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Problem 502

AMC 10/12, early questions

Geometry Difficulty 3.1 Prove it CEMC Fryer · Canada · 2018

In the diagrams shown, $ABCD$ represents a rectangular field. There are three flagpoles: $M$ on $BC$ , $P$ on $AD$ , and $Q$ on $CD$ . Paul runs along the path $A \to D \to C \to M \to A$ . Tyler runs along the path $A \to P \to Q \to C \to B \to A$ .

What is the length of $MA$ ?
What is the total distance that Tyler runs?
Paul and Tyler start running at the same time. Tyler runs at a speed of 145 m/min. Paul runs at a constant speed and finishes 1 minute after Tyler. Determine Paul’s speed, in m/min.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.