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Romanian Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Romania

In a triangle $ABC$ denote by $D$ , $E$ , respectively $F$ , the points where the angle bisectors of $\angle CAB$ , $\angle ABC$ , respectively $\angle BCA$ , meet its circumcircle.

a) Prove that the ortocenter of triangle $DEF$ coincides with the incenter of triangle $ABC$ .

b) Prove that if $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \overrightarrow{0}$ , then the triangle $ABC$ is equilateral.

Solution

a) Let $I$ be the incenter of the triangle $ABC$ . $D, E, F$ are the midpoints of the arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ .
The angle determined by the lines $AD$ and $EF$ is equal to $\frac{1}{2}(\widehat{AE} + \widehat{DF}) = \frac{1}{4}(\widehat{AB} + \widehat{BC} + \widehat{CA}) = 90^\circ$ , hence $AD \perp EF$ . Likewise, $BE \perp DF$ , and the claim follows.

b) Let $O$ be the circumcenter of the triangle $ABC$ . The given relation implies that $OA + OB + OC = OD + OE + OF$ .
Invoking Sylvester's theorem, the triangles $ABC$ and $DEF$ share the same orthocenter. Thus, in the triangle $ABC$ point $I$ is both incenter and orthocenter, yielding the conclusion.

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