Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 19, 2011"
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==Solution== | ==Solution== | ||
− | + | Let the angles of the triangles at the interior point be <math> A, B, </math> and <math> C </math>, such that <math> A+B+C=360^\circ </math>. Assume the contrary, that there are at least <math> 2 </math> acute triangles. Assume WLOG that <math> A </math> and <math> B </math> are acute or right angles, so that <math> A, B\leq90^\circ </math>. Therefore, <math> A+B\leq90^\circ+90^\circ=180^\circ </math>. Now, since <math> A+B+C=360^\circ </math>, we have <math> A+B=360^\circ-C \implies 360^\circ-C=A+B\leq180^\circ </math>, so <math> C\geq180^\circ </math>. However, this is a contradiction, since <math> C </math> must be the vertex of a triangle, and therefore cannot be more than <math> 180^\circ </math>. Therefore, there cannot be more than <math> 1 </math> acute or right angles at the interior point, and therefore there must be at least <math> 2 </math> obtuse angles, creating at least <math> 2 </math> obtuse triangles. |
Latest revision as of 21:36, 18 June 2011
Problem
AoPSWiki:Problem of the Day/June 19, 2011
Solution
Let the angles of the triangles at the interior point be and , such that . Assume the contrary, that there are at least acute triangles. Assume WLOG that and are acute or right angles, so that . Therefore, . Now, since , we have , so . However, this is a contradiction, since must be the vertex of a triangle, and therefore cannot be more than . Therefore, there cannot be more than acute or right angles at the interior point, and therefore there must be at least obtuse angles, creating at least obtuse triangles.