Edge of the triangle

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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

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Let the edges of one of the faces of the tetrahedron have lengths c7d457e388298246adb06c587bccd419ea67f7e88136a7ef6a03334a7246df9097e5bcc31ba33fd2, and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc. Let 96ab646de7704969b91c76a214126b45f2b07b25a3a59bb1293ee3f6dec19de4019a7178874219ae, and bb2c93730dbb48558bb3c4738c956c4e8f816437 be the lengths of the sides that are not adjacent to the sides with lengths c7d457e388298246adb06c587bccd419ea67f7e88136a7ef6a03334a7246df9097e5bcc31ba33fd2, and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc, respectively.

Without loss of generality, assume that 9d9107feaaa80fa6df7739a4d522aed22b76bbb0. I shall now prove that either 505b6cb9b69b8dacfb85f66770f2810dd6ee8c79 or 059367ba94e44d4ec15e909c79d6e787a4598c82, by proving that if 7b3370cc2927972ff7246adc832c469305b4efd1, then 059367ba94e44d4ec15e909c79d6e787a4598c82.

Assume that 7b3370cc2927972ff7246adc832c469305b4efd1. The triangle inequality gives us that 60be48f0aa0ab45ec2d576f416aa478902dbf9b2, so a3a59bb1293ee3f6dec19de4019a7178874219ae must be greater than 8136a7ef6a03334a7246df9097e5bcc31ba33fd2. We also have from the triangle inequality that 92517f1f9ab681652c5c0ab5831b8429b8d6a263. Therefore cd71c4d758d8475637c62803d57376ce26831897. Therefore either 505b6cb9b69b8dacfb85f66770f2810dd6ee8c79 or 059367ba94e44d4ec15e909c79d6e787a4598c82.

If 505b6cb9b69b8dacfb85f66770f2810dd6ee8c79, then the vertex where the sides of length c7d457e388298246adb06c587bccd419ea67f7e88136a7ef6a03334a7246df9097e5bcc31ba33fd2, and bb2c93730dbb48558bb3c4738c956c4e8f816437 meet satisfies the given condition. If 059367ba94e44d4ec15e909c79d6e787a4598c82, then the vertex where the sides of length c7d457e388298246adb06c587bccd419ea67f7e83372c1cb6d68cf97c2d231acc0b47b95a9ed04cc, and a3a59bb1293ee3f6dec19de4019a7178874219ae meet satisfies the given condition. This proves the statement.

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