The lengths of all six edges of a tetrahedron are integers. The lengths of five of the edges are 14, 20, 40,52 , and 70. The number of possible lengths for the sixth edge is
(A) 9
(B) 3
(C) 4
(D) 5
(E) 6
The important fact to remember to solve this problem is that if $a, b, c$ are the side lengths of a triangle, then $a+b>c, a+c>b$, and $b+c>a$. This is called the "Triangle Inequality". We can simplify this slightly by saying that if $a, b, c$ are the side lengths of a triangle with $a \leq b \leq c$, then we need $a+b>c$ to be true. (The other two inequalities are automatically satisfied since $c$ is the longest side.)

Let $x$ represent the length of the unknown edge of the tetrahedron.

We represent the tetrahedron schematically as in the diagram. We can think here of triangle $A B C$ as being the "base" of the tetrahedron, and vertex $D$ as being the "top" vertex of the tetrahedron. This diagram also emphasizes that the tetrahedron is formed by 4 triangles $A B C, A B D, A D C$, and $B D C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-255.jpg?height=374&width=377&top_left_y=1922&top_left_x=1424)

Let $B C=70$. Now $B C$ is an edge in two triangles $(A B C$ and $B D C)$ and so one of these triangles must be formed entirely from known edge lengths. Of the known edge lengths,
only 40 and 52, or 20 and 52 can form a triangle with a third side of 70, by the Triangle Inequality. So let $A C=52$ (the common edge length of these two triangles).

If $A B=40$, then the three remaining edge lengths are 14,20 and $x .14$ and 20 cannot together make a triangle with any of 40,52, or 70 . So there are no possibilities for $x$ in this case.

If $A B=20$, the three remaining edge lengths are 14,40 and $x .14$ and 40 cannot together make a triangle with 70 , so either $B D=x$ or $D C=x$.

If $D C=x$, then triangle $A B D$ must have side lengths 20, 14 and 40, which is impossible, so $B D=x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-256.jpg?height=374&width=382&top_left_y=756&top_left_x=1294)

Case 1: $A D=14$ and $D C=40$

From $\triangle A B D, 20+14>x$ and $14+x>20$, so since $x$ is an integer, $7 \leq x \leq 33$.

From $\triangle B D C, 40+70>x$ and $x+40>70$, so since $x$ is an integer, $31 \leq x \leq 109$.

Combining these conditions, $31 \leq x \leq 33$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-256.jpg?height=385&width=374&top_left_y=1176&top_left_x=1301)

Case 1: $D C=14$ and $A D=40$

From $\triangle A B D, 20+40>x$ and $20+x>40$, so since $x$ is an integer, $21 \leq x \leq 59$.

From $\triangle B D C, 14+70>x$ and $x+14>70$, so since $x$ is an integer, $57 \leq x \leq 83$.

Combining these conditions, $57 \leq x \leq 59$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-256.jpg?height=379&width=379&top_left_y=1602&top_left_x=1296)

So there are 6 possibilities for $x$.

[Note: In fact, all six of these possibilities do in fact form tetrahedra with the other 5 edge lengths. As we discovered as we were creating this question, it is possible to come up with side lengths that do form 4 triangles, but which cannot be put together to form a tetrahedron. This is a very interesting problem to work on.]

ANSWER: (E)