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Ant highway with energy loss (Posted on 2018-08-24) Difficulty: 5 of 5
An ant starts at the origin of the coordinate plane. The ant only has enough energy to walk for 10 units along either the x- or y-axis before stopping. These lines are the two ant highways.

However, if it goes off the highway, more energy is needed for the same distance. That is, the further the ant deviates, the greater the energy consumption, as represented by the factor 1+0.4d, where d is the distance to the nearest highway. For example, if the ant is 1 unit away from the nearest highway, it uses 1.4 times the energy on the highway to walk 1 meter.

What is the area of the set of points the ant can reach before getting tired and stopping?

No Solution Yet Submitted by Jer    
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possible analytical approach (thoughts) | Comment 6 of 9 |
I am still pondering how to do this analytically.

I believe I have 2 correct premises and possible way forward.

1) the formula I gave for straight line distance obtainable by Ant going off-road in a straight line is correct: I verified it with a simulation. It can be generalized for any amount of energy Ant has remaining when Ant goes off-road by scaling the formula by E_remain/10.

2) The symmetry of the X and Y highways will result in 45 degree wedges joined in mirror symmetry. I.e. solving for one wedge suffices.

Now the problem I face is the case where Ant has a 2 component trip with the first segment on-road and the second off-road. This likely (it can be shown true or false pretty easily) can get Ant to a further radius than an all off-road trip. So how to compute these larger radii? Here is the plan: for a given theta between 0 and 45, find the on-road distance at which Ant should depart from the highway to reach the maximum radius on the vector defined by theta. The resulting r's for each theta give the Ant's maximum extent in the wedge. 

Whether this maximization is tractable analytically remains to be seen.

  Posted by Steven Lord on 2018-08-26 10:48:16
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