Ants Not on a Can!
Three ants start at the blue vertex of the bottom base of a unit anticube comprised of parallel unit square bases and unit equilateral triangular faces, and come back to it. All three ants walk at the same constant speed and take the shortest possible paths. Each of them also makes one complete lap.
- Ant arrives to the opposite midpoint of the equilateral triangle, then to the red midpoint, then to the opposite vertex and back to the blue point.
- Ant also arrives to the red midpoint and back to the starting point, but along the equilateral triangular faces at the minimum path length between the blue and the red points.
- Ant goes around the circumference of the square base.
Which ant finishes first?
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1 solution
Ant A's route :
* Uphill and downhill is the height of unit equilateral triangle each
* Highland is crossing over a unit square along two opposing sides
* Lowland is crossing under a unit square from one vertex to the opposite one.
A's distance
= 2 × √[1² - (1/2)²] + 1 + √[1² + 1²]
= 2 × 0.5√3 + 1 + √2
= √3 + 1 + √2
Ant B's route :
* Uphill and downhill in one big round trip surrounding the hill, touching the top level once and returning to the bottom level opposite vertex where it started
* Uphill and downhill half rounds are mirroring each other
* Each half round is crossing 4 adjacent unit equilateral triangles arrange alternately up and down from one acute angled base vertex to the opposite end's triangle's height-base perpendicular crosspoint.
B's distance
= 2 × √[(1 + 1)² + {1² - (1 / 2)²}]
= 2 × 0.5√19
= √19
Ant Y's route :
* Only surrounding the 'hill' on the very bottom level and returning to the bottom level original vertex where it started
* The hill's base is in a unit square shape
Y's distance
= 4 × 1
= 4
==> Comparing all three distances,
4 < 1 + √2 + √3 < √19
Answer : Ant Y finishes first.