at time $t_k$ points A and B coincide for the $k^\text{th}$ time.

I've been able to determine $t_1$, but cant seem to determine the next moment. I'll describe how I've gotten $t_1$ and hopefully you can suggest how to proceed (or if I'm doing it wrong, how to go about solving for all $t$). We're given the angular velocities of the two points:

  • $v_A = 60$° ($\pi\over 3$) per minute;
  • $v_B = 42$° ($7\pi\over 30$) per minute.

We also know the starting angles of the two points (shown also on the graph):

  • $d_A = {\pi \over 2}$ and $d_B = 0$.

To calculate $t_1$, we just have to solve the following equation for $t$:

$$\left({\pi\over 2} + {\pi \over 3}\cdot t \right)= \left(0 + {7\pi\over 30}\cdot t\right) \tag{$T_1$}.$$

Calculation yields the value of $5$ for $t$, so $\color{brown}{t_1 = 5}$ minutes.

Now, another basic calculation tells us that at minute $t_1$, points A and B form an angle of $5\pi\over 3$ with respect to $OB$ (sorry, forgot to label 'O' on the graph). So, I figured that to calculate moment $t_2$, it will suffice to solve the following equation for $t$ and add $t_1$ to it:

$$\left({5\pi\over 3} + {\pi \over 3}\cdot t \right)= \left({5\pi\over 3} + {7\pi\over 30}\cdot t\right) \tag{$T_2^?$}.$$

But, of course, the first summands are canceled out, leaving us with:

$$\left({\pi \over 3}\cdot t \right)= \left({7\pi\over 30}\cdot t\right) \tag{$T_2^?$}.$$

This solution is true only for $t=0$, so clearly something went wrong with my reasoning. (Of course, $t=0 + t_1 = t_1$, which is a moment of coincidence, but it's not the moment we're looking for). I would appreciate any help with the strategy I've taken or the way I should approach it instead.

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    $\begingroup$ We use degrees, since typing those $\pi$ is a nuisance, and degree intuition is better, and fractions are unpleasant. In $k$ minutes, $A$ travels $18$ degrees more than $B$. Since $B$ starts out $90$ degrees ahead, the first time of coincidence is $5$ minutes. For the next time of meeting, $A$ has to gain $360$ more degrees, which takes $20$ minutes, so $t_2=25$. And $20$ more minutes gets us to the third time of meeting, and so on. The $k$-th time of meeting is $5+20(k-1)$. $\endgroup$ Jun 2, 2014 at 20:53
  • $\begingroup$ Thank you André! $20k-15$ it is! $\endgroup$
    – Readingtao
    Jun 2, 2014 at 20:55
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    $\begingroup$ You are welcome. You can see that viewing the problem concretely makes the answer pop out. $\endgroup$ Jun 2, 2014 at 20:57
  • $\begingroup$ It does! Thank you all for your help. $\endgroup$
    – Readingtao
    Jun 2, 2014 at 20:58
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    $\begingroup$ Please note that I misread and interchanged $A$ and $B$. It is $15+20(k-1)$. $\endgroup$ Jun 2, 2014 at 21:29

3 Answers 3

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All your thinking so far is good. What you're missing is that the next time your points meet, A will have gone around the circle an extra time. So their angles won't be equal - $A$'s will be exactly $2\pi$ more than $B$'s, accounting for the extra lap. If you solve $$ \frac{\pi}{3} \cdot t = \frac{7\pi}{30} \cdot t + 2\pi, $$ you get $t= 20$. That tells you they'll meet again another 20 minutes later, at a total time of 25 minutes after the start. Can you guess from there what time the third meeting will be? Even better - can you explain why? :)

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  • $\begingroup$ I was just thinking about the 2pi. I had the stupid thought to divide by 2pi, but this makes sense. Thank you. $\endgroup$
    – Readingtao
    Jun 2, 2014 at 20:51
  • $\begingroup$ I'll think about you question and get back to you soon. $\endgroup$
    – Readingtao
    Jun 2, 2014 at 20:53
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An easy way to do this is to solve this equation $$42t\equiv 90+60t \bmod 360$$

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    $\begingroup$ Ah! That's the same as cool papa's answer, but with degrees. I shouldn't have complicated by converting to radians. Thank you. $\endgroup$
    – Readingtao
    Jun 2, 2014 at 20:52
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    $\begingroup$ The use of modular arithmetic helps a lot here too, that way, if you see a more convoluted problem in the future where you can't intuit the number of additional laps it will take, you don't have to worry. $\endgroup$ Jun 2, 2014 at 20:53
  • $\begingroup$ This doesn't lend itself to solving the equation very well. How would you go about solving for t? Just guess and check? $\endgroup$
    – j will
    Dec 11, 2014 at 23:18
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Keep in mind that when you're calculating distance around a circle in this way you have to mod it by $2\pi$. If, for example, A and B start both straight up and A makes a revolution every minute while B makes a revolution every 2 minutes, then they'll coincide after 2 minutes, though the distance traveled is very different.

So instead, just look at the difference in distance traveled. A travels a certain distance farther than B every minute. Every time that distance is equal to a multiple of $2\pi$ they will coincide.

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  • $\begingroup$ Thank you. That is helpful. $\endgroup$
    – Readingtao
    Jun 2, 2014 at 21:06

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