## A Tetrahedron

You are given six sticks of integral lengths 1, 2, 3, 4, 5 and 6. Using these sticks, can you make a tetrahedron (4-sided, 3-D figure, with a triangle on each side)? If so, show how. If not, replace any one of the sticks with the smallest stick of integral length greater than 6 that allows you to build such a tetrahedron and show how it can be done.

#### Source: Based upon the 9/16/96 problem at the SMSU Problem Corner.

Ken's Solution:
A tetrahedron cannot be made with the sticks as given. To see this, consider the stick of length 1. Whatever side it is on has side-lengths of a>b>1. But since all lengths are integers, a>=b+1, so a triangle cannot be made with one side of length 1.

If we substitute a length-7 stick for the length-1 stick, there are five unique ways to make the tetrahedron. The constraints are that the lengths on a side with the length-2 stick must differ by only 1, and (3,4) cannot be on the same side with the 7.

[Assume you're looking down/up at the figure. These solutions assume rotations and reflections are the same:]
 ``` / \ / | \ 2 3 5 / | \ / / \ \ / 4 7 \ /_/____6____\_\ ``` ``` / \ / | \ 2 3 6 / | \ / / \ \ / 4 7 \ /_/____5____\_\ ``` ``` / \ / | \ 2 3 6 / | \ / / \ \ / 4 5 \ /_/____7____\_\ ``` ``` / \ / | \ 2 3 7 / | \ / / \ \ / 4 5 \ /_/____6____\_\ ``` ``` / \ / | \ 2 4 6 / | \ / / \ \ / 5 3 \ /_/____7____\_\ ```
Rick Simineo provided the following solution:
To solve this, one simple Geometric principle must be known: No side of a triangle may be longer than the other two sides combined. Now, then, about the tetrahedron:

Each side is part of two triangles. Starting with the largest integer, one can find ONLY two triangles that can coexist: 6,5,2 and 6,4,3. This leaves the 1, which is not long enough to make two triangles, either 1,5,4 and 1,2,3, or 1,5,2, and 1,3,4. Therefore the original integers can not be used.

Changing the offender 1 to a 7 (the smallest integer greater than 6) would give us more options. Taking the other five integers and putting the largest with the smallest of them, we get 7,6,2 with the 3,4,and 5 left over. The 3 & 4 can't make a triangle with the 7, but the 5 and 3 can. (So can the 5 & 4. No difference.) This leaves the 4 with the triangles 7,6,2, and 7,5,3. You can make the 4 fit in EITHER the set of 4,5,6 and 4,3,2 OR the set of 4,6,3 and 4,5,2, giving you the desired solution.

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