A Different Version of the Cyclists' Problem

For all who solved an infinite GP for this problem, and for all those who didn't.

$\bullet$ The total time taken by $A$ and $B$ to meet is their distance divided by their relative speed. Thus, the total time for which $A$ and $B$ cycle, is $\frac{100}{20+30}$ , that is, $2$ seconds. (LOL)

$\bullet$ In the whole of the $2$ seconds, the fly would have been flying to and fro between $A$ and $B$ . Thus, in all, the fly, with a constant speed of $39$ , would have travelled a distance of $39 \times 2 = 78$ metres.

$\bullet$ Of this $78$ meters, some of it would have been from $A$ to $B$ and some of it from $B$ to $A$ . Let's call them $X$ and $Y$ respectively, such that $X+Y=78$ .

$\bullet$ In the end, $A$ would have travelled a distance of $20\times 2=40$ metres. This means that $A$ and $B$ meet in the end at $40$ metres from $A$ .

$\bullet$ This is amazing, that, in the end, the fly is at $40$ metres from $A$ . So, since it went $X$ metres from $A$ to $B$ , and $Y$ metres from $B$ to $A$ , we get $X-Y=40$ .

$\bullet$ Kudos! We have $X+Y=78$ and $X-Y=40$ , giving us $X=59$ and $Y=19$ .

Thus the total distance the fly flew from $B$ to $A$ is $19$ metres. QED.

Cool na?

Variant of Satvik's solution:

• Let $t$ be the time the fly was flying. $20t +30t = 100 \implies t= 2$
• Hence, the collision happens 40 meters from A.
• Let $t$ be the time the fly was flying from A to B. $39t + (-39)(2-t) = 40$
• Solve for $t$ and enter then answer as $39(2-t)$

Total distance S = 100m. ${ t }_{ 1 }=\frac { 100 }{ 69 }$

Distance traveled by A and B = 50t.

Thus ${ S }_{ left,1 }=\quad S-50t\quad =S\frac { 19 }{ 69 } =\frac {1900}{69}$

and fly travels ${ S }_{ 1 } = \frac {3900}{69}$

Now, ${ t }_{ 2 }=\frac { { S }_{ left,1 } }{ 59 }$

and fly travels ${ S }_{ 2 }=\frac { 3900 }{ 69 } \times \frac { 19 }{ 59 }$

again, ${ S }_{ left,2 }=\quad S-50t\quad =S\frac { 9 }{ 59 } =\frac { 1900 }{ 69 } \times \frac { 9 }{ 59 }$

${ t }_{ 3 }=\frac { { S }_{ left,2 } }{ 69 }$

and fly travels ${ S }_{ 3 }=\frac { 3900 }{ 69 } \times \frac { 19 }{ 59 } \times \frac {9}{69}$

once it is multiplied by $\frac { 19 }{ 59 }$ and once by $\frac { 9 }{ 69 }$ then again by $\frac { 19 }{ 59 }$

thus, the distance from B to A is: ${ S }_{ 2 }+{ S }_{ 4 }+{ S }_{ 6 }...$

$= { S }_{ 1 } \times \frac { 19 }{ 59 } + { S }_{ 1 } \times \frac { 19 }{ 59 } \times \frac {9}{69} \times \frac { 19 }{ 59 }$ and so on

applying sum of infinite GP,

net distance = $\frac { { S }_{ 1 }\times \frac { 19 }{ 59 } }{ 1-\frac { 19\times 9 }{ 59\times 69 } } =\frac { { S }_{ 1 }\times \frac { 19 }{ 59 } }{ \frac { 59\times 69-19\times 9 }{ 59\times 69 } }$

$=\frac { { S }_{ 1 }\times 19 }{ \frac { 3900 }{ 69 } }$

where ${ S }_{ 1 } = \frac {3900}{69}.$

thus, distance = $\boxed { 19 }$ .