My question is as follows: a plane that has taken the shape of a pentagon is intersecting the skeleton of the cube. Or I guess we could think of it as a cross section. Points $M$ and $N$ were used to help make this construction. My question is that I want to prove

$$|AB| + |AE| \lessgtr |BC| + |CD| + |DE|$$

I put an $\lessgtr$ to represent either $>$ or $<$ because I do not recall which symbol it was. So, the idea of the problem is stating whether the sum of the 2 lengths of this pentagons sides I stated is either greater then or is less then the sum of the lengths of the other three sides in this case.

From the picture there is not much to use I feel like. One idea I had was to construct a line from $B$ to $E$ and then use the triangle inequality to represent triangle $BAE$. I was also making other triangles with this pentagon and then tried to compare all the triangle inequalities that I came up with. The problem is I do not think it got me anywhere that I could see.

Anyone provide any help? It would be greatly appreciated.

  • 1
    $\begingroup$ As M,N move toward their nearest cube corners, the pentagon approaches a triangle which seems to imply that the * should become a > sign, if it is to be one of the two. (That's not to say it's definitely >). I tried using coordinates, but it got messy since involved sums of radicals arose. $\endgroup$ – coffeemath Aug 26 '14 at 3:42

Suppose you lower the bottom face of the cube enough that it no longer intersects the plane. (That is, you replace the cube with a tall square prism.) Then the pentagon becomes a parallelogram $ABFE$, with $C$ and $D$ lying on the edges $BF$ and $FE$ respectively. Clearly $|AB|+|AE|=|BF|+|FE|$, and $|BF|+|FE|\ge|BC|+|CD|+|DE|$.

  • $\begingroup$ Wait how do you know $|BF| + |FE| \geq |BC| + |CD| + DE|$. I dont see the reasoning as to why it should be. $\endgroup$ – Nick Aug 26 '14 at 18:14
  • $\begingroup$ I don't see where your conclusion comes from though. $|BF| > |BC|$ and $|EF| > |ED|$ so we can say that $|BF| + |EF| > |BC| + |ED|$. How do you know that adding $|CD|$ to the right hand side will not change the inequality? I mean I get the triangle inequality you brought up but I dont see how it proves that by tacking on the $|CD|$ part the left hand side is still greater. $\endgroup$ – Nick Aug 26 '14 at 21:32
  • $\begingroup$ I mean I see that you have triangle $CDF$ leftover. And by the triangle inequality $|CD| \leq |CF| + |DF|$. But if its at most equal then, it would imply the result? $\endgroup$ – Nick Aug 26 '14 at 22:55
  • $\begingroup$ $|BF|+|FE| = \big(|BC|+|CF|\big)+\big(|FD|+|DE|\big) = |BC|+\big(|CF|+|FD|\big)+|DE| \ge |BC|+|CD|+|DE|$ $\endgroup$ – Rahul Aug 26 '14 at 23:45
  • $\begingroup$ thanks! it makes sense now. $\endgroup$ – Nick Aug 27 '14 at 0:58

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