Difference between revisions of "2021 WSMO Team Round Problems/Problem 5"
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==Problem== | ==Problem== | ||
Two runners are running at different speeds. The first runner runs at a consistent 12 miles per hour. The second runner runs at <math>t+4</math> miles per hour, where <math>t</math> is the number of hours that have passed. After <math>n</math> hours, the runners have run the same distance, where <math>n</math> is positive. Find <math>n</math>. | Two runners are running at different speeds. The first runner runs at a consistent 12 miles per hour. The second runner runs at <math>t+4</math> miles per hour, where <math>t</math> is the number of hours that have passed. After <math>n</math> hours, the runners have run the same distance, where <math>n</math> is positive. Find <math>n</math>. | ||
| + | |||
| + | ''Proposed by pinkpig'' | ||
==Solution== | ==Solution== | ||
| + | After <math>n</math> hours, the first runner has run <math>12n</math> miles. The second runner runs at <math>t+4</math> miles per hour, so his average speed over <math>n</math> hours is | ||
| + | <cmath>\frac{(n+4)+4}{2} = \frac{n+8}{2}.</cmath> | ||
| + | |||
| + | Thus, the second runner travels | ||
| + | <cmath>\frac{n+8}{2} \cdot n = \frac{n^2+8n}{2} \text{ miles}.</cmath> | ||
| + | |||
| + | Setting the distances equal: | ||
| + | <cmath>\frac{n^2+8n}{2} = 12n \Rightarrow n^2+8n = 24n \Rightarrow n^2 = 16n \Rightarrow n = \boxed{16}.</cmath> | ||
| + | ~pinkpig | ||
Latest revision as of 13:05, 9 September 2025
Problem
Two runners are running at different speeds. The first runner runs at a consistent 12 miles per hour. The second runner runs at 
miles per hour, where 
is the number of hours that have passed. After 
hours, the runners have run the same distance, where is positive. Find .
Proposed by pinkpig
Solution
After hours, the first runner has run 
miles. The second runner runs at miles per hour, so his average speed over hours is
![\[\frac{(n+4)+4}{2} = \frac{n+8}{2}.\]](http://latex.artofproblemsolving.com/b/7/a/b7a41abea84ad74bbc23a3ea0cca850ec766040b.png)
Thus, the second runner travels
![\[\frac{n+8}{2} \cdot n = \frac{n^2+8n}{2} \text{ miles}.\]](http://latex.artofproblemsolving.com/7/3/5/735e0fe3710a55e605b1e6a775d2d822f99b0a68.png)
Setting the distances equal:
![\[\frac{n^2+8n}{2} = 12n \Rightarrow n^2+8n = 24n \Rightarrow n^2 = 16n \Rightarrow n = \boxed{16}.\]](http://latex.artofproblemsolving.com/9/8/8/9880b5d6384b76030d2b0f6c79e567474c9eec60.png)
~pinkpig
