Patrick started walking at a constant speed along a straight road from his school to the park. One hour after Patrick left, Tanya started running at a constant speed of 2 miles per hour faster than Patrick walked, following the same straight road from the school to the park. One hour after Tanya left, JosΓ© started bicycling at a constant speed of 7 miles per hour faster than Tanya ran, following the same straight road from the school to the park. All three people arrived at the park at the same time. The distance from the school to the park is nmβ miles, where m and n are relatively prime positive integers. Find m+n.
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Suppose that Patrick walks at a speed of v miles per hour and took t hours. Then, we can make the following table (using d=rt, where d is distance, r is speed, and t is time):
PersonPatrickTanyaJoseΛβVelocity (v)vv+2v+9βTime (t)ttβ1tβ2βDistance (D)vt(v+2)(tβ1)(v+9)(tβ2)ββ
Note that these distances must all be equal. Equating Patrick and Tanya's equations gives
vt=(v+2)(tβ1)=vt+2tβvβ2
so 2tβvβ2=0, or v=2tβ2. Equating Patrick and JosΓ©'s equations gives
vt=(v+9)(tβ2)=vt+9tβ2vβ18
so 9tβ2vβ18=0, or v=4.5tβ9. Equating these gives
2tβ2=4.5tβ9βΉ2.5t=7βΉt=2.8
Thus, we get v=2tβ2=2β
2.8β2=3.6, and the distance travelled is
d=vt=25252β
for an answer of m+n=252+25=277β.
The problems on this page are the property of the MAA's American Mathematics Competitions