3 Race Runner Run in a Race Contestants A,B,C run in a race a total of 100 meters.A finishes the end point with B 10 meters behind and C 10 meters behind "B".They fight over the fairness of the race so decide to race again. A is now 10 meters behind start point, B on the start line, C 10 meters ahead of the start line. Who crosses the finish line first.
I think all cross finish at the same time. Is it correct?
 A: Let's call the length of the race $L$, and the time required by A to finish it $t$. Then $$v_A=\frac Lt\\v_B=\frac{L-10}t\\v_C=\frac{L-20}t$$
When they race for the second time, we want to calculate the time required by each contestant to get to the finish line:
$$t_A=\frac{L+10}{v_A}=\frac{L+10}L t\\t_B=\frac L{v_B}=\frac{L}{L-10}t\\t_C=\frac{L-10}{v_C}=\frac{L-10}{L-20}t$$
One can immediately see that the times might be different. For example, assuming that the track is long enough:
$$t_A-t_B=\left(\frac{L+10}L-\frac{L}{L-10}\right)t=\frac{-100}{L(L-10)}t<0$$
That means that A still arrives earlier than B.
Note Another way to think about this, for A it will take longer than $t$ to get to the finish line. In fact, after time $t$ all contestants will be 10 meters away from the finish line. Since A is faster than B, who is faster than C, the order stays the same.
A: The starting position for $B$ for a draw with $A$ can be found by comparing the ratios of the race lengths for $A$ and $B$:
$$\frac{\text{distanceA}}{\text{distanceB}}=\frac{\text{new distanceA}}{\text{new distanceB}}$$
$$\frac{L}{L-10}=\frac{L+10}{L+k}$$
$$L^2+kL=L^2-100$$
$$k=\frac{-100}{L}$$
As $k$ is always negative, $B$ must always start in front of the starting line in order to draw.
Similarly for $A$ and $C$.
