Lap times

Clement had a good idea for tackling this problem which involved the combined velocities of the two cyclists. The clue to this method is to consider the relative velocity of the two cyclists (by subtracting their individual velocities), all you need to think about then is when the relative cyclist completes a lap. Andrei made a good attempt at this question and his solution is given below. By using relative velocities there is a much neater solution than the one given here. Perhaps you might like to think about this Andrei.

I used the following notations:

• $L$ - the length of the trajectory
• $t_1$ -the time first cyclist goes round the track once
• $t_2$ -the time in which the second cyclist goes round the track once
• $x$ - the distance from the start point on the circle (smaller than one circumference)
• $d_1$ - the distance travelled by the first cyclist
• $d_2$ - the distance travelled by the second cyclist
• $n_1$ - the number of complete circumferences made by the first cyclist
• $n_2$ - the number of complete circumferences made by the second cyclist
• $v_1$ - the speed of the first cyclist
• $v_2$ - the speed of the second cyclist
• $T$ - the time up to the moment when the two cyclists will meet the first time

First I observe that $n_1$ - $n_2$ = 1

This is because the cyclist meet first time when the first travelled one circumference more that the second. I supposed that the first has a higher speed.

I can write the following relations: $$d_1 = n_1\times L + x$$ $$d_2 = n_2\times L + x $$Now, I divide both members of equations (1) and (2) by L: $${d_1 \over L} ={n_1+ {x \over L} }$$ $${d_2 \over L} ={n_2+ {x \over L} }$$ I know that the length of the track is, for each situation, the speed multiplied by the time of travel of one circumference: $${L = v_1\times t_1} \Rightarrow {v_1 = {L \over t_1}}$$ $${L = v_2\times t_2} \Rightarrow {v_2 = {L \over t_2}}$$ Each cyclist travels the same time T : $${d_1 = v_1\times T}$$ $${d_2 = v_2\times T}$$ Now, I replace $v_1$ by ${L/t_1}$ and $v_2$ by ${L/t_2}$ and divide both members by L: $${d_1 = {L \over t_1}\times T} \Rightarrow {{d_1 \over L}= {T \over t_1}}$$ $${d_2 = {L \over t_2}\times T} \Rightarrow {{d_2 \over L}= {T \over t_2}}$$ I see that (3) = (9) and (4) = (10). So: $${n_1 + {x \over L}} = {T \over t_1}$$ $${n_2 + {x \over L}} = {T \over t_2}$$ Now, I subtract (12) from (11), and I obtain an equation for T: $${T({1 \over t_1} - {1 \over t_2})} = 1$$ Now I divide both members by T, and finally I find an expression for T in terms of t_1 and t_2: $${T = {{t_1t_2} \over {t_2 - t_1}}}$$ But $t_1$ and $t_2$ are known, so I can calculate T.

Now I substitute (14) into (11) and (12): $${t_2 \over {t_2 - t_1}} = {n_1 + {x \over L}}$$ $${t_1 \over {t_2 - t_1}} = {n_2 + {x \over L}}$$ I observe that ${x \over L}$ is smaller than 1, so $n_1$ and $n_2$ are the integer parts of ${t_2 \over {t_2 - t_1}}$ and ${t_1 \over {t_2 - t_1}}$ respectively, and ${x \over L}$ is the remaining fraction, that shows at what fraction of ${x \over L}$ the cyclists meet first time}.

If ${t_2 \over {t_2 - t_1}}$ and ${t_1 \over {t_2 - t_1}}$ are integers, the cyclists meet at the start line.

I look for the condition the cyclists meet at half circumference, i.e. when ${x \over L} = {1 \over 2}$ .

From (15) and (16) I find: $${t_2 \over {t_2 - t_1}} = {n_1 + {1 \over 2}}$$ $${t_1 \over {t_2 - t_1}} = {n_2 + {1 \over 2}}$$ where $n_1$, $n_2$ $\epsilon$ N .

I have to look for pairs ($t_1$, $t_2$) that fulfill the previous two relations. One possibility is $t_2$ - $t_1$ = 2 , and both $t_1$, $t_2$ - odd:

($t_1$ , $t_2$) = (3, 5), (5, 7), (7, 9),

or the previous pairs multiplied by an integer:

(6, 10), (9, 15),

(10, 14), (15, 21),

and so on.

We received the following solution from Guy from Ransom Everglades Middle School, Miami, Fl, USA, which differs from the original published solution and looks at some alternative ideas.

In order to look at the general case, without numbers, I make $R_1$ the number of seconds it takes for the first cyclist to complete a lap, and therefore his rate in laps per second is $\frac{1}{R_1}$. Also, The rate of the second cyclist is $\frac{1}{R_2}$ laps/sec. (Arbitrarily, I take $R_1$ to be the smaller number of seconds, and therefore generate the faster rate).

I now make $k$ the difference in the number of laps travelled by the two cyclists after $t$ seconds. Obviously, if they are to meet, then $k$ must be an integer. We then obtain the equation: $\frac{t}{R_1}=\frac{t}{R_2}+k \quad (1)$

We now substitute 72 for $R_1$and 80 for $R_2$and solve for $t$, eventually obtaining: $t=720k$

To get the first time (not counting the starting line) that the cyclists meet, we set $k=1$, and get that $t = 720$ seconds. And indeed, after 720 seconds, one cyclist has done 10 laps, and one has done 9, and they meet at the starting line.

We now do the same thing with $R_1=70$ and $R_2=85$, so we get the equation:

$\frac{t}{70} = \frac{t}{85} + 1$

$\frac{17t}{1190} = \frac{14t}{1190} + 1$

$\frac{3t}{1190} = 1$

$t = \frac{1190}{3}$

So, after $\frac{1190}{3}$ seconds, the first cyclist has travelled $5 \frac{2}{3}$ laps, and the second cyclist has travelled $4 \frac{2}{3}$ laps, so they meet at the $\frac{2}{3}$ lap mark.

Now we are looking for two lap times such that they meet for the first time at the halfway lap mark. To make things easier on ourselves, we solve for t in our original equation (1) with letters: $$t = \frac{k\times R_1\times R_2}{R_2 - R_1}$$

Making $k=1$ we get:

$t = \frac{R_1\times R_2}{R_2 - R_1}$

From this equation, we can obtain formulas for the number of laps each cyclist has done, by dividing it by their lap time.

$L_1 = \frac{R_2}{R_2 - R_1}$

$L_2 = \frac{R_1}{R_2 - R_1}$

We want these $L$'s to be of the form $I + \frac{1}{2}$, where $I$ is an integer. So:

$\frac{R_2}{R_2 - R_1}= I + \frac{1}{2}$

And, with some algebra:

$\frac{R_2}{R_1} = \frac{2I + 1}{2I - 1}$

This equation is satisfied by all $R$'s who are consecutive odd numbers, or multiples thereof; i.e. keep the same ratio.

For example: (3,5); (6,10); (7,9); (21,27); etc. All these pairs meet at the halfway lap mark.