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Geometry Difficulty 5.6 AIME, harder Prove it India

Let $ABC$ be an equilateral triangle, and $D, E, F$ be points on $BC, CA, AB$ respectively. Let $\angle BAD = \alpha$ , $\angle CBE = \beta$ , $\angle ACF = \gamma$ . Suppose $\alpha + \beta + \gamma \ge 120^\circ$ . Prove that the union $\Sigma$ of triangular regions $BAD, CBE, ACF$ covers the triangle $ABC$ .

Solution

Suppose some point $P$ does not lie in any of the triangles $BAD, CBE, ACF$ . Then $\angle BAD < \angle BAP$ , $\angle CBE < \angle CBP$ and $\angle ACF < \angle ACP$ . Thus
$\alpha + \beta + \gamma < \angle BAP + \angle CBP + \angle ACP.$
If $P$ is the circumcentre of $ABC$ , then $\angle BAP = \angle CBP = \angle ACP = 30^\circ$ and thus $\alpha + \beta + \gamma < 90^\circ$ contradicting $\alpha + \beta + \gamma \ge 120^\circ$ . Thus $P$ cannot be the circumcentre of $ABC$ . Hence $PA, PB, PC$ are not all equal. We may assume $PA > PB$ .

In the above figure, triangles $PAB$ and $PLK$ are similar. If $PA > PB$ , then $PL > PK$ and hence $\angle PKL > \angle PLK = \angle PAB$ . Thus
$\begin{align*} \angle BAP + \angle CBP + \angle ACP &= \angle PLK + \angle LKC + \angle AKM \\ &< \angle PKL + \angle LKC + \angle AKM \\ &< \angle BKC = 120^\circ. \end{align*}$
It follows that if $\alpha + \beta + \gamma \ge 120^\circ$ , then no such point $P$ which lies outside all the triangles $BAD, CBE, ACF$ can exist.

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