Paths
Paths
[Blueprint of a campus. Two buildings in the upper and lower left corners, respectively, and a rectangular lawn. A road encloses the lawn, another road traverses horizontally through the center of the lawn. The character is in the lower left and the upper right corner, where it says "my apartment".] [Dashed line 1, from the lower-left along the road to the top-left corner, then to the top-right corner.] 60 seconds [Dashed line 2, from the lower-left along the road up to the center crossroads, then diagonally over the lawn to the top-right corner.] 48 seconds (80%) [Dashed line 3, diagonally from the lower-left to the top-right corner.] 44.7 seconds (74%) My apartment 1=t 2=(t*(1+√2))/3 3=(t*√5)/3 When I'm walking, I worry a lot about the efficiency of my path.

This comic centers around the consideration of what is the shortest path available to a person traveling by foot. Cueball has to travel across a rectangular distance, which has an established path around the periphery. When Cueball follows these paths, he has to walk for 60 seconds. He realizes that by ignoring the paths and taking the desire lines from corner to corner, his route will be shorter, and he calculates that he could cut up to 26% of his time.  As a result, every time he has to travel this rectangle, he worries about the extra time taken as a result of following the path. There are downfalls to this plan, however. This is convenient for Cueball but probably not for the building owner, as many rectangular lawns have delicate decorations such as flowers on them.
[Blueprint of a campus. Two buildings in the upper and lower left corners, respectively, and a rectangular lawn. A road encloses the lawn, another road traverses horizontally through the center of the lawn. The character is in the lower left and the upper right corner, where it says "my apartment".] [Dashed line 1, from the lower-left along the road to the top-left corner, then to the top-right corner.] 60 seconds [Dashed line 2, from the lower-left along the road up to the center crossroads, then diagonally over the lawn to the top-right corner.] 48 seconds (80%) [Dashed line 3, diagonally from the lower-left to the top-right corner.] 44.7 seconds (74%) My apartment 1=t 2=(t*(1+√2))/3 3=(t*√5)/3 When I'm walking, I worry a lot about the efficiency of my path.
