A triangle problem

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Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another

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In triangle e2a559986ed5a0ffc5654bd367c29dfc92913c36, let da35a0d0a8952651e752f9a0afa0662c0c143b9abe0834a58059a61a79d86701cc1292ee73a0858c474cd45035e0eea00cfc2d9355528e69dc34389db7cb244faa138b87f4b915395f6ea28ee806fbe9, and cd78384b4990cba6fbcb386d57c9c76be38a1dfa. Using the Law of Sines gives that

bf84e7b0edb0bfeec24f83f1cc28aff629df5a3a

Therefore 896121481645621b1e537c45e695aa1ddfeb6b1c. Using the Law of Cosines gives that

ad2736e984813c97a47a7e307ee456b3da59b5de

This can be simplified to ca8d8f623cfcc98b9ccd4e80810408615a53a2e7. Since c7d457e388298246adb06c587bccd419ea67f7e88136a7ef6a03334a7246df9097e5bcc31ba33fd2, and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc are positive integers, b1503e4a9c3d306f9badb922068dc9c5b0f26ef1. Note that if 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is between c7d457e388298246adb06c587bccd419ea67f7e8 and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc, then 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is relatively prime to c7d457e388298246adb06c587bccd419ea67f7e8 and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc, and 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 cannot possibly divide b87077407887cbeefafb810630ce3d7934037b0e. Therefore 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is either the least of the three consecutive integers or the greatest.

Assume that 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is the least of the three consecutive integers. Then either 09a092ecea151fb05eba1170ba2be565493cc283 or 3d2089682bad978c293caefb7f9ad8f81f32550e, depending on if fffda67433ea32215503519363a5f50652df548b or ee64770c2ba84450f104425ee501e4d9c5b0329a. If 09a092ecea151fb05eba1170ba2be565493cc283, then 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is 1 or 2. 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 couldn’t be 1, for if it was then the triangle would be degenerate. If 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is 2, then 199f2f2cab45fd9af4f2578f45a370fafe5cf1d4, but c7d457e388298246adb06c587bccd419ea67f7e8 and 3372c1cb6d68cf97c2d231acc0b47b95a9ed04cc must be 3 and 4 in some order, which means that this triangle doesn’t exist. therefore 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 cannot divide 8cd8ac52a9c6112e9ac737cba648d2c51efb97d7, and so 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 must divide 0f1e916cbacb9c9e781c35613c1006fe12aaca7f. If 3d2089682bad978c293caefb7f9ad8f81f32550e then e3da2f72c17d10f35e850b51c55178b508d2b5bd, so 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is 1, 2, or 4. Clearly 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 cannot be 1 or 2, so 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 must be 4. Therefore b2e783bfaf70566692a4e282e67d3215243b6bdf. This shows that fa5507ab5a3d1f06941dabd1c26505d646367983 and 24f7150d029fa128fa37d225f668174382d6219b, and the triangle has sides that measure 4, 5, and 6.

Now assume that 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is the greatest of the three consecutive integers. Then either b30dcdc54476aa909b5af740ecd5d1e69e110316 or ad6d5bababdbe3e0515d946d832c5a7e67091309, depending on if d3ecf802d869f203bcd82747c335d3dd50a32513 or 2a39bfcef0dfb19750221718a8ec7688ed681504b30dcdc54476aa909b5af740ecd5d1e69e110316 is absurd, so ad6d5bababdbe3e0515d946d832c5a7e67091309, and e3c4e67b5d51aadf1254f75b30e0a1f396fa45dc. Therefore 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so 8136a7ef6a03334a7246df9097e5bcc31ba33fd2 cannot be the greatest of the three consecutive integers. This shows that there is exactly one triangle with this property – and it has side lengths of 4, 5, and 6. 3bbb9e25e5be095e317771a9c5d00877ee9c8012

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