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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 $A, B,$ 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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