Let PDevyn represent Devyns position with respect to time, and let PRiley represent Rileys position with respect to time. Let tstart be the starting time of the race, and tfinish be the end of the race. Setf$($t$)=$PDevyn$($t$)$PRiley$($t$).$Note, we may assume that PDevyn and PRiley$($t$)$ are continuous on $[$tstart$,$tfinish$]$ and that they are differentiable on $($tstart$,$tfinish$).$ Hence the same is true for f$.$ Since both runners start and finish at the same place,f$($tstart$)=$PDevyn$($tstart$)$PRiley$($tstart$)$f$($tfinish$)=$PDevyn$($tfinish$)$PRiley$($tfinish$)=$and$=.$In fact, this shows us that the average rate of change off$($t$)=$PDevyn$($t$)$PRiley$($t$)$on$[$tstart$,$tfinish$]$is $.$ Hence by the mean value theorem, there is a point ctstartctfinishwith f$($c$)=.$ However,$=$f$($c$)=$PDevyn$($c$)$PRiley$($c$).$Hence at c$,$PDevyn$($c$)=$PRiley$($c$),$this means that there was a time when they were running at exactly the same velocity.