Problem 3

For each

k=1,2,3,4,5

, find necessary and sufficient conditions on

a>0

for the existence of a tetrahedron with

k

edges of length

a

and the remaining

6-k

edges of length

1

Step 2 of 5: Analyze two long edges

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In plain words

The combinatorial placement of the two special edges matters: adjacent and opposite edges provide different ranges, whose union gives the answer.

\sqrt{2-\sqrt3}<a<\sqrt{2+\sqrt3}\quad\text{(adjacent $a$-edges)},\qquad 0<a<\sqrt2\quad\text{(opposite $a$-edges)}

Detailed analysis

If the two $a$ -edges share a vertex, hinge the two unit triangles about the opposite unit edge. The two flat limits are $\sqrt{2-\sqrt3}$ and $\sqrt{2+\sqrt3}$ , so the open interval between them is attainable. If the $a$ -edges are opposite, a circle-locus argument gives $0<a<\sqrt2$ . The union is therefore $0<a<\sqrt{2+\sqrt3}$ for $k=2$ .