View source for AoPS Wiki talk:Problem of the Day/June 19, 2011

==Problem==
{{:AoPSWiki:Problem of the Day/June 19, 2011}}
==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.