85: Paths

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It's true, I think about this all the time.

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My apartment
  1. 1=t
  3. 3=(t*√5)/3






Where do the equations come from to figure out #2 & #3 - can anybody derive it? (talk) (please sign your comments with ~~~~)

The equation #2 comes from the second route. t(1+√2)/3 is how far the second path takes the guy. If each block is a unit square, the diagonal to the corner is √2 while the next part is 1. The t/3 part is making it comparable to the first one (the first one is t despite it being 3 unit squares). Equation #3 is t√(5)/3. Plugging 1, 2, and √5 into wolfram|alpha for triangle side lengths makes it a right triange, so the √5 comes from the side length (assuming unit squares) while the t/3 makes it comparable to the first one.Mulan15262 (talk) 02:36, 1 December 2014 (UTC)

Interestingly enough, if the three sides are equal in time taken (20 seconds each), the time it would take for path #2 would be 20rt2 + 20, and path three would be roughly 40, which comes out to 60, 48.28, and 40 seconds by using very simple geometry. (talk) (please sign your comments with ~~~~)

Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.Flewk (talk) 17:10, 24 December 2015 (UTC)
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