Difference between revisions of "1968 IMO Problems/Problem 4"

Latest revision as of 23:35, 18 July 2016

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. $\blacksquare$

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	[[Category:Olympiad Geometry Problems]]		[[Category:Olympiad Geometry Problems]]
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1968 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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Difference between revisions of "1968 IMO Problems/Problem 4"

Latest revision as of 23:35, 18 July 2016

Problem

Solution

See Also