It is given a rectangular parallelepiped $3\times4\times5$. Which are the farthest points from a given vertex, provided one can only walk on the surface of the parallelepiped?
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$\begingroup$ Which given vertex? Or are you after 8 answers (with some repeated)? $\endgroup$– AliCommented Jul 18, 2016 at 13:08
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$\begingroup$ Are you looking for the shortest distance between two vertices that are the furthest apart in 3-space? If not, what's to stop an answer from walking across the same surface multiple times in order to extend the length of the path? $\endgroup$– Ian MacDonaldCommented Jul 18, 2016 at 14:14
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1$\begingroup$ I am looking for the shortest distance between two points on the surface of a parallelepiped (provided that one of those points is one vertex, and that you can only move on the surface of the parallelepiped). $\endgroup$– zarCommented Jul 18, 2016 at 15:12
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1$\begingroup$ @Ali Through mirroring and rotations, all vertices would be considered equivalent don't you think? $\endgroup$– TreninCommented Jul 18, 2016 at 17:01
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2$\begingroup$ I say please re-open: this puzzle really is puzzle-like in nature, going by the 3rd of the 3 bullet-points xnor listed at "So, what makes something a math puzzle rather than math problem? I think there's a few features." in the answer here: meta.puzzling.stackexchange.com/questions/2783/… $\endgroup$– Rosie FCommented Jul 23, 2016 at 7:39
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1 Answer
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Maybe I got the terms wrong here but I think :
The farthest point in a rectangular parallelpiped from a vertex is the vertex at the other end of a space diagonal that goes through the given vertex.
The distance is $\sqrt{a^2+b^2+c^2}$
So in this case $\sqrt{3^2+4^2+5^2}$ which is about $7.07$
Let the cuboid be $x=0,\dots,5; y=0,\dots,4; z=0,\dots,3$. And let the starting vertex be $(0,0,0)$.
It might be thought that the furthest point is the opposite vertex $(5,4,3)$. The shortest route to that vertex crosses an edge of length $5$, and its length is $\sqrt{(4+3)^2+5^2}=\sqrt{74}$.
However, there is a $z_0=2\frac{10}{13}$ where, for $z\geqslant z_0$, the distance to $(5,4,z)$ by the shortest route is $\sqrt{(3+y)^2+(8-z)^2}$. For $z=z_0$, this distance $= \sqrt{76 \frac{61}{169}} \sim 8.7384750812$. (So I think this puzzle really does count as a puzzle because its correct answer is different from the one you might at first think.)
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2$\begingroup$ @Lordofdark. You got too much spare time and a wild imagination. 707...I mean LOL $\endgroup$– MariusCommented Jul 18, 2016 at 13:20
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$\begingroup$ I'm sorry, I left an important information in the keyboard: you are allowed to only walk on the surface of the parallelepiped. I will edit the question. $\endgroup$– zarCommented Jul 18, 2016 at 14:08