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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 19, 2011"

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==Solution==
 
==Solution==
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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 $A, B,$ and $C$, such that $A+B+C=360^\circ$. Assume the contrary, that there are at least $2$ acute triangles. Assume WLOG that $A$ and $B$ are acute or right angles, so that $A, B\leq90^\circ$. Therefore, $A+B\leq90^\circ+90^\circ=180^\circ$. Now, since $A+B+C=360^\circ$, we have $A+B=360^\circ-C \implies 360^\circ-C=A+B\leq180^\circ$, so $C\geq180^\circ$. However, this is a contradiction, since $C$ must be the vertex of a triangle, and therefore cannot be more than $180^\circ$. Therefore, there cannot be more than $1$ acute or right angles at the interior point, and therefore there must be at least $2$ obtuse angles, creating at least $2$ obtuse triangles.

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