1968 IMO Problems/Problem 4

Problem

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.


Solution

Let the edges of one of the faces of the tetrahedron have lengths $a$, $b$, and $c$. Let $d$, $e$, and $f$ be the lengths of the sides that are not adjacent to the sides with lengths $a$, $b$, and $c$, respectively.

Without loss of generality, assume that $\max(a,b,c,d,e,f)=a$. I shall now prove that either $b+f>a$ or $c+e>a$, by proving that if $b+f\leq a$, then $c+e>a$.

Assume that $b+f\leq a$. The triangle inequality gives us that $e+f>a$, so $e$ must be greater than $b$. We also have from the triangle inequality that $b+c>a$. Therefore $e+c>b+c>a$. Therefore either $b+f>a$ or $c+e>a$.

If $b+f>a$, then the vertex where the sides of length $a$, $b$, and $f$ meet satisfies the given condition. If $c+e>a$, then the vertex where the sides of length $a$, $c$, and $e$ meet satisfies the given condition. This proves the statement. $\blacksquare$

See Also

1968 IMO (Problems) • Resources)
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions
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