Two rectangular parallelepiped - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:10:03Z http://mathoverflow.net/feeds/question/79621 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79621/two-rectangular-parallelepiped Two rectangular parallelepiped heartwork 2011-10-31T13:00:27Z 2011-12-17T07:35:36Z <p>Prove that if we have two rectangular parallelepiped (cuboids) such that one of them is placed inside the other then the sum of the three lengths of the inner parallelepiped is at most the sum of the three lengths of the exterior parallelepiped. In 2 dimensions the problem is trivial. Does this hold in higher dimensions? There is a way to prove it in higher Euclidean dimensions?</p> http://mathoverflow.net/questions/79621/two-rectangular-parallelepiped/79627#79627 Answer by Vitali Kapovitch for Two rectangular parallelepiped Vitali Kapovitch 2011-10-31T13:39:37Z 2011-10-31T13:39:37Z <p>This is true for 3-dimensional parallelepipeds. Let the sides of the ambient parallelepiped be $A,B,C$ and of the inner one $a,b,c$. Then it's clear that $A^2+B^2+C^2\ge a^2+b^2+c^2$ since the diameter of the ambient body is obviously bigger. Also, it's easy to see that the there is a 1-Lipschitz map of $\mathbb R^3$ onto the smaller parallelepiped (just take the nearest point projection map). Being 1-Lipschitz this map does not increase 2-dimensional area and therefore, applying this to the surface of the bigger parallellepiped we get $2(AB+BC+AC)\ge 2(ab+bc+ac)$. Adding the above two inequalities we get $(A+B+C)^2\ge (a+b+c)^2$.</p> <p>I learned of this problem from <a href="http://dl.getdropbox.com/u/1577084/problems.pdf" rel="nofollow">Anton's Petrunin's list of exercises in orthodox geometry</a> (which has a lot of other cool problems btw). </p> <p>I don't know if the same holds in higher dimensions.</p> http://mathoverflow.net/questions/79621/two-rectangular-parallelepiped/79655#79655 Answer by thei for Two rectangular parallelepiped thei 2011-10-31T21:32:51Z 2011-10-31T22:05:20Z <p>This version works for all parallelepipeds, not only rectangular ones:</p> <p>If you replace each parallelepiped by all points that have distance at most $\varepsilon$ to a point in the parallelepiped, you can still place the smaller inside the bigger one. In particular, the smaller object has a smaller volume. </p> <p>We divide up the extended parallepipeds by extending the planes corresponding to the six faces. This gives the volume of the original solid in the center, parallelepipeds of height $\varepsilon$ on top of each face, partial cylinders (with slanted parallel ends) of radius $\varepsilon$ and length the corresponding edge, and partial spheres of radius $\varepsilon$ around each vertex.</p> <p>The partial spheres add up to exactly one whole sphere simply by translation. The partial cylinders corresponding to parallel edges add up to one whole cylinder by translation.</p> <p>Now let $\varepsilon$ tend to infinity (yes, really). The term with $\varepsilon^3$ comes from the sphere of radius $\varepsilon$ and does not depend on the parallelepiped at all, . The coefficient of $\varepsilon^2$ comes from the cylinders and is clearly the sum of the edges times some constant.</p> <p>So, for the inequality to be valid the sum of the edges of the smaller box must be smaller than the sum of the edges of the larger box.</p> <p>I don't see any problem with the generalization to higher dimension.</p> http://mathoverflow.net/questions/79621/two-rectangular-parallelepiped/83687#83687 Answer by heartwork for Two rectangular parallelepiped heartwork 2011-12-17T07:35:36Z 2011-12-17T07:35:36Z <p>What if we ask the same problem with tetrahedron instead of parallelepiped. Consider two tetrahedrons one placed inside the other. Prove that the sum of all 6 sides of inner tetrahedron is at most the sum of the 6 sides of exterior tetrahedron. This looks much harder!</p>