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Romanian Mathematical Olympiad - District Round

Geometry Difficulty 4.8 AIME Prove it Romania

The triangle $ABC$ has $\angle ABC = 90^\circ$ and $\angle BCA = 30^\circ$ . Let $AD$ be the bisector of the angle $\angle BAC$ , $D \in BC$ , and $BE \perp AC$ , $E \in AC$ . Denote $M$ the intersection of the lines $AD$ and $BE$ , and $P$ the midpoint of the segment $CM$ . Prove that $AC = 4 \cdot DP$ .

Solution

The $30^\circ$ angle theorem yields $AC = 2AB$ . Since $\angle CAD = \angle ACD = 30^\circ$ , the triangle $ADC$ is isosceles, with $DA = DC$ .

From $\angle ADB = \angle MBD = 60^\circ$ follows that the triangle $MBD$ is equilateral.

This shows that $MB = BD = \frac{1}{2}AD$ , hence $M$ is the midpoint of the segment $AD$ .

Denote $S$ the reflection of $D$ into $B$ . Then the triangle $ADS$ is equilateral. This yields $DS = AD = CD$ , hence $DP$ is a midline of the triangle $CSM$ . From $AB = SM$ (medians in the equilateral triangle $ADS$ ) follows that $DP = \frac{1}{2}MS = \frac{1}{2}AB = \frac{1}{4}AC$ .

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