, 2018

In the first diagram (Paul’s Path), $\triangle ABM$ is right-angled at $B$. By the Pythagorean Theorem, $MA^2=AB^2+BM^2$ or $MA^2=105^2+100^2=21\,025$ or $MA= \sqrt{21\,025}=145$ m (since $MA>0$). In the second diagram (Tyler’s Path), $AD=BC=200$ m and $DC=AB=105$ m (since $ABCD$ is a rectangle). Thus $PD=AD-AP=200-140=60$ m. Also, $DQ=DC-QC=105-60=45$ m. Since $\triangle PDQ$ is right-angled at $D$, using the Pythagorean Theorem, we get$PQ^2=PD^2+DQ^2$ or $PQ^2=60^2+45^2=5625$ or $PQ= \sqrt{5625}=75$ m (since $PQ>0$). The total distance that Tyler runs is $$\text{AP+PQ+QC+CB+BA=140+75+60+200+105=580 m .}$$ The total distance that Paul runs is $$\text{AD+DC+CM+MA=200+105+(200-100)+145=550 m .}$$ Tyler runs at a speed of 145 m/min, and so it takes Tyler $580\div145=4$ min to finish his path. Paul begins at the same time as Tyler and finishes his path 1 minute after Tyler, and so Paul takes $4+1=5$ min to finish his path. In this time, Paul runs a total distance of 550 m and so Paul’s speed is$550\div5=110$ m/min.